Undergraduate Lecture Notes in Physics
Wolfgang Demtröder
Mechanics and
Thermodynamics
Undergraduate Lecture Notes in Physics
Series editors
N. Ashby, University of Colorado, Boulder, USA
W. Brantley, Department of Physics, Furman University, Greenville, USA
M. Deady, Physics Program, Bard College, Annandale-on-Hudson, USA
M. Fowler, Dept of Physics, Univ of Virginia, Charlottesville, USA
M. Hjorth-Jensen, Dept. of Physics, University of Oslo, Oslo, Norway
M. Inglis, Earth &Space Sci, Smithtown Sci Bld, SUNY Suffolk County Community College, Long
Island, USA
H. Klose, Humboldt University, Oldenburg, Germany
H. Sherif, Department of Physics, University of Alberta, Edmonton, Canada
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Wolfgang Demtröder
Mechanics and
Thermodynamics
Wolfgang Demtröder
Kaiserslautern, Germany
demtroed@rhrk.uni-kl.de
ISSN 2192-4791
Undergraduate Lecture Notes in Physics
ISBN 978-3-319-27875-9
DOI 10.1007/978-3-319-27877-3
ISSN 2192-4805 (electronic)
ISBN 978-3-319-27877-3 (eBook)
Library of Congress Control Number: 2016944491
© Springer International Publishing Switzerland 2017
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even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations
and therefore free for general use.
The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed
to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty,
express or implied, with respect to the material contained herein or for any errors or omissions that may have been made.
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Preface
The present textbook represents the first part of a four-volume series on experimental Physics. It covers
the field of Mechanics and Thermodynamics. One of its goal is to illustrate, that the explanation of our
world and of all natural processes by Physics is always the description of models of our world, which
are formulated by theory and proved by experiments. The continuous improvement of these models
leads to a more detailled understanding of our world and of the processes that proceed in it.
The representation of this textbook starts with an introductory chapter giving a brief survey of the history and development of Physics and its present relevance for other sciences and for technology. Since
experimental Physics is based on measuring techniques and quantitative results, a section discusses
basic units, techniques for their measurements and the accuracy and possible errors of measurements.
In all further chapters the description of the real world by successively refined models is outlined. It
begins with the model of a point mass, its motion under the action of forces and its limitations. Since
the description of moving masses requires a coordinate system, the transformation of results obtained
in one system to another system moving against the first one is described. This leads to the theory
of special relativity, which is discussed in Chap. 3. The next chapter upgrades the model of point
masses to spatially extended rigid bodies, where the spatial extension of a body cannot be ignored
but influences the results. Then the deformation of bodies under the influence of forces is discussed
and phenomena caused by this deformation are explained. The existence of different phases (solid,
liquid and gaseous) and their relation with external influences such as temperature and pressure, are
discussed.
The properties of gases and liquids at rest and the effects caused by streaming gases and liquids are
outlined in Chap. 7 and 8.
Many insights in natural phenomena, in particular in the area of atomic and molecular physics could
only be explored after sufficiently good vacua could be realized. Therefore Chap. 9 discusses briefly
the most important facts of vacuum physics, such as the realization and measurement of evacuated
volumina.
Thermodynamics governs important aspects of our life. Therefore an extended chapter about definitions and measuring techniques for temperatures, heat energy and phase transitions should emphazise
the importance of thermodynamics. The three principle laws ot thermodynamics and their relevanve
for energy transformation and dissipation are discussed.
Chapter 11 deals with oscillations and waves, a subject which is closely related to acoustics and optics.
While all foregoing chapters discuss classical physics which had been developed centuries ago,
Chap. 12 covers a modern subject, namely nonlinear phenomena and chaos theory. It should give
a feeling for the fact, that most phenomena in classical physics can be described only approximately
by linear equations. A closer inspection shows that the accurate description demands nonlinear equations with surprising solutions.
A description of phenomena in physics requires some minimum mathematical knowledge. Therefore a
brief survey about vector algebra and vector analysis, about complex numbers and different coordinate
systems is provided in the last chapter.
A real understanding of the subjects covered in this textbook can be checked by solving problems,
which are given at the end of each chapter. A sketch of the solutions can be found at the end of the
book.
For further studies and a deeper insight into special subjects some selected literature is given at the
end of each chapter.
v
vi
Preface
The author hopes that this book can transfer some of his enthusiasm for the fascinating field of physics.
He is grateful for any comments and suggestions, also for hints to possible errors. Every e-mail will
be answered as soon as possible.
Several people have contributed to the realization of this book. Many thanks go the Dr. Schneider
and Ute Heuser, Springer Verlag Heidelberg, who supported and encouraged the authors over the
whole period needed for translating this book from a German version. Nadja Kroke and her team
(le-tex publishing services GmbH) did a careful job for the layout of the book and induced the author
to improve ambiguous sentences or unclear hints to equations or figures. I thank them all for their
efforts.
Last but not least I thank my wife Harriet, who showed much patience when her husband disappeared
into his office for the work on this book.
Kaiserslautern, December 2016
Wolfgang Demtröder
Contents
1
Introduction and Survey . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1.1
The Importance of Experiments . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
1.2
The Concept of Models in Physics . . . . . . . . . . . . . . . . . . . . . . . . .
3
1.3
Short Historical Review . . . . . . . . . . . . . . . . .
1.3.1 The Natural Philosophy in Ancient Times
1.3.2 The Development of Classical Physics . . .
1.3.3 Modern Physics . . . . . . . . . . . . . . . . .
1.4
The Present Conception of Our World . . . . . . . . . . . . . . . . . . . . . . 11
1.5
Relations Between Physics and Other Sciences
1.5.1 Biophysics and Medical Physics . . . . .
1.5.2 Astrophysics . . . . . . . . . . . . . . . . . .
1.5.3 Geophysics and Meteorology . . . . . .
1.5.4 Physics and Technology . . . . . . . . . .
1.5.5 Physics and Philosophy . . . . . . . . . .
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16
1.6
The Basic Units in Physics, Their Standards and Measuring Techniques
1.6.1 Length Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.6.2 Measuring Techniques for Lengths . . . . . . . . . . . . . . . . . . .
1.6.3 Time-Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.6.4 How to measure Times . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.6.5 Mass Units and Their Measurement . . . . . . . . . . . . . . . . . . .
1.6.6 Molar Quantity Unit . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.6.7 Temperature Unit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.6.8 Unit of the Electric Current . . . . . . . . . . . . . . . . . . . . . . . .
1.6.9 Unit of Luminous Intensity . . . . . . . . . . . . . . . . . . . . . . . . .
1.6.10 Unit of Angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
16
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19
20
23
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24
24
25
25
25
1.7
Systems of Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
1.8
Accuracy and Precision; Measurement Uncertainties and Errors . . . .
1.8.1 Systematic Errors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.8.2 Statistical Errors, Distribution of Experimental Values, Mean
Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.8.3 Variance and its Measure . . . . . . . . . . . . . . . . . . . . . . . . .
1.8.4 Error Distribution Law . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.8.5 Error Propagation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.8.6 Equalization Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . .
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5
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5
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7
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27
29
29
31
32
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
2
Mechanics of a Point Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
2.1
The Model of the Point Mass; Trajectories . . . . . . . . . . . . . . . . . . . . 40
2.2
Velocity and Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
vii
viii
Contents
2.3
Uniformly Accelerated Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
2.3.1 The Free Fall . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
2.3.2 Projectile Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
2.4
Motions with Non-Constant Acceleration . . . . . . . . . . . . . . . . . . . . 44
2.4.1 Uniform Circular Motion . . . . . . . . . . . . . . . . . . . . . . . . . . 44
2.4.2 Motions on Trajectories with Arbitrary Curvature . . . . . . . . . 45
2.5
Forces
2.5.1
2.5.2
2.5.3
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47
47
48
50
2.6
The Basic Equations of Mechanics . . . . . . . . . . . . . . . . . . . . . . . . .
2.6.1 The Newtonian Axioms . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6.2 Inertial and Gravitational Mass . . . . . . . . . . . . . . . . . . . . . .
2.6.3 The Equation of Motion of a Particle in Arbitrary Force Fields .
51
51
52
53
2.7
Energy Conservation Law of Mechanics . . . . . . . . . . . . .
2.7.1 Work and Power . . . . . . . . . . . . . . . . . . . . . . .
2.7.2 Path-Independent Work; Conservative Force-Fields
2.7.3 Potential Energy . . . . . . . . . . . . . . . . . . . . . . . .
2.7.4 Energy Conservation Law in Mechanics . . . . . . . .
2.7.5 Relation Between Force Field and Potential . . . . .
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59
61
62
2.8
Angular Momentum and Torque . . . . . . . . . . . . . . . . . . . . . . . . . . 63
2.9
Gravitation and the Planetary Motions . . . . . . . . . . . . . . . . . .
2.9.1 Kepler’s Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.9.2 Newton’s Law of Gravity . . . . . . . . . . . . . . . . . . . . . .
2.9.3 Planetary Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.9.4 The Effective Potential . . . . . . . . . . . . . . . . . . . . . . .
2.9.5 Gravitational Field of Extended Bodies . . . . . . . . . . . .
2.9.6 Measurements of the Gravitational Constant G . . . . . . .
2.9.7 Testing Newton’s Law of Gravity . . . . . . . . . . . . . . . . .
2.9.8 Experimental Determination of the Earth Acceleration g
.........................................
Forces as Vectors; Addition of Forces . . . . . . . . . . . . . . .
Force-Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Measurements of Forces; Discussion of the Force Concept
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64
64
66
66
68
69
71
72
74
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
3
Moving Coordinate Systems and Special Relativity . . . . . . . . . . . . . . . . . . 81
3.1
Relative Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
3.2
Inertial Systems and Galilei-Transformations . . . . . . . . . . . . . . . . . . 82
3.3
Accelerated Systems; Inertial Forces . . .
3.3.1 Rectilinear Accelerated Systems
3.3.2 Rotating Systems . . . . . . . . . .
3.3.3 Centrifugal- and Coriolis-Forces
3.3.4 Summary . . . . . . . . . . . . . . . .
3.4
The Constancy of the Velocity of Light . . . . . . . . . . . . . . . . . . . . . . 89
3.5
Lorentz-Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
3.6
Theory of Special Relativity . . . . . . . . .
3.6.1 The Problem of Simultaneity . .
3.6.2 Minkowski-Diagram . . . . . . . .
3.6.3 Lenght Scales . . . . . . . . . . . . .
3.6.4 Lorentz-Contraction of Lengths
3.6.5 Time Dilatation . . . . . . . . . . .
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83
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93
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96
Contents
3.6.6
3.6.7
The Twin-Paradox . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
Space-time Events and Causality . . . . . . . . . . . . . . . . . . . . . 99
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
4
Systems of Point Masses; Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
4.1
Fundamentals . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.1.1 Centre of Mass . . . . . . . . . . . . . . . . . . . . .
4.1.2 Reduced Mass . . . . . . . . . . . . . . . . . . . . .
4.1.3 Angular Momentum of a System of Particles
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104
104
105
105
4.2
Collisions Between Two Particles . . . . . . . . . . . . . . .
4.2.1 Basic Equations . . . . . . . . . . . . . . . . . . . . .
4.2.2 Elastic Collisions in the Lab-System . . . . . . . .
4.2.3 Elastic Collisions in the Centre-of Mass system
4.2.4 Inelastic Collisions . . . . . . . . . . . . . . . . . . . .
4.2.5 Newton-Diagrams . . . . . . . . . . . . . . . . . . . .
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107
108
109
111
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114
4.3
What Do We Learn from the Investigation of Collisions? . . . . . . . . . . 115
4.3.1 Scattering in a Spherical Symmetric Potential . . . . . . . . . . . . 115
4.3.2 Reactive Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
4.4
Collisions at Relativistic Energies . . . . . . . . . . . . . . . .
4.4.1 Relativistic Mass Increase . . . . . . . . . . . . . . . .
4.4.2 Force and Relativistic Momentum . . . . . . . . . .
4.4.3 The Relativistic Energy . . . . . . . . . . . . . . . . . .
4.4.4 Inelastic Collisions at relativistic Energies . . . . .
4.4.5 Relativistic Formulation of Energy Conservation
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119
119
120
121
122
122
4.5
Conservation Laws . . . . . . . . . . . . . . . . . . .
4.5.1 Conservation of Momentum . . . . . .
4.5.2 Energy Conservation . . . . . . . . . . . .
4.5.3 Conservation of Angular Momentum
4.5.4 Conservation Laws and Symmetries . .
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123
123
124
124
124
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Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
5
Dynamics of rigid Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129
5.1
The Model of a Rigid Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130
5.2
Center of Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130
5.3
Motion of a Rigid Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
5.4
Forces and Couple of Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132
5.5
Rotational Inertia and Rotational Energy . . . . . . . . . . . . . . . . . . . . 133
5.5.1 The Parallel Axis Theorem (Steiner’s Theorem) . . . . . . . . . . . 134
5.6
Equation of Motion for the Rotation of a Rigid Body . . . . . . . . . . . .
5.6.1 Rotation About an Axis for a Constant Torque . . . . . . . . . . .
5.6.2 Measurements of rotational inertia; Rotary Oscillations About
a Fixed Axis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.6.3 Comparison Between Translation and Rotation . . . . . . . . . . .
136
137
139
139
ix
x
Contents
5.7
Rotation About Free Axes; Spinning Top . . . . . . .
5.7.1 Inertial Tensor and Inertial Ellipsoid . . . .
5.7.2 Principal Moments of Inertia . . . . . . . . .
5.7.3 Free Rotational axes . . . . . . . . . . . . . . .
5.7.4 Euler’s Equations . . . . . . . . . . . . . . . . .
5.7.5 The Torque-free Symmetric Top . . . . . . .
5.7.6 Precession of the Symmetric Top . . . . . . .
5.7.7 Superposition of Nutation and Precession
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139
140
141
143
144
145
147
148
5.8
The Earth as Symmetric Top . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152
6
Real Solid and Liquid Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153
6.1
Atomic Model of the Different Aggregate States . . . . . . . . . . . . . . . 154
6.2
Deformable Solid Bodies . . . . . . . . . . . . . . . . .
6.2.1 Hooke’s Law . . . . . . . . . . . . . . . . . . . .
6.2.2 Transverse Contraction . . . . . . . . . . . . .
6.2.3 Shearing and Torsion Module . . . . . . . . .
6.2.4 Bending of a Balk . . . . . . . . . . . . . . . . .
6.2.5 Elastic Hysteresis; Energy of Deformation .
6.2.6 The Hardness of a Solid Body . . . . . . . . .
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155
156
157
158
159
161
162
6.3
Static Liquids; Hydrostatics . . . . . . . . . . . . . . . .
6.3.1 Free Displacement and Surfaces of Liquids
6.3.2 Static Pressure in a Liquid . . . . . . . . . . .
6.3.3 Buoyancy and Floatage . . . . . . . . . . . . .
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162
162
163
165
6.4
Phenomena at Liquid Surfaces . . . . . . .
6.4.1 Surface Tension . . . . . . . . . . . .
6.4.2 Interfaces and Adhesion Tension .
6.4.3 Capillarity . . . . . . . . . . . . . . . .
6.4.4 Summary of Section 6.4 . . . . . . .
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166
166
168
170
171
6.5
Friction Between Solid Bodies . . . . . . . . . . . .
6.5.1 Static Friction . . . . . . . . . . . . . . . . . .
6.5.2 Sliding Friction . . . . . . . . . . . . . . . . .
6.5.3 Rolling Friction . . . . . . . . . . . . . . . . .
6.5.4 Significance of Friction for Technology
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171
171
172
173
174
6.6
The Earth as Deformable Body . . . . . . . . . . . .
6.6.1 Ellipticity of the Rotating Earth . . . . . .
6.6.2 Tidal Deformations . . . . . . . . . . . . . . .
6.6.3 Consequences of the Tides . . . . . . . . . .
6.6.4 Measurements of the Earth Deformation
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174
175
175
178
179
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Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181
7
Gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183
7.1
Macroscopic Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184
7.2
Atmospheric Pressure and Barometric Formula . . . . . . . . . . . . . . . . 185
Contents
7.3
Kinetic Gas Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.3.1 The Model of the Ideal Gas . . . . . . . . . . . . . . . .
7.3.2 Basic Equations of the Kinetic Gas Theory . . . . . .
7.3.3 Mean Kinetic Energy and Absolute Temperature . .
7.3.4 Distribution Function . . . . . . . . . . . . . . . . . . . .
7.3.5 Maxwell–Boltzmann Velocity Distribution . . . . . .
7.3.6 Collision Cross Section and Mean Free Path Length
7.4
Experimental Proof of the Kinetic Gas Theory . . . . . . . . . . . . . . . . . 196
7.4.1 Molecular Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196
7.5
Transport Phenomena in Gases . . . . . . . .
7.5.1 Diffusion . . . . . . . . . . . . . . . . . .
7.5.2 Brownian Motion . . . . . . . . . . . .
7.5.3 Heat Conduction in Gases . . . . . .
7.5.4 Viscosity of Gases . . . . . . . . . . . .
7.5.5 Summary of Transport Phenomena
7.6
The Atmosphere of the Earth . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204
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188
188
189
190
190
191
195
198
198
200
201
202
203
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208
8
Liquids and Gases in Motion; Fluid Dynamics . . . . . . . . . . . . . . . . . . . . . . 209
8.1
Basic Definitions and Types of Fluid Flow . . . . . . . . . . . . . . . . . . . . 210
8.2
Euler Equation for Ideal Liquids . . . . . . . . . . . . . . . . . . . . . . . . . . . 212
8.3
Continuity Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212
8.4
Bernoulli Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213
8.5
Laminar Flow . . . . . . . . . . . . . . . . . . . . . . . . .
8.5.1 Internal Friction . . . . . . . . . . . . . . . . . .
8.5.2 Laminar Flow Between Two Parallel Walls
8.5.3 Laminar Flows in Tubes . . . . . . . . . . . . .
8.5.4 Stokes Law, Falling Ball Viscometer . . . . .
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216
216
218
219
220
8.6
Navier–Stokes Equation . . . . . . . . . . . .
8.6.1 Vortices and Circulation . . . . . . .
8.6.2 Helmholtz Vorticity Theorems . .
8.6.3 The Formation of Vortices . . . . .
8.6.4 Turbulent Flows; Flow Resistance
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220
221
222
223
224
8.7
Aerodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.7.1 The Aerodynamical Buoyancy . . . . . . . . . . . . . .
8.7.2 Relation between Dynamical and Flow Resistance
8.7.3 Forces on a flying Plane . . . . . . . . . . . . . . . . . .
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226
226
227
228
8.8
Similarity Laws; Reynolds’ Number . . . . . . . . . . . . . . . . . . . . . . . . . 228
8.9
Usage of Wind Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229
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Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235
9
Vacuum Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237
9.1
Fundamentals and Basic Concepts . . . . . . . . . . . . . . . . . . . . . . . . . 238
9.1.1 The Different Vacuum Ranges . . . . . . . . . . . . . . . . . . . . . . 238
9.1.2 Influence of the Molecules at the Walls . . . . . . . . . . . . . . . . 239
xi
xii
Contents
9.1.3
9.1.4
9.1.5
Pumping Speed and Suction Capacity of Vacuum Pumps . . . . 239
Flow Conductance of Vacuum Pipes . . . . . . . . . . . . . . . . . . 240
Accessible Final Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . 241
9.2
Generation of Vacuum . . . . . . . . . . . . . . . . . . . . .
9.2.1 Mechanical Pumps . . . . . . . . . . . . . . . . . .
9.2.2 Diffusion Pumps . . . . . . . . . . . . . . . . . . . .
9.2.3 Cryo- and Sorption-Pumps; Ion-Getter Pumps
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241
242
244
246
9.3
Measurement of Low Pressures . . . . . . . . . . . . . . . .
9.3.1 Liquid Manometers . . . . . . . . . . . . . . . . . . .
9.3.2 Membrane Manometer . . . . . . . . . . . . . . . .
9.3.3 Heat Conduction Manometers . . . . . . . . . . .
9.3.4 Ionization Gauge and Penning Vacuum Meter
9.3.5 Rotating Ball Vacuum Gauge . . . . . . . . . . . .
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247
248
248
249
249
250
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252
10
Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253
10.1
Temperature and Amount of Heat . . . . . . . . . . . . . . . . . . . . . . . . .
10.1.1 Temperature Measurements, Thermometer, and Temperature
Scales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.1.2 Thermal Expansion of Liquids and Solids . . . . . . . . . . . . . . .
10.1.3 Thermal Expansion of Gases; Gas Thermometer . . . . . . . . . .
10.1.4 Absolute Temperature Scale . . . . . . . . . . . . . . . . . . . . . . . .
10.1.5 Amount of Heat and Specific Heat Capacity . . . . . . . . . . . . .
10.1.6 Molar Volume and Avogadro Constant . . . . . . . . . . . . . . . .
10.1.7 Internal Energy and Molar Heat Capacity of Ideal Gases . . . . .
10.1.8 Specific Heat of a Gas at Constant Pressure . . . . . . . . . . . . .
10.1.9 Molecular Explanation of the Specific Heat . . . . . . . . . . . . .
10.1.10 Specific Heat Capacity of Solids . . . . . . . . . . . . . . . . . . . . . .
10.1.11 Fusion Heat and Heat of Evaporation . . . . . . . . . . . . . . . . .
254
10.2
Heat Transport . . . . . . . . . . . . . . . . .
10.2.1 Convection . . . . . . . . . . . . . .
10.2.2 Heat Conduction . . . . . . . . . .
10.2.3 The Heat Pipe . . . . . . . . . . . .
10.2.4 Methods of Thermal Insulation .
10.2.5 Thermal Radiation . . . . . . . . .
266
266
267
271
271
273
10.3
The Three Laws of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . .
10.3.1 Thermodynamic Variables . . . . . . . . . . . . . . . . . . . . . . . .
10.3.2 The First Law of Thermodynamics . . . . . . . . . . . . . . . . . . .
10.3.3 Special Processes as Examples of the First Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.3.4 The Second Law of Thermodynamics . . . . . . . . . . . . . . . . .
10.3.5 The Carnot Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.3.6 Equivalent Formulations of the Second Law . . . . . . . . . . . .
10.3.7 Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.3.8 Reversible and Irreversible Processes . . . . . . . . . . . . . . . . .
10.3.9 Free Energy and Enthalpy . . . . . . . . . . . . . . . . . . . . . . . .
10.3.10 Chemical Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.3.11 Thermodynamic Potentials; Relations Between Thermodynamic Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.3.12 Equilibrium States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.3.13 The Third Law of Thermodynamics . . . . . . . . . . . . . . . . . .
10.3.14 Thermodynamic Engines . . . . . . . . . . . . . . . . . . . . . . . . .
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254
256
258
259
260
261
261
262
263
264
265
. 279
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281
282
283
286
286
290
291
292
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292
293
294
295
Contents
10.4
Thermodynamics of Real Gases and Liquids . . .
10.4.1 Van der Waals Equation of State . . . . .
10.4.2 Matter in Different Aggregation States
10.4.3 Solutions and Mixed States . . . . . . . .
10.5
Comparison of the Different Changes of State . . . . . . . . . . . . . . . . . 309
10.6
Energy Sources and Energy Conversion .
10.6.1 Hydro-Electric Power Plants . . .
10.6.2 Tidal Power Stations . . . . . . . .
10.6.3 Wave Power Stations . . . . . . .
10.6.4 Geothermal Power Plants . . . .
10.6.5 Solar-Thermal Power Stations . .
10.6.6 Photovoltaic Power Stations . . .
10.6.7 Bio-Energy . . . . . . . . . . . . . . .
10.6.8 Energy Storage . . . . . . . . . . . .
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299
299
301
307
309
312
312
313
313
314
315
316
316
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319
11
Mechanical Oscillations and Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321
11.1
The Free Undamped Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . 322
11.2
Mathematical Notations of Oscillations . . . . . . . . . . . . . . . . . . . . . 323
11.3
Superposition of Oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324
11.3.1 One-Dimensional Superposition . . . . . . . . . . . . . . . . . . . . . 324
11.3.2 Two-dimensional Superposition; Lissajous-Figures . . . . . . . . . 327
11.4
The Free Damped Oscillator . . . . . . . .
11.4.1
< !0 , i. e. weak damping . . . .
11.4.2
> !0 , i. e. strong Damping . . .
11.4.3
D !0 (aperiodic limiting case) .
11.5
Forced Oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330
11.5.1 Stationary State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331
11.5.2 Transient State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333
11.6
Energy Balance for the Oscillation of a Point Mass . . . . . . . . . . . . . . 333
11.7
Parametric Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 334
11.8
Coupled Oscillators . . . . . . . . . . . . . . . . . . . . . . . .
11.8.1 Coupled Spring Pendulums . . . . . . . . . . . . .
11.8.2 Forced Oscillations of Two Coupled Oscillators
11.8.3 Normal Vibrations . . . . . . . . . . . . . . . . . . .
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335
335
338
339
11.9
Mechanical Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.9.1 Different Representations of Harmonic Plane Waves . . .
11.9.2 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.9.3 General Description of Arbitrary Waves; Wave-Equation
11.9.4 Different Types of Waves . . . . . . . . . . . . . . . . . . . . . .
11.9.5 Propagation of Waves in Different Media . . . . . . . . . .
11.9.6 Energy Density and Energy Transport in a Wave . . . . . .
11.9.7 Dispersion; Phase- and Group-Velocity . . . . . . . . . . . . .
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339
340
341
341
342
344
350
350
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328
329
329
330
11.10 Superposition of Waves; Interference . . . . . . . . . . . . . . . . . . . . . . . 352
11.10.1 Coherence and Interference . . . . . . . . . . . . . . . . . . . . . . . . 352
11.10.2 Superposition of Two Harmonic Waves . . . . . . . . . . . . . . . . 353
xiii
xiv
Contents
11.11 Diffraction, Reflection and Refraction of Waves .
11.11.1 Huygens’s Principle . . . . . . . . . . . . . . .
11.11.2 Diffraction at Apertures . . . . . . . . . . .
11.11.3 Summary . . . . . . . . . . . . . . . . . . . . . .
11.11.4 Reflection and Refraction of Waves . . . .
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354
355
356
358
358
11.12 Standing Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.12.1 One-Dimensional Standing Waves . . . . . . . . . . . . . . .
11.12.2 Experimental Demonstrations of Standing Waves . . . .
11.12.3 Two-dimensional Resonances of Vibrating Membranes
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359
359
360
361
11.13 Waves Generated by Moving Sources . . .
11.13.1 Doppler-Effect . . . . . . . . . . . . .
11.13.2 Wave Fronts for Moving Sources .
11.13.3 Shock Waves . . . . . . . . . . . . . .
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363
363
364
365
11.14 Acoustics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.14.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.14.2 Pressure Amplitude and Energy Density of Acoustic Waves .
11.14.3 Sound Generators . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.14.4 Sound-Detectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.14.5 Ultrasound . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.14.6 Applications of Ultrasound . . . . . . . . . . . . . . . . . . . . . .
11.14.7 Techniques of Ultrasonic Diagnosis . . . . . . . . . . . . . . . . .
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366
366
367
368
368
369
370
371
11.15 Physics of Musical Instruments . . . . . . . . . .
11.15.1 Classification of Musical Instruments
11.15.2 Chords, Musical Scale and Tuning . .
11.15.3 Physics of the Violin . . . . . . . . . . .
11.15.4 Physics of the Piano . . . . . . . . . . .
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372
372
372
374
375
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Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 376
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 378
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379
12
Nonlinear Dynamics and Chaos . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 381
12.1
Stability of Dynamical Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . 383
12.2
Logistic Growth Law; Feigenbaum-Diagram . . . . . . . . . . . . . . . . . . 386
12.3
Parametric Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 388
12.4
Population Explosion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 389
12.5
Systems with Delayed Feedback . . . . . . . . . . . . . . . . . . . . . . . . . . 390
12.6
Self-Similarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 391
12.7
Fractals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392
12.8
Mandelbrot Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393
12.9
Consequences for Our Comprehension of the Real World . . . . . . . . . 397
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 397
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 398
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 399
13
Supplement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401
13.1
Vector Algebra and Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 402
13.1.1 Definition of Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 402
13.1.2 Representation of Vectors . . . . . . . . . . . . . . . . . . . . . . . . . 402
Contents
13.1.3
13.1.4
13.1.5
13.1.6
14
Polar and Axial Vectors . . . . . . . . .
Addition and Subtraction of Vectors
Multiplication of Vectors . . . . . . . .
Differentiation of Vectors . . . . . . .
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403
403
404
405
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407
408
408
409
13.2
Coordinate Systems . . . . . . . .
13.2.1 Cartesian Coordinates .
13.2.2 Cylindrical Coordinates
13.2.3 Spherical Coordinates .
13.3
Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 410
13.3.1 Calculation rules of Complex Numbers . . . . . . . . . . . . . . . . . 410
13.3.2 Polar Representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 411
13.4
Fourier-Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 411
Solutions of the Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413
14.1
Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 414
14.2
Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 414
14.3
Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 418
14.4
Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 420
14.5
Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423
14.6
Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 425
14.7
Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 426
14.8
Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 429
14.9
Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 431
14.10 Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 433
14.11 Chapter 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 436
14.12 Chapter 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 441
Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445
xv
1.1
The Importance of Experiments . . . . . . . . . . . . . . . . . . . . . . .
2
1.2
The Concept of Models in Physics . . . . . . . . . . . . . . . . . . . . .
3
1.3
Short Historical Review . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
1.4
The Present Conception of Our World . . . . . . . . . . . . . . . . . . .
11
1.5
Relations Between Physics and Other Sciences . . . . . . . . . . . . .
14
1.6
The Basic Units in Physics, Their Standards and Measuring Techniques 16
1.7
Systems of Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
26
1.8
Accuracy and Precision; Measurement Uncertainties and Errors . . . .
27
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
34
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
35
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
35
© Springer International Publishing Switzerland 2017
W. Demtröder, Mechanics and Thermodynamics, Undergraduate Lecture Notes in Physics, DOI 10.1007/978-3-319-27877-3_1
Chapter 1
1
Introduction and Survey
1
2
1 Introduction and Survey
Chapter 1
The name “Physics” comes from the Greek (“'i” D
nature, creation, origin) which comprises, according to the definition of Aristotle (384–322 BC) the theory of the material
world in contrast to metaphysics, which deals with the world of
ideas, and which is treated in the book by Aristotle after (Greek:
meta) the discussion of physics.
Definition
The modern definition of physics is: Physics is a basic science, which deals with the fundamental building blocks of
our world and the mutual interactions between them.
The goal of research in physics is the basic understanding
of even complex bodies and their composition of smaller
elementary particles with interactions that can be categorized into only four fundamental forces. Complex events
observed in our world should be put down to simple
laws which allow not only to explain these events quantitatively but also to predict future events if their initial
conditions are known.
In other words: Physicists try to find laws and correlations for our world and the complex natural events and to
explain all observations by a few fundamental principles.
Note, however, that complex systems that are composed of
many components, often show characteristics, which cannot be
reduced to the properties of these components. The amalgamation of small particles to larger units brings about new and
unforeseen characteristics, which are based on cooperative processes. The whole is more than the sum of its parts (Heisenberg
1973, Aristotle; metaphysics VII). Examples are living biological cells, which are composed of lifeless molecules or molecules
with certain chemical properties consisting of atoms that do not
show these properties of the molecule.
The treatment of such complex systems requires new scientific
methods, which have to be developed.
This should remind enthusiastic physicists, that physics alone
might not explain everything although it has been very successful to expands the borderline of its realm farther and farther in
the course of time.
1.1
The Importance of Experiments
The more astronomically oriented observations of ancient Babylonians brought about a better knowledge of the yearly periods
of the star sky. The epicycle model of Ptolemy gave a nearly
quantitative description of the movements of the planets. However, modern Physics in the present meaning started only much
later with Galileo Galilei (1564–1642, Fig. 1.1), who performed
as the first physicist well planned experiments under defined
conditions, which could give quantitative answers to open questions. These experiments can be performed at any time under
conditions chosen by the experimentalist independent of external influences. This distinguishes them from the observations
of natural phenomena, such as thunderstorms, lightening or volcanism, which cannot be influenced. This freedom of choosing
the conditions is the great advantage of experiments, because
all perturbing external influences can be partly or even completely eliminated (e. g. air friction in experiments on free falling
bodies). This facilitates the analysis of the experimental results
considerably.
Experiments are aimed questions to nature, which yield
under defined conditions definite answers.
The goal of all experiments is to find reasons and causes for all
phenomena observed in nature, to see connections between the
manifold of observations and to categorize them under a common law. Even more ambitious is the quantitative prediction of
future experimental results, if the initial conditions of the experiments are known.
A physical law connects measurable quantities and concepts. Its clear form is a mathematical equation.
Such mathematical descriptions give a clearer insight into the
relations between different physical laws. It can reduce the manifold of experimental findings, which might seem at first glance
uncorrelated but turn out to be special cases of the same general
law that is valid in all fields of physics.
Examples
1. Based on many careful measurements of planetary orbits by Tycho de Brahe (1546–1601), Johannes Kepler
(1571–1630) could postulate his three famous laws
for the quantitative description of distances and movements of the planets. He did not find the cause for
these movements, which was discovered only later by
Isaac Newton (1642–1727) as the gravitational force
between the sun and the planets. However, Newton’s
gravitation law did not only describe the planetary orbits but all movements of bodies in gravitational fields.
The problem to unite the gravitational force with the
other forces (electromagnetic, weak and strong force)
has not yet been solved, but is the subject of intense
current research.
2. The laws of energy and momentum conservation were
only found after the analysis of many experiments in
different fields. Now they explain and unify many
experimental findings. Such a unified summary of
different physical laws and principles to a consistent
general description is called a physical theory.
J
3
Chapter 1
1.2 The Concept of Models in Physics
Figure 1.1 Left: Galileo Galilei. Right: Looking of Cardinales through Galilio’s Telescope
Its range of validity and predictive capability is checked
by experiments.
Since the formulation of a theory requires a mathematical
description, a profound knowledge of basic mathematics is indispensible for every physicist.
1.2
The Concept of Models in
Physics
The close relation between theory and experiments is illustrated
by the following consideration:
If a free falling body in a vacuum container at the surface of
the earth is observed one finds that the fall time over a definite
distance is independent of the size or form of the body and also
independent of its weight. In contrast to this result is the fall
of a body in any fluid, instead of vacuum where the form of
the body does play a role because here perturbing influences,
such as friction often cannot be neglected. Neglecting these perturbations one can replace the body by the model of a point
mass. With other words: In these experiments the falling body
behaves like a point mass, because its size does not matter. The
theory can now give a complete description of the movement of
point masses under the influence of gravitational forces and it
can predict the results of corresponding future experiments (see
Chap. 2).
Now the experimental conditions are changed: For a body
falling in water the velocity and fall time do depend on size
and weight of the body, because of friction and buoyancy. In
this case the model of a point mass is no longer valid and has to
be broadened to the model of spatially extended rigid bodies
(see Chap. 5). This model can predict and quantitatively explain
the movements of extended rigid bodies under the influence of
external forces.
If we now further extend our experimental condition and let a
massive body fall onto a deformable elastic steel plate, our rigid
body model is no longer valid but we must include in our model
the deformation of the body, This results in the model of extended deformable bodies, which describes the interaction and
the forces between different parts of the body and explains elasticity and deformation quantitatively (see Chap. 6).
The theory of phenomena in our environment is always the
description of a model, which describes the observations.
If new phenomena are discovered which are not correctly
4
1 Introduction and Survey
Chapter 1
represented by the model, it has to be broadened and refined or even completely revised.
REDUCTION due to natural laws
mathematical formulation
Simple description
The details of the model depend on the formulation of the question asked to nature and on the kind of experiments which
should be explained. Generally a single experiment tests only
certain statements of the model. If such an experiment confirms
these statements, we say, that nature behaves in this experiment
like the model predicts, i. e. nature gives the same answer to selected experiments as the model.
Since theory can in principle calculate all properties of an accepted model it often gives valuable hints, which experiments
could best test the validity of the model.
SEVERAL
MODELS
An impressive example is the development of quantum chromodynamics. This modern theory describes the substructure
of particles, which had been regarded as elementary, such as
protons, neutrons and mesons, but are really composed of still
smaller particles, the quarks. Theoretical predictions about the
possible masses of unstable particles, composed of these quarks,
which appear as resonances in the collision cross sections, allowed the experimentalists to restrict their search which is like
the search for a needle in the haystack, to the predicted energy
range, which facilitated their efforts considerably.
The model concept for the description of observations in nature
is in particular obvious in the world of microphysics (atomic,
molecular and nuclear physics), because here the particles cannot be seen with the naked eye and therefore a vivid picture
cannot be given. Attempts to transfer vivid models useful in
macrophysics to microphysics have often led to misunderstandings and wrong ideas. One example is the particle-wave dualism
for the description of microparticles (see Vol. 3).
Figure 1.2 comprises the discussion above. One example shall
illustrate the development and refinement of models in physics.
The explanation of lightning by Greek philosophers was the
god Zeus who flung flashes to the earth while he was in a furious mood. Modern models explain lightning by the separation
of positive and negative electrical charges by charged water
drops floating in turbulent air, leading to large electric voltages
between different clouds or between clouds and earth with resulting strong discharges. This modern model is based on many
detailed observations with high speed photographic instruments
and on experimental simulations of lightning in high voltage
laboratories where discharges can be observed under controlled
conditions.
The goal of sciences is the understanding of natural phenomena observed under different conditions and to categorize their
FEW BASIC
MODELS
Refined
MODEL
Natural laws,
observational facts, intuition
PROBLEM
DEFINITION
EXPERIMENT to
verify or falsify the
models
Sophisticated
but elegant
Observation of
natural
phenomena
Perception with
sensory organs
or instruments
Such a cooperation and mutual inspiration of theoretical
and experimental physics contribute in an outstanding way
to the progress in physical knowledge.
REDUCTION due to
experiments
MODEL OF
REALITY
Curiosity, reconsideration
of basic observations
REALITY
Figure 1.2 Schematic representation of the way, how scientists gain information on nature
explanations under a common law. It is assumed, that the observed reality exists independent of the observer. However, the
experiments performed in order to reproduce the observations
demand nevertheless characteristic features of the observing
subject, such as imagination for the planning of decisive experiments, an open mind for new ideas, etc. Many ideas turn
out to be wrong. They can be already excluded by comparison
with former experiments. Such ideas which do not contradict
already existing knowledge can contribute to a working hypothesis. Even such a hypothesis might be only partly correct and
has to be modified by the results of further experiments. If all
these results confirm the working hypothesis it can become a
proved theory, which allows us to summarize many observations to a general law (see Fig. 1.3).
This procedure where a theory is built up from many experimental results is called the inductive method.
In theoretical physics often a reverse procedure is chosen. The
starting point are fundamental basic equations such as Newton’s
law of gravitation or the Maxwell equations or symmetry laws.
From these general laws the outcome of possible experiments is
predicted (deductive method).
Both procedures have their justification with advantages and
drawbacks. They supplement each other.
An important aspect which one should keep in mind is summarized in the following fundamental statement:
Physics describes objective and as accurate as possible the reality of the material world. For human beings this is, however,
only a small section of the world we experience, as a specific
example illustrates: From the standpoint of physics a painting
can be described, by giving for each point .x; y/ the reflectivity
R.; x; y/, which depends on the wavelength , the spectrum of
Reality
Objective
physical law
Mathematics
Imagination
of physicist
Idea
Working hypothesis
Theory
Observation
Experiment
Results of
measurements
Subjective
interpretation
Objective knowledge about reality
Figure 1.3 Schematic diagram of gaining insight into natural phenomena
the illuminating radiation source and the angles of incidence and
observation direction. A computer which is fed with these characteristic input data can reproduce the painting very accurately.
Nevertheless this physical description lacks an essential part of
the painting, which is in the mind of the observer. When looking at the painting a human being might remember other similar
paintings which he compares with the present painting, even if
these other paintings are not present but only in the mind they
still change the subjective impression of the observer. The subject of the painting may induce cheerful or sad feelings in the
mind of the observer, it may call back remembrances of former
events or impressions which are related to this painting. All
these different influences will determine the judgement about
the painting, which therefore might be different for different observers.
All these aspects are not the realm of physics, because they
are subjective, although they are essential for the quality of the
painting as judged by human beings and they represent an important part of the “reality” as perceived by us.
These remarks should warn physicists, not to forget that our fascinating science is only competent for the description of the
material basis of our world. Although the other nonmaterial
realms are based on the material world their description and
understanding reaches far beyond physics. The question, how
living cells are built from inanimate molecules and how the
human mind is related to the structure of the brain are still pending but exciting problems, which might be solved in the future.
This is related to the question whether the human brain is more
than a highly developed computer, which is the subject of hot
discussions between the supporter of artificial intelligence and
biologists.
For more detailed discussions of these questions, the reader is
referred to the literature [1.1a–1.6].
1.3
Short Historical Review
The historical development of physics can be roughly divided
into three periods:
The natural philosophy in ancient times
The development of classical physics
The modern physics.
1.3.1
The Natural Philosophy in Ancient Times
The investigation of natural phenomena and the efforts to explain them by rational arguments started already 4000 years
ago. The astronomical observation of the Babylonian and the
Egyptian scientists were important for the prediction of annual occurrences, such as the Nile flood or the correct time
for sowing. The Greek philosophers produced many ideas for
the explanation of the observed natural phenomena. All these
ideas were treated within the framework of general philosophy.
For example, the textbook on Physics ('i ˛%o˛i& D
lectures on physics) by Aristotle contains mainly philosophical
considerations about space and time, movements of bodies and
their causes.
Probably the most important achievement of Greek philosophy
was the overcoming of the widespread mythology, where the
life of mankind was governed by a hierarchy of gods, whose
mood was not predictable and everybody had to win the liking of gods by sacrificing precious gifts to them. Most Greek
philosophers abandoned the belief, that the world was a playing
ground for gods, demons and ghosts who generated thunderstorm, floods, sunshine or disastrous droughts just according to
their mood (see Homer’s Odyssey).
The Greek philosophers believed that all natural phenomena
obeyed eternal unchanging laws which were not always obvious
5
Chapter 1
1.3 Short Historical Review
6
1 Introduction and Survey
Chapter 1
Anaxagoras (499–428 BC) was the first to postulated that the
world consists of many infinitely small different particles. The
force which keeps them together is the Nus (D world spirit).
Leucippus (489–428 BC) and his student Democritus (455–
370 BC) followed these ideas and refined this hypothesis. Democritus assumed that the world consists of atoms (˛oo& D
indivisble), very small indivisible identical particles, which
move forever in an infinite empty space. The different forms
of matter differ only by the number and arrangement of atoms
of which they are composed. This hypothesis comes close to our
present understanding of the atomic composition of the different
elements in the periodic table (see Sect. 1.4).
The doctrine of the “atomists” was declined by Plato (427–
347 BC) and Aristotle (Fig. 1.4) since it contradicted their view
of a continuous world. Since these two philosophers had such
a great reputation the atomistic theory was forgotten for nearly
2000 years.
Aristotle (384–322 BC) (Fig. 1.4) regarded nature as the forever
moving and developing universe, where at the beginning a “divine mover” was assumed who started the whole world. The
planets move apparently without obvious mover and therefore
Aristotle assumed that they do not consist of the four earthly
elements fire, air, water and soil but of a fifth “divine element”
which he called “Ether”. This ether should be massless and elastic and should penetrate the whole world, including rigid bodies.
Figure 1.4 Aristotle. With kind permission of the “Deutsches Museum”
because of the complex nature but which were independent of
men or gods. This means that it is, at least in principle, possible
to find such laws merely by human reason.
Example
A solar eclipse is no longer described by a monster that
engulfs the sun, but by the temporarily blocking of the
sunlight by the moon. This changes the solar eclipse from
an accidental event to a predictable occurrence.
J
Famous representatives of Greek philosophy were Thales from
Milet (624–546 BC), who discovered magnetism and frictional
electricity, but could not correctly explain his findings. Empedocles (495–435 BC) assumed that fire, water, air and soil formed
the four basic elements, which can mix, divide and build compositions from which all other material is composed. The
mathematical aspect of natural phenomena was introduced by
Pythagoras (572–492 BC) and his scholars who assumed that
numbers and mathematical relations between these numbers reflect the reality. They made acoustic experiments with striking
chords of different lengths and measured the resulting tones.
However, they erroneously generalized their results to other
fields such as the movement of the planets.
Archimedes (287–212 BC) studied in Alexandria, the centre of
science at that time. Later he moved to Syracuse on Sicily. He
was the greatest mathematician, physicist and technical expert
of his time. He succeeded to calculate the area and the perimeter of a circle, the surfaces of spheres, cones and cylinders
and he solved third order equations. As a physicist he determined the centre of mass for bodies of different shape, he found
the lever principle, calculated the buoyancy of bodies in water
(Archimedes’ principle), he built a planetarium and measured
star positions and proved the curvature of the sea surface. He
was famous for his technical achievements. He invented and
constructed about 40 different machines, such as the worm gear
drive, catapults, hydraulic levers for lifting ships and many machines used for warfare.
In spite of great success in many fields the Greek philosophers
could not reach natural science in the present sense, because
they did not accept the experiment as the touchstone for every
theory. They believed that an initial observation was sufficient
and that all subsequent conclusions and knowledge could be
achieved by pure thinking without further confirming or disproving experiments.
This rather speculative procedure has influenced, due to the
great impact of Aristotle’s generally accepted teaching, many
generations of philosophers for more than 1500 years. Even
when Galilei Galileo observed through his telescope the four
moons of Jupiter, most philosophers and high members of the
church did not believe him, because his observation contradicted
the theory of Aristotle, who taught that the planets were fixed
on crystal spheres moving with the planet around the earth. If
moons circled around Jupiter they had to penetrate these crystal
spheres and would smash them. Therefore, the moons should
1.3 Short Historical Review
Chapter 1
be impossible. Even when Galilei offered to the sceptics to look
through the telescope (Fig. 1.1b) many of them refused and said:
“Why should we look and be deceived by optical illusions when
we are sure about Aristotle’s statements”.
Although some inconsistencies in Aristotle’s teaching had been
found before, Galilee was the first to disprove by his observations and experiments the whole theory of the shining example
of Greek philosophy, in particular when he also advertised the
new astronomy of Copernicus, which brought him many enemies and even a trial before the catholic court.
1.3.2
The Development of Classical Physics
One may call Galileo the first physicist in the present meaning.
He tried as the first scientist to prove or disprove physical theories by specific well-planned experiments. Famous examples
are his experiments on the movement of a body with constant
acceleration under the influence of gravity. He also considered
how large the accuracy of his experimental results must be in order to decide between two different versions for the description
of such movements. He therefore did not choose the free fall
(it is often erroneously reported, that he observed bodies falling
from the Leaning tower in Pisa). This could never reach the required accuracy with the clocks available at that time. He chose
instead the sliding of a body on an inclined plane with an angle
˛ against the horizontal. Here only the fraction g sin ˛ acts on
the body and thus the acceleration is much smaller.
His astronomical observations (phases of Venus, Moons of
Jupiter) with a self-made telescope (after he had learned about
its invention by the optician Hans Lipershey (1570–1619) in
Holland) helped the Copernican model of the planets circling
around the sun instead of the earth, finally to become generally
accepted (in spite of severe discrepancies with the dogmatic of
the church and heavy oppression by the church council).
The introduction of mathematical equations to physical problems, which comprises several different observations into a
common law, was impressively demonstrated by Isaac Newton (Fig. 1.5). In his centennial book “Philosophiae Naturalis
Principia Mathematica” he summarizes all observations and
the knowledge of his time about mechanics (including celestial mechanics D astronomy) by reducing them to a few basic
principles (principle of inertia, actio D reactio, the force on a
body equals the time derivative of his momentum and the gravitational law).
Supported by progress of mathematics in the 17th century (analytical geometry, infinitesimal calculus, differential equations)
the mathematical description of physical observations becomes
more and more common. Physics emancipates from Philosophy
and develops its own framework using mathematical language
for the clear formulation of physical laws. For example classical
mechanics experiences its complete and elegant mathematical form by J. L. de-Lagrange (1736–1813) and W. R. Hamilton
(1805–1865) who reduced all laws for the movement of bodies
under arbitrary forces to a few basic equations.
7
Figure 1.5 Sir Isaac Newton. With kind permission of the “Deutsches Museum
München”
Contrary to mechanics which had developed already in the 18th
century to a closed complete theory the knowledge about the
structure of matter was very sketchy and confused. Simultaneously different hypotheses were emphasized: One taken form
the ancient Greek philosophy, where fire, water, air and soil
were assumed as the basic elements, or from the alchemists who
favoured mercury, sulphur and salt as basic building blocks of
matter.
Robert Boyle (1627–1591) realized after detailed experiments
that simple basic elements must exist, from which all materials can be composed, which however, cannot be further divided.
These elements should be separated by chemical analysis from
their composition. Boyle was able to prove that the former assumption of elements was wrong. He could, however, not yet
find the real elements.
A major breakthrough in the understanding of matter was
achieved by the first critically evaluated quantitative experiments investigating the mass changes involved in combustion
processes, published in 1772 by A. L. de Lavoisier (1743–1794).
These experiments laid the foundations of our present ideas
about the structure of matter. Lavoisier and John Dalton (1766–
1844) recognised metals as elements and postulated like Boyle
that all substances were composed of atoms. The atoms were
now, however, not just simple non-divisible particles, but had
8
1 Introduction and Survey
Chapter 1
specific characteristics which determined the properties of the
composed substance. Karl Wilhelm Scheele (1724–1786) found
that air consisted of nitrogen and oxygen.
Antoine-Laurent Lavoisier furthermore found that the mass of
a substance increased when it was burnt, if all products of the
combustion process were collected. He recognized that this
mass increase was caused by oxygen which combined with the
substance during the burning process. He formulated the law
of mass conservation for all chemical processes. Two elements
can combine in different mass ratios to form different chemical
products where the relative mass ratios always are small integer
numbers.
The British Chemist John Dalton was able to explain this law
based on the atom hypothesis.
Examples
1. For the molecules carbon monoxide and carbon dioxide the mass ratio of oxygen combining with the same
amount of carbon is 1 W 2 because in CO one oxygen
atom and in CO2 two oxygen atoms combine with one
carbon atom.
2. For the gases N2 O (Di-Nitrogen oxide), NO (nitrogen
mono oxide), N2 O3 (nitrogen trioxide), and NO2 ) nitrogen dioxide) oxygen combines with the same mass
J
of nitrogen each time in the ratio 1 W 2 W 3 W 4.
Dalton also recognized that the relative atomic weights constitute a characteristic property of chemical elements. The
further development of these ideas lead to the periodic system
of elements by Julius Lothar Meyer (1830–1895) and Dimitri
Mendelejew (1834–1907), who arranged all known elements in
a table in such a way that the elements in the same column
showed similar chemical properties, such as the alkali atoms in
the first column or the noble gases in the last column.
Why these elements had similar chemical properties was recognized only much later after the development of quantum theory.
The idea of atoms was supported by Amedeo Avogadro (1776–
1856), who proposed in 1811 that equal volumes of different
gases at equal temperature and pressure contain an equal number
of elementary particles.
A convincing experimental indication of the existence of atoms
was provided by the Brownian motion, where the random movements of small particles in gases or liquids could be directly
viewed under a microscope. This was later quantitatively
explained by Einstein, who showed that this movement was induced by collisions of the particles with atoms or molecules.
Although the atomic hypothesis scored indisputable successes
and was accepted as a working hypothesis by most chemists
and physicists, the existence of atoms as real entities was a matter of discussion among many serious scientists until the end of
the 19th century. The reason was the fact that one cannot see
atoms but had only indirect clues, derived from the macroscopic
behaviour of matter in chemical reactions. Nowadays the improvement of experimental techniques allows one to see images
of single atoms and the theoretical basis of atomic theory leaves
no doubt about the real existence of atoms and molecules.
The theory of heat began to become a quantitative science after
thermometers for the measurement of temperatures had been developed (air-thermoscope by Galilei, alcohol thermometer 1641
in Florence, mercury thermometer 1640 in Rome). The Swedish
physicist Anders Celsius (1701–1744) introduced the division
into 100 equal intervals between melting point (0 ı C) and boiling point (100 ı C) of water at normal pressure. Lord Kelvin
(1824–1907) postulated the absolute temperature, based on gas
thermometers and the general gas law. On this scale the zero
point T D 0 K D 273:15 ı C is the lowest temperature which
can be closely approached but never reached (see Chap. 10).
Denis Papin (1647–1712) investigated the process of boiling
and condensation of water vapour (Papin’s steam pressure pot).
He built the first steam engine, which James Watt (1736–1819)
later improved to reliable technical performance. The terms
amount of heat and heat capacity were introduced by the English physicist and chemist Joseph Black (1728–1799). He
discovered that during the melting process heat was absorbed
which was released again during solidification.
The more precise formulation of the theory of heat was essentially marked by establishing general laws. Robert Mayer
(1814–1878) postulated the first law of the theory of heat, which
states that for all processes the total amount of energy is conserved. Nicolas Carnot (1796–1832) started 1831 after some
initial errors a fresh successful attempt to describe the conversion of heat into mechanical energy (Carnot’s cycle process).
This was later more precisely formulated by Rudolf Clausius
(1822–1888) in the second law of heat theory.
A real understanding of heat was achieved, when the kinetic
gas theory was formulated. Here the connection between heat
properties and mechanical energy was for the first time clearly
formulated. Since the dynamical properties of molecules moving around in a gas were related to the temperature of a gas, the
heat theory was now called thermodynamics, which was formulated by several scientists (Clausius, Avogadro, Boltzmann)
(see Fig. 1.6). They proved under the assumption that gases consist of many essentially free atoms or molecules, which move
randomly around and collide with each other, that the heat energy of a gas is equivalent to the kinetic energy of these particles.
The Austrian physicist Joseph Loschmidt (1821–1895) found
that under normal pressure the gas contains the enormous number of about 3 1019 atoms per cm3 .
Optics is one of the oldest branches of physics which was already studied more than 2000 years ago where the focussing of
light by concave mirrors was used to ignite a fire. However,
only in the 17th century optical instruments and their imaging properties were studied systematically. A milestone was
the fabrication of lenses and the invention of telescopes. Willibrord Snellius (1580–1626) formulated his law of refraction
(see Vol. 2, Chap. 9), Newton found the separation of different colours when white sun light passed through a prism. The
explanation of the properties of light was the subject of hot discussions. While Newton believed that light consisted of small
particles (in our present model these are the photons) the experiments on interference and diffraction of light by Grimaldi
9
Chapter 1
1.3 Short Historical Review
Figure 1.6 Ludwig Boltzmann. With kind permission from Dr. W. Stiller Leipzig
(1618–1663), Christiaan Huygens (1629–1695), Thomas Young
(1773–1829) and Augustin Fresnel (1788–1827) decided the
dispute in favour of the wave theory of light. Melloni showed
1834 that the laws for visible light could be extended into the
infrared region and Max Felix Laue (1879–1960) and William
Bragg (1862–1942) demonstrated the wave character of X-rays,
which had been discovered by Conrad Roentgen (1845–1923),
by their famous experiments on X-ray diffraction in crystals.
The velocity of light was first estimated by Ole Rømer (1644–
1710) by astronomical observations of the appearance time of
Jupiter moons and later more precisely determined by Huygens. With measurements on earth Jean Foucault (1819–1868)
and Armand Fizeau (1819–1896) could obtain a rather accurate
value for the velocity of light.
William Gilbert (1544–1603) was called “the father of electricity”. He investigated the magnetic field of permanent magnets
and measured the magnetic field of the earth with the help of
magnetic needles. He made extensive experiments on friction
electricity and divided the different materials into electrical and
non-electrical substances. He built the first electroscop and
measured the forces between charged particles. Stephen Gray
(1670–1736) discovered the electrical conductivity of different
materials and made detailed experiments on electric induction.
He made electricity very popular by spectacular demonstrations.
Figure 1.7 James Clerk Maxwell. With kind permission from the American
Institute of Physics, Emilio Segre Visual hives, College Park MD
Charles Augustin Coulomb (1736–1806) built the first electrometer, constructed the Coulomb torsion balance and formulated
the famous Coulomb law for the forces between charged particles. Benjamin Franklin (1706–1790) recognized that lightening
is not a fire but an electrical discharge and constructed the first
lightning conductor. Luigi Galvani (1737–1798) discovered the
stimulation of nerves by electrical currents (frog’s leg experiments); and the contact voltage between different conductors,
which lead to the construction of batteries (Galvanic element).
Allessandro Volta (1745–1827) continued the experiments of
Galvani and he categorized the different metals in an electrochemical series.
Hans Christean Oersted (1777–1851) discovered the magnetic
field of an electric current. Andre Marie Ampere (1775–1836)
coined the terms “electrical current” and electrical voltage. By
many detailed experiments, he established modern electrodynamics.
Michael Faraday (1791–1867) performed basic experiments on
the relations between electric currents and magnetic fields (Faraday’s induction law). He prepared the foundations for the
development of alternating currents and their applications.
James Clerk Maxwell (1831–1879) (Fig. 1.7) summarized all
known results of former experiments by a few basic equations
(Maxwell’s equations) and gave them a general mathematical
10
1 Introduction and Survey
Chapter 1
formulation, which represents the basis for electrodynamics and
optics. Their solutions are electro-magnetic waves, which found
a brilliant confirmation by the experiments of Heinrich Hertz
(1857–1894), who showed that these waves were transversal and
propagate in space with the velocity of light.
1.3.3
Modern Physics
At the end of the 19th century, all problems in physics seemed
to be solved and many physicists believed, that a closed theory
describing all known facts could be realized in the near future.
This optimistic opinion changed, however, in a dramatic way,
induced by the following experimental findings.
The Michelson experiment (see Sect. 3.4) showed without
doubt, that the velocity of light is constant, independent of
the direction or the velocity of the observer. This result was
in sharp contrast to former concepts and induced Albert Einstein (Fig. 1.8) to formulate his theory of special relativity
(see Sect. 3.6).
Experimentally found deviations from the theoretically expected spectral intensity distribution of the thermal radiation
of hot bodies, as calculated by Stephan Boltzmann and Wilhelm Wien, could not be explained by classical physics. This
Figure 1.9 Max Planck. With kind permission of the “Deutsches Museum
München”
Figure 1.8 Albert Einstein. With kind permission of the “Deutsches Museum
München”
discrepancy led Max Planck (1858–1947) (Fig. 1.9) to the
conclusion of quantized energy of radiation fields. This bold
assumption, which could perfectly reproduce the experimental results, represented the beginning of quantum theory that
was later on imbedded in a concise mathematical framework
by Erwin Schrödinger (1887–1961) and Werner Heisenberg
(1901–1976) (see Vol. 3). The concept of energy quanta
was further supported experimentally by the photoelectric
effect, which was quantitatively explained by Einstein, who
received the Nobel Prize for his theory of the photo-effect
(not for his theory of relativity!).
New experimental techniques allowed investigating the
structure of atoms and molecules. The light emitted from
atoms or molecules could be sent through a spectrograph
and showed discrete lines, indicating that it has been emitted from discrete energy levels. Through the development of
spectral analysis by Gustav Robert Kirchhoff (1824–1887)
and Robert Bunsen (1811–1899) it was found that atoms of
a specific element emitted spectral lines with wavelengths
characteristic for this element. The results could not be explained by classical physics but needed quantum theory for
their interpretation. Today the physics of atomic electron
shells and their energy levels can be completely described
by a closed theory called quantum-electrodynamics.
1.4 The Present Conception of Our World
The properties of atomic nuclei could be only investigated after appropriate detectors had been developed. Nuclear physics
is therefore a rather new field where most of the results were
obtained in the 20th century. The substructure of atomic nuclei
and the physics of elementary particles could start after particle accelerators could be operated and many results in this field
have been achieved only recently.
Elementary Particles
The entire material world known up to now is composed of
only a few different particles. The three most important are the
electron (e ), the proton (pC ) and the neutron (n). All other
elementary particles (muons, -Mesons, Kaons, -particles
etc.) exist after their production only a very short time (10 6 –
10 15 s). They convert either spontaneously or by collisions into
other particles which finally decay into pC , e , neutrinos or photons h. Although neutrinos are stable particles they show such
a small interaction with matter that they are difficult to detect
and they therefore play no role in daily life.
Recent experiments and theoretical consideration have shown,
that the particles pC , n, mesons and hyperons, which had been
regarded as elementary, show a substructure (see Vol. 4), According to our present understanding they consists of smaller
particles, called “quarks”, which occur in 6 different species.
All building blocks of matter can be divided into two groups:
This short historical review should illustrate that many concepts
which today are taken for granted, are not as old and have been 1. the quarks, which build up the heavy particles (baryons),
accepted only after erroneous ideas and a long way of successive
such as proton, neutron, mesons and hyperons
corrections, guided by new experiments. It is worthwhile for 2. the light particles (leptons) electron, myon and neutrino.
every physicist to look into some original papers and follow the
gradual improvements of concepts and representation of results. Each of these two groups consists of three families of elementary particles, which are listed in Tab. 1.1. For each of these
More extensive literature about the historical development of particles there exists an anti-particle with equal mass but oppophysics and about bibliographies of physicists can be found in site charge. For instance the anti-particle of the electron e is the
positron eC , the proton pC has as anti-particle the anti-proton p
the references [1.6–1.14c].
and the anti-neutron has the same mass and the charge zero as
the neutron.
1.4
The Present Conception of Our
World
As the result of all experimental and theoretical investigations
our present model of the material world has been established
(Fig. 1.10). In this introduction, we will give only a short summary. The subject will be discussed more thoroughly in Vol. 3
and 4 of this textbook series.
Macroscopic bodies
(solid, liquid and gaseous)
Atoms
Molecules
Nuclei and electrons
Atomic nuclei
(protons + neutrons)
Quarks, gluons
Figure 1.10 Build up of our material world (H. J. Jodl [1.14b])
According to present theories the interaction between the particles can be described by the exchange of “interaction particles”,
which are called the quanta of the interaction field. For example the quanta of the electromagnetic field, which determine the
interaction between charged particles are the photons h .
The quanta of the strong interaction between nucleons are called
gluons. The gravitons are the quanta of the gravitational field.
Our present knowledge is that there exist only four different
kinds of interaction, which are summarized in Tab. 1.2.
An essential goal of present research is to reduce the four types
of interaction to one common force (grand unification). The reduction of the manifold of different particles to two groups of
elementary particles was in a certain sense successful, because
the classification into two groups with three families in each
group gives a rather simple arrangement. However, the number of 24 different particles together with their antiparticles is
still large and adding the 15 interaction quanta the total number
of elementary particles is 39. Whether the “grand unification”
will allow a further reduction or a simpler ordering scheme is
still an open question.
This field of research is very interesting because it ventures to
the limit where matter and energy might become indistinguishable. It is also closely related to processes occurring at the very
beginning of our universe where elementary particles and their
interaction played a major role in the extremely hot fireball during the first seconds of the big bang.
Chapter 1
This illustrates that always in the history of natural sciences new
experimental results forced physicists to revise former concepts
and to formulate new theories which, however, should include
proved earlier results. In most cases the old theories were not
completely abandoned but their validity range was restricted and
more precisely characterized. For example the classical physics
is perfectly correct for the description of the motion of macroscopic bodies or for many applications in daily life, while for
the description of the micro-world of atoms and molecules it
may completely fail and quantum theory is necessary.
11
12
1 Introduction and Survey
Chapter 1
Table 1.1 The three families of Leptons and Quarks
Leptons
Name
Electron
Symbol
e
Mass MeV=c2
0.51
Electron neutrino
āe
< 10
Myon
Myon neutrino
ā
< 10
Tau-lepton
£
1840
Tau-neutrino
ā
< 10
5
105:66
4
4
Quarks
Name
Up
Symbol
u
0
Down
d
Charge
1
1
Charm
c
0
Strange
s
1
Top
t
0
Bottom
b
Table 1.2 The four types of interaction between particles (known up to now)
and their field quanta. There are 8 gluons, 2 charged (WC and W ) W bosons,
1 neutral boson (Z0 ) and probably only 1 graviton with spin I D 2
Interaction
Strong interaction
El. magn. interaction
Weak interaction
Gravitational interaction
Field quantum
Gluons
Photons
W bosons
Z bosons
Gravitons
Rest mass MeV=c2
0
0
81;000
91;010
0
Atomic Nuclei
Protons and Neutrons can combine to larger systems, the atomic
nuclei. The smallest nucleus is the proton as the nucleus of the
hydrogen atom. The largest naturally existing nucleus is that of
the uranium atom with 92 protons and 146 neutrons. Its diameter is about 10 14 m. Besides the nuclei found in nature there
are many artificially produced nuclei, which are however, generally not stable but decay into other stable nuclei. Nearly every
atom has many isotopes with nuclei differing in the number of
neutrons. Meanwhile there is a wealth of information about the
strong attractive forces, which keep the protons and neutrons
together in spite of the repulse electrostatic force between the
positively charged protons.
Atoms
Atomic nuclei together with electrons can form stable atoms,
where for neutral atoms the number of electrons equals the number of protons. The smallest atom is the hydrogen atom, which
consists of one proton and one electron. The diameter of atoms
ranges from 5 10 11 m to 5 10 10 m and is about 10;000 times
larger than that of the nuclei, although the mass of the nuclei
is about 2000 times larger than that of the electrons. The electrons form a cloud of negative charge around the nucleus. The
electro-magnetic interaction between electrons and protons has
been investigated in detail and there is a closed theory, called
quantum electrodynamics, which describes all observed phenomena of atomic physics very well.
The chemical properties of the different atoms are completely
determined by the structure of the atomic electron shell. This is
illustrated by the periodic system of the elements (Mendelejew
1869, Meyer 1870), where the elements are arranged in rows
and columns and ordered according to the number of electrons
of the atoms (see Vol. 3). With each new row a new electron
Mass MeV=c2
300
Charge
2=3
306
1=3
450
1=3
4300
1=3
1200
1:7 105
2=3
2=3
shell starts. In each column the number of electrons in the outer
shell (valence electrons) is equal and the chemical properties of
the elements in the same column are similar. A real understanding of the periodic table could only be reached 60 years later
after the quantum theory of atomic structure had been developed.
Molecules
Two or more atoms can combine to form a molecule, where the
atoms are held together by electro-magnetic forces. The magnitude of the binding energy depends mainly on the electron
density between the nuclei. Biological molecules such as proteins or DNA-molecules may consist of several thousand atoms
and have diameters up to 0:1 µm, which is about 1000 times
larger than the hydrogen atom. Molecules form the basis of
all chemical and biological substances. The properties of these
substances depend on the kind and structure of the molecules,
such as the geometrical arrangement of the atoms forming the
molecule.
Macroscopic Structures, Liquids and Solid Substances
Under appropriate conditions many equal or different atoms
can form large macroscopic bodies which can contain a huge
number of atoms. Depending on temperature they can exist in
the solid or liquid phase. The interaction between the atoms
is in principle known (el. magn. forces) but difficult to calculate because of the enormous number of participating atoms
(1022 =cm3 ). Most theoretical treatments therefore use statistical
methods. Up to now many characteristics of macroscopic bodies
can be calculated and understood from their atomic structure but
a general exact theory of liquids and solids, which can explain
also finer details, is still not available. Therefore approximations
are used where each approximate model can describe special
features quite well but others less satisfactorily. Examples are
the band structure model, which can explain the electrical conductivity but not as well the elastic properties.
Structure and Dynamics of Our Universe
In our universe all of the constituents discussed so far are
present.
Free elementary particles (pC , n, e , photons h, also short
lived mesons in the cosmic radiation, in the atmosphere of
stars and in hot interstellar clouds, in the hot fireball during
some minutes after the big bang, of our universe).
1.4 The Present Conception of Our World
Macroscopic view
Microscopic view
(matter = system
of atoms and molecules)
ELECTRODYNAMICS
SOLIDSTATE
PHYSICS
OPTICS
Molecules as
compounds
of atoms
PHYSICS OF
LIQUIDS
AND GASES
Electro
magne
tic radia
tion
MECHANICS
Chapter 1
MATTER
MOLECULAR
PHYSICS
Single atoms consisting of electron shell nucleus
ATOMIC
PHYSICS
ION PHYSICS
Single nuclei as system of elementary particles
NUCLEAR
PHYSICS
Single particle
ELEMENTARY
PARTICLE
PHYSICS
Figure 1.11 Family tree of physics (with kind permission of Dr. H. J. Jodl) [1.14b]
Atomic nuclei in the inner part of stars, in neutron stars and
in hot gas clouds.
Atoms in atmospheres of planets and stars and in the interstellar medium.
Molecules in molecular clouds, in comet tails, in interstellar
space, in the atmospheres of cold stars and of planets.
Solid and liquid macroscopic bodies (in planets and moons,
in meteorites).
For the understanding of the origin and the development of
our universe the interactions between these particles have to be
known. Although in the early stage of the universe and later on
in the interior of stars all four kinds of interaction played a role,
gravitation is by far the most important force between celestial
bodies such as stars, planets and moons.
Systematic Hierarchy of Physics
The systematic building up principle from small to larger entities discussed so far would suggest to start studies of physics
13
with elementary particles and then proceed gradually to larger
systems. However, since the theoretical treatment of elementary
particles and nuclear physics is rather difficult, it is advisable
from the didactical point of view to go the opposite way, We
therefore start with classical physics of macroscopic bodies and
proceed then to smaller structures like atoms, molecules, nuclei
and elementary particles (see Fig. 1.11). The Physics courses
therefore start with classical mechanics and thermodynamics
(Vol. 1), continue with electrodynamics and optics (Vol. 2) and
then with a basic knowledge of quantum mechanics treat the
physics of atoms, molecules, solid and liquid states (Vol. 3) to
arrive finally at nuclear physics, elementary particle physics and
astrophysics (Vol. 4).
There exist a large number of good books on the subjects treated
in this section [1.14b–1.19], which discuss in more detail the
questions raised here. In order to gain a deeper understanding
of how all this knowledge has been achieved, a more thorough
study of basic physics, its fundamental laws and the experimental techniques, which test the developed theories, is necessary.
The present textbook will help students with such studies.
14
1 Introduction and Survey
Chapter 1
1.5
Relations Between Physics and
Other Sciences
Since physics deals with the basic elements of our material
world it represents in principle the foundations of every natural
science. However, until a few decades ago the scientific methods in chemistry, biology and medicine were more empirically
oriented. Because of the complex nature of the objects studied
in these sciences it was not possible to start the investigations
“ab initio” in order to understand the atomic structure of large
complex molecules and biological cells to say nothing of the
human body and its complex reactions as the research object in
medicine. Therefore, in former years a more phenomenological
method was preferred.
With refined experimental techniques developed in recent years
(electron microscopy, (Fig. 1.12), tunnel microscopy, x-ray
structural analysis, neutron diffraction, nuclear magnetic resonance tomography and laser spectroscopy) in many cases it
became possible to uncover the atomic structure even of complex molecules such as the DNA (Fig. 1.13). Here physics was
helpful in a twofold way: First of all physicists developed, often in cooperation with engineers, the experimental equipment
and secondly it provided the theoretical understanding for the
atomic basis of the research objects. Therefore the differences
in the research methods become less and less important and the
cooperation between researchers of different fields is rapidly increasing, indicated by the growing number of interdisciplinary
research projects. For example the essential question of the
relation between molecular structure and chemical binding is
attacked in common efforts by experimental chemists, theoretical quantum chemists and physicists. Overstated one may say
that chemistry is applied quantum theory and therefore a branch
of physics.
Due to the complex diagnostic techniques in medicine the cooperation between physicists and medical doctors has enormously
increased as will be outlined in the next section.
Figure 1.12 Scavenger Cells visualized with an electron microscope
Bases
Phosphates
Nucleotidsequence
Strand I
Strand II
Figure 1.13 Double Helix of DNA (deoxyribonucleic acid)
1.5.1
Biophysics and Medical Physics
Meanwhile biophysics has developed to an independent branch
of physics. Some of the many research projects are the physical processes in living cells, e. g. the energy balance during cell
processes, the ion transport through cell membranes, the penetration of bacteria and viruses into cells, the different steps of
photosynthesis or the visual process. The very sensitive detection techniques for the detection of single molecules, developed
in physics laboratories, allow the tracing of single laser excited
molecules on their way from outside a cell through membrane
channels into the cell interior. In particular the realization of
ultra short laser pulses down to below a femtosecond (10 15 s)
opens for the first time the possibility to view ultrafast processes
such as molecular isomerisation.
In recent years, medical physics has been established at many
universities and research institutes. The development of new
diagnostic techniques and therapy methods are based on experimental techniques invented and optimized in physics laboratories and on new insights about the interaction between radiation
and tissue. Examples of such new methods are ultrasonic diagnostics with improved spatial resolution, nuclear magnetic
resonance tomography, thermography or laser-induced cell fluorescence. One specific example is the localization of brain
tumours by optical coherence tomography and methods for
1.5 Relations Between Physics and Other Sciences
1.5.2
Chapter 1
their operation with laser techniques, which are investigated
in cooperation between laser physicists and neurosurgeons.
[1.20a–1.23b]
Astrophysics
For ages the closest relation with physics had the astronomy,
which tried to determine the positions of stars, the movement
of planets and the prediction of eclipses. Modern astronomy
goes far beyond this type of problems and looks for information about the composition of stars, conditions for their birth
and the different stages of their development. It turns out that
nearly all branches of physics are necessary in order to solve
these problems. Therefore, this part of modern astronomy is
called astrophysics. The cooperation with physicists who measure in the laboratory processes relevant for the understanding
of star atmospheres and the energy production in the interior of
stars has greatly improved our knowledge in astrophysics (see
Vol. 4). One of the results is for example, that in the universe
the same elements are present as can be found on earth and that
the same physical laws are valid as known from experiments
on earth. The correct interpretation of many astrophysical observations could only be given, because laboratory experiments
had been performed which could give unambiguous decisions
between several possible explanations of astrophysical phenomena.
The following facts have contributed essentially to the impressive progress in astronomy.
The development of new large telescopes in the optical,
near infrared and radio region, of satellites and space probes
(Fig. 1.14) and sensitive detectors.
New and deeper knowledge in the fields of atomic, nuclear and elementary particle physics, in plasma physics and
magneto-hydrodynamics.
Faster computers for the calculation of more complex models for the present composition, the birth, evolution and final
stages of stars [1.24a–1.24c].
Figure 1.14 Last inspection of the Giotto-space probe before its journey to the
comet Halley (with kind permission of the European Space Agency ESA)
its future development. The system shows a chaotic behaviour.
This astonishing feature has lead to a new branch of physical and
mathematical sciences, called chaos research (see Chap. 12).
[1.25–1.30b]
1.5.4
1.5.3
Geophysics and Meteorology
Although geophysics and meteorology have developed into
autonomous disciplines, they are completely based on fundamental physical laws. In particular, in meteorology it is evident
how important fundamental physical processes are, such as the
interaction of light with atoms and molecules, collisions between electrons, ions, atoms and molecules or light scattering
by aerosols and dust particles. Without the detailed understanding of these and other processes the complex preconditions for
the local and global climate could not be calculated within a climate model. However, it turns out, that in spite of the knowledge
of these basic processes it is often not possible to give a reliable
long term weather forecast, because already tiny changes of the
present status of the atmosphere could result in huge changes of
15
Physics and Technology
The application of physical research has pushed the development of our industrial society in a way, which can hardly be
overestimated. Examples are the inventions of the steam engine, the electromotor, research on semiconductors, which form
the basis of computers, information technology, such as the
telephone and extremely fast optical communication over glass
fibres, Lasers and their various applications, precision measuring techniques down into the nanometre range. This connection
between applied physics and technology has received new impetus through the urgent problems of energy crisis, lack of raw
materials, global warming, which have to be solved within a
limited time. Urgent problems are, for example
the development of new energy sources, such as nuclear
fusion, which demands a profound knowledge of plasma
physics under extreme conditions,
16
1 Introduction and Survey
Chapter 1
directions to the philosophical way of thinking. The essential
goal of modern physics is the understanding and the detailed description of our world and the reduction of many observations
to a few general laws. The essential point is, that the human
consciousness and the attitude against the human surroundings
are changed by this new knowledge. The fascinating question,
how cognitive faculty is received by communication with other
thinking persons and whether the structured mind which allows
to process this information to form a unique world view, had
been already formed prenatal had been extensively discussed
by the great philosopher Immanuel Kant (1724–1804) in his famous book “Kritik der reinen Vernunft”.
Figure 1.15 Hexagonal structure of a graphite surface, visualized by a tunnel
microscope (M. Müller, H. Öchsner, TU Kaiserslautern)
the optimization of wind converters,
the development of solar cells with sufficiently high efficiencies,
increasing the conversion efficiency from heat into electrical
energy,
improving the transport efficiency of energy.
Further examples are the development of reliable electrically
driven cars with new designs of batteries, hydrogen technology,
magnetically levitated railways (trans-rapid), development of “clean air cars” etc.
Of particular interest for many branches of industry is the research on new materials such as met-glasses (amorphous metals
with particular properties such as high tensile strength), compound materials or amorphous semiconductors, which have
found meanwhile numerous applications. Surface science
(Fig. 1.15) has given the basic understanding for corrosion processes, catalytic effects and the properties of thin films in optics
and for the creation of very hard surfaces of tools, which decrease the wear and tear of such tools considerably.
One should keep in mind that for densely populated countries
such as Germany, which do not have sufficient raw material at
their disposal, technological innovations and inventions of new
products as well as progress in environmental protection are essential for a better and safe life in the future. Here physicists
encounter great challenges and new ideas and a critical but pragmatic way of thinking are demanded, characteristics, which are
trained during the physics education. [1.30a–1.30b]
1.5.5
Nowadays biophysicists and neurologists try to understand by
well aimed experiments the connection between specified parts
of the brain and the storage of information which we receive
from outside. All these progress in natural sciences has influenced philosophical theories. Although the approach to this
subject is often different for philosophers and scientists, an intense discussion between the representatives of the two camps
could remove many misunderstandings and could lead to a more
extensive view of our world. If such discussions should be fruitful, both sides have to learn more about the way of thinking and
arguing of the other side. The study of physics and its way of
arguing can shape the way we are looking onto our world and
represents an essential part of our culture.
An important aspect of such cooperation is the critical evaluation of ethical questions related to scientific research, which
have found more and more concern in our society. Since the developments in physics and their applications, essentially change
our daily life, physicists have to think about the consequences of
their scientific results. The research itself is unbiased and valuefree. Ethical problems arise when the results of basic research
are applied in such a way, that society might be damaged by
such applications. For instance, the discovery of nuclear fission
by Otto Hahn could be used for peaceful applications as well as
to build an atomic bomb; lasers can be used for health treatment
in medicine or as laser weapons.
People who demand social relevance for every research projects
forget that this is a question of possible applications, which can
often not be predicted from basic research. There are many
examples where basic research was done without any ideas of
possible benefit for the public, such as the beginning of solid
state physics, low temperature physics, semiconductor research.
[1.31–1.35]
1.6
The Basic Units in Physics,
Their Standards and
Measuring Techniques
Physics and Philosophy
Since its beginning in the Greek period, physics always had a
close relation to philosophy (see Sect. 1.3). Already for the
Greek philosophers recognition in natural sciences gave new
Since any objective description of nature demands quantitative
relations between measurements of different objects, which can
be expressed by numbers, one has to define units for the results
of measurements. This means that every numerical result of a
measurement must be expressed in multiples of such units. One
needs a scale that can be compared with the measured quantity.
To measure always means to compare two quantities!
There are several possibilities for choosing units. For the length
unit for instance one may use units which are given by nature
such as foot or the distance between two atoms in a crystal;
for the time unit the time interval between two successive heart
beats, or the time between two culminations of the sun. A better choice of physical units is to use arbitrary but suitable units,
which are conveniently adapted to daily life. Such units have
to be defined by standards with which they can be always compared (calibration).
Every standard has to meet the following demands:
It must be possible to compare with sufficient accuracy
the quantity in question with the standard.
The standard must be reproducible with the demanded
accuracy.
The production and the safekeeping of the standard and
the comparison with measurable elements must be possible with justifiable expenditure.
According to these demands ulna, foot or heartbeat period are
not good standards, because they are dependent on the person
who measures them. They may change with time and are not
general constants.
The quality of a measurement is judged according to the
following aspects:
How reliable is the measurement?
Here the experimental apparatus plays an important
role, the interpretation of the experimental results by
the observer; his ability and experience (see for instance temperature estimations guided by our senses
(Chap. 10 or “optical illusions” Vol. 2)).
How accurate is the measurement, i.e. how large is the
maximum possible error of the result?
Are measurements performed under different experimental conditions reproducible?
off in space and time one certainly needs basic units for length
and time. We will see that all physical quantities can be derived
from three basic units for length, time and mass. One would
therefore need in principle only these three basic units. It turns
out, however, that it is useful to add four more basic units for
the temperature, the mole fraction of material, for the strength
of an electric current and the luminous intensity of radiation
sources, because many derived units can be simpler expressed
when these four additional units are included [1.37–1.39].
In the following we will discuss the different basic units and
also give a short outline of the historical development of this
units and their increasing accuracy. This shall illustrate how
new measurement techniques have improved the quality of a
measurement and asked for new and better standards that could
meet the demands for higher accuracy and reproducibility.
1.6.1
Length Units
As length unit the metre (m) was chosen in 1875 which was
originally meant as the 1=10;000;000 fraction of the equator
quadrant (¼ of the earth circumference). The prototype as the
primary standard was kept in Paris. In order to maintain this
normal as reproducible as possible, it was realized by the distance between two markers on a platinum-iridium rod with a
low thermal expansion coefficient. The rod was kept in a box at
0 ı C. More precise later measurements of the earth circumference showed that the metre deviated from the original definition
by about 0.02%. The comparison of length standards with this
prototype was only possible with a relative uncertainty of 10 6 .
This means that it is only possible to detect a deviation of larger
than 1=1000 mm. This does not meet modern requirements of
accuracy.
Therefore in 1960 a new length standard was defined by the
wavelength of the orange fluorescence line of a discharge
lamp filled with the krypton isotope 86 (Fig. 1.16), where
the conditions in the krypton lamp (pressure, discharge current and temperature) were fixed. The metre was defined as
1;650;763:73 . The wavelength can be measured with an
uncertainty of 10 8 , which is 100 times more accurate than the
comparison with the original metre standard in Paris.
With increasing accuracy of measurements this standard was
again abandoned and a new standard was chosen, which was
based on a completely new definition. Since time can be measured much more accurate than length, the length standard was
Of course, each physical quantity cannot be measured more
accurately than the accuracy of the normal’s measurement.
Therefore such a normal should be chosen which is so accurately defined that it does not represent a limitation for the
accuracy of the measurement. For many measurements, a stopwatch or a micrometre-screw might not be accurately enough
and should not be used as normal.
The question is now how many basic units are necessary to describe all physical quantities. Since all physical processes go
Figure 1.16 The old definition of the length unit, based on the wavelength of
a Krypton line (valid from 1960–1983)
17
Chapter 1
1.6 The Basic Units in Physics, Their Standards and Measuring Techniques
18
1 Introduction and Survey
Chapter 1
Table 1.4 Labels for different orders of magnitude of length units
Table 1.3 Range of actual lengths in our world
Object
Radius of the electron
Radius of the proton
Distance between atoms in solids
Thickness of the skin of a soap bubble
Mean distance between air molecules at 105 Pa
Radius of the earth
Distance earth–moon
Distance earth–sun
Diameter of the solar system
Distance to the nearest star
Diameter of our galaxy
Extension of the universe
Dimension/m
10 18
10 15
10 10
10 7
10 6
6 106
4 108
1:5 1011
1014
4 1016
3 1020
3 1025
related to time measurements via the velocity c of light. The
weighted average of the most precise measurement of the speed
of light in vacuum is now defined as
1 attometer
1 femtometer
1 picometer
1 nanometer
1 micrometer
1 millimeter
1 centimeter
1 dezimeter
1 kilometer
Often used units in
– atomic and nuclear physics
1 fermi D 1 femtometer
1 X-unit
1 Ångström
– astronomy:
1 astronomical unit
mean distance earth–sun
1 light year
1 parsec
D 1 am
D 1 fm
D 1 pm
D 1 nm
D 1 µm
D 1 mm
D 1 cm
D 1 dm
D 1 km
D 10 18 m
D 10 15 m
D 10 12 m
D 10 9 m
D 10 6 m
D 10 3 m
D 10 2 m
D 10 1 m
D 103 m
D 10 15 m
D 1 XU D 1:00202 10
D 1Å
D 10 10 m
D 1 ly
D 1 pc
13
m
D 1 AU
1:496 1011 m
D 9:5 1015 m
D 3 1016 m D 3:2 ly
c D 299;792;458 m=s :
Definition
This means that the speed of light is no longer a result of new
measurements but is defined as a fixed value.
1 AU is the distance to the centre of the sun, which a hypothetical body with negligible mass would have, if it moves
on a circle around the sun in 365.256 8983 days.
Definition
The length unit 1 m is now fixed by the following definition:
One metre is the length of the path that is travelled by
light in vacuum during the time interval 1=299;792;485 s.
From the relation c D between speed of light c, frequency
and wavelength of an electro-magnetic wave the wavelength
of any spectral line can now be determined from the frequency
(which can be measured with a much higher accuracy than
wavelengths) and the defined speed of light (see Sect. 1.6.2 and
1.6.4).
One light-year (1 ly) is the distance, which light travels in 1 year.
An object has a distance of one parsec (1 pc) if the astronomical
unit seen from this object appears under an angle of one second
of arc (100 ) (Fig. 1.17). The distance d of a star, where this angle
is ˛ is d D 1 AU= tan ˛. With tan 100 D 4:85 10 6 we obtain
1 pc D 2:06 105 AU D 3:2 ly :
Note: In some countries other non-metric length units are
in use: 1 inch D 2.54 cm D 0.0245 m and 1 yard (1 yd) D
0.9144 m, 1 mile (1 mi) D 1609.344 m.
However, in this textbook only SI units are used.
The order of magnitude of length-scales in physics covers the
enormous range from 10 18 m for the size of elementary particles to 10C25 m for the radius of the present universe (Tab. 1.3).
It is therefore appropriate to give metre scales in powers of ten.
For specific powers a shorthand notation is used, e. g. 10 6 m D
1 micrometer (µm); 103 m D 1 kilometer (km). These shorthand
notations are listed in Tab. 1.4.
In astronomy, the distances are very large. Therefore, appropriate units are used. The astronomical unit AU is the mean
distance between earth and sun. The new and more exact definition, adopted 1976 by the International Astronomical Union
is the following:
Figure 1.17 Definition of the astronomical units 1 AU and 1 pc
Edge-like surfaces
for measurements
of inside diameters
Set screw
Slider
Pole for depth
measurements
Vermier scale
Zero position
Fixed
Leg
Movable
Measurement: 5.4 mm
Figure 1.18 Caliper gauge with vernier scale
1.6.2
Measuring Techniques for Lengths
For measuring of lengths in daily life secondary standards are
used which are not as accurate as the primary standards but are
more readily usable. The accuracy of such standards is adapted
to the application for which they are constructed. One simple example is the sliding vernier (Fig. 1.18). Its accuracy is
based on the nonius principle. The upper scale is divided into
millimetres, the lower scale has 10 scale divisions for 9 mm,
which means that every division is 9=10 mm. For the situation
in Fig. 1.18b the division mark 9 mm on the upper scale coincides with the division mark 4 on the lower scale. The distance
D between the two fold limbs is then
D D .9
4 9=10/ mm D 5:4 mm :
The uncertainty of the measurement is about 0.1 mm.
Higher accuracies can be reached with a micrometer screw
(Fig. 1.19) where a full turn of the micrometer drum corresponds to a translation of 1 mm. If the scale on the drum is
divided into 100 divisions each division mark corresponds to
Anvil Measuring arbor
Measuring
surfaces
of carbide
Figure 1.19 Micrometer caliper
Scale cylinder
Reference line
Arbor fixing screw
Bow
Coarse
setting
0.01 mm. The shackle is thermally isolated in order to minimize
thermal expansion, With differential micrometer screws, which
have two coaxial drums turning into opposite directions, where
one drum produces a translation of 1 mm per turn, the other of
0:9 mm in the backward direction, one full turn corresponds
now to 0.1 mm. This allows an accuracy of 0:001 mm D 1 µm.
This is about the accuracy limit of mechanical devices.
More accurate length measurements are based on optical techniques. For distances below 1 m interferometric methods are
preferable (see Vol. 2) where lasers (see Vol. 3) are used as
light sources. Here distances are compared to the wavelength
of the light source. Modern interferometers reach accuracies
of =100. With a wavelength of D 500 nm an accuracy of
5 nm D 5 10 9 m can be achieved.
Larger distances can be measured via the travel time of a light
pulse. For instance the distance of the retro-reflector which the
astronauts have positioned on the moon, can be measured within
a few cm using laser pulses with 10 12 s pulse width (LIDAR
technique see Fig. 1.20). Measuring this distance from different locations on earth at different times even allows to detect
continental drifts of the earth crust plates [1.41–1.42].
For the exact location of planes, ships or land vehicles the global
positioning system GPS has been developed. Its principle is
illustrated by Fig. 1.21.
The navigator, who wants to determine his position, measures
simultaneously the phases of radio signals emitted from at least
four different satellites. The radio signals on frequencies at
1575 MHz and 1227 MHz are modulated. This allows to determine unambiguously the distances di from the receiver to the
satellites Si from the measured phase differences i . From these
four distances di the position (x; y; z) of the receiver can be determined with an uncertainty of only a few cm if relativistic effects
(see Sect. 3.6) are taken into account! In order to achieve this
accuracy, the frequencies of the radio signals must be kept stable within 10 10 . This can be realized with atomic clocks which
19
Chapter 1
1.6 The Basic Units in Physics, Their Standards and Measuring Techniques
20
1 Introduction and Survey
Chapter 1
a precise value of the distance between Earth and Venus. From
the angle between the radii Earth–Sun and Earth–Venus at the
time of the measurement the distance Earth–Sun can be obtained
by trigonometric relation in the triangle Earth–Venus–Sun and
using Kepler’s 3rd law (see Sect. 2.9).
As the result of many different measurements, which became
more and more accurate, the Astronomical Union has recommended in 2012 to take the average of these measurements as
the definition of the Astronomical Unit:
1 AUdef D 149;597;870;700 m :
1.6.3
Figure 1.20 Measurement of the distance Earth–Moon with the LIDARtechnique
14
reach a relative stability = D 10 . The exact position of
the satellites is fixed by radio signals from several stations and
receivers at selected precisely known locations on earth. The
European Space Agency has launched several satellites for the
realization of a new GPS System called Galileo with predicted
higher accuracy.
Also a more precise value of the astronomical unit 1 AU can be
obtained by measuring the travel time of short light pulses. A
radar pulse is sent from the earth to Venus where it is reflected.
The time delay between sending and receiving time is measured
for a time of closest approach of Venus to Earth. which gives
Time-Units
The unit of time is the second (1 s). Its initial definition was
1 s D 1=.60 60 24/ d D .1=86;400/ of a solar day ;
where a solar day is defined as the time between two lower culminations of the sun i.e. between two successive midnights.
When the earth rotates around its axis with the angular velocity
! one sun day is d D .2 C ˛/=!; where the additional angle ˛
is due to the revolution of the earth around the sun. On the other
hand a sidereal day (D time between two culminations of a
star) is d D 2=! and therefore shorter by 1=365 d (Fig. 1.22a).
365.25 solar days correspond to 366.25 sidereal days.
Later it was found that the period of a solar day showed periodic and erratic changes, which can amount up to 30 s per day.
(Fig. 1.22b) These changes are caused by the following effects:
A yearly period due to the non-uniform movement of the
earth on an ellipse around the sun (Fig. 1.23 and Sect. 2.9).
The velocity v2 around the perihelion (minimum distance
between earth and sun) is larger than v1 around the aphelion (maximum distance). Since the revolution of the earth
around the sun and the rotation of the earth around its axis
have the same rotation sense, a solar day is longer around the
perihelion than around the aphelion.
A half-year period due to the inclination of the earth axis
against the ecliptic (the plane of the earth’s movement around
the sun), which causes a variation of the sun culmination at a
point P on earth (Fig. 1.24).
In order to eliminate the effect of such changes on the definition
of the second, a fictive “mean sun” is defined which (seen by an
observer on earth) moves with uniform velocity (D yearly average) along the earth equator. The time between two successive
culmination points of this fictive sun defines the mean solar day
hdi. This gives the definition of the mean solar second
Figure 1.21 Principle of the Global Positioning System GPS
1 s D .1=86;400/hdi :
21
Chapter 1
1.6 The Basic Units in Physics, Their Standards and Measuring Techniques
To far distant
star
Sun
Earth
Earth´s orbit
Figure 1.22 a Difference between solar day and sidereal day, b Difference between the true and the mean solar time
Figure 1.23 Changing velocity of the earth during one revolution on its elliptical path around the sun
Figure 1.25 Definition of the tropical year
Since even the tropical year suffers in the course of time small
variations, the astronomers introduced 1960 the ephemeris
time, based on tables which give the calculated positions of sun,
moon and planets at a given time [1.24d].
Figure 1.24 Variation of the point of culmination of the sun with a half-year
period, due to the inclination of the earth axis
With the development of modern precise quartz clocks it was
found that even this mean solar day showed periodic and irregular variations due to changes of the earth’s moment of inertia
caused by melting of glaciers at the poles, falling of leaves in
autumn, volcano eruptions, earth quakes, and turbulent movements of material in the liquid part of the earth’s interior. The
deviations from the mean sun day amount up to 10 milliseconds
per day and cause a relative deviation of 10 2 =85;400 10 7
per day. Therefore the astronomers no longer use the earth rotation as a clock but rather the time span of the tropical year. This
is the revolution period of the earth around the sun between two
successive spring equinoxes, which are the intersection point of
the ecliptic and the equator plane vertical to the earth’s axis
(Fig. 1.25). This tropical year equals the annual period of the
mean sun on its way along the earth’s equator.
The astronomical definition of the second is now 1 s D
period of the tropical year 1900 divided by 31;556;925:9747.
For daily use, quartz clocks are more convenient and therefore
more useful secondary time standards. Their essential part is a
quartz rod of definitive length, which is excited by an external
electric high frequency field to length oscillations (see Vol. 2).
If the exciting frequency is tuned to the resonance frequency of
the quartz rod, the oscillation amplitude reaches a maximum.
By appropriate feedback the system becomes a stable self sustaining oscillator which does not need an external frequency
source. The relative frequency deviation of good quartz clocks
are = 10 9 . The second is then counted by the number of
oscillation periods per time. Of course, the quartz clocks need a
calibration with primary time standards.
The subdivisions of the second and longer time periods are listed
in Tab. 1.5.
A better time standard which is still valid up to now is the caesium atomic clock. Its principle is illustrated in Fig. 1.26.
22
1 Introduction and Survey
Chapter 1
Figure 1.26 Caesium atomic clock. a Experimental arrangement; b level scheme of the hyperfine-transition; c detector signal as a function of the microwave
frequency; d Definition of the second as a multiple of the oscillation period T
Table 1.5 Labelling of subdivisions of the second or of longer time intervalls
Subdivisions of second
1 millisecond
1 mikrosecond
1 nanosecond
1 picosecond
1 femtosecond
1 attosecond
Larger time units
1 hour
1 day
1 year
3
D 1 ms
D 1 µs
D 1 ns
D 1 ps
D 1 fs
D 1 as
D 10
D 10
D 10
D 10
D 10
D 10
s
s
9
s
12
s
15
s
18
s
D 1h
D 1d
D 1a
D 3:6 103 s
D 8:64 104 s
D 3:15 107 s
6
Cs-atoms evaporate through a hole in an oven into a vacuum
tank. Several apertures collimate the evaporating atoms and
form a collimated atomic beam which passes through a microwave resonator M placed between two six pole magnets A
and B. They act on atoms with a magnetic moment like an optical lens and focus the atomic beam onto the detector D where
the focusing characteristics depend on the hyperfine structure
level of the atoms. If the resonator is excited on the frequency
D .E2 E1 /=h which corresponds to the transition between
the two hyperfine levels F D 3 ! F D 4 in the S1=2 electronic
ground state of Cs (Fig. 1.26b) (see Vol. 3), the atoms can absorb the microwave radiation and are transferred from the F D 3
level into the F D 4 level. In this level they have a different
magnetic moment and are therefore defocused in the magnetic
field B. They cannot reach the detector D and the measured signal decreases (Fig. 1.26c). When the microwave frequency
is tuned over the resonance at D 9;192;631;770 s 1 a dip
in the signal S./ appears which is transferred by a feedback
circuit to the microwave generator and keeps its frequency exactly on resonance. The frequency stability of the microwave
generator is now determined by the atomic transition frequency
and serves as a very stable clock, called atomic clock. The
achieved frequency stability of modern versions of the Cs-clock
is = D 10 15 .
The new definition of the second, which is still valid today,
is: 1 s is the time interval of 9,192,631,770.0 oscillation
periods of the Cs clock.
Table 1.6 gives a survey about the time scales of some natural
phenomena, which extend from 10 23 to 10C18 s.
The new definition of the second shows that the time measurement is put down to frequency measurements. The frequency of
any oscillating system is the number of oscillation periods per
second. Its metric unit is [1 s 1 ] or [1 hertz D 1 Hz]. Larger
units are
1 kilohertz D 1 kHz D 103 s 1 ,
1 Megahertz D 1 MHz D 106 s 1 ,
1 Gigahertz D 1 GHz D 109 s 1 ,
1 Terahertz D 1 THz D 1012 s 1 .
Smaller units are
1 Millihertz D 1 mHz D 10 3 s 1 ,
1 Microhertz D 1 µHz D 10 6 s 1 .
Table 1.6 Time scales occuring in natural phenomena
Natural phenomenon
Period=s
Transit time of light over the diameter of an atomic nucleus
10 23
Revolution period of electron in the hydrogen atom
10 15
Transit time of electrons in old tv-tubes
10 7
Oszillation period of tuning fork
2:5 10 3
Time for light propagation sun–earth
5 102
1 day
8:64 104
1 year
3:15 107
Time since the first appearance of homo sapiens
2 1013
Rotational period of our galaxy
1016
Age of our earth
1:6 1017
Age of universe
5 1017
1.6.4
How to measure Times
For the measurement of times periodic processes are used with
periods as stable as possible. The number of periods between
two events gives the time interval between these events if the
time of the period is known. Devices that measure times are
called clocks.
Quartz Clocks: Modern precision clocks are quartz clocks with
a frequency instability = 10 9 . This means that they
deviate per day from the exact time by less than 10 4 s.
Atomic Clocks: For higher accuracy demands atomic clocks are
used, which are available as portable clocks (Rubidium clocks
with = 10 11 ) or as a larger apparatus fixed in the lab
e. g. the Cs clock with = 10 15 .
As world-standard Cs-clocks are used at several locations (National Institute of Standards and Technology NIST in Boulder, Colorado, Physikalisch-Technische Bundesanstalt PTB in
Braunschweig, Germany and the National Physics Laboratory
in Teddington, England) which are connected and synchronized
by radio signals. Two of such clocks differ in 1000 years by less
than 1 millisecond [1.44a–1.44b].
events, such as the rearrangement of the atomic electron shell
after excitation with fast light pulses or the dissociation of
molecules which occur within femtoseconds (1fs D 10 15 s) can
be time- resolved with special correlation techniques using ultrafast laser pulses with durations down to 10 16 s.
1.6.5
Mass Units and Their Measurement
As the third basic unit the mass unit is chosen. The mass of
a body has always a fixed value, even if its form and size is
altered or when the aggregation state (solid, liquid or gaseous)
changes as long as no material is lost during the changes. The
mass is the cause of the gravitational force and for the inertia of
a body, which means that all bodies on earth have a weight and
if they are moving, magnitude and direction of their velocity is
not changing as long as no external force acts on the body (see
Sect. 2.6).
As mass unit the kilogram is defined as the mass of a
platinum-iridium cylinder, which is kept as the primary
mass standard in Paris. (Fig. 1.27)
Initially the kilogram should have been the mass of a cubic
decimetre of water at 4 ı C (at 4 ı C water has its maximum density). Later more precise measurements showed, however, that
the mass of 1 dm3 water was smaller by 2:5 10 5 kg D 0:025 g
than the primary standard.
In Tab. 1.7 the subunits of the kilogram, which are used today,
are listed. For illustration in Tab. 1.8 some examples of masses
which exist in nature are presented.
Frequency stabilized Lasers: A helium-Neon laser with a frequency of 1014 Hz can be locked to a vibrational transition of the
CH4 molecule and reaches a stability of 0.1 Hz, which means a
relative stability = 10 15 comparable to the best atomic
clocks [1.45]. With the recently developed optical frequency
comb (see Vol. 3) stabilities = 10 16 could be achieved
[1.46]. It is therefore expected, that the Cs-standard will soon
be replaced by stabilized lasers as frequency and time standards.
The time resolution of the human eye is about 1/20 s. For the
time resolution of faster periodic events stroboscopes can be
used. These are pulsed light sources with a tuneable repetition
frequency. If the periodic events are illuminated by the light
source, a steady picture is seen, as soon as the repetition frequency equals the event frequency. If the two frequencies differ
the appearance of the event is changing in time the faster the
more the two frequencies differ.
Periodic and non-periodic fast events can be observed with high
speed cameras, which reach a time resolution down to 10 8 s;
with special streak cameras even 10 12 s can be achieved. Faster
Figure 1.27 Standard kilogram of platin-iridium, kept under vacuum in Paris
(https://en.wikipedia.org/wiki/Kilogram#International_prototype_kilogram)
23
Chapter 1
1.6 The Basic Units in Physics, Their Standards and Measuring Techniques
24
1 Introduction and Survey
Chapter 1
Table 1.7 Subdivisions and multiples of the kilogram
Unit
1 gram
1 milligram
1 microgram
1 nanogram
1 pikogram
1 ton
1 megaton
1 atomic mass unit
Denotion
D 1g
D 1 mg
D 1 µg
D 1 ng
D 1 pg
D 1 AMU
Mass/kg
10 3
10 6
10 9
10 12
10 15
103
109
1:6605402 10 27
Table 1.8 The masses of particles and bodies found in nature
Body
Electron
Proton
Uranium nucleus
Protein molecule
Bacterium
Fly
Man
Earth
Sun
Galaxy
Mass/kg
9:1 10 31
1:7 10 27
4 10 25
10 22
10 11
10 3
102
6 1024
2 1030
1042
Masses can be measured either by their inertia or they weight,
since both properties are proportional to their mass and unambiguously defined (see Sect. 2.6). The inertia of a mass is
measured by the oscillation period of a spring pendulum. Here
the mass measurement is reduced to a time measurement.
1 mol is the amount of a substance that consist of as many
particles as the number N of atoms in 0.012 kg of the carbon nuclide 12 C.
These particles can be atoms, molecules, ions or electrons.
The number N of particles per mol with the numerical value
N D 6:02 1023 =mol, is called Avogadro’s number (Amedeo
Avogadro 1776–1856).
Example
1 mol helium has a mass of 0:004 kg, 1 mol copper corresponds to 0:064 kg, one mol hydrogen gas H2 has the
J
mass 2 0:001 kg D 0:002 kg.
1.6.7
Temperature Unit
The unit of the temperature is 1 Kelvin (1 K). This unit can be
defined by the thermo-dynamic temperature scale and can be reduced to the kinetic energy of the molecules (see Sect. 10.1.4).
Because of principal considerations and also measuring techniques, which are explained in Chap. 10, the following definition was chosen:
1 Kelvin is the fraction (1=273:16) of the thermodynamic
temperature of the triple point of water.
The weight of a mass is determined by comparison with a mass
normal on a spring balance or a beam balance and therefore reduced to a length measurement. Today balances are available
with a lower detection limit of at least 10 10 kg (magnetic balance, electromagnetic balance, quartz fibre microbalance).
The triple point is that temperature where all three phases of
water (ice, liquid water and water vapour) can simultaneously
exist (Fig. 1.28).
Note: In some countries non-metrical units are used: 1 pound D
0:453 kg.
There are plans for a new definition of 1 K which is independent
on the choice of a special material (here water). It reads:
1.6.6
Molar Quantity Unit
1 Kelvin is the temperature change which corresponds to
a change .kT/ D 1:3806505 10 23 Joule of the thermal energy kT, where k D 13;806;505 10 23 J=K is the
Boltzmann constant.
As already mentioned in the beginning of this section in addition to the three basic units for length, time and mass four
further units (molar quantity, temperature, electric current and
luminous intensity of a radiation source) are introduced because
of pragmatic reasons. Strictly speaking they are not real basic
units because they can be expressed by the three basic units.
Definition
The unit of molar quantity is the mol, which is defined as
follows:
Figure 1.28 Phase diagram and triple point of water
New very accurate measurements of the Boltzmann constant
allow a much better definition of the temperature T with an uncertainty of T=T 8 10 6 .
1.6.8
Unit of the Electric Current
The unit of the electric current is 1 Ampere (1 A) (named after
Andre-Marie Ampère 1775–1836). It is defined as follows:
1 Ampere corresponds to a constant electric current
through two straight parallel infinitely long wires with
a distance of 1 m which experience a mutual force of
2 10 7 Newton per m wire length (Fig. 1.29).
The definition of the electric current unit is therefore based on
the measurement of the mechanical quantities length and force
(see Vol. 2)
1.6.9
Unit of Luminous Intensity
The luminous intensity of a radiation source is the radiation
power emitted into the solid angle 1 Sterad D 1=.4/. It could
be defined in Watt=Sterad, which gives the radiation power independent of the observing human eye. However, in order to
characterize the visual impression of the light intensity of a light
source, the spectral characteristics of the radiation must be taken
into account, because the sensitivity of the human eye depends
on the wavelength. Therefore the definition of the light intensity is adapted to the spectral sensitivity maximum of the eye at
a wavelength D 555 nm. The luminosity unit is called 1 candela (1 cd).
1 cd is the radiation power of (1/6839)W/Sterad emitted
by a source at the frequency 540 THz ( D 555 nm) into
a selected direction.
Note: 1. The luminous intensity of a source can differ for different directions.
2. The definition of the candela is related to the radiation power in Watt=Sterad, which shows that the
candela is not a basic unit.
Figure 1.29 Illustration how the unit of the electric current is defined
1.6.10
Unit of Angle
Plane angles are generally measured in degrees of arc. The full
angle of a circle is 360ı. The subdivisions are minutes of arc
(1ı D 600 ) and seconds of arc (10 D 6000 ! 1ı D 360000 ). Often
it is convenient to use dimensionless units by reducing angle
measurements to length measurements of the arc length L of a
circle, which corresponds to the angle ˛ (Fig. 1.30).
The circular measure (radian) of the angle ˛ is defined as the
ratio L=R of circular arc L and radius R of the circle. The unit
of this dimensionless quantity is 1 radian (rad) which is realized
for L D R. Since the total circumference of the circle is 2R the
angle ˛ D 360ı in the unit degrees corresponds to ˛ D 2 in
the units radian D rad.
The conversion from radians to degrees is
1 rad D
360ı
D 57:296ı D 57ı 170 4500 :
2
While the plane angle ˛ D L=R cuts the arc with length L out
of a circle with radius R the solid angle ˝ D A=R2 is the angel
of a cone that cuts the area A D ˝R2 out of a full sphere with
area 4R2 and radius R (Fig. 1.31). The dimensionless unit of
the solid angle is 1 steradian (1 sr) for which A D R2 .
Definition
1 sr is the solid angle of a cone which cuts an area A D
1 m2 out of the unit sphere with R D 1 m.
Since the total surface of a sphere is 4R2 the total solid angle
around the centre of the sphere with A D 4R2 is ˝ D 4.
The three planes xy, xz, yz through the positive coordinate axis
Cx, Cy, Cz cut a sphere around the origin (0; 0; 0) into 8 octands, The solid angle of one octand is
˝D
1
1
4 D sr :
8
2
Note: The numerical values of the units for the basic physical quantities discussed so far have been often adapted by the
International Comission for Weights and Measures (CIPM for
Figure 1.30 To the definition of the radian ˛ D L =R
25
Chapter 1
1.6 The Basic Units in Physics, Their Standards and Measuring Techniques
26
1 Introduction and Survey
Chapter 1
1.7
Systems of Units
As has been discussed in Sect. 1.6 the three basic quantities and
their units in physics are
length with the unit 1 Meter D 1 m
time with the unit 1 second D 1 s
mass with the unit 1 kilogram D 1 kg
with four additional quantities
molar quantity with the unit 1 mole D 1 mol
temperature with the unit 1 Kelvin D 1 K
electric current with the unit 1 Ampere D 1 A
radiation luminosity with the unit 1 candela D 1 cd
where these four quantities can be reduced in principle to the
three basic quantities and are therefore no real basic quantities.
All other quantities in physics can be expressed by these 3 basic
quantities with the additional 4 quantities for convenient use.
This will be shown for each derived quantity in this textbook
when the corresponding quantity is introduced.
Each physical quantity is defined by its unit and its numerical
value. For instance the speed of light is c D 2:9979 108 m/s or
the earth acceleration g D 9:81 m/s2 etc.
Figure 1.31 a To the definition of the solid angle ˝; b Illuration of the solid
angel element d˝ D dA =r 2
the French comite international des poids et measures) in order to take into account the results of new and more accurate
measurements. At present, considerations are made to reduce
all quantities to combinations of fundamental constants in order
to give them more accurate and time independent values. This
has been realized up to now only for the length unit which is
defined through the fixed speed of light and the frequency of the
Cs-clock. This might be soon generalized to all physical units
in order to get a system of time-independent values for the units
which do not need to be corrected in future times.
One example is the mass unit. There are many efforts in several
laboratories to create a better and more accurately defined mass
normal. One realistic proposal is a large silicon single crystal
in form of a polished sphere, where the atomic distances in the
crystal have been precisely measured with X-ray interferometry.
This allows the determination of the total number of atoms in the
crystal and the mass of the crystal can be related to the mass of
a silicon atom and is therefore reduced to atomic mass units and
the Avogadro constant [1.48a]. Although it has been shown, that
such a mass normal would be more accurate ( m=m 10 8 )
and would represent a durable mass standard, it has not yet been
internationally acknowledged.
Similar considerations are discussed for the temperature unit
1 K which might be reduced to the Boltzmann constant k (see
above).
In a physical equation all summands must have the same
units.
These units or the products of units are called the dimension of
a quantity. The check, whether all summands in a equation have
the same dimension is called dimensional analysis. It is a very
helpful tool to avoid errors in conversion of different systems of
units.
Each physical quantity can be expressed in different units, for
example, times in seconds, minutes or hours. The numerical
value differs for the different units. For instance the velocity
v D 10 m/s equals v D 36 km/h. In order to avoid such numerical conversions one can use a definite fixed system of units.
If the three basic units are chosen as
1 m for the length unit,
1 s for the time unit,
1 kg for the mass unit.
The system is called the mks-system. If the unit Ampere for the
electric current is added, the system is called the mksA.-system,
often named the SI-System after the French nomenclature System International d’Unites. It has the very useful advantage that
for the conversion from mechanical into electrical and magnetic
units all numerical conversion factors have the value 1. All basic
units and also the units derived from them are called SI units.
In theoretical physics often the cgs system is used, where the
basic units are 1 cm (instead of 1 m), 1 Gramm (instead of 1 kg)
and only the time unit is 1 s as in the SI-system. According to
international agreements from 1972 only the SI-system should
be used. In this textbook exclusively SI units are used.
For a more detailed representation of the subject the reader is
referred to the literature [1.37–1.39,1.50].
1.8
electron charge e and electron mass m. According to the CODATA publication of NIST the value accepted today is me D
9:10938291.40/ 10 31 kg, where the number in brackets gives
the uncertainty of the last two digits.
Accuracy and Precision;
Measurement Uncertainties and
Errors
Every measurement has in different ways uncertainties which
can be minimized by a reliable measuring equipment and careful observation of the measurement. The most important part
in the measuring process is an experienced and critical experimenter, who can judge about the reliability of his results. The
final results of an experiment must be given with error limits
which show the accuracy of the results. There are two different
kinds of possible errors: Systematic and statistical errors.
1.8.1
Systematic Errors
Most systematic errors are caused by the measuring equipment,
as for instance a wrong calibration of an instrument, ignoring of external conditions which can influence the results of
the measurement (temperature change for length measurements,
lengthening of the string of a threat pendulum by the pendulum
weight or air pressure changes for measurements of optical path
length). Recognizing such systematic errors and their elimination for precision measurements is often difficult and demands
the experience and care of the experimental physicist. Often the
influence of systematic errors on the experimental results is underestimated. This is illustrated by Fig. 1.32, which shows the
results of measurements of the electron mass during the time
from 1950 up to today with the error bars given by the authors.
Due to improved experimental techniques the error bars become
smaller and smaller in the course of time. The dashed line gives
the value that is now accepted. One can clearly see, that all the
error bars given by the authors are too small because the systematic error is much larger.
The electron mass can be only determined by a combination
of different quantities. For example, from the deflection of
electrons in magnetic fields one can only get the ratio e=m of
Relative deviation from mean value in ppm
Accuracy and Precision; Measurement Uncertainties and Errors
1.8.2
Statistical Errors, Distribution of
Experimental Values, Mean Values
Even if systematic errors have been completely eliminated, different measurements of the same quantity (for instance the
falling time of a steel ball from the same heights) do not give
the same results. The reasons are inaccurate reading of meters,
fluctuations of the measured quantity, noise of the detection system etc. The measured results show a distribution around a mean
value. The width of this distribution is a measure of the quality
of the results. It is illustrative to plot this distribution of measured values xi in a histogram (Fig. 1.33), where the area of the
rectangles represents the number ni x D ni of measurements
which have given a value within the interval from xi x=2 to
xi x=2.
The mean value x of n measurements is chosen in such a way
that the sum of the squares of the deviations (x xi ) from the
Figure 1.33 Typical histogram of the statistical distribution of measured values
xi around the mean value x
80
9,109558
40
9,109540
0
–40
9,10968
9,1085
CODATA VALUE 2010: 9,10938291(40) ×10–31kg
–80
9,1083
–120
–160
–200
9,10721
–240
1950
1960
1970
1980
Year
1990
2000
2010
Figure 1.32 Historical values of measurements of the electron mass in units of 10 31 kg, demonstrating the underestimation of measuring uncertainties. The
relative deviations m/m from the best value accepted today are plotted in units of 10 6 (ppm D parts per million)
27
Chapter 1
1.8
28
1 Introduction and Survey
Chapter 1
mean value become a minimum, i.e.
n
X
.x
SD
iD1
2
xi / D Minimum :
(1.1)
For the derivative follows:
From (1.6) we obtain by squaring
n
X
dS
D2
.x
dx
iD1
The absolute error " of theP
arithmetic mean x equals the
ei of the absolute errors of
arithmetic mean he1 i D 1n
the individual results xi .
xi / D 0 :
1
" D 2
n
2
This gives for the mean value
1 X 2
1 XX
ei ej 2
e :
C 2
n i
n i i
(1.2)
the
P arithmetic mean value of all measured results. Because
.x xi / D 0 the arithmetic mean is at the centre of the symmetric distribution, which means that the sum of the positive
deviations equals the sum of the negative ones. Contrary to this
symmetric distribution of values with statistical errors the systematic errors cause deviations in one direction.
The question is now how much the mean value deviates from the
true, but generally unknown value of the measured quantity. We
will now prove, that after elimination of all systematic errors
the arithmetic mean converges against the true value xw with
increasing number of measurements. This means:
n
1X
xw D lim
xi :
n!1 n
iD1
(1.3)
(1.4)
xi
Since for statistical errors the deviations ei and ej are uncorrelated.
The quantity
D
p
he2 i D
s
P
.xw xi /2
n
(1.8a)
is named standard deviation or root mean square deviation. It equals the square root of the squared arithmetic
mean he2 i
1X 2
1X
.xw
ei D
n
n iD1
The smaller quantity
m D
p
1
D
n
"2
D
r
sX
xi /2
1 X 2
ei
n2
.xw
xi /2
(1.8b)
(1.8c)
i
(1.5)
x:
From (1.8a)–(1.8c) we can conclude
The mean values of these errors are
X
X
hei D .1=n/
ei I he2 i D .1=n/
e2i :
m D p :
n
From (1.2) it follows
xi / D
xw D 0 :
is the mean error of the arithmetic mean x.
and the absolute error of the mean value as the difference
" D xw
n
1X
ei D xw
n!1 n
iD1
lim
n
We define the absolute error of the measured value xi as the
difference
n
1X
xD
.xw
n iD1
The double sum converges for n ! 1 towards zero because for
any fixed number j it follows from (1.3)
he2 i D
Since it is impossible to perform infinitely many measurements the true value generally remains unknown!
" D xw
(1.7)
j¤i
n
1X
xD
xi ;
n iD1
ei D xw
X 2
1 X 2
ei D 2
e
n i i
i
n
1X
n
iD1
ei :
(1.6)
(1.9)
The mean error of the arithmetic mean equals the mean
error of the individual measurements divided by the square
root of the total number n of measurements.
In the next section it will be shown that approaches a constant
value for n ! 1. Equation 1.9 then implies, that lim m D 0,
which means that the arithmetic mean x approaches the true
value xw for a sufficiently large number n of measurements.
1.8.3
Variance and its Measure
We introduce instead of the unknown deviations ei D xw xi
of the measured values from the true value xw the deviations
vi D x xi from the mean value, which contrary to ei are known
values.
According to (1.4) and (1.5) we can express the vi by the quantities ei and ".
xi
D xw
xi
D ei
":
.xw
For the standard deviation of the individual results xi we
obtain the mean deviation of the arithmetic mean value
s
P
.x xi /2
n
2
2
D
s ! D
;
(1.13)
n 1
n 1
which can be obtained from measurements and is therefore a known quantity.
Since for a finite number n of measurements the true value of
the measured quantity is generally unknown, also the absolute
errors and the mean errors and m cannot be directly determined. We will now show how and m are related to quantities
that can be directly measured.
vi D x
Accuracy and Precision; Measurement Uncertainties and Errors
(1.10)
x/
The mean square deviation of the measured values xi from the
arithmetic mean x can then be written as
1X 2
1X
vi D
.ei "/2
n i
n i
!
#
"
2" X
1 X 2
2
ei
D
ei C "
n
n i
i
For the mean deviation of the arithmetic mean (also called
standard deviation of the arithmetic means) we get
m2 D
1 X 2
ei
n i
"2 ;
because according to (1.6) " D .1=n/
with (1.8a,b,c) yields the relation
s2 D
1 X 2
ei
n i
P
"2 D 2
From the equations (1.8b), (1.9) and (1.12) it follows
s2 D
D
X
1
n
1
n2
n2
1X
n
D .n
xi /2
.xw
i
xi /2
.xw
i
1/m2 D
1
n
n
(1.11)
2 :
(1.12)
.x
n.n
xi /2
:
1/
(1.14)
For 10 measurements of the period of a pendulum the following values have been obtained:
T1 D 1:04 s; T2 D 1:01 s; T3 D 1:03 s; T4 D 0:99 s;
T5 D 0:98 s; T6 D 1:00 s; T7 D 1:01 s; T8 D 0:97 s;
T9 D 0:99 s; T10 D 0:98 s.
The arithmetic mean is T D 1:00 s. The deviations xi D
Ti T of the values Ti from the mean T are
x1 D 0:04 s; x2 D 0:01 s; x3 D 0:03 s; x4 D 0:01 s;
x5 D 0:02 s; x6 D 0:00 s; x7 D 0:01 s; x8 D 0:03 s;
x9 D 0:01 s; x10 D 0:02 s. This gives
†.Ti
ei . The comparison
m2 :
1
n
s2 ! m D
sP
Example
s2 D
D
1
hTi/2 D †x2i D 46 10
4 2
The standard deviation is then
p
D .46 10 4 =9/ D 2:26 10
s :
2
s
and the standard deviation of the arithmetic mean is
p
m D .46 10 4 =90/ D 0:715 10 2 s :
1.8.4
J
Error Distribution Law
In the histogram of Fig. 1.33 the resolution of the different measured values depends on the width xi of the rectangles. All
values within the interval xi are not distinguished and regarded
to be equal. If ni is the number of measured values within the
interval xi and k the total number of intervals xi we can write
Eq. 1.2 also as
k
xD
1X
ni xi
n iD1
with
k
X
iD1
ni D n :
(1.15)
The histogram in Fig. 1.33 can be obtained in a normalized
P form
when we plot the fraction ni =n (ni D ni =xi and n D ni /,
29
Chapter 1
1.8
30
1 Introduction and Survey
Chapter 1
Figure 1.34 Normalized statistical distribution and distribution function of
measured data
which represents the number of measured values within the unit
interval xi D 1 (Fig. 1.34). The heights of the rectangles
give these fractions. The quantity ni =n can be regarded as
the probability that the measured values fall within the interval
xi . With increasing number n of measurements we can decrease the width of the intervals xi which means that the total
number k of all intervals increases. For xi ! 0 the number
k ! 1 and ni ! 0 but P
the fraction ni =xi approaches a
finite value. The sum n D
ni xi which represents the total
number of measured values, stays of course constant. The discontinuous distribution of the histogram in Fig. 1.34 converges
against a continuous function f .x/, which is shown in Fig. 1.34
as black dashed curve. The function f .x/ is defined as
f .x/ D .1=n/ lim.ni =xi / D .1=n/ dn=dx I
which has its maximum at x D xw . The inflection points of
the curve f .x/ are at x D xw ˙ . The full width between the
inflection points where f .x/ D f .xw /=e is therefore 2. The distribution f .x/ is symmetrical around its centre at xw (Fig. 1.35).
For infinitely many measurements the arithmetic mean x becomes xw .
When the standard deviation has been determined from n measurements, the probability P./ that further measured values fall
within the interval x D xw ˙ and are therefore within the standard deviation from the true value. It is given by the integral
(1.16a)
f .x/ is the continuous distribution function. The product f .x/
dx gives the probability to find a measured value
P in the interval
from x dx=2 to x C dx=2. From (1.16a) and ni xi follows
the normalization
Z
i
h
X
(1.16b)
f .x/dx D lim .1=n/
ni xi D 1 :
This means that the probability to find a measured value somewhere within the total x-range must be of course 100% D 1,
because it has to be somewhere in this range.
R
The integral f .x/dx represents the area under the black curve
which is normalised to 1 because the ordinate in Fig. 1.34 is
given as the normalized quantity ni =n.
The standard deviation is a measure for the width of the distribution f .x/. Its square 2 gives, as for the discontinuous
distribution (1.8b), the mean square deviation of the arithmetic
mean from the true value xw , which determines the centre of the
symmetric curve f .x/
ZC1
2
2
D he i D
.xw x/2 f .x/ dx :
(1.17)
2
Figure 1.35 Error distribution function (Gaussian distribution) around the true
value xw for different standard deviations
1
The quantity is named the variance.
If only statistical errors contribute, the normalized distribution
of the measured values can be described by the normalized
Gauss-function
1
2
2
f .x/ D p
e .x xw / =2 ;
(1.18)
2 2
P .jxw
xi j / D
xZ
w C
f .x/dx :
(1.19)
xw
When inserting (1.18) the integral can be solved and yields the
numerical values
P .ei / D 0:683
P .ei 2/ D 0:954
P .ei 3/ D 0:997
.68% confidence range/
.95% confidence range/
.99:7% confidence range/ :
The results of a measurement are correctly given with the 68%
confidence range as
(1.20)
xw D x ˙ :
This means that the true value falls with a probability of 68%
within the uncertainty range from x to x C around the
arithmetic mean, if all systematic errors has been eliminated.
The relative accuracy of a measured value xw is generally given
as =x.
Cautious researchers extend the uncertainty range to ˙3 and
can than state that their published result lies with the probability
of 99.7%, which means nearly with certainty within the given
limits around the arithmetic mean. The result is then given as
xw D x ˙ 3 D x ˙ 3
s
P
.xi x/2
:
n 1
(1.21)
1.8
Accuracy and Precision; Measurement Uncertainties and Errors
xw D x ˙ m D x ˙
sP
.xi x/2
:
n.n 1/
Chapter 1
Since the arithmetic mean is more accurate than the individual
measurements often the uncertainty range is given as the standard deviation m of the arithmetic mean which is smaller than
. The result is then given as
(1.22)
Example
For our example of the measurements of the periods of a
pendulum the result would be given with the 69% confidence range as
Tw D hTi ˙ D .1:000 ˙ 0:025/ s
Figure 1.36 Error propagation for a function y D f .x /
and for the 99.7% confidence range as
Tw D hTi ˙ 3 D .1:000 ˙ 0:075/ s :
For the standard deviation m of the arithmetic mean one
gets
Tw D hTi ˙ m D .1:0000 ˙ 0:0079/ s :
The relative uncertainty of the true value is then with a
probability of 68%
Tw =Tw D 7:9 10
3
D 0:79% :
J
Remark. For statistical processes where the measured quantity is an integer number xi D ni that statistically fluctuates (for
instance the number of electrons emitted per sec by a hot cathode, or the number of decaying radioactive nuclei per sec) one
obtains instead of the Gaussian function (1.18) a Poisson distribution
f .x/ D
1.8.5
xx
e
xŠ
x
x D integer number :
(1.23)
Error Propagation
If a quantity y D f .x/ depends in some way on the measured
quantity x, the uncertainty dy is related to dx by (Fig. 1.36)
dy D
df .x/
dx :
dx
(1.24)
When the quantity x has been measured n-times its standard deviation is
sP
.Nx xi /2
x D
;
n 1
31
which results in the standard deviation of the yi values
s
sP
2
P
f .Nx/ f .xi /
.Ny yi /2
y D
D
n 1
n 1
df .x/
x :
D
dx xN
(1.25)
Often the value of a quantity, which is not directly accessible to
measurements, and its uncertainty should be known. Examples
are the density of a body which can be calculated as the ratio of
mass and volume of the body, or the acceleration of a moving
mass which is determined from measurements of distances and
times.
The question is now: What is the accuracy of a quantity f .x; y/,
if the uncertainties of the measurements of x and y are known.
Assume one has made n measurements of the quantity x from
which the uncertainty range of the arithmetic mean is determined as
s
P 2
vi
with vi D xi x
x ˙ x D x ˙
n 1
and m measurements of the quantity y with the mean
s
P 2
uk
with uk D yk y ;
y ˙ y D y ˙
m 1
one obtains the quantity
fik D f .xi ; yk / D f .x C vi ; y C uk /
@f .x; y/
D f .x; y/ C vi
@x
0
@f .x; y/
C uk
C :::
@y
0
(1.26)
by a Taylor expansion, where .@f =@x/0 is the partial derivative
for the values x; y. Often the deviations vi and uk are so small
32
1 Introduction and Survey
Chapter 1
that the higher powers in the expansion can be neglected. The
mean value of all fik is then
m
n
1 XX
1 XX
f D
f .x; y/
fik D
nm i k
n m iD1 kD1
@f
@f
C vi .x; y/ C uk .x; y/
@x
@y
X @f
1
vi
D
n m f .x; y/ C m
nm
@x
i
X @f
uk
Cn
D f .x; y/ ;
@y
k
P
because @f =@xjx;y is constant and
vi D
P
(1.27)
Figure 1.37 a Mean error of a length measurement, that consists of
two individual measurements x and y ; b Error propagation for the measurement of an area x y
2. The area A D x y of a rectangle shall be determined
for the measured side lengths x and y. The true values
of x and y are
ui D 0.
xw D x ˙ x ;
@A
.x; y/ D y ;
@x
The arithmetic mean f of all values fik equals the value
f .x; y/ of the function f .x; y/ for the arithmetic means x; y
of the measured values xi yk .
A D x y ˙ xy
q
2
2
Dxy˙
y x C x y :
In books about error calculus [1.53a–1.55] it is shown, that the
standard deviation of the derived quantity f is related to the standard deviations x and y of the measured values xi , yk by
f D
s
x2
@f
@x
2
C y2
@f
@y
2
The mean uncertainties x and y propagate to the uncertainty
f of the derived mean f .x; y/. The 68% confidence range of the
true value fw .x; y/ D f .xw ; yw / is then
fw .x; y/ D f .x; y/ ˙
s
x2
@f
@x
2
C y2
@f
@y
2
:
(1.29)
p
With the inequality a2 C b2 jaj C jbj the uncertainty (1.29)
can be also written as
ˇ
ˇ ˇ
ˇ
ˇ @f ˇ ˇ @f ˇ
ˇ
ˇ
ˇ
(1.30)
f D fw f .x; y/ ˇx ˇ C ˇy ˇˇ :
@x
@y
Examples
1. The length L is divided into two sections x and y with
L D x C y which are separately measured (Fig. 1.37a).
The final result of L is then, according to (1.27) and
(1.28) with @f =@x D @f =@y D 1,
L D xCy˙
q
x2 C y2 :
The relative error of the product A D x y
xy
D
A
(1.28)
:
This means: the mean error of a sum (or a difference)
equals the square root of the sum of squared errors of
the measured values.
yw D y ˙ y ;
@A
.x; y/ D x ;
@y
s
x
x
2
C
y
y
2
equals the Pythagorean sum of the relative errors of
the two factors x and y.
3.
y D ln x I
x D x ˙ x )
@y
D 1=x
@x
y D ln x ˙ x =x
The mean absolute error of the logarithm of a meaJ
sured value x equals the relative error of x.
1.8.6
Equalization Calculus
Up to now we have discussed the case, where the same quantity
has been measured several times and how the arithmetic means
of the different measured values and its uncertainty can be obtained. Often the problem arises that a quantity y.x/, which
depends on another quantity x shall be determined for different values of x and the question is how accurate the function
y.x/ can be determined if the measured values of x have a given
uncertainty.
Example
1. A falling mass passes during the time t the distance
d D 21 g t2 and its velocity v D g t is measured at
different times ti .
2. The change of the length L D L0 ˛ T, a long
rod with length L and thermal expansion coefficient ˛
experiences for a temperature change T, is measured
at different temperatures T.
J
In our first example distances and velocities are measured at different times. The goal of these measurements is the accurate
determination of the earth acceleration g. In the second example length changes and temperatures are measured in order to
obtain the thermal expansion coefficient ˛ as a function of temperature T.
The relation between y.x/ and x can be linear (e. g. v D g t),
but may be also a nonlinear function (e. g. a quadratic or an exponential function). Here wee will restrict the discussion to the
simplest case of linear functions, in order to illustrate the application of equalization calculus to practical problems.
This will become clear with the following example.
Accuracy and Precision; Measurement Uncertainties and Errors
This is the case if the sum of the squared deviations
reaches a minimum.
X
(1.31)
SD
.yi axi b/2
Differentiating (1.31) gives the two equations. (Note that
a and b are here the variables!)
n
X
@S
xi .yi
D 2
@a
iD1
We consider the linear function
y D ax C b
and will answer the question, how accurate the constants a
and b can be determined when y is calculated for different
measured values of x.
Solution
It is often the case that the values x can be measured more
accurately than y. For instance for the free fall of a mass
the times can be measured with electronic clocks much
more accurately than distances or velocities. In such cases
the errors of x can be neglected compared to the uncertainties of y. This reduces the problem to the situation
depicted in Fig. 1.38. The measured values y.x/ are given
by points and the standard deviation by the length of the
error bars.
The question is now, how it is possible to fit a straight line
to the experimental points in such a way that the uncertainties of the constants a and b become a minimum.
b/ D 0
(1.32a)
n
X
@S
.yi
D 2
@b
iD1
b/ D 0 :
axi
(1.32b)
Rearranging yields
X
X
X
xi yi
xi D
a
x2i C b
i
a
Example
axi
X
i
(1.33a)
i
i
xi C b n D
X
(1.33b)
yi :
i
The last equation is matched exactly for the point .x; y/
with the mean coordinates
X
X
x D .1=n/
xi I y D .1=n/
yi :
Inserting these values into (1.33b) yields after division by
the number n the relation
axCbDy :
This proves that the point .x; y/ fulfils the equation and is
located in Fig. 1.38 exactly on the red straight line.
From (1.33b) one obtains for the slope b of the straight
line
X
X
xi :
b D y ax D .1=n/
yi .a=n/
Inserting this into (1.33a) gives with the abbreviation
d D n
X
x2i
X 2
xi ;
the constants a and b as
P
P P
xi yi
n
yi
xi
aD
;
d
P 2 P P P
yi
xi
xi
xi yi
bD
:
d
(1.34a)
(1.34b)
The true constants a and b give the true values yw .xi / D
axi C b within the 68% confidence limits yi ˙ y around
the mean value y. From (1.18) and (1.19) one obtains the
probability P.yi / to find the measured value yi
Figure 1.38 Equalization calculus for the function y D ax C b , when
the values xi can be measured much more accurate than the values yi
P.yi / /
1
e
y
.yi axi b/2 =2y2
:
(1.35)
33
Chapter 1
1.8
34
1 Introduction and Survey
Chapter 1
The uncertainties of the constants a and b can be obtained
according to the error propagation rules. The results are
a2
D
n y2
d
;
b2
D
y2
P
d
x2i
:
(1.36)
The full width between the two points P.yw /=e is y
p
2.
For more information on error analysis and regression fits
see [1.53a–1.56].
J
Summary
Physics deals with the basic building blocks of our world,
their mutual interactions and the synthesis of material from
these basic particles.
The gain of knowledge is pushed by specific experiments.
Their results serve for the development of a general theory
of nature and to confirm or contradict existing theories.
Experimental physics started in the 16th century
(e. g. Galilei, Kepler) and led to a more and more refined
and extensive theory, which is, however, even today not yet
complete and consistent.
All physical quantities can be reduced to three basic quantities of length, time and mass with the basic units 1 m, 1 s, and
1 kg. For practical reasons four more basic quantities are introduced for molar mass (1 mol), temperature (1 K), electric
current (1 A) and the luminous power (1 cd).
The system of units which uses these basic 3 C 4 units is
called SI-system with the units 1 m, 1 s, 1 kg, 1 mol, 1 K, 1 A
and 1 cd.
Every measurement means the comparison of the measured
quantity with a normal (standard).
The length standard is the distance which light travels in
vacuum within a time interval of .1=299;792;458/ s. The
time standard is the transition frequency between two hyperfine levels in the Cs atom measured with the caesium
atomic clock. The present mass standard is the mass of the
platinum-iridium kilogram, kept in Paris.
Each measurement has uncertainties. One distinguishes between systematic errors and statistical errors. The mean
value of n independent measurements with measured values
xi is chosen as the arithmetic mean
n
xD
1X
xi ;
n iD1
which meets the minimum condition
n
X
.x
iD1
xi /2 D minimum:
If all systematic errors could be eliminated the distribution
of the measured values x show the statistical Gaussian distribution
f .x/ / e
.x xw /2 =2 2
;
about the most probable value, which equals the true value
xw . The half-with of the p
distribution between the points
f .xw /=e D f .xw ˙ / is 2 Within the range x D xw ˙
fall 68% of all measured values. The standard deviation of
individual measurements is
sP
.x xi /2
D
;
n 1
the standard deviation of the arithmetic means is
sP
.x xi /2
m D
:
n.n 1/
The true value xw lies with the probability of 68% within the
interval xw ˙ , with a probability of 99.7% in the interval
xw ˙ 3. The Gaussian probability distribution for the measured values xoi has a full width at half maximum of
p
x1=2 D 2 2 ln 2 D 2:35 :
Problems
1.1
The speed limit on a motorway is 120 km/h. An international commission decides to make a new definition of the hour,
such that the period of the earth rotation about its axis is only
16 h. What should be the new speed limit, if the same safety
considerations are valid?
arguments why the length of the mean solar day can change for
different years?
1.2
Assume that exact measurements had found that the diameter of the earth decreases slowly. How sure can we be, that
this is not just an increase of the length of the meter standard?
1.9
How many water molecules H2 O are included in 1 litre
water?
1.3
Discuss the following statement: “The main demand for
a length standard is that its length fluctuations are smaller than
length changes of the distances to be measured”.
1.4
Assume that the duration of the mean solar day increases
by 10 ms in 100 years due to the deceleration of the earth rotation. a) After which time would the day length be 30 hours? b)
How often would it be necessary to add a leap second in order
to maintain synchronization with the atomic clock time?
1.5
The distance to the next star (˛-Centauri) is d D 4:3
1016 m. How long is the travelling time of a light pulse from
this star to earth? Under which angle appears the distance earthsun from ˛-Centauri? If the accuracy of angular measurements
is 0:100 what is the uncertainty of the distance measurement?
1.6
A length L is seen from a point P which is 1 km (perpendicular to L) away from the centre of L, under an angle of
˛ D 1ı . How accurate can the length be determined by angle
measurements from P if the uncertainty of ˛ is 10 ?
1.7
Why does the deviation of the earth orbit from a circle
cause a variation of the solar day during the year? Give some
1.8
How many hydrogen atoms are included in 1 kg of hydrogen gas?
1.10 The radius of a uranium nucleus (A D 238) is 8:68
10 15 m. What is its mean mass density?
1.11 The fall time of a steel ball over a distance of 1 m is
measured 40 times, with an uncertainty of 0.1 s for each measurement. What is the accuracy of the arithmetic mean?
1.12 For which values of x has the error distribution function
expŒ x2 =2 fall to 0.5 and to 0.1 of its maximum value?
1.13 Assume the quantity x D 1000 has been measured with
a relative uncertainty of 10 3 and y D 30 with 3 10 3 . What is
the error of the quantity A D .x y2 /?
1.14 What is the maximum relative error of a good quartz
clock with a relative error of 10 9 after 1 year? Compare this
with an atomic clock (= D 10 14).
1.15 Determine the coefficients a and b of the straight line
y D ax C b which gives the minimum squared deviations for the
points .x; y/ D .0; 2/; .1; 3/; .2; 3/; .4; 5/ and .5; 5/. How large
is the standard deviation of a and b?
References
1.1a. R.D. Jarrard, Scientific Methods. An online book:
https://webct.utah.edu/webct/RelativeResourceManager/
288712009021/Public%20Files/sm/sm0.htm
1.1b. K. Popper, The Logic of of Scientific Discovery. (Routledgen 2002)
1.2. R. Feynman, The Character of Physical Laws (Modern
Library, 1994)
1.3. W. Heisenberg, Physics and Philosophy (Harper Perennial Modern Thoughts, 2015)
1.4. A. Franklin, Phys. Persp. 1(1), 35–53 (1999)
1.5. L. Susskind, G. Hrabovsky, The theoretical Minimum.
What you need to know to start doing Physics (Basic
Books, 2013)
1.6. J.Z. Buchwald, R. Fox (eds.), The Oxford Handbook of
the History of Physics (2014)
1.7. A. Einstein, The evolution of Physics From early Concepts to Relativity and Quanta (Touchstone, 1967)
1.8. D.C. Lindberg, E. Whitney, The Beginning of western
Sciences (University of Chicago Press, 1992)
1.9. https://en.wikipedia.org/wiki/History_of_physics
1.10. H.Th. Milhorn, H.T. Milhorn, The History of Physics
(Virtualbookworm.com publishing)
1.11a. W.H. Cropper, Great Physicists: The Life and Times
of Leading Physicists from Galileo to Hawking (Oxford University Press, 2004) https://en.wikipedia.org/
wiki/Special:BookSources/0-19-517324-4
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Chapter 1
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36
1 Introduction and Survey
Chapter 1
1.11b. John L. Heilbron, The Oxford Guide to the History of Physics and Astronomy (Oxford University Press, 2005) https://en.wikipedia.org/wiki/Special:
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1.11c. J.Z. Buchwald, I.B. Cohen (eds.), Isaac Newton’s natural
philosophy (Cambridge, Mass. and London, MIT Press,
2001)
1.12. J.T. Cushing, Philosophical Concepts in Physics: The
historical Relation between Philosophy and Physics
(Cambridge Univ. Press, 2008)
1.13a. E. Segrè, From Falling Bodies to Radio Waves: Classical Physicists and Their Discoveries (W.H. Freeman,
New York, 1984) https://en.wikipedia.org/wiki/Special:
https://www.worldcat.
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1.13b. E. Segrè, From X-Rays to Quarks: Modern Physicists and Their Discoveries (W.H. Freeman, San
Francisco, 1980) https://en.wikipedia.org/wiki/Special:
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1.13c. G. Gamov, The Great Physicists from Galileo to Einstein (Dover Publ., revised edition 2014) ISBN: 9780486257679
1.14a. P. Fara, Science, A Four Thousand Year History (Oxford
Univ. Press, 2010) ISBN: 978-0199580279
1.14b. H.J. Jodl, in: E. Lüscher, H.J. Jodl, (eds.), Physik,
Gestern, Heute, Morgen (München, Heinz Moos-Verlag
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1.14c. N.N., From Big to Small: The Hierarchy Problem in Physics (String Theory) (what-when-how – In
Depth Tutorials and Information) http://what-when-how.
com/string-theory/from-big-to-small-the-hierarchyproblem-in-physics-string-theory/
1.15. St. Weinberg, The Discovery of Subatomic Particles (Scientifique American Library Freeman Oxford, 1984)
1.16a. G. Gamov: One, Two, Three . . . Infinity. Facts and Speculations of Science (Dover Publications, 1989) ISBN:
978-0486256641
1.16b. G. Gamov: Mr. Tomkins in Paperback (Cambridge Univ.
Press, reprint 2012) ISBN: 978-1107604681
1.17. N. Manton, Symmetries, Fields and Particles. (University of Cambridge), http://www.damtp.cam.ac.uk/user/
examples/3P2.pdf
1.18. F. Close, Particle Physics: A Very Short Introduction.
(Oxford Univ. Press, 2004)
1.19. R. Oerter, The Theory of Almost Everything: The Standard Model, the Unsung Triumph of Modern Physics.
(Plume, Reprint edition, 2006)
1.20a. T. Plathotnik, E.A. Donley, U.P. Wild, Single Molecule
Spectroscopüy. Ann. Rev. Phys. Chem. 48, 181 (1997)
1.20b. N.G. Walter, Single Molecule Detection, Analysis, and
Manipulation, in Encyclopedia of Analytical Chemistry,
ed. by R.A. Meyers (John Wiley & Sons Ltd, 2008)
1.21. E. Schrödinger, What is life? (Macmillan, 1944)
1.22a. D. Goldfarb, Biophyiscs Demystified. (Mc Grawhill,
2010) ISBN: 978-0071633642
1.22b. R. Glaser, Biophysics: An Introduction. (Springer,
Berlin, Heidelberg, 2012) ISBN: 978-3642252112
1.23a. K. Nouri (ed.), Lasers in Dermatology and Medicine.
(Springer, Berlin, Heidelberg, 2012) ISBN: 978-085729-280-3
1.23b. K. Nouri (ed.), Laser Applications in Medicine International Journal for Laser Treatment and Research.
(Elsevier, Amsterdam)
1.24a. St. Hawking, The Universe in a Nutshell. (Bantam Press,
2001)
1.24b. L.M. Krauss, A Universe from Nothing. (Atria Books,
2012)
1.24c. E. Chaisson, St. McMillan, Astronomy Today. (Addison
Wesley, 2010)
1.24d. T. Dickinson, A. Schaller, T. Ferris, Night watch. A practical Guide to viewing the Universe. (Firefly Books)
1.24e. B.W. Carroll, D.A. Ostlie, An Introduction to Modern Astrophysics. (Pearson Education, 2013)
1.25. A.E. Musset, M. Aftab Khan, S. Button, Looking into
the Earth: An Introduction to Geological Geophysics,
1st edn. (Cambridge University Press, 2000) ISBN: 9780521785747
1.26a. C.D. Ahrens, Meteorology Today: Introduction to
Wheather, Climate and the Environment. (Brooks Cole,
2012) ISBN: 978-0840054999
1.26b. St.A. Ackerman, J.A. Knox, Meteorology: Understanding the Atmosphere, 4th edn. (Jones & Bartlett Learning,
2013)
1.27. P.J. Crutzen, Pure Appl. Chem. 70(7), 1319–1326 (1998)
1.28. P. Saundry, Environmental physics. (The Encyclopedia of Earth, 2011), http://www.eoearth.org/view/article/
152632
1.29a. F.K. Lutgens, E.J. Tarbuck, D.G. Tasa, Essentials of Geology, 11th edn. (Prentice Hall) ISBN: 978-0321714725
1.29b. W.S. Broecker, How to Build a Habitable Planet. (Eldigio Press, 1988)
1.30a. R.A. Müller, Physics and Technology für Future Presidents. (Princeton Univ. Press, 2010)
1.30b. M.L. Forlan, Modern Physics and Technology. (World
Scientific, Singapore, 2015)
1.31. L. Sklar, Philosophy of Physics. (Oxford University
Press, 1990) ISBN: 978-0198751380
1.32. Th. Brody, The Philosophy behind Physics. (Springer,
Berlin, Heidelberg, 1994)
1.33. St. Cole, Am. J. Sociol. 89(1) 111–139 (1983)
1.34a. B.R. Cohen, Endeavour 25(1) 8–12 (2001)
1.34b. R.M. Young, Science and Humanities in the understanding of human nature. (The Human Nature Review),
http://human-nature.com/rmyoung/papers/pap131h.html
1.35. W. Heisenberg, Physics and Beyond: Encounters and
Conversations. (Harper & Row, 1971)
1.36a. P. Becker, The new kilogram is approaching. (PTBnews, 3/2011), http://www.ptb.de/en/aktuelles/archiv/
presseinfos/pi2011/pitext/pi110127.html
1.36b. S. Hadington, The kilogram is dead, long live
the kilogram!
(The Royal Society of Chemistry,
2011), http://www.rsc.org/chemistryworld/News/2011/
October/31101101.asp
1.37. N.N., SI based Units. (Wikipedia), https://en.wikipedia.
org/wiki/SI_base_unit
1.38. N.N.,
https://www.bing.com/search?q=Phyiscal%20 1.49. V.V. Krutikov, Measurement Techniques. (Springer,
basis%20for%20SI%20units&pc=cosp&ptag=ADC890
Berlin, Heidelberg, 2013)
F4567&form=CONMHP&conlogo=CT3210127
1.50. Journal of Measurement Techniques (Springer Berlin,
Heidelberg)
1.39. B.W. Petley, The fundamental physical constants and the
1.51. K.T.V. Grattan (ed.), Measurement: Journal of Interfrontiers of measurements. (Adam Hilger, 1985)
national Measuring Confederation (IMEKO) (Elsevier,
1.40. N.N., History of the Meter. (Wikipedia), https://en.
Amsterdam), ISSN: 0263-2241
wikipedia.org/wiki/History_of_the_metre
1.41. J. Levine, Ann. Rev. Earth Planet Sci. 5 357 (1977)
1.52. L. Marton (ed.), Methods of experimental Physics.
1.42. J. Müller et al., Lunar Laser Ranging. Recent Results.
(29 Vol.) (Academic Press, New York, 1959–1996) (con(AG Symposia Series, Vol 139, Springer, 2013)
tinued as Experimental Methods in the Physical Sciences
1.43a. K.M. Borkowski, Journal of the Royal Astronomical Soup to Vol. 49, published by Elsevier Amsterdam in 2014),
ciety of Canada 85 121 (1991)
ISBN 13: 978-0-12-417011-7
1.53a. J.R. Taylor, An Introduction to Error Analysis:The Study
1.43b. http://en.wikipedia.org/wiki/Tropical_year
of Uncertainties in Physical measurements, 2nd ed.
1.44a. N.N., Atomic clock. (Wikipedia), https://en.wikipedia.
(University Science Books, 1996)
org/wiki/Atomic_clock
1.44b. N.N., Clock network. (Wikipedia), https://en.wikipedia. 1.53b. D. Roberts, Errors, discrepancies, and the nature of
physics. (The Physics Teacher, March 1983) a very useorg/wiki/Clock_network
ful introduction
1.45. A. deMarchi (ed.), Frequency Standards and Metrology.
1.54. J.W. Foremn, Data Smart: Using Data Science to trans(Springer, Berlin, Heidelberg, 1989)
form Information into Insight. (Wiley, 2013)
1.46. S.A. Diddams, T.W. Hänsch et al., Phys. Rev. Lett. 84
1.55. R.C. Sprinthall, Basic statistical Analysis, 9th ed. (Pear5102 (2000)
son, 2011)
1.47. Ch. Gaiser, Metrologia 48 382 (2011)
1.48a. N.N., Kilogram (Wikipedia), https://en.wikipedia.org/ 1.56. J. Schmuller, Statistical Analysis with Excel for Dummies, 3rd ed. (For Dummies, 2013)
wiki/Kilogram
1.48b. N.N., Mole (unit) (Wikipedia), https://en.wikipedia.org/
wiki/Mole_(unit)
37
Chapter 1
References
2.1
The Model of the Point Mass; Trajectories . . . . . . . . . . . . . . . .
40
2.2
Velocity and Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . .
41
2.3
Uniformly Accelerated Motion . . . . . . . . . . . . . . . . . . . . . . .
42
2.4
Motions with Non-Constant Acceleration . . . . . . . . . . . . . . . . .
44
2.5
Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
47
2.6
The Basic Equations of Mechanics . . . . . . . . . . . . . . . . . . . . .
51
2.7
Energy Conservation Law of Mechanics . . . . . . . . . . . . . . . . . .
56
2.8
Angular Momentum and Torque . . . . . . . . . . . . . . . . . . . . . .
63
2.9
Gravitation and the Planetary Motions . . . . . . . . . . . . . . . . . .
64
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
76
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
77
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
79
© Springer International Publishing Switzerland 2017
W. Demtröder, Mechanics and Thermodynamics, Undergraduate Lecture Notes in Physics, DOI 10.1007/978-3-319-27877-3_2
Chapter 2
2
Mechanics of a Point Mass
39
40
2 Mechanics of a Point Mass
As has been discussed in the previous chapter, the theoretical description of the physical reality often proceeds by successively
refined models which approach the reality more and more with
progressive refinement. In this chapter the motion of bodies under the influence of external forces will be depicted by the model
of point masses, which neglects the spatial form and extension
of bodies, which might influence the motion of these bodies.
Chapter 2
2.1
The Model of the Point Mass;
Trajectories
For many situations in Physics the spatial extension of bodies
is of no importance and can be neglected because only their
masses play the essential role. Examples are the motion of the
planets around the sun where their size is very small compared
with the distance to the sun. They can be described as point
masses.
The position P.t/ of a point mass in the three-dimensional space
can be described by its coordinates, which are defined if a suitable coordinate-system is chosen. These coordinates are fx; y; zg
in a Cartesian system, fr; #; 'g in a spherical coordinate system
and f%; #; zg in cylindrical coordinates (see Sect. 13.2).
The motion of a point mass is described as the change of its
coordinates with time, for example in Cartesian coordinates
The motion performed by P.t/ on its trajectory is called translation. Contrary to the point mass bodies with extended size can
also perform rotations (Chap. 5) and vibrations (Chap. 6).
Note: The model of a point mass moving on a welldefined trajectory fails in micro-physics for the motion
of atoms or elementary particles described correctly by
quantum mechanics (Vol. 3), where position and velocity cannot be precisely given simultaneously. Instead of
a precisely defined trajectory where the point mass can be
find at a specific time with certainty at a well-defined position, only probabilities P.x; y; z; t/dxdydz can be given for
finding the point mass in a volume dV D dxdydz around
the position .x; y; z/. Strictly speaking a geometrical exact trajectory does not exist in the framework of quantum
mechanics.
Examples
1. Motion on a straight line
x Dat ;
y Dbt ;
zD0:
Elimination of t gives the usual representation y D
.b=a/x of a straight line in the .x; y/-plane.
The point mass moves in the x; y-diagram on the
straight line with the slope .b=a/ (Fig. 2.2).
9
x D x.t/>
=
y D y.t/ r D r.t/ ;
>
z D z.t/ ;
where the position vector r D fx; y; zg combines the three coordinates x, y and z (Sect. 13.1).
Note: Vectors are always marked as bold letters.
The function r.t/ represents a trajectory in a three-dimensional
space, which is passed by the point mass in course of time
(Fig. 2.1). The representation r D r.t/ is called parameter representation because the coordinates of the point P.t/ depend on
the parameter t.
Figure 2.2 Motion on a straight line in the x -y plan
Motions where one of the coordinates are timeindependent constants are named planar motions,
because they are restricted to a plane (in our example
the x; y-plane)
2. Planar circular motion
We can describe this motion by the coordinates R and
' (Fig. 2.3), where R is the radius of the circle and
'.t/ the angle between the x-axis and the momentary
radius vector R.t/. From Fig. 2.3 the relations
x D R cos !t ; y D R sin !t ;
R D const ;
! D d'=dt :
can be derived. Squaring of x and y yields
x2 C y2 D R2 .cos2 !t C sin2 !t/ D R2 ;
Figure 2.1 Illustration of a trajectory
which is the equation of a circle with radius R. The
point mass m with the coordinates fx; y; 0g moves with
the angular velocity ! D d'=dt and the velocity v D
R ! on a circle in the x; y-plane.
2.2
Velocity and Acceleration
41
Figure 2.3 Circular motion
J
Note: The point mass moves relative to a chosen coordinate system (in our case a plane system with the origin at
x D y D 0). The description of this motion depends on
the choice of the reference frame (coordinate system) (see
Chap. 3).
Example
2.2
Velocity and Acceleration
For a uniformly moving point mass the position vector
rDvt
with
(2.1)
v D fvx ; vy ; vz g D const ;
increases linearly with time. This means that in equal time intervals t equal distances r are covered.
The ratio v D r=t is the velocity of the point mass. The unit
of the velocity is Œv D 1 m=s.
A motion where the magnitude and the direction of the velocity vector v is constant, i. e. does not change with time, is called
uniform rectilinear motion (Fig. 2.5). In Cartesian coordinates
with the unit vectors eO x ; eOy ; eO z , the velocity vector v can be written as
v D vx eO x C vy eOy C vz eO z
˚
or v D vx ; vy ; vz :
Equation 2.1 reads for the components of v as
x D vx t I
y D vy t I
z D vz t :
(2.1a)
Example
Uniform motion along the x-axis:
vx D v0 D const I
vy D vz D 0 ! v D fv0 ; 0; 0g :
The trajectory is the x-axis and the motion is x D v0 t. J
Figure 2.4 Part of the moon trajectory described in two different coordinate systems. a Origin in the mass centre of the moon-earth system,
located in the focal point of the ellipse; b origin in the centre of the sun.
The deviations from the elliptical path of the mass centre earth-moon
are here exaggerated in order to illustrate these deviations. In reality the
orbit of the moon around the sun is always concave, i. e. the curvature
radius always points towards the sun. The orbital plane of the moon is
inclined against that of the earth
In general the velocity will not be constant but can change with
time its magnitude as well as its direction. Let us regard a point
mass m, which is at time t in the position P1 (Fig. 2.6). Slightly
The orbital motion of the moon around the earth is approximately an ellipse if r.t/ is measured in a coordinate
Figure 2.5 Uniform motion on a straight line
Chapter 2
system with the origin in the centre of mass of the earthmoon system.(Fig. 2.4a). If one chooses, however, the
centre of the sun as origin, the trajectory is much more
complex (Fig. 2.4b), because now two motions are superimposed: the orbit around the centre of mass and the
J
motion of the centre of mass around the sun.
42
2 Mechanics of a Point Mass
We will now discuss the time dependence of the velocity v in
more detail: Let us regard a point mass with the velocity v.t/ at
the point P1 of the curve v.t/. At a slightly later time t C t the
point mass has arrived at P2 and has there generally a different
velocity v.t C t/ (Fig. 2.7). We define the mean acceleration
a as
aD
Figure 2.6 Non-uniform motion on an arbitrary trajectory in space
Chapter 2
later at the time t C t it has proceeded to the point P2 . The
ratio
!
P1 P2
r.t C t/ r.t/
r
D
D
Dv
.t C t/ t
t
t
is the average velocity v over the distance P1 P2 .
For t ! 0 the two points P1 and P2 merge together and we
define as the momentary velocity v.t/ the limiting value
r.t C t/
t!0
t
v.t/ D lim
r.t/
D
dr
D rP ;
dt
which equals the time derivative of the function r.t/. In order
to distinguish this time derivative dr=dt D rP .t/ from the spatial
derivative y0 .x/ D dy=dx the time derivative is marked by a point
instead of an apostrophe.
Since the derivative df =dx of a function f .x/ gives the slope of
the curve f .x/ at the point P.x; y/ the velocity v has at any point
the direction of the tangent (Fig. 2.6). Its magnitude is in Cartesian coordinates:
q
p
(2.2)
v D jvj D vx2 C vy2 C vz2 D xP 2 C yP 2 C zP 2 :
Examples
1. Linear accelerated motion
z D a t2 ! vz D zP D 2a t :
For a D const the velocity increases linearly with
time. For a D g=2 this describes the free fall with
the initial velocity vz .t D 0/ D 0 (see Sect. 2.3.1).
Here only the magnitude, not the direction of the velocity changes with time.
2. Uniform circular motion
9
x D R cos !t ) xP D R ! sin !t >
=
y D R sin !t ) yP D R ! cos !t
>
;
zD0
) zP D 0
p
! jvj D xP 2 C yP 2 C zP 2 D R ! :
v.t C t/
t
v.t/
:
Analogous to the definition of the momentary velocity the momentary acceleration is the limit
v.t C t/
t!0
t
a.t/ D v.t/
P D rR.t/
a.t/ D lim
v.t/
D
dv
D v.t/
P D rR .t/
dt
(2.3)
The acceleration a.t/ is the first time derivative dv=dt of the
velocity v.t/ and the second derivative d2 r=dt2 of the position
vector r.t/. a.t/ D fax ; ay ; az g is a vector and has the dimensional unit Œa D Œ1 m=s2 .
2.3
Uniformly Accelerated Motion
A motion with a D const where the magnitude and the direction
of a do not change with time is called uniformly accelerated
motion. It is described by the equation
rR .t/ D a D const :
(2.4)
Equation 2.4 is named differential equation because it is an
equation between the derivative of a function and other quantities (here the constant vector a).
The vector equation (2.4) can be written as the corresponding
three equations for the components
xR .t/ D ax
yR .t/ D ay
zR .t/ D az :
The equation of motion (2.4) is readily solvable. The velocity if
obtained by integrating (2.4) which yields:
Z
v.t/ D rP.t/ D a dt D a t C b :
(2.5)
For ! D const the magnitude of v does not change, only
its direction.
J
Figure 2.7 Definition of acceleration
2.3
The integration constant b (b is a vector with constant components) can be defined by choosing the initial conditions for the
motion. For t D 0 is rP .0/ D v.0/ D b. I. e. the constant b gives
the initial velocity v.0/ D v0 .
Uniformly Accelerated Motion
z
–v
h
z(t)
v=–gt
Further integration of (2.5) gives the trajectory r.t/
c D r.0/ D r0 :
(2.6)
y.t/ D 21 ay t2 C v0y t C y0 ;
z.t/ D
1
a
2 z
(2.6a)
2.3.2
Projectile Motion
2
t C v0z t C z0 :
One should realize the following statement:
All functions f .x/ C c with arbitrary constants c have the same
derivative y0 D f 0 .x/ because the derivative of a constant is zero.
This implies:
All functions f .x/ C c, which represent an infinite parametric
curve family, are solutions of the differential equation y0 D
f 0 .x/. Therefore infinitely many position vectors r.t/ are found
for the same velocity v.t/. Only the initial conditions select one
specific position vector.
We will illustrate this by several examples in the next sections.
2.3.1
As starting point we choose x.0/ D y.0/ D 0; z.0/ D h; and
the z-axis is again the vertical direction, while the x-axis marks
the horizontal direction, so that the trajectory for the projectile
is in the x-z-plane (Fig. 2.9). The initial velocity should be v0 D
fv0x ; 0; v0z g. The acceleration is a D f9; 0; gg. Equation 2.6
becomes then
x.t/ D v0x t ;
y.t/ D 0 ;
z.t/ D
1 2
gt
2
C v0z t C h :
The motion is therefore a superposition of a uniform straight
motion into the x-direction and a uniformly accelerated motion
into the z-direction. For v0z D 0 we obtain the special case of
the horizontal throw and for v0x D 0 the vertical throw.
Elimination of t D x=v0x yields the projectile parabola
The Free Fall
We choose the vertical direction as the z-axis. A body experiences in the gravitational field of the earth the acceleration
ax D ay D 0 ;
az D g D 9:81 m=s2 ;
z.x/ D
When a body at rest falls at time t D 0 from the height h, the
initial conditions are x.0/ D y.0/ D 0: z.0/ D h; vx .0/ D
vy .0/ D vz .0/ D 0.
v 2 sin ' cos '
v0x v0z
D 0
:
g
g
Ch:
(2.7)
The derivative gives vz .t/ D g t. The motion z.t/ plotted
p in
the z-t-plane represents a parabola (Fig. 2.8). For t D 2h=g
the body has reached the ground at z D 0. The falling time for
the distance h is
p
tfall D 2h=g ;
(2.8)
p
and the final velocity at z D 0 is vmax D 2hg.
(2.9)
(2.10)
For a given value of the initial velocity v0 the maximum of xs is
achieved for ' D 45ı . In order to calculate the projectile range
With these initial conditions the system of equations (2.6a) reduces to
1 2
gt
2
1 g 2 v0z
x C
xCh :
2
2 v0x
v0x
The value x D xs where the maximum occurs is found for
dz=dx D 0.
xS D
where the numerical value is obtained from experiments.
z.t/ D
t
Figure 2.8 Path-time function z .t / (red curve ) and velocity-time function (dotted line )
This vector-equation can be written for the 3 components
x.t/ D 12 ax t2 C v0x t C x0 ;
√2h/g
Figure 2.9 Projectile motion
Chapter 2
r.t/ D 12 at2 C v0 t C c with
43
44
2 Mechanics of a Point Mass
xw we solve (2.9) for z.xw / D 0. This gives
v0x v0z
xW D
˙
g
"
v0x v0z
g
2
2v 2
C 0x h
g
#1=2
:
(2.11)
Chapter 2
Since xw > 0 only the positive sign is possible. With the relation
vz0 vx0 D 12 v02 sin 2' we can transform (2.11) into
"
#
2gh 1=2
v0
2
sin 2' v0 C v0 C
xW D
:
(2.12)
2g
sin2 '
The optimum angle 'opt for achieving the largest throwing range
for a given initial velocity v0 is achieved when dxw =d' D 0.
This gives
1
0
B
'opt D arcsin @ q
1
2C
2gh=v02
C
A :
The scalar product of two vectors becomes zero, if either at least
one of the vectors is zero or if the two vectors are orthogonal.
Since eO t ¤ 0 and dOet =dt ¤ 0 it follows
(2.13)
For thep special case h D 0 (2.13) simplifies because of
arcsin. 2=2/ D =4 to 'opt D 45ı (see the detailed derivation of (2.13) in the solution of Problem 2.5c).
2.4
Figure 2.10 a uniform circular motion, b Illustration of the angular velocity
Motions with Non-Constant
Acceleration
While the differential equation for motions with constant acceleration is elementary integrable this might not be true for
arbitrary time dependent accelerations. We will at first treat the
simple example of the uniform circular motion, where the magnitude of the acceleration is constant but not the direction.
dOet
? eO t :
dt
This means that the acceleration a is orthogonal to the velocity
v which is collinear with eO t . The vector dOet =dt gives the angular velocity of the tangent to the circle. Since the radius vector
R is orthogonal to the vector v both vectors turn with the angular velocity ! D d'=dt. This means that the magnitude is
jdOet =dtj D !. This gives for the acceleration
aDv
dOet
D R ! 2 eO a D R! 2 rO ;
dt
(2.14)
where the unit vector ea D R=R always points into the direction towards the centre of the circle, and rO D r=jrj points into
the opposite direction.
Proof
2.4.1
Uniform Circular Motion
For the uniform circular motion equal distances are gone for
equal time intervals. This means that the magnitude of the velocity v is constant and the component a' of the acceleration
a D far ; a' g in the direction of v must be therefore zero.
The path length s on the circle arc for the angle ' is s D
R ' (Fig. 2.10a). The magnitude of the velocity is then
vD
ds
d'
DR
D R! :
dt
dt
The quantity ! D d'=dt is the angular velocity with the dimension Œ! D Œrad=s.
The acceleration is now
dv
dOet
dv
d
D .vOet / D
eO t C v
dt
dt
dt
dt
deOt
Dv
because v D const :
dt
aD
Because eO 2t D 1 ! 2Oet dOet =dt D 0.
)
(
R cos !t
rD
R sin !t
)
(
R ! sin !t
vD
R ! cos !t
(
)
R! 2 cos !t
aD
D ! 2 r D R! 2 rO :
R! 2 sin !t
J
The vector of the acceleration for the uniform circular motion
aD
R! 2 rO with
jaj D R ! 2
is called centripetal-acceleration because it points towards the
centre of the circle (Fig. 2.11).
If also the orientation of the plane in the three-dimensional space
should be defined, it is useful to define a vector ! of the angular
velocity which is vertical to the plane of motion (Fig. 2.10b) and
has the magnitude ! D j!j D d'=dt D v=R.
Motions with Non-Constant Acceleration
45
Chapter 2
2.4
Figure 2.11 Rollercoaster, where the superposition of centripetal acceleration and gravity changes along the path and inluences the feelings of the passenger
(with kind permission of Foto dpa)
2.4.2
Motions on Trajectories with Arbitrary
Curvature
The change of the magnitude of the velocity is described by at
while the change of the direction of v is described by an .
In the general case the velocity v will change its magnitude as
well as its direction with time. However, the momentary velocity v.t/ at time t is always the tangent to the trajectory in the
point P.t/, while the acceleration a.t/ can have any arbitrary direction (Fig. 2.12). The acceleration can be always composed of
two components at D dv=dt eO t along the tangent to the curve
(tangential acceleration) and an in the direction of the normal
to the tangent, i. e. perpendicular to at (normal acceleration).
For v D v eO t where eO t is the unit vector tangential to the trajectory, the acceleration a is
aD
dv
dOet
dv
D
eO t C v
D at C an :
dt
dt
dt
(2.15)
Figure 2.12 Tangential and normal acceleration
46
2 Mechanics of a Point Mass
For the small section of the curve we get
ds D %d'
d' ds
d'
1
d'
D
D
vD v:
dt
ds dt
ds
%
(2.16a)
(2.16b)
The acceleration vector becomes
v2
dv
eO t C eO n
dt
%
aD
(2.16c)
Chapter 2
Examples
Figure 2.13 a Derivation of the normal acceleration. b Local radius of curvature of a trajectory with arbitrary curvature %
1. Assume a motion on a straight line experiences the
acceleration a.x/ D b x4 .
Calculate the velocity v.x/ for the initial condition
v.0/ D v0 .
Solution
For an D 0 the trajectory is a straight line, where the body
moves with changing velocity if at ¤ 0. For at D 0 the point
mass moves with a constant velocity jvj on a curve which is determined by an .t/. For the free fall of Sect. 2.3.1 is an D 0 and
at D const., while for the uniform circular motion at D 0 and
an D const.
For the motion on trajectories with arbitrary curvature the acceleration can be obtained as follows: We choose the x-y-plane as
the plane of the two vectors v.t/ and a.t/, which implies that all
vectors have zero z-components.
dv
dv dx
dv
aD
D
D
v ;
dt
dx dt
dx
d'
d'
sin ' eO x C cos ' eOy
dt
dt
d'
D
eO n
dt
The normal acceleration is therefore
an D v
d'
eO n :
dt
We regard in Fig. 2.13b an infinitesimal section between the
points A and A0 of an arbitrary curve and approximate this
section by a circular arc AA0 with the center of curvature M.
Shortening the section AA0 more and more, i.e. the points A
and A0 converge towards the point P1 the curve section AA0 approaches more and more the circular arc with radius MP1 . The
radius % D MP1 is the radius of curvature of the curve in the
point P1 .
Zv
vdv :
v0
Inserting a and integration yields
1
b
5
x5
x50 D
1
2
v 2 .x/
Resolving this equation for v.x/ gives
v.x/ D
q
2
b
5
x5
v02 :
x50 C v02 :
2. The open parachute of a parachutist experiences, due
to air friction, a negative acceleration besides the acceleration by gravity.
a D b v2
There we get
dOet
D
dt
adx D
x0
According to Fig. 2.13a the two mutually vertical unit vectors eO t
and eO n can be composed as
eO t D cos ' eO x C sin ' eOy
eO n D cos.' C /Oex C sin.' C /Oey
2
2
D sin ' eO x C cos ' eOy
Zx
with
b D 0:3 m
1
:
a) What is his constant final velocity ve ?
b) What is the time-dependent velocity v.t/, if the
parachutist opens his parachute only after t0 D 10 s
free fall for which friction can be neglected?
Solution
a) A constant final velocity is reached, when the total acceleration becomes zero. This is the case when
p
g b ve2 D 0 ! ve D g=b D 5:7 m=s :
b) The equation of motion after the parachute is opened is
with the z-axis in the vertical direction
zR D g
b zP2 :
2.5 Forces
by the concept of forces. When a body changes its state of motion we say that a force acts upon the body.
2
With v D dz=dt and dv=dt D d z=dt we obtain
dv=dt D b
If, for instance, two bodies collide we say: Each of the two
bodies has exerted during the collision a force onto the other
body, which causes a change of the state of motion for both
bodies.
b v2 ;
which leads to the equation
Zv
v0
dv
1
D
2
g bv
g
Zv
v0
1
dv
D
v 2 =ve2
Zt
dt0 D t
t0 :
t0
We substitute v=ve D x, for x > 1 i. e. for v > ve we
get
Z
dx
1 xC1
D ln
1 x2
2 x 1
1 ve v C ve
CC :
! t t0 D
ln
2 g
v ve
For t D t0 ! v D v0 D g t0 D 98:1 m=s. This gives
for the integration constant C the value
1 ve v0 C ve
ln
2 g
v0 ve
v C ve v0 ve
1 ve
t0 D
:
ln
2 g
v ve v0 C ve
CD
!t
A body without any interaction with its surroundings (or for
which the vector sum of all forces is zero), is called a free body.
A free body does not change its state of motion. Strictly speaking there are in reality no free bodies without any interaction
(because we would not see them). However, in many cases the
interaction is so small, that we can neglect it. Examples are
atoms in a tank where a very good vacuum has been established,
or a sliding carriage on a nearly frictionless horizontal air track.
Such free bodies move uniformly on a straight trajectory. For
such cases the model of a free body is justified.
2.5.1
Forces as Vectors; Addition of Forces
Since velocity changes which are caused by forces are vectors,
also forces must be described by vectors, i. e. they are defined
by their magnitude and their direction.
Eliminating v from this equation for v yields
d ec.t t0 / C 1
with
d ec.t t0 / 1
v0 C ve
and c D 2g=ve :
dD
v0 ve
v.t/ D ve
The velocity decreases from the initial value v.t0 / D v0
at t0 exponentially to the final value ve for t D 1.
However, already after t t0 D 2ve =g D 1:16 s the
velocity has reached 96.7% of its final value.
J
2.5
Forces
We will now discuss the question, why a body performs that motion that we observe, why for instance the earth moves around
the sun on an elliptical trajectory, or why a stone in a free fall
moves on a vertical straight line to the ground.
Newton recognized that the cause for changes of a body’s velocity must be interactions of the body with its surroundings. These
can be long range interactions such as the gravitational interaction between the sun and the earth, or short range interaction
which work for example in collisions between colliding billiard
balls, or even ultrashort range strong interactions between neutrons in an atomic nucleus. All such interactions are described
Figure 2.14 Vector sum of forces. a all forces act on the same point, b equivalent representation of the vector sum
Chapter 2
2
47
48
2 Mechanics of a Point Mass
Note: When forces act on extended bodies, also the point of
origin is important (see Sect. 5.4).
Chapter 2
A force, as any vector, can be reduced to the sum of its components. This reduction depends on the chosen coordinate system.
For example in Cartesian coordinates the vector and its components are F D fFx ; Fy ; Fz g. If we choose the coordinate system
in such a way that the z-direction points into the direction of
F, the component representation becomes F D f0; 0; Fz D Fg
with F D jFj. Often the solution of a problem can be essentially simplified by choosing the optimum coordinate system
(see Sect. 2.3.2). If several forces act on a body the total force is
the vector sum of the individual forces (superposition principle)
X
FD
Fi :
2. A circular pendulum is a mass m hold by a string
which is fixed at a point P. The mass can move on
a circle in the x-y-plane while the string movement
forms the surface of a cone (Fig. 2.16). The total force
F D m g C Fel as the sum of gravitational force and
elastic force of the stressed string acting on the mass m
always points towards the centre of the circle in the xy-plane and acts as centripetal force which causes the
circular motion of m.
r
i
This vector equation is equivalent to the three equations for the
components
X
X
X
Fiz :
Fiy Fz D
Fix Fy D
Fx D
i
i
i
The addition of several vectors is illustrated in Fig. 2.14a and b.
Both ways to add vectors are P
equivalent, because the origin of
the vectors can be shifted. If
Fi D 0 the total force is zero
and the body remains in its constant state of motion (either at
rest or in a uniform motion on a straight line.
Restoring
spring force
F
m·g
Figure 2.16 Circular pendulum with the vector diagram
2.5.2
J
Force-Fields
Examples
1. A body with mass m rests on a friction-free sloped
plane (Fig. 2.15). The gravitational force can be
regarded as the vecor sum of the two forces F? perpendicular to the sloped plain and Fk parallel to this
plane. F? exerts a force onto the surface of the plane
and causes an opposite force N of equal magnitude by
the elastic response of the surface. Only the force Fk
can cause an acceleration of the body. It can be compensated by an opposite force Z in order to reach a
zero total force and keep the body at rest on the sloped
plane. This situation can be described by the equation
m g D Fk C F? D .Z C N/ :
Attractive force Z and elastic force N compensate the
gravitational force and the body remains at rest.
Often the force acting on a body depends on the location. If it is
possible to unambiguously assign to each point .x; y; z/ a force
with defined magnitude and direction the spatial force function
F.x; y; z/ is called a force-field. Its components depend on the
chosen coordinate system:
F.r/ D F.x; y; z/
F.r; #; '/
F.r; '; z/
in Cartesian coordinates, or
in spherical coordinates, or
in cylinder coordinates.
In a graphical representation the direction of the force is illustrated by “force-lines” where the force at any point .x; y; z/ is
the tangent to the force-line (Fig. 2.17).
If the force has for any point in space only a radial component
with a magnitude which depends on the distance r to the centre
r D 0 the force field is centro-symmetric and is called a central
force field. It can be written as
F D f .r/ rO ;
where rO D r=jrj is the unit vector in radial direction. The sign
of the scalar function f .r/ is: f .r/ < 0 if the forces point to the
centre and f .r/ > 0 is it points from the centre away.
Surfaces where the force field has the same magnitude are called
equipotential surfaces. (see Sect. 2.7.5)
Figure 2.15 Equilibrium of forces for a body on an inclined plane
Central force fields are spherical symmetric.
Examples
1. Central force fields
a) Gravitational force field of the earth (Fig. 2.17a)
F depends on the distance for the earth’s centre. For the idealized case that the earth can be
described by a homogeneous sphere with spherical symmetric mass distribution (see Fig. 2.9)
the gravitational force is for r > R (R D
radius of the earth)
FD G
mM
rO
r2
(M D mass of earth, m D mass of body, G D
gravitation constant, unit vector rO D r=jrj)
b) Force field of a positive electric charge Q
(Fig. 2.17b).
In the electric force field of an electric charge Q the
force on a small test charge q is
FD
2. Non-central force fields
a) Dipole force field
The force field in the surrounding of two charges
CQ and Q with equal magnitude but opposite
sign is no longer spherical symmetric. The force
on a test charge not only depends on the distance
from the centre of the two charges but also on
the angle # of the position vector against the connecting line of the two charges (Fig. 2.18). The
calculation of the force field gives (see Sect. 1.5 of
Vol. 2)
1
qQ 1
rO 1
rO2 :
F D F1 C F2 D
4"0 r12
r22
1 qQ
rO I
4"0 r2
("0 = dielectric constant see Vol. 2). The spherical symmetric force field has the same form as the
gravitational force field.
Figure 2.18 Force field of an electric dipole and the force on a negative
test charge
b) Force field of a planetary system
At each position r the gravitational forces on a test
mass exerted by the sun, the planets and
Pthe moons
superimpose. The force field F.r/ D
Fi is very
complex. It even can be zero at certain points in
space, for example at a point N between earth and
moon (neutral point) where the opposite gravitational forces from earth and moon just compensate
(Fig. 2.19).
Figure 2.17 Spherical symmetric force fields a gravitational force field
of a mass M (attractive force) and b electric force field of a positive
charge Q and repulsive force on a positive test charge
Figure 2.19 Gravitational field between earth and moon
49
Chapter 2
2.5 Forces
50
2 Mechanics of a Point Mass
Chapter 2
c) Homogeneous force field of a parallel plate capacitor
For a voltage V between the plates with a distance d the force on an electric charge Cq is F D
Cq .V=d/ eO z vertical to the plates and pointing
from the positively charged plate to the negative
one (Fig. 2.20). The force F has at any point inside
the capacitor the same magnitude and direction.
Such a force field is called homogeneous.
Within a small volume also the gravitational field
of the earth can be treated as a homogeneous force
field as long as the vertical extension z of this
volume is very small compared to the radius R of
the earth. The force on a mass is then F D m g,
where jgj D 9:81 m=s2 is the earth gravitational
acceleration which remains constant in a small volume.
Z
Comet
Fg
Figure 2.22 Interaction between sun and comet as an example for the far
distance effect of forces
displaced from its equilibrium position x0 and then released, it
performs oscillations around x0 (see Sect. 2.9.7).
+
Often forces can act on bodies without physical contact between
them. Examples are the gravitational force between sun and
earth or between sun and a comet (Fig. 2.22). In the latter case
the comet is attracted by the sun due to the gravitational force
and vice versa. Its tail is repelled because of the radiation pressure and the sun wind which is exerted by particles (protons and
electrons) emitted from the sun.
Figure 2.20 Homogeneous force field for electric charges inside a parJ
allel plate capacitor
Measurements of Forces; Discussion of
the Force Concept
Forces can be measured due to their effect on the deformation of
elastic bodies (see Chap. 6). One example is the spring balance
(Fig. 2.21). Here the elongation of a spring under the influence
of a force is measured. Its displacement x x0 from the equilibrium position x0 is proportional to the acting force
Fx D D.x
Train
–
d
2.5.3
FL
x0 / :
(2.17)
If the spring constant D D F=x is known, the determination of
the force F is reduced to a length measurement x D x x0 . The
spring constant D can be obtained from measurements of the
oscillation period of the spring balance. After a mass m has been
Figure 2.21 Spring balance for the measurement of forces
Even if there is no direct contact between two bodies we say that
a force acts on each body which causes the change of its motional state i. e. its velocity with time. Also for the investigation
of atomic collision processes the information on the forces between the colliding atoms is obtained from the observed change
of the velocities of the two collision partners (see Sect. 4.3).
Here the change of the momentum dp=dt is used to determine
the force. This explanation goes beyond the ordinary meaning
of forces as directly perceptible phenomena as for instance the
physical strength.
In all cases the force is a synonym for the interaction between
bodies. The range of distances between the interacting bodies
can reach from 10 17 m to infinity.
The question, what the real cause for this interaction is and
whether it is transferred between the interacting bodies infinitely
fast or with a finite speed can be up to now only partly answered
and is the subject of intense research but is not yet fully understood. Theoretical predictions claim a finite transfer time which
equals the speed of light. The description of the interaction between very fast moving bodies has therefore to take into account
this finite transfer time (retardation, see Sect. 3.5). For velocities which are small compared to the speed of light this effect
can be neglected (realm of non-relativistic physics).
We will now discuss more quantitatively the relations between
forces and the change of motional states of bodies.
2.6 The Basic Equations of Mechanics
The Basic Equations of
Mechanics
The mathematical description of the motion of bodies under the
influence of forces can be reduced to a few basic equations.
These equations are based on assumptions (axioms) which are
suggested by experiments. They were first postulated by Isaac
Newton in his famous multi-volume opus “Philosophiae naturalis principia mathematica” which was published in the years
1687–1726 [2.1].
Figure 2.23 Forces as cause for a change of momentum
With p D m v we can write this in the form
2.6.1
The Newtonian Axioms
FDm
For the introduction of the force model and its relation with the
state of motion of bodies Newton started from three basic assumptions which were taken from daily experience. They are
called the three Newtonian axioms (sometimes also Newton’s
three laws).
dm
dv
C
v:
dt
dt
(2.18a)
The second term describes a possible change of the mass m with
the velocity of the particle. There are many situations where
this second term becomes important, for instance when a rocket
is accelerated by the expulsion of fuel (see Sect. 2.6.3) or when
a particle is accelerated to very high velocities, comparable to
the velocity c of light, where the relativistic mass m.v/ increase
with velocity, cannot be neglected (see Sect. 4.4.1).
First Newtonian Axiom
Example
Each body remains in the state of rest or of uniform motion
on a straight line as long as no force is acting on it.
A freight train moves with the velocity v in the horizontal x-direction (Fig. 2.24). It is loaded continuously with
sand from a stationary reservoir above the train. The mass
increase per time dm=dt is assumed to be constant. When
friction can be neglected the total force onto the train is
zero. The equation of motion is then
As the measure for the state of motion of a body with mass m
we define the momentum
0 D m dv=dt C A v
pDmv :
The momentum p is a vector parallel to the velocity v and has
the dimension Œp D Œkg m s 1 . A particle on which no force
is acting is called a free particle.
(2.18b)
with m D m0 C A t. Integration yields
v
m0
ln
D ln
v0
m0 C A t
With this definition Newton’s first law can be formulated as
The momentum of a free particle is constant in time.
This means: always when a particle changes its state of motion
a force is acting on it, i. e. it interacts with other particles or it is
moving in a force field (Fig. 2.23).
Figure 2.24 Example to Eq. 2.18a
with the solution
Second Newtonian Axiom
Since we attribute a force to any change of momentum we
define the force F as
dp
FD
:
dt
(2.18)
v.t/ D v0
1
:
1 C .A=m0 / t
(2.18c)
With m0 D 1000 tons and dm=dt D A D 1 ton=s the train
velocity v.t/ D v0 .1 C 1 C 10 3 t/ 1 the velocity slows
down to v0 =2 in 1000 s.
J
Chapter 2
2.6
51
52
2 Mechanics of a Point Mass
Strand
Spring
Figure 2.25 actio D reactio for the example of gravitational forces F1 D
between two masses
F2
Chapter 2
If the mass m is constant .dm=dt D 0/ Eq. 2.18b takes the simple
form
FDma
with
aD
dv
:
dt
The unit of the force is ŒF D 1 kg m s
2
Air track
Figure 2.26 Experiment to prove the 3. Newtonian law a with two equal
spring balances, b with two equal masses on an air track
(2.18d)
D 1 Newton D 1 N.
Figure 2.27 The gravitational force F D m g of a mass m on a solid surface
is compensated by the antiparallel deformation force of the solid surface
Third Newtonian Axiom
When two bodies interact with each other but not with a
third partner the force acting on the first body has equal
magnitude but opposite direction as the force on the second body (Fig. 2.25). Newton’s formulation in Latin was
actio D reactio
F1 D F2 :
We will apply Newton’s axioms to a system of two masses m1
and m2 which interact with each other, i. e. they collide, but are
otherwise completely isolated from their surroundings. Such a
system is called a closed system.
Since there are no external forces on a closed system we can
conclude in analogy to a free particle that the total momentum
of the system remains constant:
p1 C p2 D const :
(2.19a)
Differentiating this equation yields
dp1
dp
C 2 D 0 ) F1 D F 2 :
dt
dt
(2.19b)
which are hold together by a string. If the string is burnt by
a candle, the two masses are pushed by the expanding spring to
opposite sides and slide on the air track with equal velocities,
which means that they have equal but opposite momenta. The
velocities can be accurately measured by photoelectric barriers.
Newton’s third law can be also proved for resting bodies. A
mass m resting on a solid surface acts with the gravitational
force F1 D m g on the surface which is deformed and responds
with an equal but opposite elastic force Fel D F1 D mg
(Fig. 2.27).
2.6.2
Inertial and Gravitational Mass
The property of bodies to remain in their state of motion when
left alone (i. e. when no force is acting on them) is called their
inertia. Since the accelerating force is proportional to the mass
of the body its mass can be regarded as the cause of the inertia
and is therefore called the inertial mass minertial . Newton’s second law means this inertial mass. There are many demonstration
experiments which illustrate this inertia. Assume, for example,
a glass of water standing on a sheet of paper. If the paper is
pulled suddenly away, the glass remains a rest without moving,
because of its inertia.
This axiom can be proved experimentally with two equal spring
balances (Fig. 2.26a), which are connected to each other at one
end. If one pulls at the two other ends into opposite directions
they show that on each spring balance the same force is acting.
There is another property of masses which is the gravitational
force (Fgrav D m g on the earth surface). This force is also
called the weight of the mass. Experiments measure the weight
of a mass of 1 kg as
Another experimental verification is shown in Fig. 2.26b where
a spring is compressed by two equal masses on an air track
Fgrav D 1 kg 9:81 m=s2 D 9:81 N :
2.6 The Basic Equations of Mechanics
2.6.3
53
The Equation of Motion of a Particle in
Arbitrary Force Fields
Integration of Newton’s equation of motion F D mdv=dt yields
the equations
Note: The gravitational force is always present when the mass
m is attracted by another mass M and it is proportional to the
product m M (see Sect. 2.9.2).
The question is now: Are these two properties related to the
same mass i. e. is minertial D mgrav ?
1
m
Z
Z
(2.20a)
Fdt C C1 ;
v.t/ dt C C2
Z Z
Z
1
Fdt dt C C1 dt C C2
D
m
r.t/ D
(2.20b)
For the velocity v.t/ and the position vector r.t/ with the integration constants C1 and C2 which are fixed by the initial conditions
(e.g. v.t D 0/ D v0 and r.t D 0/ D r0 ).
Whether these equations are analytical solvable depends on the
form of the force F which can be a function of position r, velocity v or time t. We will illustrate this by some examples.
Many detailed and accurate measurements for many different
masses have proved that within the relative uncertainty of 10 10
there is no measurable difference between minertial and mgrav .
Constant Forces
Starting from this experimental result Einstein has postulated
the general equivalence principle that inertial and gravitational masses are always equal.
For the most simple case of constant forces F D const, which
do not depend on time nor on the position or velocity of the
particle the integration of (2.20) immediately gives
By the following “Gedanken-experiment” he has shown, that it
doesn’t make sense to distinguish between inertial and gravitational masses:
An observer in a closed lift measures a mass m hanging on a
spring balance (Fig. 2.28). He cannot distinguish, whether the
elevator is resting in a gravitational field with the gravitational
force Fgrav D m g on the mass m (Fig. 2.28a) or whether the
elevator moves upwards with the velocity v D gt and the acceleration g in a force-free surrounding (Fig. 2.28b). Both
situations lead to the same elongation of the spring balance. Any
further experiment performed inside the closed elevator leads to
the same results for the two situations (a) and (b).
For instance when the observer in the elevator throws a ball in
the horizontal direction the trajectory of the ball is for both situations a parabola (see Fig. 2.9).
We will therefore no longer distinguish between inertial and
gravitational mass and call it simply the mass m of a body which
has the two characteristic features of inertia under acceleration
and weight in gravitational fields.
Note: The question what the mass of a body really means is up
to date not answered, although great efforts are undertaken to
solve this problem.
F D m a D const ;
v.t/ D at C C1 with C1 D v0 D v.t D 0/ ;
r.t/ D
1 2
at
2
C v 0 t C r0
with
(2.21)
r0 D r.t D 0/ :
The trajectory of the particle can be directly determined, if the
initial conditions are known. It is advisable to choose the coordinate system in such a way that the force coincides with one of
the coordinate axes.
Example
The motion of a particle under the influence of the constant force F D f0; 0; mgg pointing into the z-direction
gives the three equations for the 3 components of the force
xR D 0 ) xP D Ax
) x D Ax t C Bx
yR D 0 ) yP D Ay
) y D Ay t C By
zR D g ) zP D gt C Az ) z D 21 gt2 C Az t C Bz :
(2.22)
These equations describe every possible motion of particles under the influence of the earth gravitation in a
volume which is small compared with the dimensions of
the earth where the gravitational force can be regarded as
constant. From the many possible solutions of (2.22) the
initial conditions with fixed values of A and B select special solutions (see examples in Sect. 2.3).
J
Chapter 2
Figure 2.28 Einstein’s Gedanken-experiment for the equivalence of gravitational and inertial mass a in the homogeneous gravitational field of the earth;
b in a gravitation-free space inside an accelerated lift
v.t/ D
54
2 Mechanics of a Point Mass
p
2Rg the maximum vertical
For the initial velocity v0 !
height rmax becomes infinity and the projectile can leave the
earth. This velocity is called the escape velocity. Inserting the
numerical values for R and g gives
v0 v2 D
p
2Rg D 11:2 km=s :
(2.26a)
(escape velocity)
Chapter 2
The velocity v2 is often named the 2nd cosmic velocity while
the first cosmic velocity v1 is the velocity of a projectile which
is fired in horizontal direction and orbits around the earth on a
circle closely above the earth surface. From the relation
v12
GM
D 2 ! v1 D
R
R
Figure 2.29 Launch of a body from the earth surface
As an example of position dependent forces we choose the gravitational force
mM
rO :
r2
The acceleration a has in this central force field only a radial
component ar D G M=r2 . For vertical motions the velocity becomes in spherical coordinates v D fvr ; 0; 0g and its
magnitude is jvj D v D vr . Our problem therefore becomes
one-dimensional. From the relation
.G M=r2 /dr.
1 2
GM
v D
C C1 :
2
r
(2.23)
1 2
v
2 0
GM
1
D v02
R
2
gR ;
gR :
(2.24)
At the maximum vertical height r D rmax the velocity becomes
v.rmax / D 0 and we obtain from (2.24)
rmax D
1
1. Assume the time dependent force F D b t C c with
b D 120 N=s and c D 40 N, which points into the
x-direction, is acting on the mass m D 10 kg. For
t D 0 the mass should be at x D 5 m with a velocity
v.0/ D 6 m=s. Calculate the position x.t/.
R
:
.v02 =2Rg/
The straight motion proceeds along the x-axis. The acceleration is a D F=m and the velocity
v.t/ D
because a.R/ D g D G.M=R2 /. This gives
gR2
1 2
1
v D
C v02
2
r
2
There are many situations where the force on a particle changes
with time. One simple example is a mass hanging on a spring,
which is induced to vertical oscillations, or a comet moving on a
parabolic trajectory through the solar system. We will illustrate
the solution of the equation of motion for time dependent forces
by two numerical examples.
Solution
Let us discuss the case that a projectile is fired from the earth
surface .r D R/ upwards in vertical direction with the initial velocity v0 (Fig. 2.29). The integration constant C1 then becomes
C1 D
Note: The general case of arbitrary motion in a central force
field is treated in Sect. 2.9.
Examples
dv
dv dr
dv
aD
D
D
v ;
dt
dr dt
dr
Integration yields
(2.26b)
Time-dependent Forces
The minus-sign indicates that the attractive force points into the
direction of r.
it follows: v dv D
p
GM
D gR
R
the numerical value
p of v1 becomes (when neglecting the earth
rotation) v1 D v2 = 2 7:9 km=s.
Forces F.r/ that depend on the Position
F.r/ D G
r
(2.25)
1
m
Z
Zt
0
F./d D
b 2
c
t C t C v0x I
2m
m
b 3
c 2
t C
t C v0x t C x0
6m
2m
D .2t3 C 2t2 C 6t C 5/m with t in s :
x.t/ D
vx d D
2. What is the final velocity of a mass m initially at
rest .v.0/ D 0/ which experiences a force F.t/ D
A expŒ a2 t2 ?
2.6 The Basic Equations of Mechanics
For the limit t ! 0; m=t ! dm=dt is limt!0 .m
v=t/ D 0.
Solution
Z
Z
Since the time derivative dp=dt of the momentum equals the
force Fg D m g of gravity acting on the rocket we obtain
p
A
2 2
e a t dt D
I
2a
F dt D A
p
A
:
v1 D
2 am
J
Acceleration of a Rocket
In the example for position dependent forces we have assumed
that the projectile starts with the initial velocity v0 > 0. In
reality it starts with v0 D 0. However, the velocity v > 0 is
reached within a short distance that is very small compared with
the earth radius R. We will now study the acceleration during
the start phase of the rocket in more detail. Within this small
distance d R, which the rocket passes during its acceleration,
we can fairly assume the earth acceleration g to be constant.
During the burning phase the rocket is continuously accelerated
by the recoil momentum of the propellant hot gases (Fig. 2.30).
With v 0 we denote the velocity of the propellant gases relative to
the surface of the earth which represents our reference coordinate system, and with v the rocket velocity in this system. The
escaping gas mass per second is m=t. The momentum of the
rocket at time t is p.t/ D m v. At time t C t the mass of the
rocked has been reduced by m (which equals the mass of the
expanding gas during this time interval) and its velocity has increased by v while the gases have transported the momentum
m v 0 . The total momentum of the system rocket + gas is then
with m > 0
p.t C t/ D .m
m/.v C v/ C m v0 :
(2.27a)
During the time interval t the momentum of the system has
changed by
p D p.t C t/ p.t/
D m v C m.v0
v/
m v :
dp
dv
dm 0
Dm
C
.v
dt
dt
dt
v/ D m g :
(2.27c)
The velocity v0 of the propellant gases relative to the earth depends on the velocity v of the rocket. For jvj < jv0 j the direction
of v0 is downwards, for jvj > jv0 j it is upwards. It is therefore
better to introduce the velocity ve D v0 v of the propellant
gases relative to the rocket, which is independent of v and constant in time. This converts Eq. 2.27c into
m
dv
dm
C
ve D m g :
dt
dt
(2.27d)
With v D f0; 0; vz g, ve D f0; 0; ve g, g D f0; 0; gg this equation becomes after division by m and multiplication by dt
dv D ve
dm
dt
g dt ;
(2.27e)
Integration from t D 0 up to t D T (propellant time of the
rocket) yields
v.T/ D v0 C ve ln
m0
m
gT ;
(2.28)
where v0 D v.t D 0/.
Numerical Example
Launching of a Saturn rocket with m0 D 3 106 kg; ve D
4000 m=s, T D 100 s, v0 D 0. Final mass at t D T is
m.T/ D 106 kg, which means that the mass of the fuel is
2 106 kg. Equation 2.28 yields
v.T D 100 s/ D 0 C 4000 m=s ln 3
D 3413:5 m=s :
9:81 m=s2 100 s
J
(2.27b)
The heights z.t/ of the rocket during its burning time for constant loss of mass q D dm=dt D const is readily obtained. With
m.t/ D m0 q t, Eq. 2.28 becomes
q
t
gt I
v.t/ D v0 ve ln 1
m0
Z
1 2
q
z.t/ D v0 t ve ln 1
t dt
gt C C0 ;
m0
2
and integration yields
z.t/ D v0 t
Figure 2.30 Acceleration of a rocket
ve
Z
ln 1
q
t dt
m0
1 2
gt C C0 :
2
The integration constant is C0 D 0 (because z.0/ D 0).
Chapter 2
m v.t D 1/ D
55
56
2 Mechanics of a Point Mass
Since
R
ln xdx D x ln x
x the integration gives
z.t/ D .v0 C ve / t C ve
m0
q
t ln 1
q
t
m0
1 2
gt :
2
(2.29)
This equation illustrates that even for z D 100 km the correction
term 2z=R for g amounts only to 3%. This means for the calculation of the velocity v only a correction of 1%, because the
term g T in Eq. 2.28 represents only about 1=3v.
The integration of (2.28) is now more tedious but an approximation is still possible, if the function (2.29) is inserted for z.t/.
Numerical Example
Chapter 2
For our example above we obtain with q D 2 104 kg=s,
v0 D 0; ve D 4000 m=s T D 100 s
z.T / D .4 105 C 2 103 ln 0:33 4:9 104 m
D .400 219:7 49/ km D 131 km ;
v.T / D 4 103 ln.0:33/ 981 m=s D 3413 m=s ;
J
This example illustrates that with z.T/ R the earth acceleration does not change much and can be regarded as constant.
It further demonstrates that with a single stage the escape velocity v D 11200 m=s of the rocket cannot be achieved with
reasonable fuel masses. It is therefore necessary to use multistage rockets.
2.7
Energy Conservation Law of
Mechanics
In this section we will discuss the important terms “work”,
“power”, “kinetic and potential energy” before we can formulate the energy conservation law of mechanics.
2.7.1
Work and Power
If a point mass m proceeds along the path element r in a force
field F.r/ (Fig. 2.31), the scalar product
W D F.r/ r
Numerical Example
After the end of the burning time T1 of the first stage
the velocity of the rocket in our example is v.T1 / D
3400 m=s. The second stage starts with a mass m.T1 / D
9 105 kg (the fuel tank with m D 105 kg has been pushed
off) including m D 7 105 kg for the fuel. The burning
time is again 100 s and the final mass m.T2 / D 2:105 kg.
According to (2.28) the final velocity v is
v.T1 C T2 / D .3400 C 4000 ln.9=2/
D 8435 m=s :
The third stage starts with a velocity v D 8435 m=s and a
mass m D 1:8 105 kg (the fuel tank of the 2nd stage with
m D 2 104 kg has been pushed off). With T3 D 100 s we
obtain the final velocity
v.Tfinal / D .8400 C 4000 ln 7:2 9:8 100/ m=s
D 15;000 m=s > vescape :
is called the mechanical work, due to the action of the force F
on the point mass m.
The work is a scalar quantity!
Written in components of the vectors F and r Eq. 2.31a reads
W D Fx x C Fy y C Fz z :
9:81 100/
dm
m
g.1
2z=R/dt :
Remark. In the cgs-system the unit is ŒW D 1 dyn cm D
1 erg D 10 7 J.
J
(2.30)
(2.31b)
The unit of work is Œwork D Œforce length D 1 N m D
1 Joule D 1 J.
Note: For the second and third stage one should, strictly speaking, take into account the decrease of the earth acceleration
g with increasing z. Instead of the constant g one should
use the function g.z/ D G M=r2 with r D z C R and
M D mass of the earth. With the approximation .1 C z=R/ 2
1 2z=R one obtains instead of (2.27e) the equation
dv D ve
(2.31a)
Figure 2.31 Definition of work
If the point mass moves under the action of the force F from
point
P P1 to point P2 the total work on this path is the sum W D
Wi of the different contributions Wi D F.ri / ri which
converges in the limit ri ! 0 to the integral
WD
ZP2
(2.32a)
F dr :
P1
The integral is called line-integral or curvilinear integral. Because of the relation F dr D Fx dx C Fy dy C Fz dz it can be
reduced to a sum of simple Rieman integrals:
Z
F dr D
Zx2
Fx dx C
x1
Zy2
Fy dy C
y1
Zz2
Fz dz ;
(2.32b)
z1
which can be readily calculated if the force is known (see the
following examples). In Equation 2.32 is W > 0 for F dr > 0
i.e. if the force F has a component in the direction of the movement. In this case the mass m is accelerated. According to this
definition the work is positive if the energy of the mass m is
increased. Work which is performed by the mass on other systems decreases its energy and is therefore defined as negative
(see Sect. 2.7.3).
Energy Conservation Law of Mechanics
3. The work performed by a mountaineer against the
gravitational force (man C pack D 100 kg), who
Rclimbs up the Matterhorn2 (z D 18003 m) is2 W2 D
Fg dz D mgz D 10 9:811:810 kgm =s D
17:6 105 J 0:5 kWh.
The work is negative, because the force is antiparallel to the direction of the movement. The mountaineer
produces energy by burning his food and converts it
into potential energy thus decreasing its internal energy. The prize for the electrical equivalent of 0:5 kWh
is about 10 Cents!
4. In order to expand a coil spring one has to apply a
force F D Fr opposite to the restoring spring force
Fr D D.x x0 / which is proportional to the elongation .x x0 / of the spring from its equilibrium position
x0 . The work which has to be applied is
Z
Z
W D Fx dx D D .x x0 /dx
D 21 D.x
x0 /2 :
This is equal to the area A in Fig. 2.32a between the
x-axis and the straight line F D D.x x0 /.
If F is perpendicular to r (and therefore also to the velocity v)
the work is W D 0, because then the scalar product F dr D 0.
The work per time unit
PD
dW
dt
(2.33a)
is called the power P. Its unit is ŒP D 1 J=s D 1 Watt D 1 W.
d
PD
dt
Zt
F.r.t0 /; t0 / rP.t0 /dt0
t0
(2.33b)
D F.r.t/; t/ v.t/ D F v :
Remark. In daily life the electrical work is defined in kWh.
With 1 J D 1 Ws the relation is 1 kWh D 3:6 106 Ws.
Examples
1. Uniform circular motion under the action of a radial
constant force. Here v always points in the direction of
the tangent to the circle, but the force is always radial,
i. e. F?v. The scalar product F v D 0 and therefore
the work is zero.
2. A mass is moved with constant velocity without friction on a horizontal plane. (motion on a straight line).
The gravitational force is always perpendicular to the
motion, ! F dr D 0. The work is zero.
Figure 2.32 a Work for expanding a spring, b work of a car climbing
up a slope
5. A car (m D 1000 kg) moves with constant velocity of
48 km=h on a straight line with a slope of 5ı against
the horizontal (Fig. 2.32b). What is the work the engine has to produce within 5 min, if friction effects can
be neglected?
The force in the direction of the motion is
F D Fg sin ˛ D m g sin ˛ :
The distance which the car moves within 5 min is
s D 48 km 5=60 D 4 km D 4000 m :
57
Chapter 2
2.7
58
2 Mechanics of a Point Mass
The work is then with 1 kWh D 103 3:6 103 Ws D
3:6 106 J
In conservative force fields the work for moving a mass m
on a closed loop is zero.
W D 4 103 9:81 sin 5ı 103 N m
Wa
6
D 3:4 10 J 1 kWh :
Wb D
ZP2
F dra
P1
The power is
Chapter 2
dW
3:4 106 J
PD
D
1:13 104 W
dt
300 s
D 11:3 kW :
ZP2
D
ZP2
P1
F dra C
P1
J
D
I
F drb
ZP1
(2.34)
F drb
P2
F dr D 0 :
The work depends only on initial and final position of the
motion, not on the chosen path between them.
2.7.2
Path-Independent Work; Conservative
Force-Fields
We regard a force field F.r/ that depends only on the position
r but not on time. When a mass m is moved from point P1 to
point P2 on the path (a) (Fig. 2.33) the work necessary for this
motion is
Z
Wa D F dra :
On the path (b) it is
Wb D
Z
F drb :
In Vector-Analysis it is proved that the equivalent condition for a
conservative force field F.r/ is curl F D 0 (theorem of Stokes).
For the definition of curl F see Sect. 13.1. It is
curl F D rot F D r F
@Fz @Fy @Fx
D
;
@y
@z
@z
With other words:
Conservative Force Fields
:
1. A homogeneous force field F.r/ D f0; 0; Fz g with
Fz D const (Fig. 2.34a) is conservative because
Z
F dr D Fz dz ! W D F dr
If we move the mass from P1 to P2 and back to P1 the total
work is then zero.
D
Zz2
Fz dz D
z1
Zz1
Fz dz !
z2
I
F dr D 0 :
2. Every time-independent central force field, written
in spherical coordinates (see Sect. 13.1) as F D
fFr ; F# D 0; F' D 0g, which depends only on the
distance r from the centre r D 0 and not on the angles
# and ' is conservative.
It can be written as F.r/ D f .r/ rO , where f .r/ is a
scalar function of r (Fig. 2.34b).
Z
Figure 2.33 Path-independent work in a conservative force field
Note: Not every force field F.r/ is conservative! (see Example
below)
Examples
Force-field F
@Fx
@y
Conservative force fields are a special case of force fields F.r/
that depend only on the position r, not on time or velocity.
If for arbitrary paths (a) and (b) always Wa D Wb we name the
integral path-independent and the force field F.r/ conservative.
In conservative force fields the work necessary to move a
mass m from a point P.r1 / to a point P.r2 / is independent
of the path between the two points.
@Fz @Fy
;
@x
@x
F dr D
Zr2
r1
Fr dr D
Zr1
r2
Fr dr )
I
F dr D 0 :
Energy Conservation Law of Mechanics
Figure 2.34 Examples for conservative force fields. a Homogeneous
field, b central field
Figure 2.35 Movement in a non-conservative force field
F.r/ D y ex C x 2 ey
Non-conservative Force Fields
1. Position-dependent non-central force field
F.r/ D yex C x2 ey :
The work one has to expend for moving a body from
point P1 D f0; 0; 0g to point P2 D f2; 4; 0g is
WD
D
ZP
0
Z2
F dr D
Z2
Fx dx C
xD0
y dx C
xD0
Z4
Z4
Fy dy
yD0
x2 dy :
yD0
We choose two different paths (Fig. 2.35):
(a) along the straight line y D 2x
(b) along the parabola y D x2 .
On the path (a) is y D 2x ) x2 D .y=2/2
Z
F dra D
Z2
0
Z4 2
y
2xdx C
dy
2
On the path (b) is y D x2 .
Z
)
H
F drb D
Z2
0
2. For time-dependent force fields the integral cannot be
path-independent, because the force field varies during
the travel of the body and therefore the work expended
for the different paths is generally different.
3. If the force depends on the velocity of the body (for
instance the friction for a body moving through a
medium or on a surface, or the Lorentz-force F D
q .v B/ on a charge q moving with the velocity
v in a magnetic field B) such fields are generally not
conservative because the velocity differs generally on
the different paths. For friction forces Ff the force is
for small velocities v proportional to v.Ff v/, when
the body moves slowly through a liquid. For large velocities is Ff v 3 for example when a body moves
through turbulent air. For all friction forces heat is
produced and therefore the mechanical
energy cannot
H
be preserved. In all these cases F dr ¤ 0 (see also
Sect. 6.5)
J
Time-dependent or velocity-dependent forces are generally not conservative.
0
ˇ2
y3 ˇˇ4
16
ˇ
D x2 ˇ C ˇ D 4 C
D 28=3 ;
0
12 0
3
2
x dx C
Z4
y dy
0
8
1 ˇˇ2 1 ˇˇ4
32
D x3 ˇ C y2 ˇ D C 8 D
:
3 0 2 0
3
3
F dr ¤ 0. The force field is not conservative!
2.7.3
Potential Energy
When a body is moved in a conservative force field from a starting point P1 .r1 / to another point P2 .r2 / the work expended or
gained during this movement does not depend on the path between the two points. If P0 is a fixed point P0 and P.r/ has an
arbitrary position r the work solely depends on the initial point
P0 and the final point P.r/. It is therefore a function of P.r/
with respect to the fixed point P0 . This function is called the
potential energy Ep .P/ of the body.
The work
W D
ZP2
P1
Def
F dr D
Ep .P2 /
Ep .P1 / ;
59
Chapter 2
2.7
(2.35a)
60
2 Mechanics of a Point Mass
the potential energy Ep .P/ is then negative for F dr < 0. It
is equal to the work one has to spend in order to bring the
body from the point P to infinity. For instance the potential
energy of a mass m in the gravitational field of the earth Fg D
GMm=r2 at a distance r D R from the centre of the earth is
then
Ep .R/ D GMm=R ;
(2.35d)
Chapter 2
where G is Newton’s constant of gravity and M is the mass
of the earth (Fig. 2.36b).
3. The work which one has to spend on the body (for F dr < 0
Figure 2.36 Different possibilities to choose the zero of the potential energy:
or which can be gained from the body (for F dr > 0) when it
a Ep .z D 0/ D 0; b Ep .r D 1/ D 0
is moved from point P1 to point P2 is of course independent
of the choice of the zero point because it depends only on the
difference Ep D E.P1 / Ep .P2 / of the potential energies.
which the force F.r/ accomplishes on the body when it is moved
between two points P1 and P2 is equal to the difference of the
Examples
potential energies in these two points. For F dr > 0 the force is
directed into the direction of the motion. The potential energy
1. A body with mass m is lifted in the constant gravidifference Ep D Ep .P1 / Ep .P2 / is then negative. This means,
tational force field F D f0; 0; mgg from z D 0 to
that the mass m can deliver the work W but looses potential
z
D h, where h R (earth radius). The necessary
energy.
work to achieve this lift is
One example is the free fall in the gravitational field of the earth,
when a mass m falls from the height h with potential energy mg
Z
Zh
h to the ground with h D 0. When we lift the mass m from h D 0
m g dz
W D F dr D
to h > 0 against the gravitational force, the scalar product F dr
0
is negative and the potential energy increases (Fig. 2.36a). The
D
m
g
h
D
E
Ep .h/ :
p .0/
work spend on the body to lift it against the force results in an
increase of the potential energy. A body with a positive potential
If we choose Ep .z D 0/ D 0 the potential energy for
energy can convert this potential energy again into work. An
z D h is Ep .h/ D Cmgh (Fig. 2.37a). The work apexample is water falling down through pipes and drives a turbine
plied
to the mass m appears as potential energy.
which drives maschines and produces electricity.
Ep
Note:
1. The sign of work and potential energy difference in (2.35a)
has been chosen in such a way, that for Fdr < 0 ! W < 0
but Ep > 0, i. e. one has to spend work in order to move the
body against the force which increases its potential energy.
Work which the body can deliver to its surrounding for F
dr > 0 decreases its potential energy.
2. The defined zero Ep D 0 for the potential energy is not fixed
by the definition (2.35a). If we choose the fixed reference
point P0 as the zero point of the potential energy and define
Ep .P0 / D 0, then the absolute value of the potential energy
in point P is given by
WD
ZP
F dr D Ep .P/ :
(2.35b)
P0
For our example of the free fall we can choose h D 0 as
the reference point with Ep .0/ D 0. In many cases where
a body can be moved to very large distances from the earth
(for instance space crafts) it is more convenient to choose
r D 1 as the reference point for Ep .1/ D 0. We then have
the definition
Z1
F dr D Ep .P/ Ep .1/ D Ep .P/ ;
(2.35c)
P
Ep= m⋅g⋅h
h
F=–G⋅
F=m⋅g
0
Ep=0
a)
(r=∞)=0
⋅ ⋅∧r
mM
r2
⋅ mR⋅M
=–m⋅g⋅R
Ep= –G
r=R
b)
Figure 2.37 a Approximately homogeneous gravitational force field
as small section of the spherical field of the earth in b. The selection of
the definition Ep D 0 is Ep .z D 0/ D 0 in case a and Ep .r D 1/ D 0
for case b
2. In an attractive force field, such as the gravitational
field of the earth F D .GMm=r2 /er a mass m is
moved from r D R (earth surface) to r D 1. In this
case is F dr < 0. The necessary work is negative:
WD
Z1
GMm
rO dr D
r2
r
D
GMm
D Ep .r/ :
r
Z1
r
GMm
dr
r2
(2.35e)
2.7
F D .q1 q2 =r2 /er
2.7.4
Energy Conservation Law in Mechanics
Multiplying the Newton equation
FD m
dv
dt
scalar with the velocity v and integrating over time yields
Z
F vdt D m
Zt1
dv
v dt :
dt
(2.36)
t0
The integral on the left hand side gives with v D dr=dt
Z
F vdt D
ZP1
F dr D Ep .P0 /
The increase of kinetic energy of a body is equal to the
work supplied to this body.
R
In conservative force fields F dr is equal to the change of
potential energy. Then Eq. 2.36 states:
(2.38b)
Ep .P0 / C Ekin .P0 / D Ep .P/ C Ekin .P/ D E :
the potential energy is positive and one wins work
when the charge separation increases, while the potential energy decreases.
When a body with mass m should be moved from the
earth surface r D R to r D 1 one needs the work
W D GMm=R. With g D GM=R2 this can be written
as W D mgR.
Numerical example: With g D 9:81 m=s2 , R D
6371 km, the work to launch a mass of 100 kg is
W D 6:25 109 J D 1736 kWh.
J
Ep .P1 / ;
P0
When a body is moved in a conservative force field from
a point P0 to a point P the total mechanical energy E (sum
of potential and kinetic energy) is conserved, i. e. it has for
all positions in the force field the same amount.
Examples
1. For the free fall starting from z D h with the velocity
v.h/ D 0 we choose Ep .h D 0/ D 0. For arbitrary z
the following equations hold:
Ep .z/ D
Zz
0
mgdz D mgz :
With v D g t and s D h z D
1 2 2
g t D g.h z/ (see Sect. 2.3).
2
This gives
1 2
gt
2
Ekin .z/ D 21 mv 2 D m g .h
!
1 2
v
2
D
z/ :
The sum Ep .z/ C Ekin .z/ D mgh is independent of z
and for all z equal to the total energy E D mgh.
2. A body with mass m oscillates in the x-direction,
driven by the force F D D x. For each point of its
path the total energy is E D Ep .x/ C Ekin .x/ D const.
For x D 0 the potential energy is zero. In the upper
turning points for x D ˙xm the velocity is zero and
therefore Ekin D 0. (Fig. 2.38).
where the last equality is valid for conservative force fields.
The right hand side of (2.36) gives
m
Z
dv
v dt D m
dt
Zv1
v dv D
m 2
v
2 1
m 2
v :
2 0
v0
The expression
Ekin D mv 2 =2
(2.37)
is called the kinetic energy of a body with mass m and velocity
v D jvj.
R
The integral F dr represents the work W which is supplied to
the body. The statement of Eq. 2.36 can therefore be formulated
as:
Ekin D W :
(2.38a)
61
Figure 2.38 Example of energy conservation for a harmonic motion
The oscillation can be described by
x D xm sin !t ! v D dx=dt D xm ! cos !t :
R
The potential energy is Ep D Dxdx D 21 Dx2 D
1
Dx2m sin2 !t. The kinetic energy is Ekin D 12 mv 2 D
2
Chapter 2
Ep .r/ is negative because E.r D 1/ D 0. To raise the
mass m work has to be applied, which is converted to
the increase of potential energy (Fig. 2.37b).
For repulsive potentials (e.g. the Coulomb potential of
two positive electrical charges q1 and q2 )
Energy Conservation Law of Mechanics
62
2 Mechanics of a Point Mass
Defining the gradient of the function Ep .x; y; z/ as
1
m
2
x2m ! 2
2
cos !t. From the Newton equation F D
ma D m d2 x=dt2 we obtain by comparison with
F D Dx the relation D D m ! 2 . Inserting this
into the expression for the potential energy we get
E D Ep C Ekin D 12 mx2m ! 2 .sin2 !t C cos2 !t/
D 21 m x2m ! 2 ;
Chapter 2
which is independent of x.
2.7.5
J
Relation Between Force Field and
Potential
If a body in a conservative force field is moved from the point P
by an infinitesimal small distance r to a neighbouring point P0
(Fig. 2.39) the potential energy changes by the amount
Ep D
@Ep
@Ep
@Ep
x C
y C
z ;
@x
@y
@z
(2.39)
The movement of the body from P to P0 requires the work
(2.40)
0
where F is an average of F.P/ and F.P /. The comparison between (2.39) and (2.40) yields
Fx D
Fz D
(2.42)
F D gradEp D r Ep ;
(2.41a)
where the symbol r D nabla (r has the form of an old Egyptian
string instrument called nabla) is an abbreviation to make the
equation more simple to write.
The potential energy Ep of a body with mass m in the gravitational field of a mass M depends on both masses. However, for
m M (for instance a mass m in the gravitational field of the
earth with M m) the small contribution of m to the gravitational field can be neglected. In such cases it is possible to
define a function V.P/ for each point P, called the gravitational
potential
Def
V.P/ D lim
1
Ep .P/
m
I
(2.42a)
which is the potential energy pro unit mass m in the limit of
m ! 0 in the gravitational field of M. V.P/ is a scalar function
which depends only on the position of P and on the mass M that
generates the gravitational field.
The gravitational potential of the earth is for instance
V.r/ D G ME =r ;
The gravitational field strength is defined as
Since this equation holds for arbitrary paths, i. e. arbitrary values
of x, y, z it follows that
@Ep
I
@y
;
where r is the distance from the centre of the earth.
Fr D Fx x C Fy y C Fz z
@Ep
@Ep
@Ep
D
x
y
z :
@x
@y
@z
@Ep
I Fy D
@x
@Ep
:
@z
the relations (2.41) for the components of F can be combined
into the vector equation
m!0
where the partial derivative @E=@x means that for the differentiation of the function E.x; y; z/ the two other variables are kept
fixed (see Sect. 13.1.6).
W D F r D Ep ;
@Ep @Ep @Ep
;
;
gradEp D
@x @y @z
Def
(2.41)
gradV :
(2.43)
FG D m G :
(2.44)
GD
The force on a mass m is then
For the gravitational field of a spherical symmetric mass M one
obtains
GDG
M
rO ;
r2
(2.43a)
and for the force on a body with mass m in this field Newton’s
gravitational law
FG D G
Figure 2.39 Relation between force and potential
mM
rO :
r2
(2.44a)
These definitions are completely equivalent to their pendants in
electrostatics: The electrical potential of an electric charge Q
and the Coulomb law (see Vol. 2, Sect. 1.3).
2.8 Angular Momentum and Torque
Angular Momentum and Torque
Assume a point mass moving with the momentum p D m v
on an arbitrary path r D r.t/ (Fig. 2.40). We define its angular
momentum L with respect to the coordinate origin r D 0 as the
vector product
(2.45)
L D .r p/ D m .r v/ :
Note, that L is perpendicular to r and v!
In Cartesian coordinates L has the components (see Sect. 13.4)
Lx D ypz
Lz D xpy
zpy I
ypx :
Ly D zpx
xpz I
L D m Œr .vr C v' /
D m .r v' / because r vr D 0 :
The value of L is
d'
dt
because jr v' j D r2
jLj D L D m r v sin.r; v/ D m r v D m r2 ! :
sin.r; v/ D 1
because r ? v :
(2.47)
For the uniform circular motion is r D constant and v D
constant ! L D constant.
(2.46)
If the body moves in a plane but on an arbitrarily curved path
we can compose the velocity in any point of the path of a radial
component vr k r and a tangential component v' ? r using
polar coordinates r and ' (Fig. 2.40). This gives the relations:
jLj D m r2
circle centre perpendicular to the circular plane, i. e. into
the direction of the angular velocity vector ! (Fig. 2.41).
d'
D r2 ! :
dt
These equations describe the following facts:
For planar motions the angular momentum L always
points into the direction of the plane-normal perpendicular to the plane (Fig. 2.40). The vector product (r v)
forms a right-handed screw.
Figure 2.41 Constant angular momentum of the uniform circular moJ
tion
Differentiating (2.45) with respect to time we obtain
dL
dp
dr
D
p C r
dt
dt
dt
D .v p/ C .r pP / D .r pP /; because v k p ;
dL
dp
D .r F/; because F D
:
dt
dt
(2.48)
The vector product
D D .r F/
(2.49)
is the torque of the force around the origin r D 0 acting on the
mass m at the position r. Equation 2.48 can then be written as
When the angular momentum is constant, the motion proceeds
in a plane perpendicular to the angular momentum vector.
dL
DD:
dt
(2.49a)
The change of the angular momentum L with time is equal
to the torque D.
In other words: If the torque on a mass is zero, its angular momentum remains constant.
Figure 2.40 Angular momentum L referred to an arbitrarily chosen origin 0
for a plain motion of a point mass m
Example
For the uniform circular motion the constant angular momentum points into the direction of the axis through the
Note the equivalence between linear momentum p and angular
momentum L:
dp
D F;
dt
dL
DD;
dt
p D constant for F D 0 and L D constant for D D 0.
(2.50)
In central force fields F.r/ D f .r/ rO the torque D D r F D 0
because F k r. Therefore the angular momentum is constant
Chapter 2
2.8
63
64
2 Mechanics of a Point Mass
Figure 2.42 Illustration of angular momentum of a body moving on a straight
line with respect to a point P which does not lie on the straight line
Chapter 2
for all motions in a central force field. This implies that all
trajectories are in a plane, perpendicular to the angular momentum vector.
Note: Angular momentum and torque are always defined with
respect to a selected point (for instance the origin of the coordinate system). Even a body moving on a straight line can have
an angular momentum with respect to a point, which is not on
the straight line.
In Fig. 2.42 the amount L of the angular momentum L of the
mass m moving with the constant velocity v on a straight line is
with respect to the point P
L D m r v sin # D m b v
where b (called the impact parameter) is the perpendicular distance of P from the straight line.
2.9
Figure 2.43 Tycho de Brahe (1546–1601) (with kind permission of “Deutsches
Museum”)
Gravitation and the Planetary
Motions
In the previous section we have learned that in central force
fields the angular momentum L is constant in time. The motion of a body therefore proceeds in a plane perpendicular to L.
The orientation of the plane is determined by the initial conditions (for instance by the initial velocity v0 ) and is then fixed for
all times. The most prominent example are the motions of the
planets in the central gravitational field of the sun which we will
now discuss.
2.9.1
Kepler’s Laws
Based on accurate measurements of planetary motions (in particular the motion of Mars) by Tycho de Brahe (Fig. 2.43)
Johannes Kepler (Fig. 2.44) could show, that the heliocentric
model of Copernicus allowed a much simpler explanation of the
observations than the old geocentric model of Ptolemy where
the earth was the centre and the planets moved around the earth
in complex trajectories (epicycles).
Kepler assumed at first circular trajectories because such motions seemed to him as perfect in harmony with God’s creation.
However, this assumption led to small inconsistencies between
Figure 2.44 Johannes Kepler (1571–1630) (with kind permission of
“Deutsches Museum”)
2.9 Gravitation and the Planetary Motions
65
Chapter 2
Figure 2.46 Kepler’s first law
ϕ
Figure 2.45 Initial model of Kepler illustrating the location of the planets at
the corners of regular geometric figures (with kind permission of Prof. Dr. Ron
Bienek)
Figure 2.47 Kepler’s second law. S: sun, ®: center of ellipse
calculated and observed motions of the planets which exceeded
the error limits of the observations. After a long search with
several unsuccessful models (for instance a model where the
planets were located at the corners of symmetric figures which
rotate around a centre (Fig. 2.45). Kepler finally arrived at his
famous three laws which were published in his books: Astronomia Nova (1609) and Harmonices Mundi Libri V (1619).
Kepler’s first law
The planets move on elliptical trajectories with the sun in
one of the focal points (Fig. 2.46).
the triangle SP1 P2 is then
dA D
1
2
jr vj D 21 jrj jvj sin ˛ D
1
2
jLj
:
m
(2.51)
Kepler’s second law therefore states that the angular momentum
of the planet is constant. Kepler’s first law postulates that the
motion of the planets proceeds in a plane. Since the angular
momentum is perpendicular to this plane it follows that also the
direction of L is constant.
Kepler’s second law
The radius vector from the sun to the planet sweeps out in
equal time intervals equal areas (Fig. 2.47).
Kepler’s third law
The squares of the full revolution times Ti of the different
planets have the same ratio as the cubes of the large half
axis ai of the elliptical paths.
T12 =T22 D a31 =a32
or
Ti2 =a3i D constant ;
where the constant is the same for all planets.
The 2. Kepler’s law tells us that the areas Ai in Fig. 2.47 is for
equal time intervals t always the same, i. e. the area A1 D
SP.t1 /P.t1 C t/ D A2 D SP.t2 /P.t2 C t/. For sufficiently
small time intervals dt we can approximate the arc length ds D
P1 P2 D vdt in Fig. 2.48b by the straight line P1 P2 . The area of
Figure 2.48 Kepler’s second law as conservation of angular momentum.
a schematic representation of the equal area law. b calculation of the area
covered by the radius vector in the time interval dt
66
2 Mechanics of a Point Mass
2.9.2
Newton’s Law of Gravity
Newton came to the conclusion that the free fall of a body as
well as the motion of the planets have a common cause: the
gravitational attraction between two masses. In order to find a
quantitative formulation of the gravitational force he started his
considerations with Kepler’s laws. Since the angular momentum
of the planetary motion is constant the force field has to be a
central force field
Chapter 2
F.r/ D f .r/ rO :
The gravitational force which acts on a body with mass m at the
surface of the earth with mass M (which is equal to its weight)
is proportional to m. According to the principle actio D reactio
and also because of symmetry principles the equal but opposite
force acting on M should be also proportional to the mass M of
the earth (Fig. 2.25). It is therefore reasonable to postulate that
the gravitational force is proportional to the product m M of the
two masses. We therefore can write for the force between two
masses m1 and m2
and it demands special very sensitive detection techniques in
order to measure it. The gravitational constant G can be determined from such experiments in the lab. Among all physical
constant it is that with the largest uncertainty. Therefore many
efforts are undertaken to determine G with new laser techniques
which should improve the accuracy [2.5a–2.5b]. The present
accepted numerical value is
G D 6:67384.80/ 10
11
N m2 =kg2
with a relative uncertainty of 1:2 10 4 .
Note: The gravitational force is always attractive, never repulsive! This differs from the static electric forces between two
charges Q1 and Q2
F.r/ Q1 Q2 =r2C ;
which can be attractive or repulsive, depending on the sign of
the charges Qi .
(2.52a)
Fg D G m1 m2 f .r/ rO :
The proportionality factor G is the Newtonian gravitational constant.
2.9.3
The function f .r/ can be determined from Kepler’s third law.
Since (2.52a) must be also valid for circular orbits we obtain for
the motion of a planet with mass m around the sun with mass
Mˇ the equation
Since the gravitational force field is conservative the sum of potential and kinetic energy of a planet is constant. Because it is a
central field also the angular momentum L D r p is constant.
This can be used to determine the orbit of a planet which proceeds in a plane with constant orientation perpendicular to L.
We use polar coordinates r and ' with the centre of the sun as
coordinate origin (Fig. 2.49).
G m Mˇ f .r/ D m ! 2 r ;
(2.52b)
because the gravitational force is the centripetal force which
causes the circular motion of the planet with the angular velocity
! D v=r. The revolution period of the planet is T D 2=!. For
the orbits of two different planets Kepler’s third law postulates:
T 2 =r3 D const :
With ! D 2=T this gives ! 2 r3 D const or ! 2 r 3 .
Inserting this into (2.52b) yields f .r/ r 2 .
Planetary Orbits
The kinetic energy is
m 2
m 2
v D
vr C v'2
2
2
m 2
D
rP C r2 'P 2 :
2
(2.53)
L D mr2 'P D const :
(2.54)
Ekin D
The amount L D jLj of the angular momentum L is
We then obtain Newton’s law of gravity
Fg .r/ D G
m Mˇ
rO :
r2
(2.52c)
The minus sign indicates that the force is attractive.
The gravitational force
F.r/ D G
m1 m2
rO
r2
acts not only between sun and planets but also between arbitrary
masses m1 and m2 separated by the distance r. However, the
force between masses realized in the laboratory is very small
Figure 2.49 Elliptical orbit in Cartesian and in polar coordinates
2.9 Gravitation and the Planetary Motions
Conservation of energy demands
largest distance (Aphelion) from the sun the derivative dr=dt D
0. Inserting this into (2.56) gives
L2
m 2
D E D const ;
rP C
2
2mr2
(2.55)
where E and L2 are temporally constant. Resolving (2.55) for
dr=dt gives
s
2
dr
L2
:
(2.56)
D
E Ep
dt
m
2mr2
For the angular variable '.t/ one gets from (2.54)
(2.57)
Division of (2.57) by (2.56) yields
2
E
m
Ep
L2
2mr2
1=2
;
integration gives
Z
d' D '
D
L
m
'0
Z
r2
q
2
m
dr
E
Ep
L2 =.2mr2/
:
(2.58)
This allows to get the polar representation of the orbit in the
following way:
With Ep D G M m=r the integral in (2.58) belongs to the type
of elliptical integrals with the solution for the initial condition
'.0/ D '0 D 0 (see integral compilation [2.6a–2.6b]):
' D arccos p
L2 =r
Gm2 M
.Gm2 M/2 C 2mE L2
!
:
(2.59)
With the abbreviations
aD
GmM
2E
and
"D
s
1C
2EL2
G2 m3 M 2
;
(2.59a)
The solutions of this equation are
rmin;max D
2 2 2
1=2
G m M
L2
GmM
C
:
˙
2E
4E2
2mE
(2.61)
a) E < 0.
For E < 0 is the constant a D GmM=.2E/ > 0 and " < 1.
The orbit is an ellipse with the major axis a and the excentricity ". This can be readily seen from (2.60), when the
transformation D r cos ' and D r sin ' to Cartesian
coordinates with the origin in the focal point S is applied.
This gives
p
a 1 "2
(2.61a)
" D 2 C 2 :
When we shift the origin f0; 0g from S into the centre of the
ellipse with the transformation x D C a" and y D we
obtain from (2.61a) the well-known equation for an ellipse
in Cartesian coordinates
x2
y2
C
D1
a2
b2
with
b2 D a2 1
"2 :
(2.61b)
For the special case " D 0 ) a D b the orbit becomes a
circle with r D const. From (2.54) it follows because of L D
const that d'=dt D const the planet proceeds with uniform
velocity around the central mass M.
For a negative total energy E < 0 the planet proceeds
on an elliptical orbit (Kepler’s first law).
b) E D 0.
For E D 0 one immediately obtains from (2.59)
rD
Eq. 2.59 can be written as
L2
D0:
2m r2
We distinguish between three cases:
L
d'
:
D
dt
mr2
d'
L
D
dr
mr2
GmM
r
E
L2
:
Gm2 M.1 C cos '/
(2.62)
This is the equation of a parabola [2.6a, 2.6b] with the minimum distance rmin D L2 =.2Gm2 M/ from the focal point for
' D 0.
c) E > 0.
Solving for r gives
Since in (2.61) the distance r has to be positive .r > 0/ for
a 1 "2
E > 0 only the positive sign before the square root is possirD
:
(2.60)
1 C " cos '
ble. Therefore only one rmin exists and the orbit extends until
infinity .r D 1/. For E > 0 ) " > 0 (see (2.59a)). The
This is the equation of a conic section (ellipse, hyperbola or
orbit is a hyperbola.
parabola) in polar coordinates with the origin in the focal point
S Œ2W6. The minimum distance rmin D a.1 "/ is obtained In Tab. 2.1 the relevant numerical data for all planets of our sofor cos ' D C1, the maximum distance rmax D a.1 C "/ for lar system are compiled, where the earth moon is included for
cos ' D 1. For the shortest distance (perihelion) and the comparison.
' D arccos
a.1
"2 /
"r
r
:
(2.59b)
Chapter 2
Ep C
67
68
2 Mechanics of a Point Mass
Table 2.1 Numerical values for the orbits of all planets in our solar system. The earth moon is included for comparison
Name
Symbol
Large semi axis a of orbit
In AU
Chapter 2
Mercury
Venus
Earth
Mars
Jupiter
Saturn
Uranus
Neptun
Earth moon
¡
¢
£
¤
¥
¦
§
¨
ª
In
In light
106 km travel time t
0.39
57.9
3:2 min
0.72
108.2
6:0 min
1.00
149.6
8:3 min
1.52
227.9
12:7 min
5.20
778.3
43:2 min
9.54
1427
1:3 h
19.18
2870
2:7 h
30.06
4496
4:2 h
0.00257
0.384
1:3 s
Revolution Mean
Numerical Inclination
period T
velocity excentricity of orbit
In km s 1
88 d
225 d
1:00 a
1:9 a
11:9 a
29:46 a
84 a
165 a
27:32 d
47.9
35.0
29.8
24.1
13.1
9.6
6.8
5.4
1.02
7:0ı
3:4ı
–
1:8ı
1:3ı
2:5ı
0:8ı
1:8ı
5:1ı
0.206
0.007
0.017
0.093
0.048
0.056
0.047
0.009
0.055
Distance from earth
Minimum Maximum
in AU
in AU
0.53
1.47
0.27
1.73
–
–
0.38
2.67
3.93
6.46
7.97
11.08
17.31
21.12
28.80
31.33
356410 km 406740 km
replaced by
aD
GM
D
2E
GmM 2
:
2E .m C M/
4. For the accurate calculation of the planetary orbits one has to
take into account the interactions between the planets. Because of the small deviations from a central force field the
angular momentum is no longer constant but shows slight
changes with time.
5. Most of the comets have been formed within our solar system. They therefore have a negative total energy E < 0 and
Figure 2.50 Inclination angles of the orbital planes for the different planets
move on elongated elliptical orbits with a b.
against the earth ecliptic
Remark.
1. Pluto is since 2006 no longer a planet but is now listed
according to a decision of the International Astronomical
Union in the group of dwarf planets. To this group also belong Ceres, Eris and about 200 additional dwarf planets in
the Kuiper belt far beyond the orbit of Neptune.
2. The orientation of the orbital plane of a planet depends on
the initial conditions when the solar system was created from
a rotating gas cloud [2.7]. Since these initial conditions
were different for the different planets the orbital planes
are slightly inclined against each other (Fig. 2.50). Furthermore the gravitational interaction between the planets is
small compared to the interaction with the sun, but not completely negligible. This disturbes the central force field and
leads over longer time periods to a change of the orientation
of the orbital planes.
3. For more accurate calculations (which are necessary for astronomical predictions) one has to take into account that the
sun is not exactly located in a focal point of the ellipse. Because Mˇ is not infinite, the sun and the planets move around
the common centre of mass, which is, however, not far away
from the focal point because Mˇ m [2.8]. For more accurate calculations one has to replace the mass m of a planet
by the reduced mass D m Mˇ =.m C Mˇ / (see Sect. 4.1)
where Mˇ is 700
P times larger than the mass of all planets
.Mˇ 700 mi /. The constant a in Eq. 2.60 has to be
2.9.4
The Effective Potential
The radial motion of a body in a central force field, i. e. the solution of Eq. 2.56, can be illustrated by the introduction of the
effective potential.
We decompose the kinetic energy in (2.53) into a radial part
.m=2/Pr2 which represents the kinetic energy of the radial motion, and an angular part 21 m r2 .d'=dt/2 which stands for the
kinetic energy of the tangential motion at a fixed distance r. The
second part can be expressed by the angular momentum L
tan
Ekin
D
L2
1 2 2
mr 'P D
2
2mr2
(2.63)
(see (2.55)). Since for a given constant L this part depends only
on r but not on the angle ' or on the radial velocity rP , it is added
to the potential energy Ep , which also depends only on r. The
sum
L2
Epeff D Ep .r/ C
(2.64)
2mr2
is the effective potential energy. Often the effective potential
Vpeff D Epeff =m
is introduced which is the potential energy per mass unit. The
part L2 =.2 m r2 / is called the centrifugal potential energy
2.9 Gravitation and the Planetary Motions
2.9.5
Figure 2.51 Effective potential energy Epeff .r / as the sum of potential energy
and centrifugal energy
Gravitational Field of Extended Bodies
In the preceding sections we have discussed the gravitational
field generated by point masses. We have neglected the spatial
extension of the masses and have assumed that the total mass
is concentrated in the centre of each body. This approximation
is justified for astronomical situations because the distance between celestial objects is very large compared to their diameter.
Example
2 2
and L =.2 m r / the centrifugal potential, while the radial part
Ep .r/=m is the radial potential.
The kinetic energy of the radial motion is then
rad
Ekin
D 21 mPr2 D E
Epeff ;
(2.65)
where E is the constant total energy.
In the gravitational force field is
Epeff D G
mM
L2
:
C
r
2mr2
(2.66)
Both parts are depicted in Fig. 2.51. The centrifugal-term Ez
decreases with increasing r as 1=r2 and is for large r negligible
while for small values of r it can overcompensate the negative
radial part to make the total energy positive.
The minimum of Epeff is obtained from dEpeff =dr D 0. This gives
r0 D
L2
:
Gm2 M
The radius of the sun is Rˇ D 7 108 m, the mean distance
sun–earth is r D 1:5 1011 m, i. e. larger by the factor 210!
J
(2.67)
The kinetic energy of the radial motion Ekin .r/ D E Epeff .r/
at the distance r from the centre is indicated in Fig. 2.51 as the
vertical distance between the horizontal line E D constant and
the effective potential energy. The body can only reach those
intervals r D rmin rmax of r where E Epeff > 0.
We will now calculate the influence of the spatial mass distribution on the gravitational field. We start with the field of a hollow
sphere in a point P outside the sphere (Fig. 2.52). The hollow
sphere should have the radius a and the wall-thickness da a.
A disc with the thickness dx cuts a circular ring with the breadth
ds D dx= sin # and the diameter 2y. The mass of this ring (thickness da and breadth ds) is for a homogeneous mass density %
dM D 2y% ds da
D 2a % dx da
because y D a sin # :
All mass elements dM of this ring have the same distance to the
point P. Therefore the potential energy of a small probe mass m
in the gravitational field generated by dM is
dEp D G m dM=r D G m 2a % da dx=r :
These intervals depend on the total energy E, as is illustrated in
Fig. 2.51.
> 0. (horizontal line 1)
E < 0 but Erad
kin
The body moves between the points A.rmin / and B.rmax /.
They correspond to the radii r D a.1 ˙ "/ for the motion
of planets on an ellipse around the sun.
E < 0 but Erad
D 0 (horizontal line 2)
kin
The orbital path has a constant radius r0 , which means it is a
circle. In the diagram of Fig. 2.51 the body always remains
at the point M in the minimum of Epeff .
Figure 2.52 Potential and gravitational field-strength of a hollow sphere
Chapter 2
rad
E > 0 and Ekin
< jEpeff .r D 1/j (horizontal line 3)
The body has the minimum value of r in the point C, where
rad
D 0. It can reach r D 1. Its orbit is a hyperbola.
Ekin
ED0
rad
D Epeff . The body reaches
From (2.65) it follows that Ekin
the minimum distance rmin in the point D on the curve E.r/.
rad
D 0 and Epeff D 0. It can reach r D 1, where
Here is Ekin
rad
Ekin D 0. The orbit is a parabola.
D
A
2
69
70
2 Mechanics of a Point Mass
The gravitational force in the inner volume of the hollow sphere
is then
F D gradEp D 0
for
R<a:
(2.72)
In the inner volume of the hollow sphere there is no gravitational
field. The force on a test mass m is zero. The contributions
from the different parts of the hollow sphere cancel each other.
In Fig. 2.53 the potential energy Ep .R/ and the force F.R/ are
shown inside and outside of the hollow sphere.
Chapter 2
Figure 2.53 Potential energy of a sample mass m and gravitational field
strength in the gravitational field of a hollow sphere with mass M
The gravitational field of the total mass M is obtained by integrating over x from x D a to x D Ca.
ZCa
dx
Ep D 2%Gma da
:
(2.68)
r
xD a
From Fig. 2.52 the relations
2
2
r D y C .R
2
2
Da CR
2
2
2
2
x/ D y C x C R
2RxI
2Rx
r dr D R dx
can be verified. This yields
Ep D
2%a da m
G
R
ZR
a
dr
rDRCa
D G
(2.69)
mM
;
R
because M D 4a2 %da is the mass of the hollow sphere.
The gravitational force on the mass m is
FG D gradEp
dEp O
mM O
D
R:
RD G
dR
R2
(2.70)
The gravitational field of a hollow sphere with mass M is outside
the sphere exactly the same as if the mass M is concentrated in
the centre of the sphere (Fig. 2.53).
For R < a the calculation proceeds in the same way. Only the
upper limit of the integration changes. For x D Ca the limit
becomes r D a R as can be seen from Fig. 2.52. With
rDa
Z R
dr D 2R
rDaCR
the potential energy becomes
Ep D
mM
D const for R a :
G
a
A homogeneous full sphere can be composed of many concentric hollow spheres. Its mass is
ZR0
MD
% 4a2da :
aD0
For a test mass outside the sphere .R > R0 / we obtain from
(2.69)
Ep D G
D G
4
%m
R
ZR0
0
a2 da D G
4 3
R %m
3R 0
(2.71a)
mM
:
R
For a point inside the sphere .R < R0 / we perform the integration in two steps over the ranges 0 a R and R a R0 .
From the Eqs. 2.71 and 2.71a the potential energy can be derived
as
2 R
3
Z 2
ZR0
a
da
C
a da5
Ep D 4%Gm 4
R
(2.73)
aDR
aD0
2
R
1 2 1 2
D 4%Gm
C R0
R I
3
2
2
since M D .4=3/ %R30 this becomes
Ep D
GMm 2
R
2R30
3R20 :
(2.74)
The physical meaning of the two steps for the integration is
the following: For a test mass in the point P(R) only the mass
elements of the sphere with r R contribute to the total gravitational force while the contributions of all mass elements with
r R exactly cancel each other. The second term in (2.73) gives
a constant part to the potential energy and therefore no contribution to the force. From (2.71) and (2.74) one obtains the force
(Fig. 2.54 lower part)
Mm
rO for R R0
R2
GMm
ROr for R R0 :
R30
FD G
FD
(2.75)
Remark. The earth is not a sphere with homogeneous density
(2.71) 1. Because it is an oblate spheroid due to the rotation of the
earth which deforms the plastic earth crust [2.10].
2.9 Gravitation and the Planetary Motions
71
North pole
South pole
Figure 2.56 The shape of the earth as geoid. The deviation of the geoid from
a spheroid with .a b /=a D 1=298:25 (dotted curve ) is shown 80 000 times
exaggerated. Even the geoid gives only the approximate shape of the real earth
Figure 2.54 Potential energy Ep and gravitational force F of a sample mass m
in the gravitational field of a full sphere with mass M
3. The mass distribution is not exactly spherical symmetric.
The gravitational field of the earth is therefore not exactly
a central force field. This implies that the angular momentum of a satellite, orbiting around the earth is not really
2. Because the density increases towards the centre. Therefore
constant. Measurements of the change of the orbital plane
the mass M.R/ inside a sphere with radius R < R0 increases
with time (the position r.t/ of a satellite can be determined
n
with R only as R (with n < 3, Fig. 2.55). The earth accelwith RADAR techniques with an uncertainty of a few cm!)
eration g measured in a deep well therefore decreases with
allows the determination of the mass distribution %.#; '/ in
rq .q < 1/ [2.11].
the earth [2.9a, 2.9b].
4. The equipotential surfaces of the earth form a geoid
(Fig. 2.56). One of these surfaces, which coincides with the
average surface of the oceans is defined as the normal zero
surface. All heights on earth are given with respect to this
surface.
2.9.6
Measurements of the Gravitational
Constant G
Measurements of the planetary motions allow only the determination of the product GMˇ of gravitational constant G and mass
of the sun. The absolute value of G has to be measured by laboratory experiments. Such experiments were at first performed
1797 by Henry Cavendish and later on repeated by several
scientists with increased accuracy [2.12a–2.14], where Lorand
Eötvös (1848–1919) was especially of high repute because of
his very careful and extensive precision experiments [2.2].
Most of these experiments use a torsion balance (Fig. 2.57). A
light rod (1) with length 2L and two small lead balls with equal
masses m hangs on a thin wire. Two large masses M1 D M2 D
M are placed on a rotatable rod (2), which can be turned to the
two positions (a) or (b). Due to the gravitational force between
m and M the light rod (1) is clockwise turned for the position (a)
and counter-clockwise for the position (b) by an angle ' where
the retro-driving torque
Figure 2.55 a Radial cut through the earth showing the different layers. b radial density function %.r /
Dr D
d4
G
'
2
16l
(2.76)
Chapter 2
Equator
72
2 Mechanics of a Point Mass
Mirror
z
Mirror
Uper turning
point
Atomic cloud
Tungsten mass
upper position
Laser
Scale
Atomic cloud
Chapter 2
Lower position
Vacuum tube
Atomic trap
Laser beam
Figure 2.57 Eötvös’ torsion balance for measuring Newton’s gravitational
constant G
of the twisted wire just compensates the torque with the amount
2L Fg generated by the gravitational forces
Fg D G
mM
16 2 2 3 3
D
G
% R1 R2 :
r2
9r2
(2.77)
Here G is the torsion module of the wire, d its diameter and
l its length, % the mass density of the spheres, R1 and R2 their
radii and r the distance between their centres. In the equilibrium
position, where the two torques cancel, we have the condition
Dr D 2L Fg . This gives for the gravitational constant
GD
9G r2 .d=2/4
' :
64 l L %2 R31 R32
(2.78)
In order to maximize the force Fg , the density % should be as
high as possible, because the distance r between the masses m
and M cannot be smaller than rmin D R1 CR2 . The measurement
of ' is performed by placing a mirror at the turning point of the
rod with the masses m, which reflects a laser beam by an angle
2'. On a far distant scale the deflection of the laser spot is a
measure for the angle '.
The most accurate measurement proceeds as follows: The
masses M are turned into the position (a). The system now performs oscillations around the new equilibrium position '1 which
can be determined as the mean of the turning points of the oscillations. Now the masses M are turned into the new position
(b). Again oscillations start around the new equilibrium position '2 , which is determined in the same way. The difference
' D '1 '2 than gives according to (2.78) the gravitational
constant G.
Equation 2.78 tells us, that the diameter d of the wire should be
as small as possible. New materials, such as graphite composites, have a large tear strength. They can carry the masses m
even for small values of d. This increases the sensitivity.
Figure 2.58 Atom interferometer for the measurement of the Newtonian gravitational constant G [2.13b]
In recent years new methods for measuring G have been developed. Most of them are based on optical techniques. We will
just discuss one of them: A collimated beam of very cold atoms
(laser-cooled to T < 1 µK) is sent upwards through an evacuated
tube (Fig. 2.58). At the heights z D h where 21 mv 2 D mgh they
reach their turning point where they fall down again. A large
tungsten mass surrounds the tube and can be shifted upwards or
downwards. Above the mass the atoms experience during their
upwards motion an acceleration .g C g/ due to the gravitational attraction by the earth .g/ and the mass .g/. Below the
mass their acceleration is .g g/. These accelerations are
measured via atom interferometry [2.13b].
Figure 2.59 gives the results of many experiments in the course
of time, using different measuring techniques. This illustrates,
that the error bars are still large but the differences between the
results of many experiments are even larger, indicating the underestimation of systematic errors. The value accepted today
G D 6:67384.80/ m3 kg
1
s
2
is the weighted average of the different measurements where
the number in the brackets give the standard deviation (see
Sect. 1.8.2). The relative error is 1:2 10 4 which illustrates
that among all universal constants G is the one with the largest
uncertainty.
2.9.7
Testing Newton’s Law of Gravity
In order to test the validity of the 1=r2 dependence of the gravitational force (2.52) several precision experiments have been
performed [2.13d]. An interesting proposal by Stacey [2.17] is
based on the following principle: In the vertical tunnel within
2.9 Gravitation and the Planetary Motions
NIST-82
torsion balance
TR&D-96
torsion balance
LANL-97
torsion balance
73
CODATA 1998
UWash-00
torsion balance
BIPm-01
torsion balance
UWup-02
simple pendulum
MSL-03
torsion balance
HUST-05
torsion balance
UZur-06
beam balance
Chapter 2
CODATA 2002
CODATA 2006
HUST-09
torsion balance
JILA-10
simple pendulum
CODATA 2010
BIPM-13
torsion balance
THIS WORK atom interferometry
6.665
6.670
6.675
6.680
G (10–11m3kg–1s–2)
Figure 2.59 Results of different measurements of the Newtonian gravitational constant G [2.13b]
a large water reservoir a sensitive gravitation-balance is placed,
where two masses m are hold at different heights, one above the
water level and one below (Fig. 2.60). When the water level is
lowered by h, the change of the gravitational force differs for
the two masses. For the lower mass it increases by
ıFG D G m 2% h
(2.78a)
because the water above the mass decreases, while the water
below the mass stays constant. For the upper mass the force
decreases because the distance between the mass and the water
surface increases (see problem 2.34).
There is still an open question concerning the exact validity
of the r 2 dependence in Newton’s gravitation law over astro-
Upper
waterlevel
Lower
nomical distances. Astronomical observations of the rotation of
galaxies showed, that the visible mass distribution in the galaxy
could not explain the differential rotation !.R/ as the function
of the distance R from the galaxy centre, if Newton’s law is assumed to be valid. There are two different explanations of this
discrepancy: Either the 1=r2 dependence of FG is not correct
over large distances, or there exists invisible matter (dark matter) which interacts with the visible matter only by gravitation
and therefore changes the gravitational force of the visible matter.
Such very difficult precision experiments have a great importance for testing fundamental physical laws. There are many
efforts to develop theories which reduce the four fundamental
forces (see Tab. 1.2) to a common origin and to understand more
deeply the difference between energy and matter. One example
of such precision experiments are tests of possible differences
between gravitational and inertial mass as has been performed
by Eötvös 1922 and Dicke 1960 and many other scientists.
Here the inertial mass is measured for different materials by the
oscillation period of a gravitational torsion balance [2.18a]. The
results obtained up to now show that the ratio min =mg of inertial
mass to gravitational mass does not differ from 1 within the error
limits. For two different materials A and B a possible difference
.A; B/ D Œmin =mg A
Figure 2.60 Possible method for measuring the 1=r 2 dependence of the gravitational force
Œmin =mg B < 10
12
must be very small and lies below the detection limit of 10
with the presently achievable accuracy.
12
74
2 Mechanics of a Point Mass
Table 2.2 Mass and mean density of sun, planets and the earth-moon
Planet
Chapter 2
Sun
Mercury
Venus
Earth
Mars
Jupiter
Saturn
Uranus
Neptun
Moon
Symbol
¡
¢
£
¤
¥
¦
§
¨
ª
Mass=earth mass
3:33 105
0:0558
0:8150
1:0
0:1074
317:826
95:147
14:54
17:23
0:0123
Mean density %
in 103 kg=m3
1.41
5.42
5.25
5.52
3.94
1.314
0.69
1.19
1.66
3.34
From the revolution period T D 2=! of a satellite around the
earth (e.g. the moon or an artificial satellite) the mass M of the
earth can be determined. For a circular motion the gravitational
force is equal to the centripetal force
m ! 2 r D G mM=r2 :
With the known gravitational constant G and the measured distance r of the satellite from the earth centre the mass of the earth
is obtained from
2
3
M D ! r =G :
The experimental value is
M D 5:974 1024 kg :
From measurements of the gravity acceleration g on the earth
surface the equation
m g D G m M=R2
yields the earth radius R. From M and R the mean density % D
3M=.4R3 / can be derived.
A comparison of the densities of the different planets (Tab. 2.2)
illustrates that the inner planets (Mercure, Venus, Earth and
Mars) formed of rocks have comparable densities around % D
5 g=cm3 , while the outer gas planets and the sun have much
lower densities. These differences give hints to the formation
process of our solar system [2.7] (see Volume 4).
Figure 2.61 Measuring the free fall acceleration g with a pendulum
with L this device is called a mathematical pendulum. The motion of the pendulum under the influence of gravity can be best
calculated when the force Fg D m g is decomposed into the
two components (Fig. 2.61):
a radial component Fr in the direction of the string, which
generates in the string an equal but opposite restoring force.
Since the total force component in this direction is zero, it
does not contribute to the acceleration.
a tangential component Ft D m g sin ' which causes a
tangential acceleration at D g sin '.
The pendulum represents an example of a position dependent
force which is not a central force. The angular momentum
is therefore not preserved. However, if the initial velocity for
' ¤ 0 lies in the plane of the components Fr and Ft the motion
remains in this plane. It can be therefore described by planar
polar coordinates z and '. The equation of motion reads
m g sin ' D m L ' :
Expanding sin ' into a Taylor-series
sin ' D '
'3
'7
C
3Š
7Š
:
The higher order terms can be neglected for small elongations
'. For example is for ' D 10ı D 0:17 rad the term ' 3 =3Š D
8:210 4 which means that the second term is already smaller by
the factor 208 than the first term. The error in the approximation
sin ' ' is for ' D 10ı only < 0:5%.
The equation of motion (2.79a) is then in the approximation
sin ' '
'R D .g=L/' :
2.9.8
Experimental Determination of the Earth
Acceleration g
The most accurate determination of g can be performed by measuring the oscillation period of a pendulum. This pendulum
consist of a sphere with the mass m suspended by a string with
length L (measured between suspension point A and the centre
C of the sphere). If the mas of the string is negligibly small
compared to m and the radius R of the sphere small compared
(2.79a)
With the initial condition '.0/ D 0 the solution is
p
'.t/ D A sin. g=L t/ :
(2.79b)
(2.80)
The pendulum performs a periodic oscillation with the oscillation period
p
(2.81)
T D 2 L=g :
Measuring the time for 100 periods with an uncertainty of 0:1 s
allows the determination of T with an error of 10 3 s. The
2.9 Gravitation and the Planetary Motions
75
largest uncertainty comes from the measurement of the length
L. The errors for L and T in the determination of
gD
4 2 L
T2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ g ˇ
ˇ
ˇ
ˇ 2 ˇ T ˇ C L :
ˇ T ˇ
ˇ g ˇ
L
Example
T=T D 5 10 5 , L=L D 10
g=g D 1:1 10 3 .
3
Figure 2.63 Dependence of the oscillation period on the deflection of the pendulum
for L D 1 m. )
J
For a more accurate solution of (2.79a) we use the energy conservation law (see Sect. 2.7), which saves one integration. From
Fig. 2.62 we see that
Ekin D 21 m v 2 D 21 mL2 ' 2 :
m 2 2
L 'P C mgL.1
2
cos '0 / :
Where '0 is the angle at the turning point where Ekin D 0. Solving for ' gives
d'
D
dt
2g.cos '
cos '0 /
L
:
Integration yields
s
L
2g
Z'0
'D0
d'
D
p
cos ' cos '0
(2.83)
k D sin.'0 =2/ ;
s
L
1
T.'0 / D 2
1 C '02 C :
g
16
cos '/
E D Ep C Ekin D
r
0
which can be solved by a Taylor expansion of the integrand [2.18b]. The result is
The constant total energy is
D mgL.1
=2
Z
p
d
T D 4 L=g
p
1 k2 sin2
with
cos '/
Ep D m g L .1
With the substitution sin D sin.'=2/= sin.'0 =2/ the integral
can be reduced to an elliptical integral
ZT=4
dt D T=4 :
(2.82)
tD0
(2.84)
For the accurate determination of T the oscillation period is
measured as a function of the elongation '0 and the measured
values are extrapolated towards '0 D 0 (Fig. 2.63).
If the shape of the earth is approximated by a spheroid the dependence of g on the latitude ˇ D 90ı # can be approximated
by the formula
g.ˇ/ ge 1 C 0:0053024 sin2 ˇ
5:8 106 sin2 2ˇ
(2.85)
where ge D g.ˇ D 0/ D 9:780327 m=s2 is the earth acceleration at the equator. This formula takes into account, that g is
diminished by the centrifugal acceleration of the rotating earth
which depends on ˇ (see Sect. 3.2). Because of the inhomogeneous mass distribution of the earth additional local changes of
g appear which are not considered in (2.85).
Instead of the pendulum nowadays modern gravimeters are used
for the determination of g. These are sensitive spring balances
which had been calibrated with a precision pendulum. The
restoring force F D D.x x0 / is determined by measuring
the displacement from the equilibrium position by a calibrated
mass m and gets the local variation of the earth acceleration g
according to [2.19]
m g D D.x
Figure 2.62 Illustration of the integration of the pendulum equation based on
the energy conservation
x0 / :
Recently two identical satellites were launched which orbit
around the earth on identical paths with an angle distance '.
Chapter 2
give a total error of g according to
76
2 Mechanics of a Point Mass
This distance can be measured very accurately (within a few
millimetres) by the time laser pulses need to travel from one
satellite to the other and back. Local variations of the gravity
cause a different local acceleration which changes the distance
d D R ' between the satellites. This allows the determination
of even tiny changes of the gravity force [2.20a, 2.20b, 2.20c].
Chapter 2
Summary
A body with mass m can be described by the model of a point
mass as long as its spatial extensions are small compared to
its distance to other bodies.
The motion of a body is described by a trajectory r.t/, which
the body traverses in the course of time. Its momentary velocity is v.t/ D rP D dr=dt and its acceleration is a.t/ D
dv=dt D d2 r=dt2 .
Motions with a.t/ D 0 are called uniform straight-line motions. Magnitude and direction of the velocity are constant.
For the uniform circular motion the magnitude ja.t/j is constant, but the direction of a.t/ changes uniformly with the
angular velocity !.
A force acting on a freely movable body causes an acceleration and therefore a change of its state of motion.
A body is in an equilibrium state if the vector sum of all acting forces is zero. In this case it does not change its state of
motion.
The state of motion of a body with mass m and velocity v is
defined by the momentum p D m v.
The force F acting on a body is defined as F D dp=dt
(2. Newton’s law).
For two bodies with masses m1 and m2 which interact with
each other but not with other bodies the 3. Newtonian law is
valid: F1 D F2 (F1 is the force acting on m1 , F2 acting on
m2 /.
The work executed by the force F.r/ on a bodyR moving along
the trajectory r.t/ is the scalar quantity W D F.r/dr.
Force fields where the work depends only on the initial point
P1 and the final point P2 but not on the choice of the path
between P1 and P2 are called conservative. For such fields
is rot F D 0. All central force fields are conservative.
To each point P in a conservative forceRfield a potential energy Ep .P/ can be attributed. The work F.r/dr D E.P1 /
E.P2 / executed on a body to move it from P1 to P2 is equal
to the difference of the potential energies in P1 and P2 . The
choice of the point of zero energy is arbitrarily. Often one
chooses E.r D 0/ D 0 or E.r D 1/ D 0.
The potential energy E.P/ and the force F.r/ in a conservative force field are related by F.r/ D gradEp .
The kinetic energy of a mass m moving with the velocity v is
Ekin D 12 mv 2 .
In a conservative force field the total energy E D Ep C Ekin
is constant (law of energy conservation).
The angular momentum of a mass m with momentum p, referred to the origin of the coordinate system is L D r p D
m .r v/. The torque acting on a body in a force field F.r/
is D D r F. It is D D dL=dt.
All planets of our solar system move in the central force field
F.r/ D G .m M=r2 /Or of the sun. Therefore their angular momentum is constant. Their motion is planar. Their
trajectories are ellipses with the sun in one focal point.
The gravitational field of extended bodies depends on the
mass distribution. For spherical symmetric mass distributions with radius R the force field outside the body (r > R)
is exactly that of a point mass, inside the body (r < R) the
force F.r/ increases for homogeneous distributions linearly
with r from F D 0 at the centre r D 0 to the maximum value
at r D R.
The free fall acceleration g of a body with mass m equals the
gravitational field strength G D F=m at the surface r D R of
the earth with mass M. With Newton’s law of gravity g can
be expressed as g D G .M=R2 /Or (G D gravitational constant). It can bepdetermined from the measured oscillation
period T D 2 L=g of a pendulum with length L, or with
gravitational balances.
Problems
77
Problems
velocities and accelerations in the points B and C of the circular path with radius R? What is the maximum ratio R=h to
prevent that the body falls down in B? How large is then the
velocity v.B/?
Chapter 2
2.1
A car drives on a road behind a foregoing truck (length
of 25 m) with a constant safety distance of 40 m and a constant
velocity of 80 km=h. As soon as the driver can foresee a free
distance of 300 m he starts to overtake. Therefore he accelerates
with a D 1:3 m=s2 until he reaches a velocity of v D 100 km=h.
Can he safely overtake? How long are time and path length of
the overtaking procedure if he considers the same safety distance after the overtaking? Draw for illustration a diagram for
s.t/ and v.t/.
2.2
A car drives half of a distance x with the velocity v1 D
80 km=h and the second half with v2 D 40 km=h. Estimate and
calculate the mean velocity hvi as the function of v1 and v2 .
Make the same consideration if x1 D 1=3x and x2 D 2=3x.
2.3
A body moves with constant acceleration along the xaxis. It passes the origin x D 0 with v D 6 cm=s. 2 s later it
arrives at x D 10 cm. Calculate magnitude and direction of the
acceleration.
Figure 2.64 Looping path (Probl. 2.8 b)
2.9
How large is the escape velocity
2.4
An electron is emitted from the cathode with a velocity a) of the moon (d D 384 000 km) in the gravitational field of
the earth?
v0 and experiences in an electric field over a distance of 4 cm a
constant acceleration a D 3 1014 m=s2 , reaching a velocity of b) of a body on the surface of the moon in the gravitational field
of the moon?
7 106 m=s. How large was v0 ?
2.5
A body is thrown from a height h D 15 m with an initial 2.10 What is the minimum fuel mass of a one stage rocket
with a payload of 500 kg for a horizontal launch at the equator
velocity v0 D 5 m=s
to bring the rocket to the first escape velocity of v1 D 7:9 km=s
a) upwards,
when the velocity of the propellant gas relative to the rocket is
b) downwards.
ve D 4:5 km=h
Calculate for both cases the time until it reaches the ground.
a) in the east direction
c) Derive Eq. 2.13.
b) in the west direction?
2.6
Give examples where both the magnitude and the direction of the acceleration are constant but the body moves 2.11 Check the energy conservation law for the examples
nevertheless not on a straight line. Which conditions must be given in the text. Show, that (2.26) follows directly from the
condition Ekin Ep , i. e. 21 mv 2 m g R.
fulfilled for a straight line?
2.7
A car crashes with a velocity of 100 km=h against a thick
tree. From which heights must it fall down in order to experience the same velocity when reaching the ground? Compare this
with two equal cars with velocities of 100 km=h crashing head
on against each other.
2.8
a) A body moves with constant angular velocity ! D 3 rad=s
on a vertical circle in the x-z-plane with radius R D 1 m in
the gravity field F D f0; 0; gg of the earth. How large are
its velocities at the lowest and the highest point on the circle?
How large is the difference between the two values? Could
you relate this to the potential energy?
b) A body starts with v0 D 0 from the point A.z D h/ in
Fig. 2.64 on the frictionless looping path. How large are
2.12 A rocket to the moon is launched from a point at the
equator. How much energy is saved compared to a vertical
launch, when it is shot in the eastern direction under 30ı against
the horizontal?
2.13 A wooden cylinder (radius r D 0:1 m, heights h D
0:6 m) is vertically immersed in water with 2=3 of its length
which is its equilibrium position. Which work has to be performed when it is pulled out of the water? How is the situation
if the cylinder lies horizontally in the water? How deep does it
immerse?
2.14 A body with mass m D 0:8 kg is vertically thrown upwards. In the heights h D 10 m its kinetic energy is 200 J. What
is the maximum heights it can reach?
78
2 Mechanics of a Point Mass
Chapter 2
2.15 A spiral spring of steel with length L0 D 0:8 m is ex- 2.26 A vertical straight tunnel is cut through the earth bepanded by the force F D 20 N to a length L D 0:85 m. Which tween opposite points A to B on the earth surface.
work is needed to expand the spring to twice its initial length, if a) Show that without friction a body released in A performs a
the force is always proportional to the expansion L D L L0 ?
harmonic oscillation between A and B.
b) What is the oscillation period?
2.16 What is the minimum initial velocity of a body at a ver- c) Compare this value with the period of a satellite, which cirtical launch from the earth when it should reach the moon?
cles around the earth closely above the surface.
d) A straight tunnel is cut between London and New York.
2.17 What is the distance of a geo-stationary satellite from
What is the travel time of a train without friction and extra
the centre of the earth? Which energy is needed to launch it?
driving force (besides gravity) which starts in London with
How accurate has its distance to the earth centre be stabilized in
the velocity v0 D 0? How much does the time change, if
v0 D 40 m=s?
order to maintain its position relative to a point on earth within
0:1 km=d?
2.27 Calculate the distance earth-moon from the period of
2.18 What is the change of potential, kinetic and total energy revolution of the moon T D 27 d (mass of the earth is M D
of a satellite when its radius r on a stable circular orbit around 6 1024 kg).
the earth centre is changed? What is the ratio Ekin =Ep ? Does
it depend on r? Express the total energy E by m, g, r and the 2.28 Saturn has a mass M D 5:7 1026 kg and a mean density
mass ME of the earth. Are these quantities sufficient or are more of 0:71 g=cm3 . How large is the gravitational acceleration on its
needed?
surface?
2.19 Prove, that the force F D m g sin ' et for the mathematical pendulum is conservative and that for arbitrary values
of ' conservation of energy Ekin C Ep D const holds.
2.29 How large is the relative change of the gravity acceleration g between a point on the earth surface and a point with
h D 160 km above the surface?
2.20 Assume one is able to measure the length L D 10 m of 2.30 How large is the change g of the earth acceleration due
a pendulum within 0:1 mm and the period T within 10 ms. How to the attraction by
many oscillation periods have to be measured in order to equal- a) the moon and
ize the contribution of L and T to the accuracy of g? How b) the sun?
large is then the uncertainty of g?
Compare the two changes and discuss them. How large is the
relative change g=g?
2.21 How much accuracy is gained for the determination of G
with the gravity balance if the large masses M are increased by 2.31 Two spheres made of lead with masses m1 D m2 D
a factor of 10? How accurate has the measurement of the angle 20 kg are suspended by two thin wires with length L D 100 m
' to be in order to determine G with an accuracy of 10 4 ? Give where the suspension points are 0:2 m apart. What is the dissome physical reasons for the limits of the accuracy of '.
tance between the centres of the spheres, when the gravitational
field of the earth is assumed to be spherical symmetric?
2.22 The comet Halley has a period of 76 years. His small- a) without
est distance to the sun is 0.59 AU. How large is its maximum b) with the gravitational force between the two masses.
distance to the sun and what is the eccentricity of its elliptical
orbit? Hint: Look for a relation between T and rmin D a.1 "/ 2.32 Based on the energy conservation law determine the
and rmax D a.1 C "/.
velocity of the earth in its closest distance from the sun (Perihelion) and for the largest distance (aphelion). How large is
2.23 Assume that the gravity acceleration at the equator of the difference v to the mean velocity? Discuss the relation
a rotating planet is 11:6 m=s2 , the centripetal acceleration a D between the eccentricity of the elliptical orbit and v.
0:3 m=s2 and the escape velocity for a vertical launch 23:6 m=s.
At the heights h D 5000 km above the surface is g D 8:0 m=s2 . 2.33 A satellite orbiting around the earth has the velocity
What are the radius R and the mass M of the planet. How fast is vA D 5 km=s in the aphelion and vP D 7 km=s in its perihelion. How large are minor and major half axes of its elliptical
it rotating? Which planet meets these requirements?
orbit?
2.24 The gravitational force exerted by the sun onto the moon
is about twice as large as that exerted by the earth. Why is the 2.34 Prove the equation (2.78a).
moon still circling around the earth and has not escaped?
2.25 Which oscillation period would a pendulum have on the
moon, if its period on the earth is 1 s?
References
79
2.1. F. Cajori (ed.), Sir Isaac Newton’s Mathematical Principles of Natural Philosophy and His System of the World.
(Principia). (University of California Press, Berkely,
1962)
2.2. R.V. Eötvös, D. Pekart, F. Fekete, Ann. Phys. 68 11
(1922)
2.3. A.P. French, Special Relativity Theory. (MIT Interoductor Series Norton & Company, 1968)
2.4. A. Beer, P. Beer (ed.) Kepler. Four Hundred Years. (Pergamon, Oxford, 1975)
2.5a. J. Stuhler, M. Fattori, T. Petelski, G.M. Tino, J. Opt. B:
Quantum Semiclass. Opt. 5 75 (2003)
2.5b. A. Bertoldi et al., Europ Phys. Journal D 40, 271 (2006)
2.6a. B.R. Martin, C. Shaw Mathematics for Physicists.
Manchester Physics Series. (Wiley, London, 2015)
2.6b. A. Jeffrey, D. Zwillinger, Table of Integrals, Series and
Products. 8th ed. (Elsevier Oxford, 2014)
2.7. M.A. Seeds, D. Backman, Foundations of Astronomy,
13th ed. (Cangage Learning, 2015)
2.8. E. Chaisson, St. McMillan, A Beginners Guide to the
Universe. 7th ed (Benjamin Cummings Publ. Comp.,
2012)
2.9a. W.M. Kaula, Satellite Measurements of the Earth’s Gravity Field. in Methods of Experimental Physics. ed. by
R. Celotta, J. Levine, Ch.G. Sammis, Th.L. Henyey,
Vol. 24, part B (Academic Press, San Diego, 1987), p.163
2.9b. Ch. Hirt et al., Geophys. Research Lett. 40(16), 4279
(2013)
2.9c. http://en.wikipedia.org/wiki/Gravity_of_Earth
2.10. R.H. Rapp, F. Sanso (ed.) Determination of the Geoid.
(Springer, Berlin, Heidelberg, 1991)
2.11. C.M.R. Fowler The Solid Earth: An Introduction to
Global Geophysics, 2nd ed. (Cambridge Univ. Press,
Cambridge, 2004)
2.12a. https://en.wikipedia.org/wiki/Gravitational_constant
2.12b. B. Fixler, G.T. Foster, J.M. McGuirk, M.A. Kasevich,
Science 315(5808), 74 (2007)
2.13a. H.V. Parks, Phys. Rev. Lett. 105, 110801 (2010)
2.13b. G. Rosi et al., Nature 510, 518 (2014)
2.13c. C. Moskowitz, Puzzling Measurement of “Big G” Gravitational Constant. Scientific American, Sept. 18, 2013
2.13d. T. Quinn, Nature 408, 919 (2000)
2.14. P.J. Mohr, B.N. Taylor, Rev. Mod. Physics 80, 633 (2008)
2.15. NIST: Reference on constants, units and uncertainties.
http://physics.nist.gov/cuu/
2.16a. C.C. Speake, T.M. Niebauer et al., Phys. Rev. Lett. 65,
1967 (1990)
2.16b. C.W. Misner, K.S. Thorne, J.A. Wheeler, Gravitation.
(Freeman, San Franscisco, 1973)
2.17. F. Stacey, G. Tuck, Phys. World 1(12), 29 (1988)
2.18a. C.B. Braginski, V.I. Panov, Sov. Phys. JETP 34, 464
(1971)
2.18b. https://en.wikipedia.org/wiki/Elliptic_Intergral
2.19. R. Celotta, J. Levine, Ch.G. Sammis, Th.L. Henyey (ed.),
Methods od experimental Physics, Vol. 24 (Academic
Press, San Diego, 1987)
2.20a. https://en.wikipedia.org/wiki/Geographic_information_
system
2.20b. http://www.stevenswater.com/telemetry_com/leo_info.
aspx
2.20c. Ch. Hwang, C.K. Shum, J. Li (ed.), Satellite Altimetry
for Geodesy, Geophysics and Oceanography. (Springer,
Berlin, Heidelberg, 2004)
Chapter 2
References
3.1
Relative Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
82
3.2
Inertial Systems and Galilei-Transformations . . . . . . . . . . . . . . . .
82
3.3
Accelerated Systems; Inertial Forces . . . . . . . . . . . . . . . . . . . . .
83
3.4
The Constancy of the Velocity of Light . . . . . . . . . . . . . . . . . . .
89
3.5
Lorentz-Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . .
90
3.6
Theory of Special Relativity . . . . . . . . . . . . . . . . . . . . . . . . . .
92
Chapter 3
3
Moving Coordinate Systems
and Special Relativity
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
© Springer International Publishing Switzerland 2017
W. Demtröder, Mechanics and Thermodynamics, Undergraduate Lecture Notes in Physics, DOI 10.1007/978-3-319-27877-3_3
81
82
3 Moving Coordinate Systems and Special Relativity
For the description of the location and the velocity of a body in
a three-dimensional space one needs a coordinate system where
the position vectors r.t/ and its time derivative dr=dt D v.t/ are
defined. Of course are all physical processes independent of the
choice of the coordinate system. However, their mathematical
formulation can be much simpler in a suitable coordinate system
than in other systems. It is therefore essential to choose that system which allows the optimum description of a process and to
find the transformation equations to change from one to another
coordinate system.
For example is the coordinate system connected with the earth
which moves around the sun, the best choice for the description of measurements on earth. For astronomical observations
the results of such measurements must be transformed into a
galactic coordinate system which has its origin in the galactic
centre and moves with the rotating galaxy, in order to eliminate
the complex motion of the earth relative to the galactic centre.
For coordinate systems at rest these transformations impose no
problems. The situation is different for systems which move
against each other.
Chapter 3
In this chapter we will discuss question which arise for transformations between moving coordinate systems when physical
processes are described in different systems. It turns out that
many concepts derived from daily life experience which were
taken for granted, had to be revised. The mathematical framework for such revisions is the special relativity theory developed
by Albert Einstein, which will be briefly treated in this chapter.
3.1
Figure 3.1 Definition of the relative distance
3.2
Inertial Systems and
Galilei-Transformations
Two observers B and B0 sit in the origins O and O0 of two coordinate systems S.x; y; z/ and S0 .x0 ; y0 ; z0 / which move against each
other with the constant velocity u (Fig. 3.2). Both observers
measure the motion of an object A. which has the position vector r.x; y; z/ in the system S and r0 .x0 ; y0 ; z0 / in the system S0 .
As can be erived from Fig. 3.2 it is
r0 D r
which can be written for the components as
8̂
x0 .t/ D x.t/
ˆ
ˆ
<y0 .t/ D y.t/
0
ˆ
ˆ z .t/ D z.t/
:̂ 0
t Dt
Relative Motion
An observer, sitting in the origin O of a coordinate system looks
at two objects A and B with the coordinates rA and rB and the
relative distance
rAB D rA
(3.1)
rB ;
which move with the velocities
vA D
drA
dt
(3.3)
ut ;
9
ux t >
>
>
uy t =
;
uz t >
>
>
;
(3.3a)
where t D t0 means that both observers use synchronized equal
clocks for their time measurements. This is not obvious and is
generally not true if the velocity u approaches the velocity of
light (see Sect. 3.4). For the velocity of A the two observers find
vD
dr
dt
and v0 D
dr0
:
dt
(3.4)
From (3.3) follows
and vB D
v0 D v
drB
dt
u:
(3.5)
relative to the coordinate system O (Fig. 3.1). The velocity of A
relative to B is then
vAB D
drAB
D vA
dt
vB ;
(3.2a)
vA D vAB :
(3.2b)
while the velocity of B relative to A
vBA D vB
This illustrates that position vector and velocity do depend on
the reference system.
Figure 3.2 The coordinates of a point A, described in two different systems O
and O 0 which move against each other with the constant velocity u
3.3 Accelerated Systems; Inertial Forces
83
In other words: An observer siting in a train who does not look
out through the window cannot decide by arbitrary many experiments whether he sits in a train at rest or in a train moving
against another reference system with constant velocity.
3.3
Accelerated Systems; Inertial
Forces
Figure 3.3 Description of the free fall in two different inertial systems
a0 D
dv0
dv
D
Da:
dt
dt
(3.6)
Both observers in the systems which move with constant velocity u against each other, measure the same value for the
acceleration a. Because the force on a body with mass m is
F D m a both observers come to the same conclusion about
the force acting on A and find the same relations for dynamical
processes in the two systems.
The observer in an accelerated system can, however, ascertain
that his system moves accelerated against another system. If
he takes into account this acceleration he comes to the same
conclusions about physical laws for the observed motion of the
body A as an observer in an inertial system.
We will discuss this for two different accelerated motions:
0
Such systems which move with a constant relative velocity u a) rectilinear motion of S against S with constant acceleration
0
b) rotation of S against S around the common origin 0 D 00 .
against each other are named inertial systems.
Between the quantities r, v and t for the motion of an object A
measured in two different inertial systems the Galilei transformations pertain
r D r0 C ut ;
v D v 0 C u ) a D a0
t D t0 ;
and
0
FDF ;
(3.7)
where u D juj c is the constant velocity of S against S0 .
Because of F D F0 both observers measure the same forces and
derive identical physical laws. This can be illustrated by the
example of the free fall observed in the two systems S and S0
moving with the velocity u D ux in the x-direction against each
other (Fig. 3.3):
A body A which is released at the heights z D h falls down
in the system S0 along the z0 -axis (x0 D y0 D 0), which moves
with the velocity u against the z-axis in the system S. For the
observer O0 in S0 the motion of A appears as vertical free fall.
For the observer O in S the body A starts at z D z0 D h with
the velocity v.h/ D u in the x-direction, which bends down into
the z-direction because of the gravitation. The trajectory of A
is for O a parabola (horizontal throw see Sect. 2.3.2). However,
both observes measure the same fall acceleration g D f0; 0; gg
and the same fall times. They derive the same law for the free
fall.
All inertial systems are equivalent for the description of
physical laws.
Remark. In the following sections we will always assume that
the observers O and O0 sit in the origins 0 and 00 of the systems
S and S0 .
The discussion of the description of physical processes in accelerated coordinate systems leads to the introduction of special
forces (inertial forces), which are often confusing students.
Therefore these forces will be discussed as vivid as possible in
order to illustrate that these forces are no real forces but are only
necessary, when the observer in the accelerated system does not
take into account the acceleration of his system.
3.3.1
Rectilinear Accelerated Systems
If the origin 00 of the system S0 moves along the x-axis of S
with the time dependent velocity u.t/ D u0 C a t.a D ax ex
with ax D du=dt D d2 x=dt2 / against S, only the magnitude of
the velocity changes not its direction (Fig. 3.4). An example is
an observer in a train accelerating on a straight track.
For a body A with the coordinates .x0 ; y0 ; z0 / in the system S0
the observer in S measures the coordinates x D u0 t C 21 at2 C x0 ,
y D y0 , z D z0 , if for t D 0 the two origins of S and S0 coincide
and the relative velocity u.t/ between S and S0 at time t D 0
is u0 . The velocity of A is then v 0 D fvx0 ; vy0 ; vz0 g for O0 and
v D fvx D u0 C a t C vx0 , vy D vy0 , vz D vz0 g for O.
The description of different situations by O (sitting in a system
S at rest) and O0 (sitting in the accelerated System S0 ) shall be
illustrated by three examples. Note that S0 is no inertial system!
Chapter 3
The acceleration a of A can be derived from (3.5) as
If the two observers sit in two systems which move against each
other with a velocity u.t/ changing with time resulting in an
acceleration a D dv=dt they measure for the motion of a body
A relative to their system different accelerations and therefore
conclude that different forces act on A.
84
3 Moving Coordinate Systems and Special Relativity
Figure 3.4 Coordinates of point A in a system S with origin O and a system
S 0 with origin O 0 , that moves against O with the acceleration a in x -direction
Examples
Chapter 3
1. The observer O0 is sitting on a carriage at a fixed table with plain tabletop. On the tabletop rests a ball A
without friction (Fig. 3.5a). If the system S0 is accelerated to the left (i. e. in x-direction), both observers O
(in the system S at rest) and O0 see that the ball moves
accelerated towards O0 . Both O and O0 make the same
observation but interpret this in a different way:
O0 says: The ball moves accelerated towards me.
Therefore a force F D m a must act on the ball.
O says: The system S0 moves with the acceleration a
to the left, while the ball does not participate in the
acceleration and stays at rest. This means: Not the
ball is accelerated towards O0 , but O0 is accelerated
towards the resting ball. Therefore no force is acting
on A.
a)
b)
Figure 3.5 a A freely movable ball; b a ball fixed to a spring balance,
both on a table that is accelerated into the x -direction with constant
acceleration a
Note: If O0 knows that his system S0 is accelerated, he
also knows that the ball must stay at rest, because it
is frictionless and therefore not linked with the table,
which means that it will not participate in the motion
of the table. In order to explain his observation of the
acceleration a of the ball he introduces a force F D m
a which he calls fictitious force (often named pseudoforce), because he knows that this is not a real force
but merely the description of a virtual acceleration a
of the ball when its motion is described in a reference
system which itself is accelerated with the acceleration
a. Often the notation “inertial force” is used in order
to point to the inertial mass of the ball which prevents
it to follow the acceleration of the table.
2. The observer O0 connects the ball with an elastic
spring scale and holds the other end with his hand
(Fig. 3.5b). If the system S0 is now accelerated with the
acceleration a to the left O0 observes that the spring
is compressed. The spring balance measures the force
F1 D m a. He must apply an equal but opposite
force F2 D Cm a in order to keep the ball at rest.
O0 says: The total force F D F1 C F2 acting on the
ball is zero in accordance with my observation that the
ball rests.
The observer O in the rest system S says: Since the
ball is now connected with the table in S0 it participates
in the acceleration a of S0 . The observer O0 has to
apply the force F D m a in order to transfer the
same acceleration a to the ball as the system S0 and
to keep the ball at rest relative to the system S0 .
3. A mass m in an elevator is suspended by a spring
balance (Fig. 3.6). If the elevator moves with the acceleration a D f0; 0; ag downwards (Fig. 3.6a) the
spring balance measures the force F D m.g a/, if the
elevator moves upwards with the acceleration Ca the
balance measures F D m.g C a/ where g D f0; 0; gg
is the earth acceleration. The observer O0 , sitting in
the elevator, says: The body is at rest. Therefore the
total force acting on it must be zero. The total force
F D F1 C F2 C F3 (Fig. 3.6c) is the sum of
D the weight of the mass m
opposite force of
F2 D m.g a/ D
the spring balance
D inertial force
F3 D m a
F1 D m g
:
O0 must introduce the inertial force F3 in order to explain his observation.
The observer O outside the elevator at rest says: The
body with mass m is connected with the elevator. It
therefore participates in the acceleration of the elevator. This demands the force F D m a. The total force
acting on the body is the sum of its weight F1 D m g
and the restoring force F2 D m .g a/ of the
spring balance. Which gives, as expected the total
force F D m g m .g a/ D m a.
If the suspension cable of the elevator is ruptured and
the elevator goes down in a free fall itsPacceleration is
a D g. For O0 the total force remains Fi D 0 while
for O the total force becomes F D m g.
These examples illustrate, that the inertial forces are
introduced only for measurements in accelerated coordinate systems if the acceleration of the system is
not taken into account. They are therefore also called
fictitious forces or pseudo-forces. A transformation to
an inertial system lets all pseudo-forces vanish. This
3.3 Accelerated Systems; Inertial Forces
85
means an observer O in an inertial system does not
need any pseudo-force for the explanation of the observed physical processes.
Figure 3.7 A system S 0 , that rotates around the axis ! against S . Both systems have the same origin O D O 0
Note: r D r0 means that we regard the same vector in both
systems with the same magnitude but different components.
Figure 3.6 Elevator experiment. Description of the forces acting on
a mass m , that hangs on a spring balance in an elevator accelerated
downwards in a and upwards in b. In c the forces are listed as observed
by O 0 in the elevator (left hand side ) and by O at rest outside the elevator
(right hand side )
J
3.3.2
Rotating Systems
We regard two coordinate systems S.x; y; z/ and S0 .x0 ; y0 ; z0 / with
the unit vectors eO x , eOy , eO z and eO x0 , eOy0 , eO z0 of the coordinate axes
and a common origin 0 D 00 . S0 rotates against S with the constant angular velocity ! D f!x ; !y ; !z g around 0 D 00 (Fig. 3.7).
S0 is therefore no inertial system. We assume that for all times
0 D 00 .
A point A should have at time t in the system S the position
vector
r.t/ D x.t/ eO x C y.t/ eOy C z.t/ eO z
dx
dy
dz
eO x C eOy C eO z :
dt
dt
dt
The endpoints of the unit vectors eO x0 , eOy0 , eO z0 perform a circular motion with the angular velocity ! around 0 D 00 . Their
velocity is then
(3.8)
(3.9)
u D .! eO x0 /x0 C .! eOy0 /y0 C .! eO z0 /z0
D ! .Oex0 x0 C eOy0 y0 C eO z0 z0 /
D ! r0 D ! r; because r r0 :
0
The same point A has in the system S the position vector
r0 .t/ D r.t/ D x0 eO x0 C y0 eOy0 C z0 eO z0 :
dr0
dy0
dz0
dx0
(3.11)
D
eO x0 C
eOy0 C
eO z0 :
dt
dt
dt
dt
However, if the observer O in the inertial system S describes the
velocity of A in the coordinates of S0 . he knows that the axis of
S0 are rotating and therefore not constant in time. He therefore
must write:
0
dx
dy0
dz0
0 0 0
v.x ; y ; z / D
eO x0 C
eOy0 C
eO z0
dt
dt
dt
dOex0
dOey0
dOez0
(3.12)
C x0
C y0
C z0
dt
dt
dt
D v0 C u :
v0 .t/ D
dOex0
dOey0
dOez0
D ! eO x0 I
D ! eOy0 I
D ! eO 0z : (3.13)
dt
dt
dt
Inserting this into (3.12) the second term in (3.12) becomes
and the velocity
v.t/ D
If the observer O0 does not take into account that his system
rotates, he will define as the velocity of A in his system
(3.10)
We therefore get the transformation between the velocity v of
the point A measured by O in the system S and the velocity v 0
measured by O0 in the system S0
v D v0 C .! r/ :
(3.14)
Chapter 3
y
86
3 Moving Coordinate Systems and Special Relativity
Note: v 0 is the velocity measured by O0 , if he does not take into
account, that his system S0 rotates with the angular velocity !,
while v in (3.9) is the velocity in the resting system S and v
in (3.14) the velocity of A measured by O but expressed in the
coordinates of the rotating system S0 .
The acceleration a can be obtained by differentiating (3.14). The
result is
dv
dv0
dr
aD
;
(3.15)
D
C !
dt
dt
dt
because we have assumed that ! D const. The observer O0 gets
the result for a, expressed in the coordinates of his system S0 :
dvy0
dv 0
dv0
dv 0
D eO x0 x C eOy0
C eO z0 z
dt
dt
dt
dt
dOey0 0
dOez0 0
dOex0 0
(3.16)
C
vx C
vy C
vz
dt
dt
dt
D a0 C .! v0 / ;
0
0
Chapter 3
where a is again the acceleration of A measured by O in the
system S0 . We therefore obtain with (3.15)
aD
dv
D a0 C .! v0 / C .! v/ :
dt
a0 D a C 2.v0 !/ C ! .r !/
D a C aC C acf :
(3.18)
While the observer in his resting system S measures the acceleration a D dv=dt, the observer O0 in his rotating system S0 has
to add additional terms for the acceleration in order to describe
the same motion of A. These are
the Coriolis-acceleration
aC D 2.v0 !/ ;
(3.19a)
the centrifugal acceleration
acf D ! .r !/ :
Figure 3.8 Centrifugal- and Coriolis-force acting on a mass m in A.x ; y ; z D
0/ described in a system S 0 , that rotates with constant angular velocity ! around
the z -axis
3.3.3
Inserting for v the expression (3.14) we finally obtain from
(3.15)
a D a0 C 2.! v0 / C ! .! r/ ;
(3.17)
and for a0
azf
(3.20a)
Special Cases: If the point A moves parallel to the rotation
axis we have v k ! and therefore the Coriolis acceleration becomes aC D 0. The Coriolis acceleration appears only, if v0 has
a component perpendicular to !. When we choose the z-axis
as the direction of ! (Fig. 3.8), both the Coriolis acceleration
aC and the centrifugal acceleration acf lie in the x-y-plane. The
centrifugal acceleration points outwards in the radial direction.
The direction of the Coriolis acceleration depends on the direction of the velocity v 0 in the coordinate system .x0 ; y0 ; z0 /. Since
the vz0 -component does not contribute to aC only the projection
v? D fvx0 ; vy0 g is responsible for the determination of the vector
aC D ! fvy0 ; vx0 ; 0g :
The vector aC is perpendicular to v? as can be immediately seen
when forming the scalar product aC v0? .
Centrifugal- and Coriolis-Forces
According to Newton’s laws accelerations are caused by forces.
Therefore the observer O0 , who measures in his rotating system S0 additional accelerations has to introduce additional forces
based on the equation F D m a These are the Coriolis force
FC D 2 m .v0 !/ ;
(3.19b)
and the centrifugal force
Fcf D m ! .r !/ :
(3.20b)
Both forces are inertial forces or virtual forces because they are
not real forces due to interactions between bodies. They have
only to be introduced if the rotation of the coordinate axes of
the rotating system S0 are not taken into account. If the same
motion of the body A are described in an inertial system S or in
the rotating system S0 where the rotation of the coordinate axes
in considered, these forces do not appear.
We will illustrate these important facts by some examples.
Examples
1. A mass m is attached to one end of a string with length
L while the other end is connected to the end of a bar
with length d which rotates with the angular velocity
! around a vertical axis fixed to the centre of a rotating
table (Fig. 3.9). In the equilibrium position the string
forms an angle ˛ against the vertical direction, where
˛ depends on !, d and L. The observer O in the resting frame S and the observer O0 sitting on the rotating
table describe their observations as follows:
O says: Since m moves with constant angular velocity
! on a circle with radius r D d C L sin ˛ a centripetal
force Fcp D m ! 2 r acts on m which is the vector
sum of its weight m g and the restoring force Fr of the
string (Fig. 3.9a).
O0 says: Since m is resting in P
my system S0 the total
force on m must be zero, i. e.
Fi D 0. The vector
sum mgCFr has to be compensated by the centrifugal
force Fcf D Cm! 2 r (Fig. 3.9b). He has to introduce
the virtual force Fcf if he does not take into account
the rotation of his system.
Figure 3.9 Forces on a rotating string pendulum, described by the observer O at rest and O 0 rotating with the pendulum
2. In a satellite, circling around the earth with constant
angular velocity ! experiments are performed concerning the “weightlessness” (Fig. 3.10). For example
an astronaut can freely float in his satellite without
touching the walls.
The observer O0 in the satellite (i. e. the astronaut)
says: I know that the gravity force
Fg D .GmM=r2 /Oer
acts on me, where r is the distance to the centre of the
earth. It is compensated by the opposite centrifugal
force
Fcf D Cm! 2 r eO r :
The total force acting on me is zero and therefore I can
freely float.
Note: The state of the astronaut should be better called
“force-free” instead of “weightlessness”.
The observe O in a resting system S (for example the
galactic coordinate system) says: The gravity force Fg
acts as centripetal force on both the satellite and the astronaut. Both are therefore forced to move on a circle
around the earth. The acceleration a D .GM=r2 /Oer
is the same for the astronaut and the satellite and the
difference of the accelerations is zero. Therefore the
astronaut can freely float in his satellite.
Note: Both observes can describe consistently the situation of the astronaut, however the observer S0 has to
introduce the inertial force Fcf if he does not take into
account the accelerated motion of his space ship.
3. A sled moves with constant velocity v on a linear track
and writes with a pen on a rotating disc (Fig. 3.11).
The marked line on the rotating disc is curved where
the curvature depends on the velocity v of the sled, the
perpendicular distance d of the track from the centre
of the disc and the angular velocity ! of the rotating
disc. The two observers O and O0 describe the observed curve as follows:
O says: The sled moves with constant velocity on a
straight line, as can be seen from the marked line outside the disc. Therefore no force is acting on the sled
and its acceleration is zero. The curved path marked
on the disc is due to the fact that the disc is rotating.
O0 says: I observe a curved path. Therefore a force
has to act on the sled. By experiments with different
values of v, ! and d he finds:
For d D 0 is ja0 j / v 0 !; a ? v0 and a ? !.
For d ¤ 0 is a D c1 ! C c2 ! 2 with c1 / v and c2 / r,
where r is the distance of the sled from the centre of
the disc. The quantitative analysis of his measurements gives the result:
a0 D 2.v0 !/ C ! .r !/ ;
which is consistent with (3.18) and shows that the acceleration of the sled measured by O0 is the sum of
centrifugal and Coriolis accelerations.
Glider
with pen
M
Earth
Mass M
Figure 3.10 Force-free conditions in a satellite orbiting around the
earth
Figure 3.11 Experimental illustration of the inertial forces. A glider,
moving on a straight line above a rotating disc writes with a pen its path
on the rotating disc which appears as a curved trajectory
87
Chapter 3
3.3 Accelerated Systems; Inertial Forces
88
3 Moving Coordinate Systems and Special Relativity
Chapter 3
This example illustrates clearly that the two accelerations and the corresponding forces are only virtual,
because the sled moves in fact with constant velocity on a straight line and therefore experiences no real
forces.
4. A hollow sphere filled with sand hangs on a string
which is connected to a fixed suspension point and
swings in the fixed x-z-plane of a resting system S,
driven by the gravity force Fg D m g with g D
f0; 0; gg.
Below the swinging pendulum is a rotating table in the
x-y-plane which rotates with the angular velocity !
around a vertical axis through the minimum position
of the pendulum.
If the sand flows through a small hole in the hollow
sphere it draws for ! D 0 a straight line on the table while for ! ¤ 0 a rosette-like figure is drawn
(Fig. 3.12) with a curvature which depends on the ratio of oscillation period T1 of the pendulum to rotation
period T2 of the rotating table.
The two observers give the following explanations:
O says: The x-z-oscillation plane remains constant
because the driving force Fg D m g sin ˛ (see
Sect. 2.9.7) lies always in the x-z-plane and therefore
the motion must stay in this plane. The projection of
the trajectory onto the x-y-plane should be a straight
line. The curved trajectory drawn on the rotating table is caused by the rotation and not by an additional
force.
O0 says sitting on the rotating table: I see a curved path
which must be caused by forces, which depend on !,
v 0 and r. Its form can be explained by the centrifugal and the Coriolis forces. My careful measurements
prove that the paths is due to the action of the total acceleration a0 D acf C aC in accordance with Eq. 3.18.
Figure 3.12 Apparent trajectory written on a rotating disc by a sand
pendulum than oscillates in a constant plane
5. Foucault pendulum. Since our earth is a rotating system, the path drawn by a linearly swinging pendulum
onto the ground must show curved lines as discussed
in example 4). However, because of the slow earth rotation (! D 7:3 10 5 s 1 ) the curvature is very small.
Using a pendulum with a large length L and a corresponding large oscillation period T the rotation of
the earth under the linearly swinging pendulum could
be first demonstrated 1850 by Leon Foucault (1819–
1868) who used a copper ball (m D 28 kg) suspended
by a 67 m long string (T D 16:4 s). The turn of the
oscillation plane against the rotating ground occurs
with the angular velocity !s D ! sin ' where ' is
the geographic latitude of the pendulum (Fig. 3.13).
in Kaiserslautern with ' D 49ı the pendulum plane
turns in 1 h by 11ı 320 , which can be readily measured.
Using shadow projection of the pendulum string defining the oscillation plane this turn can be quantitatively
measured within a physics lecture.
Figure 3.13 Explanation of the turning plane of oscillation of the
Foucault-pendulums on the surface the rotating earth
6. An impressive demonstration of the Coriolis force is
provided by the motion of cloud formations around
a low pressure region as for instance realized by tornados or typhoons (Fig. 3.14). For an observer on
the rotating earth looking from above onto the ground
the wind does not blow radially into the low pressure region but rotates on the northern hemisphere
anticlockwise around it, on the southern hemisphere
clockwise. Around a high pressure region the rotation is clockwise on the northern hemisphere and
anticlockwise on the southern.
Note: If a small balloon which floats in the air is used
as indicator of the wind flow an observer on earth
would see the balloon moving on one of the lines in
Fig. 3.14. An observer O at a fixed position outside
the earth, would however see, that the balloon moves
on a straight line radially into the centre of the deep
pressure region or out from the centre of a high pressure region. These centres are fixed at a point on earth
and rotate with the earth.
3.4 The Constancy of the Velocity of Light
3.3.4
89
Summary
Inertial forces (virtual forces) have to be introduced, if the motion of bodies are described in accelerated coordinate systems.
These forces are not caused by real interactions between bodies
but only reflect the acceleration of the coordinate system. They
do not appear if the same motion is described in an inertial system.
In rotating systems with a fixed centre the inertial forces are centrifugal and Coriolis forces. In systems with arbitrarily changing
velocities further inertial forces have to be introduced.
The Constancy of the Velocity of
Light
We consider a body A which has the velocity v measured in
the system S but the velocity v0 in a system S0 , which moves
itself with the velocity u against the resting system S. According
to the Galilei transformations the different velocities are related
through the vector sum (Fig. 3.15)
v D v0 C u :
(3.21a)
Therefore one might suggest, that also the velocity of light,
emitted from a light source which is fixed in a system S0 moving
with the velocity u against the system S, should be measured in
the system S as the vector sum
c D c0 C u ;
(3.21b)
where c0 is the velocity measured by O0 in his system S0 . This
means that the observer O should measure the velocity c1 D
c0 C u if c0 and u have the same direction, and c2 D c0 u if they
have opposite directions.
Very careful measurements performed 1881 by Albert Abraham Michelson and Edward Morley [3.2a, 3.3] and later on by
many other researchers [3.4a, 3.4b] produced evidence that the
velocity of light is independent of the relative velocity u between source and observer. For example measurements of the
velocity of light from a star at different times of the year always
Figure 3.14 Stream-lines of the air around a deep-pressure region.
a on the northern hemisphere; b on the southern hemisphere. On the
northern hemisphere the Coriolis force acts (seen in the wind direction
from above) in the right direction against the radial force of the pressure
gradient, on the southern hemisphere in the left direction. c Satellite
photo of the “death-hurricane” north of Hawaii (with kind permission of
J
NASA photo HP 133)
Figure 3.15 Galilei transformations of velocities in two inertial systems
Chapter 3
3.4
90
3 Moving Coordinate Systems and Special Relativity
Figure 3.16 Experimental possibility to prove the constant velocity of light, by
measuring the velocity of light from a distant star at two different days with a
time interval of half a year, when the earth on its way around the sun moves
towards the star and away from it
Chapter 3
brought the same result although the earth moved with a velocity of 30 km=s at one time of the year against the star and half
a year later away from the star (Fig. 3.16). This result was very
surprising and brought about many discussions but induced the
formulation of the theory of special relativity by Albert Einstein.
According to these unambiguous experimental results we must
conclude:
The velocity of light is the same in all inertial systems,
independent of their velocity against the light source.
The Galilei-transformations (3.7) which appear very plausible
apparently fail for very large velocities. It turns out, that in particular the assumption t D t0 in Eq. 3.3a needs a critical revision.
It must be precisely defined what “simultaneity” means for two
events at different locations. The question is: How does one
measure the times of two events at different locations?
To illustrate this point we regard in Fig. 3.17 two systems S and
S0 where light pulses are emitted from the points A and B in the
system S and from A0 and B0 in the system S0 . If the two systems
do not move against each other (Fig. 3.17a) the situation is clear:
The observers O and O0 measure the arrival time of the two light
pulses in O resp. O0 and can decide, whether the pulses had
been sent from A and B resp. from A0 and B0 simultaneously
or at different times. For the first case they arrive in O or O0
simultaneously. Both observers come to the same result.
Figure 3.17 Illustration of the problem of measurements of simultaneous
events in A and B resp. in A0 and B0 in two different systems: a which are
at rest, b which move against each other
The situation is more difficult, if S0 moves with the velocity vx
against S (Fig. 3.17b). We assume, that at time t D 0 the origins
of both systems coincide, i. e. O D O0 and therefore also A and
A0 as well as B and B0 coincide. If now at t D 0 two pulses are
emitted form A and B in S and from A0 and B0 in S0 the observer
O in the rest frame measures their arrival times in O. During
the light travel time t for the pulses from A or B the system S0
has moved over the distance x D vx t to the right side in
Fig. 3.17b. The pulses from B0 therefore arrive earlier in O0 than
those from A0 . Therefore O0 concludes that the pulses from B0
had been sent earlier than those from A0 .
Now we will take the standpoint of O0 , who assumes that his
system S0 is at rest and that S moves with the velocity vx to the
left in Fig. 3.17b. He now defines the Simultaneity of the events
in A0 and B0 if he receives the light pulses at O0 simultaneously.
Now the pulses from A arrive for O earlier that those from B.
This illustrates that the definition of simultaneity depends on the
system in which the pulses are measured. The reason for this
ambiguity is the finite velocity of light. If this velocity would
be infinite, the problem of simultaneity would not exist because
then the travel time for the signals from the two points A and B
would be always zero.
The question is now: what are the true equations for the transformation between different inertial systems?
3.5
Lorentz-Transformations
We regard two inertial systems S.x; y; z/ and S0 .x0 ; y0 ; z0 / with
parallel axes and with O.t D 0/ D O0 .t0 D 0/ which move
with the constant velocity v D fvx ; 0; 0g against each other in
the x-direction (Fig. 3.18). Assume, that a short light pulse is
emitted at t D 0 from O D O0 . The observer O measures that
the pulse has reached the point A after a time t. He describes his
observation by the equation
rD ct
or: x2 C y2 C z2 D c2 t2 :
(3.22a)
The observer O0 in S0 measures that the pulse has arrived in A
after the time t0 . he therefore postulates:
r0 D c t 0
or: x0 2 C y0 2 C z0 2 D c2 t0 2 :
(3.22b)
Figure 3.18 Schematic diagram for deriving the Lorentz-transformations
3.5 Lorentz-Transformations
x.O0 / D v t
for x0 D 0 :
Since the coordinate x0 refers to the system S0 the transformation
to the coordinates x.A/ of the point A, expressed in the system
S must depend on the argument (x v t). We make the ansatz
x0 D k.x
(3.23)
v t/ ;
where the constant k has to be determined. At time t D 0 the
two observers were at the same place x D x0 D 0 and have
simultaneously started their clocks, i. e. t.x D 0/ D t0 .x0 D
0/ D 0. However, the time measurements for t > 0 are not
necessarily the same for the two observers, because they are no
longer at the same place but move against each other with the
velocity v. The simplest transformation between t and t0 is a
linear transformation
t0 D a.t
(3.24)
bx/ ;
where the constants a and b have again to be determined. Inserting (3.23) and (3.24) into (3.22b) yields with y D y0 and
z D z0
k2 x2 2vxt C v 2 t2 C y2 C z2
D c2 a2 t2 2bxt C b2 x2 :
Rearrangement gives
2
k
b2 a2 c2 x2 2 k2 v ba2 c2 xt C y2 C z2
D a2 k2 v 2 =c2 c2 t2 :
Inserting the expressions for a, b and k into (3.23) and (3.24)
gives the special Lorentz-Transformations
x
x D p
1
vt
v 2 =c2
t
t0 D p
;
0
y Dy;
vx=c2
1
v 2 =c2
0
z Dz
For v D 10 km=s .36 000 km=h/ is v=c 3 10 5 and
.1 v 2 =c2 / 1=2 1 C 21 v 2 =c2 D 1 C 10 10 . The difference between Galilei and Lorentz transformations is then
only 5 10 10 and therefore smaller than the experimental
J
uncertainty.
With the abbreviation D .1 v 2 =c2 / 1=2 the Lorentz transformations can be written in the clear form
x0 D .x
y0 D y
z0 D z
t0 D .t
(3.26)
;
between the coordinates .x; y; z/ and .x0 ; y0 ; z0 / of two inertial
systems which move against each other with the constant velocity v D fv; 0; 0g. These equations were first formulated 1890 by
Hendrik Lorentz [3.5]. They show, that for v c the Lorentz
transformations converge towards
the Galilei transformations
p
(because for v 2 =c2 1 ! 1 v 2 =c2 1), which are therefore a special approximation for small velocities v:
x D .x0 C vt0 /
y D y0
z D z0
vt/
vx=c2 /
(3.26a)
:
t D .t0 C vx0 =c2 /
Note: The Lorentz transformations have, compared to the
Galilei transformations, only one additional assumption: The
constancy of the velocity of light and its independence of the
special inertial system, which was used in (3.22a,b), where both
observers anticipate the same value of the velocity of light.
We will now discuss, how the velocity u of a body A, measured
in the system S transforms according to (3.26) into the velocity
u0 of A, measured by O0 in S0 .
For O pertains:
ux D
This has to be identical with (3.22a) for all times t and all locations x. Therefore the coefficients of x and t have to be identical.
This gives the equations
9
k2 b2 a2 c2 D 1>
=
aDkD p 12 2
1 v =c
(3.25)
)
k2 v ba2 c2 D 0
2
>
b
D
v=c
:
;
2
2 2 2
a
k v =c D 1
0
Example
dx
I
dt
uy D
while for O0 applies
0
u D
fu0x ; uy0 ; u0z g
dy
I
dt
uz D
dz
;
dt
dx0 dy0 dz0
D
;
;
dt0 dt0 dt0
(3.27)
:
Using (3.26) and considering that x D x.t/ depends on t, we get:
dx0
dx0 dt
dx0 . dt0
u0x D 0 D
0 D
dt
dt dt
dt dt
dx
v
v
u
x
D dt v dx D
vux :
1
1 c2 dt
c2
Solving for ux gives the back-transformation
u0x C v
ux D
:
1 C u0x v=c2
(3.28a)
In the same way one obtains
uy0 D
uy
I
.1 vux =c2 /
uy D
u0z D
uz
I
.1 vux =c2 /
uz D
uy0
1 C vu0x =c2
u0z
1 C vu0x =c2
;
:
(3.28b)
(3.28c)
These equations demonstrate that the velocity components uy
and uz perpendicular to the velocity v D vx of S0 against S
Chapter 3
Both observers know about the result of the Michelson experiment. They therefore assume the same velocity of light c. The
coordinate x of the origin O0 measured in the system S is
91
92
3 Moving Coordinate Systems and Special Relativity
transforms differently from the component ux parallel to vx . For
vx u c2 one obtains again the Galilei transformations.
If the body A moves parallel to the velocity v i. e. parallel to the
x-axis and therefore also to the x0 -axis, we have uy D uz D 0 )
u D ux , the Lorentz transformations simplify to
u0 D
For u D c we get
u0 D
u
1
v
:
vu=c2
c v
c;
1 v=c
(3.28d)
(3.28e)
which means that O and O0 measure the same value for the light
velocity in accordance with the results of the Michelson experiment.
3.6
Theory of Special Relativity
Chapter 3
Starting with the results of the Michelson experiment and the
Lorentz transformations Einstein developed 1905 his theory of
special relativity [3.5–3.8], which is based on the following postulates:
All inertial systems are equivalent for all physical laws
The velocity of light in vacuum has the same value in all
inertial systems independent of the motion of the observer or
the light source.
Figure 3.19 Illustration of the different results in a space-time diagram, when
measuring simultaneous events in two different systems that move against each
other
Now we regard the same situation in the system S0 which moves
with the velocity v D vx against S (Fig. 3.19b). The points A, B
and C should rest in the system S0 , they therefore move with the
velocity vx against the system S and pass in the x-t-diagram of S
inclined straight lines with the inclination angle ˛2 and the slope
tan ˛2 D dt=dx D 1=v. The light pulses travel with the same
velocity c as in S. At time t D 0 both systems S and S0 should
coincide. The light pulse, emitted from B at t D 0 now reaches
the two points A and C for the observer O not simultaneously
but in A at t D t1 and in C at t D t2 , which correspond with the
intersection points A01 and C01 in Fig. 3.19b. The reason is that
A propagates towards the light pulse but C from it away.
For the comparison of measurements of the same event by two
observers in two different systems S and S0 , which move against
each other the time definition plays an important role. The
Lorentz-transformations (3.26) show that also the time has to
be transformed when changing from S to S0 . We will therefore
at first discuss the relativity of simultaneity. The presentation in Sect. 3.6 follows in parts the recommendable book by
French [3.8].
Since for the observer O0 in S0 the points A, B and C are resting
in S0 the events A01 and C01 (we define as event the arrival of the
light pulse in the point A01 or C01 ) has to occur simultaneously,
equivalent to the situation for S in Fig. 3.19a because all inertial
systems at rest are completely equivalent. In the x0 -t0 -diagram
the line through the points A01 and C01 has to be a line t0 D const
i. e. it must be parallel to the x0 -axis (Fig. 3.20). One has to
choose for the moving system S0 other x0 - and t0 -axes which are
inclined against the x- and t-axes of the system S. The x0 -axis
(t0 D 0) and the t0 -axis (x0 D 0) are generally not perpendicular
to each other.
3.6.1
One obtains the t0 -axis in the following way: If O0 moves with
the velocity v D vx against O he propagates in the system S0
along the axis x0 D 0 which is the t0 -axis (because he is in his
The Problem of Simultaneity
We will now treat the problem of simultaneity in different inertial systems in some more detail. We regard three points A, B
and C which rest in the system S and have equal distances, i. e.
AB D BC D x. In an x-t-coordinate-system with rectangular axis. For t D 0 the three points are located on the x-axis
(Fig. 3.19a). In the x-t-diagram the points A, B and C proceed
in the course of time on vertical straight lines since they are
fixed in the system S and have therefore constant positions x.
At time t D 0 a light pulse is emitted from point B. The light
pulse, however proceeds on an inclined straight line with an inclination angle ˛1 with tan ˛1 D t1 =x. This line intersects the
vertical position lines of A and C at the points A1 and C1 . The
connecting line through A1 and C1 is the horizontal line t D t1 .
Since the light travels with the same speed c in all directions the
pulses reach the points A and C at the same time t1 D x=c.
Figure 3.20 a Space axis and time axis in a moving inertial system S 0 are
inclined by the angle ˛ against the axes in a system S at rest. b Definition of
the velocity u of a point A in the two systems S and S 0
3.6 Theory of Special Relativity
93
system S0 always resting at the origin x0 D 0). In the system S
this axis is x D v t which is inclined against the t-axis x D 0 by
the angle ˛ with tan ˛ D v=c. The slope of the t0 -axis against
the x-axis in the system S is dt=dx D 1=v.
Any event E is completely defined by its coordinates .x; t/ in S
or .x0 ; t0 / in S0 .
Note, however, that for the same event E the spatial and time
coordinates .xE ; tE / for O in S are different from .x0E ; tE0 / for O0
in S0 (Fig. 3.20)
We regard a point mass A which moves with the velocity ux
against O and with u0x against O0 . Its velocity is determined by
O and O0 by measuring the coordinates x1 .t1 / and x2 .t2 / in S
resp. x01 .t10 / and x02 .t20 / in S0 (Fig. 3.20b).
O obtains: ux D
x2
t2
x1
;
t1
O0 obtains: u0x D
x02
t20
x01
:
t10
The velocity ux is represented in S by the reciprocal slope
x=t D ux of the straight line A1 A2 , In S0 however by
u0x D x0 =t0 . One can see already from, Fig. 3.20b that
ux ¤ u0x , which is quantitatively described by Eq. 3.28.
3.6.2
Minkowski-Diagram
The relativity of observations and their dependence of the reference system can be illustrated by space-time-diagrams as shown
in Fig. 3.20. Each physical event which occurs at the location r D fx; y; zg at time t can be represented by a point in
the four-dimensional space-time fx; y; z; tg. For simplicity we
will restrict the following to one spatial dimension x and the
relative motion of S0 against S should occur only in the xdirection. Then the four-dimensional representation reduces to
Figure 3.22 Minkowski diagram of the axes .x ; t / of a system S at rest, of the
axes .x 0 ; t 0 / of a system S 0 that moves with the velocity v against S and of the
axes .x 00 ; t 00 / of a system S 00 , moving with v against S
a two-dimensional one. Furthermore the time axis t is changed
to c t in order to have the same physical dimension [m] for both
axes. Such a depiction is called Minkowski-diagram (Fig. 3.21).
A body A at rest propagates in an orthogonal .x; ct/ diagram
on a vertical line while a body B with the constant velocity v
relative to O propagates on a sloped straight line with the slope
c t=x D c=v. A light pulse which is emitted from x D 0
at t D 0 and propagates with the velocity c into the x-direction
traverses on a straight line with the inclination of 45ı against the
x-axis because the slope is tan ˛ D c=c D 1. It is represented
by the diagonal in an orthogonal .x; ct/-diagram. Such lines for
moving bodies or for light pulses are called world lines or spacetime-lines, which can be also curved. Two events A and B occur
in the system S simultaneously, if their points in the Minkowskidiagram lie on the line t D t1 parallel to the x-axis (Fig. 3.19).
The ct0 -axis in S0 is the world line of O0 .
We had already discussed in the preceding section that the axis
of two inertial systems S and S0 , which move against each
other with the constant velocity vx are inclined against each
other. If the x- and the ct-axes in system S are orthogonal
the ct0 -axis has the slope tan ˛ D c=vx against the x-axis.
Also the x0 -axis is inclined against the x-axis. According to
the Lorentz-transformations the relation t0 D 0 ) t D v x=c2
must be satisfied (Fig. 3.22). Its slope against the x-axis is therefore dt0 =dx D tan ˇ 0 D v=c. The angle between the x0 and the
ct0 axes is D ˛-ˇ 0 D arctan.c=v/ arctan.v=c/.
For illustration also a third system S00 is shown in Fig. 3.22,
which moves with the velocity v D vx against S. The slope of
the x00 -axis against the x-axis is now tan ˇ 00 D vx =c. the angle
between ct00 -axis and ct-axis is also ˇ 00 . The ct00 -axis forms an
angle ı D 2.ˇ 0 C ˇ 00 / C > 90ı against the x00 -axis.
3.6.3
Figure 3.21 Minkowski diagram showing the world lines of a point A resting
in the system, of a point B moving in the system with the velocity u and a light
pulse emitted at time t D o from the origin
Lenght Scales
Not only the inclination of the axis but also their scaling is different in the systems S, S0 and S00 . Since the velocity of light
is the same in all inertial systems (which implies c D dx=dt D
Chapter 3
For each observer the simultaneity of two events at different spatial points depends on the coordinate system in
which the events are described.
94
3 Moving Coordinate Systems and Special Relativity
Figure 3.24 A yardstick with length O A that rests in the system S , appears
shortened in the system S 0 moving against S
This shows that length standards in different inertial systems
can be in fact different. If O in his system S measures distances
in another system S0 , moving against S, he uses a larger scale,
which means that the length of distances appears shorter.
Chapter 3
Figure 3.23 Illustration of the invariant s 2
3.6.4
c0 D dx0 =dt0 D const) the quantity
s2 D .ct/2
x2 D .ct0 /2
x0 2
(3.29)
must be equal in all inertial systems. This can be also seen from
the Lorentz-equations (3.26). The quantity s2 is therefore invariant under transformation between different inertial systems. For
s2 D 0 the world-line x D ˙ct of a light pulse is obtained. For
the motion of a body with velocity v < c starting at t D 0 and
x D 0 it follows x2 .t/ < .ct/2 ) s2 > 0.
In the .x; ct/-diagram no points with x2 > .ct/2 .s2 < 0/ can be
reached by signals emitted by O at t D 0. The area in Fig. 3.23
with s2 < 0 is non-accessible, while all points with s2 > 0 can
be reached by such signals.
Such an invariant quantity like s2 can be used to fix the scale
length in Minkowski-diagrams. If we allow also imaginary values of s, the square s2 can be also negative. For s2 D 1 we
obtain from (3.29) for all inertial systems (i. e. for S as well as
for S0 ) the hyperbola
x2 .ct/2 D x02 .ct0 /2 D 1 ;
Lorentz-Contraction of Lengths
One of the surprising results of the Lorentz-transformations is
the contraction of the length of bodies in Systems S0 moving
against the observer in a rest frame S. In the foregoing section
we have already indicated that this contraction is caused by the
change of the length scale L0 and that it can be ascribed to the
problem of simultaneity.
Assume a rod with the endpoints P01 and P02 rests in the moving system S0 . The coordinates x01 and x02 therefore move in the
course of time on straight lines parallel to the t0 -axis (Fig. 3.25).
The observer O0 measures at time t10 the length
L0 D P01 P02 D x02 .t10 /
x01 .t10 / :
For the observer O in S the rod resting in S0 moves with the
system S0 with the velocity v in the x-direction. In order to determine the length of the rod, O has to measure the endpoints x1
which is drawn in Fig. 3.23. It intersects in the system S the
x-axis .t D 0/ in the point A at x D 1. This defines the scale
length L D 1 for the system S.
Also in the system S0 is x0 D 1 for t0 D 0, which gives the scale
length L0 D 1 for the observer O0 . However, for the observer O
in S the length L0 appears as L ¤ 1 as can be seen from Fig. 3.24
where L D OA but L0 D 0B. Each observer measures for the
length in his own system another value than for the length in
a system moving against his system. This seems very strange
but is a consequence of the problem of simultaneity, because
in order to measure the lengths L and L0 , O has to measure the
endpoints 0 and A or 0 and B simultaneously, i. e. at the same
time t, while O0 measures them at the same time t0 .
Figure 3.25 Graphical illustration of the Lorentz-contraction of a yard stick
with length L resting in the moving system S 0 , when O in S measures the length
L 0 , expressed in the Minkowski diagram of S 0
3.6 Theory of Special Relativity
95
and x2 simultaneously, i. e. for t D t1 . These endpoints are for
t D t1 at the intersection points P1 .t1 / D P01 and P2 .t1 / ¤ P02 of
the world-lines x01 .t/ and x02 .t/ with the horizontal line t D t1 in
Fig. 3.25. For O is therefore the length of the rod
L D P1 P2 D x2
x1 ;
where x1 and x2 are the vertical projections of P1 and P2 onto
the x-axis t D 0 (Fig. 3.25).
x01 D .x1
) x02
) L0 D
vt1 /I
x01 D .x2
x02 D .x2
x1 /
L ) L < L0 I
for
vt2 /
t1 D t2
because
(3.30)
>1:
The lengths of a moving rod seems for an observer to be
shorter than that of the same rod at rest.
The contraction does not depend on the sign of the velocity
v D ˙vx .
The contraction is really relative as can be seen from the
following example: Two rods should have the same length
L1 D L2 if both are resting in the same system S. Now L2 is
brought into a moving system S0 where it rests relative to the
origin O0 of S0 . For the observer O the length L2 seems to be
shorter then L1 but for O0 L1 seems to be shorter than L2 . This
implies that the Lorentz contraction is symmetric. This is no
contradiction, because the different length measurements are
due to the different observations of simultaneity as has been
discussed before.
Each observer can only make statements of events and times
with respect to his own system S. If he transfers measurements of events in moving systems S0 to his own system S,
he has to take into account the relative velocity of S0 against
S and must use the Lorentz transformations. Then O and O0
come to the same results.
Note that both observers O and O0 come to consistent results for measurements in their own system and in the
other system which moves against their own system, if
they use consequently the Lorentz-transformations.
The answer to the often discussed question whether there is a
“real contraction” depends on the definition of “real”. The only
information we can get about the length of the rod is based on
measurements of the distance between its endpoints. For rods
moving against the observer the locations of the two endpoints
have to be measured simultaneously, which gives the results discussed above.
The relativity of the contraction can be visualized in the
Minkowski diagram of Fig. 3.26. We regard again two iden-
Figure 3.26 Relativity of the Lorentz contraction: a The yardstick O A D 1
rests in S , b the yardstick O B D 1 rests in the moving system S 0
tical yardsticks with the scale L D 1, which rest in the system
S resp. S0 . The yardstick in S has for the observer O the endpoints O and A with the distance OA D 1. The world line for
O is the ct-axis x D 0 and for A the parallel vertical line x D 1.
In Fig. 3.26a also the world line x D c t of a light pulse and
the hyperbola x2 c2 t2 D 1 are drawn. The intersection of the
hyperbola with the x-axis t D 0 defines the scale L D 1, in the
system S.
How is the situation in the system S0 ? The world line of A intersects the x0 -axis ct0 D 0 in the point A0 . Therefore the distance
OA0 is for O the length L0 D 1 of the yardstick. However, for O0
in his system S0 the length of the yardstick is x0 D 1 given by the
distance OB0 where B0 is the intersection point of the parabola
x02 .ct0 /2 D 1. For O is the length of the moving yardstick
therefore smaller than for O0 , who regards the stick resting in
his system.
Note that the parabola is the same in both systems S and S0 (see
Sect. 3.6.3).
For O0 is the scale of O which he measures as OA0 shorter than
his own scale OB0 this means that it appears for O0 shorter.
Now we take a scale OB0 which rests in the system S0 and has
there the length x0 D 1 because B0 is the intersection point of the
parabola x02 .ct/2 D 1 with the x0 -axis ct0 D 0 (Fig. 3.26b).
The world line of O0 is the ct0 -axis x0 D 0 and that of the point
B the line through B0 parallel to the ct0 axis. This line intersects
the x-axis in the point B. The observer O measures the length
of the scale x0 D 1 as the distance OB which is shorter than the
distance OA with x D 1. Now the scale x0 D 1 of the observer
O0 is shorter for the observer O.
This illustrates that the length contraction is due to the
different prolongation of the scale which is caused by the
different simultaneity for measurements of the endpoints by
O and O0 .
Note that both observers O and O0 come to contradictionfree statements concerning measurements in their own
system and in the other system if they use the Lorentztransformations.
Chapter 3
Since x0 differs from x the two lengths L and L0 are different. Because the scale lengths s and s0 are different one
cannot directly geometrically compare the length of the rod
measured in S and S0 from Fig. 3.25, but has to use the Lorentztransformations.
96
3 Moving Coordinate Systems and Special Relativity
3.6.5
Time Dilatation
We regard a clock, which rests in the origin O of system S. We
assume that this clock sends two light pulses at times t and tCt
with a time delay t of the second pulse. An observers O at the
location x0 in the system S receives the light pulses at times t1
and t2 , at the event points A and B of his world line x D x0 D
const (Fig. 3.27). For O the time interval between the two pulses
is
t D AB D t2
t1 :
An observers O0 sitting at x0 D x00 in the system S0 which moves
with the velocity v against S receives the light pulses at the intersection points A0 and B0 of his world line x0 D x00 with the
two axes x0 D ct0 and x0 D c.t0 C t0 / which are observed at
times t10 and t20 measured with his clock in S0 .
Chapter 3
The observer O in S knows, that these times t10 and t20 are
transformed into his measured times t1 and t2 by the Lorentz
transformations
t10 D
t1
v x0 0
t2 D
c2
t2
v x0
:
c2
According to these equations he determines the time difference
in the moving system S0 as
t0 D t20
t10 D
t :
(3.31)
Since D .1 v 2 =c2 / 1=2 > 1 the observer O at rest measures for the moving system S0 a longer time interval t0
between the two pulses than the moving observer O0 . Because the clock resting in S moves for the observer S0 he
measures, that this clock runs slower than his own clock. This
can be expressed by: Moving clocks run slower. Equivalent to the length contraction also the time dilatation is caused
by the different observations of simultaneity in the systems
S and S0 . This effect increases with increasing velocity v
and reaches essential values only for velocities v close to
Figure 3.28 Measurement of the lifetime of relativistic muons with two detectors at different heights h1 and h2 above sea level
the velocity c of light. However this time dilatation can be
measured with very precise clocks already for smaller velocities. For example, if two clocks are synchronized in Paris
and transported by a fast plane (such as the concorde with
v D 2400 km=h D 667 m=s ) D 1 C 8:9 10 12 ) to New
York the difference between t and t0 during a flight time of
3 hours is 8:9 10 9 s D 8:9 ns.
A much more precise measurement of the time dilatation can be
obtained with faster moving clocks. Examples of such fast moving clocks are fast elementary particle such as electrons, protons
or muons which move with velocities v c.
The cosmic radiation (electrons and protons with very high energy) produce in the upper earth atmosphere at collisions with
the atomic nuclei of the atmospheric molecules muons with
velocities v c which reach nearly the velocity of light. Part
of these muons reach the earth surface, while part of them decay during their flight through the atmosphere according to the
scheme
! e C ā C āe
(3.32)
into an electron and two neutrinos (see Vol. 4). The lifetime of decelerated resting muons can be precisely measured
as D 5 10 6 s.
In order to measure the lifetime 0 of fast flying muons the rate
of muons, incident onto a detector is measured at different altitudes above sea level, for instance at the altitude h D h1 on the
top of a mountain and at h D h2 at the bottom of the mountain
(Fig. 3.28). For a mean decay time 0 of the muons moving with
the velocity v the relative fraction dN=N decays during the time
interval dt= 0
dN D a .N= 0 /dt :
Integration yields
N.h2/ D N.h1 / e
Figure 3.27 Minkowski diagram for illustration of the time dilatation. Two
signals with the time difference t D t2 t1 in the resting system S reach the
moving observer O 0 in S 0 with the time difference t 0 D t
t= 0
with
t D
h2
h1
v
;
where the factor a < 1 takes into account the scattering of
muons by the atmospheric molecules. This factor can be calculated from known scattering data. The often repeated measurements clearly gave essentially higher lifetimes 0 D 45 10 6 s
3.6 Theory of Special Relativity
of the moving muons than D 5 10 6 s for muons at rest.
From 0 D follows D 9 ! v D 0:994c. The muons
have a velocity v which is smaller than c by only 6 ‰.
Meanwhile many short-lived elementary particles can be produced and accelerated to high velocities. Comparing the
lifetimes of these particles at rest and while moving, unambiguously confirms the time dilatation postulated by the Lorentz
transformations.
The relativistic time dilatation can be illustrated by an
“gedanken-experiment” proposed by Einstein using a light pulse
clock (Fig. 3.29). The system consists of a box with length L.
On one side a flash lamp is mounted at the point A and on the
opposite end a mirror M1 . The flash lamp emits a short light
pulse and starts a clock. The light pulse reflected by M1 is received by a detector which stops the clock. The time interval
t0 D 2L=c is used as time scale in the system S in which the
light clock rests.
Now we let the system S move with the velocity v relative to a
system S0 in a direction perpendicular to the length L. For the
observer O0 in S0 the light pulse now travels from A to B and
is reflected to C. With AN D NC D v t=2 it follows from
Fig. 3.29
"
#1=2
t0 2
2
AB C BC D 2 L C v
D c t0
2
(3.33)
2L
0
) t D
:
.c2 v 2 /1=2
The observer O measures t D 2L=c. The comparison between
t and t0 gives:
t
D t ;
(3.34)
t0 D
.1 v 2 =c2 /1=2
which turns out to be identical with (3.31).
3.6.6
The Twin-Paradox
No other problem of special relativity has aroused so many
controversial discussions as the twin paradox (often called the
Figure 3.30 Minkowski diagram for the twin paradox
clock-paradox), discussed by Einstein in his first paper 1905
about relativity. It deals with the following situation:
Two clocks which are synchronized show equal time intervals
when sitting in the same system S at rest. One of the clocks
is taken by O0 on a fast moving spacecraft and returns after the
travel time T (measured by O in S) back to the other clock which
always had stayed in the system S. A comparison of the two
clocks shows that the moved clock is delayed, that means that it
shows a smaller value T 0 than T [3.8–3.11].
This “gedanken-experiment” has meanwhile be realized and the
time dilatation has been fully verified (see previous section). For
manned space missions this means that an astronaut A after his
return to earth after a longer journey through space is younger
than his twin brother B who has stayed at home. The “gained”
time span is, however, for velocities of spacecrafts which can
be realized up to now, very small and therefore insignificant.
Nevertheless an understanding of the twin paradox is of principal significance because it illustrates the meaning of relativity,
which is often used in a popular but wrong way.
We have discussed in the previous section that the time dilatation is relative, i. e. for each of the two observers O and O0 the
time scale of the other seems to be prolonged. Why is it then
possible to decide unambiguously that A and not B is younger
after his return?
The essential point is that A is not strictly in an inertial system,
even if he moves with constant velocity, because at his returning point he changes the system from one that moves with the
velocity Cv into one that moves with v against B. This shows
that the measurements of A and B are not equivalent.
In order to simplify the discussion we will categorize the journey
of A into three sections, which are illustrated in the Minkowski
diagram of the resting system S of B in Fig. 3.30.
A starts his journey from x D 0 at time t D t0 D 0, reaches
in a negligibly small time interval his final speed v until he
arrives at his point of return P1 .xr ; T=2/ after the time t1 D
T=2.
At time t1 D T=2 he decelerates to v D 0 and accelerates
again to v. This should all happen within a time interval
which is negligibly small compared with the travel time T.
Astronaut A flies with v2 D v back to B and reaches B in
x D 0 at t2 D T.
Chapter 3
Figure 3.29 Einstein’s “light-clock” for illustration of the time dilatation
97
98
3 Moving Coordinate Systems and Special Relativity
While the world line of B in Fig. 3.30 is the vertical line x D 0,
A follows the line x D v t ! ct D .c=v/x until the point
of return P1 from where he travels on the line x D xr v.t
T=2/ ) ct D .c=v/.xr x/ C cT=2, until the point P2 .0; T/
where he meets with B.
From (3.29) we obtain for the invariant
ds2 D c2 dt2
dx2 D c2 dt02
dx02 :
This yields the different travel times for A and B: For B is always dx D 0. We therefore get for the total distance s in the
Minkowski diagram:
Z
Z
s D ds D c dt D c T :
Chapter 3
For the moving astronaut A the resting observer B measures on
the way OP1 : dx D v dt ! ds2 D c2 dt2 v 2 dt2 , which gives
for the total path
Z
Z
p
cT
cT 0
D
;
ds D c2 v 2 dt D
2
2
and on the way P1 P2 back: dx D vdt:
Z
Z
p
cT
cT 0
2
2
v
dt D
D
:
ds D c
2
2
The total travel time measured by B in S for the system S0 of his
twin A is then T 0 D T= < T. This result can be also explained
by the Lorentz contraction: For A is the path L shortened by
the factor . Therefore the travel time T for A is shorter by the
factor since A as well as B measure the same velocity v of A
relative to B.
The asymmetry of the problem can be well illustrated by regarding light pulses sent at constant intervals by A to B and by B to
A. Both the observer B and the astronaut A send these light signals at the frequency f0 measured with their clocks. The sum of
the sent pulses at a frequency f0 D 1=s gives the total travel time
in seconds (Fig. 3.31).
Figure 3.32 Doppler-effect of the signal frequency illustrated in the Minkowski
diagram
While A moves away from B both observers receive the pulses
at a lower frequncy f1 because each successive pulse has to travel
a longer way than the preceding pulse. The asymmetry occurs at
the turning point P1 . While A on his way back now receives the
pulses with a higher frequency f2 directly after he turns around,
B receives the pulses from A with the frequency f2 only at times
t c x0 after the return time. He receives the same total
number N 0 of pulses as has been sent by A but he receives for a
longer time signals with lower frequency than A. Therefore he
measures a longer travel time for A than A himself.
This is illustrated in Fig. 3.31 for v D 0:6c. B sends during
the travel time of A altogether 20 pulses, which are all received
by A. While B sends his signals at constant time intervals 0 ,
A receives them on the outbound trip with larger time intervals
1 i. e. lower frequency and on the way back with shorter time
intervals 2 , i. e. higher frequency. Measured in the system S0
of A the travel time T 0 is shorter by the factor . The astronaut
A sends during this time only N 0 D N= D 16 pulses which
are all received by B. Since B receives the signal sent by A at
the return time only delayed, he receives signals with the larger
interval 1 (lower frequency) for a longer time and only after the
time tr0 C L=2c the signals with the shorter intervals 2 .
This is further illustrated by the Minkowski diagram of Fig. 3.32
which explains the relativistic Doppler-effect. The astronaut A
is at time t D 0 at the point .x D 0; t D 0/ in the Minkowski
diagram. He moves, measured by B on the line
xDvt ! ct D
Figure 3.31 Illustration of the twin paradox, using the signals sent and received by A and B
c
x:
v
The observer B sitting always at x D 0 sends light pulses at a
repetition frequency f0 . A pulse sent by B at time t0 travels in
the .x; ct/-diagram on lines with a 45ı slope which intersects the
world line of A in the point .x1 ; t1 /, where it is received by A.
The next pulse is sent by B at time t D t0 C D t0 C 1=f0
and reaches A at .x2 ; t2 /. According to Fig. 3.31 the following
3.6 Theory of Special Relativity
99
Table 3.1 Measurement of multiple physical quantities of resident and traveler (according to [3.7])
Measurement of B (resident)
Total travel time
TD
Total number of sent signals
f T D
Reversal time of A
tu D
2L
v
L
v
T0 D
2fL
v
C
D
Travel time after reversal
L
c
fL
v .1
D Lv .1 ˇ/
1=2
f 00 t2 D f 1Cˇ
vL .1
1 ˇ
Number of received signals with frequency
1=2
f 00 D f 1Cˇ
1 ˇ
D
fL
.1
v
N D f 0 tu C f 00 t2 D
D
2fL
v
Conclusion regarding the time measured by the other
T0 D
2L
v
relations apply:
D .1
2fL
v .1
ˇ2 /
L
v
D
t20 D
ˇ/
L
v
1=2
1 ˇ
1Cˇ
fL
v .1
D
f 00 t20 D f
D
ˇ 2 /1=2
2fL
v
f 0 tu0 D f
ˇ 2 /1=2
Total number of received signals
N D f 0 tu C f 00 t2 ; N 0 D f 0 tu0 C f 00 t20
ˇ D v=c;
tu0 D
ˇ 2 /1=2
L
c
L
v
t2 D
2L
v
f T0 D
D vL .1 C ˇ/
1=2
1 ˇ
Lv .1 C ˇ/
f 0 tu D f 1Cˇ
Number
of received signals
with frequency
1=2
1 ˇ
0
0
f f D f 1Cˇ
Measurement of A (traveler)
1=2
L
v
ˇ/
1
ˇ2
1=2
L
1
v .1 ˇ 2 /1=2
fL
.1
v
1Cˇ
1 ˇ
1=2
.1
ˇ 2 /1=2
C ˇ/
N 0 D f 0 tu0 C f 00 t20 D
TD
2fL
v
2L
v
their systems. This shows that there are no contradictions in
the description of the twin paradox. Observer B knows, that
the travel time T 0 measured by A is shorter than the time measured by himself because A is sitting in a moving system, and A
knows that B measures in his resting system a longer time.
x1 D c .t1 t0 / D x0 C v t1
x2 D c .t2 t0 / D x0 C v t2 :
Subtraction of the first from the second equation yields
c
I
c v
t1 D
t2
x1 D
x2
vc
:
c v
3.6.7
Figure 3.32 illustrates that for A the time intervals 0 are longer
on the outward flight than on the return flight. Astronaut A measures in his system S0 according to the Lorentz transformations
0 D t20
D
With
D .1
0 D
t10 D
.1 C ˇ/ ;
ˇ2 /
h
.t2
1=2
1Cˇ
1 ˇ
i
v
.x2 x1 /
2
c
with ˇ D v=c :
t1 /
) f0 D
1
D f0
0
1 ˇ
1Cˇ
1=2
Since the speed of light is the upper limit for all velocities with
which signals can be transmitted from one space-time point
.x1 ; t1 / to another point .x2 ; t2 / all space-time events can be classified into those which can be connected by signals and those
which cannot. In the first case an event in .x2 ; t2 / can be caused
by an event in .x1 ; t1 /.
In the Minkowski diagram of Fig. 3.33 the two diagonal lines
x D ˙c t are the worldliness of light signals passing through
this becomes
1=2
Space-time Events and Causality
:
Astronaut A measures therefore on the outward flight the
smaller repetition frequency f1 which is smaller than f0
by the factor Œ.1 ˇ/=.1 C ˇ/1=2 and on the return flight
with the velocity v he measures the higher repetition rate
f2 D Œ.1 C ˇ/=.1 ˇ/1=2 f0 .
In Tab. 3.1 the different measurements of A and B are summarized. The table shows again, that the total number of pulses sent
by B is equal to the number received by A but different from the
number sent by A. The last line in Tab. 3.1 makes clear, that B
can conclude the travel time measured by A from the number
of pulses received from A and vice versa can A conclude the
time measured by B. Both observers are therefore in complete agreement in spite of the different times measured in
Figure 3.33 Two-dimensional Minkowski diagram with the shaded areas for
past and future and the white areas for non-accessible space-time point .r; t /
Chapter 3
Physical quantity
100
3 Moving Coordinate Systems and Special Relativity
the point .x D 0; t D 0/. These world lines divided the spacetime into different regions: All regions with .x; t > 0/ with
jxjj ct represent the future seen from .x D 0; t D 0/. They
can be reached by signals sent from .0; 0/, while the region with
.x; t < 0/ form the past.
This can be also expressed in the following way: All events
in space-time point .x; t/ can be causally connected with each
other, i. e. an event in .x2 ; t2 / can be caused by an event in .x1 ; t1 /
if both points lie in the red shaded regions in Fig. 3.33. this
means signals can be transferred between these points and interactions between bodies in these points are possible. For instance
the event A can influence the event B in Fig. 3.33 but not the
event C.
An observer in the space-time point .x; t/ with jxj jctj can
never receive a signal from points in the white regions with
jxj > jctj. We call these regions therefore “elsewhere”.
In a three-dimensional space-time diagram .x; y; ct/ the surfaces x2 C y2 D c2 t2 form a cone called the light-cone. Past
and future are inside the cone. “elsewhere” is outside. In a
four-dimensional space-time diagram .x; y; z; ct/ this light cone
becomes a hyper-surface.
Very well written introductions to the special relativity and its
consequences without excessive Mathematics, which are also
understandable to undergraduate students can be found in [3.7–
3.11].
Summary
Chapter 3
For the description of motions one needs a coordinate system. Coordinate systems in which the Newtonian Laws can
be formulated in the form, discussed in Sect. 2.6 arte called
inertial systems. Each coordinate system which moves with
constant velocity v against another inertial system is also an
inertial system.
The transformation of coordinates .x; y; z/, of time t and of
velocity v and therefore also of the equation of motion from
one to another inertial system is described by the Lorentz
transformations. They are based on the constancy of the
speed of light c, confirmed by experiments, which is independent of the chosen inertial system and has the same
value in all inertial systems. For small velocities v c the
Lorentz transformations approach the classical Galilei transformations.
The description of motions in accelerated systems demand
additional accelerations, which are caused by “inertial or
virtual” forces. In a rotating system with constant angular
velocity these are the Coriolis force FC D 2 m.v0 !/ which
depends on the velocity v 0 of a body relative to the rotating
system, and the centrifugal force Fcf D m ! .r !/ which
is independent of v 0 .
The theory of special relativity is based on the Lorentz transformations and discusses the physical effects following from
these equations when the motion of a body is described in
two different inertial systems which move against each other
with constant velocity v. An essential point is the correct
definition of simultaneity of two events. Many statements of
special relativity can be illustrated by space-time diagrams
.x; ct/ (Minkowski diagrams), as for instance the lengthcontraction or the time-dilatation. Such diagrams show that
these effects are relative and symmetric, which means that
each observers measures the lengths in a system moving
against his system contracted and the time prolonged. The
description of the two observers O and O0 are different but
consistent. There is no contradiction.
For the twin-paradox an asymmetry occurs, because the astronaut A changes its inertial system at the point of return. It
is therefore possible to attribute the time dilation unambiguously to one of the observers.
The statements of special relativity have been fully confirmed by numerous experiments.
Problems
From a point A on the earth equator a bullet is shot in
3.1
An elevator with a cabin heights of 2:50 m is accelerated 3.2
with constant acceleration a D 1 m=s2 starting with v D 0 at horizontal direction with the velocity v D 200 m=s.
a) in the north direction
t D 0. After 3 s a ball is released from the ceiling.
b) in the north-east direction 45ı against the equator
a) At which time reaches it the bottom of the cabin?
b) Which distance in the resting system of the elevator well has c) In the north-west direction 135ı against the equator
What are the trajectories in the three cases described in the systhe ball passed?
c) Which velocity has the ball at the time of the bounce with tem of the rotating earth?
the bottom in the system of the cabin and in the system of
the elevator well?
References
101
3.7
A plane disc rotates with a constant angular velocity
! D 2 10 s 1 around an axis through the centre of the disc
perpendicular to the disc plane. At time t D 0 a ball is
launched with the velocity v D fvr ; v' g with vr D 10 m=s,
v' D 5 m=s (measured in the resting system) starting from the
point A .r D 0:1 m; ' D 0ı /. At which point .r; '/ does the ball
reach the edge of the disc?
3.8
A bullet with mass m D 1 kg is shot with the velocity
v D 7 km=s from a point A on the earth surface with the geographical latitude ' D 45ı into the east direction. How large
are centrifugal and Coriolis force directly after the launch? At
which latitude is its impact?
3.13 At January 1st 2010 the astronaut A starts with the constant velocity v D 0:8c to our next star ˛-Centauri, with a
distance of 4 light years from earth. After arriving at the star,
A immediately returns and flies back with v D 0:8c and reaches
the earth according to the measurement of B on earth at the 1st
of January 2020. A and B had agreed to send a signal on each
New Year’s Day. Show that B sends 10 signals, but A only 6.
How many signals does A receive on his outbound trip and how
many on his return trip?
3.14 Astronaut A starts at t D 0 his trip to the star Sirius (distance 8.61 light years) with the velocity v1 D 0:8c. One year
later B starts with the velocity v2 D 0:9c to the same star. At
which time does B overtake A, measured
3.9
Two inertial systems S and S0 move against each other a) in the system of A,
with the velocity v D vx D c=3. A body A moves in the system b) of B and
S with the velocity u D fux D 0:5c; uy D 0:1c; uz D 0g. What c) of an observer C who stayed at home?
At which distance from C measured in the system of C does this
is the velocity vector u0 in the system S0 when using
occur?
References
3.1.
3.2a.
3.2b.
3.3.
3.4a.
https://en.wikipedia.org/wiki/Fictitious_force
A.A. Michelson, E.W. Morley, Am. J. Sci. 34, 333 (1887)
R. Shankland, Am. J. Phys. 32, 16 (1964)
A.A. Michelson, Studies in Optics. (Chicago Press, 1927)
A. Brillet, J.L. Hall, in Laser Spectroscopy IV, Proc. 4th
Int. Conf. Rottach Egern, Germany, June 11–15 1979.
(Springer Series Opt. Sci. Vol 21, Springer, Berlin, Heidelberg, 1979)
3.4b. W. Rowley et al., Opt. and Quant. Electr. 8, 1 (1976)
3.5. A. Einstein, H.A. Lorentz, H. Minkowski, H. Weyl, The
principles of relativity. (Denver, New York, 1958)
3.6. R. Resnik, Introduction to Special Relativity. (Wiley,
1968)
3.7. E.F. Taylor, J.A. Wheeler, Spacetime Physics: Introduction to Special Relativity, 2nd ed. (W.H. Freeman &
Company, 1992)
3.8. A.P. French, Special Relativity Theory. (W.W. Norton,
1968)
3.9. D.H. Frisch, H.J. Smith, Am. J. Phys. 31, 342 (1963)
3.10. N.M. Woodhouse, Special Relativity. (Springer, Berlin,
Heidelberg, 2007)
3.11. C. Christodoulides, The Special Theory of Relativity:
Theory, Verification and Applications. (Springer, Berlin,
Heidelberg, 2016)
Chapter 3
3.3
A ball hanging on a 10 m long string is deflected from a) the Galilei transformations and
its vertical position and rotates around the vertical axis with b) the Lorentz transformations?
! D 2 0:2 s 1 . What is the angle of the string against the How large is the error of a) compared to b)?
vertical and what is the velocity v of the ball?
3.10 A meter scale moves with the velocity v D 2:8 108 m=s
3.4
In the edge region of a typhoon over Japan (geographical passing an observer B at rest. Which length is B measuring?
latitude ' D 40ı ) the horizontally circulating air has a velocity
of 120 km=h. What is the radius of curvature r of the path of the 3.11 A space ship flies with constant velocity v to the planet
Neptune and reaches Neptune at its closest approach to earth.
air in this region?
How large must be the velocity v if the travel time, measured by
3.5
A fast train .m D 3 106 kg/ drives from Cologne to the astronaut is 1 day? How long is then the travel time meaBasel with a velocity of v D 200 km=h exactly in north-south sured by an observer on earth?
direction passing 48ı latitude. How large is the Coriolis force
3.12 Light pulses are sent simultaneously from the two endacting on the rail? Into which direction is it acting?
points A and B of a rod at rest. Where should an observer O
3.6
A body with mass m D 5 kg is connected to a string with sit in order to receive the pulses simultaneously? Is the answer
different when A, B and O moves with the constant velocity v?
L D 1 m and rotates
At which point in the system S an observer O0 moving with a
a) in a horizontal plane around a vertical axis
velocity vx against S receives the pulses simultaneously if he
b) in a vertical plane around a horizontal axis
At which angular velocity breaks the string in the cases a) and knows that the pulses has been sent in the system S simultaneously from A and B?
b) when the maximum tension force of the string is 1000 N?
Systems of Point Masses;
Collisions
4.1
Fundamentals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104
4.2
Collisions Between Two Particles . . . . . . . . . . . . . . . . . . . . . . 107
4.3
What Do We Learn from the Investigation of Collisions? . . . . . . . . . 115
4.4
Collisions at Relativistic Energies . . . . . . . . . . . . . . . . . . . . . . 119
4.5
Conservation Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
4
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126
Chapter 4
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
© Springer International Publishing Switzerland 2017
W. Demtröder, Mechanics and Thermodynamics, Undergraduate Lecture Notes in Physics, DOI 10.1007/978-3-319-27877-3_4
103
104
4 Systems of Point Masses; Collisions
In the preceding chapters we have discussed the motion of a
single particle and its trajectory under the influence of external
forces. In this chapter we will deal with systems of many particles, where besides possible external forces also interactions
between the particles play an important role.
4.1
Fundamentals
RS
At first we introduce several expressions and definitions of fundamental terms and notations for systems of many particles.
Figure 4.1 Definition of center of mass
4.1.1
Centre of Mass
We consider N point masses with position vectors ri and define
as the centre of mass the point with the position vector
P
mi ri
1 X
RS D Pi
mi ri ;
(4.1)
D
M i
i mi
where M D
P
With the acceleration of the centre of mass aS D dvS =dt we
obtain
(4.5)
F D MaS :
mi is the total mass of all N particles (Fig. 4.1).
When the masses mi move with the velocities vi D dri =dt we
define the velocity vS of the centre of mass as
Chapter 4
1 X
dRS
D
mi vi :
vS D
dt
M i
(4.2a)
With the momenta pi D mP
i v i (4.2a) can be also expressed by
the total momentum P D pi as
P D MvS :
(4.2b)
If no external forces are acting on the particles, we need to regard only internal forces, i. e. interactions between the particles.
Such a system without external forces is called a closed system.
P P
From the Newtonian law Fik D Fki it follows: i k¤i Fik D
0. In a closed system the vector sum of all forces is zero.
P
With Fi D k¤i Fik and Fi D dpi =dt the total momentum of the
system
X
PD
pi D const :
(4.3)
Since P is the momentum of the centre of mass we can state:
The centre of mass of a closed system moves with constant
momentum. This implies that its velocity does not change.
If an external total force F ¤ 0 acts onto the system we can
write
dP
d X
FD
;
(4.4)
pi D
dt
dt
The centre of mass of an arbitrary system of particles
movesP
in the same way as a body with the total mass
M D
mi would move under the action of the external
force F.
Often it is useful to choose a coordinate system with the centre
of mass as origin, which moves with the velocity vS of the centre
of mass against the fixed laboratory system. Such a system is
called the centre of mass system (CM-system).
The position vectors ri in the lab-system are related to the position vectors riS in the CM-system (Fig. 4.1) by
(4.6a)
ri D riS C RS :
Inserting into (4.1) gives
X
i
mi riS D
D
X
mi .ri
RS /
i
X
mi ri
X
mi riS D 0
RS
i
X
i
mi D 0 ;
(4.6b)
This implies that in the CM-system
P the position vector RS of the
centre-of-mass is RS D .1=M/ mi riS D 0.
4.1 Fundamentals
which can be verified by differentiation of (4.6a). For the momenta we therefore get
i
mi viS D
X
i
piS D 0 :
For a closed system of two masses m1 and m2 the total kinetic
energy in the lab-system is
Ekin D
D
Ekin D Ekin C 21 MVS2 :
.S/
(4.7b)
Reduced Mass
We consider two particles with masses m1 and m2 which interact
with each other due to the forces F12 D F21 . Without other
external forces the equations of motion read:
dv2
F21
:
D
dt
m2
where v12 D v1
v2 / D
1
1
C
m1
m2
F12 ;
(4.10)
2
2
viS
mi vi2
1
MVS2 :
2
(4.11a)
is the difference of Ekin in the lab-system and the kinetic energy
of the CM.
P
Inserting vS D .1=M/ mi vi gives with (4.9)
1 2
v :
2 12
(4.11b)
The kinetic energy of a closed system of two particles in the
CM-system equals the kinetic energy of a single parrtivle with
the reduced mass which moves with the relative velocity v12 .
This important relations can be summarized as:
The relative motion of two particles under the influence of
their mutual interaction F12 D F21 can be reduced to the
motion of a single particle with the reduced mass driven
by the force F12 .
This is illustrated in Fig. 4.2 where two masses m1 D m and
m2 D 1:5m move around their centre of mass S which moves
itself with the velocity VS . An example of such a system is a
double-star system, where two stars with different masses circulate around their common CM (see Vol. 4).
(4.8a)
4.1.3
(4.8b)
We consider two point masses m1 and m2 with their mutual interaction forces
Subtraction yields
d
.v1
dt
2
i
X
1
.S/
The total motion of the closed system can be divided into a uniform motion of S with the constant velocity VS and a relative
motion of the two particles against S.
dv1
F12
I
D
dt
m1
D
X mi
Ekin D
In the Lab-system the kinetic energy of a closed system
.S/
can be written as the sum of Ekin in the CM-system plus
the kinetic energy of the total mass M concentrated in the
center of mass S (translational energy of the system).
dv12
:
dt
S
The kinetic energy Ekin
of the two particles in the CM-system
(4.7a)
C .m1 v1S C m2 v2S / V S :
(4.9)
This means: For the relative motion of the two particles the
equation of motion is completely analogous to Newton’s equation (2.18a) for a single particle with the mass . This shows
the usefulness of defining the reduced mass.
.S/ Def
The last term is zero because p1S C p2S D 0 and we obtain:
4.1.2
F12 D
Ekin D
C 12 .m1 C m2 / VS2
m1 m2
;
m1 C m2
and rewrite Eq. 4.8b we get
(4.6d)
The sum of all momenta in the CM-system is always zero.
1
m v 2 C 12 m2 v22
2 1 1
2
2
1
C m2 v2S
m1 v1S
2
D
(4.6c)
vi D viS C V S ;
X
Introducing the reduced mass
v2 is the relative velocity of the two particles.
Angular Momentum of a System of
Particles
F12 D F21
Chapter 4
The relation between the velocity vi in the lab-system and viS in
the CM-system is
105
106
4 Systems of Point Masses; Collisions
VS
μ
Figure 4.3 Torques acting on a system of two masses under the influence of
external forces
F12
r12
v12
Figure 4.2 a Velocity VS of the CM of a system of two masses with velocities
vi ; b Reduction of the relative motion of two masses mi to the motion of a single
particle with the reduced mass under the action of the force F12
and the external forces F1 acting on m1 and F2 acting on m2 .
The torques on the two masses with respect to the origin 0 of
the coordinate system are
For the special case that no external forces are present the torque
is zero and therefore the angular momentum L is constant.
The total angular momentum of a closed system of particles is constant.
Chapter 4
D1 D r1 .F1 C F12 / ;
D2 D r2 .F2 C F21 / ;
Using CM-coordinates we can divide the angular momentum
(4.13) according to (4.6a)
and the total torque of the system is then (Fig. 4.3)
D D .r1 F1 / C .r2 F2 / C .r1
The time derivative of the total angular momentum of a
system of particles referred to an arbitrary point is equal
to the total torque exerted onto the system and referred to
the same point.
r2 / F12 :
Since the direction of the internal forces F12 D F21 lies in the
direction of the connecting line r12 D .r1 r2 / the last term
vanishes and the total torque
D D .r1 F1 / C .r2 F2 /
(4.12)
becomes the vector sum of the torques on the individual particles. Without external forces the total torque on the system is
zero!
The total angular momentum L of the system with respect to the
origin 0 is
L D .r1 p1 / C .r2 p2 / ;
(4.13)
and we obtain, analogous to the Eq. 2.48 for a single particle:
dL
D .r1 F1 / C .r2 F2 / D D :
dt
L D m1 .r1S C RS / .v1S C V S /
C m2 .r2S C RS / .v2S C V S / :
For many particles this reads
X
LD
mi .riS C RS / .viS C V S /
i
D M .rS V S / C
C
X
i
X
i
mi .riS viS /
mi .RS viS / C
X
i
mi .riS V S / :
P
P
The terms i mi .RS vS / and i mi .riS vS / are zero according
to (4.6d) and (4.6b) and it follows:
X
mi .riS viS / :
(4.14a)
L D M .rS V S / C
i
The first term
(4.14)
The derivation of these equations and the situations discussed
for a system of two particles can be readily generalized to a
system of many particles. This gives the important statement:
L0S D M .RS V S /
(4.15a)
is the angular momentum of the total mass contracted in the CM
referred to the origin of the coordinate system. The second term
gives the total angular momentum referred to the CM.
4.2 Collisions Between Two Particles
For
P a system of two particles we can transform LS because of
i mi v iS D 0 into
LS D
X
D .r1S
LiS D .r1S p1S / C .r2S p2S /
r2S / p1S D r12 v12 ;
107
around the CM. Although there is no analytical
solution for the exact orbit, very good numerical
approximations have been developed [4.1b].
(4.15b)
(with piS D v12 ). This follows from (4.6d) and (4.10). We can
therefore state:
The angular momentum LS of a system of two particles
in the CM is equal to the angular momentum of a single
particle with the reduced mass and the position vector
r12 D r1 r2 .
1. The relative motion of the earth-moon system around
their common center of mass S (Fig. 4.4) can be reduced to the motion of a single body with reduced
mass D mE mMo =.mE C mMo / 0:99mMo in
the central gravitational force field between earth and
moon around the centre M of the earth. The centre of mass is located inside the earth 4552 km away
from the centre M because the mass of the moon
mMo 0:01mE is small compared with the earth mass.
In the CM-system earth and moon describe nearly circular elliptical orbits around the common CM with
radii
Figure 4.4 a Motion of the moon in the CM-system earth–moon.
b Motion of the moon and the CM in the galactic coordinate system
where the sun also moves
2. The hydrogen atom is a two-body system of an electron with mass me and proton with mass mp . Because
mp D 1836me the reduced mass is D 0:99946me
me . In a classical picture proton and electron circulate
around the CM. With the mean distance rpe between
proton and electron the CM lies .1=1836/rpe from the
centre of the proton. The motion of the two particles
can be separated into the translation of the CM with
the velocity VS and the motion of a particle with mass
with the relative velocity vpe around the CM. The
total kinetic energy of the H-atom is then:
rE D .mMo =.mE C mMo // rEMo 0:01rEMo
Ekin D
and
2
:
mp C me VS2 C 12 vpe
For velocities of the H-atom which correspond to
thermal energies at room temperature the first term
. 0:03 eV/ is very small compared to the second
J
term of the “internal” energy ( 10 eV).
rMo D .mE =.mE C mMo // rEMo 0:99rEMo ;
where rEMo is the distance between earth and moon.
In a coordinate system which is referred to the centre
of our galaxy the lunar orbit is a complicated curve,
shown in Fig. 4.4b where the deviations from the path
of the CM are exaggerated in order to elucidate the
situation. This complicated motion can be composed
of
a) the motion of the moon around the CM of the
earth-moon system
b) the motion of the CM around the centre of mass of
the solar system, which
P is located inside the sun,
because Mˇ > 103 mPlanets .
c) the motion of the CM of the solar system around
the centre of our galaxy.
d) The exact calculation of the lunar orbit has to take
into account the simultaneous gravitational attraction of the moon by the earth and the sun, which
changes with time because of the changing relative
position of the three bodies. Because of this “perturbation” the lunar orbit is not exactly an ellipse
1
2
4.2
Collisions Between Two Particles
This section is of great importance for the understanding of
many phenomena in Atomic and Nuclear Physics, because an
essential part of our knowledge about the structure and dynamics of atoms and nuclei arises from investigations of collision
processes.
When two particles approach each other they are deflected due
to the interaction forces between them. The deflection occurs in the whole spatial range where the forces are noticeable
(Fig. 4.5). Due to this interaction both particles change their
momentum and often also their energy. However, conservation
laws demand that momentum and energy of the total system are
always preserved.
Chapter 4
Examples
108
4 Systems of Point Masses; Collisions
where p0i is the momentum of particle i after the collision and
U is that part of the initial energy that had been converted
into internal energy of one or both of the collision partners
and is therefore missing in the kinetic energy after the collision
(U < 0). If internal energy of the colliding partners has been
transferred into kinetic energy we get U > 0.
Figure 4.5 Schematic illustration of a collision with the asymptotic scattering
angles 1 an 2
The Eq. 4.16 and 4.17 describe the collision process completely
in that sense, that relations between magnitude and direction of
the individual momenta of the particles after the collision can be
determined, if they are known before the collision.
Depending on the magnitude of U we distinguish between three
cases:
The exact form of the trajectory of the particles inside the interaction zone can be determined only if the exact interaction
potential is known. However, it is possible to make definite
statements about magnitude and direction of the particle momenta after the collision in a great distance from the interaction
zone. These statements are based solely on the conservation of
momentum and energy. We will illustrate this in more detail in
the following section.
4.2.1
Basic Equations
Chapter 4
Although the total energy of the two colliding partners is preserved during collisions, part of the translational energy is often
converted into other forms of energy, as for instance potential energy or thermal energy. From (4.3) it follows however,
that the total momentum of the collision partners is always retained.
The basic equations for collision processes between two particles with velocities v which are small compared to the velocity
c of light (non-relativistic collisions) can be written as:
conservation of momentum (Fig. 4.6)
p01
C
p02
D p1 C p2
During reactive collisions (for instance during chemical reactions or in high energy collisions) new particles can be produced
and the masses of the collision partners may change. An example is the reaction
H2 C Cl2 ! HCl C HCl :
These reactive collisions are treated later.
Note:
(4.16)
conservation of energy
p02
p02
p21
p2
2
1
C 2 CU
0 C
0 D
2m1
2m2
2m1
2m2
U D 0, elastic collisions. The total kinetic energy is preserved, while the kinetic energy of the individual particles
generally changes.
U < 0, inelastic collisions. The total kinetic energy after the
collision is smaller than before. Part of the initial kinetic energy has been converted into internal energy of the collision
partners.
U > 0, superelastic collisions (sometimes called collisions
of the second kind). At least one of the collision partners had
internal energy before the collision which was transferred
into kinetic energy during the collision.
The kinetic energy after the collision is larger than before the
collision.
(4.17)
Figure 4.6 Conservation of total momentum at the collision of two particles
While the kinetic energy is only preserved in elastic
collisions the total momentum is preserved for all kinds
of collisions (Fig. 4.6).
Inelastic, super-elastic and reactive collisions can only
occur, if at least one of the collision partners has an
internal structure. This means that it must consist of
at least two particles, which are bound together. Examples are atoms (consisting of nuclei and electrons)
or nuclei (consisting of protons and neutrons). Part
of the kinetic energy of the collision partners then
can be transferred into the increase of the internal energy (potential or kinetic energy of the constituents).
For collision partners consisting of many particles (for
example solids) the increase of kinetic energy of the
constituents can be defined as an increase of the temperature (see Sect. 7.3) which is then called “thermal
energy” (see Sect. 10.1).
4.2 Collisions Between Two Particles
Elastic Collisions in the Lab-System
The description of collision processes can be essentially simplified when the appropriate coordinate system is chosen. For
many situations one of the collision partners, for instance m2 , is
at rest before the collision. We choose its position as the origin
of our coordinate system, which is fixed relative to the laboratory system. In this system is therefore p2 D 0 (Fig. 4.7).
We assume that the masses do not change during the collision
(m1 D m01 , m2 D m02 ). With U D 0 for elastic collisions we
obtain from (4.16) and (4.17)
p1 D p01 C p02 D p0 ;
(4.16a)
p21
p02
p02
D 1 C 2 :
2m1
2m1
2m2
(4.17a)
We choose the direction of the initial momentum p1 as the xdirection (Fig. 4.8) ) p1 D fp1 ; 0; 0g. The angular momentum
L D r p points into the z-direction. Because the angular
momentum is constant the motion of the collision partners is
restricted to the x-y-plane. The endpoint of the vector p02 is the
point P.x; y/. From Fig. 4.8 we derive the relations:
.p1
Figure 4.8 Illustration of (4.18)
impinging particle is reached, when p01 is the tangent to the circle. For m1 > m2 ! p1 D m1 v1 > 2v1 , which means that
jp1 j > 2r. The magnitude of the momentum of the impinging
particle is larger than the diameter of the circle. From Fig. 4.9
we can then conclude the relation
sin 1max D
.m1
m2
v1
D
:
D
/v1
m1
m1
(4.19)
Examples
1. m1 D 1:1m2 ) D 0:52m2 ) sin 1max D 0:91
) 1max D 65ı :
x2 C y2 D p02
2;
x/2 C y2 D p02
1 :
2. m1 D 2m2 ) D 0:67m2 ) sin 1max D 0:5
) 1max D 30ı :
Inserting into (4.17a) yields
3. m1 D 100m2 ) D 0:99m2
) 1max D 0:6ı :
p21
.p1 x/2 C y2
x2 C y2
D
C
:
2m1
2m1
2m2
J
Rearranging gives with the reduced mass D m1 m2 =.m1 Cm2 /
the equation
.x
v1 /2 C y2 D .v1 /2
(4.18)
of a circle in the x-y-plane with the radius r D v1 and the
centre M D fv1 ; 0g. This implies that the endpoints of all
possible vectors p02 which fulfil energy-and momentum conservation have to lie on the circle around M, if they start from the
origin f0; 0g (Fig. 4.9).
Special Case: Central Collisions
If p02 has the same direction as p1 the deflection angle becomes
2 D 0 (central or collinear collision). All vectors p1 , p01 and p2
are collinear and coincide in Fig. 4.9 with the x-axis. We obtain
The angles 1 and 2 are the deflection angles of the two collisions partners. The maximum deflectionangle 1max of the
Figure 4.7 Collision of a particle with mass m1 and momentum p1 with a mass
m2 at rest, drawn in the Lab-system
Figure 4.9 Momentum diagram of elastic collisions for m1 > m2 . All possible
endpoints of the vector p02 are located on the circle with radius v1 around M
Chapter 4
4.2.2
109
110
4 Systems of Point Masses; Collisions
from Fig. 4.9:
Special cases of non-collinear collisions
p1 D 2v1 C p01
)m1 v10 D m1 v1
)v10 D
2
We will now discuss the general case of non-collinear collisions
and illustrate it by some important special cases of the mass ratio
m1 =m2 .
m1 m2
v1
m1 C m2
m1 D m2 D m
m1 m2
v1 I
m1 C m2
(4.20)
p02 D 2v1
)v20 D 2
2m1
v1 D
v1 :
m2
m1 C m2
The momentum of the pushed particle gets its maximum value
p02 D 2v1 for collinear collisions.
Also the kinetic energy, transferred from m1 to m2 during a
collinear elastic collision reaches its maximum value
Ekin D
p02
2m21 m2
2
max
Ekin
D
v2
2m2
.m1 C m2 /2 1
max
Ekin
D4
D 12 m:
Equation 4.18 gives for the radius of the circle in Fig. 4.9
r D 21 mv1 , which implies that the momentum p1 D mv1 of
the incident particle is the diameter of the circle (Fig. 4.11). For
non-collinear collisions the momenta p01 and p02 after the collision are perpendicular to each other according to the theorem of
Thales. For the deflection angles it follows 1 C 2 D =2.
The paths of the two particles are perpendicular to each
other after the non-collinear collision, i. e. p01 ? p02 .
Example
(4.21)
m1 m2
42
E1 D
E1 ;
2
M
m1 m2
which equals the fraction 42 =.m1 m2 / of the initial energy E1 of
the impinging mass m1 . In Fig. 4.10 this maximum transferred
fraction is shown as a function of the mass ratio m1 =m2 .
v20
v10
)
Chapter 4
D v1 . The two masses
D 0 and
For m1 D m2 it is
exchange their momentum during the collision, i. e. after the collision m1 is at rest and m2 moves with the momentum p02 D p1 .
For equal masses m1 D m2 the energy of the incident particle is completely transferred to the resting mass m2 during
a collinear collision.
For the deceleration of neutrons in nuclear reactors a
material with many hydrogen atoms is the best choice.
Because the protons have nearly the same mass as neutrons.
J
m1 m2
)
m1
The radius of the circle in Fig. 4.9 becomes for the limiting case
m1 =m2 ! 0 equal to the momentum p1 D m1 v1 of the incident particle (Fig. 4.12a). The magnitude of p1 does not change
during the collision .jp1j D jp01 j/ but all directions of p01 are
possible. The scattering angle 1 can take all values in the range
1 C.
The maximum momentum transfer onto m2 is
jp02 jmax D 2r D 2p1 :
The maximum transferred energy is
∆Emax
kin
E1
=
4m1m2
(m1+m2)2
=
4
(m1/m2)+2+1/(m1/m2)
max
Ekin
D
.2p1/2
4p21 m1
m1
D
D 4 E1 :
2m2
2m1 m2
m2
1,0
0,5
0
01
5
10
15
20
(m1/m2)
Figure 4.10 Maximum energy transfer E D E E10 for a collinear elastic
collison of a particle with mass m1 onto a mass m2 at rest for different ratios
m1 =m2
Figure 4.11 Elastic collision between particles of equal mass
(4.22)
4.2 Collisions Between Two Particles
111
cident mass m1 is according to (4.19)
sin ' D
m2
:
m1
Example
In collisions of ˛-particles (helium nuclei) with electrons
at most the fraction E1 D 0:00054E1 of the initial
energy E1 can be transferred to the electron. The maximum deflection angle of the ˛-particles is ' sin ' D
1:36 10 4 rad D 0:480 . When ˛-particles pass through
matter the electron shell of the atoms contributes to the
deflection only a tiny part. Most of the deflection is
caused by the atomic nuclei (see Rutherford scattering in
Vol. 3).
J
In collisions of a small mass m1 against a large mass m2
the maximal transferred fraction of the initial kinetic energy is 4.m1 =m2 /.
Examples
Elastic Collisions in the Centre-of Mass
system
When none of the collision partners is resting, the description
of the collision process is often simpler in the CM-system than
in the lab-system. Since, however, the observation of the process always occurs in the lab-system the measured results must
be transformed into the CM-system in order to compare them
with the predictions calculated in the CM-system. The relations
between position vectors and velocities in the two systems is
illustrated in Fig. 4.13b and the results are compiled in Tab. 4.1.
1. Impact of a particle onto a solid wall.
max
m2 D 1 ) Ekin
D0
but: p02 D 2p1 :
During the elastic collision of a particle with a solid
wall the particle is elastically reflected and p01 D p1 .
Therefore twice the initial momentum is transferred to
the wall but no energy!
2. Collision of an electron with a proton at rest.
m1 D m2 =1836. The maximal transferred energy
max
D
occurs in central collisions and is then Ekin
J
4.m1 =m2 /E1 D 0:00218 E1 .
m1 m2
)
RS
m2 .
In this case the radius of the circle in Fig. 4.9 is r D m2 v1
(Fig. 4.12b). For central collisions is
m2 v20 D 2r D 2m2 v1 ) v20 D 2v1 ;
and the transferred energy is
Ekin D
m2
m2 0 2
v D 4 E1 :
2 2
m1
VS
(4.23)
For non-collinear collisions the energy transferred to m2 is
smaller. The maximum deflection angle ' D 1max of the in-
Figure 4.13 Graphical representation of the relations between a Lab- and CMcoordinates and b Lab- and CM-velocities
Chapter 4
4.2.3
Figure 4.12 Elastic collision for m1 m2 (a) and m1 m2 (b)
112
4 Systems of Point Masses; Collisions
Table 4.1 Compilation of quantities relevant for collisions in the lab-system
and the CM-system
M D m1 C m2 D total mass
D
m1 m2
m1 Cm2
RS D
VS D
D reduced mass
1
.m1 r1
M
1
.m1 v1
M
C m2 r2 / D position vector of CM
C m2 v2 / D velocity of CM
r12 D r1 r2 D relative distance
v12 D v1 v2 D relative velocity
RS D position vector of i-th particle in the CM-system
riS D ri
viS D vi
V S D velocity of i-th particle in the CM-system
Figure 4.14 In the CM-system an elastic collision is represented by a turn of
the momentum vectors without changing their length
piS D mi viS D momentum of i-th particle in the CM-system
P
piS D 0
Note: in order to distinguish the deflection angles in the labsystem written as capital letters, from those in the CM we will
label all deflection angles in the CM-system by lower case letters #.
i D deflection angle of i-th particle in the lab-system
#i D deflection angle of i-th particle in the CM-system
Note: We will denote the center of mass with the index S.
Since the total momentum in the CM-system is always zero, we
can write for two particles 1 and 2
p1S D p2S
and
p01S D p02S :
Chapter 4
The sum of the momenta of the collision partners before
the collision and after the collision is in the CM-system
always zero.
From the energy conservation (4.17) it therefore follows:
1
2
1
1
C
m1
m2
p02
1S D
1
2
1
1
C
m1
m2
p21S C U ;
Example
Deceleration of neutrons (mass m1 , velocity v1 ) in elastic
collisions by atomic nuclei (mass m2 ) at rest. The CMvelocity is
V S D V 0S D
v1
m1 v1
D
m1 C m2
1CA
A D m2 =m1 :
Av1
I
1CA
v1
D 0 VS D
I
1CA
D v01 C v2S D v01 V S :
v1S D v1
v2S
VS D
energy conservation in the S-system
p02
p2
1S
D 1S C U :
2
2
(4.24)
S
For elastic collisions .U D 0/ in the CM-system is p21S D p02
1S
and p22S D p02
2S . This means:
In the CM-system each collision partner retains in elastic
collisions its kinetic energy.
In the CM-system the result of an elastic collision is merely a
turn of the momentum vectors which are always pointing into
the opposite direction (Fig. 4.14).
J
The velocity of the two particles in the CM-system is according
to Fig. 4.15
v01S
which can be written when using the reduced mass
with
Figure 4.15 Determination of energy transfer at elastic collisions
4.2 Collisions Between Two Particles
02
0
v102 D v1S
C VS2 C 2v1S
vS cos #1 ;
where #1 is the angle between v01S and V S . Because V S k v1S
0
and v1S , i. e. the
it follows that #1 is also the angle between v1S
deflection angle of m1 in the CM-system. Inserting the relations
above for VS we obtain
02
v1S
D v12
2
A C 2A cos #1 C 1
:
.1 C A/2
The ratio of the kinetic energies of the neutron after and before
the collision is then
0
Ekin
A2 C 2A cos #1 C 1
v 02
:
D 12 D
Ekin
.1 C A/2
v1
For central collisions is #1 D and the ratio becomes
0 central
Ekin
A 1 2
D
:
Ekin
AC1
0
For the transferred energy E D Ekin
From (4.26) it follows that only in cases where the two collision
partners have equal but opposite momenta (m1 v1 D m2 v2 )
V S D 0) the total kinetic energy can be converted into internal
energy. The two particles then stick together and are at rest,
their total momentum is zero before and after the collision. For
all other collisions jUj < jUmax j. Therefore the general rule is:
2
For all inelastic collisions not more than E D 21 v12
of the initial kinetic energy can be converted into internal
energy. At least the proportion
1
.m1
2
Ekin we then obtain
4A
4m1 m2
E
D
D
:
Ekin
.A C 1/2
.m1 C m2 /2
For m1 D m2 the transferred energy E=E takes on its maximum value E=E D 1. This means that the neutron can transfer
its total kinetic energy if it suffers a central collision with a proton.
4.2.4
In a completely inelastic collision, where the two particles
stick together after the collision, just the kinetic energy
of the two particles in the CM-system is converted into
internal energy of one or both collision partners.
Examples
1. A glider with mass m1 on an air-track hits a second
glider with mass m2 at rest .v2 D 0/. The two colliding ends are covered with plasticine, which causes
the two gliders to stick together after the collision and
they move with the CM-velocity
VS D
For inelastic collisions part of the initial kinetic energy is transferred into internal energy U of the collision partners. Such
collisions are only possible, if at least one of the partners has
a variable internal sub-structure, which means that it has to be
composed of two or more particles. For point masses there are
no inelastic collisions!
For inelastic collisions momentum conservation remains valid
(4.16) and also energy conservation with U < 0 (4.17). In the
limiting case of maximal inelastic collisions the two collision
partners stick together after the collision and move with the CMvelocity.
m1 v1 C m2 v2
VS D
:
(4.25)
m1 C m2
From (4.17) and (4.25) we obtain for the maximum fraction of
the kinetic energy, which is transferred into internal energy
1
1
.m1 C m2 / VS2
m1 v12 C m2 v22
2
2
1 2
1 m1 m2
.v1 v2 /2 D
D
v ;
2 m1 C m2
2 12
(4.26a)
which is identical to the kinetic energy of the two particles in
the CM-system (see (4.11b)).
(4.26b)
of the CM-motion remains as kinetic energy of the collision partners.
Inelastic Collisions
UD
C m2 /VS2 D 21 MVS2
m1
v1 :
m1 C m2
The kinetic energy after the collision is
0
Ekin
D
m21
m1 C m2 2
VS D
v2 ;
2
2.m1 C m2 / 1
and the energy converted into the plasticine energy is
0
U D Ekin
Ekin D
For m1 D m2 this gives
UD
m2
Ekin :
m1 C m2
1
E
2 kin
:
2. A neutron n with velocity v1 impinges on a proton p
at rest. This produces a deuteron d D np.
nCp !d:
Because of m1 D m2 the deuteron moves with the
CM-velocity VS D 12 v1 and has therefore half of the
0
D Ekin of the incident neuinitial kinetic energy Ekin
tron. The other half is converted into internal energy
of the deuteron, which is excited into a higher energy
J
state, that can decay by emission of -radiation.
Chapter 4
0
VS cos #1
Squaring gives with v01S V S D v1S
113
114
4 Systems of Point Masses; Collisions
Summarizing the results: In inelastic collisions of particles with
equal masses where one collision partner is at rest at most half
of the kinetic energy of the incident particle can be converted
into internal energy
jUj jUmax j D
1m 2
v :
22 1
(4.27a)
The amount Umax U remains as kinetic energy of the collision partners in addition to the kinetic energy 12 MVS2 of the
CM-motion.
electron can be converted into excitation energy of the
Hg-atoms.
2. A heavy particle with mass m1 D 100m2 collides with
a particle of mass m2 . Now D 0:99m2 and U D
.0:99=100/m2v12 =2. This implies that only about 1%
of the kinetic energy is converted into internal energy
U.
J
4.2.5
Special Cases
If a particle with mass m1 suffers a totally inelastic collision
with a wall .m2 m1 ! m1 / it remains adsorbed
at the wall and transfers its kinetic energy completely to the
0
D 0/.
wall, which heats up. .U D Ekin ; Ekin
If two equal masses collide head-on with p1 D p2 the total
momentum after the collision must be zero. With v12 D v22 D
v 2 the increase of internal energy is
UD
1
.m1 C m2 /v 2 ;
2
as in the first case the total kinetic energy is converted into
internal energy. These two special cases are illustrated in
Fig. 4.16 and compared with the corresponding elastic collisions.
Examples
Chapter 4
1. Collisional excitation of mercury atoms by electron impact (Franck-Hertz-Experiment). Because of
mHg me the reduced mass m me . From (4.26) we
can conclude that nearly the total kinetic energy of the
Newton-Diagrams
The measurements of deflection angles at collisions between
atoms or molecules is performed in the laboratory-system. The
determination of the interaction potential derived from these
deflection angles is, however, much easier in the CM-system.
The relations between the relevant parameters in the two systems (velocities, deflection angles, energy transfer) for arbitrary
elastic or inelastic collisions can be visualized with the help of
Newton diagrams, which connects the velocities in the lab- system with those in the CM-system (Fig. 4.17). The parameters
used in the following are listed in Tab. 4.1.
With the relations
r1 D RS C .m2 =M/ r12 and
r2 D RS .m1 =M/ r12 ;
v1 D V S C .m2 =M/ v12 and
v2 D V S .m1 =M/ v12 ;
(4.29)
which can be derived from Fig. 4.13, the kinetic energy can be
separated into the two parts
Ekin D 12 m1 v12 C 21 m2 v22 D 12 MVS2 C
ƒ‚
…
„
„ƒ‚…
Ekin in lab frame
Figure 4.16 Comparison of elastic and completely inelastic collisions: a particle against a wall, b collision between two equal masses
(4.28)
Ekin of
CM-motion
2
1
v12
2
„ƒ‚…
:
Ekin of relative motion
in the CM-system
(4.30)
Figure 4.17 Newton diagram of elastic collision between two particles
4.3 What Do We Learn from the Investigation of Collisions?
115
Figure 4.19 Illustration of impact parameter
Since for elastic collisions the kinetic energy of each partner in
the CM-system is preserved, the vector of the relative velocity
v12 retains its magnitude but turns around the centre of mass
S where the end of the vector describes a circle with the radii
v1S D .m2 =M/v12 and v2S D .m1 =M/v12 . The deflection angles
#1 in the CM-system can be determined graphically from the
deflection angles 1 in the lab-system.
In particular the maximum deflection angle 1max can be determined readily. It appears when v10 is the tangent to the Newton
circle.
For inelastic collisions (Fig. 4.18) part of the kinetic energy
1
0
2
is converted into excitation energy, which means that v12
v12
2
becomes smaller. However, still the centre-of-mass S divides
the connecting line between the endpoints of the vectors v1 and
0
v2 in the ratio m1 =m2 of the two masses. The endpoints of v12
are now located on a circle with smaller radius (dashed circles
in Fig. 4.18).
For both elastic and inelastic collisions the range of possible
deflection angles and the maximum deflection angles can be determined from the Newton diagrams. Therefore such diagrams
are very useful for the planning of experiments, because they
tell us, in which deflection ranges one must look for scattered
particles for given initial conditions [4.2].
4.3
What Do We Learn from the
Investigation of Collisions?
Since F.r/ D r Ep the force F is a measure for the potential
energy Ep .r/ of the interaction between A and B which depends
on the distance r between A and B. The deflection of A therefore depends on the impact parameter b in Fig. 4.19, which is a
measure of the closest approach between A and B. It is defined
in the following way:
For large distances between A and B the force F is negligible and
the incident particle A will follow a straight line. If there would
be no interaction between A and B the incident particle A would
follow this straight line and pass B at the closest distance b. This
line is parallel to the straight line through B where B is resting in
the origin of the coordinate system. To each impact parameter
b belongs a certain deflection angle in the lab-system resp. #
in the CM-system., which depends on the interaction potential
V.r/ between A and B.
4.3.1
Scattering in a Spherical Symmetric
Potential
In Sect. 4.1.3 it was shown, that the relative motion of two particles around the CM caused by the mutual interaction force F.r/
can be reduced to the motion of a single particle with the reduced mass in the spherical symmetric potential with its origin
at the position of one of the two particles (usually the one with
the larger mass). If the force F.r/ is known, the deflection angle
# in the CM-system can be determined from Eq. 4.3 and the re-
The deflection of a particle A during the collision with another
particle B is due to the momentum transfer
p D
ZC1
F dt ;
(4.31)
1
which is caused by the force F acting between A and B while
passing by each other. The momentum change p experienced
Figure 4.20 Relation between momentum change p and deflection angle #
in the CM-system
Chapter 4
Figure 4.18 Newton-diagram of inelastic collisions between two particles
by A is, of course, compensated by the change p experienced
by B, because the total momentum has to be conserved.
116
4 Systems of Point Masses; Collisions
lation sin.#=2/ D 21 p
(Fig. 4.20). In Sect. 4.2.5 it was shown
pA
how the angle measured in the lab-system can be transformed
into the angle # in the CM-system.
The deflection of particles in a potential is called potential
scattering. We will illustrate this potential scattering and its
treatment by some examples.
Examples
1. Collision of two hard spheres with radii r1 and r2
(Fig. 4.21).
If the impact parameter b is larger the sum r1 C r2
no collision takes place. The hard sphere A moves on
a straight line and passes B without deflection. For
b r1 C r2 the colliding partner A is reflected at the
surface of B (Fig. 4.21a). In order to determine the
deflection angle # of A we decompose the momentum
p1 of A into a component pr parallel to the connecting
line M1 M2 at the touch of the two spheres and a component p t in the tangential direction perpendicular to
pr (Fig. 4.21b). We assume the surface of the spheres
as frictionless. Then no rotation of the spheres can be
excited and the component p t does not change during
the collision. For the component pr we can conclude
from (4.20) for central collisions
p0r D
m1 m2
p :
m1 C m2 r
(4.32a)
From Fig. 4.21b one can deduce from b D r1 C r2 for
D 2˛ the dependence of the deflection angle on the
impact parameter b
.b/ D 2 arccos
b
:
r1 C r2
(4.32b)
For b D 0 is D i. e. A is reflected back., for
b > r1 C r2 is D 0. The function .b/ is called
deflection function. Its curve depends on the interaction potential .V.r//. For the collision of two hard
spheres the potential is a step function (Fig. 4.22a) and
the deflection function is the monotonic curved shown
in Fig. 4.22b for m2 m1 .
For the general case of arbitrary ratios m1 =m2 we obtain from Fig. 4.21c:
D˛Cˇ
with pr =p t D tan ˛ it follows
tan ˇ D p0r =pr D .m1
m2 /=.m1 C m2 / :
While (4.32b) is only strictly valid for m2 D 1, the
deflection function
#.b/ D arccos
b
r1 C r2
(4.32c)
in the CM-system is correct for arbitrary ratios m1 =m2 .
Chapter 4
Figure 4.22 a Potential V .r / for hard spheres; b deflection function
for hard sphere collisions
Figure 4.21 Determination of the deflection function for collisions
between two hard spheres. a Definition of impact parameter, b decomposition of impact momentum, c momentum vector addition
For m2 m1 this gives p0r D pr . In this case is in
Fig. 4.21c ˇ D ˛ and the deflection angle becomes
D 2˛.
2. Scattering of a particle in a potential V.r/ / 1=r.
This important case applies for instance for the
Coulomb-scattering of charged particles (electrons
or ˛-particles) on atomic nuclei (see Vol. 3) or the
Kepler-orbit of comets in the gravitational field of the
sun.
For a potential V.r/ with the potential energy Ep D
a=r the force between the interacting particles A and
B with masses m1 and m2 is
a
(4.33)
F D grad Ep D 2 rO :
r
For a > 0 a repulsion between the particles occur, for
a < 0 an attraction. The angular momentum in the
CM-system is according to (4.15)
LDrv
with
D
m1 m2
;
m1 C m2
4.3 What Do We Learn from the Investigation of Collisions?
where r is the distance between A and B and v is the
relative velocity.
Since L is in the central potential temporally constant
the orbit of the particle remains in the plane ?L, which
we choose as the x-y-plane (Fig. 4.23). The particle A
is incident parallel to the x-axis with an impact parameter b and the initial velocity v0 . It is convenient to use
polar coordinates for the description of its trajectory.
The magnitude of L is then
L D jr vj D r2
d'
D v0 b ;
dt
(4.34)
where the last term describes the angular momentum
of A for large distances r ! 1 referred to the particle B which sits at the origin r D 0. It should be
emphasized that we use the CM-system for our description. In the lab-system B does not stay at r D 0
but moves around the common centre of mass. We
compose the force F.r/ of the components Fx and Fy .
For the deflection of A only the component Fy is responsible. From Fig. 4.23 we see that
dvy
a sin '
D
:
r2
dt
(4.35)
With the equation .1 C cos #/= sin # D cot.#=2/ the
relation between deflection angle # and impact parameter b for scattering in the potential with energy
Ep D a=r becomes
v02
#
2Ekin
cot
bD
b:
(4.37a)
D
2
a
a
The ratio a=b gives the potential energy of the interaction between the particles A and B at a distance r D b.
Inserting this into (4.37a) gives the result
#
2Ekin
:
(4.37b)
D
cot
2
Ep .b/
The deflection angle # in the cm-system is determined
by the ratio of twice the kinetic energy and the potential energy at a distance r D b between the interacting
particles. The deflection function #.b/ is shown in
Fig. 4.24. For b D 0 is cot.#=2/ D 1 ! # D .
The particle A is scattered back into the x-direction.
The turning point which is the closest approach r0 can
be obtained from Ekin D Ep ! v02 =2 D a=r0 . This
gives r0 D 2a=.v02/.
Chapter 4
Fy D
A (r)
Figure 4.23 Scattering of a particle in the potential V .r / with r D
distance AB
From (4.34) and (4.35) we obtain
dvy
a sin ' d'
D
:
dt
v0 b dt
(4.36)
The total deflection of A during its path through the
potential V.r/ is obtained by integration over the
whole range from r D 1 to r D 1.
For A. 1/ we have vy D 0 and ' D 0, for A.C1/
is vy D v0 sin # with # D 'max . For the elastic
potential scattering the magnitude v0 of the velocity
remains constant. Therefore the integration of (4.36)
yields
Z #
Z
a
sin ' d'
dvy D
v0 b
0
! v0 sin # D
a
.1 C cos #/ :
v0 b
117
Figure 4.24 Deflection function #.b / (a) and trajectories of a particle
in a potential V .r / / 1=r for different impact parameters but constant
initial energy. Each deflection angle corresponds to a different ratio f D
2Ekin =Epot .b /. ➀: # D ; ➁: # D 43 ; ➂: # D 105ı , f D 0:76; ➃:
# D 60ı , f D 1:7; ➄: # D 30ı , f D 3:7
For the gravitational potential is a D Gm1 m2 (see
(2.52) and we get from (4.37b) with M D m1 C m2 the
result
v02
#
cot
b:
(4.37c)
D
2
GM
The deflection angle depends only on the masses, the
initial velocity v0 and the impact parameter b. For a
118
4 Systems of Point Masses; Collisions
comet is m1 m2 D Mˇ . The total mass M is then
with a very good approximation M D Mˇ . The mass
of the comet does not affect the deflection angle.
According to (2.60) the trajectories of the particle m1
for E D Ekin C Ep > 0 are hyperbolas. In Fig. 4.24b
some of these hyperbolas are shown for a repulsive potential .a > 0/ and different impact parameters b. For
the interaction between two positively charged particles with charges q1 and q2 is a D .1=4"0/ q1 q2
J
(see Vol. 2).
4.3.2
Reactive Collisions
Reactive collisions provide the basis of all chemical reactions.
A simple example is the reaction
A C BC ! ABC ! AB C C ;
(4.38)
where an atom A with the velocity vA collides with a molecule
BC (velocity vBC ), forms a complex ABC, which can decay into
the fragments AB C C (Fig. 4.25).
Chapter 4
Momentum conservation is also valid for reactive collisions.
The momentum of the right side in equation (4.38) must be
therefore the same as on the left side. The kinetic energy is,
however, in general not conserved because part of this energy
may be converted into internal energy .U < 0/. In cases where
the reactants on the left side are already excited, this internal
energy may be also transferred to kinetic energy (U > 0, superelastic collisions). The measurement of velocities and deflection
angles after the collision gives information about the energy balance of the reaction and the interaction potential between the
reactants, if the initial conditions are known. The potential is
in general no longer spherical symmetric but depends on the
spatial orientation of the molecule BC against the momentum
direction of A. The reaction probability can differ considerably
for collinear collisions, (Fig. 4.26a) from that for non-collinear
collisions where the internuclear axis of BC is inclined by the
angle ˛ against the momentum direction of A (Fig. 4.26b).
Figure 4.25 Schematic representation of a reactive collision, where a collision
complex is formed that decays again
Figure 4.26 Collinear (a) and noncollinear (b) collision, where angular momentum of the relative motion is transferred to rotational angular momentum of
the molecule BC
Often the reactants A and BC fly in two perpendicular collimated molecular beams. In the intersection volume of the two
beams the reactants collide. For this arrangement the directions
of the reactants are known and their velocities can be selected
by velocity selectors which interrupt the beams (see Sect. 7.4.1).
The initial conditions are then well known (apart from the often
unknown internal energies).
Note, that the masses are generally not constant for reactive
collisions, because the reduced mass .A C BC/ before the collision differs in general from .AB C C/ after the collision.
If the kinetic energy E2 of the reaction products is smaller than
the kinetic energy E1 of the reactants, the reaction is called endotherm. One has to put energy into the system in order to make
the reaction possible. If energy is released in the reaction it is
called exotherm. In this case the kinetic energy of the reaction
products is larger than that of the reactants. Measurements of the
velocities of reactants and reaction products can decide which
type applies to the investigated reaction.
The energy balance is illustrated by the potential diagram of
Fig. 4.27. Often the reactants have to overcome a potential barriers in order to start the reaction. In this case a minimum initial
energy is necessary even for exothermic reactions.
The heights of the potential barrier and with it the reaction
probability depends on the internal energy (vibrational- rota-
Figure 4.27 Diagram of potential energy as function of the reaction coordinate
for a endothermic and b exothermic reactive collisions
4.4
Collisions at Relativistic Energies
119
tional or even electronic energy) of the reaction partners. For
the measurement of these internal energy several spectroscopic
techniques have been developed which allow to determine the
excitation state of the partners involved in the reaction.
An ideal experiments should allow to measure all relevant parameters of a collision process such as the internal energies, the
deflection angles and the velocities of all particles. Such modern
techniques are discussed in Vol. 3.
Collisions at Relativistic Energies
Up to now we have used the Newtonian laws (energy- and momentum conservation) for the description of collision processes
and we have assumed that the masses of the reaction partners are
constant (besides in reactive collisions). This is justified as long
as the velocities of the collision partners are small compared
with the velocity c of light (see Chap. 3).
For the investigation of interactions between elementary particles and atomic nuclei, higher energies of the collision partners
are required. Such energies, where the velocity of particles
comes close to the velocity of light can be realized in particle
accelerators and storage rings (see Vol. 4). We will now discuss,
how the rules governing collisions at relativistic energies (the
domain of high energy physics) must be formulated.
4.4.1
*
*
Figure 4.28 Grazing elastic collision between A and B at relativistic velocities.
a In the system S has A a large and B a small velocity since Vy Vx . b In the
sytem S , that moves relativ to S the situation is reversed
Relativistic Mass Increase
We regard two particles A and B which have equal masses m1 D
m2 D m, if they are at rest. We assume that A and B move in a
system S with velocities
v1 D fvx1 ; vy1 g and
which implies that the roles of A and B are just interchanged.
According to Eq. 3.28 for the transformation of velocities when
changing from system S to S the observer O in S measures
the velocity component
v2 D f0; vy2 g
vy D
against each other where vy1 D vy2 (Fig. 4.28a).
The particle B should suffer an elastic striking collision with A
such that during the collision the x-component vx1 of A remains
constant but the y-component is reversed. The velocity of A after
0
the collision
qis then v1 D fvx1 ; vy1 g. The magnitude of its veloc-
2
2
is therefore also preserved. Because the
C vy1
ity jv1 j D vx1
momentum must be constant the velocity of B after the collision
must be
v02 D f0; vy2 g :
We assume that vy vx . For the magnitudes of the velocities
this implies
2
2 .1=2/
v1 D .vx1
C vy1
/
vx1
and
v2 v1 :
Now we describe this collision between A and B in a system S
which moves against S with the velocity v D vx1 into the xdirection (Fig. 4.28b). In this system we get for the velocities of
A and B:
vx1
D0
but vx2
D
vx1 ;
1
vy =
:
vx v=c2
(4.39)
0
Since in the system S for the particle A holds: vx1 D vx1
¤ 0,
0
for B, however, vx2 D vx2 D 0 the observer O measures for the
two particles different y-components of the velocities
D
vy1
vy1 =
vy1 =
D
D vy1 ;
1 vx1 v=c2
1 v 2 =c2
since v vx1 ;
vy2 =
vy2
D
D vy2 = ;
1 vx2 v=c2
(4.40a)
(4.40b)
while for the observer O in S the velocity component of A is vy1
and for B it is vy2 .
In both inertial systems S and S the conservation of total momentum holds, since the physical laws are independent of the
chosen inertial system (see Sect. 3.2). This yields for the ycomponent of the total momentum
mA vy1
mB vy2 D mA vy1
mB vy2
D0:
(4.41)
Chapter 4
4.4
120
4 Systems of Point Masses; Collisions
For mA D mA and mB D mB this condition cannot be fulfilled,
i. e. the conservation of momentum would fail, because accord
ing to (4.40) vy1 =vy1
. We are therefore
is different from vy2 =vy2
forced (if we will not give up the well proved conservation of
momentum) to assume that the mass of a particle is changing
with its velocity. For the limiting case vx1 vy1 0 we can
write:
vA vx1 D v I vA 0 ;
vB 0 I vB vx1 v :
We therefore get with m.v D 0/ D m0 for (4.41)
m.v/vy1 C m0 vy2 D 0 ;
m0 vy1
C m.v/vy2
D0;
(4.42a)
(4.42b)
with (4.40) this gives
2
vy2 vy1
.m.v//
D
D
vy2
vy1
m20
) m.v/ D m0 D p
1
m0
v 2 =c2
4.4.2
Force and Relativistic Momentum
The work, which has to be spent for the acceleration of a mass
is used with increasing velocity more and more for the increase
of the mass and less for the increase of the velocity.
The Newton-equation (2.18) between force and momentum with
the inclusion of the relativistic mass increase (4.43) is
!
vm0
d
d
dp
p
FD
D .mv/ D
dt
dt
dt
1 v 2 =c2
!
(4.44a)
m0
d
D
v C ma :
p
dt 1 v 2 =c2
This gives with d=dt D .dv=dt/ .d=dv/
2
FD
:
(4.43)
The mass m.v/ of a moving particle increases with its velocity
v. the mass m0 D m.v D 0/ is called its rest-mass. This mass
increase is noticeable only for large velocities [4.3].
Chapter 4
Examples
1. For v D 0:01c ) m D 1:00005m0. The relative mass
increase m=m D .m m0 /=m0 5 10 5 .
2. For v D 0:9c ) m D 2:29m0 .
3. For v D 0:99c ) m D 7m0 .
J
In Fig. 4.29 the increase of the mass m.v/ is plotted against the
normalized velocity v=c. This illustrates also, that the maximum velocity of a particle with m0 ¤ 0 is always smaller than
the velocity of light because for v ! c it follows from (4.43)
m.v/ ! 1.
m0 v=c2 a
v C ma
v 2 =c2 /3=2
2
v
v2
D 3 m0 a 2 eO v C 1
eO a ;
c
c2
.1
(4.44b)
where eO v and eO a are unit vectors in the direction of v and a.
These equations show that for large velocities v the force F is
no longer parallel to the acceleration a but has a component in
the direction of v. For v c the first term in (4.44b) can be
neglected and we obtain the classical Newton equation F D ma.
In order to keep the Newton equation F D dp=dt the relativistic
momentum
m0 v
p.v/ D m.v/ v D
(4.45a)
has to be used which has the magnitude
p D ˇ m0 c
.ˇ D v=c/ :
(4.45b)
For the relativistic momentum the conservation law (4.41) is fulfilled for all velocities.
We will now discuss, how the components of the force are transformed for a transition from a system S where a particle has a
velocity v and a mass m D m0 into a system S which moves
with the velocity U D Cv against S. In S is therefore v D 0
and m D m0 . We choose the axes of the coordinate system such
that v D fvx ; 0; 0g. It follows then from (4.44) with eO v D eO a
Fx D
dpx
D
dt
3
m0 ax :
(4.45c)
In S the x-component of the acceleration a becomes
ax D
3
ax
as can be seen from (3.26) and (3.28). Therefore the component
of the force in the system S becomes
Figure 4.29 Dependence of a mass m on the ratio v=c
Fx D m0 ax D
3
m0 ax Fx :
(4.45d)
4.4
We obtain the remarkable result that the x-components in the
two systems which move against each other in the x-direction,
are equal!
This no longer true for the components perpendicular to the relative motion of the two systems, because we get for vy vx the
result
2
Fy D m0 ay D
m0 ay D
Fy ;
Fx
D
Fy
2 ax
:
ay
(4.45f)
This shows again that for ¤ 1 the force F is no longer parallel
to the acceleration a as in the nonrelativistic case.
4.4.3
121
of the box. For v c the light pulse reaches the right wall of
the box at a time t2 D L=c and is absorbed by the wall. This
transfers the momentum p D C.E=c/Oe to the right wall. The
total momentum transferred to the box is therefore zero and the
box is again at rest. However, during the time t2 the box has
moved to the right by a distance
(4.45e)
and therefore obtain for the ratio
Collisions at Relativistic Energies
x D v t2 D
E
L
:
Mc2
(4.46)
Since the box plus light pulse represent a closed system where
no external forces act onto the system, the centre of mass cannot have moved. The CM of the box certainly has moved by x
into the x-direction. Therefore the light pulse must have transported mass into the Cx-direction in order to leave the CM of
the total system (box + light pulse) at rest. If we attribute a mass
m to the light pulse with energy E the CM of the total system
stays at rest, if
mL
The Relativistic Energy
M x D 0 :
(4.47)
This gives with (4.46) the result
In classical mechanics the kinetic energy of a particle
is different in diverse inertial systems which move against each
other, because the velocity v is different.
In order to obey energy conservation when changing from one
system to the other the total energy of a particle has to be defined
in such a way, that it is Lorentz-invariant. i. e. that it does not
change for transformations into different inertial systems (see
Sect. 3.3).
We will at first present an intuitively accessible description,
which is based on a “Gedanken-experiment” of Einstein. We regard in Fig. 4.30 a box with length L and mass M. We assume,
that at the time t1 D 0 a light pulse with energy E is emitted
from the left side of the box which travels with the velocity of
light c to the right. According to results of classical physics the
momentum of the light pulse is p D .E=c/Oe (see Vol. 2). Because of the conservation of momentum the left wall and with
it the total box suffers a recoil p into the left direction. This
results in a velocity
v D p=M D .E=Mc/Oe
(4.48a)
According to this consideration each mass m is correlated
to the energy E D mc2 . Mass and energy are proportional
to each other.
When we insert the rest mass m0 from (4.43) we obtain from
(4.48a) the energy of a mass m that moves with the velocity v
ED p
1
m0 c2
v 2 =c2
D m0 c2 C .m
m0 /c2 :
(4.48b)
This energy E can be composed of two parts:
The rest energy m0 c2 which a particle at rest must have, and the
kinetic energy
Ekin D .m.v/
m0 /c2 ;
(4.49a)
which is here described as the increase of its mass m.v/. If we
expand the square root in (4.48b) according to
p
1
1
v 2 =c2
D1C
1 v2
3 v4
C
C ;
2 c2
8 c4
the kinetic energy becomes
Ekin D
1
3 v4
m0 v 2 C m0 2 C :
2
8 c
(4.49b)
In the limiting case v c we can neglect the higher order terms
in (4.49b) and obtain the classical result
Ekin D 12 mv 2 :
Figure 4.30 Einstein’s “Gedanken-experiment” illustrating E D mc 2
This shows that the classical expression for the kinetic energy is
an approximation for v c. Since in daily life the condition
Chapter 4
Ekin D
m D E=c2 ) E D mc2 :
1
mv 2
2
122
4 Systems of Point Masses; Collisions
v c is always fulfilled, the relativistic expression is important
only for cases where v approaches c as in high energy physics
or astrophysics.
B has in S according to the relativistic addition of velocities
(3.28) the velocity
v2 D
Squaring (4.48b) and multiplying both sides with c2 gives
E2 D
m20 c6
D m20 c4 C m2 c2 v 2 :
c2 v 2
(4.50)
Inserting (4.45) for the relativistic momentum yields
E2 D m20 c4 C p2 c2 :
This gives the relativistic relation between total energy E
and momentum p
q
E D c m20 c2 C p2 :
(4.51)
For v c the square root can be expanded and gives the result
Ekin D E
m0 c2
p2
1
D m0 v 2
2m
2
with the classical momentum p D m0 v.
Chapter 4
Inelastic Collisions at relativistic
Energies
The relativistic energy and its conservation can be illustrated
by the example of a collinear completely inelastic collision
(Fig. 4.31). We regard two particles A and B with equal masses
m which fly against each other width velocities v1 D fv1 ; 0; 0g
and v2 D f v1 ; 0; 0g, measured in the system S. In a completely inelastic collision their total kinetic energy is converted
into internal energy of the collision partners (see Sect. 4.2.4).
After the collision they form a compound particle AB with the
velocity v D 0 (Fig. 4.31 upper part).
In a system S , which moves with the velocity v D v1 against
S the particle A has the velocity v1 D 0, the compound particle
AB which rests in S has in S the velocity u D v1 . The particle
v2 v
2v
D
;
v2 v=c2
1 C v 2 =c2
(4.52)
where v2 D v1 is the velocity of B in S and v D v1 is the
velocity of S against S.
The conservation of momentum demands for the collision described in S
m v2 v2 D Mu D Mv1 ;
(4.53)
where M is the mass of the compound AB with the velocity u D
v1 measured in S .
Conservation of energy requires, when dividing by c2
m v2 C m0 D M :
(4.54)
Inserting from (4.53) the relation M D m.v2 /v2 =v into (4.54)
we obtain
m v2
v
D
:
(4.55)
m0
v2 C v
Equation 4.52 gives the relation between v and v2
vD
4.4.4
1
"
c2
1
C
1
v2
v22
c2
1=2 #
I
(4.56)
inserting this into (4.55) gives the mass ratio
m v2
D 1
m0
v22
c2
1=2
D
and therefore again the general relation
m.v/ D p
1
m0
v 2 =c2
v2 ;
D .v/m0 ;
(4.57)
(4.58)
which has been already derived in Sect. 4.4.1.
4.4.5
Relativistic Formulation of Energy
Conservation
In order to show, that the relativistic energy E D m c2 is
conserved we must discuss the relativistic formulation of the
Newton equation F D dp=dt. Thereto we replace the classical
position vector r by the Lorentz four-vector
R D xOex C yOey C zOez C ictOe t D r C ictOe t ;
Figure 4.31 Description of a collinear completely inelastic collision in two different inertial systems S and S
(4.59)
defined in the four-dimensional Minkowski space .x; y; z; ict/
(see Sect. 3.6.2), where the unit vector eO t is perpendicular to
the three spatial axes. From (4.59) one can derive that R2 D
r2 c2 t2 .
4.5 Conservation Laws
This gives the total differential
dR2 D dx2 C dy2 C dz2
c2 dt2 D
c2 d 2 ;
where we have used as abbreviation the differential
r
1
d D dt2
.dx2 C dy2 C dz2 /
c2
s
v2
D dt 1
D dt=
c2
(4.60)
The quantity dW represents the work, performed on the particle
with mass m. For conservative forces F D grad V, which have
a potential, is dW equal to the change of the potential energy Ep .
Integration of (4.67) over the time yields
Ep C mc2 D const D E ;
(4.61)
which corresponds to the classical energy conservation (2.38) if
Ekin is replaced by mc2 .
For a particle with the velocity v we can write
of the “eigen-time” , which approaches the classical time differential dt for v c.
The differentiation of (4.59) gives the four-vector of the velocity
dx
dy
dz
dt
dR
D
eO x C eOy C eO z C ic eO t
d
d
d
d
d
v C icOe t
:
D p
1 v 2 =c2
dR
v C icOe t
D m0 p
:
d
1 v 2 =c2
(4.63)
(4.64)
Using these definitions we can derive the relativistic energy conservation law. We multiply (4.64) with dR=d
dR
F
d
d2 R
dR
D m0
d 2
d
2
m0 d dR
:
D
2 d d
(4.65)
According to (4.60) is .dR=d/2 D c2 D const. Therefore the
right side of (4.65) is zero!
dR
D0:
F
d
(4.66)
This can be separated in a spatial and a temporal part, which
gives
1
dr
d 2
F
D0
mc
1 v 2 =c2
dt dt
) d mc2 D F dr D dW :
1
.v=c/2
:
Ep C p
1
m0 c2
v 2 =c2
DE:
(4.68b)
Conservation Laws
In the foregoing sections we have discussed, that there are physical quantities which are conserved in closed systems, i. e. they
do not change in the course of time.
As a reminder, please note,
F D
m0 c2
The equation (4.68a) of energy conservation then becomes
4.5
In analogy to the Newton equation F D dp=dt we define the
four-force (also called the Minkowski-force)
d2 R
dP
D m0 2
d
d
d
d
D
.m v/ C ic .mOe t /
dt
dt
d
D
F C i .mcOe t / :
dt
mc2 D m0 .v/ c2 D p
(4.62)
The four-momentum is defined as
P D m0
(4.68a)
(4.67)
that a closed system is a system which has no interaction
with the outside, i. e. there are no external forces acting
on the particles of the system, although the particles may
interact with each other.
Such conserved quantities are the total momentum p, the total
energy E and the angular momentum L of a closed system. The
conservation of these quantities is, because of its great importance, formulated in special conservation laws, which shall be
summarized and generalized in the following sections.
4.5.1
Conservation of Momentum
For a single free particle (no forces acting on it) the momentum
conservation reads:
The momentum p D m v of a free particle is constant in
time.
This is identical with the Newton postulate (Sect. 2.6).
Generalized for a system of particles this reads:
Chapter 4
123
124
4 Systems of Point Masses; Collisions
The total momentum of a closed system of particles which
may interact with each other, does not change with time.
This can be also formulated as: If the vector sum of all external
forces acting on a system of particles is zero, the total momentum of the system does not change with time. According to the
3. Newton’s axiom actio D reactio the vector sum of all internal
forces is anyway zero.
Note that the momentum of the individual particles can indeed
change!
4.5.2
Energy Conservation
We have seen in Sect. 2.7 that in conservative force fields the
sum of kinetic anf potential energy is constant. This energy conservation can be generalized to a system of particles and also
further types of energy (internal energy, thermal energy, mass
energy E D mc2 ) can be included. The law of energy conservation in the general form is then:
The total energy of a closed system is constant in time,
where the different forms of energy can be completely or
partially converted into each other
Chapter 4
For instance the kinetic energy of a particle can be converted
into thermal energy at a collision with the wall, or the mass energy of electron and positron can be converted into radiation
energy if the two particles collide.
4.5.3
Conservation of Angular Momentum
If the vector sum of all torques Di which act on a system of
particles is zero, the total angular momentum L of the system
P
remains constant. This follows from the relation dL=dt D Di .
Note: For the definition of the angular momentum
LD
X
.ri pi / ;
the reference point (generally the origin of the coordinate system from which the position vectors ri start) has to be defined.
P
Since for a closed system Di D 0 the conservation of angular
momentum can be also formulated as
In a closed system the total angular momentum L remains
constant in time.
4.5.4
Conservation Laws and Symmetries
A more detailed investigation of the real causes of the conservation laws reveals that these laws are based on symmetry
properties of space and time [4.9]. In order to prove this, we
introduce the Lagrange function L
L.ri ; vi / D
N
X
mi
vi2
2
i
D Ekin
Epot .r1 ; r2 : : : rN /
(4.69)
Ep
of a closed system with N particles, which represents the difference of kinetic and potential energy. From (4.69) the relations
@L
D mi vi D pi
@vi
@Ep
@L
D
D Fi
@ri
@ri
(4.70a)
(4.70b)
follow immediately. This gives the equation of motion Fi D
mi dvi =dt in the general form
d
dt
@L
@vi
D
@L
:
@ri
(4.71)
Note: The Lagrange equation (4.71) can be derived quite general from a fundamental variation principle, called the principle
of minimum action [4.8].
This principle also gives the definite justification for the following statements and their explanation.
1. The conservation of momentum is due to the homogeneity of space.
This homogeneity of space guarantees that the properties of a
closed system do not change when all particles are shifted by an
amount ", which means that their position vectors changes from
r to r C ". Because of the homogeneity all masses and velocities
remain unchanged.
The Lagrange function in a homogeneous space does not depend
on the position vectors ri i. e.
X
@L=@ri D 0 :
From (4.71) we can conclude
X d @L
d X @L
D
D0
dt @vi
dt i @vi
i
X @L
X
)
D
pi D p D const :
@vi
(4.72)
Summary
125
2. The conservation of energy follows from the homogeneity of time.
The homogeneity of time implies that the Lagrange function L
does not explicitly depend on time. Which means that @L=@t D
0.
The total derivation of L is
3N
X @L
X @L
dL
xPi C
xRi :
D
dt
@xi
@xPi
iD1
Figure 4.32 Definition of ı' and ı r
which means that E is constant in time.
For the change ı L D 0 of the Lagrange function L we obtain
ıL D
X @L
@L
ıri C
ıv D 0 :
@r
@v
i
i
i
With the relations
and @L=@ri D Fi D
@L=@vi D pi
Finally the conservation of angular momentum follows from
the isotropy of space, which means that no specific direction
in space is preferred.
we can write (4.75) in the form
This isotropy implies that an arbitrary rotation of a closed system does not change the mechanical properties of the system.
In particular the Lagrange function should not change when the
system rotates by an angle @'.
i
We introduce the vector ı' with the magnitude ı' and the direction of the rotation axis. The change of the position vector ri
of the point P is (Fig. 4.32)
ıri D ı' ri :
(4.74a)
The velocity of P is then changing by
ıvi D ı' vi :
(4.75)
X
dpi
dt
pPi .ı' ri / C pi .ı' vi / D 0
) ı'
"
X
i
..ri pP i / C .vi pi //
#
(4.76)
d X
D ı'
.ri pi / D 0 :
dt
Since this must hold for arbitrary values of ı' it follows
X
.ri pi / D L D const :
(4.77)
(4.74b)
Summary
The centre of mass of a system of N point masses mi with the
position vectors ri has the position vector
1 X
1 X
rS D P
mi ri D
mi ri :
mi
M
The coordinate system with the CM as origin is called the
centre-of-mass system
The vector sum of all momenta mi vi of the masses mi in the
CM-system is always zero.
The reduced mass of two masses m1 and m2 is defined as
D
m1 m2
:
m1 C m2
The relative motion of two particles with the mutual interaction forces F12 D F21 can be reduced to the motion of a
single particle with the reduced mass which moves with
the velocity v12 D v1 v2 around the centre of m1 .
Chapter 4
If we replace according to (4.71) @L=@xi by d=dt.@L=@Oe/ we
obtain
X d @L
X d @L X @L
dL
C
xRi D
xPi
D
xPi
dt
dt @xPi
@xPi
dt @xPi
X
d
@L
(4.73)
)
L D0
xPi
dt
@xPi
X @L
L D E D const ;
)
xPi
@xPi
126
4 Systems of Point Masses; Collisions
A system of particles with masses mi , where no external
forces are present is called a closed system. The total momentum and the total angular momentum of a closed system
are always constant, i. e. they do not change with time (conservation laws for momentum and angular momentum).
In elastic collisions between two particles the total kinetic
energy and the total momentum are conserved. For inelastic collisions the total momentum is also conserved but part
of the initial kinetic energy is transferred into internal energy (e. g. potential energy or kinetic energy of the building
blocks of composed collision partners). Inelastic collisions
can only occur if at least one of the collision partners has a
substructure, i. e. is compound of smaller entities.
While for elastic collisions in the lab-system the kinetic energies Ei of the individual partners change (although the total
energy is conserved), in the CM-system also the Ei are conserved.
2
In inelastic collisions only the kinetic energy 12 v12
of the
relative motion can be transferred into internal energy. At
least the part 21 MvS2 of the CM-motion must be preserved as
kinetic energy of the collision partners.
The collision between two particles with masses m1 and m2
can be reduced in the CM-system to the scattering of a single
particle with reduced mass
D
m1 m2
m1 C m2
by a particle with mass m1 fixed in the CM. This can be
also described by the scattering in a potential depending on
the interaction force between the two particles.
The deflection angle ' of the particle in the CM-system depends on the impact parameter b, the reduced mass , the
initial kinetic energy 12 v02 and the radial dependence of the
interaction potential.
The evaluation of collisions at relativistic velocities v demands the consideration of the relativistic mass increase.
Then also energy and momentum conservation remain valid.
The conservation laws for energy, momentum and angular
momentum can be ascribed to general symmetry principles,
as the homogeneity of space and time and the isotropy of
space.
Problems
Chapter 4
4.1
Two particles with masses m1 D m and m2 D 3m suffer c) Which fraction of the initial kinetic energy has been converted into internal energy? Calculate this fraction in the
a central collision. What are their velocities v10 and v20 after the
lab-system and the CM-system.
collision if the two particles had equal but opposite velocities
v1 D v2 before the collision
4.5
A mass m1 D 1 kg with a velocity v1 D 4 m=s collides
a) For a completely elastic collision
with a p
mass m2 D 2 kg. After the collision m1 moves with
b) For a completely inelastic collision?
v10 D p 8 m=s under an angle of 45ı against v1 and m2 with
4.2
A wooden block with mass m1 D 1 kg hangs on a wire v 0 D 2 m=s under an angle of 45ı
2
with length L D 1 m. A bullet with mass m2 D 20 g is shot with a) What was the velocity v2 ?
3
the velocity v D 10 m=s into the block and sticks there. What b) Which fraction of the initial kinetic energy has been conis the maximum deflection angle of the block?
verted into internal energy in the lab-system and the CMsystem?
4.3
A proton with the velocity v1 collides elastically with a c) How large are the deflection angles #1 and #2 in the CMdeuteron (nucleus consisting of proton and neutron) at rest. Afsystem?
ter the collision the deuteron flies under the angle of 45ı against
v1 . Determine
4.6
Two cuboids with masses m1 D 1 kg and m2 < m1 slide
a) the deflection angle 1 of the proton
frictionless on an air-track, which is blocked on both sides by a
b) the CM-velocity
vertical barrier (Fig. 4.33). Initially m1 is at rest and m2 moves
c) the velocities v10 and v20 of proton and deuteron after the col- with constant velocity v2 D 0:5 m=s to the left. After the collilision.
sion with m1 the mass m2 is reflected to the right, collides with
the barrier .m D 1/ and slides again to the left. We assume
4.4
A particle with mass m1 D 2 kg has the velocity v1 D that all collisions are completely elastic.
f3Oex C 2Oey eO z g m=s. It collides completely inelastic with a par- a) What is the ratio m1 =m2 if the two masses finally move to
ticle of mass m2 D 3 kg and velocity v2 D f 2Oex C 2Oey C 4Oez g.
the left with equal velocities?
Determine
b) How large should m2 be in order to catch m1 before it reaches
a) The kinetic energies of the two particle before the collision
the left barrier?
in the lab-system and the CM-system.
c) Where collide the two masses at the second collision for
b) Velocity and kinetic energy of the compound particle after
m2 D 0:5 kg?
the collision.
References
Figure 4.33 Illustrating Probl. 4.6
127
4.8
An elevator ascends with constant velocity v D 2 m=s.
When its ceiling is still 30 m below the upper point A of the lift
shaft a ball is released from A which falls freely down and hits
elastically the ceiling of the elevator, from where it is elastically
reflected upwards.
a) Where does it hit the elevator ceiling?
b) What is its maximum height after the reflection?
c) Where does it hit the elevator ceiling a second time?
4.7
A steel ball with mass m1 D 1 kg hangs on a wire
with L D 1 m, vertically above the left edge of a resting mass 4.9
An ˛-particle (nucleus of the He-atom) hits with the vem2 D 5 kg which can slide without friction on a horizontal air- locity v1 an oxygen nucleus at rest .m2 D 4m1 /. The ˛-particle
track. (Fig. 4.34). The steel ball with the wire is lifted by an is deflected by 64ı , the oxygen nucleus by 51ı against v1 . The
angle ' D 90ı from the vertical into the horizontal position and collision is completely elastic.
then released. It collides elastically with the glider. What is the a) What is the ratio v 0 =v 0 of the velocities after the collision?
2
1
maximum angle ' of m1 after the collision?
b) What is the ratio of the kinetic energies after the collision?
4.10 A particle has in a system S a kinetic energy of 6 GeV.
and the momentum P D 6 GeV=c. What is its energy in a system S0 , where its momentum is measured as 5 GeV=c? What is
the relative velocity of S0 against S?
Figure 4.34 Illustration of Probl. 4.7
4.1a. http://berkeleyscience.com/pm.htm
4.1b. A. Tan, Theoryof Orbital Motion. (World Scientific Publ.,
2008)
4.2. R. D. Levine, Moleclar Reaction Dynamics. (Cambridge
Univ. Press, 2005)
4.3. A. Einstein, N. Calder, Relativity: The special and the
general Theory.
4.4a. E.F. Taylor, J.A. Wheeler, Spacetime Physics: Introduction to Special Relativity, 2nd ed., (W. H. Freeman &
Company, 1992)
4.4b. W. Rindler, Introduction to Special Relativity, 2nd ed.,
(Oxford University Press, 1991)
4.5. R.M. Dreizler, C.S. Lüdde, Theoretical Mechanics.
(Springer, Berlin, Heidelberg, 2010)
4.6. A.P. French, Special Relativity. (W.W. Norton & Co, 1968)
4.7. A. Das, The Special Theory of Relativity. (Springer,
Berlin, Heidelberg, 1996)
4.8. Goldstein, Classical Mechanics. (Addison Wesley, 2001)
4.9. J. Schwichtenberg, Physics from Symmetry. (Springer,
Heidelberg, 2015)
Chapter 4
References
Dynamics of rigid Bodies
5.1
The Model of a Rigid Body . . . . . . . . . . . . . . . . . . . . . . . . . 130
5.2
Center of Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130
5.3
Motion of a Rigid Body . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
5.4
Forces and Couple of Forces . . . . . . . . . . . . . . . . . . . . . . . . 132
5.5
Rotational Inertia and Rotational Energy . . . . . . . . . . . . . . . . . 133
5.6
Equation of Motion for the Rotation of a Rigid Body . . . . . . . . . . 136
5.7
Rotation About Free Axes; Spinning Top . . . . . . . . . . . . . . . . . 139
5.8
The Earth as Symmetric Top . . . . . . . . . . . . . . . . . . . . . . . . . 149
5
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151
Chapter 5
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152
© Springer International Publishing Switzerland 2017
W. Demtröder, Mechanics and Thermodynamics, Undergraduate Lecture Notes in Physics, DOI 10.1007/978-3-319-27877-3_5
129
130
5 Dynamics of rigid Bodies
Up to now we have discussed idealized bodies where their
spatial extension could be neglected and they were therefore
adequately described by the model of a point mass. We have
investigated their motion under the influence of forces and
have presented besides Newton’s laws fundamental conservation laws for linear momentum, energy and angular momentum.
All phenomena found in nature which are due to the spatial extension of bodies demand for their explanation an extension of
our model. Besides the translation of point masses, discussed so
far, we have to take into account the fact that extended bodies
can also rotate around fixed or free axes.
At first, we will restrict ourselves to the motion of free extended
bodies under the influence of forces. The motion of single
volume elements of an extended body against each other, that
results in a deformation of the body will be discussed in the next
chapter. Such still idealized extended bodies that do not change
their form, are called rigid bodies.
5.1
The Model of a Rigid Body
We can partition an extended rigid body with volume V and total mass M into many small volume elements Vi with masses
mi which are rigidly bound together (Fig. 5.1). We can regard theses mass elements mi as point masses and treat them
according to the rules discussed in Chap. 2.
The total body can then be composed of these volume elements:
VD
N
X
MD
Vi ;
iD1
N
X
mi :
iD1
We name the ratio
Œ% D kg=m3
% D m=V I
(5.1)
Chapter 5
the mass density of the volume element V. The total mass can
then be expressed as
N
X
MD
%i Vi :
(5.2)
iD1
If the volume elements V become smaller and smaller, their
number N correspondingly larger, the sums converge for the
limiting case V ! 0 to volume integrals [5.1]
N
X
V D lim
Vi !0
N!1 iD1
MD
Z
Vi D
Z
V
dV I
(5.3)
%dV ;
V
where the volume integral stands for the three-dimensional integral
2 0
1 3
Zz2 Zy2 Zx2
V D 4 @ dxA dy5 dz
(5.4)
z1
y1
x1
Figure 5.1 Partition of a spatially extended body into small volume elements
Vi
for the example of a cuboid, while for a spherical volume with
radius R and a volume element dV D r2 sin # dr d# d' (see
Sect. 13.2) the integral can be written as
VD
ZR Z Z2
r2 sin # dr d# d' :
(5.5)
rD0 #D0 'D0
The mass density %.x; y; z/ can generally depend on the location
.x; y; z/. For homogeneous bodies % is constant for all points of
the body and we can extract % out of the integral. The mass M
of the body can then be expressed as
Z
M D % dV D %V :
(5.6)
V
5.2
Center of Mass
As has been shown in the previous chapter the position vector
rS of the CM of a system with N particles at the positions ri
(Fig. 5.2) is
PN
ri mi
rS D PiD1
N
iD1 mi
(5.7)
N
1 X
ri %.ri /Vi :
D
M iD1
For the limiting case V ! 0 and N ! 1 this becomes
Z
1
rS D
r dm
M
V
Z
(5.8a)
1
D
r%.r/ dV :
M
V
5.3
Motion of a Rigid Body
131
This becomes with z D rcos # and dV D r2 drsin #d#d'
zS D
1
V
ZR Z=2 Z2
r3 cos # sin # dr d# d'
(5.9)
rD0 #D0 'D0
D
3
R:
8
J
Figure 5.2 Definition of the Center of Mass of an extended body
This corresponds to the three equations for the components
Z
1
xS D
x%.x; y; z/ dV ;
M
V
Z
1
yS D
y%.x; y; z/ dV ;
M
V
Z
1
z%.x; y; z/ dV :
zS D
M
V
For homogeneous bodies (% D const) (5.8a) simplifies to
Z
1
r dV :
(5.8b)
rS D
V
V
5.3
Motion of a Rigid Body
The center points Pi of the volume elements dVi are defined by
their position vectors ri , the CM by rS . The vector
riS D ri
rS
points from the center of mass S to the point Pi (Fig. 5.4). The
vector
driS =dt D viS D vi
vS
(5.10)
gives the relative velocity of Pi with respect to the CM.
In a rigid body all distances are fixed, i.e. jriS j D const. Differ2
D const gives
entiation of riS
2riS vS D 0 ;
Center of Mass of a homogeneous hemisphere.
If the center of the sphere is at the origin .x D y D z D 0/
(Fig. 5.3) symmetry arguments require xS D yS D 0. For
% D const we obtain from (5.8b)
Z
Z
1
1
zS D
z% dV D
z dV :
M
V
V
which implies that the vector of the relative velocity is perpendicular to the position vector. This can be written as (see
Sect. 2.4)
viS D .! riS / ;
(5.11)
where ! is the angular velocity of Pi rotating about an axis
through the CM perpendicular to the velocity vector viS .
V
Figure 5.3 Center of Mass of a hemisphere
Figure 5.4 The motion of a rigid spatially extended body
Chapter 5
Example
132
5 Dynamics of rigid Bodies
For the general motion of the rigid body the velocity of the point
Pi
vi D vS C .! riS /
(5.12)
can be composed of two contributions: The translational velocity vS of the CM and the rotation .! riS / of Pi around the CM.
Since the consideration discussed above is valid for an arbitrary
point Pi we can make the general statement:
The motion of an extended rigid body can always be composed of the translation of its CM and a rotation of the
body about its CM.
Note: The rotational axis is not necessarily constant but can
change its direction in the course of time, even when no external
forces act onto the body (see Sect. 5.7).
The Eq. 5.10 and 5.11 are based on the condition rik2 D const
for a rigid body. They are no longer valid, if deformations of the
body occur, because then vibrational motions of Pi against the
CM can be present as additional movements.
The complete characterization of the motion of a free rigid body
demands 6 time-dependent parameters: The position coordinates
rS .t/ D fxS .t/; yS .t/; zS .t/g
for the description of the CM-motion and three angular coordinates for the description of the rotation of the rigid body about
its CM.
The free rigid body has six degrees of freedom for its motion.
Figure 5.5 The two forces F1 and F2 have equal magnitudes but different
points of application P1 and P2 . They effect different motions of the body
on the same point S and since F2 .S/ C F3 .S/ D 0 they cancel
each other.
Now we combine the two antiparallel forces F1 and F3 with
equal magnitude (Fig. 5.6) which form a couple of forces, but
regard at first the remaining single force F2 , which acts on the
center of mass S. This force causes a translation of the CM. The
couple of forces brings about a torque
DS D .riS F1 / ;
(5.13a)
referred to the center of mass S. Since F1 C F3 D 0 this couple
of forces does not cause an acceleration of the CM. It induces,
however, a rotation of the body around S. Summarizing we can
make the general statement:
A force F acting on an arbitrary point P ¤ S of an
extended body causes an acceleration of the CM and a rotation of the body about the center of mass S.
An extended rigid body initially at rest suffers an accelerated translation of its center of mass S and a rotation
around S when a force F acts on a point P ¤ S.
In this chapter we will investigate such motions in more detail.
Chapter 5
If one point of the body (for example the CM) is kept fixed the
body can still rotate about this point but cannot perform a translation. The number of degrees of freedom then reduces to three,
namely the three rotational degrees of freedom. If the body rotates around a fixed axis, only one degree of freedom is left,
namely the one-dimensional rotation described by the angle '.
5.4
At first we will restrict the treatment to the special case where
the body rotates around a space-fixed axis. The motion is then
restricted to a rotation which has only one degree of freedom.
Forces and Couple of Forces
While a force F acting on a point mass is unambiguously defined
when its magnitude and direction is given, for forces acting on
an extended body the point of origin P has to be added (Fig. 5.5).
We will investigate the change of motion which an extended
body suffers under the action of a force F.Pi / with its origin
in an arbitrary point Pi . We can simplify the treatment when
we add two antiparallel forces F2 .S/ and F3 .S/ D F2 .S/ with
equal magnitude which both act on the center of mass S and
therefore do not affect the motion of the body, because they act
Figure 5.6 Decomposition of a force F1 into a couple of forces F1 F3 and
a force F2 that attacks at the Center of Mass
5.5
Ft
Rotational Inertia and Rotational Energy
133
Figure 5.9 Principle of beam balance
r
Figure 5.7 Rotation of a rigid body about a fixed axis A , induced by a force
F attacking at the point P . Decomposition of the projection of F onto the x -y plane into a normal and a tangential component
We choose the direction of the rotation axis A as the z-direction
and the x-y-plane through the point P.r/ where the force F acts
on the body, which has the distance r from the rotation axis
(Fig. 5.7). We can then decompose the force into the three components Fz k A, the radial component Fr k r and the tangential
component F t ? r and ? A. Fz is perpendicular to the x-y-plane
and the other two components are in the x-y-plane.
The torque exerted by the force F onto the body is
D D .r F/ D .r F t / C .r Fz / ;
since .r Fn / D 0.
The first term causes a torque about the z-axis, and therefore an
acceleration of the rotation about the z-axis. The second term
would change the direction of the rotation axis. If this axis is
fixed by axle bearings the torque only acts on the bearings and
does not lead to a change of motion.
If the rotation axis intersects the center of mass S (Fig. 5.8)
which we choose as the origin of our coordinate system, the
torque exerted by the weight Fw D M g of the body is zero, as
can be seen by the following derivation:
The torque with respect to the rotation axis caused by the gravitational force mi g on the mass element mi is Di D
.riS mi g/. The torque exerted by the weight of the whole
body is then
Z
Z
D D .r g/ dm D g r dm
(5.13b)
V
V
D
.g MrS / D 0 ;
because the center of mass S is the origin and therefore is rS D 0.
If a body can rotate around an axis through the CM it is
always at equilibrium, independent on the space orientation of the axis because the torque exerted by its weight is
always zero.
All beam balances are based
P on this principle (Fig. 5.9). The
balance is at equilibrium if Di D 0, which means
r1 F1 C r2 F2 D 0 :
This is the equilibrium condition for a balance as two-armed
lever.
5.5
Rotational Inertia and Rotational
Energy
We consider an extended body which rotates about a fixed axis
A with the angular velocity ! (Fig. 5.10). The mass element
mi with the distance ri? D jri j from the axis A has the velocity
vi D ri !. Its kinetic energy is then
Ekin .mi / D 12 mi vi2 D 21 mi ri2 ! 2 :
Figure 5.8 Contribution of the mass element m to the torque about an axis
through the Center of Mass, due to the weight of m
Figure 5.10 Definition of moment of inertia
Chapter 5
P
134
5 Dynamics of rigid Bodies
The summation over all mass elements gives the total rotational
energy of the body
!
Z
N
1X
1
2
2
2
Erot D lim
mi ri? ! D ! 2 r?
dm : (5.14)
N!1
2
2
iD1
m !0
i
The expression
Def
ID
Z
V
2
r?
dm D
Z
2
r?
% dV
(5.15)
V
is called the rotational inertia (often also moment of inertia) of
the rotating body referred to the axis A. With this definition we
obtain for the rotational energy
Erot D 21 I! 2 :
(5.16a)
The angular momentum of mi with respect to the axis A is
2
Li .mi / D ri? .mi vi / D ri?
mi ! ;
(5.17a)
which gives the total angular momentum of the body as
LDI! :
(5.17b)
Replacing in (5.16a) ! 2 by L2 =I 2 we obtain for the rotational
energy
1
L2
Erot D I! 2 D
:
(5.16b)
2
2I
The rotational inertia I is a measure for the mass distribution
in an extended body relative to the rotational axis. For geometrically simple bodies with homogeneous mass distribution
% D const the rotational inertia I can be readily calculated, as is
illustrated in the following examples. For bodies with a complex
geometrical structure I has to be measured (see below).
Chapter 5
If the rotational axis A intersects the center of mass S.r D 0/,
the rotational inertia can be written as
Z
IS D % r2 dV :
The rotational inertia is always defined with respect to a
definite rotational axis and depends on the location of this
axis with relative to the CM.
5.5.1
Figure 5.11 Steiner’s rule: The drawing shows the plane through dm , perpendicular to the axis A
The Parallel Axis Theorem (Steiner’s
Theorem)
If a body rotates about an axis B which is parallel to the axis A
through the CM, the rotational inertia with respect to the axis B
can be readily calculated, if it is known with respect to the axis
A. If the distance between the two axes is a (Fig. 5.11) we can
write
Z
Z
IB D r2 dm D .rmS C a/2 dm
Z
Z
Z
2
D rmS
dm C 2a rmS dm C a2 dm :
R
According to (5.8) is rmS dm D RS M D 0, because the center
of mass S is the origin of the coordinate system and therefore is
RCM D RS D 0.
We then obtain
IB D IS C a2 M :
(5.18)
Equation 5.18 is called the parallel axis theorem or Steiner’s
theorem. It states:
The inertial moment of a body rotating around an axis B is
equal to the sum of the inertial moment with respect to an
axis A through the center of mass S with a distance a from
the axis B plus the moment of inertia of the total mass M
concentrated in S with respect to B.
This illustrates that it is sufficient to determine the rotational
inertia with respect to an axis A through S. With (5.18) we can
then obtain the rotational inertia with respect to any axis parallel
to A.
In the following we will give examples for the calculation of I
for homogeneous bodies with different geometrical structures.
Example
1. Thin disc. We assume the height h in the z-direction is
small compared to the extension of the body in the xand y-directions.
a) Rotational axis in the z-direction:
Z
Iz D % .x2 C y2 /dV :
b) Rotational axis in the x-direction:
Z
Z
Ix D % .y2 C z2 /dV % y2 dV ;
because jzj h=2 ymax .
5.5
Rotational Inertia and Rotational Energy
135
Figure 5.12 Moment of inertia of a thin disc with arbitrary shape (a),
circular disc (b)
Figure 5.13 Moment of inertia of a hollow cylinder
Iy D %
Z
.x2 C z2 /dV %
Z
x2 dV :
This shows that with the approximation z x; y one
obtains
(5.19)
Iz Ix C Iy :
For plane bodies (for example a triatomic molecule)
the rotational inertia for the rotation around an axis
through the CM perpendicular to the plane is equal to
the sum of the two other moments of inertia.
For the case of a thin circular disc we obaqin from
(5.19) because of the rotational symmetry (Fig. 5.12b)
Ix D Iy D
1
I
2 z
:
For the homogeneous circular disc with radius R it is
not difficult to calculate Iz :
Iz D %
Z
r2 dV D 2h%
Z
r3 dr D % h R4 =2 ;
because dV D 2r h dr. With M D % R2 h this
gives
Iz D 12 MR2 :
(5.20a)
2. Hollow cylinder with height h, outer radius R and wall
thickness d R (Fig. 5.13). Rotation about the z-axis
as symmetry axis:
Iz D %
Z
2
r dV D 2h%
V
ZR
Remark. A simpler derivation starts with the relation
for the total mass of the hollow cylinder with outer
radius r2 and inner radius r1 :
M D % h r22 r12 :
Z
Z
Iz D r2 dm D 2% h r3 dr D 12 % h r24 r14
D 21 % h r22 r12 r22 C r12
21 M 2R2 D MR2 :
3. Full cylinder with radius R and height h.
Iz D 2%h
D
ZR
0
r3 dr D
h%R4
2
(5.20c)
M 2
R ;
2
which, of course, concurs with (5.20a).
4. Thin rod (length L diameter (Fig. 5.14))
a) Rotation about the vertical axis a through the center of mass S.
IS D %
D
Z
2
x dV D %A
CL=2
Z
x2 dx
L=2
(5.21a)
1
1
%AL3 D
ML2 :
12
12
r3 dr ;
R d
with dV D 2 R h dr and d R one obtains
Iz D h % R4
.R
d/4 2%hR3 d :
This gives with M D 2 r % d h
Iz D M R2 :
(5.20b)
Figure 5.14 Rotation of a thin rod with arbitrary cross section about
an axis a through the Center of Mass and about an axis b at one end of
the rod
Chapter 5
c) Rotational axis in the y-direction:
136
5 Dynamics of rigid Bodies
b) Rotation about the vertical axis b through an end
point of the rod. According to the parallel axis theorem (5.18) the moment of inertia is
2
L
2
1
1
1
2
D
ML C ML2 D ML2 :
12
4
3
Ib D IS C M
(5.21b)
This result could have been obtained also directly
from
Ib D %A
ZL
0
x2 dx D
1
1
%AL3 D ML2 :
3
3
5. Diatomic molecule. Because of the small electron
mass .me 1=1836mp/ the electrons do not contribute essentially to the moment of inertia when the
molecule rotates around an axis A through the CM
perpendicular to the inter-nuclear axis (Fig. 5.15). Because the diameter of the nuclei .d 10 14 m/ is
very small compared with the inter-nuclear distance
R . 10 10 m/ we can treat the nuclei as point masses
and obtain
ISA D m1 r12 C m2 r22 :
Figure 5.16 Derivation of the moment of inertia of a sphere
6. Moment of inertia for a homogeneous sphere. Because of the spherical symmetry the moment of inertia
is independent of the direction of the rotational axis
through the center of the sphere. The moment of inertia of a mass element m with a distance a from the
rotation axis (Fig. 5.16) is dI D a2 dm. For the whole
sphere we obtain
Z
IS D % a2 dV
with dV D r2 sin #drd#d' and a D r sin #.
(5.22a)
With the inter-nuclear distance R D r1 C r2 and the
reduced mass D m1 m2 =.m1 Cm2 / (5.22a) becomes
with r1 =r2 D m2 =m1
2
ISA D R :
IS D %
Chapter 5
Erot D L2 =2IB
for a given angular momentum L becomes much larger
than for the rotation around A (see Chap. 11 and
Vol. 3).
r4 sin3 # dr d# d'
rD0 #D0 'D0
1
D %R5 2
5
(5.22b)
When the molecule rotates around its inter-nuclear
axis B the nuclei lie on the rotational axis and do not
contribute to the moment of inertia. Now the electrons
provide the major contribution. Because of the small
electron mass the moment of inertia is now very small
and the rotational energy
ZR Z Z2
Z
sin3 # d#
(5.23)
#D0
4
2
2
D %R2 R3 D MR2 :
5
3
5
J
These examples with their rotational inertia are compiled in
Tab. 5.1.
5.6
Equation of Motion for the
Rotation of a Rigid Body
For the rotation of a rigid body around a space-fixed axis the
angular momentum of a mass element mi is:
2
Li D .ri? pi / D mi .ri? vi / D mi ri?
!;
Figure 5.15 Moment of inertia of a diatomic molecule
(5.24)
where the velocity vi is perpendicular to the rotation axis (pointing into the direction of ! which is the z-direction) and to the
radius r.
5.6 Equation of Motion for the Rotation of a Rigid Body
137
Table 5.1 Moments of inertia of some symmetric bodies that rotate about a
symmetry axis
Geometrical figure
Realization
Moment of inertia
Thin disc
1
MR2
2
Hollow cylinder
with thin wall
MR2
Figure 5.17 Torque acting on the rotation about a fixed axis induced by a force
F attacking at the point P
Full cylinder
1
MR2
2
The integration overall mass elements yields
1
ML2
12
Thin rod L r
DDI
Homogeneous sphere
(5.27)
where ' is the angle of r against the x-axis.
2
MR2
5
Hollow sphere
with thin wall
d!
d2 '
DI 2 ;
dt
dt
2
2
3 MR
5.6.1
Diatomic molecule
Ix D
1
M.b2
12
C c2 /
Iy D
1
M.a2
12
C c2 /
Iz D
1
2
12 M.a
C b2 /
ID
m1 m2
m1 Cm2
For a constant torque which does not change with time the integration of (5.27) yields the equation of rotation analogous to
(2.6) for the translation of a body
'D
D 2
t C At C B :
2I
(5.28a)
The integration constants A and B are specified by the initial
conditions '.0/ D '0 and d'.0/=dt D !0 . This gives for
(5.28a)
R2
'D
D 2
t C !0 t C ' 0 :
2I
(5.28b)
The time-derivative of (5.24) is
dLi
dvi
D mi ri?
D .ri? F t / D Dik ;
dt
dt
Example
(5.25)
where Dik D ri? Fi is the component of the torque Di parallel
to the rotation axis A. The other components Fz and Fr of the
force F are compensated by elastic forces of the mounting of the
fixed axis A (Fig. 5.17).
For the magnitude Di D jDik j of the not compensated torque we
obtain from (5.24) and (5.25)
2
Di D mi ri?
d!
:
dt
(5.26)
1. A full cylinder, a hollow cylinder and a ball with equal
masses M and equal radii r role down an inclined
plane. All three bodies start at the same time. Who
will win the race? The question can be answered experimentally as demonstration experiment during the
lecture and arises always astonishment.
Solution: The rotation takes place around the momentary rotation axis which is the contact line between
the body and the plane (Fig. 5.18). The torque acting on the body is D D M g sin ˛ where ˛ is the
inclination angel of the plane. The rotational inertia is
Chapter 5
Cuboid
Rotation About an Axis for a Constant
Torque
138
5 Dynamics of rigid Bodies
according to the parallel axis theorem I D IS C Mr2 .
Equation 5.27 then becomes
(5.29)
Mgr sin ˛ D IS C Mr2 !P :
The translational acceleration a D d2 s=dt2 of the center of mass S is equal to the acceleration r d!=dt of
the perimeter which rolls on the inclined plane. This
gives the relation
d2 s
Mgr sin ˛
D r!P D r
dt2
IS C Mr2
(5.30)
g sin ˛
D
D
a
:
1 C IS =Mr2
Compare this with the acceleration of a body which
slides frictionless down the plane without rolling. In
this case the acceleration is a t D g sin ˛.
For the rolling body part of the potential energy is
converted into rotational energy and only the rest is
available for translational energy. The translational acceleration is reduced by the factor b D .1 C IS =Mr2 /,
which depends on the moment of inertia IS of the
rolling body. The race is therefore won by the body
with the smallest moment of inertia. According to
Sect. 5.5.1 these are:
Ball:
Full cylinder:
2. Maxwell’s Wheel. A cylindrical disc with radius R,
mass M and rotational inertia IS D 12 MR2 is centered
on a thin axis through S with radius r R. The disc
hangs on a strand which is wrapped around the axis
(Fig. 5.19). The mass of the axis should be negligible compared with the mass M of the disc. When the
wheel is released it will roll down on the strand under
the influence of the torque D D r Mg and will move
down with the acceleration
d2 '
rD
D
dt2
I
r2 Mg
D 1
MR2 C Mr2
2
g
:
D
1 C R2 =2r2
aDr
The acceleration g is therefore reduced by the factor
.1 C 12 R2 =r2 /. This allows to observe this small acceleration when performing the experiment. After the
CM of the wheel has travelled the distance h the total
potential energy Mgh has been converted into kinetic
energy Mgh D Ekin D Etrans C Erot where
b D 7=5 ! a D 5=7 g sin ˛ ;
b D 3=2 ! a D 2=3 g sin ˛ ;
Hollow cylinder: b D 2
! a D 1=2 g sin ˛ :
Therefore the ball wins the race barely before the full
cylinder, while the hollow cylinder arrives last. It is
instructive to consider the situation from another point
of view: When the body has travelled the distance s
from the starting point on the inclined plane the loss
of potential energy is Epot D M g h D M g s sin ˛
which is converted into kinetic energy Ekin D Etrans C
Erot D 21 .Mv 2 C ! 2 IS / D 12 Mv 2 .1 C IS =Mv 2 /. This
gives for the translational velocity
Chapter 5
v2 D
2gs sin ˛
:
1 C IS =Mr2
Differentiation yields with d.v 2 /=dt D 2va the result
(5.30) for the acceleration a.
Figure 5.18 Rolling cylinder on an inclined plane
1
1
2
Mvtrans
D Mr2 ! 2
2
2
2r2
D Mgh 2
R C 2r2
Etrans D
and
Erot D
1 2
I!
2
D Mgh
R2
:
R2 C 2r2
Hint: The result is obtained from the relations r2 ! 2 D
v 2t and v 2t D g2 T 2 =.1 C 21 R2 =r2 /2 with the fall time
T D Œ.2h=g/.1 C 12 R2 =r2 /1=2 . By far the larger fraction 1=.1 C 2r2 =R2 / of the total energy is converted
into rotational energy.
At the lowest point of its path where the strand is completely unwound, the wheel continues to rotate in the
same sense (why?) and the strand winds up again,
which causes the wheel to rise nearly up to the starting
point. Because of frictional losses it does not completely reach its original starting heights.
Figure 5.19 Maxwell’s wheel
J
5.7
Measurements of rotational inertia;
Rotary Oscillations About a Fixed Axis
The experimental determination of inertial moments for bodies
with arbitrary form uses a rotary table consisting of a circular
disc with a concentric axis which can turn in fixed ball bearings
(Fig. 5.20). A coil spring with one end attached to the axis and
the other end to the mounting is bent by the turn of the table and
causes by its tension a restoring torque which is proportional to
the displacement angle ' from the equilibrium position ' D 0
(see Sect. 6.2)
D D Dr ' :
(5.31)
The proportionality factor Dr is called torsional rigidity. Its
value depends on the rigidity of the spring. The equation of
motion (5.27) for this case is
I0 'R D
Dr ' ;
(5.32a)
where I0 is the inertial moment of the rotary table. We have
neglected any friction. The solution of the differential equation
(5.32a) is with the initial condition '.0/ D 0
' D a sin
p
Dr =I0 t :
(5.32b)
Once deflected from its equilibrium position the rotary table performs a harmonic oscillation with the oscillation period
p
T0 D 2 I0 =Dr :
(5.32c)
If a circular disc with known mass M and radius R is placed
concentrically on the table, the moment of inertia increases to
I D I0 C 21 MR2 and the oscillation period becomes
T1 D 2
q
I0 C 12 MR2 =Dr :
(5.32d)
From the difference T12 T02 D 2 2 R2 =Dr the torsional rigidity
Dr can be determined. Now a body A with arbitrary form can be
placed on the table. The total moment of inertia IA now depends
on the location of A with respect to the center of the table. The
measured oscillation period
p
T D 2 .I0 C IA / =Dr
(5.32e)
allows the determination of IA . With the parallel axis theorem
(5.18) the moment of inertia IS of A with respect to its center of
mass S is IS D IA Ma2 , where a is the distance between the
center of the rotary table and the CM of A.
5.6.3
Comparison Between Translation and
Rotation
Table 5.2 shows a comparison between corresponding quantities
for the description of translation of a point mass and rotation of
an extended body. Note the analogous notation for momentum,
angular momentum, energy and power, if the mass m is replaced
by the moment of inertia I.
5.7
Rotation About Free Axes;
Spinning Top
Up to now we have discussed only rotations of rigid bodies
about space-fixed axes. Even for the example of the cylinder
rolling down the inclined plane the direction of the rotational
axis remained constant although it performed a translation.
In the present section we will deal with the more complex situation that a body can rotate about a free axis, which might change
its direction in space. We will treat at first the case that no external forces act on the body and then discuss the cases where
external torques are present.
Such rigid bodies rotating about free axes are called spinning
tops or gyroscopes.
Table 5.2 Comparison of corresponding quantities for rotation and translation
Figure 5.20 Turntable with cut through the ball bearing
139
Translation
Length L
Mass m
Velocity v
Momentum p D m v
Force F
dp
FD
dt
m
Ekin D v 2
2
Restoring force
FD Dx
Period of a linear
oszillation
p
T D 2 m=D
Rotation
Angle '
Moment of inertia I
Angular velocity !
Angular momentum L D I !
Torque D D r F
dL
DD
dt
I
Erot D ! 2
2
Restoring torque
D D Dr '
Period of torsional
oszillation
p
T D 2 I=Dr
Chapter 5
5.6.2
Rotation About Free Axes; Spinning Top
140
5 Dynamics of rigid Bodies
For the general motion one must take into account the translation of the CM (which can be always treated separately) and
the rotation around the CM. If the motion is discussed in the
CM-system where the CM is at rest, one has to regard only the
rotation about the CM. We will see, that the space-orientation
of free axes generally changes with time and the motion of an
arbitrary point of the rigid body might perform a complicated
trajectory.
In order to calculate the motion about free axes we have to determine the dependence of the moment of inertia on the direction
of the rotation axis, which, however, always should intersect the
CM.
5.7.1
Inertial Tensor and Inertial Ellipsoid
When a rigid body rotates with the angular velocity ! around an
axis through the center of mass S with arbitrary space orientation
(Fig. 5.21) the mass element mi moving with the velocity vi D
! ri has the angular momentum
Li D mi .ri vi / D mi .ri .! ri // ;
(5.33a)
using the vector relation (see Sect. 13.1.5.4)
A .B C/ D .A C/B
this can be transformed into
Li D mi r2i !
.A B/C ;
.ri !/ ri :
where the coefficients Iik are abbreviations for the expressions
Ixx D
Z
x2 dm
Z
xy dm
r2
Ixy D Iyx D
Z
2
r
Iyy D
y2 dm
Z
yz dm
Iyz D Izy D
Z
2
Izz D
r
(5.35a)
z2 dm
Z
xz dm :
Ixz D Izx D
Equation 5.34b can be readily checked when inserting the relations r2 D x2 C y2 C z2 and r ! D x!x C y!y C z!z into (5.34a)
and using (5.35a). The components Iik can be written in form of
the matrix
0
Ixx
B
e
I D @Iyx
Izx
Ixy
Iyy
Izy
1
Ixz
C
Iyz A ;
Izz
(5.35b)
which allows to write Eq. 5.34b in the vector form
(5.33b)
The total angular momentum of the rigid body is then obtained
by integration over all mass elements. This gives
Z
2
r ! .r !/ r dm :
(5.34a)
LD
Chapter 5
This vector equation corresponds to the three equations for the
components
Lx D Ixx !x C Ixy !y C Ixz !z
Ly D Iyx !x C Iyy !y C Iyz !z
(5.34b)
Lz D Izx !x C Izy !y C Izz !z ;
0 1 0
Lx
Ixx
B C B
D
L
@ y A @Iyx
Lz
Izx
Ixy
Iyy
Izy
10 1
!x
Ixz
CB C
Iyz A @!y A :
Izz
!z
(5.34c)
This can be shortened to
L De
I!:
(5.34d)
The mathematical term for I is a tensor of rank two, which
is called inertial tensor. The diagonal elements of I give the
moments of inertia for rotation axes in the direction of the coordinate axes x; y; z.
To illustrate the advantage of introducing this inertial tensor we
will at first calculate the rotational energy of the body for a
rotation about an arbitrary axis !. For a mass element mi
(Fig. 5.21) the rotational energy is
1
mi vi2
2
Figure 5.21 Rotation of a body about an arbitrary axis through the Center of
Mass
D 21 mi .! ri / .! ri /
D 21 mi !2 r2i .! ri /2 ;
where the right hand side follows from the vector relation
.A B/ .A B/ D A2 B2 .A B/2 .
5.7
Rotation About Free Axes; Spinning Top
141
The spatial integration over all mass elements gives the rotational energy of the whole rigid body
Z
1
.! r/2 dm
2
Z
!x2 C !y2 C !z2 2
x C y2 C z2 dm
D
2
Z
2
1
!x x C !y y C !z z dm
2
1 2
!x Ixx C !y2 Iyy C !z2 Izz
D
2
C !x !y Ixy C !x !z Ixz C !y !z Iyz ;
!2
2
Z
r2 dm
(5.36)
Figure 5.22 Inertial ellipsoid
where the definitions (5.35b) have been used. Within the tensor
notation (5.36) can be written as
Erot D 12 !T IQ ! ;
which explicitly means
Erot
0
Ixx
1
B
D .!x !y !z / @Iyx
2
Izx
Ixy
Iyy
Izy
The moment of inertia I! D k=R2 for a rotation about an arbitrary axis ! D f!x ; !y ; !z g is proportional to 1=R2 where
R is the distance from the center of the ellipsoid to its surface
(Fig. 5.22). With this notation one can say that the scalar value
I of the moment of inertia as a function of the spatial orientation
.˛; ˇ; / of the rotation axis represents the inertial ellipsoid.
10 1
!x
Ixz
CB C
Iyz A @!y A :
Izz
!z
This shows that for arbitrary orientations of the rotation
axis all elements of the inertial tensor can contribute to
the rotational energy.
When the rotation axis ! forms the angles ˛; ˇ;
ordinate axes the components of ! are
!x D ! cos ˛ ;
!y D ! cos ˇ ;
with the co-
!z D ! cos
an ellipsoid, with axes which depend on the coefficients Iik .
Since I / M R2m the constant k D M R4m has the dimension
Œk D kg m4 . Its value depends on the mass M of the rigid body
and the mass distribution relative to the center of mass S which
is expressed by a mean distance Rm .
:
When the rotational energy is written in the form of Eq. 5.16 as
5.7.2
Principal Moments of Inertia
We introduce a coordinate system .; ; / which is generated
by three orthogonal vectors , and with axes which point
into the directions of the principal axes a, b and c of the inertial
ellipsoid (Fig.
p 5.23). Their magnitude is normalized when dividing by k. In this coordinate system the ellipsoid equation
(5.37) becomes with R2 I D 1
Erot D 21 I! 2 ;
2 Ia C 2 Ib C 2 Ic D 1 :
the comparison with (5.36) yields for the scalar moment of inertia
In this principal axis coordinate system all off-diagonal elements Iik with i ¤ k of the inertial tensor I are zero and the
I D cos2 ˛ Ixx C cos2 ˇ Iyy C cos2 Izz
C 2 cos ˛ cos ˇ Ixy C 2 cos ˛ cos Ixz
C 2 cos ˇ cos Iyz :
(5.37a)
When we introduce a vector R in the direction of the rotation
axis with the components x D R cos ˛; y D R cos ˇ; z D
R cos Eq. 5.37a can be written as
R2 I D x2 Ixx C y2 Iyy C z2 Izz
C 2xyIxy C 2xzIxz C 2yzIyz :
(5.37b)
This is a quadratic equation in x, y, and z with constant coefficients Iik . All points .x; y; z/ for which R2 I D const are located
on an ellipsoid because (5.37) describes for R2 I D k D const
Figure 5.23 Definition of principal axes of inertia
(5.38)
Chapter 5
Erot D
142
5 Dynamics of rigid Bodies
tensor becomes a diagonal tensor
2
Ia
6
e
I D 40
0
3
0
7
05 :
Ic
0
Ib
0
(5.39)
Mathematically such a principal axes transformation can be performed by a diagonalization of the corresponding matrix [5.2].
The principal inertia moments Ia , Ib , Ic (i.e. the moments of inertia for rotations about the principal axes a, b, c) are the solutions
of the determinant equation
ˇ
ˇI
ˇ xx I
ˇ
ˇ Iyx
ˇ
ˇ Izx
Ixy
Iyy I
Izy
Ixz
Iyz
Izz
ˇ
ˇ
ˇ
ˇ
ˇD0:
ˇ
Iˇ
(5.40)
Note, that generally the principal moments of inertia do
not concur with the elements Ixx , Iyy , Izz , because all elements of the tensor can change under the principal axes
transformation.
According to international agreements [5.3] the assignment of
the principal moments follows the definition:
The moment of inertia for a rotation about an arbitrary axis with
direction angels ˛, ˇ, ', against the x, y, z, axis is (Fig. 5.24)
2
With the principal moments of inertia the angular momentum
and the rotational energy can be written as
L D fLa ; Lb ; Lc g D f!a Ia ; !b Ib ; !c Ic g ;
1 2
! Ia C !b2 Ib C !c2 Ic
2 a
L2
L2
L2
D a C b C c :
2Ia
2Ib
2Ic
Erot D
Ia Ib Ic :
2
Figure 5.25 Examples of asymmetric tops
2
I D Ia cos ˛ C Ib cos ˇ C Ic cos
:
(5.41)
This equation corresponds to (5.37a) since all off-diagonal elements are zero. The principal axes transformation has made the
expression for the general moment of inertia I simpler.
(5.42)
(5.43)
If all three principal moments are different .Ia ¤ Ib ¤ Ic ¤ Ia /
the body is called an asymmetric top.
Example: A cuboid with three different side lengths a, b, c
(Fig. 5.25b) or the NO2 molecule (Fig. 5.25a).
If two principal moments of inertia are equal the body is called
a symmetric top.
Chapter 5
Example: All bodies with rotational symmetry (circular cylinder
linear molecules but also quadratic cuboids).
Every body with rotational symmetry is a symmetric top,
but a symmetric top has not necessarily a rotational symmetry (for example a quadratic post). The inertial ellipsoid
of a symmetric top is, however, always rotationally symmetric.
We distinguish between
Figure 5.24 Inertial moment about an arbitrary axis
Prolate symmetric tops (Fig. 5.26a) with Ia < Ib D Ic . The
inertial ellipsoid is a stretched rotational ellipsoid where the
diameter along the symmetry axis z is larger than the diameter in the x-y-plane (Fig. 5.27a).
Oblate symmetric tops (Fig. 5.26b) with Ia D Ib < Ic .
The inertial ellipsoid is a squeezed rotational ellipsoid (disc,
Fig. 5.27b).
5.7
Rotation About Free Axes; Spinning Top
143
For
Prolate top
Oblate top
Figure 5.26 Examples of symmetric tops: a prolate and b oblate symmetric
top
Figure 5.28 Angular momentum L and rotational axis are generally not parallel. This is illustrated in a .x ; y ; z / coordinate system that coincides with the
principal axes .a ; b ; c /
the body rotates about one of its principal inertial axes, which
implies that only one of the three components !x , !y , !z is
not zero.
For a symmetric top is L k ! if the body rotates around an
arbitrary axis perpendicular to the symmetry axis.
For an asymmetric top the angular momentum L and the
rotational axis ! D f!x ; !y ; !z g generally point into different directions, because the components Ix , Iy , Iz , in
Eq. 5.42 are different, except if the body rotates aboud
one of its principal axes.
When all three principal moments of inertia are equal, the body
is a spherical top, because in this case its inertial ellipsoid is a
sphere.
Examples: A ball or a cube.
5.7.3
Free Rotational axes
The Eq. 5.42 and Fig. 5.28 give the following important information: Angular momentum L and rotational axis ! point for
all bodies with free axes (where the rotation axis is not fixed by
mountings) only then into the same direction if at least one of
the following conditions is fulfilled.
Ia D Ib D Ic (spherical top) or
The principal axes of a body are therefore also called free axes
because the body can freely rotate about them even if they are
not fixed by mountings.
The experiment shows, however, that a stable rotation is only
possible about the axes of the smallest and the largest moment
of inertia. For the rotation about a free axis of the intermediate
moment of inertia any tiny perturbation makes the motion instable and the body finally flips into a rotation about one of the
other two principal axes.
Examples
1. A cuboid with Ia < Ib < Ic is suspended by a thread
(Fig. 5.29) and can be induced to rotations about the
thread by a small motor which twists the thread. The
cuboid rotates stable if the thread direction coincides
with the axis of the inertial moments Ia or Ic . If it
is suspended in a way that the thread direction coincides with the axis of Ib the cuboid flips for faster
rotations into a rotation around the axis b, as shown in
Fig. 5.29c), it rotates then no longer about the thread
but around the dashed line in Fig. 5.29c), which is a
free axis.
Chapter 5
Figure 5.27 Inertial ellipsoids of a prolate and b oblate symmetric top
Since without external torque the angular momentum L is constant and has a constant orientation in space the body has in
these three cases a space fixed rotational axis and rotates around
this constant axis with constant angular velocity !. Its rotational motion is then identical to the rotation about axes with
fixed mountings (see Sect. 5.6)
144
5 Dynamics of rigid Bodies
Figure 5.31 Stable flight of a discus
5.7.4
J
Euler’s Equations
For an arbitrary orientation of the rotation axis angular momentum L and rotation axis ! are no longer parallel. The motion of
the body is now more complicated. In order to investigate this
motion as seen by an observer sitting in a space-fixed inertial
coordinate system S, we have to describe it in this system S.
Figure 5.29 Rotation of a cuboid about free axes: a stable rotation
about the axis of maximum moment of inertia; b instable rotation about
the axis of median moment of inertia, which jumps into a rotation about
the axis c of maximum moment of inertia (c)
2. A closed chain hangs on a thread and is induced to rotations by a motor (Fig. 5.30). Due to the centrifugal
force the chain widens to a circle which orientates itself in a horizontal plane, because in this position the
rotation takes place about the axis of the maximum
inertial moment and therefore the rotational energy
Erot L2 =2I becomes a minimum. In this stable rotational mode the rotation axis does not coincide with
the direction of the thread.
Chapter 5
Figure 5.30 Rotation of a chain about the axis of maximum moment
of inertia
3. A thrown discus flies stable as long as the rotation proceeds about the symmetry axis (axis of the maximum
moment of inertia) (Fig. 5.31).
The time derivative dL=dt of the angular momentum is equal to
the torque D acting on the body.
dL
DD:
(5.44)
dt S
A coordinate system K where the axes are the principal axes
of the body, which is therefore rigidly connected to the rotating
body rotates with the angular velocity ! against the system S.
In this system the time derivative of L is (see Sect. 3.3.2)
dL
dL
D
.! L/ ;
(5.45)
dt K
dt S
which gives the vector equation for the torque D
dL
DD
C .! L/ :
dt K
(5.46)
This equation corresponds formally to (3.14) if we replace L by
r. Note, that in (5.46) dL=dt is the derivative of L in the body
fixed principal axes system K, while ! is the angular velocity
in the space-fixed system S. If (5.46) is written for the three
components in the direction of the principal axes one obtains
for example for the axis a the relation
dL
C .! L/a
Da D
dt a
d
D .Ia !a / C .!b Lc !c Lb /
dt
d!a
D Ia
C !b Ic !c !c Ib !b ;
dt
where Da is the component of the torque in the direction of the
principal axis a.
Similar equations can be derived for the other two components.
This leads to the Euler-equations
5.7
d!a
C .Ic
dt
d!b
Ib
C .Ia
dt
d!c
Ic
C .Ib
dt
Ia
145
Ib / !c !b D Da
Ic / !a !c D Db
(5.47)
Ia / !b !a D Dc :
The Torque-free Symmetric Top
A symmetric top has two equal principal moments of inertia. If
the symmetry axis of its inertial ellipsoid is the axis c we have
Ia D Ib ¤ Ic . For rotational symmetric bodies the symmetry
axis is also called the figure axis. For a bicycle wheel as symmetric top this is the visible wheel axis (Fig. 5.32). Without any
external torque .D D 0/ the magnitude and the direction of the
angular momentum L is constant. Such a top with D D 0 is
called force-free top although it should be called more correctly
torque-free top.
When the top rotates about its figure axis, L and ! coincide
with this axis. The top rotates as if its axis would be hold by a
stable mounting (see Sect. 5.6). If, however, ! points into an
arbitrary direction which does not coincide with the figure axis
the motion becomes complicated.
For the description of this motion one has to distinguish between
three axes (Fig. 5.33a):
The space-fixed angular momentum axis L
The momentary (not space-fixed) rotation axis !
The figure axis of the symmetric top, which is only spacefixed, if L coincides with this axis.
We can win a qualitative picture for the motion of the figure
axis by the following consideration: For D D 0 the angular
momentum L and the rotational energy Erot are both constant.
Figure 5.32 Bicycle wheel as symmetric top
Figure 5.33 Figure axis c, angular momentum L and momentary rotation
axis !: a Decomposition of ! and L into the components parallel and perpendicular to the figure axis of the symmetric top. b Nutation cone of LE and
!E
Then we obtain from Eq. 5.43
L2x C Ly2 C L2z D const D C1 ;
L2a
Ia
C
L2b
Ib
C
L2c
Ic
D const D C2 :
(5.48a)
(5.48b)
In a space-fixed coordinate system with the axes Lx , Ly , Lz
(5.48a) represents the equation of a sphere. Equation 5.48b
describes an ellipsoid in the principal axes system. Since
the components of the space-fixed vector L must obey both
equations, the endpoint of L can only move on the curve of intersection between sphere and ellipsoid (Fig. 5.34). Since the
ellipsoid is determined by the principal axes system of the top,
i.e. rotates with the top, while L is space-fixed, the top and
therefore also its inertial ellipsoid move in such a way, that the
endpoint of L always remains on the curve of intersection. This
causes a nutation of the figure axis and the momentary rotation
axis ! about the space-fixed axis L (Fig. 5.33b).
While the figure axis can be seen straight forward, the momentary rotation axis can be made visible by an experimental trick:
A circular disc with red, black and white circular segments is
Figure 5.34 The top of the angular momentum vector moves on the intersection curve of angular momentum sphere and energy ellipsoid
Chapter 5
5.7.5
Rotation About Free Axes; Spinning Top
146
5 Dynamics of rigid Bodies
Figure 5.36 Nutation cone, herpolhode cone and polhode cone for a the prolate, b the oblate top
nutation. The solutions show, that the magnitude ! D j!j is
constant in the body-fixed system as well as in the lab-system,
because ! 2 D !a2 C !b2 C !c2 D A2 C C2 D const. However,
the individual components !a and !b can change and therefore
the direction of !.
We separate ! into a component
!c parallel to the figure axis
p
and a component !? D .!a2 C !b2 / D A perpendicular to the
figure axis (Fig. 5.33a) . Squaring (5.42) yields then
L D Ia !? C Ic !c :
Figure 5.35 Visualization of the momentary rotational axis
Chapter 5
centered on the peak of the figure axis (Fig. 5.35). When the
top rotates the three colors blur to an olive-brown mixed color.
Only at the intersection point with the momentary rotation axis
one can see the color of the specific segment which wanders
slowly from red over black to white which indicates the motion
of the momentary rotation axis.
In order to calculate the motion of the figure axis and the momentary rotation axis more quantitatively we apply the Euler
equations (5.47) to the special case D D 0 and Ia D Ib . They
simplify to
!P a C ˝!b D 0 ;
!P b ˝!a D 0 ;
(5.49)
!P c D 0 ;
with the abbreviation ˝ D ..Ic
this system of equations are
Ia /=Ia /!c . The solutions of
˝a D A cos ˝t ; ˝b D A sin ˝t ;
˝c D C with A; C D const ;
(5.50)
as can be proved by inserting (5.50) into (5.49). While ! is the
angular frequency of the spinning top, ˝ is the frequency of the
(5.51)
The figure axis forms a constant angle ˛ against the space-fixed
angular momentum axis with
q
2
2
Ia !a C !b
Ia A
Ia !?
D
D
:
tan ˛ D
Ic !c
Ic
!c
Ic !c
This means that the figure axis migrates on a cone with the full
aperture angle 2˛ around the space-fixed axis L (Fig. 5.33b and
5.36). This cone is called nutation-cone.
The vector ! with its constant magnitude forms the constant
angle ˇ with the figure axis where sin ˇ D !? =! D A=!c . The
momentary rotation axis ! is also wandering on a cone with
the opening angel 2.ˇ ˛/, called herpolhode cone around
the space fixed axis of L. This common motion of figure axis
and momentary rotation axis without external torque is called
nutation.
The common motion of figure axis and momentary rotation axis
can be illustrated by two cones (nutation cone and herpolhode
cone) with different opening angles centered around the spacefixed axis L. A third cone (polhode cone) centered around the
nutating figure axis touches the space fixed herpolhode cone
along the momentary rotation axis and rolls on the outer surface (prolate top Fig. 5.36a) or the inner surface (oblate top
Fig. 5.36b) of the herpolhode cone. The contact line shows the
momentary rotation line !. The apex of the three cones lies in
the center of mass of the nutating body.
5.7
5.7.6
Rotation About Free Axes; Spinning Top
147
Precession of the Symmetric Top
If an external torque D acts on the body the angular momentum
is no longer space-fixed, because of D D dL=dt. Depending on
the direction of D relative to the figure axis the direction and
also the magnitude of L changes with time. At first we will
discuss the simplest case where the body rotates with the angular
velocity ! around its figure axis c and all three axis L, ! and c
coincide. In this case there is no nutation and for D D 0 the
body would rotate with ! D const about the space-fixed figure
axis.
If the top is not supported in its CM, the gravitational force generates a torque
Figure 5.38 Precession of a spinning top
DDrmg;
where r is the vector from the support point to the CM. If the which shows that ! is independent of the space orientation of
p
symmetric top spins with the angular momentum L the torque is the figure axis and depends only on the angular momentum L
perpendicular to L and therefore changes only its direction but and the torque D.
not its magnitude (Fig. 5.37). During the time interval dt the
direction of L changes by the angle d' and we can derive from The general treatment of precession has to take into account the
three vectors (Fig. 5.39)
Fig. 5.37
d'
dL
1. The angular velocity !F about the figure axis
D jLj
:
jdLj D jLj d' ! D D
dt
dt
2. The angular velocity !p of precession around the vertical zaxis
The angular momentum axis and with it the coincidental figure
3.
The
total angular velocity ! D !F C !p
axis rotate with the angular velocity
According to Fig. 5.39 we get the relations:
(5.52)
about an axis perpendicular to the plane of D and L where we
have assumed that !p !. This motion is called precession.
If the figure axis forms the angle ˛ against the vertical direction
the magnitude of the torque is D D m g sin ˛. The change dL
of the angular momentum L is now for dL L (Fig. 5.38)
dL D jLj sin ˛ d' :
For !p ! therefore the equation for the precession frequency
becomes
mgr
mgr sin ˛
!p D
D
;
(5.53)
I! sin ˛
I!
!F D ! e with e D fsin cos '; sin sin '; cos g
!p D 'P f0; 0; 1g
(5.53a)
! D f! sin cos '; ! sin sin '; ! cos C 'g
P :
We separate ! into a component !k parallel and !? perpendicular to the figure axis e.
!k D e .! C 'P cos /
!? D e .! e/
D 'P sin f cos cos ';
(5.53b)
cos sin '; sin g :
Chapter 5
d'
D
D
!p D
D
D
dt
L
I!
θ
Figure 5.37 The gravitational force causes a torque acting on a top, that is not
supported in the Center of Mass
Figure 5.39 Illustration of the equation of motion for a spinning top
148
5 Dynamics of rigid Bodies
The total angular momentum L is
L D Ik !k C mrs2 C I? !?
D Ik e.! C 'P cos /
C I? C mrs2
'P sin f cos cos ';
(5.53c)
cos sin '; sin g ;
where Ik is the moment of inertia for a rotation about the figure
axis and I? about an axis perpendicular to the figure axis.
Because !, d'=dt and do not change with time the time
derivative of (5.53c) is
dL
D Ik .! C 'P cos / eP
dt
I? C mrs2 'P 2 cos sin f sin '; cos '; 0g (5.53d)
D Ik sin .! C 'P cos /'P
I? C mrs2 'P 2 sin cos nO :
where n D f sin '; cos '; 0g is the unit vector in the direction
of the torque D. With dL=dt D D D m g rs sin n we obtain
the equation
!p Ik ! C !p2 cos .Ik
I? / D mgrs :
(5.53e)
which has two solutions for the precession frequency !p . The
difference between the two solutions depends on the difference
Ik I? of the two moments of inertia [5.4].
5.7.7
Superposition of Nutation and
Precession
Chapter 5
In the general case the top does not rotate about its figure axis.
Without external torque the top would perform a nutation around
the space-fixed angular momentum axis L. With an external
torque the angular momentum axis is no longer constant but precesses with the angular velocity !p around an axis through the
underpinning point A parallel to the external force (Fig. 5.39)
while the figure axis performs a nutation around the precessing
axis L. With this combination of precession and nutation the
end of the figure axis describes a complicated path (Fig. 5.40).
The exact form of this trajectory depends on the ratio of nutation
frequency ˝ to the precession frequency !p .
For the demonstration of nutation and precession a special bearing of the top is useful called a gimbal mounting where the figure
axis can be turned into arbitrary directions and the top is always
“torque-free” (Fig. 5.41). This can be realized if the figure axis
is mounted by ball bearings in a frame which can freely rotate
around an axis perpendicular to the figure axis. The mounting
of this axis can again rotate about a vertical axis. If the system is
turned around the vertical axis, the figure axis of the top diverts
from its horizontal direction. Reversal of the turning direction
also reverses the direction of this diversion. If a short push is
applied perpendicular to the top axis, the angular momentum
Figure 5.40 Path of figure axis when precession and mutation are superimposed
axis is forced into another direction and the top starts to nutate.
When a mass m is attached to the first frame, a torque acts on
the top which starts to precess around a vertical axis.
Before the invention of the GPS system the precession of the top
was used for navigation purposes (gyro compass). Its function
is explained in Fig. 5.42, where a rotating disc is suspended in
a mounting B which can turn around a vertical axis a through
the suspension point A. The top axis KA can freely turn only in
a horizontal plane. The center of mass lies below the point A.
Different from a torque-free top rotating around its figure axis
where the figure axis and the angular momentum axis coincide,
for the gyro compass the suspension axis through the point A
is rigidly connected with the earth and participates in the earth
rotation with the angular velocity !. Therefore a torque is acting
on the gyro perpendicular to the drawing plane. The gyro turns
around the axis a until the figure axis is parallel to the rotation
axis of the earth and points into the south-north direction. Now
angular momentum axis L and the forced rotation axis !E are
parallel (Fig. 5.42b), and the torque forcing the turn of the figure
axis becomes zero.
This property can be experimentally studied with the gimbal
mount by simulating the earth rotation by the rotation of the
outer mounting in Fig. 5.41. The figure axis then turns into the
vertical position.
Figure 5.41 Gimbal mount of a symmetric top
5.8 The Earth as Symmetric Top
149
Figure 5.43 The earth as symmetric top. The arrows indicate the difference
forces F1 F2 ath the Centers of Mass S1 and S2 of the two opposite sides of
the bulge
At the earth equator !E and L are parallel. This is not the case
for a point P on another circle of latitude because L has to lie
in a horizontal plane. However, also in this case the gyro adjusts itself in such a way that the component L parallel to !E
becomes maximum. The vector L becomes the tangent to the
circle of longitude, points therefore again to the north. Only at
the two poles of the earth the gyro fails, because here L is always
perpendicular to !E .
5.8
The Earth as Symmetric Top
In a good approximation the earth can be described by a clinched
rotational ellipsoid, i.e. an oblate symmetric top with Ia D Ib <
Ic . The equator diameter is with 12 756 km by about 43 km
larger than the pole diameter with 12 713 km. This clinch is
caused by the centrifugal force due to the rotation of the earth
(see Sect. 6.6). For the following considerations we will compose this oblate ellipsoid by a sphere plus additional bulges
which have their maximum thickness at the equator (red area
in Fig. 5.43).
Because of the inclination of the earth axis .' D 90ı 23:5ı D
66:5ı / against the ecliptic (orbital plane of the earth’s motion
around the sun) the two centers of mass S1 of the bulge towards
the sun and S2 of the bulge opposite to the sun are located above
and below the ecliptic (Fig. 5.43) in contrast to the center of
mass S of the sphere which lies in the ecliptic. While for the
Besides the gravitational force between earth and sun the attraction between moon and earth must be taken into account. The
calculation is here more complicated because the orbital plane
of the moon around the earth is inclined by an angle of 5:1ı
against the ecliptic. The calculation shows that the influence on
the earth is of the same order of magnitude than that of the sun.
Altogether both torques cause the lunar-solar precession where
the earth axis propagates on a cone with an opening angle of
2 23:5ı by an angle ' 5000 per year which gives a precession period of about 25 750 years for ' D 2 (Platonic year).
Within a Platonic year the cone is once circulated. The elongation of the earth axis describes a circle on the celestial sphere
around the ecliptic pole (Fig. 5.44).
Remark.
This precession causes a turn of the intersecting line between ecliptic and equatorial plane by 360ı within
25 850 years. This shifts the vernal equinox (where day and
night both last 12 hours) by about 5000 per year. It causes furthermore a shift of the signs of the zodiac between their naming
2000 years ago and today by about one month. For example the
real constellation of the Gemini (twins) coincides in our times
with the sign of the zodiac Cancer. This is unknown to many astrological oriented people who come into trouble if they should
explain whether the real stars or the signs of the zodiac are responsible for the fate of a person.
The precession of the earth axis described above is not uniform
because of the following reasons:
Chapter 5
Figure 5.42 Gyro-compass: a mount, b direction of L and !E at the equator
and for higher lattitudes
mass mE of the sphere concentrated in S the centripetal force
F1 D GmE Mˇ =r2 due to the gravitational attraction between
earth and sun is just compensated by the equal but opposite centrifugal force F2 D mE vE2 =r this is no longer true for the centers
of mass S1 and S2 of the bulges. Since S1 is closer to the sun the
centripetal force predominates while for S2 the centrifugal force
prevails. Since the net forces for S1 and S2 are antiparallel they
form a couple of forces which act as a torque on the earth and
cause the earth axis to precess (solar precession).
150
5 Dynamics of rigid Bodies
Figure 5.44 Due to the precession of the earth its axis traverses a circle on the
celestial sphere around the ecliptic pole. In 1950 it pointed towards the pole star
Because of the inclination of the earth axis the torque exerted
by the sun changes periodically during the year (Fig. 5.45).
It is maximal on December 22nd and June 21st and minimal
at March 21st and September 23rd
The torque exerted by the moon changes because the inclination of the moon’s orbital plane changes with a period of
9.3 years.
Also the other planets cause a small torque acting on the
earth. Because the relative distances to the earth change in
time, this causes a tiny variation of the precession.
The motion of the earth around the sun proceeds on an elliptical path and therefore the distance r between earth and sun
changes periodically. It is minimum in December and maximum in June (Fig. 5.45). Therefore the gravitational force
acting on the earth changes correspondingly.
Astronomers call these short-period fluctuations of the precession nutations although they are strictly speaking no nutations
but perturbations of the precession.
Chapter 5
There are real torque-free nutations superimposed on the complicated precession. They are caused by the fact, that the figure
axis of the earth and the rotation axis do not exactly coincide (Fig. 5.46). The figure axes (south-north pole intersection)
Figure 5.45 Position of the earth axis during the revolution of the earth about
the sun. Note, that the direction of the angular momentum does not change
during the year
Figure 5.46 Nutation of the earth axis
therefore nutates around the precessing angular momentum axis
with a measure period of about 303 days. On the other hand the
nutation period is
2 Ia
Tnut D
:
(5.54)
! Ic Ia
From the measured nutation period one can therefore determine
the difference Ic Ia of the inertial moments [5.5a, 5.5b].
Since the earth is no rigid body the mass distribution and therefore the inertial moments can change, for instance by volcanic
eruptions or by convective currents in the liquid interior of the
earth [5.6a, 5.6b]. This causes small fluctuations of the nutation.
In Fig. 5.47 the wandering of the north-pole of the rotational axis
during the year 1957 is shown.
The above discussion has shown, that the more precise measurements have been made, the more different influences on the
motion of the earth axis have to be taken into account. Even
today there are discussions about the best model for the earth
motion [5.7, 5.8].
Figure 5.47 Migration of the north pole of the earth’s rotation axis during the
year 1957 about the average position of the period 1900–1905. One second of
arc corresponds to about 30 m
Problems
151
Summary
The model of the extended rigid body neglects all internal
motions (Deformations and vibrations). The center of mass
S has the coordinates
Z
Z
1
1
rS D
r%.r/dV D
rdV for % D const :
M
V
V
The motion of a free rigid body can be always composed of
a translation of the center of mass S with the velocity vs and
a rotation of the body around S with the angular velocity !.
The motion of the extended body has therefore 6 degrees of
freedom.
For the motion of an extended body not only magnitude and
direction of the force acting on the body are important but
also the point of action on the body.
An arbitrary force acting on an extended body can always
be composed of a force acting on the center of mass S
(translational acceleration) and a couple of forces causing an
accelerated rotation.
The moment of inertia (rotational inertia) for Ra rotation about
2
%dV where
an axis through the center of mass S is Is D r?
r? is the distance of the volume element dV from the rotation
axis. The moment of inertia for a rotation around an arbitrary axis with a distance a from the parallel axis through S
is I D Is C Ma2 (parallel axis theorem or Steiner’s theorem).
The kinetic energy of the rotational motion is Erot D 21 I! 2 .
The equation of motion for a body rotating about a spacefixed axis is Dk D I d!=dt, where Dk is the component of
the torque parallel to the rotation axis.
The moment of inertia Is depends on the direction of the rotation axis relative to a selected axis of the body. It can be
described by a tensor. The directions of the axes with the
maximum and the minimum inertial moment determine the
principal axes system. In this system the tensor is diagonal.
The diagonal elements are the principal moments of inertia.
If two of the principal moments are equal, the body is a symmetric top. If all three are equal the body is a spherical top.
Angular momentum L and angular velocity ! are related by
L D I !, where I is the inertial tensor, which is diagonal
in the principal axes system. In the general case L and ! are
not parallel.
If the body rotates about a principal axis, L and ! are parallel
and without external torque their directions are space-fixed.
For an arbitrary direction of ! the momentary rotation axis !
nutates around the angular momentum axis which is spacefixed without external torque.
Under the action of an external torque the angular momentum axis L precedes around the external force and in addition
the momentary rotation axis nutates around L. The relation
between L and D is D D dL=dt.
The general motion of a top is completely described by the
Euler-equations.
The earth can be approximately described by a symmetric
top, which rotates about the axis of its maximum moment
of inertia. The vector sum of the gravity forces exerted by
the sun, the moon and the planets results in a torque which
causes a periodic precession of the earth axis with a period
of 25 850 years. In addition changes of the mass distribution in the earth cause a small difference between symmetry
axis and momentary rotation axis. Therefor the earth axis
performs an irregular nutation around the symmetry axis.
5.1
Determine the center of mass of a homogeneous sector
of a sphere with radius R and opening angle ˛.
5.3
A cylindrical disc with radius R and mass M rotates
with ! D 2 10 s 1 about the symmetry axis (R D 10 cm,
M D 0:1 kg).
5.2
What are moment of inertia, angular momentum and ro- a) Calculate the angular momentum L and the rotational energy
tational energy of our earth
Erot .
a) If the density % is constant for the whole earth
b) a bug with m D 10 g falls vertical down onto the edge of the
b) If for r R=2 the homogeneous density %1 is twice the dendisc and holds itself tight. What is the change of L and Erot ?
sity %2 for r > R=2?
c) The bug now creeps slowly in radial direction to the center
c) By how much would the angular velocity of the earth change,
of the disc. How large are now !.r/, I.r/ and Erot .r/ as a
if all people on earth (n D 5109 with m D 70 kg each) would
function of the distance r from the center r D 0?
gather at the equator and would start at the same time to run
into the east direction with an acceleration a D 2 m=s2 ?
Chapter 5
Problems
152
5 Dynamics of rigid Bodies
5.4
The mass density % of a circular cylinder (radius R,
height H) increases in the radial direction as %.r/ D %0 .1 C
.r=R/2 /.
a) How large is its inertial moment for the rotation about the
symmetry axis for R D 10 cm and %0 D 2 kg=dm3?
b) How long does it take for the cylinder to roll down an inclined plane with ˛ D 10ı from h D 1 m to h D 0?
5.5
Calculate the rotational energy of the Na3 -molecule
composed of 3 Na atoms (m D 23 AMU) which form an isosceles triangle with the apex angle ˛ D 79ı and a side length of
d D 0:32 nm when it rotates around
p the three principal axes
with the angular momentum L D l.l C 1/ „. Determine at
first the three axis and the center of mass.
5.6
A wooden rod with mass M D 1 kg and a length l D
0:4 m, which is initially at rest, can freely rotate about a vertical
axis through the center of mass. The end of the rod is hit by
a bullet (m D 0:01 kg) with the velocity v D 200 m=s, which
moves in the horizontal plane perpendicular to the rod and to the
rotation axis and which gets stuck in the wood.
What are the angular velocity ! and the rotational energy Erot of
the rod after the collision? Which fraction of the kinetic energy
of the bullet has been converted to heat?
5.7
A homogeneous circular disc with mass m and radius R
rotates with constant velocity ! around a fixed axis through the
center of mass S perpendicular to the disc plane. At the time
t D 0 a torque D D D0 e at starts to act on the disc. What
is the time dependence !.t/ of the angular velocity? Numerical
example: !0 D 10 s 1 , m D 2 kg, R D 10 cm, a D 0:1 s 1 ,
D0 D 0:2 Nm.
5.8
A full cylinder and a hollow cylinder with a thin wall
and equal outer diameters roll with equal angular velocity !0 on
a horizontal plane and then role up an inclined plane. At which
height h do they return? (Friction should be neglected), numerical example: R D 0:1 m, !0 D 15 s 1 .
References
5.1. Ph. Dennery, A. Krzwicki, Mathematics for Physicists.
(Dover Publications, 1996)
5.2. J.C. Kolecki, An Introduction to Tensors for Students of
Physics and Engineering.
5.3. https://en.wikipedia.org/wiki/Moment_of_inertia#
Principal_moments_of_inertia
5.4. W. Winn, Introduction to Understandable Physics. (Authort House, 2010)
5.5a. K. Lambrecht, The earth’s variable Rotation. Geophysical
Causes and Consequences. (Cambridge University Press,
Cambridge, 1980)
5.5b. W.H. Munk, G.J. MacDonald, The Rotation of the Earth. A
Geophysical Discussion. (Cambridge Monographs, 2010)
5.6a. St. Marshak, Earth: Portrait of a Planet, 4th ed.
(W.W. Norton & Company, 2011)
5.6b. I. Jackson, The Earth’s Mantle. (Cambridge University
Press, 2000)
5.7. V. Dehant, P.M. Mathews, Precession, Nutation and Wobble of the Earth. (Cambridge University Press, 2015)
5.8. A. Fothergill, D. Attenborough, Planet Earth: As You’ve
Never Seen it Before. (BBC Books, 2006)
Chapter 5
Real Solid and Liquid Bodies
6.1
Atomic Model of the Different Aggregate States . . . . . . . . . . . . . 154
6.2
Deformable Solid Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . 155
6.3
Static Liquids; Hydrostatics . . . . . . . . . . . . . . . . . . . . . . . . . . 162
6.4
Phenomena at Liquid Surfaces . . . . . . . . . . . . . . . . . . . . . . . . 166
6.5
Friction Between Solid Bodies . . . . . . . . . . . . . . . . . . . . . . . . 171
6.6
The Earth as Deformable Body . . . . . . . . . . . . . . . . . . . . . . . . 174
6
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181
Chapter 6
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181
© Springer International Publishing Switzerland 2017
W. Demtröder, Mechanics and Thermodynamics, Undergraduate Lecture Notes in Physics, DOI 10.1007/978-3-319-27877-3_6
153
154
6 Real Solid and Liquid Bodies
In this chapter, we will proceed with the stepwise refinement of
our “model of reality”. We will take into account the experimental fact that extended bodies can change their shape under
the influence of external forces. We will also discuss the important question why and under which conditions real bodies can
exist in different aggregation states as solids, liquids or gases.
We will see, that an atomic model, which considers the different
interactions between the atoms, can at least qualitatively explain
all observed phenomena. For a quantitative description, a more
profound knowledge about the atomic structure is demanded.
The quantitative calculation of the detailed characteristics of
solids or liquids is still not trivial, even with fast computers
and sophisticated programs, because of the enormous number
(1023 =kg) of atoms involved. Here symmetry considerations
are helpful to facilitate the description.
If all physical characteristics of an extended body (density, elasticity, hardness etc.) are constant within the body, we call it
a homogeneous body. Are they also independent of the direction the body is isotropic. A liquid metal is an example of
an isotropic and homogeneous body while a NaCl-crystal (table
salt) is homogeneous but not isotropic.
6.1
Atomic Model of the Different
Aggregate States
Many experiments have proved that all macroscopic bodies are
composed of atoms or molecules (see Vol. 3). Between two
atoms, which consist of a positively charged small nucleus and
a negatively charged extended electron cloud, attractive as well
as repulsive interactions can occur. The superposition of all
these interactions results in a force F.r/ and a potential energy
Ep .r/ which depend on the distance r between the interacting
atoms and which are qualitatively depicted in Fig. 6.1. At the
equilibrium distance r0 , the potential energy Ep .r/ shows a minimum and the force F.r/ D grad Ep becomes zero. For shorter
distances r < r0 the repulsive forces dominate and for larger distances r > r0 the attractive forces. For both cases, the potential
energy increases. When an atom A is surrounded by many other
atoms Ai at distances ri the total force F acting on A is the vector
sum of all individual forces Fi :
FD
X
Figure 6.1 Qualitative dependence of potential energy Ep .r / and force F .r /
between two atoms as a function of distance r between the nuclei of two adjacent atoms
vectors
ri D n1i a C n2i b C n3i c ;
(6.1)
where the n˛i are integers and the basis vectors a; b; c define the
unit cell (or elementary cell) in the crystalline solid. They are
marked as red vectors in Fig. 6.2. Their magnitudes and directions define the crystal structure of the solid. The forces between
the atoms can be modelled by elastic springs (Fig. 6.3) where
the restoring force constants ki can be different in the different directions. At the absolute temperature T the atoms vibrate
about their equilibrium position r0 . Their mean kinetic energy
is hEkin i D .1=2/kT per degree of freedom (see Sect. 7.3) where
the equilibrium positions correspond to the minimum of the potential energy Ep in Fig. 6.1. For temperatures far below the
melting temperature, hEkin i is small compared to the magnitude
jEp .r0 /j of the potential energy at the equilibrium position which
means that the atoms cannot leave their equilibrium positions.
If the temperature rises above the melting temperature the
mean kinetic energy hEkin i becomes larger than the binding energy EB D Ep .r0 /. The atoms cannot be kept any
Fi .ri / :
The resulting potential energy Ep of atom A depends on the spatial distribution of the surrounding atoms Ai and is related to the
force F by F D grad Ep .
Chapter 6
In crystalline solids the atoms are arranged in regular lattices
(Fig. 6.2) while in amorphous solids they sit on more or less
statistically distributed sites. Examples for the first cases are
NaCl-crystals, or noble gas crystals at low temperatures, while
examples for amorphous solids are glasses or amorphous silicon, which is used for solar cells.
When we place the atom A in a crystalline solid at the origin
r D 0 of our coordinate system the atoms Ai have the position
Figure 6.2 Regular structure of a solid crystal. The basis vectors a, b, c form
the elementary cell with volume VE D .a b/ c. The position vector of the
point A is rA D 2a C b C c
6.2 Deformable Solid Bodies
155
Figure 6.3 Spring model of a solid crystal. The restoring force is for an isotropic
crystal equal for all three directions a, b, c, for an unisotropic crystal they differ
longer on their positions ri but can diffuse around. The
crystalline solid melts and converts to the liquid state.
Also in the liquid state the minimum of the potential energy remains at the mean distance hr0 i between the atoms. This means
that the density in the liquid state is not very different from that
in the solid state. However, now a single atom is no longer
bound to a fixed position but can move freely within the liquid. Nevertheless, there is still a certain order. If one plots the
probability W.r/ that an atom A occupies a position with the
distance r from its neighbors (Fig. 6.4) a pronounced maximum
is found at r D r0 which is close to the minimum distance r0
in the crystalline solid. Similar to the amorphous solid the liquid has a short range order, while a crystalline solid has a long
range order, because it is possible to assign a definite position
r D n1 a C n2 b C n3 c (with integers ni ) to each atom regardless
how far away it is (Fig. 6.2). While the crystalline solid can be
described by the spring model of Fig. 6.3, many features of liquids can be modelled by the string model of Fig. 6.5. Here the
atoms are connected by strings of constant lengths where the directions can be arbitrarily altered. The balls in this mechanical
model can move similar to the atoms in a liquid.
A further increase of the temperature above the boiling point
makes the mean kinetic energy large compared to the magnitude
of the potential energy. The potential energy is then negligible
Figure 6.5 Atomic model of a liquid. The balls are connected with each other
by strings. The model illustrates the free mobility of each atom
and the atoms can move freely. They form a gas that fills the total accessible volume. The interaction energy is only noticeable
during collisions of the atoms with each other.
The mean distance hri between the atoms or molecules and
therefore also the density % D M=V of the gas with total mass M
depends on the volume V which is accessible to the N D M=m
molecules with mass m. At normal pressure p D 1 bar the density of the gas is about three orders of magnitude smaller than
that of solids or liquids.
Examples
The density of air at p D 1 bar and T D 300 K. 20 ı C/
is % D 1:24 kg=m3 , while the density of water is about
103 kg=m3 and that of lead is 11:3 103 kg=m3.
J
The considerations above show that the aggregation state of material depends on the ratio hEkin i=Ep and therefore on the temperature and on the binding energy of the atoms or molecules of
the body.
P(r)
We will now discuss the most important characteristic features
of the different aggregation states in a phenomenological manner. The more detailed treatment is given in Vol. 3.
6.2
r0
r
Figure 6.4 Probability P .r / that an atom A1 in a liquid has the distance r to
an arbitrary other atom A2
Deformable Solid Bodies
External forces can change the shape of solid bodies. If the
body returns to its original shape after the exposure to the external forces we call it elastic. For a plastic body the deformation
remains.
Chapter 6
P(r)
156
6 Real Solid and Liquid Bodies
6.2.1
Hooke’s Law
If a pulling force acts onto the end face of an elastic rod with
length L and cross section A, which is hold tight at the other end
x D 0 (Fig. 6.6) the length L is prolonged by L. The linear
relation between the magnitude F D jFj of the force and the
prolongation L
F D E A L=L
(6.2)
is called Hooke’s law, which is valid for sufficiently small relative length changes L=L. The proportional factor E is the
elastic modulus with the dimension N=m2 . For technical applications often the dimension kN=mm2 D 109 N=m2 is used.
Table 6.1 gives numerical values for some materials.
For materials with a large elastic modulus E one needs
a large force to achieve a given relative change of length
L=L. With other words: Materials with a large value of
E show for a given force a small relative length change.
Introducing the tensile stress (D pulling force=cross section A)
D F=A
and the relative stretch or strain " D L=L Hooke’s law can be
written in the clearer form
DE":
(6.2a)
For sufficiently small relative stretches ", tensile stress and strain
are proportional. In this proportional range the distances between neighboring atoms vary only within a small range around
r0 (Fig. 6.1) where the interatomic force F.r/ D a r is approximately a linear function of the distance r and the potential
energy Ep .r/ can be approximated by a parabola.
Note: This linear relation is only an approximation for small
values of ". For larger " nonlinear forces appear that cannot be
neglected.
Table 6.1 Elastic constants of some solid materials. E D elastic modulus;
G D modulus of shear; K D compressibility modulus; D inverse contraction
number D Poisson number; [6.1]
Material
Aluminium
Cast iron
Ferrite steel
Stainless steel
Copper
Tungsten
Lead
Fused silica
Water ice ( 4 ı C)
E Œ109 N=m2 G Œ109 N=m2
71
26
64–181
25–71
108–212
42–83
200
80
125
46
407
158
19
7
75
32
10
3.6
K Œ109 N=m2
74
48–137
82–161
167
139
323
53
38
9
0.34
0.28
0.28
0.3
0.35
0.29
0.44
0.17
0.33
and choosing the minimum of Ep .r/ as Ep .r0 / D 0, the first
two members of the Taylor series (6.3a) vanish because also
@Ep =@rjrD0 D 0. This reduces (6.3a) to
2
@ Ep
1
Ep .r/ D .r r0 /2
2
@r2 rDr0
(6.3b)
3
@ Ep
1
C .r r0 /3
C
:
:
:
:
6
@r3 rDr0
For small elongations .r r0 / all higher order terms with powers n 3 can be neglected and (6.3b) gives for the force
F D grad Ep the linear relation of Hooke’s law. For larger
elongations, however, the higher order terms can no longer be
neglected and must be taken into account.
Surpassing the linear range at the point P in Fig. 6.7 the relative
stretch " increases faster than the tensile stress (Fig. 6.7). The
material is still elastic until the point F, i.e. it returns nearly
to its initial length after the stress is released. Above the yield
point F, internal shifts of atomic layers (lattice planes) occur
(Fig. 6.8). The body becomes malleable and the plastic flow
starts. Permanent changes of shape remain after the termination
of the external force. While for the elastic stretch the distances
r between the atoms increases linearly by r .L=L/r0 the
Expanding Ep .r/ into a Taylor series around the equilibrium position r0
1
X
.r r0 /n @n Ep
Ep .r/ D
(6.3a)
nŠ
@rn rDr0
nD0
Figure 6.7 Relative length change " of a body caused by an external tensile
force . Beyond the point P the linear elongation changes to a nonlinear one.
The point F marks the yield point, the point Z the tear point
Chapter 6
Figure 6.6 A rod fixed at x D 0 expands under the action of a force FE by
L D F L =.E A /
Figure 6.8 Model of plastic flow of a solid explained by shift of atomic planes
6.2 Deformable Solid Bodies
157
Figure 6.9 Atomic model of elastic expansion of a solid. The atom A moves
in the potential well but does not leave it
flow process can be achieved by a shift of the atomic planes
against each other, as illustrated in Fig. 6.8.
This can be made clear by Fig. 6.9 which shows the potential energy of an atom A in an atomic plane. For small length changes
all atoms remain within their potential well. For larger changes
the atomic plane is shifted against the adjacent plane. The atom
can move from one minimum into the next only if the external
pulling force is sufficiently large to lift all atoms of this plane
over the potential hills. Since such a shift changes the distance
between atoms only slightly, the minima in Fig. 6.9 are much
shallower than the minimum of the potential energy between
two atoms in Fig. 6.1. The barrier height and the modulation
period of Ep .L/ depends for an anisotropic crystal on the direction of the pulling force relative to the crystal axes.
In a real crystal lattice, defects and dislocations are present
which influence the flow process and can increase or decrease
the renitence against stretches and shifts of crystal planes against
each other.
Transverse Contraction
When a rod is stretched by an external pulling force, not only
the length L in the direction of the force is prolonged but also
the cross section decreases (Fig. 6.10). For a rod with length L
and quadratic cross section d 2 the change V of its volume V
under a length stretch L > 0 and d < 0 is
V D .d C d/2 .L C L/
d2 L
D d L C 2L dd
C Ld 2 C 2ddL C Ld 2 :
For small deformations (L L and d d), the terms in
the bracket can be neglected, because they converge quadratic
or even cubic towards zero for L ! 0. This reduces the above
equation to
The quantity
Def
D
d
d
L
L
Since a pulling force increases the volume (V > 0), we obtain
for the condition < 0:5. According to Hooke’s law (6.2a)
is L=L D =E. Inserting this into (6.6a) gives
V
D .1
V
E
2/ :
(6.6b)
If a pressure instead of a tensile stress is exerted onto the end
faces of a rod, L and V become negative but d positive because the rod is compressed in the length direction which causes
an increase of its cross section. The resulting relative volume
change can be obtained from (6.6b) when is replaced by the
pressure p.
In both cases is > 0 because for the pulling force is L > 0
and d < 0 while in case of a pressure is L < 0 and d >
0, which means that the ratio in the bracket in (6.6a) does not
change its sign.
If the body is exerted to an isotropic pressure p D , which
acts onto all sides of the body, the resulting volume change can
be obtained by the following consideration.
2
V
L
d
C2
:
V
L
d
is called the transverse contraction ratio because it is the ratio
of transverse contraction to longitudinal elongation. The relative
volume change is then expressed as
V
L
2d=d
D
1C
D " .1 2/ :
(6.6a)
V
L
L=L
(6.4)
The pressure acting on the end faces d2 decreases the length L
by L D L p=E, the pressure acting on the sides decreases
the transverse edge length by d D d p=E. However, because of the transverse action on the elongation this transverse
contraction increases the length by L D C L p=E. Taking
both effects in account the length L under the action of a uniform
pressure p changes by
L D
.L p=E/ .1
2/ :
In a similar way the transverse dimension d is changed by
(6.5)
d D
.d p=E/ .1
2/ :
(6.7)
Chapter 6
6.2.2
Figure 6.10 Transverse contraction under the influence of longitudinal tensile
stress
158
6 Real Solid and Liquid Bodies
Since L L and d d the higher order terms in the
expansion of V=V D L=L C 2d=d can be neglected and
we obtain
V
L
2d
D
C
D
V
L
d
3p
.1
E
2/ :
(6.8)
Rearrangement of (6.10) yields
E=3K D 1
2 :
(6.12b)
The division of (6.12a) by (6.12b) gives
Introducing the compressibility modulus K by the equation
pD
V
K
V
(6.9)
and the coefficient of compressibility D 1=K, Eq. 6.8 can be
written as
1
3
D
D .1 2/ :
(6.10)
K
E
This gives the relations between compressibility modulus K, coefficient of compressibility , elastic modulus E and transverse
contraction number (Poisson number) .
6.2.3
2G=3K D
Shearing and Torsion Module
1 2
:
1C
(6.12c)
Example
Torsion of a wire: We assume a force F that acts tangential on a cylinder with radius R and length L and which
causes a torsion of the cylinder (Fig. 6.12). We subdivide
the cylinder in thin radial cylindrical shucks between the
radii r and r Cdr and in axial strips with the angular width
ı'. If the upper end of the cylinder twists under the action of a torsional force F by the angle ' the prismatic
column marked in red in Fig. 6.12 experiences a shear by
the angle ˛. For r ' L one finds ˛ D r '=L.
A shear force F is a force, which acts on a body parallel to a
plane surface A (Fig. 6.11). The shearing stress
D F=A
is the tangential shearing force F per unit surface area A. The
result of the action of a shearing stress is a tilt of the axis of
the cuboid in Fig. 6.11 by an angle ˛. For sufficiently small
tilting angles ˛, the experiments prove that the tilting angle ˛ is
proportional to the applied shearing stress.
T DG˛ :
(6.11)
The constant G is called modulus of shear (or modulus of torsion).
Since the restoring forces under deformations of an elastic body
are due to interatomic forces, all elastic constants E, , K and G
must be related to each other.
As can be proved [6.2] for isotropic bodies the following relation holds:
E=2G D 1 C :
(6.12a)
Figure 6.12 Torsion of a circular cylinder
The shearing stress necessary to achieve this torsion is
according to (6.11)
D G r '=L :
Chapter 6
Since all surface elements of the upper annulus with the
area 2r dr are twisted by the same angle ' against their
position for D 0 the amount of the force dF necessary
for the shear of the whole cylindrical shuck is
dF D 2r dr D 2r2 dr ' g=L
and the corresponding torque is
Figure 6.11 Shearing of a cube under the action of shearing stress
dD D r dF D 2 r3 dr G '=L :
6.2 Deformable Solid Bodies
159
The torsion of the whole cylinder with radius R by the
angle ' is then accomplished by the torque
2G'
DD
L
ZR
0
r3 dr D
R4
G ' :
2 L
(6.13)
At equilibrium, the retro-driving torque, due to the elastic
twist of the cylinder, must be equal to the external torque.
This gives for the retro-driving torque
D D
Dr '
with Dr D
R4
G
:
2 L
(6.14)
The constant Dr , which depends on the shear modulus
G and gives the torque per unit angle, is called restoring
torque.
If a body with the moment of inertia I with respect to the
symmetry axis, is fixed to the lower end of a wire this
torsional pendulum performs rotary oscillations after the
wire has been twisted (see Sect. 5.6.2) with the oscillation
period
s
s
I
2 2L I
D 2
:
(6.15)
T D 2
Dr
R
G
Such a torsional pendulum is a very sensitive device
for measurements of small torques. Examples are the
Eötvös’s torsional pendulum for the measurement of
Newton’s gravitational constant (see Sect. 2.9.6), Coulomb’s torsional pendulum for the measurement of the
electric force between charges (Vol. 2, Chap. 1) and many
modifications of Galvanometers for the measurement of
J
small electric currents (Vol. 2, Chap. 2).
Figure 6.13 Bending of a rod which is clamped at one end
While the length of the central curve does not change by the
bending (neutral filament) the length `.z/ of a layer in the upper
half of the rod (z > 0) increases by the amount
`.z/ D z ' D z `=r :
A corresponding layer in the lower half .z < 0/ is shortened by
this amount. In order to achieve such an increase of the length `
a pulling force per unit cross section (tensile stress)
D E `=` D z E=r
has to be applied, while for the layer in the lower half .z < 0/ a
corresponding pressure
pD D
jzj E=r
is necessary. The force on a rectangular element with width b,
heights dz and distance z from the neutral filament at z D 0 is
then
bE
dF D bdz D
zdz :
(6.16a)
r
The force causes a torque
6.2.4
Bending of a Balk
dDy D
bE 2
z dz
r
(6.16b)
in the y-direction.
If a rectangular rod with cross section A D d b is clamped
at x D 0 and a force F0 is acting in the z-direction on the
other end at x D L, the bending of the rod can be approximately
described by approximating a short curved section of the rod
by a circle. When the central dashed curve in Fig. 6.14 has the
radius of curvature r, the length of the upper edge of the rod
section is .r C d=2/', that of the lower edge is .r d=2/'.
Figure 6.14 a Definition of the neutral filament at z D 0. b Illustration of
(6.17)
Chapter 6
For technical constructions (buildings, bridges, etc.) the bending of balks under the influence of suspended weights represents
an important problem and can decide about the stability of the
construction. We will illustrate the problem with a simple example, where a rod with a rectangular cross section A D d b
is clamped at one end while a force acts on the other free end
(Fig. 6.13). The calculation of such bending for arbitrary bodies
is very complicated and can be accomplished only numerically.
160
6 Real Solid and Liquid Bodies
Integration of this infinitesimal torque dDy over the total height
d of the rod gives
Dy D
Cd=2
Z
bE
r
z2 dz D
Ed 3 b
:
12r
(6.17)
d=2
This torque is caused by the vertical force F0 at the end x D L
of the rod. On the other hand the torque induced by the force F0
on a selected part of the rod at the position x is
Dy D F0 .L
x/
with
F0 D jF0 j :
(6.18)
The equilibrium position of the bent rod is determined by the
condition that the restoring torque of the elastic material (6.17)
must just compensate the torque (6.18). This yields the curvature 1=r of the rod at the distance x from the fixed end at x D 0:
12F0
.L
Ed 3 b
1=r D
x/ :
(6.19)
The neutral filament which is without a force the horizontal
straight line z D 0, becomes the bent curve z D z.x/. As is
shown in books on differential geometry the relation between
the curvature 1=r and the function z D z.x/ is
1=r D
z00 .x/
Œ1 C
z0 .x/2 3=2
;
where z0 .x/ D dz=dx and z00 .x/ D d2 z=dx2 . For small curvatures
is z0 .x/ 1 and therefore 1=r can be approximated by 1=r
z00 .x/. Integration of the equation
z00 .x/ D a .L
x/
with a D 12F0 =Ed 3 b
derived from (6.17) and (6.18), gives with the boundary conditions z.0/ D 0 and z0 .0/ D 0 the function of the neutral filament
of the strained rod
a
a 3
z.x/ D Lx2
x with a < 0 :
2
6
L3
F0
E d3 b
(6.20)
compared to z.L/ D 0 for the straight rod. The bend of the rod
s D z.L/ is also called pitch of deflection sag.
Chapter 6
The bend of a rectangular rod with length L and thickness
d is proportional to L3 and to 1=d 3 .
For x D 0 (at the clamped end of the rod) the curvature 1=r D
z00 .0/ D a L becomes maximum. The tensile stress at the upper
edge of the rod .z D Cd=2/ is
max D
Ed
12F0 L
D
:
2r
2d 2 b
As soon as max exceeds the fracture stress of the rod material,
the rod starts to notch at the upper edge at z D Cd=2 and x D 0
and the rod cracks.
Remark.
The bend of rods with arbitrary cross section
R
A D dydz can be treated in a similar way if one introduces
the geometrical moment of inertia (second moment of area)
Def
IF D
“
z2 dydz ;
(6.22a)
where z is the direction of the acting force F. For the rod with
rectangular cross section A D d b (Fig. 6.15b) we get
IF D
Cd=2
Z
Cb=2
Z
z2 dydz D
1 3
d b:
12
(6.22b)
zD d=2 yD b=2
The maximum deflection smax (pitch of deflection sag) is, in accordance with (6.20),
The free end of the rod at x D L bends by
smax D z.L/ D 4
Figure 6.15 Geometrical moments of inertia for some selected cross sections.
IFy : bending about the y -axis; IFz : about the z -axis
(6.21)
smax D
L3
F:
3E IF
(6.23)
For a rod with circular cross section (radius R) (Fig. 6.15e) we
get
1
IF D R4
4
and therefore
smax D
4L3
F:
3ER4
(6.24)
For a double T-beam (Fig. 6.15d) is
IF D
1 3
b1 d1
12
b2 d23 :
(6.25)
6.2 Deformable Solid Bodies
161
Figure 6.16 Bending of a rod, which is clamped at both ends
Figure 6.17 Mechanical hysteresis curve
smax D
1 L3
F :
4E d 3 b
with cross section A increases by L the necessary work is
ZL
ZL
W D FdL D A dL
(6.26)
0
Note, that here the sag is smaller by a factor 16! compared
to the rod which is fixed only at one end (because of the L3
dependence). The force is now distributed onto the two halves
of the rod with L=2 each.
6.2.5
Elastic Hysteresis; Energy of
Deformation
When a rod without deformation is exposed to an external tensile stress between the end faces the relative stretch " D L=L
follows the curve OA in Fig. 6.17. For small values of " the
curve ."/ is linear until a point is reached where the deformation is no longer reversible and ."/ rises slower than linear. The
point A in Fig. 6.17 is already in the irreversible region. This
means that the curve ."/ does not return on the same curve
when the stress is released but arrives at the point B for D 0.
This phenomenon is called elastic hysteresis, because the stressfree state of the body depends on its past history (the Greek word
hysteresis means: lagging behind i.e. the length change lags behind the applied stress).
When the body in the state B is exposed to an external pressure
p D onto the two end faces the curve ."/ reaches the point
C where it is also nonlinear. Releasing the pressure " does not
become zero for D 0 but arrives at the point D in Fig. 6.17,
which corresponds in the atomic model of Fig. 6.1 to an interatomic distance r < r0 .
Under a periodic change between stretch and compression the
function ."/ passes through the closed loop ABCDA, which
is called the elastic hysteresis loop. During a roundtrip one
has to expend work against the interatomic forces because the
interatomic distances r are periodically increased (stress) and
decreased (compression). When the length L of a quadratic rod
D
Z"
0
0
A Ld" D V
Z"
(6.27)
d" :
0
R
The integral d" represents the work per unit volume, necessary for the relative length change ".
In the region where Hooke’s law is valid (linear region of ."/
is D E " and the work for the elastic length change L
Welast D 12 E V "2 :
(6.28)
Returning to the original stress-free state this energy is again released. The hysteresis curve simplifies to a straight line through
the origin D " D 0.
Example
Elongation and compression of an elastic spiral spring
during the oscillation of a mass m that hangs on the
spring. During the oscillation with small amplitude with
the linear region of Hooke’s law, the potential energy of
the spring and the kinetic energy of the mass m are periodically converted into each other (see Example 2 in
Sect. 2.7.4 and Sect. 11.6). The total energy, however,
J
is always conserved.
This is no longer true for Rthe nonlinear part of the curve ."/ in
Fig. 6.17. Here the work d" has to be put into the system in
order to proceed from the point O to the point A. This work is
equal to the area under the curve OA.
R However, after releasing
the tensile stress, only the work d" that equals the area
under the curve AB can be regained. The rest is converted into
thermal energy, due to the non-elastic deformation of the body.
Altogether the net work per unit volume, put into the system
during a roundtrip along the curve ABCDA, is given by the area
enclosed by this hysteresis curve in Fig. 6.17.
Chapter 6
A beam with length L, supported on both ends by fixed bearings,
suffers by a force F acting at the midpoint x D L=2 (Fig. 6.16)
the maximum sag
162
6 Real Solid and Liquid Bodies
very small forces and are merely caused by friction or surface
effects.
Table 6.2 Hardness scale according to Mohs
Selected materials
as measurement standards
1.
2.
3.
4.
5.
Tallow
Gypsum
Calcite
Fluorite
Apatite
6.
7.
8.
9.
10.
Feldspar
Quartz
Topaz
Corundum
Diamond
Examples
Aluminium
Lead
Chromium
Iron
Graphite
Tungsten
2.3–2.9
1.5
8
3.5–4.5
1
7
At first we will discuss the simplified model of an ideal liquid,
where surface effects and friction are neglected. For the static
case of a liquid at rest, friction does not occur anyway. Surface
effects will be treated in Sect. 6.4 and the influence of friction
for streaming liquids is discussed in Chap. 8.
6.3.1
Figure 6.18 Hardness test according to Brinell
6.2.6
The Hardness of a Solid Body
The hardness of a body is a measure for the resistance, which
the body sets against a penetration of another body. Depending
on the measuring technique, there are some slightly different
hardness values. The scratch-method introduced 1820 by Mohs,
defines a body A as harder than a body B if it is possible to
scratch B with A. The hardness scale of Mohr is based on this
definition. Here the hardness scale is divided into 10 degrees of
hardness based on 10 selected minerals, listed in Tab. 6.2.
The scratch method measures in fact mainly the hardness of
the surface. This surface hardness is of particular importance
for technical applications, because the attrition of tools or of
axes and bearings depends on the surface hardness. Therefore,
several techniques have been developed for the enhancement of
the surface hardness. One example is the transformation of the
surface layers of a crystalline solid into an amorphous state by
irradiation with a powerful laser. Another example is the cover
of solid tools, e.g. drills or steel mills, with a thin layer of a hard
material such as carbon-nitride NC or titanium Ti.
For measuring the hardness of a body, often a technique is used
which had been proposed by Brinell in 1900. Here a hardened
steel ball with diameter D is pressed vertically with a constant
force F D a D2 into the sample (Fig. 6.18). The diameter d of
the resulting circular notch in the sample gives the penetration
depth, which is a measure for the Brinell-hardness.
Chapter 6
6.3
Static Liquids; Hydrostatics
In order to achieve a change of the shape of solid bodies substantial forces are required, even if the volume of the body does
not change (for example for a shear or a torsion). Although similarly large forces are necessary for a compression of liquids,
a mere deformation of liquids at constant volume requires only
Free Displacement and Surfaces of
Liquids
For ideal liquids without friction, there is no force necessary to
deform a given liquid volume. In the atomic model this means:
While in solid bodies the atoms can vibrate around fixed equilibrium positions, which do not change much under moderate
external forces, the atoms or molecules in liquids can freely
move around within the given liquid volume, determined by the
solid container (Fig. 6.5). In the macroscopic model this free
movement can be expressed by the statement:
The shear modulus of an ideal liquid is zero.
This implies that at the surface of an ideal liquid no tangential
forces can be present, because they would immediately deform
the liquid until the forces disappear and a minimum energy is
achieved. This force-free condition represents a stable state of
the liquid.
The surface of an ideal liquid is always perpendicular to the total
external force.
Examples
1. If only gravity acts onto a liquid, the surface of the
liquid forms always a horizontal plane (Fig. 6.19a)
2. The surface of a liquid in a cylinder which rotates
about a vertical axis (Fig. 6.19b) forms a surface
where the total force composed of gravity m g in the
z-direction and centrifugal force m! 2 r in the radial direction points perpendicular to the surface. The
slope of the intersection curve z.r/ of the surface at the
point A in Fig. 6.19b is
tan ˛ D
m! 2 r
! 2r
D
:
mg
g
On the other hand the slope of the curve z.r/ is given
by tan ˛ D dz=dr. Integration yields
z.r/ D
!2
g
Z
rdr D
!2 2
r CC :
2g
6.3
Static Liquids; Hydrostatics
163
For z.0/ D z0 is C D z0 and we get
z.r/ D
!2 2
r C z0 :
2g
(6.29)
The surface forms a rotational paraboloid with its axis
coincident with the rotation axis.
Figure 6.21 Relation between the pressure inside a volume element dV and
the forces acting onto the sides of dV
In an analogous way we obtain the force components in the other
directions
@p
@p
dV and Fz D
dV :
Fy D
@y
@z
Figure 6.19 a Horizontal liquid surface in a container at rest. b SurJ
face as rotational paraboloid in a rotating container
We can condense these three equations into the vector equation
F D grad p dV :
6.3.2
Static Pressure in a Liquid
Any external force acts only vertically on the surface of a liquid. If a container with a liquid is closed by a movable piston
with surface A, onto which a vertical force F acts (Fig. 6.20) we
define the pressure onto the liquid as
p D F=A;
6.3.2.1
with
F D jFj :
Forces onto a Liquid Volume Element
We consider an arbitrary cuboid volume element dV D dxdydz
inside the liquid (Fig. 6.21). We assume that a pressure p acts
in x-direction onto the left side dy dz of the cuboid. Then a
pressure
p C @p=@x dx
(6.30)
Because of the free mobility of any volume element inside the
liquid the total force onto a volume element at rest must be zero.
This implies that grad p D 0.
The pressure inside the whole liquid is constant as long as
no unisotropic forces act onto the liquid.
For a static liquid the same pressure acts onto all surface elements of the container!
This can be experimentally demonstrated by the simple device
shown in Fig. 6.22. A spherical container with small holes in
several directions of the x-y-plane is filled with dyed water and
placed above a blotting paper in the plane z D 0. When a piston
Figure 6.20 The force F , acting on a piston with area A generates a pressure
p D F =A in the liquid
Figure 6.22 Demonstration of the isotropic pressure in a liquid. When the
piston is moved the pressure p increases and the dyed water splashes through
holes onto a white paper below the device, where the spots form a circle around
the center, indication equal pressure
Chapter 6
acts onto the opposite side. The resulting force on the volume
element is then
@p
@p
Fx D p dydz
p C dx dydz D
dV :
@x
@x
164
6 Real Solid and Liquid Bodies
Figure 6.23 Hydraulic press (forces are drawn not to scale)
Figure 6.24 Pressure p .z / in an incompressible liquid in the gravity field of
the earth, as a function of height z above ground
is pressed to increase the pressure the water streams out of the
holes and all water filaments follow projectile trajectories. Their
points of impact on the blotting paper in the plane z D 0 form
a circle which proves that they all had the same initial velocity,
i.e. they stream out driven by the same pressure.
Application: Hydraulic press (Fig. 6.23)
Two cylinders with cross sections A1 and A2 A1 that are connected with each other, and are therefore at the same pressure
p, are filled with a liquid. Applying the force F1 D p A1 on
a piston in the narrow cylinder causes a force F2 D pA2 D
.A2 =A1 / F1 F1 acting on a piston in the large cylinder which
presses a sample against a fixed mounting. For demonstration
experiments, large stones can be cracked by this device. For
a displacement x2 of the large piston, the small piston has to
move by the much larger amount x1 D .A2 =A1 / x1, because
the volume V2 D A2 x2 D V1 D A1 x1 transferred from
the small to the large cylinder must be of course equal.
6.3.2.2
Then it follows from (6.31) for the pressure p.z/ in a liquid with
total heights H (Fig. 6.24)
p.z/ D % g .H
z/ :
The SI unit for the pressure is 1 Pascal D 1 Pa D 1 N=m2 , which
corresponds to 10 5 bar.
Examples
1. A water column of 10 m heights causes a hydrostatic
pressure of p D % g h D 9:81 104 Pa D 0:981 bar D
1 atmosphere. At an ocean depth of 10:000 m (Philippine rift) the hydrostatic pressure is 108 Pa (about
1000 atm). The total force onto the outer surface of a
hollow steel sphere of an aquanaut with 3 m diameter
is at this depth F D 2:8 109 N.
Hydrostatic Pressure
Taking into account that every volume element V of a liquid
has a weight % g V in the gravity field of the earth, even without external force a pressure onto the bottom of the container is
present due to the weight of the liquid above the bottom. For a
height z D H of the liquid the hydrostatic pressure at the bottom
with area A is with dV D A dz
p.z D 0/ D
ZH
0
%gA
dz D % g H ;
A
(6.31)
if we assume that the density % is independent of the pressure p.
For real liquids, there is a small change of % with the pressure p.
A measure for this dependence is the compressibility
Def
Chapter 6
D
1 @V
;
V @p
(6.32)
which describes the relative volume change V=V for a change
p of the pressure.
For liquids is very small (for example for water is D
5 10 10 m2 =N/. This shows that the density % of a liquid
changes only by a tiny amount with pressure and in most cases
the density %.p/ D %0 can be assumed to be constant.
Figure 6.25 a Water pressure acting onto a dam wall; b Additional
support by the mountain walls for a curved dam wall
2. The total force F onto the river dam with length L
caused by the water with heights H can be obtained
by integration over all contributions F.z/dz onto the
surface elements L dz of the dam.
Z
Z
F D L p.z/dz D % g L .H z/dz
D 12 % g L H 2
This force can be partly supported by choosing a
curved dam where part of the force are balanced by
the mountain walls (Fig. 6.25b). The thickness of the
dam decreases with z in order to take into account the
J
decreasing hydrostatic pressure (Fig. 6.25a).
6.3
Static Liquids; Hydrostatics
165
Figure 6.26 River dam of the river Eder, Germany. The bending of the dam towards the water side conducts part of the waterpressure against the mountain sides
(see Sect. 6.3). With kind permission of Cramers Kunstverlag, Dortmund
6.3.3
Buoyancy and Floatage
If we immerse a cuboid with basic area A and volume V D Ah
into a liquid with density %L the pressure difference between
bottom and top surface is (Fig. 6.28)
Figure 6.27 Hydrostatic paradoxon. The pressure onto the bottom is equal for
all containers filled up to the same height H
p D %L g h :
This results in an upwards directed buoyancy force
Since the hydrostatic pressure at the bottom of a liquid container
depends only on the height H of the liquid but not on the shape
of the container, the pressure at the bottom is identical for all
four containers shown in Fig. 6.27, although the total mass of the
liquid and therefore also its weight is different. This hydrostatic
paradox leads to the following astonishing but true fact: When a
hollow cube with a volume 1 m3 is filled completely with water,
the hydrostatic pressure at the bottom is 0:1 bar. If now a thin
tube with 1 cm2 cross section but 10 m height is put through a
small hole in the top wall of the cube and filled with water the
pressure in the cube rises to 1 bar. Although the additional mass
of water is only 10 3 of the water in the cube the pressure rises
by a factor of 10.
FB D %L g A h D GL ;
which is equal to the weight GL of the liquid displaced by the
body, but has the opposite direction.
This can be formulated as Archimedes’ Principle:
A body immersed in a liquid looses seemingly as much of
its weight as the weight of the displaced liquid.
This principle illustrated for the example of a cuboid, is valid for
any body with arbitrary shape as can be seen from the following
consideration:
Due to the hydrostatic pressure p D %fl g .H z/ at the height
z in a liquid with total height H the force on a volume element
dV is
F D grad p dV D .@p=@z/Oez dV D %L g dV eO z
D %L g dV :
The buoyancy force on the whole body immersed in the liquid
is then
Z
FB D g %L dV D GL :
(6.33)
Chapter 6
Equation 6.31 tells us that the pressure p at the upper surface
of a liquid volume element V D A h with height h is
smaller than at the bottom of this element by the amount %
g h. This results in an upwards force F D A % g h,
which is just compensated by the weight G D M g D % g
V D % g A h of the volume element. The total force on an
arbitrary volume element V inside a homogeneous liquid in a
homogeneous gravity force field is therefore zero.
166
6 Real Solid and Liquid Bodies
Buoyancy
FB
h1
p1
FB
h2
SB
SB
p2
FB
FB
D
Figure 6.28 Axiom of Archimedes and buoyancy
If the density %b of a body is smaller than the density %L of
the liquid, the buoyancy force becomes larger than the weight
Gb of the body and the body floats on the surface of the liquid.
Only part of the body immerses while the other part is above the
liquid surface. Equilibrium is reached if the buoyancy (it is the
weight GL of the displaced liquid) just cancels the weight Gb of
the total body.
Example
The density of ice is %i D 0:95 kg=dm3, the density of
salty sea-water at 0 ı C is %L D 1:05 kg=dm3. Therefore,
about 10% of the volume of an iceberg stick out of the
ocean surface, 90% are under water.
J
Remark. Of course, the buoyancy is also present in gases.
However, because of the much smaller density of gases the
buoyancy force is correspondingly smaller. A body in a gas atmosphere loses (seemingly) as much of its weight as the weight
of the displaced gas. This is the basis for balloon flights (see
Sect. 7.2 and Fig. 7.6.
For the stability of a floating ship it is important that in case
of heeling induced by waves there is always a restoring torque
which brings the ship back into its vertical position. This stability criterion can be quantitatively formulated in the following
way:
Chapter 6
We consider the torque generated by the gravity force Gg and
the buoyancy FB for a ship in an oblique position (Fig. 6.29).
The two forces form a couple of forces (Sect. 5.4) which cause
a torque about the center of mass SK . The point of origin for
the gravity force Gg is the center of mass SK of the ship, while
the point of origin for the buoyancy FB D Gg is the center of
mass SB of the displaced water. The symmetry plane of the ship,
indicated in Fig. 6.29b by the dashed line, intersects the vertical
direction of the buoyancy in the point M, called the meta-center.
The vector r gives the distance between M and SK . As long as
M lies above SK the resulting torque
D D .r GK / D .r FB / ;
SB
SB
D
d)
Figure 6.29 Stability of a floating body. a equilibrium position, b tilting below
the critical angle, c vector diagram of stable and unstable heeling
which has in Fig. 6.29c the counter-clockwise direction, brings
the ship back into the vertical stable position. If the slope becomes so large that M comes below SK (Fig. 6.29d) the resulting
torque acts into the clockwise direction and it brings the ship
into a larger slope. It overturns and sinks. It is therefore advantageous for the stability to have the center of mass SK as low
as possible. This can be achieved by putting heavy masses at
the bottom of the ship. In case of container ships, the cargo is
loaded on top of the ship which decreases the stability. These
ships have therefore a double mantel at the bottom where the interspace is filled with water, in order to bring the center of mass
down.
6.4
Phenomena at Liquid Surfaces
We will now upgrade our simple model of the ideal liquid in order to introduce effects which occur at surfaces of real liquids
and which are not present in ideal liquids. While inside a liquid the resulting time-averaged force on an arbitrary molecule,
exerted by all other molecules, is zero, (this allows the free relocatability of each molecule), this is no longer true for molecules
at the surface of liquids (Fig. 6.30) which are only attracted by
molecules in a half sphere inside the liquid. Therefore a residual force FR remains, which attracts the molecules towards the
interior of the liquid.
6.4.1
Surface Tension
If a molecule is brought from the inside of a liquid to the surface,
energy has to be supplied to move the molecule against the residual force FR . A molecule at the surface has therefore a higher
6.4
Phenomena at Liquid Surfaces
167
Figure 6.30 Resulting force on a molecule by all other surrounding molecules
inside a liquid and at the surface of a liquid
Figure 6.32 Measurement of surface tension by liftig an immersed metal ring
on a spring balance. It is immersed into a glass container filled
with a liquid. When the container is lowered or the metal ring is
uplifted, the lower rim of the ring emerges more and more out
of the liquid, carrying a cylindrical liquid lamella. With soapy
water more than 10 cm heights of the lamella can be reached.
The spring balance measures the force
F D 4 r ;
Figure 6.31 Determination of surface tension by measuring the force on a
sliding straight wire, that extends a liquid skin
energy than a molecule inside the liquid. In order to enlarge the
surface by an amount A molecules have to be transferred to
the surface which needs the energy W. The ratio
"D
W
I
A
Œ" D
J
m2
(6.34)
is the specific surface energy. The value of " depends on the
binding forces between the molecules of the liquid. It can be
measured with the equipment shown in Fig. 6.31. Between the
two sides of a U-shaped frame a horizontal wire with length
L can be shifted vertically. When the system is dipped into a
liquid, a liquid lamella is formed with the surface area (on both
sides) A D 2L s. For moving the horizontal wire by s, the
force F is necessary. One has to supply the energy
W D F s D " A D " 2 L s :
because the lamella has two surfaces, inside and outside of the
cylindrical membrane. The work necessary to lift the lamella up
to the height h is
W D 4 r h :
Example
Surface tension and pressure in a soap bubble (Fig. 6.33).
Because of its surface tension, the bubble tries to reduce
its surface. This increases the pressure inside the bubble.
Equilibrium is reached, if the work against the increasing pressure during the decrease r of the bubble radius
is equal to the work gained by the reduction A of the
surface area A
" 2 4.r2
.r
r/2/ D 4 r2 p :
(6.35)
D":
Surface tension and specific surface energy " are identical.
The surface tension can be impressively demonstrated by the apparatus shown in Fig. 6.32. A metal strip bent into a circle hangs
Manometer
Figure 6.33 Measurement of overpressure p in a soap bubble,
caused by surface tension
Chapter 6
The restoring force F, which is directed tangential to the surface
of the lamella, produces a tensile strain D F=2L per length
unit which is called surface tension. According to (6.35) is
168
6 Real Solid and Liquid Bodies
The sign of "ik can be obtained by the following considerations:
2
Neglecting the term with .r/ gives the excess pressure
p D 4"=r ;
(6.36)
which shows that p decreases with increasing radius r.
This can be demonstrated by the equipment in Fig. 6.34.
The lower ends of the tubes 1 and 2 are immersed into
soapy water and then lifted again. With open valves 1
and 2 but closed valve 3, two bubbles with different sizes
can be produced by blowing air into the corresponding
filling tubes. Now valves 1 and 2 are closed and valve
3 is opened. The smaller bubble starts to shrink and the
larger one inflates. This continues until the smaller bubble
completely disappears. It’s like in daily life. The powerful
people (larger ones) increase their power at the cost of the
little guys.
1
2
3
For stable interfaces between liquid and gas "ik has to be positive. Otherwise the liquid phase would be transferred into
the gas phase because energy would be gained, i.e. the liquid
would vaporize.
Also for stable interfaces between two different liquids "ik
must be positive. Otherwise the two liquids would intermix
and the interface would disappear.
For the interface between liquid and solid phases the sign
of "ik depends on the materials of the two phases. If the
molecules ML in the liquid are attracted more strongly by the
molecules Ms in the solid, than by neighboring molecules
in the liquid, is "ik < 0. If the attracting forces between
molecules ML are stronger than between ML and Ms is "ik >
0.
Also between a solid surface and a gas an interface energy
can occur, because the gas molecules can be attracted by the
solid surface (adhesion) or they can be repelled, depending
on the gas and the solid material.
We will illustrate these points by some examples: In Fig. 6.35
is the surface of a liquid 2 against the gas phase 3 close to a
vertical solid wall 1 depicted. Here the surface tensions 1;2 ;
1;3 and 2;3 tangential to the corresponding surfaces have to be
considered. We regard a line element dl perpendicular to the
plane of the drawing through the point A, where all three phases
are in contact with each other. The force parallel to the solid
surface is Fks D .1;2 1;3 /dl and Fkl D 2;3 dl is parallel to the
liquid surface. The resulting force causes a change of the liquid
Figure 6.34 Demonstration of overpressure p .r / which increases
with decreasing radius r of a soap bubble
For liquids with positive surface energy each liquid
with a given volume tries to minimize its surface
area.
This can be demonstrated by adding drop wise mercury
through a pipette into a bowl filled with diluted sulfur
acid. At first many small mercury droplets are formed
which, however, soon merge into a single larger drop. J
Liquid film
6.4.2
Interfaces and Adhesion Tension
Chapter 6
Up to now, we have only discussed surfaces of liquids as boundaries between liquid and gaseous phases. Often interfaces
between different liquids or between liquid and solid bodies can
occur. Analogue to the surface tension we define the boundary
tension ik (identical with the specific interface energy "ik / as
the energy that has to be spend (or is gained) when the interface
between the phases i and k is increased by 1 m2 .
Figure 6.35 Formation of a contact angle of a liquid surface with a vertical
solid wall. a Concave liquid surface for water-glass (1;3 > 1;2 ); b convex
surface of Hg-glass (1;3 < 1;2 ); c complete wetting for 1;3 1;2 > 2;3
6.4
Phenomena at Liquid Surfaces
169
surface, which would be a horizontal plane under the action of
gravity without surface tension.
If we neglect the small change of the gravitational force due
to the change of the surface, which is very small compared to
the forces caused by surface tension, we have the equilibrium
condition that in the point A the vector sum of all forces must be
zero. For the vertical component parallel to the solid wall this
implies:
1;2 C 2;3 cos '
1;3 D 0 :
(6.37)
The horizontal component 2;3 sin ' causes an imperceptibly
small deformation of the solid wall. This induces a restoring
deformation force which is opposite to the force 2;3 and has
the same magnitude and therefore compensates it. The wetting
angle ' can be obtained from the condition
cos ' D
1;3 1;2
:
2;3
(6.37a)
It has a definite value only for j1;3 1;2 j 2;3 . We distinguish
the following cases:
If external forces are present, such as gravitational or inertial
forces in accelerated systems, the vector sum of all forces is in
general not zero. However, the liquid surface reacts always in
such a way, that the resultant force is perpendicular to the liquid surface, i.e. its tangential component is always zero. This is
illustrated in Fig. 6.36 for the cases of a concave and a convex
curvature of the liquid surface close to the solid wall where besides the gravitational force also the attractive force F4 between
liquid and solid surfaces is taken into account.
For a liquid in a container the total force is compensated by the
restoring elastic force of the container wall.
For two non-mixable liquids 1 and 2 (for example a fat drop on
water) the angles '1 and '2 in Fig. 6.37 adjust in such a way that
the equilibrium condition
1;3 D 2;3 cos '2 C 1;2 cos '1
(6.38)
is fulfilled. This shows that a droplet of the liquid 2 can be only
formed, if 1;3 < 2;3 C 1;2 . Otherwise the droplet would be
spread out by the surface tension 1;3 until it forms a thin film,
which covers the surface of liquid 1.
Figure 6.36 The vector sum of all forces acting onto a liquid sureface must
be always vertical to the surface, for non-wetting liquids. a Concave, b convex
curved surface
Figure 6.37 Formation of a liquid drop on the surface of another liquid
Example
For the interfaces water–oil–air the numerical values of
the surface tensions are:
1:3 (water-air) D 72:5 10
1;2 (water-oil) D 46 10
2;3 (oil-air) D 32 10
3
3
3
J=m2
J=m2
J=m2 :
This shows that 1;3 > 2;3 C 1;2 . Therefore, oil cannot
J
form droplets on a water surface.
Chapter 6
1;3 > 1;2 ! cos ' > 0 ! ' < 90ı .
The liquid forms close to the solid wall a concave surface,
which forms an acute angle ' with the wall (Fig. 6.35a). It is
energetically favorable to increase the interface liquid-solid
at the cost of the interface solid–gas.
Example: Interfaces water–glas–air.
1:3 < 1;2 ! cos ' < 0 ! ' > 90ı .
The liquid forms close to the solid wall a convex surface
(Fig. 6.35b).
Example: interfaces mercury–glas–air.
For j1;3 1;2 j > 2;3 Eq. 6.37 cannot been fulfilled for any
angle '. In this case a force component parallel to the solid
surface is uncompensated. It pulls the liquid along the solid
surface until the whole surface is covered by a liquid film
(Fig. 6.35c). The interface solid–gas disappears completely.
170
6 Real Solid and Liquid Bodies
Fatty acid
Talcum
Figure 6.38 Formation of a mono-molecular layer of a fatty acid on a liquid
surface covered with a talcum powder layer
Figure 6.40 a Capillary rise of a wetting liquid, b derivation of the rise height
On the other hand, the surface energy changes by (see Fig. 6.36)
dEsurface D 2r dh.13 12 /
D 2r dh 23 cos ' ;
Figure 6.39 Fatty acid molecules forming a mono-molecular layer on a water
surface are oriented due to the attractive for one end of the interaction with the
water molecules and a repulsive interaction for the other end
If an oil drop is brought onto a water surface, it will spread out
to form a mono-molecular layer of oil which covers the whole
water surface if sufficient oil is contained in the drop. Otherwise, the oil film forms a cohesive insula of this mono-molecular
film. This can be demonstrated by the following experiment
(Fig. 6.38): Onto a water surface, powdered with talc, a droplet
of fatty acid is supplied through a pipette. The droplet immediately spreads out and displaces the talc layer. The fatty acid
molecules are oriented in such a way, that the attractive force
with the water molecules becomes maximum (Fig. 6.39). The
atomic groups COOH, which are directed against the water surface, are called hydrophilic while the groups on the opposite side
of the molecule, which are repelled by the water molecules, are
called hydrophobic. The interaction with the water molecules
causes a displacement of the charges in the fatty acid molecules
while the water molecules, which are electric dipoles, are orientated in such a way, that their positive pole is directed toward the
negative pole of the induced dipole molecules of the fatty acid
(see Vol. 2, Chap. 2).
6.4.3
where Eq. 6.37a has been used. At equilibrium is dEp C
dEsurface D 0. This gives the resulting height
h D 223 cos '=.r g %/
D 2 cos '=.r g %/ :
For completely wetting liquids .1;3 > 1;2 C 2;3 / is ' D 0.
The complete inner surface of the capillary tube is covered by
a thin liquid film and the capillary rise becomes according to
(6.40)
hD
2
:
rg%
(6.40a)
For non-wetting liquids .1;3 < 1;2 / the liquid surface inside
the capillary is convex. This convex curvature causes a force,
which is directed downwards and leads to a capillary depression
(Fig. 6.41). The depression height h is again given by (6.40),
where now cos ' D .1;3 1;2 /=2;3 < 0.
The capillary rise offers an experimental method for the measurement of absolute values of surface tensions. Instead of
Chapter 6
When a capillary tube is dipped into a wetting liquid .1;3 >
1;2 /, the wetting liquid rises in the capillary tube up to the
height h above the liquid surface (Fig. 6.40). This observation
can be explained as follows: If a liquid column in the capillary
with radius r is lifted up to the height h .h r/ the potential
energy is increased by
(6.39a)
(6.40)
The wetting angle ' is determined by Eq. 6.37). The surface
tension 2;3 D is the surface tension of the liquid against air,
introduced in Sect. 6.4.1.
Capillarity
dEp D m g dh D r2 g % h dh :
(6.39b)
Figure 6.41 Capillary depression
6.5 Friction Between Solid Bodies
171
zero. In daily life and for technical problems they play a very
important role. Without friction we would not be able to walk
nor cars could run. Also most technical processes of machine
work on material, such as drilling, milling or cutting would not
be possible without friction. On the other hand, often friction
needs to be minimized in order to avoid energy dissipation.
We will therefore discuss the basic principles of friction phenomena in more detail.
6.5.1
Figure 6.42 Demonstration of capillary rise h .d / / 1=d of a liquid confined
between two wedged plane walls with the wegde angle 2˛
capillary tubes one can also use two parallel plates with the distance d. A liquid between these plates has the capillary rise
hD
2 1
:
%g d
(6.41)
The dependence h.d/ can be demonstrated by two nearly parallel plates, which are slightly inclined against each other by a
small angle ˛ (Fig. 6.42). Since the distance d.x/ D 2x tan ˛
increases linearly with x the height h.x/ / 1=x is a hyperbola.
Static Friction
A body with a plane base (for example a cuboid) rests on a horizontal plane table. In order to move it across the table we must
apply a force in the horizontal direction, which can be measured
with a spring balance (Fig. 6.43a). The experiment shows that
in spite of the applied force the body with mass m does not
move until the force exceeds a definite value Fs . When the body
is turned over (Fig. 6.43b) so that now another surface with a
different area touches the table, this critical force Fs does not
change in spite of the different surface area in contact with the
table. However, if the body is pressed by an additional force
against the table, the critical pulling force Fs increases. The experiments show, that Fs is proportional to the total vertical force
FN exerted by the body on the table and on the roughness of the
two surfaces in contact.
The amount of this static friction force is
Summary of Section 6.4
The many different phenomena at the boundaries of liquids can
be all quantitatively explained by the magnitude of the surface
tensions or surface energies. We can make the following statements:
At each point of a stable liquid surface the total force is always perpendicular to the surface, its tangential component
is zero.
The boundary of a liquid with a given volume always approaches that shape that has the minimum surface area.
A bent convex liquid surface with radius of curvature r produces an inward pressure, that is proportional to 1=r and to
the surface tension.
6.5
Fs D s FN :
(6.42)
The static friction coefficient s depends on the materials of the
bodies in contact and on the texture of the two surfaces.
The static friction can be explained in a simple model (Fig. 6.44)
by the roughness of the two surfaces in contact. Even a polished
plane surface is not an ideal plane but shows microscopic deviations from the ideal plane, which may be caused by lattice
defects, shifts of atomic planes etc. The envelope of this microroughness gives the macroscopic deviations caused by imperfect
polishing or grinding. A measure for these deviations is the
mean quadratic deviation hz2 .x; y/i from the ideal plane z D 0.
Since one measures generally not single points but surface elements dx dy, the function z.x; y/ is averaged over the surface
Friction Between Solid Bodies
If two moving extended bodies touch each other, additional
forces occur which depend on the properties of the two surfaces.
Examples are a metal block sliding on a plane base, or a wheel
rotating around an axis. These forces are due to the interaction
between the atoms or molecules in the outer layers of the two
bodies. This interaction is reinforced by surface irregularities
and deformations, caused by the contact between the two bodies. These forces are called friction forces. For point masses
they can be completely neglected because their surface area is
Chapter 6
6.4.4
Figure 6.43 Measurement of static friction with a spring balance
172
6 Real Solid and Liquid Bodies
2. A component F? D m g cos ˛ perpendicular to the inclined
plane, which is compensated by the opposite restoring force
of the elastic deformation of the plane.
The body starts to slide downwards as soon as Fk becomes larger
than the static friction force Fs D s F? D s m g cos ˛.
This gives the condition for the coefficient s
s D
Fk .˛max /
D tan ˛max :
F? .˛max /
(6.43)
If ˛ is increased beyond ˛max the body performs an accelerated
sliding motion. This indicates that the sliding friction force is
smaller than the static friction force.
Figure 6.44 Schematic model of the surface roughness as the cause of friction.
a micro roughness (exagerated) and macroscopic coarseness; b static friction
caused by interlocking of two rough surfaces
elements dx dy and this average depends on the spatial resolution of the analyzing instrument i.e. on the size of the resolved
elements dx dy. With modern surface analysis, using tunnelmicroscopes (see Vol. 3) even the roughness on an atomic scale
can be spatially resolved.
The two surfaces in contact interlock each other due to the force
that presses them together (Fig. 6.44b) and the force Fs is necessary to release this interlocking. This can be achieved by
breaking away the “hills” of the rough surface, or by lifting the
body over these hills.
A possible way to determine experimentally the coefficient of
static friction uses the inclined plane with a variable inclination
angle ˛ in Fig. 6.45. The angle ˛ is continuously increased until
the body B with mass m starts to slide down for ˛ D ˛max .
The weight force G D m g can be decomposed into two components:
1. A component Fk D m g sin ˛ parallel to the inclined plane
6.5.2
Sliding Friction
When the body in Fig. 6.43 is moved by a force jFj > jFs j
the sliding motion is accelerated. In order to reach a uniform
motion of a sliding body with constant velocity, where the total
force is zero, one needs only the smaller force jFsl j < jFs j.
Analogue to the static friction force, the sliding friction force
Fsl is proportional to the force FN normal to the surface of the
table on which it slides.
Fsl D sl FN :
(6.44)
The coefficient of sliding friction sl depends again on the material of body and basis, but also on the relative velocity. It is,
however, always smaller than the coefficient of static friction.
This can be explained by the simplified model of the two surfaces in contact, shown schematically in Fig. 6.44, where the
roughness of the surfaces has been exaggerated. If the two bodies are at rest the peaks and the valleys of the micro-mountains
interlock. This allows a minimum distance between the two attracting surfaces resulting in a minimum energy. At the sliding
motion the two surfaces move above the peaks and the mean distance between the surfaces is larger. During the sliding motion,
parts of the peaks are ablated. This results in an attrition of the
surfaces.
The sliding motion dissipates energy, even for a horizontal motion. If the body is moved by the distance x, the necessary
work is W D Fsl x, which is converted into heat.
Chapter 6
Experiments show that the sling friction force increases with
the relative velocity. The reason is that with increasing velocity
more material of the two surfaces is ablated. The power P D
dW=dt, necessary to maintain the velocity v of a sliding motion,
increases with v n where n > 1.
Figure 6.45 Measurement of coefficient of static friction with the inclined
plane
Note: The friction between a moving body and the surrounding
air has different reasons. If the body moves through air at rest,
a thin layer of air close to the surface of the body sticks at the
surface and is therefore accelerated by the moving body. This
requires the energy 1=2mL v 2 where mL is the mass of the air
layer that also increases with the velocity v.
6.5 Friction Between Solid Bodies
6.5.3
173
Rolling Friction
When a round body rolls over a surface, also friction forces FR
occur which are caused by the interaction between the atoms
of the bodies at the line of contact. Furthermore, the base is
deformed by the weight of the round body (Fig. 6.46), which
leads to deformation forces. For the rolling of a round body
with constant angular velocity, a torque around the contact line
is necessary that just compensates the opposite torque of the
rolling friction. Around the depression of the base at the line
of contact, bulges are formed, which have to be overcome when
the body rolls.
Figure 6.47 Measurement of rolling friction with the inclined plane
The experiments tell us that the torque, necessary for keeping a
constant angular velocity, is proportional to the force FN normal
to the surface of the base
technical realizations of different ball bearings and axial bearings are shown. In Tab. 6.3, the friction coefficients for some
materials are listed.
(6.45)
Remark. For skating or tobogganing the snow melts under the
runners because of heat conduction from the warmer skates and
due to the heat produced by friction. The water film under the
vats reduces the friction considerably. Often one finds the explanation that the pressure exerted by the weight of the skater is
the main reason for melting. This effect plays, however, only a
minor part, as can be calculated from the known decrease of the
melting point with increasing pressure. (see Sect. 10.4.2.4). The
much smaller sliding friction between solid surface and liquids
is also utilized by applying lubricants between the two surfaces,
for instance between a rotating axis and its fixed support or between the moving pistons of a car engine and the cylinders. The
oil film reduces the friction by about two orders of magnitude.
DR D R FN ;
where the coefficient R of rolling friction has the dimension of
a length in contrast to the dimensionless coefficients s and sl .
Similar to the measurement of s the coefficient R can be measured with an inclined plane (Fig. 6.47). A circular cylinder with
mass m and radius r does not roll down the inclined plane, if the
inclination angle ˛ is smaller than a critical angle ˛R , which
is, however, smaller than the angle ˛max measured for the static
friction in Fig. 6.45.
For this critical angle ˛R is the counter-clockwise torque DG D
m g r sin ˛R around the contact line just equal to the clockwise
torque DR D R FN D R m g cos ˛R . This yields
R D r tan ˛R :
(6.46)
The rolling friction is proportional to the radius of the round
body. The rolling friction is much smaller than the sliding friction, because the surface irregularities, shown in Fig. 6.44, are
partly overrun. Therefore, the invention of the wheel was a great
progress for humankind. The comparison of the frictional forces
for sliding and rolling gives with (6.44) and (6.45) the ratio
Fs
Fs
s
D
:
Dr
FR
FR =r
R
(6.47)
Chapter 6
The much smaller rolling friction is utilized by ball bearings,
which reduce the friction of rotating axes compared to the sliding friction without these ball bearings. In Fig. 6.48, some
Figure 6.46 Deformation of a surface around the contact line
Figure 6.48 Ball bearings. a scheme of a radial grooves bearing; b realization;
c axial groove bearing
174
6 Real Solid and Liquid Bodies
Table 6.3 Coefficients of static, sliding and rolling friction of some materials
in contact with each other. The values strongly depend of the characteristics of
the surfaces. They therefore differ for different authors
Interacting materials
Steel–Steel
H
0.5–0.8
G
0.4
R =r
0.05
Steel with oil film
0.08
0.06
0.03–0.1
Al–Al
1.1
0.8–1.0
Steel–Wood
0.5
0.2–0.5
Wood–Wood
0.6
0.3
Diamond–Diamond
0.1
0.08
Glass–Glass
0.9–1.0
0.4
– dry
1.2
1.05
– wet without waterfilm
0.6
0.4
0.5
Rubber-tar seal
In Fig. 6.49a is an axis shown that rotates with the angular
velocity !. A circular ring with area A D .r22 r12 / is
welded to the axis and exerts a force FN and a pressure
p D FN =A onto the support base. The sliding friction
causes a torque D on the rotating axis, which has to be
compensated by an opposite torque supplied by an external force.
On the red annulus in the lower part of Fig. 6.49a acts
the force dFN D 2 r dr p, which causes the torque
dD D r dFs D s p 2 r2 dr. Integrating over all
annuli gives the total torque
DD
Zr2
r1
6.5.4
Significance of Friction for Technology
dD D
2
sl p r23
3
r13 :
(6.48)
Friction plays an outstanding role for many technical problems.
In some cases it should be as large as possible (for example for
clutches in cars or other machinery). The rolling friction for car
tires should be as small as possible, but the static friction and
the sliding friction should be as large as possible.
The friction consumes the power P D D !, which is converted into heat. This dissipated power is proportional to
the coefficient s of sliding friction, to the contact pressure p and the angular velocity !. If the annulus with area
A is supported by ball bearings (Fig. 6.49b), the torque
caused by friction decreases by some orders of magniJ
tude.
For many sliding or rotating parts of machinery, friction is damaging. It causes increased energy consumption and a destruction
of the sliding surfaces (attrition). For such cases, it is therefore
necessary to minimize friction. This can be achieved either by
reducing the sliding friction by liquid films or air buffers or by
using ball bearings. Because of its importance, meanwhile a
whole branch of science called tribology works on problems of
friction [6.4].
Another solution uses the mounting in Fig. 6.49a but now with
a liquid film between the contacting surfaces. Often air is blown
with high pressure between the two surfaces and an air buffer
supports the rotating annulus.. This allows one to realize an
extremely low friction. Examples are very fast rotating turbomolecular vacuum pumps (see Sect. 9.2.1.3), where the rotating
blades are supported by the air blow.
Example
6.6
The Earth as Deformable Body
At the end of this chapter, we will apply the results of the foregoing sections to the interesting example of our earth, which can
be deformed by several forces acting on it. In addition friction
plays an important role for phenomena such as the tides or the
differential rotation of the inner parts of the earth. Since the
earth is composed of solid material as well as of liquid phases,
it gives a good example of a realistic and more complicated deformable body.
Chapter 6
Figure 6.49 Rotating axis a without ball bearing, b with ball bearing
Our earth is not a rigid homogeneous sphere. It shows an inhomogeneous radial density profile r.r/ (Fig. 6.50), which is
determined by the pressure profile p.r/, but also by the chemical
composition, which changes with the radius r. Furthermore, the
different solid and liquid phases in the interior of the earth contribute to the inhomogeneous profile. The central region with
r < 1000 km is a solid kernel of heavy elements (iron, nickel),
while for r > 1000 km hot liquid phases of metals are predominant, covered by a relatively thin solid crust, consisting of large
plates, which float on the liquid material. The earth is therefore not a rigid body but can be deformed by centrifugal forces,
6.6 The Earth as Deformable Body
175
Figure 6.50 Radial density profile of the earth
caused by the earth rotation, and by gravitational, forces due to
the attraction by the sun and the moon. These deformations are
partly elastic (tides of the earth crust) or plastic (= inelastic). In
the latter case, the deformed material does not come back to its
original location after the force ends and a permanent change
of the shape remains. The shift of the continental plates or the
eruption of volcanos with the formation of new islands or mountains are examples of non-elastic deformations.
6.6.1
FG D
G
m M.r/
rO ;
r2
(6.49b)
where M.r/ is the mass of that part of the earth inside the radius
r. Because of the plastic deformation the mass element m
shifts until the total force F acting on it, is zero.
F D FG C Fcf C FR
Ellipticity of the Rotating Earth
The rotation of the earth with the angular velocity ! D
2=day D 7:3 10 5 s 1 causes a centrifugal force on a mass
element m with the distance a from the rotation axis
Fcf D m a ! 2 eO cf
with the unit vector eO cf perpendicular to !. This force acts in
addition to the gravitational force
(6.49a)
is the sum of gravity force FG , centrifugal force Fcf and restoring force FR . For a homogeneous earth this would result in
a rotational ellipsoid with the major diameter in the equatorial
plane
2a D 12 756:3 km ;
and with a minor diameter in the direction of the rotational axis
of
2b D 12 713:5 km :
b/=a of this rotational ellipsoid is " D
Because of the inhomogeneous mass distribution the shape of
the rotating earth deviates slightly from this rotational ellipsoid
but forms a nearly pear-shaped pattern called geoid (Fig. 2.56).
The surface of this geoid is the zero-surface for all geodetic measurements. This means: all measurements of elevations z are
related to this zero surface z D 0 [6.5].
6.6.2
Figure 6.51 Deformation of the rotating earth due to centrifugal force
Tidal Deformations
Induced by the additional forces of the gravitational attraction
by the sun and the moon the earth surface deforms in a characteristic time-dependent way. This deformation is maximum for
Chapter 6
The ellipticity " D .a
3:353 10 3 .
176
6 Real Solid and Liquid Bodies
Figure 6.53 Only for the center M of the earth are gravitational attraction by
the moon and centrifugal force of the earth–moon rotation about S equal but
opposite and cancel each other
Figure 6.52 The rotation of earth and moon about their common center of
mass S causes all points of the earth to rotate about the center S , that moves
with the revolution of the moon. This is shown, without the daily rotation of the
earth about its axis, for three different positions of the moon
the oceans (low tides and high tides) since for liquids the restoring elastic force is zero. However, it also appears with smaller
elongation in the solid crust of the earth.
The deformation of the earth and the resulting tides have three
causes:
a) The centrifugal distortion due to the motion of earth and
moon about their common center of mass.
b) The gravitational force, effected by the masses of moon and
sun.
c) The centrifugal distortion due to the rotation of the earth
around its axis.
In order to understand this tidal deformation we discuss at first
the simplified model of the deformation of the non-rotating earth
and neglect the gravitational attraction by the sun and the revolution of the earth around the sun. We restrict the discussion
therefore to the influence of the moon on the non-rotating earth.
Under the mutual gravitational attraction
FG D
G
ME MMo
rO 0 ;
r02
(6.50)
Chapter 6
earth and moon move around their common center of mass S
(also called bari-center) which lies still inside the earth (about
0.75 of the earth radius from the center). The distance between
the centers of earth and moon is r0 . During a moon-period of
27.3 days the center M of the earth moves on a circle with radius
0:75R around the baricenter S, which always lies on the line
ME -MMo . All arbitrary points Pi in the earth move around S on
circles with radii Pi S. However, the center of mass S has
no fixed position inside the earth but moves during one moon
period inside the earth on a circle with radius 0:75RE around the
center M of the earth, because the space-fixed center of mass S
lies always on the line between earth-center and moon center.
The motion of the non-rotating earth as extended body, described in the coordinate system of the earth, is therefore not a
rotation about a fixed axis but rather a shift since the space-fixed
point S. has not a fixed location inside the earth. The revolution
of the moon and the earth about S with the angular velocity ˝
causes therefore for all points of the non-rotating earth the same
centrifugal force
Fcf D m˝ 2 RS D m˝ 2 0:75R :
(6.51)
On the other hand, the gravitational attraction between earth and
moon is different for the different points of the earth because
of their different distance from the moon center. For the earth
center M it is
ME MMo
FG D G
rO0
(6.52)
r2
with r D r0 . Here gravitational force and centrifugal force just
compensate each other.
Fcf D ME ˝ 2 0:75R rO0 D FG .r0 / :
The total force in M is zero (Fig. 6.53). This is no longer true
for other points P because the distances to the moon are different and therefore the gravitational force differs while for the
non-rotting earth the centrifugal force is the same for all points
P. For example the gravitational force in the points A and B in
Fig. 6.54 is
m MMo
FG .rA / D G
rO0 ;
.r0 C R/2
(6.53)
m MMo
FG .rB / D G
r
O
:
0
.r0 R/2
Compared with the gravitational force FG .r0 / in M the force
differences are F.rA / D FG .rA / FG .r0 / and F.rB / D
FG .rB / FG .r0 / which point in the direction of the connecting line earth–moon. The magnitude of these differences can
be obtained from (6.52) and (6.53). Because R r0 , we can
approximate .1 C R=r0 / 2 1 2R=r0 and we get:
1
m MMo
1 rO 0
F.rA / D G
.1 C R=r0 /2
r02
2m MMo
R rO 0
r03
R
D 2FG .r0 / rO 0 ;
r0
R
F.rB / D C2FG .r0 / rO 0 ;
r0
G
(6.54)
6.6 The Earth as Deformable Body
177
Figure 6.54 Deformation of the earth by the tides (exagerated). The arrows
give magnitude and direction of the tidal forces
points to the center of the moon (Fig. 6.54b), while the centrifugal force is directed as for all points of the earth in the direction
of r0 and is anti-collinear to FG , while the magnitude of both
forces are equal, i.e. FG .r0 / D Fcf . (Note, that we regard
a non-rotating earth and Fcf is only due to the revolution of
earth and moon
p of mass S). With
p around the common center
cos ˛ D r0 = .r02 C R2 / and sin ˛ D R= .r02 C R2 / the resulting residual force is
1
0
r03
1
C
B .r2 C R2 /3=2
0
C
F.rC / D Fcf C FG D FG .r0 / B
2
A
@
r0 R
2
(6.56)
.r0 C R2 /3=2
!
R 23 .R=r0 /
FG .r0 /
;
r0
1
because R r0 F.rC / points nearly into the y-directions to
the center of the earth. it therefore decreases the curvature of
the earth surface (Fig. 6.57b) which causes low tide. Its amount
m MMo
F.rC / D jFG .rC / FG .r0 /j G
R
r03
1
R
D FG .r0 /
D F.rA /
r0
2
(6.57)
is smaller by the factor 1=2 than in the points A and B. For all
other points of the earth surface the resulting forces F have a
radial as well as a tangential component. The tangential component causes an acceleration of the ocean water towards the
points A or B. The borderline between the different tangential
directions lies in Fig. 6.54a left of the line CD. where the xcomponent of FG is
FGr D C 32 FG .r0 /.R=r0 / :
(6.58)
From (6.54) and (6.56) one can infer, that the maximum tide
force depends on the ratio MMo =r3 . If the numerical values for r
and MMo are inserted one obtains MMo =r3 D 1:34 10 3 kg=m3
and a tide acceleration of a1 D F=m D 1:1 10 6 m=s2 . This
leads to a deformation of the solid earth crust of up to 0:5 m.
3
Since Msun =rsun
D 6:6 10 4 kg=m3 the effect of the sun on the
tides is only about half of that of the moon and one obtains for
the contribution to the tide-acceleration a2 D 5:6 10 7 m=s2 . If
sun and moon stand both on a line through the center of the earth
(this is the case for full moon and for new moon) the actions of
moon and sun add (spring-tide). If sun and moon are in quadrature (the connecting lines sun–earth and moon–earth intersect in
the earth center under 90ı (Fig. 6.55)) the effects subtract (neap
tide).
Up to now, we have neglected the daily rotation of the earth. It
brings about two effects:
An additional centrifugal force, which causes the deformation of the earth into a oblate symmetric top (see Sect. 6.6.1)
The deformation, which amounts to about 21 km at the equator, is very much larger than that caused by the moon but it
is equal for all points on the same latitude and is not time
dependent in contrast to the tides caused by the moon.
Chapter 6
The difference F.rB / is directed from M to the center of the
moon, while F.rA / has the opposite direction. Both differences result in a convex curved deformation of the earth surface,
as shown exaggerated in Fig. 6.54. For a mass in the points C or
D the gravitational force caused by the moon
m MMo
rO D fFx ; Fy g
FG .rC / D G 2
r0 C R2
!
(6.55)
r02
cos ˛
D FG .r0 / 2
sin ˛
r0 C R2
Figure 6.55 Spring tide and neap tide caused by addition or subtraction of the
gravitational forces by moon and sun
178
6 Real Solid and Liquid Bodies
Figure 6.56 Influence of the inclination of the orbital plane of the moon on
the periodical variation of the tidal elevation
When the revolution of the moon around the earth is ignored,
the two tide maxima at the points A and B in Fig. 6.54 and the
low tides in the point C and D would travel around the earth
in 24 hours. At a fixed point one would experience every
12 hours a high tide and a low tide. The deformation of the
solid crust is about 0:5 m, that of the ocean away from the
coast about 1 m. Tide amplitudes up to 15 m are observed at
the coast and in particular in narrow bays. They are generated
by nonlinear effects during the propagation of tidal waves.
For a more accurate description of the tides the revolution of the
moon has to be taken into account. It demands the following
corrections of our simple model:
Chapter 6
The moon moves around the earth–moon-center of mass S in
27.5 days with the same direction as the rotation of the earth.
Therefore, the round-trip time of the tides is 24:87 h instead
of 24 h.
The plane of the moon’s revolution is inclined against the
equator plane (Fig. 6.56). An observer in the point A experiences a higher tide amplitude than an observer in B
12:4 h later. This can be seen as follows: The centrifugal force (6.51) caused by the revolution of the earth–moon
system around S is in A parallel to the gravitational force
FGE caused by the mass of the earth. The resulting force
F D FGE C FGM C Fcf is perpendicular to the earth surface.
The total force has to include the centrifugal force FcE caused
by the rotation of the earth, which is perpendicular to the rotation axis of the earth. In the point B the centrifugal force
Fcf has the same direction than in A, but the gravitational
force FGE has a nearly opposite direction and therefore the
vector sum of the two forces is in B smaller than in A. The
force FcE has for both points the same direction because they
are located on the same circle of latitude (Fig. 6.56). The
tide amplitudes show an amplitude modulation with a period
of about 12:4 h. The modulation index depends on the geographical latitude.
The motion of the moon changes the relative positions of the
interacting sun, moon and earth. Therefore, the vector sum
of the tide forces show also a periodic modulation.
These considerations illustrate that the total tide amplitude is determined by the superposition of many effects and is therefore
a complicated function of time (Fig. 6.57). It can be measured
with various techniques. One of them uses the time variation
of the gravitational acceleration g which depends on the ge-
Figure 6.57 Time dependent course of the tidal elevation at a fixed point on
the earth surface, measured as the corresponding variation g of the earth
acceleration g
ographic location and is affected by the tides. Another very
sensitive interferometric technique measure the local deformation of the earth crust (see Sect. 6.6.4).
6.6.3
Consequences of the Tides
With the tides of the oceans as well as with the periodic deformations of the earth crust, friction occurs which causes a partial
transfer of kinetic energy into heat. This lost kinetic energy
slows down the rotation of the earth and causes an increase of
the rotation period by 90 ns per day. Within 106 years this prolongs the duration of the day by 0:5 min (see Probl. 1.4).
The gravitational force between earth and moon causes of
course also deformations on the moon. Accurate measurements
have proved that the shape of the moon is an ellipsoid with
the major axis pointing towards the earth. The general opinion is that in former times the moon also rotated around its axis.
This rotation was, however, in the course of many million years
slowed down by friction until the moon no longer rotates and
shows always the same side to the earth.
The tidal friction of earth and moon has the following interesting
effect: The total angular momentum of the earth–moon system
is constant in time because the system moves in the central force
field of the sun (the additional non-central forces due to interactions with the other planets are negligible). Since the rotation of
the earth around its axis slows down and its angular momentum
I ! decreases, the orbital angular momentum of the earth–moon
system
jLEM j D r vrel D IEM ˝
(r D distance earth–moon, vrel D relative velocity of the moon
against the earth, IEM D inertial moment of the earth–moon
system and ˝ D angular velocity of the rotating earth–moon
system) has to increase. The moon is accelerated by the tidal
wave running around the earth. This can be seen as follows
(Fig. 6.58): The earth rotating with the angular velocity ! ˝
accelerates the tidal waves due to the friction forces: This
acceleration brings the tidal maximum slightly ahead of the connecting line between the centers of earth and moon. Due to the
slightly increased gravitational force, the moon is accelerated
while the earth rotation decreases. The larger kinetic energy
6.6 The Earth as Deformable Body
Well
Condensor
plates
C ∝ 1/d
of the moon increases its total energy .Ekin C Epot / and therefore also its distance to the earth. In former times the moon
was closer to the earth. The nowadays generally accepted theory [6.7a, 6.7b] assumes that the moon was part of the earth but
has been catapulted out of the earth by the impact of a heavy
asteroid some billion years ago (see Vol. 4).
6.6.4
Measurements of the Earth Deformation
The deformation of the earth by tidal effects can be measured
with different techniques. We will shortly discuss three of them:
6.6.4.1
Changes of the Gravitational Force
According to (6.54) the additional gravitational force caused by
the moon in the points A and B in Fig. 6.54 is
FG
2mMMo
R:
3
rMo
(6.59)
In the gravity meter shown in Fig. 6.59 a mass m is suspended by
a spring in such a way, that a small change FG D m g due to
the corresponding change of g causes a large vertical deflection
of the arrow on the scale. This is achieved by a sloped mounting
of the spring with length L and a restoring force Fr D D L.
With the slope angle ˛, the vertical deflection z causes a length
increase of the spring L D z sin ˛ (Fig. 6.59b) and a change
of the restoring force Fr D m g sin ˛.
tension spring
compression spring
α
α
Figure 6.59 Measurement of the gravitational force with a special spring balance
Wire
Ball
Figure 6.60 Measurement of the deviation of g from the vertical direction
The device measures the periodic changes of FG with a period of 24:87 h from which the tidal amplitudes can be inferred
(Fig. 6.57). Because of the different contributing effects, g.t/
follows a complicated curve.
The experimental arrangement of Fig. 6.60 allows to measure
the deviation from the vertical direction of the earth acceleration
g. Without external perturbation, g would point nearly to the
earth center (only for a spherical mass distribution it would point
exactly to the center). The additional gravitational force exerted by the moon causes a slight deviation from this direction.
The maximum angular deviation, which depends on the latitude,
amounts only to about 2:1 10 6 rad.D 0:400 /, the measurement
must be sufficiently accurate. This required accuracy can be
reached with a pendulum [6.9]. A metal ball suspended on a
long wire in a well is connected with one plate of a charged capacitor, while the other plate is fixed on the wall of the well. Any
deviation of the pendulum from the vertical direction changes
the distance between the two plates and therefore the voltage of
the capacitor (see Vol. 2, Sect. 5).
6.6.4.2 Measurements of the Earth Deformation
Here the change L of the length L between two points A1 and
A2 connected with the earth ground is measured. Figure 6.61
illustrates the method. A very sensitive Laser interferometer is
located in a gold mine deep in the ground in order to eliminate
acoustic noise from the surroundings. The two mirrors of the
laser resonator are mounted on the ground base at the points A1
and A2 separated by the distance L. The optical frequency of the
laser L D m c=.2L/ is determined by the length L of the resonator and the large integer m 1. If the length L changes due
to the deformation of the earth crust, the laser frequency changes
accordingly. This frequency change can be measured very accurately, when the laser beam is superimposed on a detector with
the output beam of a reference laser with stabilized frequency r .
The difference frequency L r in the radio-frequency range
can be counted by a digital frequency counter. Existing devices
have resonator lengths of 100 m up to several km. They can
measure deformations of the earth crust of less than 10 9 m (see
Vol. 2, Sect. 10.4). This sensitivity is sufficient to measure the
deformation of the ground base in the Rocky Mountains caused
by the tidal waves of the Pacific Ocean [6.10].
Chapter 6
Figure 6.58 Deceleration of the earth rotation and acceleration of the orbital
velocity of the moon by the tidal friction
179
180
6 Real Solid and Liquid Bodies
Mirror
Mirror
Laser
Photodetector
Frequency
counter
νR – ν L
Gold mine
Reference laser
with fixed
frequency νR
Figure 6.61 Laser interferometer for the measurement of the deformation of the earth crust
Summary
Elastic bodies show restoring forces for any deformation of
their shape. For sufficiently small deformations, these forces
are proportional to the elongation from the equilibrium position.
For a relative length increase " D L=L of a body with
length L, constant cross section and elastic modulus E one
needs a tensile stress D E " (Hooke’s law).
A length change L of a rod with length L and quadratic
cross section A D d 2 caused by the tensile stress is accompanied by a change A of its cross section. The relative
change of the volume V
V
D .1 2/
V
E
is determined by the elastic modulus E and the transvers contraction ratio D .d=d/=.L=L/.
Exposed to isotropic pressure p the relative volume change
V=V D p of a body is determined by the compressibility D .3=E/ .1 2/.
A force F acting tangentially on a wall of a body causes a
shear of the body. For a cuboid with the side area d 2 the shear
angle ˛ is related to the shear stress D F=d 2 by D G ˛
where G is the modulus of shear.
A rod with length L and cross section d b is fixed at one end.
The vertical force F acting on the other end causes a bending
s D .4L3 F/=.E d 3 b/ ;
Chapter 6
which is proportional to the third power of the length L and
the vertical width d.
Beyond the linear range of Hooke’s law plastic deformations
occur. If a periodical tensile stress acts on a rod with length
L, a closed hysteresis curve ."/ is traversed. The area enclosed by this curve represents the energy that is transformed
into heat for every cycle.
Inside a liquid the same pressure is present for all volume
elements with the same distance h from the surface. The
hydrostatic pressure p.z/ D p0 C % g .h z/ at this height
increases linearly with the height .h z/ of the liquid with
density % above the layer at z. At the upper surface z D H
of a liquid with total height H the pressure is p0 (for example
the barometric pressure of the air above the surface).
Each solid body with mass m and density %s experience in
a liquid a buoyant force FB which is equal but opposite to
the weight FG of the liquid volume displaced by the solid
body. If jFjB > m g the body floats at the liquid surface, for
jFB j D m g the body can float at any height in the liquid.
Because of the attractive forces between the molecules of a
liquid, energy is required to bring molecules from the interior
to the surface. The energy, necessary to increase the surface
by 1 m2 , is the specific surface energy. It is equal to the specific surface tension.
The shape of the surface of a liquid in a container depends
on the different surface tensions for the boundaries between
container wall and liquid, liquid and air, container wall and
air and on the gravity force. It always takes that form, for
which the energy is minimum.
Because of the surface tension a liquid can rise in a capillary
(wetting liquid) or descend (non-wetting liquid).
When two bodies come into touch, friction forces appear,
which are different for a relative velocity zero (static friction)
or for a relative motion (sliding friction). The smallest friction is found, when a circular body rolls on a plane base. The
quantitative description uses friction coefficients , which
depend on the materials of the two bodies. Generally it is
js j > jsl j > jR j, where s is the coefficient for static
friction, sl for sliding friction and R for rolling friction. A
liquid film between the two solid bodies reduces the friction
considerably.
The earth is a deformable ellipsoid which is permanently deformed by its rotation and periodically by the gravitational
forces exerted by moon and sun, which cause tidal effects.
The periodic deformations are partly non-elastic and the friction transfers part of the rotational energy into heat. This
causes a slowdown of the earth rotation and a prolongation of
the day. Conservation of the total angular momentum leads
to an increase of the distance earth–moon.
References
181
Problems
a) How deep does it immerse?
6.1
What is the change L of a steel rope with L D 9 km,
b) What is the location of center of mass and metacenter?
a) which hangs in a vertical well?
b) What is the maximum length of the rope before its rupture? c) What is the maximum angle of its symmetry axis against the
vertical direction before it becomes unstable?
c) How large is L when the rope is lowered from a ship into
the ocean? (E D 2 1011 N=m2 ; %steel D 7:7 103 kg=m3,
6.7
Which energy has to be spent in order to lift a full cube
%ocean D 1:03 103 kg=m3)
of steel from the bottom of a swimming pool with the water
6.2
A steel beam with L D 10 m is clamped at one end. A depth of 4 m to a position where the lower side of the cube is at
force F D 103 N acts on the other end in vertical z-direction. the surface of the water?
How large is the bending of this end
Which force was necessary to separate the two hemia) for a rectangular cross section d b with d D z D 0:1 m; 6.8
spheres
in the demonstration experiment by Guericke in Magdeb D y D d=2?
b) for a double T-profile (Fig. 6.15d) with b1 D d1 D 0:1 m, burg with a diameter of 0:6 m, when the pressure difference
between inside and outside was p D 90 kPa? Guericke had
b2 D d2 D 0:05 m?
used 16 horses. What should have been done in order to sepa6.3
The deep ocean aquanaut Picard reached in his spherical rate the hemi-spheres already with 8 horses?
steel submarine a depth of 10:000 m in the Philippine trench.
In order to verify that a gold bar is really made of gold
How large are pressure and total force exerted on the sphere? 6.9
What is the volume change V=V caused by the pressure
.%gold D 19:3 kg=dm3 / a goldsmith measures its weight in air
a) for a hollow sphere with wall thickness of 0:2 m?
and when totally immersed in water. Which ratio of the two
b) for a full sphere?
values is obtained
a) for a 100% gold bar?
6.4
A turbine drives a generator connected to a steel shaft b) for a 20% admixture of copper .% D 8:9 kg=dm3/?
with length L and diameter D. By which angle ˛ are the two c) What is the minimum required accuracy of the measureends of the shaft twisted if the power P D 300 kW is transferred
ments for unambiguously distinguishing between the two
at a frequency ! D 2 25 s 1
cases? What is the accuracy if an admixture of 1% of copper
a) for a steel shaft as full cylinder with D D 0:1 m; L D 20 m?
should be detected?
b) for a hollow cylinder with D1 D 5 cm and D2 D 10 cm?
6.10 A round cylinder of wood (L D 1 m, d D 0:2 m,
6.5
What is the density of water with a compressibility D % D 525 kg=m3 ) is floating in water. How deep does it immerse
4:8 10 10 m2 =N at a depth of 10:000 m?
a) in a horizontal position?
b) if a steel ball with m D 1 kg is attached to one end in order
6.6
A hollow steel cube .% D 7:8 103 kg=m3/ with edge
to bring it into a vertical floating position?
length a D 1 m and a wall thickness of d D 0:02 m and with an
open upper side floats on water.
6.1. W.D. Callister, D.G. Rethwisch, Material Science and Engineering. An Introduction. (Wiley, 2013)
6.2. Deformation, ed. by M. Hazewinkel Encyclopedia of
Mathematics. (Springer, 2001)
6.3. S. Lipschitz, Schaum’s Outline of mathematical Handbook of Formulas. (McGraw Hill, 2012)
6.4. M. Abramowitz, I. Stergun, Handbook of Mathematical
Functions. (Martino Fine Books)
6.5. R.H. Rapp, F. Sanso (ed.), Determination of the Geoid.
(Springer, Berlin, Heidelberg, 1991)
6.6a. V.M. Lyathker, Tidal Power. Harnessing Energy from Water Currents. (Wiley Scrivener, 2014)
6.6b. L. Peppas, Ocean, Tidal and Wave energy. (Crabtree Publ.
Comp., 2008)
6.6c. https://en.wikipedia.org/wiki/List_of_tidal_power_
stations
6.6d. https://en.wikipedia.org/wiki/Tidal_power
6.7a. http://novan.com/earth.htm
6.7b. P. Brosche, H. Schuh, Surveys in Geophysics 19, 417
(1998)
Chapter 6
References
182
6 Real Solid and Liquid Bodies
6.8a. D. Flannagan, The Dynamic Earth. (Freeman, 1983)
6.8b. G.M.R. Fowler, The Solid Earth. An Introduction to
Global Geophysics. (Cambridge Univ. Press, 2004)
6.9. D. Wolf, M. Santoyo, J. Fernadez (ed.), Deformation and
Gravity Change. (Birkhäuser, 2013)
6.10. J. Levine, J.L. Hall, J. Geophys. Res. 77, 2595 (1972)
6.11. A.E. Musset, M.A. Khan, Looking into the Earth: An Introduction to Geological Geophysics. (Cambridge Univ.
Press, 2000)
6.12. M. Goldsmith, M.A. Garlick, Earth: The Life on our
Planet. (Kingfisher, 2011)
Chapter 6
7.1
Macroscopic Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184
7.2
Atmospheric Pressure and Barometric Formula . . . . . . . . . . . . . . 185
7.3
Kinetic Gas Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188
7.4
Experimental Proof of the Kinetic Gas Theory . . . . . . . . . . . . . . . 196
7.5
Transport Phenomena in Gases . . . . . . . . . . . . . . . . . . . . . . . 198
7.6
The Atmosphere of the Earth . . . . . . . . . . . . . . . . . . . . . . . . . 204
7
Chapter 7
Gases
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208
© Springer International Publishing Switzerland 2017
W. Demtröder, Mechanics and Thermodynamics, Undergraduate Lecture Notes in Physics, DOI 10.1007/978-3-319-27877-3_7
183
184
7 Gases
Chapter 7
Different from solid or liquid bodies, which change their volume only slightly under the action of external forces, gases can
be expanded readily. They occupy any volume that is offered to
them. Under the action of external pressure their volume can be
reduced by orders of magnitude up to a certain limit. The reason
for these differences is their much smaller density. At atmospheric pressure, the gas density is smaller than that of solids or
liquids by about three orders of magnitude. The mean distance
between the atoms or molecules is therefore about ten times
larger. This has the consequence that their mean kinetic energy is much larger than the mean potential energy of the mutual
attraction or repulsion, while for liquid and solid bodies both energies are nearly equal at room temperature (see Sect. 6.1).
In this chapter we present at first the macroscopic properties of
gases before we will discuss in more detail the atomic explanation of the observed macroscopic phenomena. The atomic
fundamentals, developed already in the 19th century as kinetic
gas theory, was one of the most powerful supports for the existence of atoms and their relevance as constituents of matter.
7.1
Macroscopic Model
For a constant temperature the density % of a gas is proportional to its pressure.
The volume V of an enclosed gas can be changed by a movable
piston due to a variable pressure p (Fig. 7.1). For the relation
between V and p at a constant temperature T the experiment
gives the result
p V D const
(Boyle–Mariotte’s law) :
(7.1)
From V D const =p we obtain by differentiation
dV
D
dp
const
D
p2
V
:
p
1 @V
;
V @p
Œ D
V
p
)
m2
:
N
D
(7.3a)
1
:
p
(7.3b)
A gas is easier to compress for smaller pressures. For a total
mass M of a gas in a volume V the density is % D M=V. For an
enclosed gas its total mass M is constant. Its density % is then
inversely proportional to its volume V. Inserting V D M=% into
(7.1) gives
pD
const
%
M
i.e. p / % :
This can be seen from the general gas equation p V D N k T
(see Sect. 10.3), where N is the total number of molecules in
the volume V, which is independent of the boundaries of the
volume V. Since n D N=V is the density of molecules with
mass m, which is proportional to the mass density % D n m=V
we obtain p / %.
The gas pressure p with the unit Newton per square meter
For a constant temperature it follows
dV
@V
D
D
@p
dp
Remark. This is also valid for non-enclosed gases, as for
example in the free atmosphere.
(7.2)
As measure for the compressibility of a gas we define the quantity
D
Figure 7.1 A movable piston changes volume V and pressure p of an enclosed
gas volume. a Principle, b demonstration of the Boyle–Marriotte law
(7.4)
Œp D
N
D Pascal D Pa
m2
can be measured with different techniques (see Sect. 9.3). A
simple method uses a mercury manometer (Fig. 7.2). In the
container with volume V is a gas under the pressure p. The
left branch of a U-shaped tube filled partly with liquid mercury
is closed at the upper end, while the right branch is connected
with the container. If the upper valve in Fig. 7.2 is open, the
pressure p causes a difference h of the mercury heights in the
two branches. Equilibrium is reached if the pressure caused by
the gravity just compensates the pressure p in the container. This
gives the relation
% g h D p
p0 ;
where p0 is the vapour pressure of mercury above the mercury
surface in the left branch.
Historical the pressure difference corresponding to h D 1 mm
between the two mercury columns is named 1 torr, in honour of
the Italian physicist Torricelli. The unit used nowadays in the
SI-system is 1 Pascal. The following relations hold:
185
7.2 Atmospheric Pressure and Barometric Formula
Unit
Figure 7.2 Measurement of gas pressure with a mercury manometer
1 Pa D 1 N=m2
1 standard atmosphere D 1 atm D 101 325 Pa
1 torr D .1=760/ standard atmosphere D 133:3 Pa :
7.2
Atmospheric Pressure and
Barometric Formula
Similar to liquids also in gases a static pressure is present due
to the weight of the air. It can be measured with the Torricelli
U-tube in Fig. 7.3, filled partly with mercury. The left branch
is closed at the upper end, while the right branch is open. In
the left branch above the liquid mercury surface is the small
vapour pressure p0 of mercury (about 10 3 torr D 0:13 Pa at
room temperature) which can be neglected. Due to the atmo-
1 Pascal
Athmosphere Definition
abbreviation
1 Pa
1 N=m2
1 Hektopascal
1 hPa
102 N=m2
102 Pa
1 Bar
1 bar
105 N=m2
105 Pa
3
bar
Conversion
–
102 Pa
1 Millibar
1 mbar
10
1 Torricelli
1 Torr
1 mm hg
133:32 Pa
1 physical atmosphere
1 atm
760 Torr
101 325 Pa
1 technical atmosphere
1 at
1 kp=cm2
9:8 104 Pa
spheric pressure p the mercury in the right branch is depressed
by h D p=.% g/.
The pressure of the earth atmosphere at sea level h D 0 at
normal weather conditions is 101 325 Pa and is called normal
pressure or standard pressure which is often given in the unit 1
standard atmosphere (1 atm). This pressure causes in the mercury manometer of Fig. 7.3 a height difference of 760 mm.
In meteorology the hecto-pascal .1 hPa D 100 Pa/ is a commonly used unit. In Tab. 7.1 the conversion factors for some
pressure units are listed [7.1].
The weight of the air column above an area A in the height h
decreases with increasing h (Fig. 7.4). Changing the position of
A from h to h C dh decreases the weight by % g A dh and
therefore the pressure p decreases as
dp D % g dh :
(7.5)
In the case of liquids the density is independent of the height because of the small compressibility. The solution of (7.5) shows
the linear dependence p D % g h of the pressure on the height
h (Fig. 7.5b). This is no longer true for gases, where the density is proportional to the pressure and therefore depends on the
heights in the atmosphere. From (7.4) we obtain for a constant
temperature T
%0
p0
p
D
D const ) % D
p:
%
%0
p0
Inserting this into (7.5) gives
dp D
Figure 7.3 Torricelli manometer for measuring the barometric air pressure
%0
gp dh :
p0
Figure 7.4 Derivation of barometric Eq. 7.6
(7.5a)
Chapter 7
Table 7.1 Units of pressure
186
7 Gases
Chapter 7
The pressure decrease of an isothermal atmosphere follows the
exponential law (7.6a) contrary to the linear decrease in liquids.
The atmosphere has therefore no sharp boundary!
Note: The real earth atmosphere is not isothermal! The temperature decreases with increasing height (see Sect. 7.6). The
pressure p.h/ is therefore slightly different from (7.6). Nevertheless is (7.6) a useful approximation, which is sufficiently
accurate for many applications.
Analogue to the situation in liquids the Archimedes’ principle
of buoyancy is valid for bodies in air.
A body in air experiences a buoyancy force, that has the
opposite direction but the equal amount as the weight of
the displaced air.
Figure 7.5 Comparison of pressure dependence p .s / in the earth atmosphere
and in a water column of 10 m height
This principle is the basis for balloon flights. A balloon can only
rise in air, if its weight (balloon + car + passengers) is smaller
than the buoyancy force. For a balloon with total mass M and
total volume V this gives the condition
M g < V %air g :
Integration yields
ln p D
%0
gh C C :
p0
(7.5b)
With p.h D 0/ D p0 the integration constant C becomes C D
ln p0 . Solving for p gives the barometric formula
p D p0 e
%0 gh=p0
(7.6a)
:
The balloon must therefore contain a gas with a smaller density than the surrounding air. Generally helium is used, since
hydrogen is too dangerous due to its possible explosion.
Another solution is the hot-air balloon, where a burner blows
hot air into the balloon shell. The density % D p=kT is inversely
proportional to the temperature T (Fig. 7.6).
Example
Note, that the ratio %0 =p0 depends on the temperature T.
The atmospheric pressure of an isothermal atmosphere
(T D constant, independent of h) decreases exponential with
the height h (Fig. 7.5a).
Because % D .%0 =p0 / p the density % follows the same formula
% D %0 e
%0 gh=p0
(7.6b)
:
Inserting the numerical values of the atmosphere (%0
1:24 kg=m3 and p0 D 1013 hPa) into (7.6a) yields
p D p0 e
h=8:33 km
:
D
(7.6c)
For h D 8:33 km the pressure p has decreased to p0 =e
373 hPa.
The height h1=2 where the pressure has dropped to .1=2/p0 is obtained from exp.g h1=2%0 =p0 / D 2 which gives h1=2 D 5:77 km.
On a mountain with an elevation of 5:77 km the barometric
pressure sinks to half of its value at h D 0.
The density % at a temperature T D 80 ı C D 350 K is
smaller than air at room temperature T D 20 ı C D 290 K
by the factor 290=350 D 0:83. For a balloon volume of
3000 m3 the buoyancy force is then FB D g V % D
8:270 N. The maximum mass of balloon + passengers is
then 843 kg. The mass of the balloon is about 100 kg,
that of the burner with propane supply about 200 kg. This
leaves a maximum weight for the passengers of 543 kg.
J
Since the density of the earth atmosphere decreases exponentially with the height h, also the buoyancy decreases with h. This
limits the maximum altitude of weather- and research-balloons,
which should rise up to very high altitudes in order to investigate the upper part of our atmosphere. One uses extremely large
balloon volumes which are at the ground only partly filled with
helium but blow up with increasing altitude because of the decreasing air pressure (Probl. 7.18) (Fig. 7.7).
187
Chapter 7
7.2 Atmospheric Pressure and Barometric Formula
Figure 7.6 Lift of the first manned Montgolfiere. At the 21st of November 1783 the hot-air-balloon, named after its inventor Montgolfiere, started in the garden
of the castle Muette near Paris with two ballonists and landed safely after 25 min at a distance of 10 km from the starting point. The balloon was constructed with
thin branches of a willow tree, that stabilized the envelope of thin fabric, covered with coulorfully painted paper. The air inside the ballon was heated by a coal fire
in the center of the lower opening. With kind permission of Deutsches Museum München
188
7 Gases
Chapter 7
Figure 7.7 Start of a helium-filled research balloon for the investigation of the higher stratosphere. The balloon is filled only with a low He-pressure. With
increasing height the external pressure decreases and the balloon inflates, increasing its volume and the buoancy (SSC/DLR) (http://www.eskp.de/turmhoheforschungsballons-messen-ozonschicht/)
7.3
Kinetic Gas Theory
The kinetic gas theory, which was developed by Boltzmann,
Maxwell and Clausius in the second half of the 19th century,
attributes all observed macroscopic properties of gases to the
motion of atoms and their collisions with each other and the
wall. Its success has essentially contributed to the acceptance
of the atomic hypothesis (see Vol. 3, Chap. 2). The exact
theoretical description requires a more detailed and advanced
mathematical model. We will therefore restrict the treatment to
a simplified model, which, however, describes the essential basic ideas and the experimental findings correctly.
7.3.1
d > 2r0 the atoms do not interact. The interaction potential
in this model is therefore (Fig. 7.8)
Epot .r/ D 0
Epot .r/ D 1
for jrj > 2r0
for jrj 2r0 :
Such a gas model is called ideal gas, if r0 is very small against
the mean distance hdi between the atoms. This means that the
atomic volume is negligible compared to the volume V of the
gas container. In this model the atoms can be treated as point
masses (see Sect. 2.1).
The Model of the Ideal Gas
The most simple gas model is based on the following assumptions: The gas consists of atoms or molecules which can be
described by rigid balls with radius r0 . They move with statistically distributed velocities inside the gas container. Collisions
with each other or with the walls are governed by the laws of
energy-and momentum conservation. The collisions are completely elastic. Any interaction between the balls only occurs
during collisions (direct touch of two balls). For distances
Figure 7.8 Interaction potential between two rigid balls with radius r0
7.3 Kinetic Gas Theory
Chapter 7
Example
At a pressure of 1 bar and room temperature (T D 300 K)
1 cm3 of a gas contains about 3 1019 molecules. Their
mean relative distance is hdi D 3 nm. For helium atoms
is r0 0:05 nm, which implies r0 =hdi D 0:017 1.
Helium at a pressure of 1 bar can be therefore regarded as
J
ideal gas.
The pressure that the gas exerts onto the wall is caused by the
momentum transfer of the atoms to the wall. The force F acting
on the area A of the wall during collision of the atoms with the
wall is equal to the momentum transfer per second to the area A.
The pressure p D F=A is then
d momentum transfer to area A
pD
:
(7.7)
dt
area A
If for instance N dt atoms with mass m hit the wall within the
time interval dt with a velocity v in the direction of the surface
normal, the momentum transfer per second for elastic collisions
is 2N m v and the pressure onto the wall is p D 2N m v=A.
7.3.2
Basic Equations of the Kinetic Gas
Theory
We will at first regard the atoms as point masses and only take
into account their translational energy. The discussion of rotationally or vibrationally excited molecules demands a farther
reaching discussion which will be postponed to Sect. 10.2
For N molecules in a volume V the number density is n D N=V.
At first we regard only that part nx of all molecules per cm3 in
a cubical volume, which move with the velocity vx into the xdirection (Fig. 7.9). Within the time interval t
Z D nx vx A t
molecules hit the surface area A. These are just those molecules
in the cuboid with length vx t and cross section A, illustrated
in Fig. 7.9. Each of these molecules transfers the momentum
Figure 7.10 Momentum transfer for elastic collisions with a wall
px D 2 m vx . The force onto the area A is then F D Z
px =t D 2Z m vx =t and the pressure onto the wall is
p D F=A D 2m nx vx2 :
(7.8)
When molecules with the velocity v D fvx ; vy ; vz g do not
move vertically to the wall, still only the momentum 2m vx is
transferred, because the tangential components do not transfer
momentum to the wall (Fig. 7.10).
Not all molecules have the same velocity. Under stationary conditions the velocities of the molecules are isotropically
distributed, which means that each direction has the same probability. Since the pressure of a gas is isotropic, the mean
momentum transfer must be equal into all directions. The mean
square value of the velocities is
Z
1
hvx2 i D vx2 D
(7.9)
N.vx / vx2 dvx D vy2 D vz2 ;
N
where N.vx /dvx is the number of molecules in the volume V
with the velocity components vx in the interval from vx to vx C
dvx . Because on the average half of the molecules move into
the Cx-direction and the other half into the x-direction, the
pressure, exerted by all molecules with a number density n D
N=V in the x-direction on the wall in the y-z-plane is given by
p D 21 n 2mvx2 D n m vx2 :
(7.10)
With v 2 D vx2 C vy2 C vz2 it follows from (7.9)
vx2 D vy2 D vz2 D 13 v 2 :
(7.11)
This gives with (7.10)
pD
2 m
2
1
m nv 2 D n v 2 D n Ekin ;
3
3
2
3
(7.12a)
where Ekin D .m=2/ n v 2 is the mean kinetic energy per
molecule. With n D N=V this can be written as
p V D 32 N 21 mv 2 ;
(7.12b)
where N D n V is the total number of molecules in the volume
V.
Figure 7.9 Derivation of Eq. 7.8
189
Remark. We use for the average values of a quantity A the notations hAi
as well as A.
190
7 Gases
Chapter 7
7.3.3
Mean Kinetic Energy and Absolute
Temperature
All experiments give the result that the product p V depends
solely on the temperature and is for constant temperature constant (Boyle–Mariotte’s law). This implies also that the mean
kinetic energy Ekin D .m=2/v 2 depends on the temperature. It
turns out that it is convenient to define an absolute temperature
which is proportional to Ekin .
The absolute temperature T (with the unit 1 Kelvin D 1 K) is
defined by the relation
m 2
3
v D kT ;
(7.13)
2
2
where k D 1:38054 10
23
J=K is the Boltzmann constant.
With this definition Eq. 7.12 becomes the general gas-equation
pV DNkT ;
(7.14)
which represents a generalization of Boyle–Mariotte’s law (7.1)
and which reduces to (7.1) for T D const.
Each molecule can move into three directions x, y, and z. This
means it has three degrees of freedom for its translation. Collisions with other molecules change direction and magnitude of
its velocity. In the time-average all directions are equally probable. We therefore obtain for the mean square velocities the
relations
˝ ˛
˝ ˛
˝ ˛
˝ 2˛
vx t D vy2 t D vz2 t D 31 v 2 t D 13 v 2 :
The mean kinetic energy of a molecule at the temperature T is
then
Ekin D 21 kT per degree of freedom :
Remark. In statistical physics it is proved [7.2] that in a closed
system of many mutually interacting particles at thermal equilibrium the time average hAi t of a physical quantity A is equal
to the ensemble average
1 X
AD
Ai ;
N
At sufficiently high temperatures also vibrations of molecules
can be excited (see Sect. 10.3) which contribute to the total energy.
Equipartition Law
In a gas that is kept sufficiently long at a constant temperature T the energy of the atoms or molecules is uniformly
distributed by collisions over all degrees of freedom.
Therefore each molecule has on the average the energy
Ekin D f .1=2/kT where f is the number of degrees of
freedom, accessible to the molecule.
7.3.4
Distribution Function
After the descriptive discussion of the relation between mean
kinetic energy and the pressure of a gas for the special case of
a cuboid container we will now give a more quantitative representation for the general case of an arbitrary volume. For this
purpose the velocity distribution has to be defined in a quantitative way. This can be achieved with the distribution function
f .u/ (see Sect. 1.8), which describes how the quantity u is distributed among the different molecules. For u D vx we obtain
f .vx /dvx D
N.vx /dvx
N
ZC1
with N D
N.vx /dvx :
(7.15)
1
The quantity f .vx /dvx q gives the fraction of all molecules with a
velocity component vx in the interval vx to vx Cdvx . The number
of all particles within the interval vx to vx C dvx is then
N.vx /dvx D N f .vx /dvx :
(7.15a)
The number of molecules with vx u is then
N.vx u/ D N
Z1
f .vx /dvx :
(7.15b)
vx Du
averaged over all particles of the ensemble and determined at
a fixed time (ergoden hypothesis). This is however, only true
under certain conditions, which have to be proved for each case.
The “ergoden-theory” is a current field of research in mathematics and statistical physics.
From (7.15) we obtain the normalization condition
Real molecules can rotate and vibrate. The energy of these degrees of freedom have to be taken into account in addition to the
translational energy. The number of degrees of freedom therefore becomes larger. For example diatomic molecules can rotate
around two axes perpendicular to the molecular axis. This gives
the additional energy Erot D L2 =2I, where L is the angular momentum of the rotation and I the inertial moment (see Sect. 5.5).
For u D jvj D v the quantity f .v/dv gives the fraction of all
molecules with velocity amounts between v and v C dv. The
normalization is now
Z1
f .v/dv D 1 :
(7.16a)
ZC1
ZC1
1
N.vx /dvx D 1 :
f .vx /dvx D
N
(7.16)
1
1
0
sin ϑ
Unit sphere
(r = 1)
sin ϑ dφ
dΩ = dϑ· sin ϑ· dφ
F
ϑ
ϑ
dA·cos ϑ
1
ϑ
dA
dϑ
φ
dφ
Figure 7.11 Illustration of Eq. 7.17 and 7.18. Because of the small area dA and the solid angle d˝ the velocity vectors of particles hitting dA within the solid
angle d˝ are approximately parallel
We consider a surface element dA, which is hit by molecules
from the upper half space (Fig. 7.11). Within the time interval
t the number Z of molecules within the velocity interval from
v to v C dv coming from the angular range d˝ around the angle
# against the surface normal and impinging on dA is
Z D n f .v/dv dA cos # vt
d˝
:
4
(7.17)
Z2 Z=2
Z1
v 2 f .v/dv
vD0
(7.19)
cos2 # sin # d# d' :
'D0 #D0
This can be seen as follows: The product n f .v/dv gives the
particle density within the velocity interval dv. Within the time
interval t all molecules up to a distance v t from the surface
element dA can reach dA. From all molecules with isotropic
velocity distribution only the fraction d˝=4 reaches the effective area dA cos # within the solid angle d˝. The momentum
change jpj of a molecule at the impact on dA is
jpj D 2 m v cos # :
The momentum transfer of Z molecules per sec is then
ptotal =dt D Z jpj=dt :
the pressure p can be obtained from (7.17)–(7.18) as
2n m
jpjtotal
D
pD
dA t
4
The first integral gives the quadratic means v 2 . The second double integral can be analytically solved and has the solution 2=3.
This gives finally the pressure p onto the area dA
p D 13 n m v 2 ;
in accordance with (7.12a). In order to calculate the mean
square v 2 we must determine the distribution function f .v/. This
will be the task of the next section.
7.3.5
(7.17a)
Maxwell–Boltzmann Velocity
Distribution
Integration over all velocities v and over all impact angles #
yields the total momentum transfer per sec which is equal to the
product p dA of pressure p acting onto dA and the area dA.
The decrease of the air density with increasing height in our
atmosphere (barometric formula) discussed in the previous section can be explained by the velocity distribution f .v/ of the air
molecules.
Remark. The bold vector p denotes the momentum while the
scalar quantity p indicates the pressure. Although both quantities are labelled with the same letter (this is in agreement with
the general convention), there should be no confusion, because
it is clear from the text, which of the two quantities is meant.
With the solid angle
If we extend the exponent in (7.6b) with the volume V0 of a
gas with mass M D %0 V0 and insert for p0 V0 the general
gas-equation 7.14 Eq. 7.6b becomes
d˝ D
r d# r sin # d'
D d# sin # d' ;
r2
(7.18)
% D %0 e
.Mgh/=.NkT/
(7.20a)
:
For the number density n D %=m of gas molecules with mass m
we obtain with m D M=N
n.h/ D n0 e
.mgh/=.kT/
191
Chapter 7
7.3 Kinetic Gas Theory
D n0 e
Ep =kT
:
(7.20b)
192
7 Gases
Chapter 7
We will assume in the following an isothermal atmosphere
where collisions can be neglected, although collisions are responsible for the equipartition of the total energy onto all
molecules and therefore for establishing a temperature. However, this does not influence the validity of the following
derivation where we select a subgroup of all molecules with a
velocity component vz in the Cz-direction which fly upwards.
Figure 7.12 Only molecules with initial velocites Vz .h D 0/ > u reach the
height z D h
The exponent in (7.20b) represents the ratio of potential energy Ep D mgh at the height h above ground and twice the
mean kinetic energy Ekin D .1=2/kT per degree of freedom
of a molecule, due to its thermal motion at the temperature
T. We regard here at first only a one-dimensional motion in
the z-direction, which is described by the distribution of the zcomponents vz of the velocity v.
The barometric formula gives for an isothermal atmosphere the
ratio of particle densities in layers z at different heights h. If
the molecules had no kinetic energy, they would all rest on the
earth surface and form a solid layer, i.e. the earth atmosphere
would disappear.
Let us first assume the atmosphere had been built up by
molecules that start from the ground z D 0 upwards with the
velocity vz D u. They reach the height h, given by
1
m u2 D m g h
2
before they fall down in the earth gravitational field (Fig. 7.12).
In fact the primary earth atmosphere had been formed by
molecules outgassing through volcanos.
Since we have assumed an isothermal atmosphere, the velocity distribution at z D h must be the same as for z D 0. This
means, there are at z D h also molecules with vz D u which
move upwards until their kinetic energy is just cancelled by the
increase of the potential energy. The density of these molecules
is, however, smaller at z D h than at z D 0 because the number density n.h/ decreases exponentially with h according to the
barometric formula (7.20b). The specific choice of the group of
molecules starting from z D 0 does not constrict the following
argumentation.
The number N>u .z D 0/ of molecules that start from a unit
area at z D 0 with velocity components vz > u is equal to the
number N>0 .z D h/q of molecules that fly through a unit area in
the plane z D h with velocities vz > 0.
N>u .z D 0/ D N>0 .z D h/ :
(7.20c)
The number N.vz / of molecules that pass per unit time with the
velocity vz through a unit surface (flux density) is given by the
product
N.vz / D n.vz / vz
of number density n.vz / and velocity vz (Fig. 7.13).
For an isothermal atmosphere with the constant temperature T
the mean square velocity v 2 and the distribution function f .vz /
must be independent of h, since they depend only on the temperature.
The flux density N>0 .z D h/ can be expressed as
N0 .z D h/ D n.h/
Z1
vz f .vz /dvz :
(7.21a)
vz D0
It is smaller than the flux density
N>0 .z D 0/ D n.0/
Z1
vz f .vz /dvz :
(7.21b)
0
Since the two integrals in the two equations are equal, it follows
with (7.20c)
Nvz >u .0/
Nv >0 .z D h/
n.h/
D z
D
:
Nvz >0 .0/
Nvz >0 .0/
n.0/
(7.21c)
Remark.
1. Some readers have argued that the decrease of v.z/ with
increasing z contradicts the assumption of an isothermal
atmosphere. This is, however, not true, because the temperature is determined by the velocity distribution of all
molecules but not only by that of an arbitrarily selected subgroup.
2. In the real atmosphere the temperature decreases with increasing z. The main reason for that is the decreasing heat
flow from the ground into the atmosphere and the decreasing absorption of the infrared radiation emitted by the earth
surface.
Figure 7.13 Number of particles per unit volume n .vz /dvz within the intervall
u vz u C du
From the definition of the distribution function f .vz / in (7.15b)
and (7.15) it follows for the flux densities
Nvz 0 .z D 0/ D n0
Z1
vz f .vz /dvz
Z1
vz f .vz /dvz ;
vz D0
Nvz u .z D 0/ D n0
(7.22)
vz Du
with (7.20b) and the relation m g h D .1=2/m v 2 we obtain
Z1
m 2
2 u =kT
vz f .vz /dvz D C1 e
(7.23)
;
u
where C1 is a constant which depends on the temperature T.
Differentiating both sides with respect to the lower limit u yields
on the left side the negative integrand for vz D u:
mu
C1 e
kT
m 2
) f .u/ D C2 e 2 u =kT
u f .u/ D
m 2
2 u =kT
The constant C2 can be obtained from the normalization condition
Z
f .u/du D 1
and the integration
.m=2kT/.1=2/.
e
x2
dx D
p
.
This gives C2 D
Replacing u by vz , finally yields the distribution function
f .vz / D
r
The probability to find a molecule with the velocity v D
fvx ; vy ; vz g is equal to the product of the probabilities for vx , vy
and vz . One therefore obtains for the distribution function
.3=2/
m
2
f .vx ; vy ; vz / D
e .mv /=.2kT / :
(7.25)
2kT
In many cases only the magnitude jvj of the velocity is of interest, where the direction can be arbitrary. The heads of all
velocity arrows with a length between v and v C dv are located
within a spherical shell with the volume 4v 2 dv. Therefore the
integration
Z
f .vx ; vy ; vz / dvx dvy dvz
vx ;vy ;vz
over all values of v D fvx ; vy ; vz g within this spherical shell
gives for the number density n.v/dv of all molecules per unit
volume with velocities between v and v C dv the result
m
D const :
kT
with C2 D C1
R
the volume, no direction for the motion of the molecules is preferred, all directions are equally probable as has been discussed
in Sect. 7.3.2. The distributions of the velocity components are
equal for all three components vx , vy , vz and are described by
(7.24).
n.v/dv D n
m
2kT
.3=2/
4v 2 e
mv 2 =2kT
dv :
(7.26)
This is the Maxwell–Boltzmann velocity distribution
(Fig. 7.15). The normalized distribution function is then
f .v/ D n.v/=n, where n is the total number density of molecules
with any velocity.
Note
m
e
2kT
m 2
2 vz =kT
:
(7.24)
This is a symmetric Gauss-distribution illustrated in Fig. 7.14.
For a gas in a closed volume V, where the mean kinetic energy
is large compared to the difference of potential energies inside
Figure 7.14 Distribution function f .vz / of the velocity component vz
Contrary to the symmetric distribution for the velocity
components, which extends from 1 to 1, the distribution for the velocity magnitude is restricted to the range
v 0, because there are no negative velocities. The distribution is therefore asymmetric. Because of the factor v 2
it is also not symmetric around a mean value v.
Figure 7.15 Maxwell–Boltzmann velocity distribution n .v/dv with most probable velocity
p vw , mean velocity hvi and the square root of the mean velocity
square
v2
193
Chapter 7
7.3 Kinetic Gas Theory
194
7 Gases
Chapter 7
Remark. In unidirectional molecular beams generally not the
molecular density n.v/ is measured but rather the flux density
N D n.v/v, for instance by a detector which measures the number of particles reaching the detector per unit time. The velocity
distribution of the flux density N.v/ D n v f .v/ in a collimated
molecular beam where all molecules fly within a small angular
cone around the x-direction differs from n.v/ in (7.26) by the
additional factor v. The prefactor of f .v/ is therefore v 3 instead
of v 2 .
Table 7.2 Mean values of thermal velocities
Quantity
Symbol
Most probable velocity
vw
Mean velocity
Square root
of mean velocity square
v
p
hv 2 i
Mathematical expression
q
q
q
2kT
m
8kT
m
D
3kT
m
D
p2
q
vw
3
v
2 w
The maximum of the distribution (7.26) appears at the most
probable velocity vw . With the condition dn.v/=dvjmp D 0 one
obtains from (7.26) the value
r
2kT
:
(7.27)
vw D
m
The mean velocity v is defined by
vD
Z1
0
v f .v/dv
D 4
m
2kT
Integration yields
vD
.3=2/ Z1
0
r
v3 e
mv 2 =.2kT/
2vw
8kT
D p :
m
dv :
(7.28)
Figure 7.16 Velocity distribution of N2 -molecules at two different temperatures. The area under the two curves represent the total number of particles per
unit volume. For a closed gas system the two areas are equal
(7.29)
numerical values of the velocities v, vw and .v 2 /.1=2/ change by
a factor of 2 because they are proportional to the square root of
the temperature (see also Tab. 7.2).
Finally we get for the mean square v 2
v2
D
Z1
0
v 2 f .v/dv D
3kT
:
m
This gives for the mean energy of a particle with three translational degrees of freedom the result
m 2
3
1
v D kT D f kT ;
2
2
2
which has been already used in Sect. 7.3.2.
The sequence of magnitudes for the three special velocities is
p
vw < v < hv 2 i :
With the most probable velocity vw (7.27) Eq. 7.26 can be written as
4v 2
2
n.v/dv D n 3 p e mv =2kT dv
vw
(7.30)
4v 2
2
v 2 =vw
dv :
Dn 3 p e
vw
The velocity distribution depends strongly on the temperature
T. In Fig. 7.16 two distributions are shown for two different
temperatures T1 and T2 which are related by T1 =T2 D 1 W 4. The
Example
The density of nitrogen gas N2 at room temperature T D
300 K and at a pressure of 1 bar is
%.N2 / D 1:12 kg=m3; m.n2 / D 4:67 10
26
kg :
This gives
n D %=m D 2:4 1025 N2 molecules=m3
D 2:4 1019 N2 molecules=cm3 :
The numerical values for the velocities can be calculated
from (7.27) and (7.29) as
vw D 422 m=sI
v D 476 m=sI
p
hv 2 i D 517 m=s :
The mean kinetic energy of a molecule is Ekin D
.3=2/kT D 6:21 10 21 J the energy density of all
molecules per cm3 is n Ekin D n .3=2/kT D 0:15 J=cm3 .
J
7.3.6
around the centre of A2 are deflected from their straight path.
This area is called the collisional cross section (see Sect. 4.3).
Collision Cross Section and Mean Free
Path Length
The model of the ideal gas describes the gas particles by small
rigid spheres with a radius ri that is small compared to the average distance d between the spheres. A collision takes place if
the spheres touch each other, i.e. if d .r1 C r2 /.
We define the impact parameter b for the collision between two
particles A1 and A2 as the distance between two straight lines
(Fig. 7.17):
1. The path of the centre of A1 without any interaction.
2. The straight line through the centre of A2 parallel to line 1.
(see also Sect. 4.3).
In this model a collision takes place if b .r1 C r2 /. At the
collision the closest distance between the centres of A1 and A2
is d D r1 C r2 . All particles A1 for which their centre passes
through the circular area
D .r1 C r2 /2
(7.31a)
If a beam of particles A1 passes in the x-direction through a gas
with n particles A2 per cm3 at a sufficiently small number density
n (the mean distance d should be large compared to .r1 C r2 / the
probability that a particle A1 suffers a collision during the path
length x is given by the quotient
P
n x A
D
D n x
(7.31b)
A
A
P
where
is the sum of the cross sections of all atoms A2 in
the volume V D A x and A is the total cross section of the
incident beam (Fig. 7.18).
If N particles impinge per sec onto the area A of the volume
V D A x the fraction N=N that suffers a collision after a
path length x is
N
D n x:
(7.32a)
N
In its differential form this reads
dN
D
N
n dx :
(7.32b)
The negative sign should indicate that the particle flux N decreases because the collisions deflect the particles out of the
x-direction.
Integration of (7.32b) gives the particle flux
N.x/ D N0 e
n x
;
(7.33)
after a path length x through the collision volume.
Figure 7.17 Impact parameter b and collision cross section for collisions of
hard spheres with radiii r1 and r2
The path length which a particle A1 passes on the average
without a collision is
Z1
dN.x/
1
D
xj
jdx
N0
dx
0
(7.34a)
Z1
1
D n x e n x dx D
;
n
0
∆x
N0·
N0
N
Incident particles
as point masses
N0 /e
A
σ
Scattering volume
of gas particles
Total cross section
(circles) as fraction
of total area
Figure 7.18 Illustration of collision cross section and mean free path length [7.10]
195
Chapter 7
7.3 Kinetic Gas Theory
196
7 Gases
Chapter 7
where jdN.x/=dxj dx gives the number of collisions on the path
interval dx. The probability of a collision in the interval dx is
then jdN.x/=N0j.
different gas pressures and temperatures. We will here discuss
only a few of them that are based on molecular beams or on
transport phenomena in gases such as diffusion, viscosity and
heat conduction in gases.
The mean free path represents that path length after
which the number of particles in the incident beam has decreased to 1=e of its initial value.
7.4.1
The quantity D 1=.n / is called mean free path.
The average time interval between two successive collisions
can then be defined as
1
D
D
:
(7.34b)
hvi
nhvi
If both particles A1 and A2 move with velocities v1 and v2 the
to be replaced by the mean
mean velocity hvi in (7.34b) has p
relative velocity v D v1 v2 D 2v 2 . For collisions in a gas
at the temperature Tone then obtains instead of (7.34b) the mean
free collision time
1
D
p
:
(7.34c)
n 2v 2
Examples
1. At atmospheric pressure p D 105 Pa the number
density of molecules in the atmosphere is n 3
1019 cm 3 . For the elastic collision cross section D
45 10 16 cm2 the mean free path is
D
1
7 10
n
6
cm D 70 nm :
With the mean velocity hvi D 475 m=s at T D 300 K
the mean flight time between two collisions becomes
r
m
D 1:1 10
D
D p
2
6kT
2hv i
10
When atoms or molecules effuse out of a reservoir (pressure p,
volume V, temperature T) through a small hole A into a vacuum chamber they fly on straight paths until they hit the wall, if
their mean free path is longer than the dimensions D of the
chamber (Fig. 7.19). The pressure in the vacuum chamber must
be low enough which can be reached with diffusion pumps (see
Sect. 9.2).
Placing a slit B with width b at a distance d from A, only
molecules can be transmitted by the slit, that fly into the angular range j#j < " with tan " D b=2d around the x-axis.
After passing the slit they form a collimated molecular beam.
For not too high pressures p in the reservoir the molecules in
the beam follow a modified Maxwell–Boltzmann-distribution
N.v/ D nv f .v/ and the angular distribution N.#/ / N0 cos #
shows a cosine dependence. The angular distribution can be
measured with a detector slewing over the angular range #.
With a velocity selector (Fig. 7.20) subgroups of molecules with
velocities within the range vs .1=2/v < v < vs C .1=2/v
can be selected.
Such a selector consist in principal of two metallic circular discs
with radius R at a distance a, each having a slit with width S D
R '. The two slits are twisted against each other by S D
R '. When the discs rotate with the angular velocity ! only
s:
This means that nitrogen molecules in a gas under
normal conditions (p D 105 Pa, T D 300 K) suffer
1:34 1010 collisions per second!
2. In an evacuated container with a residual pressure
of p D 10 4 Pa (10 9 bar) the density is n D 3
1010 cm 3 . Now the mean free path is 70 m and
therefore large compared with the dimensions of the
container. Collisions between molecules are seldom
and the molecules fly on straight lines until they hit
the walls of the vacuum container.
J
7.4
Molecular Beams
Figure 7.19 Schematic depiction of a molecular beam apparatus
Experimental Proof of the
Kinetic Gas Theory
There are many experimental methods to prove the statements
of kinetic gas theory and to measure important quantities such as
velocity distribution, collision cross sections, mean free path at
Figure 7.20 Principle of mechanical velocity selector
7.4 Experimental Proof of the Kinetic Gas Theory
vD
!a
'
Chapter 7
molecules with velocity v can pass both slits, if their flight time
T D a=v between the two slits equals the time T2 D R '=.R
!/ D '=!. Their velocity is then
(7.35)
For a slit width S D R ' with ' ' the transmitted
velocity interval is
'
v D v
:
(7.36)
'
Varying the angular velocity ! of the discs one can select any
velocity subgroup with a velocities up to vmax D !max a=',
where !max is the maximum value that can be technically realized. This allows one to measure the velocity distribution of the
molecules in the beam.
Example
' D 20ı D 0:35 rad; a D 10 cm. If molecules with
v D 400 m=s should be selected, the velocity selector has
to rotate with ! D 1:4 103 s 1 which corresponds to
J
13 370 rpm.
Figure 7.21 Velocity selector with 6 rotating discs. a Principle of selecting a
velocity class with particle flux N .v/dv, b arrangement for measuring collision
cross sections .v/ by detecting the number N .v/ D N0 e n .v/L that have
passed the scattering chamber with length L
Cooling
system
Temperature
control
Small heat leak
When the density n.v/ of molecules in the molecular beam
follows the Maxwell–Boltzmann-distribution (7.30) the flux
N.v/ D v n.v/ is
N.v/ D n.v/ v D n
4v 3
p e
vw3
mv 2 =2kT
:
Molecules
(7.37)
The number of molecules with velocities in the interval from
v to v C dv, passing per sec through the area of 1 cm2 is then
N.v/ dv.
Remark. For the velocity selector in the above example also
molecules with v D 21 m=s would be transmitted for ' D 20ı C
360ı i.e. at the next full turn. In order to prevent this ambiguity
one has to use at least a third disc in the middle between the two
discs with a slit tilted by '=2. Generally many discs are used
with many slits (Fig. 7.21) in order to transmit more molecules
per sec. For q discs at a distance a=q the tilt of the slits between
two successive discs should be '=q.
For the detection of the transmitted molecules several different
detectors have been developed.
Bolometer (Fig. 7.22). This is a small semiconductor plate
cooled down to very low temperatures with a small heat capacity C and a small heat conductivity G to its surrounding.
The molecules impinging onto the cooled surface transfer
their kinetic energy Ekin D .1=2/mv 2 to the semiconductor. This increases its temperature by T D N.v/ Ekin =G
(see Sect. 10.2.2). The temperature increase results in a
change R D .@R=@T/T of the electric resistance R of
the semiconductor. This can be measured by the corresponding change of the electric current I D U0 =.R C R0 / that
flows through the circuit of semiconductor and external resistor R0 in series with R, if a constant voltage U0 is applied
197
Figure 7.22 Schematic design of a bolometer for measuring the flux N .v/ of
neutral molecules
(see Vol. 2). With such a device, it is possible to measure
a transferred power as small as 10 14 W! This corresponds
to a minimum rate of N.v/dv D 2:8 106 =s molecules with
v D 400 m=s.
Ionization detector (Fig. 7.23). The neutral molecules are
ionized by electron impact (see Vol. 3). The ions with charge
q D Ce are collected on an electrode at negative voltage.
For a flux N of neutral particles the electric output current of
the detector is I D N e where 1 is the ionization
probability of each neutral molecule.
Langmuir–Taylor detector. This is a heated wire where all
neutral particles with ionization energies smaller than the
work function of the hot wire are ionized if they hit the wire.
In this case, energy is gained if the electron is transferred
from the molecule to the metal wire. The ions are extracted
by an electric field and collected on a detector, for instance a
Faraday cup.
Modern methods for the measurement of the velocity distribution are based on laser-spectroscopic techniques (see
Vol. 3).
When the pressure p in the reservoir is increased, the mean free
path of the molecules in the hole A of the reservoir becomes
198
7 Gases
Chapter 7
also mass is transported. If molecules A in a sub-volume V1
can move into a volume V2 where molecules B are present, the
two species mix with each other until both sorts are uniformly
distributed over the whole volume (diffusion).
There are mainly 3 such transport phenomena:
Diffusion (mass transport),
heat conduction (energy transport),
gas flow with viscosity (momentum transport).
They occur always when local differences (gradients) of density, temperature or flow velocities are present. The important
point is that all these transport phenomena can be explained by
the kinetic gas theory. The experimental investigation of these
macroscopic processes gives information about the size of the
gas molecules and their mutual interactions (see Vol. 3).
Figure 7.23 Principle of ionization detector
7.5.1
z
Figure 7.24 Narrowing of the velocity distribution N .v/ in a supersonic beam
smaller than the dimensions of the hole. In this case the particles suffer collisions during their expansion into the vacuum.
Because the faster particles hit the slower particles ahead of
them and they transfer part of their kinetic energy, they become
slower and the slow particles faster. This implies that the velocity distribution becomes narrower (Fig. 7.24). If the mean
velocity exceeds the local velocity of sound, a supersonic beam
is formed. Its velocity distribution is described by
N.v/ D Cv 3 e
m.u v/2 =2kTt
:
Diffusion
When a bottle of an intensively smelling substance (for example
a pleasantly smelling perfume or the badly smelling hydrogen
sulphide H2 S) is opened the odour can be soon sensed in the
whole room. The molecules escaping out of the bottle must
travel in a short time over several meters through the air at atmospheric pressure in spite of the very small free mean path of
D 10–100 nm! This migration of molecules A through a gas
of molecules B resulting in a uniform spatial distribution of both
sorts A and B is called diffusion.
The diffusion is illustrated by Fig. 7.25, where a volume V
is divided by a thin wall into two parts, each containing only
molecules of one type. After removing the separating wall the
two types A and B of the molecules mix and fill the whole volume with spatially uniform concentration.
Diffusion is a net transport of particles from a region with
higher concentration to a region with lower concentration.
(7.38)
The width of the distribution around the mean velocity u D hvi
can be characterized by a translational temperature T t which is
a measure for the relative velocities of the particles in the beam.
With increasing pressure p the temperature T t decreases, which
means that the relative velocities decrease. Translational temperatures below 1 K have been realized, where all particles have
nearly the same velocity u.
7.5
Transport Phenomena in Gases
Because molecules in a gas can freely move around within the
gas container many transport processes can occur. When the
molecules collide with each other or with the wall of the container energy and momentum can be transferred. In a gas flow
Figure 7.25 Diffusion of two differrent particles with densities nA .t / and nB .t /
after opening a hole in the dividing wall at t D 0
Transport Phenomena in Gases
199
Chapter 7
7.5
Figure 7.27 Illustration of the derivation of the diffusion coefficient
2. The probability PC that A flies after its last collision at xC D
x0 C cos # with the velocity v under an angle # against
the x-direction through the plane x D x0 from right to left.
Figure 7.26 a Density gradient driving the diffusion of particles A through a
gas of particles B . At time t D 0 a hole is opened in the dividing wall and the
particles A diffuse into the volume V2 . b On the left side NA particles A per sec
are continuously supplied, which are pumped away on the right side. The curves
show the distribution nA .x / at different times ti
The examples above illustrate that diffusion always takes place
if density gradients are present. Diffusion reduces these gradients until a uniform spatial distribution is reached, where the
density gradients are zero (Fig. 7.26a) unless external conditions
maintain a stationary density gradient. This can, for example,
be realized if particles A are continuously supplied to the left
volume in Fig. 7.25 and particles A and B are simultaneously
removed on the right side, thus maintaining a concentration gradient (Fig. 7.26b).
We will now discuss diffusion in a more quantitative way. We
assume the density nA .x/ of particles A to be constant in the yand z-direction but to vary in the x-direction (Fig. 7.27). The
thermal velocities of the particles A are isotropic. This means
that the following two probabilities are equal:
1. The probability P that A flies after its last collision at x D
x0 cos # ( D mean free path) with a velocity v under an
angle # against the Cx-direction and passes the plane x D x0
from left to right.
We have to take into account that the velocities are not equal
for all molecules but follow a Maxwell–Boltzmann distribution
with the distribution function f .v/. The directions of their velocities are randomly distributed. With the density nC .x/ left of
the plane x D x0 the flux dNC of particles within the velocity interval from v to v C dv that pass in the time interval dt from left
to right under the angle # within the solid angle d˝ the plane
dA at x D x0 is
dNC .v/ D nC f .v/dv v dt dA cos # d˝=4 ;
(7.39)
because the volume from where they come, is dV D dA cos #
v dt (Fig. 7.28).
A corresponding equation is obtained for dN .v/. The question
is now which densities nC and n have to be used?
The particles start from the position x D x0 ˙ x .x
cos #/ of their last collision. There are the densities
dn
:
dx
dn
x
:
dx
nC D n0 C x
n D n0
(7.40)
On the average is x D cos #. We define the vertical particle
flux density by the vector
jD
dN
eO
dA dt
where dN is the number of particles which pass the area dA in
the plane x D x0 during the time interval dt. For that part of the
net flux of particles with the velocity v in the interval dv within
the solid angle d˝
1 dNC .v/ dN .v/
dj.v/dv D
dv
(7.41)
dA
dt
dt
200
7 Gases
Chapter 7
Figure 7.28 The number of particles passing during the time interval dt
through the area dA inclined under the angle # against the x -direction is
dn D n f .v/ v dv dt dA cos #d#=4
we obtain from (7.39) with d˝ D sin # d# d'
djx .v/dv D 2f .v/v dv
cos2 # sin # d#d' dn
:
4
dx
(7.41a)
Integration over ' gives the factor 2, over # .0 # =2/
the factor 1=3, while the integration over all velocities gives the
mean velocity
Z
v D v f .v/dv :
(7.42)
Figure 7.29 Demonstration of the fact that light particles diffuse faster than
heavy particles
One obtains finally for the total mean particle flux in the xdirection
baker is removed. Now the pressure inside the cylinder drops
below the external pressure until it finally approaches the external pressure.
Fick’s Law:
jx D
v dn
dn
D D
:
3
dx
dx
(7.43)
For the general three-dimensional case this modifies to the vector equation
jD
D grad n:
The particle flux j due to diffusion is equal to the product of
diffusion coefficient D and concentration gradient grad n.
The explanation is the larger diffusion velocity of the lighter helium atoms. It diffuses faster into the cylinder than the air in the
cylinder diffuses out. The total pressure therefore increases, until the air has diffused out and the pressure difference between
inside and outside becomes zero. After the baker has been removed there is no longer helium outside. Therefore, the helium
diffuses from the inside of the cylinder to the outside faster than
the air diffuses the opposite way. Therefore the pressure inside
drops at first below the outside pressure.
The diffusion coefficient
D D 13 v
(7.44)
is proportional to the product of mean free path and mean
velocity v. Using (7.34) and (7.28) we can express the diffusion
coefficient
r
v
8kT
1
DD
D
(7.45)
3
n 9m
for the diffusion of particles A through a gas of particles B with
density n by the collision cross section and the mass m of
particles A. This shows that heavy particles diffuse slower than
light ones.
This can be demonstrated by the experiment shown in Fig. 7.29.
A porous cylinder of clay shows in air the same pressure inside
and outside. At time t1 a baker is put over the cylinder and helium is blown into the baker. The pressure gauge shows at first
a higher pressure inside the cylinder, which gradually decreases
until it becomes equal to the pressure outside. At time t2 the
7.5.2
Brownian Motion
The diffusion of a microscopic particle A suffering collisions
with molecules B in a gas at atmospheric pressure can be observed through a microscope if the mass of A is much larger
than that of the molecules B. This can be realized with cigarette
smoke particles that have diameters of about 0:1 µm. Illuminating the diffusion chamber, makes the particles A visible (even if
their diameter is small compared to the wavelength of the illuminating light) because they scatter the incident light and appear
in the microscope as bright spots (see Vol. 2). This particle A
which contains still about 105 molecules, performs a random
walk through the gas with particles B (Fig. 7.30a). When observing the motion of A with sufficient local and time resolution
the motion appears on short straight lines between successive
collisions that abruptly change the direction of the motion. The
length of these straight lines is the mean free path length of the
particle A. The directions of the straight lines are statistically
T1
Transport Phenomena in Gases
<
T2
d
v1
P1
P2
dA1
v2
Figure 7.30 a Random path of a particle (Brownian Motion) induced by collisions with air molecules. b Histogram of the distribution of the mean free path
between successive collisions
distributed. Such a random walk can be mathematically simulated, if the lengths and the directions are generated by a random
generator. Figure 7.30b shows a histogram N.Li / for the lengths
Li of the straight lines. Each bar indicates how often the length
Li is observed within the interval L to L C L. The analytical
curve as envelope of the different bars gives the probability of
finding the length L. It has the form
W.L/ D a e
L=
:
(7.46)
The random motion of microscopic particles in liquid solutions
was first discovered 1827 by the English botanist Robert Brown
who believed initially that he observed small living microbes
until he realized that the motion was completely irregular and
should be attributed to lifeless particles. Magnified by a microscope and put on a large screen it allows a fascinating view into
the micro-world, accessible to a large auditorium.
The mathematical description first given by Albert Einstein will
be treated in Vol. 3.
The Brownian motion can be simulated by moving pucks on an
air bearing stage, where a large number of small discs are kept
in motion by vibrating wires at the 4 edges of the stage. A larger
puck with a small light bulb moves in between the small discs
and its random motion is detected by a video camera. In fact,
the random path in Fig. 7.30a has been obtained in this way.
7.5.3
Heat Conduction in Gases
Heat conduction is also based on the motion of molecules which
transfer during collisions part of their kinetic energy to the collision partner. This results in a transport of energy from regions
with higher temperature to those of lower temperature. The
mechanism of heat conduction is different for gases, liquids and
solid bodies (see Sect. 10.2). In solids the atoms are fixed to
definite positions while they can freely move in gases.
We start the discussion of heat conduction in gases with a gas
between two parallel plates at a distance d (Fig. 7.31) that are
kept at different temperatures T1 and T2 . The transport of heat
n1 > n2
Figure 7.31 Heat conduction in gases. The distance d is small compared to
the extensions of the plates. A pressure sensor measures the different pressures
p1 and p2 at different sides of the sensor
energy from the hotter to the cooler plate depends on the ratio
=d of free mean path to plate separation d.
At low gas pressures is > d and the molecules can fly between
the two plates without suffering collisions in the gas volume.
Molecules that leave the plate 1 have the mean kinetic energy
Ek1 D 12 m v 2 D 23 kT1
For an isotropic distribution of the velocity directions of the
molecules with a density n
Z D n cos # A
Z
vf .v/dv t d˝=4
molecules leaving plate 1 with velocities v within the solid angle
d˝ D sin # d# d' under the angle # against the surface normal
reach within the time interval t the surface element A on
plate 2 (see Fig. 7.28).
Integration over all possible velocities yields
n vA t
ZD
4
D
Z=2
Z2
sin # cos # d#
d'
0
n
vA with
4
'D0
vD
Z1
(7.47)
vf .v/ dv :
0
We will assume that every molecule impinging on plate 2 will
stay there for a short time, adapt the temperature of plate 2 and
desorbs again. The surface element A of plate 1 loses energy
because of the desorbing molecules
dW1
A D
dt
Z1
A U1 ;
t
201
Chapter 7
7.5
(7.48a)
202
7 Gases
Chapter 7
where dW=dt is the energy loss per unit surface and unit time
and U1 D .f =2/kT1 is the energy of a molecule with f degrees of freedom (kinetic, rotational and vibrational energy, see
Sect. 10.3). On the other hand the surface element wins the energy
dW2
Z2
A D
A U2
dt
t
with
U2 D
1
f kT2
2
(7.48b)
by molecules coming from plate 2 and impinging on plate 1.
Under stationary conditions is Z1 D Z2 .
Therefore the net energy flow is
dW
A1 D A1 .T2
dt
with
T1 /
nvkf
D
8
(7.49a)
The heat conduction (energy flux per unit time) jW D dW=dt
in gases at low pressures ( d) is proportional to the temperature difference between the walls and to the density n of
the gas molecules.
Since v / m 1=2 heavy molecules have a lower heat conduction
than light molecules. Because of their larger number f of degrees of freedom molecules transport more energy than atoms.
Since the pressure p D n k T is the same within the whole
volume between the plates the molecular densities n1 and n2 in
front of the plates differ according to
(7.49b)
The gas density is therefore lower in regions with higher temperatures.
In order to decrease the heat conduction, the gas density has to
be low, i.e. the space between the plates should be evacuated.
The thermos bottle is an example, where low heat conduction is
realized in order to keep a liquid for a longer time on a nearly
constant temperature.
If the mean free path is low compared to the plate separation ( d) the molecules often collide on their way between
the plates. The heat energy of the hotter plate is no longer
directly transported by molecules to the cooler plate but transferred within a path x during collisions to other molecules.
Therefore a temperature gradient dT=dx appears in the gas volume. The energy flow per unit time through the unit surface
element of the plane x D x0 between the plates is similar to the
discussion in Sect. 7.5.1
dW
1
d
D .v n U/
dt
3
dx
1
f
dv
D n kT
:
3
2
dx
With
dv=dx D
p
8k=m dT
dv dT
p
D
dT dx
dx
2 T
this can be written as
dW
dT
D
:
dt
dx
(7.50a)
(7.50b)
The constant
D
The constant is called heat transfer coefficient. It has the unit
Œ D 1 J s 1 m 2 K 1 .
n1
T2
D
:
n2
T1
Here the relation n U D .1=2/n f kT of Eq. 7.49 has been
used.
1
f n kv
12
is the heat conductivity, with the unit
Œ D 1
J
:
smK
The heat conductivity is for d independent of the gas
density n because, according to (7.34), is D 1=.n / and
therefore
1 f kv
D
:
(7.50c)
12
7.5.4
Viscosity of Gases
As has been discussed in the previous sections diffusion and
heat conduction can be ascribed to mass- and energy transport
by molecules. They are accomplished by the thermal motion
of molecules at local variations of density (diffusion) and temperature (heat conduction). Diffusion and heat conduction also
occur if the gas P
as a whole is at rest, i.e. if the macroscopic
momentum P D pi D 0.
When in addition to the thermal motion of the molecules a
macroscopic flow of the whole gas volume occurs (gas current),
further phenomena appear as for instance friction (viscosity), if
the flow velocity varies locally (Sect. 8.3). Also the viscosity is
related to the thermal motion of the molecules as can be illustrated by the following example:
We consider a gas, which flows into the y-direction with a flow
velocity u.x/ that varies in the x-direction (Fig. 7.32). An example is the air flow over a lake with the water surface at the plane
x D 0. Layers of the streaming air close to the water surface are
retarded by friction with the water surface and have therefore a
smaller flow velocity than higher layers.
We select a layer between the planes x D x0 ˙x=2 (Fig. 7.32b).
The velocity of the gas molecules is a superposition of their
thermal velocities with the flow velocity. Because of their thermal velocities, the gas molecules pass from their layer x D
x0 ˙ x=2 into adjacent layers and collide there with other
molecules. Since the y-component of their velocity is higher
than that of molecules in layers x > x0 they transfer part of
7.5.5
Transport Phenomena in Gases
Summary of Transport Phenomena
Diffusion, heat conduction and internal friction can be explained
by the thermal motion of the molecules and the exchange of energy and momentum during collisions. All these phenomena can
be described by the kinetic gas theory. Therefore from measurement of the macroscopic quantities diffusion coefficient D, heat
conductivity and viscosity coefficient , which describe averages over the quantities of the individual molecules, information
can be obtained about the microscopic quantities mean free path
, collision cross section D 1=n and mean velocity v of
the molecules.
In order to recall and clarify the relations between the three
transport coefficients they are here again arranged in a compact
form.
At high gas pressures ( d): Diffusion coefficient
D D 13 v D 13 v=.n / :
(7.53a)
Energy transport through heat conduction between two plates
at a distance d and temperature difference T:
dW
D T
dt
d
(7.53b)
with the heat conductivity
Figure 7.32 Viscosity of gases. The gas streams into the y -direction with a
flow velocity u .x /, that decreases with increasing x . a velocity uy .x /; b molecular model of viscosity
D
The factor is the coefficient of internal friction or coefficient
of viscosity. A consideration similar to the derivation of the
diffusion coefficient (see Sect. 7.5.1) yields with (7.34)
D 31 n m v :
(7.52)
The momentum transfer in gas flows with a velocity gradient
du=dx is proportional to the particle density n and the mean thermal velocity v D .8kT=n m/1=2 , to the mean free path and
the flow velocity gradient du=dx.
It decreases with decreasing , i.e. with increasing collision
cross section. Heavy particles cause a higher viscosity, because
of their higher momentum transfer.
(7.53c)
which is independent of the gas pressure, because D / .1=n/.
Viscosity coefficient:
their momentum to these molecules and increase their mean ycomponent. The amount of the transferred momentum depends
on the difference between the components vy .x/, i.e. on the gradient du=dx of the flow velocities.
The transport of momentum occurs in the x-direction of decreasing u.x/. When we define the momentum flux as the momentum
transfer through the unit area in the plane x D x0 , the viscosity
law for gases with locally varying flow velocities can be written
as
du
jp D
:
(7.51)
dx
1 f kv
1
D Df kn
12
4
D 31 nmv D n m D
(7.53d)
At low pressures ( d): The energy transport becomes
proportional to the gas density
dW
D T
dt
nvkf
3
3
D
D
D n
8
2
2
(7.53e)
In Tab. 7.3 the transport coefficients of some gases are compiled.
Table 7.3 Selfdiffusion coefficient D , heat conductivity , and coefficient of
viscosity of some gases at p D 105 Pa and T D 20 ı C
Gas
He
Ne
Ar
Xe
H2
N2
O2
D=m2 =s
1:0 10 4
4:5 10
5
6:0 10
6
1:6 10
5
1:3 10
4
2:4 10
5
1:81 10
=J m 1 s 1 K
1:5 10 2
5
4:6 10
2
0:5 10
2
2:6 10
2
1:7 10
2
1:7 10
1
2:0 10
2
1
=Pa s
1:5 10 5
3:0 10
5
2:1 10
5
1:7 10
5
2:0 10
5
8:0 10
6
2:0 10
5
203
Chapter 7
7.5
204
7 Gases
Chapter 7
7.6
The Atmosphere of the Earth
Our atmosphere consists of a mixture of molecular and atomic
gases. In the lower part it contains also water vapour, aerosols
and dust particles. Its composition is listed in Tab. 7.4. The
density gradient dn=dz, described by Eq. 7.6 causes diffusion
which tries to establish a uniform density. However, the gravity
acts against this tendency and results in the exponential density
function n.z/. Stationary equilibrium is reached, when for all
values of z the upwards directed diffusion current jD is just compensated by the downwards directed current jg of particles in the
gravity field of the earth
(7.54)
jD .z/ C jg .z/ D 0 :
Integration yields again the barometric formula
The diffusion current is according to (7.43)
jD D D
With n D n0 e
.mgz=kT/
n D n0 e
dn
:
dz
) dn=dz D .mg=kT/ n one obtains
jD D CD
mg
n:
kT
(7.54a)
Opposite to the gravitational force is the friction force acting on
the falling molecules. This results in a constant fall velocity vg
and therefore a constant particle current
(7.55)
jg D n vg :
Since jg D jD we obtain for the constant sink velocity
vg D jD =n D
Figure 7.33 Stationary density distribution n .z / in the isothermal atmosphere,
caused by the superposition of upwards diffusion and downward particle flux in
the gravitational field of the earth
mg
D :
kT
(7.56)
It depends on the diffusion constant D and therefore according
to (7.44) on the mean free path .
mgz=.kT/
:
(7.58)
The exponential decrease of the density in the isothermal
atmosphere is due to the common action of diffusion and
gravitational force, namely the compensation of the upwards
diffusion due to the density gradient and the downwards
motion of the molecules due to the gravitation (Fig. 7.33).
The concentration ni .z/ of molecules with mass mi therefore
depends on their mass mi . For the different molecular components in the earth atmosphere a different z-dependence appears
(Fig. 7.34). The density of the heavier components should
decrease more rapidly with increasing height than the lighter
component. However, the measurements show that the composition of the atmosphere up to altitudes of about 30 km does not
change much with the altitude. This has the following reason:
The atmosphere is not isothermal but the temperature changes
with increasing altitude (Fig. 7.35). These temperature differences causes pressure differences and strong upwards and
downwards air currents which mix the different layers of the
From (7.54) and (7.56) we obtain
D
nmg
dn
D
D :
dz
kT
(7.57)
Lighter
component
m2
Table 7.4 Gas composition of the earth atmosphere
Component
Nitrogen N2
Volume %
78.084
Oxygen O2
20.947
Argon Ar
0.934
Carbon-Dioxyd CO2
0.032
Neon Ne
0.0018
Helium He
Methane CH4
Krypton Kr
Hydrogen H2
trace gases (e.g. SO2 , O3 , NO2 )
Heavier
component
m1
5:2 10
4
1:1 10
4
2 10
4
5 10
5
< 5 10
4
Figure 7.34 Density distribution of two molecular components with different
masses in an atmosphere in the gravity field of the earth, if the atmosphere is
governed by diffusion
The Atmosphere of the Earth
205
Chapter 7
7.6
a)
b)
Figure 7.35 Temperature trend T .h / in the earth atmosphere. a Measured dependence on a logarithmic scale; b Temperature trend on a linear scale. The black
curve gives the dependence T .h / of the standard atmosphere [7.10]
atmosphere. The temperature and its dependence on z is determined by the locally varying net energy flux into the different
layers.
There are at first the absorption of the sun radiation which heats
up the atmosphere. Furthermore the absorption of the infrared
radiation emitted by the earth surface contributes to the energy
flux. The radiation scattered back by the atmosphere diminishes
the energy of the different parts of the atmosphere in a different
way [7.5–7.8].
In order to standardize the description of the atmosphere, a
model atmosphere was defined which serves as the standard atmosphere [7.7]. It is divided into different layers (Fig. 7.35).
Within these layers the temperature T.h/ is a defined function
which should be close to the measured values. Between these
layers small regions appear where the temperature is nearly constant. They are called “pauses”.
The troposphere extends from the ground to altitudes of about
8–12 km, where the upper limit depends on the season of the
year. In the troposphere our weather takes place. The temperature decreases with increasing height with a slope dT=dz
6ı =km from a mean value T.z D 0/ D 17 ı C to T.z D
12 km/ D 52 ı C. The nearly linear temperature decrease is
caused by the heat transport from the earth surface into the atmosphere (convection, heat conduction and infrared radiation)
which decreases nearly linear with increasing z.
Above the troposphere lies a thin layer, the tropopause, where
the temperature stays nearly constant. The range between 10
to 50 km altitudes is the stratosphere. In the lower part of the
stratosphere the temperature is nearly constant. With increasing
altitude it increases up to 0 ı C. The reason for the temperature
increase is the ozone layer between 30–50 km, which contains
O3 -molecules that absorb the UV-radiation from the sun and are
excited into higher energy states. The excited ozone molecules
collide with other atmospheric molecules and transfer their excitation energy into kinetic energy of the collision partners thus
raising the temperature.
Above the stratosphere lies the stratopause, followed by the
mesosphere between 50–80 km altitude. Here the temperature decreases with increasing altitude down to 93 ı C and the
molecular composition changes. With increasing altitude the
lighter elements prevail as shown in Fig. 7.34. Because of the
lower density the collision rate is much lower than in the troposphere and the mean free path is several kilometres. Therefore
air currents are less effective in mixing the different layers.
The mesopause separates the mesosphere form the higher thermosphere, (85 km until 500–800 km) where the temperature
rises up to 1700 ı C. The temperature rise is due to collisions
with high energy particles (electrons and protons) from the sun
(sun wind) which also cause the polar light phenomena (aurora
polaris). The density decreases down to 10 6 Pa D 10 11 bar. In
spite of the low density, friction effects are-non negligible. The
international space station ISS in about 350 km above ground
flies through the thermosphere and it has to be lifted from time
to time, because it loses kinetic energy and therefore altitude due
to friction. The extreme ultraviolet radiation and the solar wind
cause dissociation and ionization of the atmospheric molecules.
Part of the components in the thermosphere are therefore ions.
The thermosphere is part of a larger range in the atmosphere
called the ionosphere which extends far into the space around
206
7 Gases
Chapter 7
the earth. It is no longer spherical symmetric because the magnetic field of the earth influences the path of charged particles
(see Vol. 2) which move on spiral paths around the magnetic
field lines. The variation of the temperature T.h/ with increasing height h implicates that density n.h/ and pressure p.h/ are
no longer strictly proportional (Fig. 7.36).
The evaporation and condensation of water plays an important
role in the lower atmosphere. Also the spurious concentrations
of molecules with dipole moment such as OH, CO, CO2 , NH4
etc. and dust particles and other aerosols have a pronounced
influence on the weather conditions in the troposphere. The
Chemistry of the atmosphere based on reaction of the different species by collisions with each other, is a subject of intense
research and there are still many open questions [7.3a–7.5].
Figure 7.36 Dependence of pressure p .h / and density n .h / on a logarithmic
scale [7.11]
Summary
For a constant temperature the pressure p of a gas in a
closed but variable volume V obeys the Boyle–Mariotte law:
p V D const
The air pressure in an isothermal atmosphere decreases
exponential with the altitude h above ground, due to the gravitational force.
For the particle density n.h/ holds:
N.h/ D n.0/ expŒ mgh=kT :
Without mixing effects in the atmosphere the concentration
of particles with larger mass m therefore decreases faster
with h than for those with lighter mass.
The real earth atmosphere in not isothermal. Due to upwards and downwards air currents the different layers of the
atmosphere are mixed which leads to an equilibrium of the
concentrations of different masses.
The kinetic gas theory explains the macroscopic features
of gases such as pressure and temperature by the average
momentum and the kinetic energy of the gas molecules.
With the Boltzmann constant k the mean kinetic energy of
molecules with mass m is related to the temperature T by
.1=2/mv 2 D .3=2/kT.
The velocity distribution n.v/ of gas molecules at thermal equilibrium is the Maxwell–Boltzmann distribution
n.v/dv D v 2 expŒ .1=2/mv 2 =kTdv for the magnitude
v D jvj of the velocity. The distribution n.vi /, .i D x; y; z/ of
the velocity components vi is a Gaussian function, symmetric to vi D 0.
These distributions can be experimentally determined in
molecular beams using mechanical velocity selectors. The
molecular beams are formed by expanding a gas from a reservoir through a small hole into the vacuum, where the mean
free path is longer than the dimensions of the vacuum chamber. The beam can be collimated by small apertures which
transmit only molecules with small transverse velocities.
Always when gradients of concentrations in a gas exist,
diffusion processes occur which try to equalize the concentrations. The mean diffusion particle flux jD D D grad n
is proportional to the gradient of the particle density n. The
diffusion constant D depends on the kind of particles. Diffusion causes a mass transport from regions of high particle
density n to those of low density.
If velocity gradient in a gas flow appear, viscosity causes momentum transfer from particles with higher flow velocity to
those with lower velocity.
If temperature gradients appear in a gas, energy is transported
by diffusing molecules from regions of higher temperature to
those of lower temperature. For a one-dimensional temperature gradient dT=dx the transferred heat power is dW=dt D
dT=dx. The heat conductivity depends on the particle
density n, the mean velocity v, and the mean free path .
The density distribution n.h/ in the atmosphere is determined
by the common action of gravitational attraction of the air
molecules by the earth and the diffusion current from regions
with higher density to those with lower density. In the real
earth atmosphere furthermore vertical and horizontal air currents occur caused by local heat sources due to absorption of
sun radiation and infrared radiation from the earth surface.
This convection leads to a mixing of different layers in the
lower atmosphere.
Problems
7.1
What would be the density distribution in the atmo- 7.11 In a container is 0:1 kg helium at p D 105 Pa and T D
sphere, if the dependence of the gravitational force on the 300 K. Calculate
altitude is taken into account?
a) the number of He-atoms,
b) the meanP
free path ,
7.2
At which altitude exists, according to (7.6), a pressure c) the sum
Si of all path lengths Si which is passed by all
of 1 mbar, if the constant value T D 300 K is assumed for the
molecules in 1 s. Give this sum in the units m and light years.
temperature T.h/?
7.12 The rotating disc of a velocity selector with a slit allows
7.3
Calculate from (7.6) the pressure at h D 100 km and the N2 molecules with a Maxwellian distribution at T D 500 K to
density n for T D 250 K.
pass for a time interval t D 10 3 s. A detector at 1 m distance
from the disc measures the time distribution of the molecules.
7.4
A balloon with V D 3000 m3 floats at h D 1000 m and What is the half width of this distribution?
a temperature of 20 ı C. What is the maximum weight of balloon with ballast mass and passengers (without the weight of 7.13 What is the minimum velocity of a helium atom at
the filling gas) if one uses as filling gas
100 km above ground for leaving the earth into space? At which
a) helium
temperature would half of the N2 -molecules above 100 km altib) hydrogen gas H2
tude escape into space?
at a pressure equal to the external air pressure.
(%air D 1:293 kg=m3, %He D 0:1785 kg=m3, %H2 D 0:09 kg=m3 7.14 The exhaust gases of a factory escaping out of a 50 m
at T D 20 ı C and p D 105 Pa).
high smokestack have the density % D 0:85 kg=m3. How large
is the pressure difference at the base of the smokestack to that
7.5
A shop for diving equipment offers for measuring the of the surrounding air with %air D 1:29 kg=m3?
diving depth a glass tube with movable piston that compresses a
gas volume V D A x. Down to which depth is the uncertainty 7.15 Up to which volume a children’s balloon (m = 10g) has
of the device z 1 m if the piston edge can be read with an to be blown and filled with helium at a pressure of 1.5 bar, in
accuracy of 1 mm and x.p0 / D 0:2 m.
order to let it float in air?
7.6
Which fraction of all gas molecules has a free path that 7.16 In the centre of the sun the density of protons and elecis larger than
trons is estimated as n D 5 1029 =m3 at a temperature of
a) the mean free path
1:5 107 K.
b) 2?
a) What is the mean kinetic energy of electrons and protons?
Compare this with the ionization energy of the hydrogen
7.7
Calculate the probability that N2 -molecules in a gas at
atom (Eion D 13:6 eV).
T D 300 K have velocities within the interval 900 m=s v b) What are the mean velocities?
1000 m=s. What is the total number N.v/ of molecules with ve- c) How large is the pressure?
locities within this interval in a volume V D 1 m3 at T D 300 K
and p D 105 Pa?
7.17 Determine the total mass of the earth atmosphere from
the pressure p D 1 atm D 1013 hPa the atmosphere exerts onto
7.8
What is the thickness z of an isothermal atmospheric the earth surface.
layer at T D 280 K between the altitudes z1 and z2 with p.z1 / D
1000 hPa and p.z2/ D 900 hPa?
7.18 A research balloon has without filling a mass m D
300 kg. How large must be the volume of helium inside the
7.9
What is the square root of the mean square relative ve- balloon to let it rise up if the helium pressure at any height is
locities between two gas molecules
always 0.1 bar higher than that of the surrounding air? (T.h D
a) for a Maxwell distribution
0/ D 300 K, T.h D 20 km/ D 217 K)
b) if the magnitudes of all velocities are equal but the directions
uniformly distributed?
7.19 What would be the height of the earth atmosphere
a) if the atmosphere is compressed with a pressure at the up7.10 The mean free path in a gas at p D 105 Pa and
per edge (assumed to be sharp) of 10 atm at a temperature of
T D 20 ı C is for argon atoms Ar D 1 10 7 m and for N2
300 K?
molecules N2 D 2:7 10 7 m.
b) at T D 0 K where all gases are solidified?
a) What are the collision cross sections Ar and N2 ?
b) How large are the mean times between two successive collisions?
207
Chapter 7
Problems
208
7 Gases
Chapter 7
References
7.1. L.F. Reichl, A Modern Course in Statistical Physics. (Wiley VCH, Weinheim, 2016)
7.2. M. Scott Shell, Thermodynamics and Statistical Mechanics. (Cambridge Series in Chemical Engineering, 2015)
7.3a. R.P. Wayne, Chemistry of the Atmosphere. (Oxford Science Publ., Oxford Univ. Press, 1991)
7.3b. S. Kshudiram, The Earth Atmosphere. (Springer, Berlin,
Heidelberg, 2008)
7.4. St.E. Manahan, Enviromental Chemistry. (CRC Press,
2009)
7.5. T.E. Graedel, P. Crutzen, Atmospheric Change: An Earth
System Perspective. (Freeman and Co., 1993)
7.6. J.M. Wallace, P.V. Hobbs, Atmospheric Science. (Academic Press, 2006)
7.7. US Standard Atmosphere. (Government Printing Office, Washington, 1976) https://en.wikipedia.org/wiki/U.
S._Standard_Atmosphere
7.8. C.D. Ahrens, Essentials of Metereology. (Thomson
Broodge, 2011)
7.9a. St.A. Ackerman, J.A. Knox, Metereology: Understanding
the Atmosphere. (Jones & Bartlett Learning, 2011)
7.9b. Ch. Taylor-Butler, Metereology: The Study of Weather.
(Scholastic, 2012)
7.10. P. Staub, Physics. (TU Wien)
7.11. https://en.wikipedia.org/wiki/Atmosphere_of_Earth
8.1
Basic Definitions and Types of Fluid Flow . . . . . . . . . . . . . . . . . 210
8.2
Euler Equation for Ideal Liquids . . . . . . . . . . . . . . . . . . . . . . 212
8.3
Continuity Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212
8.4
Bernoulli Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213
8.5
Laminar Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216
8.6
Navier–Stokes Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 220
8.7
Aerodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226
8.8
Similarity Laws; Reynolds’ Number . . . . . . . . . . . . . . . . . . . . 228
8.9
Usage of Wind Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229
8
Chapter 8
Liquids and Gases in Motion;
Fluid Dynamics
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235
© Springer International Publishing Switzerland 2017
W. Demtröder, Mechanics and Thermodynamics, Undergraduate Lecture Notes in Physics, DOI 10.1007/978-3-319-27877-3_8
209
210
8 Liquids and Gases in Motion; Fluid Dynamics
Up to now we have only considered
P liquids and gases at rest
where the total momentum P D
pi D 0, although the momenta pi of the individual molecules, because of their thermal
motion, are not zero but show a Maxwellian distribution with
directions uniformly spread over all directions.
Chapter 8
In this chapter, we will discuss phenomena that occur for
streaming liquids and gases. Their detailed investigation has
led to a special research area, the hydrodynamics resp. aerodynamics which are treated in more detail in special textbooks [8.1a–8.3b].
The macroscopic treatment of fluids in motion generally neglects the thermal motion of the individual molecules but
considers only the average motion of a volume element V,
which can depend on the position r D fx; y; zg. Since even for
very small volume elements with dimensions in the mm-range
V still contains about 1015 molecules, the averaging is justified. The main difference between streaming fluids and gases is
the density, which differs by about 3 orders of magnitude. This
is closely related to the incompressibility of liquids while gases
can be readily compressed. For streaming liquids, the density %
is constant in time, while the gas density can vary with time and
position.
A complete description of the macroscopic motion of liquids
and gases demands the knowledge of all forces acting on a volume element V with the mass m D % V. These forces
have different underlying causes:
pressure differences between different local positions induce
forces Fp D grad p V on a volume element V.
The gravity force Fg D m g D % g V leads for fluid
flows with a vertical component to acceleration of V.
If the flow velocity u depends on the position r this results
in friction forces Ff between layers of the fluid flow with
different values of u.
Charged particles in streaming fluids experience additional
forces by external electric or magnetic fields (Lorentz force
see Vol. 2). Such forces play an important role in stars and
in laboratory plasmas. They are therefore extensively investigated in Plasmaphysics and Astrophysics. We will discuss
them here, however, no longer, because their treatment is the
subject of magneto-hydrodynamics and it would exceed the
frame of the present textbook.
8.1
Basic Definitions
and Types of Fluid Flow
The motion of the whole liquid is known, if it is possible to
define the flow velocity u.r; t/ of an arbitrary volume element
dV at every location r and at any time t (Fig. 8.1). All values
u.r; t0 / for a given time t0 form the velocity field (also named
flow field) which can change with time. If u.r/ does not depend
on time, the velocity field is stationary.
For a stationary flow the velocity u.r/ is at any position r temporally constant. It can, however, differ for different locations r
(Fig. 8.1b).
The location curve r.t/, which is traversed by a volume element
V (e.g. visualized by a small piece of cork) is called its streamline or stream filament (Fig. 8.1). The density of streamlines is
the number of streamlines passing per second through an area of
1 m2 . All streamlines passing through the area A form a stream
tube. Since the liquid is always moving along the stream lines
no liquid leaks out of the walls of a stream tube.
For a stationary flow the path r.t/ of a volume element dV follows the curve u.r/ of the flow field. For non-stationary flows
(@u=@t ¤ 0), this is generally not the case as is illustrated in
Fig. 8.2, where the curve u.r; t1 / of the velocity field at time t1
extends from P1 via P2 to P3 . However, when the volume element dV has arrived in P2 at the time t1 C t, the velocity field
has changed meanwhile and the volume element follows now
the curve u.r; t1 C t/ from P2 to P4 .
Since the different forces in (8.1) generally have different directions and furthermore the friction force depends on the velocity
gradient the motion of dV depends on the relative contribution
The Newtonian equation for the motion of a mass element
m D % V in motion is then
F D Fp C Fg C Ff D mRr
du
D % V
;
dt
(8.1)
where u D dr=dt is the flow velocity of the volume element V.
Before we try to solve this equation we will discuss at first some
basic definitions and features of fluids in motion.
Figure 8.1 a Stream line, stream tube and flow velocity u .r; t /; b Momentary
condition of a flow field (velocity field)
8.1 Basic Definitions and Types of Fluid Flow
211
Chapter 8
Figure 8.2 In a nonstationary flow the path of a particle does not necessarily
follow a streamline u .r ; t /
Figure 8.3 Example of a laminar flow
Figure 8.5 Laminar flow from left to right around different obstacles, photographed with the streamline device of Fig. 8.4
Figure 8.4 Streamline apparatus. a Angle view, b Side view
of the different forces. Liquids where the friction forces are negligible compared to the other forces are called ideal liquids. If
the frictional forces are large compared to all other forces we
have the limiting case of viscous liquids.
Examples for the first case are the flow of liquid helium through
a pipe or of air along the smooth wing of an airplane, while
the second case is realized by the flow of honey or molasses
out of a sloped glass container onto a slice of bread or the slow
flow of heavy oil through pipelines. The real liquids and gases
are located between these two limiting cases. A flow where the
stream lines stay side by side without mixing is called a laminar flow (Fig. 8.3). Laminar flows are always realized if the
frictional forces are dominant. They can be demonstrated with
the streamline generator. This is an apparatus where the bot-
Figure 8.6 A laminar flow coming from left becomes turbulent after impinging
on a plate
tom of two containers has narrow slits arranged in such a way,
that the liquid from each container streams alternately through
every second slit downwards between two parallel glass plates
(Fig. 8.4). When one container is filled with red dyed water and
the other with black tinted water the stream lines are alternately
black and red. With such a demonstration apparatus the stream
line conditions with different obstacles in the flow can be readily shown to a large auditorium, if projected onto a large screen
(Fig. 8.5).
Turbulent flows are generated by friction between the wall and
the peripheral layer of the flow if the internal friction of the
flow is smaller than the accelerating forces. Vortices are formed
which can intermixe the stream lines completely (Fig. 8.6).
212
8 Liquids and Gases in Motion; Fluid Dynamics
8.2
Euler Equation for Ideal Liquids
8.3
Continuity Equation
A volume element dV with the flow velocity u.r; t/ passes during the time interval dt a path length dr D udt. Starting from
the position r it reaches the position r C u dt at time t C dt and
has there the velocity
We consider a liquid volume dV D A dx, which flows in
x-direction through a pipe with variable cross section A.x/
(Fig. 8.7a). Its mass is dM D % dV D % A dx. Through
the cross section A1 flows per time unit the mass
u C du D u.r C u dt; t C dt/ :
dx
dM
D %A1
D %A1 ux1 :
dt
dt
(8.2)
Chapter 8
Even for stationary flows, the velocity can change with position.
For example, a liquid flowing through a pipe increases its velocity when the pipe cross section decreases (Fig. 8.3). The stream
line density is there increased. For nonstationary flows the velocity changes also with time even at the same location, because
@u=dt ¤ 0.
We define the substantial acceleration of a volume element dV
as the total change of its velocity u D fux ; uy ; uz g when dV
passes during the time interval dt from the position r to r C dr.
This total acceleration has two contributions:
1. the temporal change @u=@t at the same position
2. the change of u when dV passes from r to rCdr. This change
is per second .@u=@r/ .@r=@t/.
This can be written in components as
dux
@ux
@ux dx
@ux dy
@ux dz
D
C
C
C
dt
@t
@x dt
@y dt
@z dt
(8.3a)
with corresponding equations for the other components uy and
uz .
In vector form this reads with ux D dx=dt, uy D dy=dt, uz D
dz=dt
@u
du
D
C .u r/ u :
(8.3b)
dt
@t
(8.5)
We assume, that at the position x D x0 the cross section A
changes to A2 . For incompressible liquids % remains constant.
Since the liquid cannot escape through the side walls the mass
flowing per time unit through A2 must be equal to that flowing
through A1 . This gives the equation
%A1 ux1 D %A2 ux2 )
ux1
A2
:
D
ux2
A1
(8.6)
Through the narrow part of the pipe the liquid flows faster than
through a wide part. The product
jD%u
(8.7)
is called mass flow density. The product I D j A is the total
mass flow and gives the mass flowing per unit time through the
cross section A.
Equation 8.6 can then be written as I D const. The total mass
flow through a pipe is the same at every position in the pipe.
This statement about the conservation of the total mass flow can
be formulated in a more general way: The volume V contains at
Here u ru is the scalar product of the vector u and the tensor
0 @u
@u
@u 1
x
x
B @x
B
B @uy
ru D B
B @x
B
@ @uz
@y
@uy
@y
@uz
@y
@x
x
@z C
C
@uy C
C :
@z C
C
@uz A
@z
The substantial acceleration is composed of the time derivative
@u=@t of the velocity at a fixed position r and the convection
acceleration .u r/u. The first contribution is only nonzero for
nonstationary flows, the second only if the velocity changes with
the position r.
The equation of motion for an ideal liquid (frictional forces are
negligible) which experiences the accelerating forces of gravity
Fg D mg and pressure gradient Fp D grad pdV is the Euler
equation
du
@u
1
D
C .u r/ u D g
grad p :
(8.4)
dt
@t
%
This is the basic equation for the motion of ideal liquids, which
was already postulated by L. Euler in 1755.
Figure 8.7 Illustration of the continuity equation: a in a pipe with changing
diameter; b in a volume V with surface S with a mass flow dM =dt through V ;
c with a source inside a volume V
8.4
MD
Z
8.4
% dV :
The mass per volume changes with time if mass flows out of
the volume or into the volume. The mass flowing per second
through its surface S is
Z
Z
@M
(8.9)
D % u dS D j dS ;
@t
S
where the normal vector dS is perpendicular to the surface element dA.
According to Gauß’ law (see textbooks on vector analysis
e.g. [8.8]) the surface integral can be converted into a volume
integral over the volume V enclosed by the surface S.
Z
Z
(8.10)
% u dS D div.% u/dV ;
V
S
and we obtain from (8.8)–(8.10) for a constant volume V the
relation
Z
Z
Z
@
@%
% dV D
dV D div.%u/dV :
(8.11)
@t
@t
V
Since this must be valid for arbitrary volumes this gives the continuity equation
@%
C div.%u/ D 0 ;
(8.12)
@t
which states that for any mass flow the total mass is conserved,
i.e. mass is neither produced nor annihilated.
For a constant volume element dV (8.12) can be written as
div.% u/dV D
@%
dV D
@t
Bernoulli Equation
(8.8)
V
S
213
@
.dM/ :
@t
(8.12a)
The expression div.% u/ dV gives the mass that escapes per
second out of the volume element dV. Therefore div.% u/ is
called the source strength per unit volume. A source which
delivers the mass dM=dt per sec leads to a mass flow div.% u/
per sec through the surface surrounding the source (Fig. 8.7c).
The continuity equation (8.12) is valid for liquids as well as for
gases. For incompressible liquids is @%=@t D 0 and % is furthermore spatially constant. The equation of continuity simplifies
then to
(continuity equation for
(8.13a)
div.u/ D 0 incompressible liquids) :
If a liquid or a gas flows in x-direction through a pipe with variable cross section A.x/ the flow velocity is larger at locations
with smaller cross section (continuity equation). The volume
elements therefore have to be accelerated and have a higher kinetic energy than at places with larger cross section. This results
in a decrease of the pressure p. This can be seen as follows:
In order to transport the volume element dV1 D A1 x1 in the
wider part of the pipe through the cross section A1 it has to be
shifted by x1 against the pressure p1 (Fig. 8.8). This demands
the work
W1 D F1 x1 D p1 A1 x1
(8.14a)
D p1 V1 :
In the narrow part of the pipe is V2 D A2 x2 and the work
necessary to shift V2 by x2 against the pressure p2 is
W2 D p2 A2 x2
D p2 V2 :
The kinetic energy of the volume elements is
Ekin D 12 M u2 D 12 % u2 V :
For ideal liquids (frictional forces are negligible) the sum of
potential and kinetic energy has to be constant (energy conservation). This gives the equation
p1 V1 C 12 %u21 V1 D p2 V2 C 21 %u22 V2 :
p1 C 12 %u21 D p2 C 12 %u22 :
(8.16)
For a frictionless incompressible liquid flowing through a horizontal pipe with variable cross section (Fig. 8.9) we obtain for a
stationary flow from (8.16) the Bernoulli Equation
p C 12 %u2 D p0 D const :
(8.17)
The constant p0 is the total pressure which is reached at locations with u D 0. the quantity ps D .%=2/u2 D p0 p
is the dynamic stagnation pressure (ram pressure), while
p D p0 ps is the static pressure of the flowing liquid.
(8.13b)
In pipes with constant cross section A the liquid flows only into
one direction which we choose as the x-direction. Then uy D
uz D 0 and (8.13b) becomes @ux =@x D 0 ) ux D const.
(8.15)
For incompressible liquid is % D constant and therefore V1 D
V2 D V. This simplifies (8.15) to
For the three components this equation reads
@ux
@uy
@uz
C
C
D0:
@x
@y
@z
(8.14b)
Figure 8.8 Illustration of Bernoulli-equation
Chapter 8
time t the liquid mass
Bernoulli Equation
214
8 Liquids and Gases in Motion; Fluid Dynamics
Chapter 8
Figure 8.10 Measurement of pressure conditions in flows. a Measurement of
static pressure; b measurement of total pressure p0 with Pitot tube and pressure
manometer; c measurement of p0 with a stand pipe; d measurement of stagnation pressure ps D p0 p as difference of total pressure and static pressure
Figure 8.9 Demonstration of Bernoulli equation by pressure measurements in
stand pipes. The pressure difference is p D % g h . a For ideal liquids
without friction; b for real liquids with friction. The liquid streams from left to
right
The Bernoulli equation can be demonstrated with the arrangement shown in Fig. 8.9, where dyed water flows through a
horizontal glass tube with variable cross section and vertical
stand pipes. The rise h in the vertical stand tube gives the static
pressure p D % g h. At the narrow parts of the horizontal pipe
the flow velocity is larger and therefore pressure and height h are
smaller. In Fig. 8.9a the situation for an ideal frictionless liquid
with p.x/ D constant for constant cross section is shown, while
Fig. 8.9b illustrates the influence of friction on the pressure p.x/.
For tubes with constant cross section a linear decrease of p.x/ is
observed.
The three quantities p, p0 and ps can be measured at arbitrary
locations in the flow with the devices shown in Fig. 8.10a–d.
With a pressure gauge, shown in Fig. 8.10a which has a small
hole in the sidewall of a tube, the liquid flow, streaming around
the tube creates a static pressure inside the tube, which is monitored by a pressure manometer. The pitot-tube (Fig. 8.10b and
c) has a hole at the end of the tube. If the tube is aligned parallel
to the stream lines the flow velocity at the head of the tube is
u D 0, i.e. the measured pressure is the total pressure p0 . It
can be measured either with a manometer (Fig. 8.10b), or with a
vertical stand pipe (Fig. 8.10c). With a combination of pressure
gauge and Pitot tube (Fig. 8.10d) the pressure p0 is measured
at the head of the horizontal tube while a hole in the sidewalls
monitors the pressure p. The difference ps D p0 p is shown as
the difference of the heights of mercury in the U-shaped lower
part of the device.
For liquid flows in inclined pipes the difference of potential energies Epot D %ghV of a volume element V at different
Figure 8.11 Flow of a liquid through an inclined pipe
heights h has to be taken into account. If the flow, for instance,
is directed in the x-z-plane (Fig. 8.11) the height is h D z.x/ and
we obtain from (8.17) the general equation
p C %gz.x/ C 21 %u2 .x/ D const D p0 :
(8.18)
For an ideal incompressible liquid % is constant within the whole
pipe. If the cross section of the pipe is constant also the flow
velocity u is constant throughout the whole pipe. If p C % g z
p0 the flow ceases and u D 0 in the whole pipe.
Note: Although the Bernoulli equation (8.17) has been derived
for incompressible liquids the equation allows to obtain also the
pressure change of gases for laminar flows at not too high flow
velocities. For example, inserting for air flows the numerical
values p0 D 1 bar, u D 100 m=s, % D 1:293 kg=m3 into the
equation
p0
p D 21 % u2
one obtains p D 0:935p0, i. e. a pressure decrease of 6.5%
and therefore also a corresponding decrease of the density %.
However, if the flow velocity approaches the velocity of sound
(c D 340 m=s) the change of the density becomes so large that
the condition of incompressibility is even approximately not fulfilled.
Bernoulli Equation
Figure 8.13 Vaporizer
Figure 8.12 Hydrodynamic paradox: a Two curved aluminium plates, which
can swing around a yoke, are pressed together when blowing air between them;
b the lower circular plate is attracted to the upper plate when air is blown
through the pipe
The Bernoulli equation can be demonstrated by many simple experiments which often astound the auditorium. One example is
the hydrodynamic paradox. Two curved aluminum plates are
hanging on a U-shaped wire bar (Fig. 8.12a). If one blows air
between the two plates they move towards each other, contrary
to the expectation that they will be pushed away from each other.
When air is blown through a vertical pipe fixed on one end to a
circular disc S1 with a hole (Fig. 8.12b) a second disc S2 below
the fixed disc is lifted to the upper disc by the air streaming between the two discs. The distance d between the two discs with
area A must be below a critical value where the flow velocity u
of the air is sufficiently large to cause an attractive force
F D A.p0
Figure 8.14 Water jet pump
pressure. The air from the surrounding diffuses into the region
with reduced pressure where it penetrates into the water jet and
is transported out of the container into the outer space A, thus
evacuating the container. With such a device reduced pressures
down to 30 mbar can be achieved.
Undesirable effects of the Bernoulli theorem are the unroofing
of houses under the action of typhoons (Fig. 8.15). When wind
p/ D 21 % u2 A m g
between the two discs which can balance the weight m g of the
lower disc.
The Bernoulli theorem is used for many practical applications.
Examples are the vaporizer or the spray bottle (Fig. 8.13) where
air streams out of a narrow nozzle and generates a reduced
pressure, which sucks the liquid out of the bottle into the air
stream. Here it is nebulized. Another example is the water
jet vacuum pump (Fig. 8.14). Here water streams with a large
velocity through a narrow nozzle where it generates a reduced
Figure 8.15 A strong wind can unroof a house due to the reduced pressure
above the roof
215
Chapter 8
8.4
216
8 Liquids and Gases in Motion; Fluid Dynamics
Chapter 8
Figure 8.16 Aerodynamic lift at a wing profile due to the higher velocity around the upper side of the profile
blows with the flow velocity u.x/ over the roof of a house, the
pressure difference p D p0 p results in an upwards directed
force
Z
Z
F D Ly p.x/dx D Ly 12 %u2 .x/dx
on the roof, where Ly is the length of the roof in y-direction. The
pressure difference depends on the flow velocity u.x/ which is
maximum at the top of the roof, where the stream lines have the
highest density.
The Bernoulli equation is the basis of the aerodynamic lift force
and therefore important for the whole aviation. In Fig. 8.16 the
profile of an airplane wing is shown with the stream lines of air
flowing below and above the wing. For the asymmetric profile
the air flows faster above than below the wing. This cause, according to (8.17) for a wing area A and the air density %a a lift
force
F D .p2 p1 / A D 21 %L u22 u21 A :
Remark.
Since air at high flow velocities is compressible
and therefore cannot be treated as ideal liquid, the situation for
a plane is more complex because the flow velocity of the air
relative to the flying plane is very large. Besides friction forces
turbulence and density changes play an important role for the
calculation of the upwards lift (see Sect. 8.6).
8.5
Laminar Flow
Laminar flows (Fig. 8.3) are always realized when the frictional
forces exceed the accelerating forces. Therefore, we will at
first discuss the internal friction in liquids and gases and then
illustrate the importance of laminar flow by several practical examples.
8.5.1
Figure 8.17 Internal friction of liquids. a A slab is pulled with the velocity ux
through a viscose liquid. It takes along boundary layers of the liquid. b Velocity
profile and thickness D of the boundary layer
will be dragged with the moving plate due to the static friction
between liquid and plate surfaces. These layers transfer part of
their momentum %L uz dV to the neighbouring liquid layers. This
can be demonstrated by the experiment shown in Fig. 8.17a: In
a glass trough is a viscous liquid, for example glycerine. The
left part of the liquid is dyed. When an immersed plate is slowly
pulled through the liquid with the velocity u0 into the z-direction
one can see that the liquid layers adjacent to the plate surfaces
stick to the surfaces and are dragged with the velocity u0 . Perpendicular to the plate surfaces a velocity gradient is present
(Fig. 8.17b). As has been discussed in Sect. 7.5, this gradient is
due to the thermal motion of the liquid molecules, which penetrate by about a mean free path into the neighbouring layers
and transfer part of their momentum. This causes a velocity gradient du=dx perpendicular to the velocity of the plate.
In Sect. 7.5.4 it was shown, that the momentum transferred per
second and unit area between neighbouring layers is jp D
duz =dx. Since the time derivative of the momentum is equal to
the acting force we obtain for the force between adjacent layers
Internal Friction
Assume a plane sheet with area A in the y-z-plane is pulled
through a liquid with the velocity u0 into the horizontal direction (which we choose in Fig. 8.17 as the z-direction). The liquid
layers at x D x0 ˙ dx adjacent to the two plate surfaces at x D x0
F DA
du
;
dx
(8.19)
which appears when the plate is pulled with the constant velocity u0 through the liquid, where A is the total surface of the plate
(both sides!). This force must just compensate the friction force
8.5
Table 8.1 Dynamical viscosities of some liquids and gases at a temperature
T D 20 ı C
=.mPa s/
1.002
0.65
1.20
1480.0
660
1.55
1:8 10 2
1:9 10 2
%
u dm D
2
2
ZD
2u20 .1
0
jxj=D/2 A dx
(8.22)
The boundary layer has therefore the order of magnitude
du
:
dx
(8.20)
The dynamic viscosity depends strongly on the temperature,
as can be seen from Tab. 8.2. For liquid helium a superfluid
phase exists at temperatures below 2:17 K, where D 0 Pa
s [8.5].
The distance D where the liquid is dragged by the moving plate
is called fluid dynamic boundary layer. Its value can be obtained by the following consideration: In order to move the plate
by its length L against the frictional force Ff one has to accomplish the work
ˇ ˇ
ˇ du ˇ
u0
Ff L D AL ˇˇ ˇˇ D AL
;
dx
D
(8.21)
where we have assumed that a linear velocity gradient du=dx D
u0 =D is valid (Fig. 8.17b). The liquid layer with a mass dm
and a velocity u has the kinetic energy dEkin D .1=2/dm u2 .
With the constraint u.x D ˙D/ D 0 the velocity of the layer is
u D u0 .1 jxj=D/. Altogether the kinetic energy of all dragged
Table 8.2 Temperature dependence of the dynamical viscosity .T / of water
and glycerine
0
C20
C40
C60
C80
C100
Z
Due to friction part of the spent work is converted into heat.
Therefore we get Ekin < Wf . This yields with (8.21) the relation
3L 1=2
D<
:
(8.23)
%u0
In Tab. 8.1 the numerical values of for some liquids are compiled. They should be compared with the data for gases in
Tab. 7.3.
T=ı C
1
D
2
1
D A%Du20 :
3
The factor is the dynamic viscosity. It has the dimension Œ D
N s=m2 D Pa s. In the older literature often the unit Poise D
P D g cm 1 s 1 is used. The conversion is 1 P D 0:1 Pa s;
1 centipoise D 1 cP D 10 3 Pa s.
Wf D
layers is
Ekin
opposite to the direction of u0
Ff D A
Viscosity .T/=.mPa s/
Water
Glycerine
1.792
12 100
1.002
1480
0.653
238
0.466
81
0.355
31.8
0.282
14.8
217
D
s
L
:
%u0
(8.24)
The boundary layer can only develop if the distance d to the
container walls is larger than D. For d < D the static friction
between the liquid and the wall forces the velocity u.d/ D 0,
the dragged boundary layer becomes smaller and the velocity
gradient larger.
For the derivation of the general friction force on a volume element dV D dx dy dz we choose a liquid flowing into the
z-direction with an arbitrary velocity gradient
@uz @uz @uz
grad uz D
;
;
:
@x @y @z
We regard in Fig. 8.18 at first a flow that has only a gradient
@uz =@x in x-direction (@uz =@y D @uz =@z D 0). The flow velocity
uz .x/ can be expanded into a Taylor series
uz .x0 C dx/ D uz .x0 / C
@uz
dx C : : : ;
@x
(8.25)
which we truncate after the linear term.
The liquid layer between x D x0 and x D x0 C dx experiences a
friction force dFf per surface element dA D dydz. If @uz =dx > 0
this force is decelerating for the surface layer at x D x0 because
here the neighbouring layer at x D x0 dx is slower but it is
accelerating for the surface layer at x D x0 C dx, because here
the adjacent layer is faster (Fig. 8.18b). The net tangential force
is therefore
.ıFf /z D dFf .x0 C dx/ dFf .x0 /
"
@uz
D dydz
@x xDx0 Cdx
@uz
@x
xDx0
#
:
Inserting the derivatives from (8.25) yields for the bracket the
expression .@2 u=@x2 / dx and therefore for the net force onto the
volume element dV due to the velocity gradient @uz =@x
.ıFf /z D dx dy dz
@2 uz
@2 uz
D
dV
:
@x2
@x2
(8.26)
Chapter 8
Substance
Water
Benzene
Ethanol
Glycerine
Heavy fuel oil
Mercure
Air (105 Pa)
Helium (105 Pa)
Laminar Flow
218
8 Liquids and Gases in Motion; Fluid Dynamics
For arbitrary flow velocities u D fux ; uy ; uz g (8.26b) can be generalized to
Z
Ff D u dV ;
(8.26c)
V
R
this is equivalent to the three equations .Ff /i D ui dV for
the components i D x; y; z.
Chapter 8
8.5.2
Laminar Flow Between Two Parallel
Walls
In order to maintain a stationary flow with constant velocity into
the z-direction between two walls at x D d and x D Cd one
has to apply a force opposite to the friction force Ff which just
compensates Ff . This force can be, for instance, caused by a
pressure difference between the planes z D z0 and z D Cz0 .
In the following we assume that the pressure is constant in a
plane z D constant, i.e. independent of x and y.
We consider in Fig. 8.19 a volume element dV D dx dy dz
with the width dy D b in y-direction and the height dz. At its
end faces z D z1 and z D z1 C dz the pressure forces
Figure 8.18 Derivation of friction force acting on a volume element dx dy dz
in a flow with homogeneous velocity profile
A similar result is obtained for the velocity gradient @uz =@y in
y-direction.
dF1 D b dx p.z1 /
and dF2 D b dx p.z1 C dz/
are effective. They result in a total force onto the volume element dV
dp
dFz D b dx dz :
(8.27)
dz
For compressible media, e.g. for gases, a velocity gradient
@uz =@z can also appear for a flow into the z-direction if the density changes with z, while for incompressible media @uz =@z ¤ 0
only if the velocity changes, e.g. in tubes with variable cross
section.
From (8.25)–(8.26) we finally obtain for the total friction force
onto the volume element dV in case of a laminar flow with the
velocity uz the expression
2
@ uz
@2 uz
@2 uz
:
(8.26a)
.dFf /z D dV
C
C
@x2
@y2
@z2
The first two terms cause tangential forces (shear forces, see
Sect. 6.2.3), the third term, which is only nonzero for compressible media causes a normal force onto the surface element dxdy.
With the Laplace operator
D
@2
@2
@2
C 2 C 2
2
@x
@y
@z
(see Sect. 13.1.6) the total friction force onto the volume element dV, which moves with the velocity u D f0; 0; uz .x; y; z/g
can be written as
.dFf /z D uz dV :
(8.26b)
Figure 8.19 Laminar flow between two parallel walls
8.5
Laminar Flow
219
This pressure force compensates the friction force
.dFf /z D dVuz D dy dx dz
d2 uz
;
dx2
if the condition
1 dp
duz
)
D
dz
dx
x dp
C C1
dz
is fulfilled. The integration constant C1 D .duz =dx/xD0 gives the
slope of the velocity profile u.x/ at x D 0.
Integration yields
Figure 8.20 a Derivation of Hagen–Poiseuille law; b velocity profile of a laminar flow in a cylindrical tube
2
uz D
x dp
C C1 x C C2 ;
2 dz
(8.28)
Integration over r yields
since p and dp=dz do not depend on x.
For a liquid streaming between two parallel walls at x D d and
x D Cd symmetry arguments demand .du=dx/xD0 D C1 D 0.
At x D ˙d the static friction between the liquid and the walls
causes u.x D ˙d/ D 0. This gives for the integration constant
C2
d 2 dp
C2 D
:
2 dz
We then obtain for the velocity profile the parabola
u.x/ D
1 dp 2
.d
2 dz
x2 / ;
(8.29a)
with the crest at x D 0 midway between the two walls. If the
friction between the liquid and the walls is not high enough
(u.˙d/ ¤ 0), we get instead of (8.29a) the more general equation
1 dp 2
u.x/ D
x2 / C ud :
(8.29b)
.d
2 dz
u.r/ D
ZR
r
p1 p2
p1 p2 2
r dr D
R
2L
4L
r2 :
(8.30)
This velocity profile is a rotational paraboloid. It can be vividly
demonstrated by the flow of coloured glycerine through a vertical pipe (Fig. 8.20b).
The total liquid volume flowing per second through the plane
z D constant of the hollow cylinder with radii between r1 and
r1 C dr shown in Fig. 8.20a is according to (8.30)
d
2r dr .R2
.V.r// D 2r dr u D
dt
4L
r2 /
.p1
p2 / :
The total volume streaming during the time t through the pipe is
V Dt
ZR
2r u dr
rD0
4
(8.31)
R .p1 p2 /
R4 p
D
tD
t:
8L
8L
8.5.3
Laminar Flows in Tubes
The flow of liquids in cylindrical tubes plays an important role
for many technical applications (water pipes, oil pipelines), and
also in medicine (blood flow through veins). It is therefore
worthwhile to study this problem in more detail.
We assume, as in the previous example, a pressure difference
p1 p2 between the planes z D 0 and z D L in a cylindrical
pipe with radius R (Fig. 8.20) which maintains a stationary flow.
Symmetry reasons demand that the flow velocity can only depend on the distance r from the cylinder axis. For a coaxial small
cylinder with radius r the same reasoning as in the previous section gives for the condition “friction force must compensate the
pressure force”
2r L
du
D r2 .p1
dr
p2 / :
The ratio p=L D @p=@z is the linear pressure gradient along
the tube. The total volumetric flowrate (volume per second)
through the pipe is then
Hagen–Poiseuille Law
dV
R4
R4 @p
D
p D
dt
8L
8 @z
(8.32)
Note the strong dependence of dV=dt from the radius R of the
pipe .R4 Š/.
The human body utilizes this dependence for the regulation of
the blood flow by adjusting the cross section area of the veins.
Chapter 8
d2 uz
D
dx2
220
8 Liquids and Gases in Motion; Fluid Dynamics
8.5.4
Stokes Law, Falling Ball Viscometer
When a ball with radius R is dropped with the initial velocity
u D 0 into a liquid one observes at first an acceleration of the
ball due to the gravity force and after a short falling distance a
constant velocity. For this uniform motion the friction force Ff
which increases with increasing velocity just cancels the gravity
force
Fg D meff g D .%k
%Fl / 34 R3 g
(8.33)
Figure 8.22 Viscosimeter with sinking ball and photoelectric barrier
Chapter 8
diminished by the buoyancy (Fig. 8.21).
Experiments with different liquids and balls with different radii
prove that the friction force is proportional to the viscosity of
the liquid, to the radius R of the ball and to its velocity u. For
radii still small compared to the diameter of the container one
finds
Stokes Law
Ff D 6 R u0 :
(8.34a)
Example
For steel balls .% D 7:8 kg=dm3/ with radius R D
0:1 cm falling in glycerine (% D 1260 kg=m3 ) and D
1:48 Pa s, the stationary velocity becomes u0 D 1 cm=s.
In this case the second term is 3:2 10 3 1. For
R D 1 cm, however, u0 D 1 m=s and the second term
becomes 2:5 > 1 and cannot be neglected.
The Stokes Law (8.34a) therefore is correct only for sufficiently small products R u0 of ball radius and final
J
velocity u0 .
The stationary final velocity u0 is obtained for Ff C Fg D 0
u0 D
2 R2
g .%K
9
%Fl / :
(8.35)
Measuring u0 allows the determination of the viscosity , if the
densities of liquid and ball and the ball radius R are known
(Falling ball viscometer Fig. 8.22). According to (8.35) the ratio
u0 =R2 should be independent of the ball radius. This is indeed
observed for small radii R.
Stokes Law (8.34a) can be derived also theoretically. A more
detailed calculation [8.1a, 8.1b, 8.7] shows that (8.34) is only
an approximation. The exact expression for the friction force,
derived by Oseen, is
3%Fl R u0
Ff D 6R u0 1 C
:
(8.34b)
8
The second term in the bracket is for small radii R small compared to 1 and can be neglected.
8.6
Navier–Stokes Equation
In the previous sections we have discussed the different forces
acting onto a volume element dV in a streaming liquid. We can
now present the general equation of motion for a real viscous
streaming liquid. With the different contributions
dFf D u dV
dFp D grad p dV
dFg D D % g dV
(friction force)
(pressure force)
(gravity force)
to the total force and the substantial acceleration (8.3)
du
@u
D
C .u r/
dt
@t
we obtain the Navier–Stokes equation
@
%
C u r u D grad p C % g C u :
@t
(8.36a)
For ideal liquids with D 0 this reduces to the special case of
the Euler equation (8.4). The friction term u expands the
Euler equation, which is a first order differential equation, to
a second order differential equation, which is more difficult to
solve.
Figure 8.21 Uniform sink speed u0 of a ball in a viscose liquid
On the right hand side of (8.36a) the forces are listed and on the
left hand side the motion induced by these forces, which we will
now discuss in more detail.
8.6 Navier–Stokes Equation
The first term @u=@t gives the time derivative of the velocity at
a fixed location. The second term describes the change of the
velocity of dV while it moves from the position r to r C dr.
Using the vector relation
1
2
grad u2
.u rot u/ ;
Circulation
(8.36b)
that is deduced in textbooks on Vector Analysis [8.8, 8.9] (see
also Sect. 13.1.6) we see that this spatial change of u can be
composed of two contributions: The first term gives the change
of the amount of u, the second term the change of the direction
of u. This second term gives rise to vortices in the liquid, which
we will discuss next.
8.6.1
Vortices and Circulation
Circulation
flow
Circulation
flow
When a liquid streams around a circular obstacle one observes
for small velocities the streamline picture of laminar flow,
shown in Fig. 8.5. If the velocity is increased above a critical
velocity uc , which depends on the viscosity of the liquid, vortices appear behind the obstacle (Fig. 8.23). Such vortices can
be made visible by small cork pieces floating on the liquid and
moving along the streamlines. One observes that in a region
around the centre of the vortex the liquid rotates like a rigid
body. The rotational velocity
Kernel of vortex
Figure 8.24 Kernel of vortex and circulation region
uD!r
increases linear with the distance r from the centre and all particles have the same angular velocity !. This region r < rk is
called the vortex kernel (Fig. 8.24). Inserting small cork pieces
with a fixed direction arrow to the surface of the liquid it becomes apparent that they turn once around their own axis while
circulating around the vortex. (Fig. 8.25) as expected for a rigid
rotation.
Outside of the vortex kernel .r > rk / the angular velocity
! decreases with increasing distance r. The particles do no
longer rotate about their axis but keep their spatial orientation
(Fig. 8.25). This region of the vortex is called the circulation.
Here a deformation of the volume elements during the rotation
takes place (Fig. 8.26).
Figure 8.25 Orientation of cork pieces: a inside the vortex kernel (circular
motion with turning orientation), b in the circulation region (non turning orientation)
We can describe the vortex by the vortex vector
˝ D 12 rot u :
(8.37)
Figure 8.26 Deformation of a plane element in the circulation region outside
the vortex kernel
Figure 8.23 Generation of vortices in a turbulent flow around a circular obstacle
The amount of ˝ gives the angular velocity ! inside the vortex
kernel (see below). Magnitude and direction of ˝ in a vortex are
generally not constant. They change because the vortex is not
necessarily fixed in space but moves with the flowing liquid to
Chapter 8
.u r/ u D
Kernel
221
222
8 Liquids and Gases in Motion; Fluid Dynamics
Vertex
tube
Vertex
lines
Chapter 8
Figure 8.28 Explanation of circulation and its relation with rot u
Figure 8.27 Vortex lines and tube
other locations and furthermore the energy of the vortex changes
because of friction and with it the magnitude of ˝ changes.
The curves which coincide at every place with the direction of
˝ are called vortex lines. When, for instance, the particles
move on circles in the x-y-plane the vector ˝ points into the
z-direction. All lines parallel to the z-direction inside the kernel
with x2 C y2 rk2 are vortex lines (Fig. 8.27). All vortex lines
through the vortex area A form the vortex tube.
For a quantitative description of torques based on the Navier–
Stokes equation we have to study the rotational part .u rot u/
in (8.36a,b). At first we must understand, that the term rot u
describes the rotation of moving particles. We therefore regard
in Fig. 8.28 the tangential velocity components along the edge of
the surface element dx dy. As a measure of the torque strength
of the flow through the area A we define the circulation
I
Z D u ds
(8.38a)
along the edge of the surface in the counterclockwise direction.
Our surface element dx dy contributes the share
@uy
dZ D ux dx C uy C
dx dy
@x
@ux
ux C
dy dx uy dy
(8.38b)
@y
@uy @ux
D
dx dy D .rot u/z dx dy
@x
@y
to the circulation, because the z-component of rot u D r u is
defined as .r u/z D .@uy =@x @ux =@y/.
Analogous relations are obtained for the x- and y-components.
From these relations one obtains by integration the Stokes’ theorem
I
Z
u ds D rot u dA ;
(8.38c)
A
which states: “The surface integral over rot u equals the path
integral along the border of the surface element”.
For a circular current of a liquid around a centre the circulation
at a distance r from the centre is
I
Z D u ds D 2ru.r/ :
(8.38d)
The average amount ˝ of the vortex vector ˝ D 12 rot u that
points into the direction perpendicular to the surface is, according to Stokes’ theorem
I
Z
1
1
˝D
uds
jrot uj dA D
2A
2r2
u
2ru
D
D ;
2
2r
r
(8.38e)
where A is the area of the torque kernel.
Since the torque kernel rotates like a solid body, ˝ must be
independent of r. As illustrated in Fig. 8.24 the velocity u D
r ˝ increases linear with r.
The average ˝ D Z=2A of the magnitude of the torque vector
gives the circulation per surface unit and therefore the torque
strength per surface unit.
8.6.2
Helmholtz Vorticity Theorems
For an ideal liquid . D 0/ the Navier–Stokes equation (8.36a)
without external fields (gravity is neglected ! g D 0) can be
transformed into an equation that illustrates certain conservation laws. This was first recognized in 1858 by Hermann von
Helmholtz.
On both sides of (8.36a) we apply the differential operator rot,
divide by the density % and obtain from (8.36b) and (8.37) with
rot grad p D r rp D 0 the equation (see Probl. 8.11)
@˝
C r .˝ u/ D 0 :
@t
(8.39)
Together with the equation of continuity div u D 0 (8.13) for
incompressible media this equation determines completely and
for all times the velocity field of an ideal streaming liquid. This
means: If the quantities ˝ and u are given at a certain time
(8.39) describes their future development unambiguously.
For example: If for t D t0 the vortex vector ˝ for the total liquid
is ˝ D 0, it follows from (8.39) @˝=@t D 0. This means: If
an ideal liquid without vortices is set into motion it will stay
vortex-free for all times.
8.6 Navier–Stokes Equation
223
Figure 8.29 Deformation of a circular vortex during its flow with conservation
of total mass and angular momentum
If there are vortices in a liquid, it follows
˝D
1
2
rot u ) div ˝ D r .r u/ 0 :
(8.40)
This means: Inside an ideal liquid, there are no sources or sinks
for the vortex lines. They are either closed lines or they end at
the boundary of the liquid, for instance at the walls of the liquid
tube.
Inside an ideal liquid the vortex strength Z D 2˝ A is
constant in time. Vortices cannot be generated nor vanish.
The constancy of Z in a frictionless liquid is equivalent to the
conservation of angular momentum of the mass circulating in a
vortex. Because of D 0 no tangential forces can act and the
pressure forces have only radial components. Therefore, there
is no torque and the angular momentum has to be constant.
These conservation laws can be summarized by the following
model: Vortices move like solid but strongly deformable bodies
through a liquid or a gas. Without friction, their total mass and
their angular momentum remain constant although the angular
velocity and the radius of a vortex can change. This is illustrated
in Fig. 8.29 by a cylindrical vortex. The constancy of the angular
momentum L D I ! (see Sect. 5.5) with the moment of inertia
I D .1=2/Mr2 results in the equation
M1 r12 ˝1 D M2 r22 ˝2 :
Since M1 D M2 and the vortex area A D r2 this gives
A1 ˝1 D A2 ˝2 :
This means the vortex strength is constant.
8.6.3
The Formation of Vortices
In the previous section we have seen, that friction is essential for
the formation of vortices. On the other hand, it was discussed
in Sect. 8.5, that liquids with large friction show a laminar flow
where no vortices occur. Vortices must be therefore formed in
liquids with small viscosity where at certain places, e.g. at the
boundaries with walls, the friction has maxima. Here velocity
gradients occur between adjacent liquid layers because of the
static friction between these liquid layers and the wall. These
velocity gradients produce, due to friction, tangential forces,
which give rise to vortices.
When such boundary layers show small irregularities as shown
exaggerated in Fig. 8.30, the adjacent stream lines are deformed.
At the narrow positions the stream lines are compressed and the
flow velocity u increases. According to the Bernoulli equation,
a pressure gradient p develops which further increases the irregularities. Finally, an unstable condition arises which results
in the formation of vortices.
We will illustrate this vortex formation for the example of a flow
around a circular cylinder (Fig. 8.31). For sufficiently small flow
velocities u the influence of friction is small and a laminar flow
occurs (Fig. 8.5 and 8.31a). At the stagnation point S1 on the
forefront of the cylinder, the flow velocity is zero and according
to (8.17) the pressure equals the total pressure p0 . From S1 the
liquid moves along the upper side of the cylinder and is accelerated until it reaches the point P, where the velocity reaches
its maximum and the pressure its minimum. The acceleration is
caused by the pressure difference p D p0 .S1 / p.P/. At the
stagnation point S2 at the backside of the cylinder the velocity
becomes zero again, because the opposite pressure difference
decelerates the flow and brings the velocity down to zero.
When the flow velocity is increased the velocity gradient between the wall and the adjacent liquid layers also increases. This
increases the friction which is proportional to the velocity gradient. The liquid volume elements do not reach their full velocity
in the point P and therefore reach the velocity v D 0 already in
the point W before S2 (Fig. 8.31b and 8.32). The pressure force
caused by the pressure gradient between S2 and W now accelerates the volume elements into the opposite direction against
the flow velocity of the liquid layers farther away from the wall.
There are two opposite forces acting on the liquid layers close
to the wall (Fig. 8.32):
a the friction force due to the friction between the liquid layers close to the wall and the layers farther away which have
different velocities,
b the force due to the pressure gradient.
These two forces exert a torque onto the liquid layers which
cause a rotation. On each side of the cylinder a vortex is created. The two vortices have an opposite direction of rotation
Chapter 8
Figure 8.30 Generation of vortices by instabilities at boundaries between liquid layers with different velocities
224
8 Liquids and Gases in Motion; Fluid Dynamics
Chapter 8
Figure 8.31 a Laminar flow for small velocities around a circular cylinder. b Generation of vortices behind a circular cylinder for large velocities. c Pressure and
velocity behaviour for u < uc and u > uc
Figure 8.32 Illustration of torque necessary for the generation of vortices
Figure 8.34 Generation of smoke vortex by beating a membrane at the backside of a box filled with smoke
(Fig. 8.31b), i.e. the vortex vector ˝1 points into the direction
into the drawing plane while ˝2 points out of this plane.
These vortices can be visualized by dyed streamlines produced
with the apparatus shown in Fig. 8.4.
Vortices can be also produced at the end of a circular tube,
through which a liquid flows with sufficently high velocity
(Fig. 8.33).
A nice demonstration experiment where vortices are produced
in air mixed with cigarette smoke, is shown in Fig. 8.34. A
box with a thin membrane at one side and a hole with 20–
30 cm diameter on the opposite side is filled with cigarette
smoke. Beating the membrane with a flat hand, produces a sudden pressure increase inside the box and drives the air-smoke
Figure 8.33 Generation of vortices at the end edge of a tube
mixture through the hole out of the box. At the edges of the hole
vortices are produced which travel through the open air and can
be readily seen by a large auditorium. These vortices can extinguish a candle flame, several meters away from the box. The
vortices in air move nearly like a solid body through the air at
atmospheric pressure. Without vortices a pure pressure wave
would not be able to extinguish the candle flame because its intensity decreases with the distance d from the box as 1=d 2 (see
Sect. 11.9).
8.6.4
Turbulent Flows; Flow Resistance
The curls shown in Fig. 8.31 behind an obstacle, do not stay
at the location of their generation but move with the streaming
liquid due to internal friction. At the original location new vortices can now emerge, which again detach from the surface of
the immersed body and follow the liquid flow. This leads to
the formation of a “Karman vortex street” (Fig. 8.35). It turns
out that the two vortices of a vortex pair do not detach simultaneously but alternatively from the upper and the lower side of
the obstacle. In the vortex street therefore the vortices with opposite angular momentum are shifted against each other. Car
8.6 Navier–Stokes Equation
225
Figure 8.35 Karman vortex street
Chapter 8
drivers can experience such a vortex street, when driving behind
a fast truck, where they can feel the alternating directions of the
transverse gust of winds. Behind a starting jet plane the vortex
street can extend over several kilometres. Therefore there must
be always a minimum safety distance between starting planes.
The rotational energy Erot D .I=2/ ˝ 2 (I D inertial moment),
necessary for the generation of vortices has to come from the kinetic energy of the liquid flow. The flow velocity must therefore
decrease when vortices are formed.
In a laminar friction-free flow the flow velocity u in the point
S2 in Fig. 8.31 is zero and in S2 the same stagnation pressure
p0 appears as in S1 . In a turbulent flow the velocity behind the
obstacle is not zero and therefore, according to the Bernoulli
theorem the pressure is lower than p0 , causing a pressure difference between the regions before and behind the obstacle. This
results in a force F D p A on the obstacle with the cross section A in the direction of the flow. In order to keep the body at
a fixed place, an opposite force has to be applied in addition to
the force against the friction force.
Figure 8.36 Flow resistance coefficents cw for different shapes of obstacles
The pressure difference at S2 is according to Bernoulli’s theorem p / .1=2/% u2 . Therefore the force due to the pressure
difference can be written as
FD D cD
% 2
u A;
2
(8.41a)
where the dimensionless constant cD is the pressure drag coefficient. It depends on the form of the body (Fig. 8.36). This
force adds to the friction force that is also present for laminar
flows. According to the Hagen–Poiseuille law (8.31) the friction causes a pressure loss pf (see Fig. 8.9b). The Bernoulli
equation for a viscose liquid flowing through a horizontal tube,
has to be augmented to
p1 C 21 %u21 D p2 C pf C 12 %u22 I
pf < 0 :
The pressure difference pf depends on the square of the velocity u. We can write the total resistance force
Ftotal D Ff C FD D 21 cw % u2 A :
(8.41b)
The proportional factor cw is called flow resistance coefficient.
It depends analogue to cD on the form of the body in the flow. In
Fig. 8.36 the values of cw for Air flows at atmospheric pressure
are compiled for some profiles. This figure illustrates that the
streamlined profile has the smallest flow resistance coefficient.
Bodies with edges on the side of the incoming flow have larger
flow resistance coefficients than spherical profiles.
Figure 8.37 Experimental arrangement for the measurement of flow resistance
By means of the stagnation pressure ps D .1=2/% u2 Eq. 8.41b
can be written as
(8.41c)
Fw D cw ps A :
Experimental values of cw can be measured with the arrangement shown in Fig. 8.37. The body to be measured is suspended
by a bar that can turn around a horizontal axis. A fan blows air
against the body. Due to its flow resistance the body is pressed
to the right, thus expanding a spring balance on the other side of
the bar. The torque exerted by the flow resistance of the body
acting on the lever arm with length a is Fw a where
Fw D
1
2
cw % u2 A ;
while the opposite torque of the spring balance is Fs b. The force
Fs measured with the spring balance is a measure of the flow
resistance Fw and allows the determination of the coefficient cw .
226
8 Liquids and Gases in Motion; Fluid Dynamics
8.7
Aerodynamics
The knowledge of all forces that are present when air streams
around bodies with different shapes is very important not only
for aviation but also for the utilization of wind energy and the
optimization of car profiles. In this section only one aspect will
be discussed, namely the aerodynamic buoyancy (lift) and its
relation to the flow resistance of different body profiles. For a
more extensive treatment, the reader is referred to the special
literature [8.11a, 8.11b]
Chapter 8
8.7.1
Flight direction
Figure 8.39 Demonstration of Magnus effect in air
The Aerodynamical Buoyancy
In addition to the force on bodies in streaming media that acts in
the direction of the flow also a force perpendicular to the streamlines can occur. We will illustrate this by two examples: In
Fig. 8.38a we consider a laminar stream that flows around a circular cylinder. Because of symmetry reasons there could be no
net force perpendicular to the current and only a force in the direction of the stream can occur which is caused by the friction
between the flowing medium and the surface of the cylinder.
However, if the cylinder rotates clockwise the relative velocity
between surface and flowing medium is smaller at the upper side
than at the lower side. This leads to a different friction on the
two sides causing a net force upwards. This can be seen as follows: Due to friction a layer of the flowing medium close to
the surface is dragged into the direction of the rotation causing
a circulation of the layers close to the surface, which is partly
transferred to adjacent layers (Fig. 8.38b).
Figure 8.40 Generation of dynamical lift of a wing profile. a Without circulation; b With sole circulation; c Superposition of a and b
The superposition of this circulation and the laminar flow leads
to an increase of the flow velocity on the upper side and a decrease on the lower side, resulting in the streamlines shown in
Fig. 8.38c. The Bernoulli equation (8.17) tells us that this difference of the velocities results in a net upwards force F D
F1 F2 with jF1 j > jF2 j. This effect was first discovered by
Magnus and was used for the propulsion of ships. The Magnus
Effect can be demonstrated in Physics lectures with a hollow
cylinder of cardboard that can be brought into fast rotation by a
thin ribbon around the cylinder, which is fast pulled (Fig. 8.39).
The cylinder moves then against the pulling direction and rises
upwards because of the Magnus effect until its rotation is slowed
down due to friction and then slowly sinks down.
For bodies with asymmetric profiles in a flowing medium a perpendicular net force occurs even without rotation of the body
(dynamical buoyancy). It is again explained by the superposition of a circulation and the laminar flow. In this case, however,
the circulation is not caused by rotation but by the formation of
vortices. We will discuss this for the example of a wing profile
(Fig. 8.40).
Figure 8.38 Magnus effect: a laminar flow around a circular cylinder, b circulation around a rotating cylinder in a liquid at rest, c streamlines around a
rotation cylinder in an airflow as a superposition of a and c
For a laminar flow around the asymmetric wing profile the layers
of the flow medium close to the surface of the wing are decelerated due to friction. Because the path along the surface is
Figure 8.41 Distribution of lift force along lower and upper surface of a wing
profile
longer at the upper side than at the lower side, the streaming
medium arrives at the point P1 at the upper side with lower velocity than at P2 at the lower surface. The stagnation point S2 at
the backside is at the upper side behind P1 . Behind the profile
a large velocity gradient grad u occurs between neighbouring
layers of the streaming medium. If this gradient surpasses a limiting value, which depends on the velocity u and the viscosity
of the medium, a vortex develops behind the wing profile.
This can be demonstrated, when the profile is moved with
increasing velocity through air or a liquid at rest. Above a critical velocity uc the generation of a vortex is observed (starting
vortex). Since the total angular momentum of the streaming
medium must be conserved, the angular momentum of this vortex has to be compensated by a circulation around the total
profile with opposite direction of rotation. (Fig. 8.40b). The superposition with the laminar flow leads, analogous to Fig. 8.38c,
to an increase of the velocity above the wing profile and a decrease below the wing (Fig. 8.40c). According to the Bernoulli
equation (8.17) this generates an upward force with the amount
FL D p A D cL
% 2
u1
2
which is called the aerodynamical lift.
u22 A ;
Figure 8.42 Arrangement for simultaneous measurements of flow resistance
FD and lift force FL
Figure 8.42 shows a device (two-component balance) that allows the simultaneous measurement of the resistance force FD
and the lift force FL for different model profiles.
It turns out that both forces (lift and drag) depend on the angle
˛ of the profile relative to the laminar flow (Fig. 8.43). Even a
flat plank shows for a certain range of ˛ a lift force, which is,
however, smaller than for a wing profile. The two curves cD .˛/
and cL .˛/ can be plotted in a polar diagram (Fig. 8.44) (polar
profile) which illustrates the relation between cD and cL for all
possible angles of attack. The optimum angel ˛ is chosen such
that the flow resistance is as small as possible, but the lift force
is still high enough. If ˛ is too large, vortices are generated at
the upper side of the wing profile which decrease the flow velocity drastically and therefore reduce the force which can even
become negative.
(8.42)
The lift coefficient cL depends on the shape of the profile. With
pressure probes, the pressure distribution along the wing profile
can be measured. Figure 8.41 shows a typical pressure distribution (difference p to the pressure in the surrounding air) along
the upper and lower side of a wing profile, where the length of
the arrows indicates the magnitude of p [8.12].
8.7.2
Relation between Dynamical and Flow
Resistance
The Eq. 8.41 and 8.42 show that the flow resistance FD and the
FL are both proportional to the kinetic energy per unit volume
of the medium streaming around the profile, where the proportionality constants cD and cL both depend on the shape of the
profile and the smoothness of its surface.
Figure 8.43 Dependence of flow resistance coefficient cD and lift coefficient
cL on the angle of attack ˛ of a wing profile
227
Chapter 8
8.7 Aerodynamics
228
8 Liquids and Gases in Motion; Fluid Dynamics
Figure 8.46 Forces at the ascent of a motor plane
Chapter 8
Figure 8.44 Polar diagram of a modern wing profile with small flow resistance
coefficient
8.7.3
Forces on a flying Plane
At first we will discuss the flight without motor (glider). For
a stationary flight of a glider with constant velocity v the total
force on the glider (including gravity) must be zero. The total
force is the vector sum of the lift force FL , the flow resistance
force FD (Fig. 8.45) which depend on the flow velocity u D v
of the air streaming around the glider, and the gravity force m g.
A stable flight is only possible, if the glider flies on a declining
path with the glide angle . From the condition F D 0 we obtain
with FD D jFD j and FL D jFL j
tan
D
FD
FL
and
sin
D
FD
:
mg
(8.43a)
The ratio FD =FL is called glide ratio. In order to realize a small
glide angle, the force FL should be as large as possible. As can
be seen from Fig. 8.44 there is a lower limit for the glide ratio.
Modern gliders reach glide ratios of 1=50. This implies that
a glider can reach 100 km flight distance without thermal lift,
when it starts from a height of 2 km. If the glide angle is made
larger by operating the elevation unit, the velocity v of the glider
becomes larger, when is made smaller, the velocity decreases
until the uplift breaks down and the glider becomes unstable.
Without an experienced pilot, this might lead to a crash down.
Figure 8.45 Forces at the gliding flight
When the air locally heats up (for instance above a hot ground
or above chimneys of power stations) the air expands, its density
decreases and it rises upwards (thermal lift, see Sect. 6.3). This
gives an additional vertical component to the flow velocity of
the air relative to the glider. In this case the glide angle can
become negative, i.e. the glider rises upwards.
For planes with an engine (Fig. 8.46) an additional pulling force
is produced by the propeller (or a corresponding propulsive
force for jet planes). A climb is only possible, if the pulling
force FZ is larger than the magnitude of the opposite drag FD .
For the flight with constant velocity v at constant height the
pulling force must just compensate the drag .FZ D FD /. The
angle of climb, is given by the ratio
tan
D
FD
FZ
FL
:
(8.43b)
For FZ < FD the angle becomes negative and the plane can
fly with constant velocity only on a continuous descent.
8.8
Similarity Laws;
Reynolds’ Number
In Sect. 8.6.3 we have seen that vortices are caused by friction
in the layers between liquid flow and walls. Although friction
inside the liquid flow is small compared to that at the walls, it
essentially influences the behaviour of the liquid flow, because
this internal friction acts on the surface layers and starts turbulent flow.
Such boundary conditions are not included in the Navier–Stokes
equation, because this equation describes the motion of an infinitesimal volume element and its motion under the influence
of the different forces. It does not contain the special geometry
of the flow pipe. Its geometry, however, plays an important role
for the characteristics of the flow. It can be inserted into (8.36a)
as special boundary conditions, but a reliable solution demands
the knowledge of all details of such boundary conditions, which
is often missing. Therefore generally experimental solutions are
preferred which are obtained in the following way:
In hydro-and aero-dynamics the flow conditions for the motion
of large objects (ships, airplanes) is studied with small models
that have a similar but scaled down geometry. With such model
8.9 Usage of Wind Energy
We normalize all length dimensions by a unit length L, all times
by a unit time T and all velocities by L=T. We therefore define
new values of length, time and velocity:
l0 D l=L I
t0 D t=T I
u0 D u=.L=T/ D u T=L:
(8.44a)
This gives for the gradient r 0 and the pressure p0
r0 D r L I
p0 D p .T=L/2 =% ;
(8.44b)
where r 0 D L .@=@x; @=@y; @=@z/, L0 , t0 , u0 and p0 are dimensionless quantities.
With these normalized quantities the Navier–Stokes equation
becomes
@u0
1 0 0
C .u0 r 0 / u0 D r 0 p0 C
(8.45)
u ;
@t0
Re
with the dimensionless Reynolds’-Number
Re D
% L2
%UL
D
;
T
with
UD
L
:
T
(8.46)
The quantity U D L=T is the flow velocity averaged over the
length L. For ideal liquids is D 0 ! Re D 1. Here the
following statement can be made:
Flows of ideal liquids in geometrical similar containers for
which (8.44) is valid, are described by the same equation (8.45)
with the same boundary conditions. This means: At corresponding positions r0 and times t0 one obtains the same dimensionless
quantities p0 and u0 in (8.45). Even non-stationary flows have the
same progression within time intervals that are proportional to
the container dimensions and inversely proportional to the flow
velocity u.
For viscous liquids with ¤ 0 this is only valid if the Reynolds
number Re has the same value. Flows of viscous liquids are
only similar if the Reynolds number Re has the same value and
the flow proceeds in containers with similar geometrical dimensions.
We will illustrate the physical meaning of the Reynolds number Re. When we multiply numerator and denuminator in the
fraction (8.46) by L2 U we obtain
% L3 U 2
2Ekin
:
Re D
D
2
L U
Wf
(8.47)
The flow is laminar. Turbulent flows occur above a critical
Reynolds’ number Rec .
Experimental findings give for water flows in circular pipes with
diameter d the critical Reynolds’ Number
Rec D % d Uc = D 2300 :
For prevention of turbulent flows the normalized flow velocity
must always obey the condition U < Uc ! Re < Rec . If Re is
only slightly smaller than Rec vortices are formed, which, however, have diameters that are smaller than the flow pipe diameter.
Their rotational energy is small compared to the kinetic energy
of the laminar flow and they therefore do not impede the flow
very much. Only for Re Rec their rotational energy becomes
comparable to the friction energy and macroscopic vortices are
generated. The flow becomes completely turbulent.
8.9
Usage of Wind Energy
The kinetic energy of streaming air can be utilized for the generation of electric power by wind energy converters. This had been
already realized for many centuries by wind mills for grinding
grain or for pumping water.
Modern wind converters generally have three rotor blades
(coloured pictures 3 and 4). According to new insight in the
flow conditions of air around the rotor blades, the shape of
these rotors is formed in a complicated way in order to optimize
the conversion efficiency of wind energy into mechanic rotation energy of the rotor, which is then further converted through
transmission gears and electric generators into electric power.
Most of the wind energy converters produce alternating current,
which is then rectified and again converted by dc-ac converters
into alternating current. This is necessary in order to synchronize the phase of the ac-current with that of the countrywide
network.
The kinetic energy of a volume element dV of the airflow moving with the velocity v is
Ekin D 21 mv 2 D 12 %v 2 dV :
(8.48)
The air volume impinging per second onto the vertical area A is
dV D v A. The maximum power (energy per second) of the air
flow hitting the area A is then
PW D 21 % v 3 A :
(8.49)
The numerator gives twice the kinetic energy of a volume
element L3 , which moves with the velocity U, while the denominator is the friction energy Wf , which is dissipated when the
volume element L3 moves with the velocity U over a distance L.
In reality, only a fraction of the power can be converted into
rotational energy of the wind converter. Firstly the wind is not
completely decelerated to v D 0, because for v D 0 a tailback
of air would build up behind the wind converter which would
impede the air flow to the converter. Secondly, friction losses
diminish the kinetic flow energy and rise the temperature.
For small Reynolds’ numbers Ekin Wf , which implies that the
accelerating forces are small compared to the friction forces.
If the velocity of the air inflow is v1 , it is decelerated to v < v1
because of the stagnation at the rotor blades, where the pressure
Chapter 8
experiments in wind channels or in small liquid flow chambers,
the optimum shape of a wing or a hulk can be found. In order to
obtain realistic results, not only the shape of the model must be
a true scaled down version of the true object, but also the flow
conditions must be accordingly scaled down in a correct way.
How this can be achieved, will be shortly outlined:
229
230
8 Liquids and Gases in Motion; Fluid Dynamics
The power, transferred to the wind converter is then
Pw D F v D v12
A
v22 %v D a PW :
2
(8.53)
Inserting Pw from (8.49) gives for the conversion factor a the
value a D .v1 C v2 / .v12 v22 /=2v13 < 1. With a given initial
velocity v1 the maximum transferred power Pw .v2 / is reached
for d.Pw /=dv2 D 0. This gives with v D .v1 C v2 /=2 the
condition
Chapter 8
2v2 .v1 C v2 /% A=4 C .v12
v22 /% A=4 D 0 ;
which yields v2 D 13 v1 .
For the efficiency factor a one obtains a D 0:59. This means that
without any other losses at most 59% of the initial wind energy
can be converted into rotational energy of the wind converter!
Example
Figure 8.47 Schematic illustration of velocity and pressure conditions for a
rotor blade at rest in an air flow [8.13]
increases from the initial value p0 to p1 > p0 (Fig. 8.47). At the
backside of the rotor the pressure sinks to p2 < p0 . Behind the
rotor the pressure increases again to p0 and the airflow velocity
is down to v2 < v. Only after a larger distance behind the rotor
the wind velocity rises again to its initial value v D v1 .
According to the Bernoulli equation is
p1 p2 D % v12 v22 =2 :
The force acting on the rotor blades with area A is
F D .p1 p2 /A D % v12 v22 A=2 :
(8.50)
(8.51)
On the other hand this force can be written as
F D .v1
v2 /d=dt.mv/ D .v1
v2 /%vA :
(8.52)
The comparison between (8.51) and (8.52) shows that v D .v1 C
v2 /=2.
Figure 8.48 Offshore Windpark in the North Sea
v1 D 10 m=s, v2 D 4 m=s ! v D 7 m=s and a D 0:588.
A typical wind converter has rotor blades with L D 50 m
length and deliver several Megawatt electric power. At
the rotational frequency f D 1=s the velocity of the rotor
ends is already 300 m=s D 1080 km=h, which is close to
J
the limit of tensile strength of the blade material.
Note, that the power transferred to the wind converter is proportional to the third power of the initial wind velocity. This
means that already small changes of the wind velocity will
cause large changes of the power available from wind converters. Modern wind converters can operate at wind velocities
between 4 m=s and 25 m=s. For smaller velocities the transferred power is too small for a profitable operation. For higher
velocities v > 25 m=s the converters are shut down because of
possible destruction.
The efficiency of the energy conversion is reduced by several
losses. Firstly there are friction losses between different air layers with different velocities. They correspond to the friction
losses in the Navier–Stokes equation. Furthermore there are
mechanical losses of the rotating blades and the transmission
gear. Finally the losses in the electric generator have to be considered.
231
Chapter 8
8.9 Usage of Wind Energy
Figure 8.49 Windfarm Krummhörn. Rotor span width is 30 m, the nominal electric power output is 300 kW per wind converter. (With kind permission of EWE
corporation, Oldenbourg)
The available electric power is then
Pelectric D a air mech electric Pw :
For our considerations about the wind velocities before and behind the wind converter we have assumed that the rotor blades
are at rest. Because of their rotation the relative velocity between initial wind velocity and rotor velocity is smaller and the
wind does no longer impinge perpendicular to the blades. This
gives not only a smaller effective area Aeff < A but also a smaller
value of the transferred energy.
Total installed electric power per year [GW]
The power delivered by a wind converter, averaged over one
year, is only between 10% and 40% of the installed power
depending on the wind conditions at the converter location.
The highest efficiency is reached for offshore wind converters
(Fig. 8.48), because here the wind blows continuously and has
generally a higher velocity than above undulating solid ground.
For wind converters on solid ground the height should be as
large as technically possible, because the wind velocity at 100 m
altitude is much higher than directly above ground (Fig. 8.49).
In Tab. 8.3 the total installed electric power of wind converters is listed for the countries with the highest usage of wind
energy and in Fig. 8.50 the impressive increase of worldwide
annually new installed electric power from wind converters is
illustrated.
Now we will discuss the energy conversion of wind converters
in more detail: The forces driving the rotor blades can be composed of the flow resistance force and the Bernoulli-force. Their
ratio depends on the shape of the blades and on their angle of attack ˛. This is similar to the situation for air flowing around a
wing profile of an air plane (see Sect. 8.7.2 and Figs. 8.41 and
8.46). The pressure dependence p.x/ along a wing profile at rest
is shown for the upper and lower side of the profile in Fig. 8.51.
The pressure difference generates a lift force and a torque about
an axis in x-direction. Depending on the orientation of the pro-
30
25
20
15
10
5
0
1990 91 92 93 94 95 96 97 98 99 2000 01 02 03 04 05 06 07 08 09 10 11 12 13
Figure 8.50 Growth of worldwide annually installed eletric power of wind converters in GW
232
8 Liquids and Gases in Motion; Fluid Dynamics
Table 8.3 Installed electric power of wind converters for different countries
(2014)
Chapter 8
Ranking
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Country
China
USA
Germany
Spain
India
UK
Canada
France
Italy
Brazil
Sweden
Portugal
Denmark
Poland
Australia
Turkey
Rumania
Netherland
Japan
Mexico
Worldwide
Europe
Power in GW
114.763
65.879
39.165
22.987
22.465
12.440
9.694
9.285
8.663
5.939
5.425
4.914
4.845
3.834
3.806
3.763
2.954
2.805
2.789
2.381
369.553
133.969
Figure 8.51 Pressure variation along the lower and upper surface of a rotor
profile at rest, with air flowing around the profile. The rotation axis lies above
the drawing plane [8.14]
file against the direction of wind flow, the lift force as well as the
flow resistance force can be used for driving the rotor blades.
When the rotor blade rotates with the angular velocity !, the
velocity vB .r/ of the section of the blade at a distance r from the
rotation axis adds to the wind velocity v to an effective velocity
veff D v C vB .r/ (Fig. 8.52). The angle of attack ˛ against
the direction of veff must be chosen in such a way (Fig. 8.43),
that the optimum force can be used. Since veff changes with r
the profile of the blade must change with r. With increasing r
the blade must become slimmer and the direction of the profile
changes. The whole blade is therefore twisted (Fig. 8.53) in
order to reach for all sections of the rotating blade the optimum
usage of the lift force.
Figure 8.52 Velocities and forces on the rotating rotor. The rotation axis points
into the direction of v and is above the drawing plane [8.13]
Figure 8.53 Rotor blade of a fast rotating wind converter. The red areas represent the rotor profile at different distances r from the rotation axis. In this
drawing they are turned by 90ı into the drawing plane. Also the wind direction
is turned. The wind comes really from above the drawing plane
Summary
233
Summary
F D % V
du
;
dt
where u is the flow velocity of the volume element V with
the mass density %.
In a stationary flow u.r/ is at every position r constant in time
but can vary for different positions ri .
Frictionless liquids .Ff D 0/ are called ideal liquids. For
them the Euler equation
@u
C .u r/u D g
@t
1
grad p
%
describes the motion of the liquid.
The continuity equation
@%
C div.% u/ D 0
@t
describes the conservation of mass for a flowing medium.
For incompressible media .% D const/ the continuity equation reduces to div u D 0.
For frictionless incompressible flowing media the Bernoulliequation
p C 12 % u2 D const
represents the energy conservation Ep C Ekin D E D const.
The pressure decreases with increasing flow velocity u.
The Bernoulli equation is the basic equation for the explanation of the dynamical buoyancy and therefore also for
aviation.
For flow velocities u below a critical value uc laminar flows
are observed, while for u > uc turbulent flows occur. This
critical value uc is determined by the Reynolds number Re D
2Ekin =Wf which gives the ratio of kinetic energy to the friction energy of a volume element V D L3 when V is
shifted by L.
For laminar flows where the inertial forces are small compared to the friction forces no turbulence occurs and the
stream lines are not swirled.
For a laminar flow through a tube with circular cross section
R2 the volumetric flow rate
QD
R4
grad p
8
flowing per second through the tube is proportional to R4
grad p but inversely proportional to the viscosity .
A ball with radius r moving with the velocity u through a
medium with viscosity experiences a friction force
Ff D 6r u ;
that is proportional to its velocity u.
The complete description of a flowing medium is provided
by the Navier–Stokes equation (8.36a) which reduces for
ideal liquids . D 0/ to the Euler equation. The Navier–
Stokes equation describes also turbulent flows, but for the
general case no analytical solutions exist and the equation
can be solved only numerically.
For the generation and the decay of vortices friction is necessary. Vortices are generally generated at boundaries (walls
and solid obstacles in the liquid flow).
The flow resistance of a body in a streaming medium is described by the resisting force FD D cD % 12 u2 A. It depends
on the cross section A of the body and its drag coefficient cD
which is determined by the geometrical shape of the body.
The force is proportional to the kinetic energy per volume
element V of the streaming medium. In laminar flows, FD
is much smaller than in turbulent flows.
The aero-dynamical buoyancy is caused by the difference of
the flow velocities above and below the body. This difference is influenced by the geometrical shape of the body and
can be explained by the superposition of a laminar flow and
turbulent effects (circulation).
Chapter 8
The motion of particles of a flowing medium (liquids or
gases) is determined by the total force F D Fg C Fp C Ff
which is the vector sum of gravity force, pressure force and
friction force. The equation of motion is
234
8 Liquids and Gases in Motion; Fluid Dynamics
Problems
Chapter 8
8.1
Estimate the force that a horizontal wind with a velocity of 100 km=h (density of air D 1:225 kg=m3 ) exerts (% D
1:225 kg=m3; cD D 1:2)
a) on a vertical square wall of 100 m2 area
b) on a saddle roof with 100 m2 area and length L D 6 m and a
cross section that forms an isosceles triangle with ˛ D 150ı.
8.2
Why can an airplane fly “on the head” during flight
shows, although it should experience according to Fig. 8.41 a
negative buoyancy?
through a pipe with length L and radius r R at the height
z D 0?
8.6
A pressure gauge as shown in Fig. 8.10c is placed into
flowing water. The water in the stand pipe rises by 15 cm. The
measurement with the device of Fig. 8.10a shows a pressure of
p D 10 mbar. How large is the flow velocity?
8.7
A funnel with the opening angle ˛ D 60ı is filled with
water up to the height H. The water can flow into a storage vessel with volume V through a horizontal pipe at the bottom of the
8.3
Why do the streamlines not intermix in a laminar flow funnel with length L and inner diameter d.
although the molecules could penetrate a mean free path into a) What is the height H.t/ in the funnel as a function of time?
b) What is the total flow mass M.t/?
the adjacent layers?
c) After which time is the funnel empty for H D 30 cm,
Hint: Estimate the magnitude of in a liquid.
d D 0:5 cm, L D 20 cm, and D 1:002 mPa s?
8.4
Prove the relation (8.36b) using the component represen- d) After which time is the storage vessel with a volume V D
4 litre full, if the water in the funnel is always kept at the
tation.
height H by supplying continuously water?
8.5
A cylinder is filled with a liquid up to the height H. The
8.8
A water reservoir has at h below the water surface a
liquid can flow out through a pipe at height h (Fig. 8.54)
drain pipe with inner diameter d D 0:5 cm and length L D 1 m
which is inclined by the angel ˛ below the horizontal.
a) How much water flows per second through the pipe for a
laminar flow with D 10 3 Pa s and h D 0:1 m?
b) Above which angle ˛ the flow becomes turbulent if the critical Reynolds number is 2300?
8.9
What is the minimum diameter of a horizontal tube with
L D 100 m to allow a laminar flow of water of 1 l s 1 from a
vessel with a water level 20 m above the horizontal tube?
8.10 What is the vertical path z.t/ of a ball with radius r
falling through glycerine . D 1480 mPa s/ if it immerses at
t D 0 and z D 0 into the glycerine with the initial velocity
a) Calculate for an ideal liquid (no friction) the position x.H/ v0 D 2 m=s
where the outflowing liquid hits the ground and the velocity a) for r D 2 mm,
vx .H/ and vz .H/ for z D 0. Compare this result with the b) for r D 10 mm?
velocity of a free falling body starting from z D H.
b) What is the function z.t/ of the liquid surface in the cylin- 8.11 Derive the Helmholtz equation (8.39) starting from
der with radius R for a liquid with the viscosity streaming (8.36a).
Figure 8.54 To Probl. 8.5
References
235
8.1a. G. Birkhoff, Hydrodynamics. (Princeton Univ. Press,
2015)
8.1b. E. Guyon, J.P. Hulin, Physical Hydrodynamics. (Oxford
Univ. Press, 2012)
8.2. J.E.A. John, Th.G. Keith, Gasdynamics. (Prentice Hall,
2006)
8.3a. R.K. Bansal, A Textbook on Fluid Mechanics. (Firewall
Media, 2005)
8.3b. G.K. Batchelor, Introduction to Fluid Mechanics. (Cambridge Univ. Press, 2000)
8.4a. J.P. Freidberg, Ideal Magnetohydrodynamics. (Plenum
Press, 1987)
8.4b. P.A. Davidson, An Introduction to Magnetohydrodynamics. (Cambridge University Press, Cambridge, England,
2001)
8.5. F. Pobell, Matter and Methods at Low Termperatures.
(Springer, Berlin, Heidelberg, 1992)
8.6. D. Even, N. Shurter, E. Gundersen, Applied Physics. (Pentice Hall, 2010)
8.7. A.E. Dunstan, The Viscosity of Liquids. (Hard Press Publ.,
2013)
8.8. L. Brand, Vector Analysis. (Dover Publications, 2006)
8.9. K. Weltner, S. John, Mathematics for Physicists and Engineers. (Springer Berlin, Heidelberg, 2014)
8.10. G.B. Arfken, H.J. Weber, Mathematical Methods for
Physicists. A Comprehensive Guide. (Elsevier Ltd., Oxford, 2012)
8.11a. L.J. Clancy, Aerodynamics. (Pitman, London, 1978)
8.11b. J.D. Anderson, Fundamentals of Aerodynamics. (McGrawHill, 2012)
8.12. D. Pigott, Understanding Gliding. (A&C Black, London)
8.13. M. Diesendorf, GreenhouseSolutions with sustainable Energy. (University of New South Wales, 2007)
8.14. J.F. Manwell, J.G. McGowan, A.L. Rogers, Wind Energy Explained: Theory, Design and Applications. (Wiley,
2010)
8.15. R. Ferry, E. Monoian, A Field Guide to Renewable Energy
Technologies. (Society for Cultural Exchange, 2012)
8.16a. T. Burton et al., Wind Energy Handbook. (Wiley, 2001)
8.16b. N. Walker, Generating Wind Power. (Crabtree Publ.,
2007)
8.17. The European Wind Energy Association, Wind Energy:
The Facts. A Guide to the Technology, Economics and Future of Wind Power. (Routledge, 2009)
Chapter 8
References
Vacuum Physics
9.1
Fundamentals and Basic Concepts . . . . . . . . . . . . . . . . . . . . . 238
9.2
Generation of Vacuum . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241
9.3
Measurement of Low Pressures . . . . . . . . . . . . . . . . . . . . . . . 247
9
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251
Chapter 9
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252
© Springer International Publishing Switzerland 2017
W. Demtröder, Mechanics and Thermodynamics, Undergraduate Lecture Notes in Physics, DOI 10.1007/978-3-319-27877-3_9
237
238
9 Vacuum Physics
The importance of vacuum physics for the development of modern physics and technology can be hardly overestimated. Only
after the realization of a sufficiently low vacuum, many experiments in atomic, molecular and nuclear physics became
possible. These experiments have essentially contributed to the
understanding of the micro-structure of matter, of electrons and
nuclei as the building blocks and of the internal structure of
atoms and nuclei. Based on the results of these experiments
the quantum theory of matter could be successfully developed
(see Vol. 3).
Without vacuum technology, the manufacturing of semiconductor elements and integrated circuits would have been impossible
and therefore we would be still without computers.
Besides for basic research vacuum technology is nowadays used
as indispensable tool in many technical applications, which
reach from vacuum melting of special metal alloys over the production of thin optical films to the dry freezing of food. It is
therefore essential for every physics student to study at least
some basic facts of vacuum physics and technology.
Chapter 9
In this chapter we will discuss, after a summary of the most
important fundamentals, some techniques for the generation of
vacuum and the measurement of low pressures. More detailed
presentations can be found in [9.1, 9.2, 9.3].
9.1
Fundamentals and Basic
Concepts
Vacuum is produced in a container, when most of the gases
or vapours have been removed and the pressure p in the volume V becomes small compared to the atmospheric pressure
p0 1 bar. Devices, that can achieve such a reduction of the
pressure, are called vacuum pumps, because they pump part
of the gases or vapours in the volume V into other containers
or into the open air (Fig. 9.1). The achieved pressure, given in
the unit Pascal .1 Pa D 1 N=m2 D 10 2 hPa/ or often quoted
in millibar (1 mbar D 1 hPa D 100 Pa) (see Tab. 7.1) depends
essentially on the type of vacuum pumps used for the evacuation. At low pressures .p < 10 4 hPa/ the walls of the vacuum
container and the gas molecules attached to the walls play an
important role for further evacuation because their outgasing essentially influence the achievable vacuum pressure.
9.1.1
The Different Vacuum Ranges
We distinguish four different vacuum ranges, depending on the
lower pressure limit of the achievable vacuum.
Low vacuum
1 hPa < p < 1000 hPa D 1 bar
Medium vacuum
10 3 hPa < p < 1 hPa
Figure 9.1 Schematic drawing of a vacuum apparatus
High vacuum
10 7 hPa < p < 10
Ultrahigh Vacuum
p < 10 7 hPa.
3
hPa
The best vacuum, achievable today is about 10
13
hPa.
In order to give an impression how empty an evacuated container really is, Tab. 9.1 compiles the number of gas molecules
per m3 for different pressures. It is illustrative to compare these
numbers with the number nw of molecules sitting per m2 in
a monomolecular layer on the surface of the container walls.
With a mean distance of 0:3 nm between the molecules of a
monomolecular layer we get nw D 1019 =m2 . A cubic vacuum container with V D 1 m3 has a wall surface of 6 m2 . At
a pressure of 2 10 5 hPa the number of molecules on the wall
therefore equals the number of all molecules in the volume of
the container. For pressures below 10 5 hPa the number of
molecules on the wall therefore exceeds the number in the volume and in order to reach a much lower vacuum the walls have
to be outgassed by heating.
Generally there are different gases (N2 , O2 , He, Ar with pressures pi ) and vapours (water, oil or other liquids with saturation
Table 9.1 Particle density n of air molecules, mean free path and particle
flux density onto the container surface for different pressures at room temperature
p=hPa
103
1
10 3
10 6
10 9
n=m 3
2:5 1025
2:5 1022
2:5 1019
2:5 1016
2:5 1013
=m
6 10 8
6 10 5
6 10 2
60
6 104
=m 2 s
3 1027
3 1024
3 1021
3 1018
3 1015
1
9.1
Fundamentals and Basic Concepts
239
pressures psi ) in the vacuum container. The total pressure
X
(9.1)
pD
.pi C psi /
is then the sum of all partial pressures. The saturation pressure
which adjusts itself at the equilibrium between liquid and vapour
depends on the temperature (see Sect. 10.4.2).
9.1.2
Influence of the Molecules at the Walls
The number of molecules, hitting per sec an area of 1 m2 of the
walls of a vacuum container (particle flux density ˚, last column
of Tab. 9.1) depends on the particle density n in the evacuated
volume V and on the mean thermal velocity v (see Sect. 7.3).
A molecule with the velocity v D fvx ; vy vz g with the distance z
from the surface can reach the surface within the time t z=vz
as long as z is smaller than the mean free path (Fig. 9.2). For a
mean particle density n the number of wall collisions per second
of molecules in the upper half volume is
nAv
ZD
4
Z=2
Z2
sin # cos # d# d' :
0
(9.2a)
0
The first integral gives the value 1=2, the second gives 2. The
particle flux density ˚ D Z=A onto the unit area of the wall
surface is then
˚ D .1=4/n v :
(9.2b)
The numerical values in Tab. 9.1 show, that at a pressure of
p D 3 10 6 hPa and a mean velocity v D 500 m=s nearly
as many molecules hit the surface per second as are contained
in a mono-molecular layer on the surface. If all impinging
molecules would stick at the surface a clean surface would be
covered within 1 s with a monomolecular layer. This illustrates
that a really clean surface can be only realized at very low pressures (ultrahigh vacuum) and if the molecules do not stick on
the surface. This can be achieved, when the surface is heated,
which causes all impinging molecules to leave the surface immediately.
With decreasing temperature the evaporation decreases and the
inner wall of a vacuum chamber is therefore at low temperatures
always covered by a layer of adsorbed molecules. An equilibrium adjusts itself which depends on the temperature of the
surface, on the density n in the chamber and on the molecular
Figure 9.2 Illustration of the collision rate with the wall
species, where the rates of adsorbing and desorbing molecules
become equal.
Our example above has shown that at pressures p 10 3 hPa
the number of molecules adsorbed on the wall becomes larger
than the number in the evacuated volume. When a vacuum chamber is evacuated, the pressure in the chamber below
10 3 hPa will be at first essentially determined by the rate of
molecule desorbing from the wall, until the desorbing rate becomes smaller than the pumping rate that removes the molecules
out of the vacuum chamber.
9.1.3
Pumping Speed and Suction Capacity of
Vacuum Pumps
When a vacuum chamber is evacuated the gas in the chamber
has to pass through an opening and through pipes in order to
reach the vacuum pump. The volume flow rate of the pipe (often given in the unit litre per second D l=s or cubic meter per
hour D m3 =h) is the gas volume that flows per sec through a
cross section of the pipe at a given pressure p and temperature
T.
Note, that the molecular density n D N=V decreases with the
pressure according to
pV
p
p V D NkT ! N D
!nD
:
(9.3)
kT
kT
Therefore even for a constant volume flow rate the number
of molecules passing per second through the cross section decreases with p! This means for equal volume flow rates dV=dt,
the number dN=dt of molecules pumped out of the vacuum
chamber depends on pressure p and temperature T.
The suction capacity
SV D
dV
given in Œl=s or in Œm3 =h
dt
(9.4)
of a vacuum pump is defined as the volume flow rate dV=dt at
the suction intake of the pump.
The total mass flow of molecules with mass m
dM
dV
m
dV
D%
D
p
;
dt
dt
kT
dt
(9.5)
Chapter 9
For planning an experiment in the vacuum chamber the mean
free path of the molecules is of great importance. It determines the collision probability between the molecules in the
chamber (see Sect. 7.3). Table 9.1 shows that in the fine vacuum
range is small compared with the dimensions of commonly
used vacuum chambers. Collisions can be therefore not neglected. On the other hand, in the high vacuum range for
p < 10 6 hPa, becomes large compared with the dimensions
of the chamber and the molecules fly freely through the chamber
without suffering collisions until they hit a wall.
240
9 Vacuum Physics
that is pumped per second out of the vacuum chamber is the
mass suction capacity. It depends on the pressure and the volume flow rate through the chamber opening and the pipes.
strongly depends on the pressure. Here CS D LS is named
the volume flow conductance. The different pressure ranges are
characterized by the Knudsen number
Manufacturers of pumps generally give the suction capacity or
pumping speed of a pump in the unit
SL D p
dV
;
dt
ŒSL D hPa l=s
(9.6)
as the product of pressure p and volume flow rate dV=dt.
Example
With a suction capacity SV D 500 l=s about
1022 molecules per sec are pumped out of a vacuum chamber at room temperature and p D 1 hPa. At the lower
pressure p D 10 6 hPa these are only 1016 molecules per
second at the same value of SV .
Chapter 9
The suction capacity is for the first case SL D
500 hPa l=s (corresponding to 50 Watt) while for the secJ
ond case SL is only 5 10 4 hPa l=s .50 µW/.
9.1.4
Flow Conductance of Vacuum Pipes
The dimensions of vacuum pipes play an important role for the
design of a vacuum apparatus. The mass flow
dM
D Lm .p2
dt
p1 /
dV
D LS .p2
dt
p1 /
(9.7b)
is used with the unit ŒhPa m =h. Because of p V D N kT !
p D .%=m/kT with m D M=N D mass of one molecule the
coefficient of volume flow conductance LS can be related to the
mass flow conductance by
(9.7c)
LS depends on the mass m of the molecules, on the mean free
path (because / 1=p) and on the geometry of the vacuum
pipes. For simple geometries it can be calculated. For complex
geometries it must be determined experimentally. The values
are compiled in special tables [9.1].
The gas flow through pipes
dV
p
D CS
dt
p
(9.8)
which gives the ratio of mean free path and the diameter d
of openings or pipes. According to the magnitude of Kn we
distinguish between three ranges:
Range of laminar gas flow (for Reynolds numbers Re <
2200) or turbulent flow (for Re > 2200) which occurs for
Kn 1. Here is d.
Range of Knudsen flow (also called transition range) where
Kn 1 and d.
Range of free molecular flow where Kn 1 and d.
In the range Kn 1 the gas flow is essentially governed by
collisions between the gas molecules, which means that the viscosity plays an important role. The flow can be described by
hydro-dynamical models (see Chap. 8). Depending on the magnitude of the Reynolds number Re and the viscosity the flow
is laminar for Re < 2200 or turbulent for Re > 2200. Under
the conditions relevant for most vacuum systems the Reynolds
number is generally smaller than 2200 and the flow is therefore
laminar.
In the range Kn 1 collisions between the molecules can be
neglected. The viscosity does no longer influence the gas flow
and collisions with the wall determine the suction capacity. The
flow conductance becomes independent of the pressure.
We will illustrate these conditions by some examples:
Examples
3
kT
LS D
Lm :
m
;
d
(9.7a)
through a vacuum pipe is proportional to the pressure difference
.p2 p1 / between entrance and exit of the pipe. The proportionality factor Lm is the coefficient of mass flow conductance given
in the unit Œ1 m s. Generally the pumping speed
p
Kn D
(9.7d)
1. Volume flow conductance CS of a circular opening
with diameter d in the range of molecular flow .
d/. According to Eq. 9.2 the number of molecules
passing per sec through the hole with area A D d2 =4
is
Z D 14 A n v ;
with p V D N kT and Z D dN=dt we obtain for the
volume gas flow through the hole
dV
1
1
n
D A kT v D A v;
dt
4
p
4
since n D N=V D p=kT.
Since n p the volume flow dV=dt becomes independent of pressure. Inserting numerical values of v for
air at T D 300 K gives dV=dt D 11:6 A in l=s if A is
given in cm2 . A circular opening with d D 10 cm has
therefore at low pressures . d/ the volume flow
conductance CS D 900 l=s.
2. Flow through a pipe with length L and diameter d in
the range of laminar flow . d/. The pressures at
9.2 Generation of Vacuum
241
Desorption rate
the two ends of the pipe are p1 and p2 . According to
the Hagen-Poisseuille-law (8.32)
Leak rate
p2 / ;
(9.9)
we obtain for d D 5 cm, L D 1 m, p1 D 2 hPa, p2 D 0,
air D 0:018 mPa s, the numerical value p dV=dt D
170 Pam2 =s. According to (9.7) the volume flow conductance then becomes CS D 0:85 m3 =s D 850 l=s.
At a lower pressure of 10 1 hPa, where D 0:06 cm
(Tab. 9.1) which is still smaller than d the flow conductance decreases according to (9.7) and (9.9) to
CS D 42 l=s. Equation 9.9 is in this range, however, only approximately valid and the accurate value
is CS D 80 l=s.
In the range of molecular flow . > d/ CS converges with decreasing pressure towards the value
CS D 16 l=s.
J
The reciprocal of the flow conductance
RS D 1=CS
(9.10)
is the flow resistance. Completely analogue to the electrical resistance in electricity the flow resistance of consecutive flow
pipes is the sum of the individual resistances, while for flow
pipes in parallel arrangements the individual flow conductances
add to the total conductance, as can be immediately seen from
(9.7).
9.1.5
Pumping speed
Figure 9.3 The achievable final pressure is determined by the compensation
of pumping speed by leak rate C desorption rate
The final pressure achieved in the vacuum chamber is determined by the pumping speed, the leak rate and the total rate of
desorbing molecules (Fig. 9.3). At the final pressure the pumping speed just equals the sum of leak rate and desorbing gas rate.
SLeff .p/ D
dGL
dGd
C
;
dt
dt
(9.13)
where SLeff is the effective pumping speed at the outlet opening
of the chamber to the pumping pipes. It is equal to the pumping
speed of the pump minus the flow conductance of the vacuum
lines between chamber and pump.
The attainable final pressure pf results then from (9.13) with
(9.6) and (9.12):
pf D
dGd =dt C dGL =dt
;
SV
(9.14)
where SV D dVp =dt is the effective suction capacity at the exit
of the vacuum chamber.
Accessible Final Pressure
Every vacuum chamber has openings that allow access to the
experimental setup in its inside. They are closed by flange seals.
However, there are always leaks which are often difficult to find
and to close. Through these leaks molecules can penetrate from
the outside into the vacuum chamber. We define the gas rate
dGL =dt D p0 dVL =dt (p0 D atmospheric pressure) which penetrates through all leaks into the vacuum chamber as the leak
rate. It is given in the same units hPa l=s as the pumping speed
defined in (9.6).
Example
For a suction capacity SV D 103 l=s, a leak rate of
10 4 hPa l=s and a desorbing gas rate of 10 3 hPa l=s
a final pressure of pf D 1:1 10 6 hPa can be reached.
After heating the walls the desorbing rate sinks below the
leak rate and a final pressure of pf D 10 7 hPa can be
J
achieved.
As has been previously discussed molecules can also be desorbed from the inner walls and delivered into the volume of the
vacuum chamber. This leads without pumping to a pressure increase p.
9.2
For the rate dNd =dt of desorbing molecules we obtain with p
V D N kT the pressure increase per second
In order to remove gas particles out of the vacuum chamber vacuum pumps are used. The different types can be divided into
three classes (Tab. 9.2):
dp
kT dNd
D
:
dt
V dt
(9.11)
The total rate of desorbed gas is
dGd
dp
dNd
DV
D kT
:
dt
dt
dt
(9.12)
Generation of Vacuum
Mechanical pumps,
Diffusion pumps (fluid acceleration vacuum pump),
Cryo pumps and sorption pumps.
We will briefly discuss these three classes. In Fig. 9.4 the pressure ranges are compiled where the different pumps can be used.
Chapter 9
dV
d 4 p1 C p2
p
D
.p1
dt
128 L
2
242
9 Vacuum Physics
Table 9.2 Classification of the most important types of vacuum pumps
Mechanical
pumps
Rotary vane pumps
Roots pumps
Fuel acceleration
pumps
Liquid jet pumps
Vapor jet pumps
Turbopumps
Diffusion pumps
Condensation pumps,
sorption pumps
Cool traps
Kryo pumps,
sorption pumps
ion getterpumps
Rotary vane pump
Roots pump
Steam jet pump
Diffusion pump
Turbo-molecular pump
Kryo-pump
Figure 9.5 The demonstration experiment by Otto von Guericke. Engraving by
Caspar Schott
Mass spectrometer
Friction ball manometer
Chapter 9
Capacitor membrane mano.
McLoid mano.
Ionization manometer
Heat conduction mano.
Penning mano.
Membrane mano.
Figure 9.4 Pressure ranges a of the differrent types of vacuum pumps, b of
pressure detectors
9.2.1
Mechanical Pumps
Already around 1600 Galileo Galilei has produced a low vacuum in a container by using a movable piston. More detailed
experiments were performed in 1643 by Evangelista Torricelli,
who was Galileo’ successor in Florence. In honour of Torricelli
the unit of pressure has been named torr (1 torr is the pressure of
1 mm mercury column D 133:3 Pa). The unit torr has been used
for several centuries before the SI unit 1 Pa was introduced.
Spectacular experiments with evacuated spheres were performed 1645 by Otto von Guericke, the major of the German
city Magdeburg. He put two hemi-spheres together, sealed them
up with leather gaskets and evacuated the interior. This pressed
the two hemi-spheres tightly together. In order to demonstrate
the force on the hemi-spheres due to the external pressure he
roped 8 horses in on each side who tried unsuccessfully to separate the hemi-spheres. The large auditorium was very much
astonished that 16 horses could not separate the hemi-spheres
although they could be readily separated after the evacuated
sphere was filled again with air at the external pressure. Superstitious people believed in a ghost inside the sphere. An
engraving of Caspar Schott illustrates this spectacular exper-
Figure 9.6 Ancient mechanical vacuum pump
iment (Fig. 9.5). At that time the evacuation with piston air
pumps (Fig. 9.6) was tedious, because the seals were imperfect.
Nowadays the mechanical pumps are mainly rotary vane pumps,
Roots pumps or turbo-molecular pumps, which are all driven by
electro motors.
9.2 Generation of Vacuum
9.2.1.1
243
Rotary Vane Pumps
The basic principle of rotary vane pumps is schematically illustrated in Fig. 9.7. An eccentrically mounted rotor R1 rotates in a
cylindrical bore with an inlet S1 from the vacuum chamber and
an outlet A1 into the open air at atmospheric pressure. The rotor
has a slit in which two sliders are pushed by a coil spring against
the wall of the bore. When the rotor rotates in the direction of
the arrow the right side of the sliders sucks the gas from S1 and
drives it during half of a rotation period towards A1 . This is repeated every half turn thus continuously evacuating the vacuum
chamber behind S1 .
In order to obtain lower final pressures the outlet A1 can be
connected to a second pump (Fig. 9.8), which produces in A1 al-
Figure 9.8 Two stage rotary vane pump [9.1]. With kind permission of Leybold
GmbH
ready a pressure of 10 1 hPa, thus reducing the back-streaming
considerably. This leads to a final pressure of permanent gases
in S1 of about 10 3 to 10 4 hPa. However, now the saturation
vapour pressure of the pump oil (ps 10 3 hPa at T D 350 K)
is the limiting factor for the final pressure. Using a cool trap
between S1 and the vacuum chamber can reduce the saturation
pressure and realizes an oil-free vacuum in the chamber.
Typical pumping speeds of such rotary van pumps reach from
1 m3 =h for small pumps to 60 m3 =h for larger ones. To prevent
back streaming of atmospheric pressure from A to S in case of
an accidental standstill of the pump a blocking valve V is built
in at A2 .
9.2.1.2 Roots-Pump
The principle of a roots pump is shown in Fig. 9.9. Two symmetrically shaped rotors R1 and R2 rotate with opposite directions
about two axes. They are arranged in such a way that their surfaces nearly touch each other. The gap width between the two
rotors and between the rotors and the wall are only a few tenth
of a millimetre. For the momentary situation shown in Fig. 9.9
the gas volume V1 enclosed by the left rotor R2 is compressed
and pushed to the outlet A when the rotor rotates counterclockwise. A quarter of a full turn later the oppositely turning rotor
R1 pushes gas from S to a similar enclosed volume on the right
side and presses it to A. Since the rotors do not touch, there is no
Figure 9.7 Principle operation of a rotating vane pump [9.1]. With kind permission of Leybold GmbH
Figure 9.9 Principle operation of roots pump [9.1]. With kind permission of
Leybold GmbH
Chapter 9
For single-stage pumps the outlet A1 is connected to the open
air (or an exhaust gas line) and the pressure p in A1 equals the
atmospheric pressure. Due to the pressure difference between
A1 and S1 always some gas can flow back from A1 to S1 because
the slider in the rotor does not completely seal the connection
between A1 and S1 . This limits the attainable final pressure in
S1 . In order to keep the leak rate as small as possible the pump
is filled with oil which forms a film between slider and wall
and not only gives a better seal but also acts as lubricant that
prevents jamming of the rotor. With such single-stage pumps
final pressures of 10 1 –10 2 hPa are reached.
244
9 Vacuum Physics
material abrasion and roots pumps can rotate with high angular
velocities, thus increasing the pumping speed. The disadvantage
of the gap between the rotors is the backstream of gas from A
to S. With decreasing pressure in A the flow resistance of the
gaps with width d becomes larger as soon as d. Therefore
the pressure in A should be lowered by a one-stage rotary vane
pump. Roots pumps need a forepump. Large roots pumps reach
pumping speeds of up to 105 m3 =h.
9.2.1.3
Turbo-Molecular Pumps
The turbo-molecular pump, developed 1958 by W. Becker [9.5]
is based on the principle that molecules hitting a fast moving
surface, gain momentum in the direction of the surface motion
(Fig. 9.10).
Chapter 9
The turbo pump consists of a staple of fast rotating rotors with
many blades (Fig. 9.11). Assume a gas molecule M with the
thermal velocity v impinges on a blade of the rotor which has the
same temperature T as the gas. For a resting rotor the molecule
M would desorb from the blade after a short time with the velocity v0 which has about the same magnitude as v, (jvj jv0 j)
while its directions are distributed around the surface normal.
If the rotor blade moves with the velocity u the total velocity of
the desorbing molecules is the vector sum v D v0 C u. Due
Figure 9.11 Rotor of a turbo pump. With kind permission of CIT Vacuum
Technique
to the direction of u to the left in Fig. 9.10, the number of impinging molecules from the left half space is larger than that of
molecules from the right one. Because of the inclined blades the
velocities v 0 are preferentially directed into the downward direction. The rotating blades therefore transport molecules from the
upper space (inlet of the pump) to the lower space (outlet of the
pump).
This is illustrated in Fig. 9.10b. If the rotor blade moves with
the velocity u to the left, molecules moving with the velocity
v 0 < u from the upper space 1 can hit the blade only on the left
side.
9.2.2
Figure 9.10 a Momentum transfer at the reflection of molecules M at a fast
moving surface; b basic principle of turbo pump
Diffusion Pumps
Diffusion pumps are used for the generation of high- and
ultrahigh-vacuum. Their principle is shown in Fig. 9.12. A
pumping fluid 2 (oil or mercury) is evaporated by the heater at
the bottom of the pump. The vapour rises in the inner part of
the pump and leaves it at the upper end through nozzles where
it gains supersonic speed forming fast vapour jets, which are
directed downwards. Molecules from the vacuum chamber diffuse into the vapour jets and are pushed downwards by collisions
Figure 9.13 a Cooled baffle for reduction of oil return flow. b Liquid nitrogen
condensation trap
Figure 9.12 Operation principle of a diffusion pump. 1 D Heating, 2 D boiling region, 3 D pump body, 4 D water cooling, 5 D high vacuum side, 6 D
particles from high vacuum side, 7 D vapor jet, 8 D pre vacuum tube, A –D D
nozzles for vapor jets. With kind permission of Leybold GmbH [9.1]
therefore liquid nitrogen cool traps above the diffusion pump
are necessary in order to obtain a better high vacuum. Oildiffusion pumps operate with special oils that have saturation
pressures below 10 7 hPa. Therefore nowadays mainly oil diffusion pumps are used.
with the vapour molecules. Since the vapour jets are initially
free of gas, the diffusion rate into the jet is higher than out of the
jet.
In order to prevent oil molecules from reaching the vacuum container a cooled baffle is mounted above the pump (Fig. 9.13a)
which blocks the direct way of the molecules. Another solution
is a liquid nitrogen trap (Fig.9.13b) where every oil molecule
on its way to the vacuum container hits at least one cooled wall
where the molecules are adsorbed.
After diffusion into the vapor jet the gas molecules experience
by collisions with the jet molecules an additional momentum
downwards into the direction of the jet. They come into lower
regions where they diffuse into the lower jets where they experience more collisions and are transported farther downwards.
Finally they reach the outlet of the diffusion pump where they
are pumped away by a mechanical pump. A pressure ratio
pi =p0 10 7 between the pressure pi at the input and p0 at
the outlet of the diffusion pump can be reached. When the forepump maintains a pressure p0 D 10 2 hPa a pressure as low as
10 9 hPa can be realized at the high vacuum side.
The pumping speed of modern vacuum diffusion pumps ranges
from 60 l=s (for a small pump with 20 cm heights) to 50 000 l=s
(about 4–5 m high). Diffusion pumps are the favorite types of
high vacuum pumps. In Fig. 9.14 the pumping speed of medium
sized diffusion pumps as a function of the pressure on the high
vacuum side is compared with the performance of a turbo pump.
Important for the optimum performance of a pumping system is
The hot vapor jets hit the cooled wall of the pump where they
condense and flow as liquid film down to the heater. Here they
are again vaporized. In order to form oil vapor jets the free mean
path must be sufficiently large, i.e. the pressure sufficiently
low. Diffusion pumps therefore can operate only at pressures
below 10 2 –10 3 hPa. They do need a forepump, which generates the necessary minimum starting pressure.
The total pressure at the high vacuum side of the diffusion
pump is the sum of all partial pressures, including the saturation
pressure of the pump fuel. For mercury as fuel the saturation
pressure is 10 3 hPa at room temperature. For mercury pumps
Figure 9.14 Pumping speed S .p / of different types of pumps
245
Chapter 9
9.2 Generation of Vacuum
246
9 Vacuum Physics
the choice of the best forepump, which should be always able to
maintain a pressure below 10 2 hPa on the high pressure side of
the diffusion pump.
Example
A diffusion pump with a pumping speed of 2000 l=s
should maintain a pressure of 10 5 hPa in a container into
which continuously a gas streams. A gas volume of 2000 l
at p D 10 5 hPa corresponds to a volume of 2 l at a pressure of 10 2 hPa. Therefore the forepump must have at
least a pumping speed of 2 l=s D 7:2 m3 =h. Since the vacuum pipe between forepump and diffusion pump reduces
the pumping speed, a forepump with a pumping speed of
12 m3 =h should be used.
J
9.2.3
Chapter 9
Cryo- and Sorption-Pumps;
Ion-Getter Pumps
A cryopump consists essentially of one or several cooled surfaces inside the vacuum container. All gases or vapors with
condensation temperatures above the temperature of the surfaces condense and are adsorbed as liquids or solids on the
surfaces. Liquid nitrogen cooltraps therefore can condense all
gases and vapors except hydrogen and helium which need liquid
helium traps. Most cryo-pumps use closed cycle lquid helium
cooling systems (Fig. 9.15), which reach temperatures down to
about T D 10 K.
The achievable final pressure is determined by the equilibrium
of the rate of molecules impinging onto the cold surface and the
rate of evaporating molecules. The latter is determined by the
vapor pressure of the component with the lowest evaporation
temperature.
The impinging molecules have a mean velocity
p
v Tw which depends on the temperature Tw of the walls of
the vacuum chamber,
while the mean velocity of the evaporating
p
molecules v Tc depends on the lower temperature Tc of the
cold surface.
The rate of molecules impinging onto the area A is
ZD
1
n vw A ;
4
(9.15)
which equals the desorbing
p rate under equilibrium conditions.
With p D nkT and v T we obtain the partial pressure of the
i-th vapor component in the container
p
pe .i/ D ps .i/ Tw =Tc ;
(9.16a)
where ps is the saturation pressure. The attainable final pressure
is then
X
p
ptotal D
pe .i/ Tw =Tc :
(9.16b)
Cryo-pumps need a forepump which lowers the pressure in the
container down to about 10 3 hPa, because for p > 10 3 hPa the
mean free path is smaller than the dimensions of the container
and the heat conduction from the cold surfaces to the wall of the
container becomes too large (see Sect. 7.5). Furthermore at low
pressures the layer of condensed gases becomes too thick which
lowers the heat conduction from the surface of this layer to the
cooling body and increases the temperature of the surface.
The pumping speed of a cold surface Ac at a pressure p in the
vacuum container is according to (9.2) and (9.15)
ps p
1
Ls D Ac v 1
Tw =Tc
4
p
(9.17)
1 pe
1
D Ac v ˛
;
4
p
where ˛ 1 is the sticking propability
P of an impinging
ps .i/ is the sum of
molecule on the cold surface, and ps D
the saturation pressures of all vapor components at the temperature Tc of the cold surfaces.
Example
v D 400 m=s, ˛ D 1, Pe p, A D 1 cm2 ! Ls D
10 l=s, i.e. the cold surface has a maximum pumping
speed of 10 l=s per cm2 .
J
The growth rate d=dt of the adsorbed layer with thickness
.t/ on the cold surface depends on the density n D N=V of
molecules with mass m in the container and on their mean velocity v D .8kT=m/.1=2/ at the temperature T.
According to Fig. 7.28 the number of molecules with mass m
hitting per sec the area dA is
dZ D
With v D
sec
Figure 9.15 Principle of cryo pump with closed cooling cycle
1
n vdA :
4
(9.18a)
p
8kT=m we get the mass increase of the layer per
dM
1
D dZ m D n m v dA
dt
4
1 p
D n 8kTm= :
4
(9.18b)
9.3 Measurement of Low Pressures
247
Figure 9.16 Setup of a sorption pump. 1: inlet connector; 2: degasing
connector; 3: mechanical support; 4: pump body; 5: heat conduction sheets;
6: adsorption material [9.1]
The mass M of the layer with thickness and density % is M D
% dA.
This gives finally the growth rate of the layer
d
np
kTm=2
D
dt
%
(9.18c)
Example
For N2 -molecules at a pressure of 10
rate is 5 µm=h.
5
hPa the growth
J
The adsorbed layer should not be too thick, because the heat
conduction becomes worse with increasing thickness and the
surface temperature of the layer increases. This increases the
evaporation rate of the adsorbed molecules.
The adsorbing surface can be greatly increased by using molecular sieves (zeolites D alkali-aluminum silicate)). They consist
of small balls with many fine pores, into which the molecules
can diffuse and are then adsorbed. The effective surface of Zeolith is about 103 m2 per gramm. For Zeolith the diameter of the
pores is about 1:3 nm. For typical sizes of 0:5 nm for molecules
1 g Zeolith can adsorb about 2:5 1021 molecules in a monolayer. This corresponds to a gas volume of 10 000 l at a pressure
of 10 2 hPa.
The adsorption of the molecular sieves depends strongly on the
temperature. They can be therefore used at low temperatures
(liquid nitrogen temperature D 78 K) as a cryopump and later
on they can be degassed at higher temperatures and used again
as pumps. Such a sorption pump is shown schematically in
Fig. 9.16.
Another solution for high vacuum pumps are ion-getter pumps.
In a gas discharge ions are produced which are accelerated onto
the cathode. Here they sputter the cathode material (e.g. Titanium) which is adsorbed on cold surfaces, where already a layer
of condensed gases has been formed. The titanium atoms form a
film, that covers the layer of adsorbed atoms and burries it completely. A new fresh metal surface is formed where further gas
molecules can be adsorbed. Since the vapor pressure of titanium
is very low, even at room temperature very low pressures can be
obtained.
Such ion-getter pumps (chemical getter pumps) are useful for
the generation of oil-free ultrahigh vacuum (p < 10 6 hPa). In
Fig. 9.17 a possible realization is shown. A titanium wire is
heated by direct electric current or by electron bombardment.
The sputtered titanium atoms are ionized by collisions with electrons and are accelerated onto the cooled walls, which are kept
at ground potential. Here they push the adsorbed molecules
deeper into the wall and burry them under a metallic film of
neutral titanium atoms. A sputter rate of 5 mg=min represents at
p D 10 6 hPa a pumping speed of 3000 l=s.
9.3
Measurement of Low Pressures
For the measurement of pressures a variety of different measuring techniques and instruments have been developed. We will
present only a small selection. Table 9.3 compiles some of these
devices suitable for the different pressure ranges.
Table 9.3 Pressure ranges of different pressure measuring devices
Device
Pressure range/mbar
Liquid manometer
0:1–103
Mechanical spring vacuum meter
1–103
Membrane manometer
1–103
Capacity manometer
10 4 –103
Heat conduction manometer
10 3 –1
Heat conduction manometer with control 10 3 –100
feedback
McLeod manometer
10 6 –10 1
Penning ionization manometer
10 7 –10 3
Ionization manometer
10 12 –10 3
Friction manometer
10 7 –10 1
Chapter 9
Figure 9.17 Principle of getter ion pump [9.1]
248
9 Vacuum Physics
9.3.1
Liquid Manometers
Liquid manometers (Fig. 7.3) are simple devices for pressure
measurements, that have been already used in 1643 by Torricelli. Here the height difference h of a liquid with density %
in the two legs of a U-shaped tube are measured. The pressure
difference p D p2 p1 between the two ends of the U-tube is
then
p D % g h :
(9.19)
Example
With oil (% D 900 kg=m3) a pressure difference p D
1 hPa causes a height difference h D 11:3 mm. For
mercury (% D 13;546 kg=m3) one obtains h D 1 mm
J
for p D 1:33 hPa D 1 torr.
Chapter 9
When one leg of the U-tube is closed and evacuated (Fig. 7.2)
the volume above the liquid is filed with the vapour of the liquid
with the vapour pressure ps .T/, that depends on the temperature
T. The height difference is then
1
h D
.p
%g
1
ps /
p
%g
for ps p ;
(9.19a)
which gives directly the pressure p above the open leg of the
U-tube.
The accuracy and sensitivity of liquid manometers can be
considerably increased with a device, developed by McLeod
(Fig. 9.18), which is based on the Boyle–Mariotte Law (see
Sect. 7.1). At the beginning of the measurement, the container B
is lowered until the liquid level in the left leg is at h1 . The pressure p1 above h1 is the pressure in the vacuum chamber. Now B
is lifted again until the liquid level rises above the point z. The
volume V D V0 C Vc of G and the capillary above G is now separated from the vacuum chamber. The container is lifted up to a
height where the liquid level in the very left tube (pressure p1 )
is by h higher than in the capillary above G where the higher
pressure p2 is present due to the compression of the closed volume V to the much smaller volume Vc D r2 x. According to
the Boyle–Marriot law we obtain
p1 .V0 C Vc / D p2 r2 x :
The measured height difference of the liquid between the left
tube and the capillary
p2 p1
V0
p1
L
D
C
1
(9.20)
h D
%g
% g r2 x
x
yields the pressure p1 in the vacuum chamber, after the volumes
V0 , Vc D r2 L and the length x of the gas-filled part of the capillary have been determined. in case of mercury one has to take
into account the capillary depression to mercury (see Sect. 6.4).
9.3.2
Membrane Manometer
To measure the pressure in the low vacuum range robust and
simple membrane manometers are available (Fig. 9.19b). A thin
membrane separates the vacuum from the upper part at atmospheric pressure. A wire is connected at one end with the centre
of the membrane and at the other end with a hand that can rotate
around a fixed axis. Due to the pressure difference, the membrane sags and turns the hand by an angle that is proportional to
the pressure difference, which can be read on a calibrated scale.
Another realization (Fig. 9.19a) uses a bent thin hollow tube
that is connected to the vacuum chamber. The bending radius is
Flexible evacuated
tube acting as spring
Fixed
rotation axis
Scale
Membrane
a)
Figure 9.18 Principle of McLeod vacuum meter
To vacuum
container
b)
To vacuum
container
Figure 9.19 Two designs of robust and compact pressure detectors. a Spring
pressure gauge; b membrane pressure gauge
9.3 Measurement of Low Pressures
249
Figure 9.20 Membrane capacitor vacuum gauge. A thin membrane M provides together with two curved fixed plates two capacities C1 and C2 , which are
arranged in a bridge circuit (see Vol. 2, Chap. 2). They are fed by two identical
ac-voltage sources
For lower pressures in the high vacuum range (p < 10 5 hPa)
capacitance membrane manometers (Fig. 9.20) can be used.
Here a thin membrane which separates the vacuum chamber
from a chamber with a fixed reference pressure. It forms one
electrode of two electric capacitors C1 and C2 , When the pressure in the vacuum chamber decreases the membrane bends to
the left side and decreases the electrode separation of C1 but
increases that of C2 , thus increasing the capacitance of C1 and
decreases that of C2 . This changes their AC resistance in an opposite direction, which can be measured with an electric bridge
arrangement where the two capacitors are charged by two identical AC voltage supplies (see Vol. 2. Chap. 1).
9.3.3
Heat Conduction Manometers
As has been shown in Sect. 7.5 the heat conduction of a gas in
the pressure range where the mean free path is larger than
the dimensions of the vacuum chamber, is proportional to the
pressure p. This fact is used in the heat conduction manometer (Fig. 9.21) for measuring pressures. A filament of length
L, heated by an electric current I, is clamped between to yokes
along the axis of a small cylindrical tube. Its temperature Td is
determined by the supplied electric power I 2 R and the power
loss due to heat conduction.
dW
D 2r L .Td
dt
Tw / ;
(9.21)
which is given by the surface 2r L of the filament, the heat
conduction of the gas and the temperature difference T D
.Td Tw / between filament and wall (see Sect. 7.5.3). Stationary
conditions are established, when the supplied power equals the
power loss. This yields
I 2 R D 2r L T :
(9.21a)
The coefficient of heat transfer
D n v k f =8 D v p f =8T ;
(9.21b)
Figure 9.21 Heat conduction vacuum gauge. a Mechanical design; b electric
circuit
(7.49a) is proportional to the gas density n D p=kT and therefore
to the pressure p and to the degrees of freedom f of the gas
molecules. The electric resistance then becomes
R.Td / D
2r L p v f .Td
4.Td C Tw /
Tw /
:
(9.21c)
It depends on the temperature Td . It can be measured with an
electric bridge (Fig. 9.21b) (see Vol. 2, Sect. 2.4.3) and yields
the wanted pressure measurement.
Since the heat conduction in the low vacuum range ( d)
is independent of the pressure (see Tab. 9.1) heat conduction
manometers can be used only in the medium vacuum range
(1 10 3 hPa), for instance between diffusion pump and backing pump. For pressures below 10 3 hPa the heat conduction
through the gas becomes smaller than other heat leaks (for
example through the yokes of the filament). Therefore the
accuracy of pressure measurements decreases strongly below
p D 10 3 hPa.
9.3.4
Ionization Gauge and Penning Vacuum
Meter
The vacuum meters that are used most often in the high vacuum range are the ionization gauge (Fig. 9.22) and the Penning
vacuum meter (Fig. 9.23). The ionization gauge consists of a
heated filament as cathode K emitting electrons that are accelerated onto the anode A. On their way from K to A, they collide
with gas molecules and ionize them (see Vol. 3). When the free
mean path of the electrons is larger than the distance K–A the
number Nion of produced ions is proportional to the density n
of the gas molecules in the manometer and therefore also to the
pressure p D n kT. It is
Nion D Nel
X
i
ni ˛i .Eel / ;
(9.22)
Chapter 9
dependent on the pressure. When it changes the upper end of the
tube moves a hand which indicates the pressure on a calibrated
scale.
250
9 Vacuum Physics
9.3.5
Ion collector
To
vacuum
container
a)
b)
Figure 9.22 Ionization vacuum meter. a Schematic principle; b design of
Bayard–Alpert tube
Rotating Ball Vacuum Gauge
The principle of this vacuum gauge is based on the deceleration
of a rotating ball due to friction with the rest gas molecules. A
small steel hollow sphere is contact-free hold in its position by
a magnetic field (Fig. 9.24). A rotating magnetic field produced
by special coils is superimposed onto the static magnetic field. It
causes the ball to rotate with an angular velocity of about ! D
2 400 s 1 . After shut off the rotating field, the ball rotates
freely and is only decelerated by friction due to collisions with
the gas molecules. The slowing down time depends on the rate
of collisions and therefore on the gas pressure.
The angular momentum of the rotating ball is
L D I! D
8
2
MR2 ! D
%R5 ! :
5
15
(9.23)
The retarding collisions produce a mean torque
Chapter 9
DD
Figure 9.23 Penning vacuum gauge
where ni is the partial density of molecules of type i and ˛i .Eel /
the ionization probability, which depends on the energy Eel of
the electrons. The positive ions are collected on a wire D at a
negative potential against the anode.
The minimum of still detectable pressure is limited by several
factors: Firstly, the ion current decreases with decreasing pressure, which demands good current amplifiers. Secondly, the
electrons impinging onto the anode generate X-rays that can
release electrons from the ion collector. Their number is independent of the pressure and form an underground current that
overlaps the wanted signal current.
dL
D I !P ;
dt
(9.24)
onto the ball which is proportional to the gas pressure. The decrease rate of the angular velocity ! is then
d!
D
D
Da! p:
dt
I
(9.25)
The proportionality factor a depends on the radius R of the ball,
on the density % and on the mean molecular velocity v. After calibrating the system the factor a can be measured. The
pressure p, which is proportional to the density % can then be
determined from the relative deceleration d!=dt=!. The accuracy of the measurement is about p=p D 3%. Therefore the
rotating ball gauge is the most accurate vacuum meter in the
vacuum range 0:1–10 7 hPa [9.6].
For more detailed and recent information on modern techniques
of vacuum physics, the reader is referred to the literature [9.7].
Typical pressure ranges where the ionization gauge can be used
are 10 3 hPa p 10 12 hPa, where for the lower pressures
special designs have been developed which minimize the underground current (Bayard–Alpert tube Fig. 9.22b).
Instead of the thermionic emission of electrons from a heated
filament at higher voltages ( 1000 V) also a cold electron
emission between two metal plates can be realized. Since the
ionization probability is small at such high electron energies,
the ionization path must be enlarged. This is achieved by a permanent magnet with a magnetic field B that forces the electrons
on spiral paths until they reach the anode (Fig. 9.23).
These Penning manometers are robust but not as accurate as the
ionization gauges. They can be used in the vacuum range from
10 3 to 10 7 hPa.
Figure 9.24 Section through the gauge head of a friction vacuum gauge. 1 D
Steel ball, 2 D gauge tube with one open end, which is welded to the flange
7, 3 D permanent magnet, 4 D stabilization coils, 5 D four driving coils, 6 D
horizontal position detector. With kind permission of Leybold GmbH
Problems
251
A volume V is called evacuated if the total gas pressure in V
is small compared to the atmospheric pressure.
The different vacuum ranges are:
Low vacuum
(1 hPa p 103 hPa)
Medium vacuum
(10 3 hPa p 1 hPa)
High vacuum
(10 7 hPa p 10 3 hPa)
Ultrahigh vacuum (p 10 7 hPa)
Vacuum is generated with vacuum pumps. The most important types are mechanical pumps (rotary vane pumps)
and roots-pumps, (which are used as fore pumps for the
generation of fine vacuum), turbo-molecular pumps for the
generation of oil-free high- and ultrahigh vacuum, oil- and
mercury diffusion pumps, cryo-pumps and ion getter pumps
for the generation of ultrahigh vacuum.
The gas pressure in a vacuum chamber can be measured with
one of the following devices:
liquid barometer
(0:1 hPa p 103 hPa)
membrane manometer
(p 1 hPa)
heat conduction manometer (p 10 3 hPa)
capacitance manometer
(p 10 5 hPa)
friction vacuum manometer (p 10 7 hPa)
ionization gauge
(p 10 12 hPa)
The suction capability SV D dV=dt is the gas volume flow
through the suction connection of a pump. Often the product
SL D p SV of pressure and suction capability is called the
pumping speed.
The vacuum lines (tubes and pump connectors) between vacuum chamber and pump reduce the total suction capability.
Their flow conductance LS D p dV=.p2 p1 / should be
as high as possible, in order to make the pressure difference
between entrance and exit of the vacuum line small.
The achievable final pressure in the vacuum chamber is determined by the pumping speed of the pump, by the leak rate
and the desorption rate of molecules from the inner walls of
the chamber.
Problems
9.1
A vacuum chamber is connected to the outside at atmospheric pressure through a capillary tube with length L D 10 cm
and 0:5 mm inner diameter. What should be the effective suction
capacity of the vacuum pump in order to maintain a pressure of
10 3 hPa?
9.2
Which force was necessary to separate the two hemispheres of Guericke’s demonstration experiment, when the diameter of the spheres was 60 cm and the inner pressure 100 hPa?
9.3
In a cubic vacuum chamber with a volume V D 0:4 m3
a pressure of p D 10 5 hPa is maintained. What are the particle
density n, the mean free path and the mean time between
two successive collisions between particles at room temperature? How large is the ratio Z1 =Z2 of the rate Z1 for mutual
collision between particles to the rate Z2 for collision of particles with the walls? How large is the total mean path length that
a particle traverses within 1 s, and what is the sum of the path
lengths of all particle in the chamber?
9.4
Assume, the vacuum chamber of Probl. 9.3 is operated
under ultrahigh vacuum and the inner walls are free from all adsorbed molecules. At t D 0 oxygen is let in until the pressure
rises to 10 7 hPa. How long does it take until the walls are cov-
ered by a monomolecular layer, if each oxygen molecule covers
an area of 0:15 0:2 nm2 and its sticking probability is 1?
9.5
A vacuum chamber should be evacuated down to a pressure of 10 6 hPa using a diffusion pump with the effective
pumping speed of 3000 l=s. What is the minimum effective
pumping speed of the mechanical fore pump in order to maintain a vacuum of 0:1 hPa at the outlet of the diffusion pump?
9.6
The ionization cross section of nitrogen molecules N2
for collisions with electrons of 100 eV energy is D 1
10 18 cm2 . How large is for an electron current of 10 mA the ion
current at a pressure of 10 7 hPa in the ionization gauge when
the path length of the electrons is 2 cm?
9.7
Through the heated filament of a thermal conductivity
gauge flows the electric current I D U=R.T/ at a constant
voltage U. The heating power under vacuum conditions is
Pel D U 2 =R0 . What is the dissipation power due to heat conduction in a cylindrical chamber with diameter of 2 cm at a gas
pressure of p D 10 2 hPa when the temperature of the filament
is T1 D 450 K and that of the wall is T2 D 300 K? (The length of
the filament is 5 cm, its diameter 0:5 mm, the distance filamentwall is 1 cm). Which fraction of the electric energy Eel D U I
is dissipated by heat conduction if U D 0:5 V and I D 2 A?
Chapter 9
Summary
252
9 Vacuum Physics
9.8
The total angular momentum transfer onto a ball at rest
in a gas at thermal equilibrium is zero. Why is the rotating ball
in a Langmuir friction gauge slowed down? Estimate the torque
that the gas molecules transfer to a ball with a radius of 1 cm
rotating with the angular velocity ! D 2 400 s 1 at a temperature of T D 300 K and a pressure of p D 10 3 hPa. How long
does it take until ! has decreased by 1%?
References
9.1. D.M. Hoffman, B. Singh, J.H. Thomas, Handbook of Vacuum Science and Technology. (Academic Press, 1998)
9.2. N.S. Harris, Modern Vacuum Practice, reprint 1997 (MacGraw Huill Publ., 1989)
9.3. J.F. Lafferty, Foundations of Vacuum Science and Technology. (John Wiley and Sons, 1998)
9.4. Ph. Dnielson, A Users Guide to Vacuum Technology. (John
Wiley and Sons, 1989)
9.5. https://en.wikipedia.org/wiki/Turbomolecular_pump
9.6. J.F. O’Hanion, A User’s Guide to Vacuum Technology.
(John Wiley & Sons, 2005), p. 385
9.7. K. Jouston (ed.), Handbook of Vacuum Technology. (Wiley
VCH, Weinheim, 2008)
9.8. J.F. O’Hanion, A User’s Guide to Vacuum Technology. 3rd
ed. (John Wiley and Sons, 2003)
9.9. N. Yoshimura, Vacuum Technology. (Springer, 2014)
Chapter 9
Thermodynamics
10.1
Temperature and Amount of Heat . . . . . . . . . . . . . . . . . . . . . 254
10.2
Heat Transport . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266
10.3
The Three Laws of Thermodynamics . . . . . . . . . . . . . . . . . . . . 279
10.4
Thermodynamics of Real Gases and Liquids . . . . . . . . . . . . . . . 299
10.5
Comparison of the Different Changes of State . . . . . . . . . . . . . . 309
10.6
Energy Sources and Energy Conversion . . . . . . . . . . . . . . . . . . 309
10
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318
Chapter 10
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319
© Springer International Publishing Switzerland 2017
W. Demtröder, Mechanics and Thermodynamics, Undergraduate Lecture Notes in Physics, DOI 10.1007/978-3-319-27877-3_10
253
254
10 Thermodynamics
The insight, that heat is just one of several forms of energy
and can be explained by a mechanical model, is today common
knowledge, but it is only about 170 years old. The physician
Julius Robert Mayer (1814–1878) formulated in 1842 his ideas
about the energy conservation for the conversion of mechanical
energy into heat, and he could already give a numerical value
for the thermal energy equivalent (Sect. 10.1.5). However, only
after the development of the kinetic gas theory (see Sect. 7.3)
the microscopic explanation of heat of a macroscopic body as
the total energy (kinetic plus potential energy) of all molecules
of the body was possible. As has been explained in Sect. 7.3, a
measure of the mean kinetic energy of all particles with mass m
in a gas volume, which have three degrees of freedom for their
motion is the absolute temperature
TD
1 2 m 2
v :
k 3 2
(10.1)
With this definition of the temperature all macroscopic phenomena and the general laws derived from them (Boyle.-Marriott,
general gas law) could be reduced to microscopic models describing matter as composed of atoms and molecules.
Chapter 10
In this chapter, we will discuss in more detail the measurement
of temperature, the definition of temperature scales, the experimental findings of energy transport and conversion, of material
changes with temperature such as thermal expansion and phase
transitions. An important subject is the formulation of basic
laws of thermodynamics which can be regarded as a summary
of many experimental results. We try to explain all macroscopic
phenomena as far as possible by microscopic models, where,
however, some explanations need a deeper knowledge of atomic
physics, which will be imparted in volume 3 of this textbook
series.
At the end of this chapter a short excursion to the thermodynamics of real gases and liquids is presented, which might be helpful
for the explanation of many phenomena observed in nature.
10.1
Temperature and
Amount of Heat
The definition of the absolute temperature, given in (10.1), is
for most practical applications of temperature measurements not
very helpful. One has to use measuring techniques that are reliable, accurate and easy to handle.
Qualitative information about the temperature can be already
obtained with the heat sensibility of our body. Our skin has
sensors that inform us whether a body is cold or hot. This sensing is, however, not very accurate and depends on the previous
experience, as the following experiment illustrates: Three containers with (1) hot water, (2) lukewarm water and (3) cold water
are placed side by side. Dipping a finger at first into (1) and then
into (2) the lukewarm water seem to be cold, but dipping at first
into (3) and then into (2) the same lukewarm water seems to be
warm.
Figure 10.1 a Liquid thermometer; b thermo couple
This demonstrates that the sensing of our body is not reliable
and cannot be used for quantitative measurements. In order to
measure temperatures and the amount of heat, measuring instruments and techniques have to be developed that are for practical
applications sufficiently easy to handle and which give reliable
and reproducible results.
10.1.1
Temperature Measurements,
Thermometer, and Temperature Scales
For the measurement of temperatures in principal all physical
effects can be used that depend on the temperature. These are
for instance:
The geometrical dimensions of solid, liquid or gaseous bodies, which generally increase with the temperature. Metal
wires become longer, liquid or gas volumes expand with increasing temperature at constant pressure.
The electrical resistance of a body changes with the temperature T. For metals it increases with T for semiconductors it
decreases (see Vol. 2, Chap. 2)
The electric contact potential difference between two different metals in contact changes with temperature (Thermovoltage see Vol. 2, Sect. 2.9)
The radiation power emitted by a hot body increases with T 4
and can be used for the measurement of the temperature of
remote bodies such as stars (Radiation Pyrometer see Vol. 2
Chap. 12).
Devices for the measurement of temperatures are called thermometers (Tab. 10.1). For daily practice the expansion of
liquids are generally used (liquid thermometer, Fig. 10.1a) or
the change of the contact voltage (thermo-elements, Fig. 10.1b).
For the quantitative specification of a temperature, numerical
values for fixed temperatures have to be defined that can be accurately reproduced under readily realizable external conditions
(temperature fix points). Furthermore, a temperature scale has
to be defined. This has been historically realized in different
ways.
10.1 Temperature and Amount of Heat
255
Table 10.1 The mostly used thermometers
Thermometer type
Liquid thermometers:
Mercury
Alcohol
Pentane mixture
Solid state thermometers:
Metal rod
Bimetal
Resistance thermometers:
Metal wire
Semiconductor
Thermo couple:
Fe-CuNi (iron-constantan)
Ni-CrNi
Ni-CrNi
Pt-PtRh
W-WMo
Pyrometer
38 to C800
110 to C210
200 to C30
150 to C1000
dependent on specific metal
150 to C500
250 to C1000
273 to C400
200 to C760
270 to C1000
200 to C1370
50 to C1700
200 to C3000
C800 to C3000
Measuring principle
Error limits
Thermal expansion of liquid in glas capillary
depending on scale division 0:1–1 ı C
Thermal expansion of metals
1–2% of scale range
Length expansion difference
Dependent on model
Temperature dependence of electric resistance
0:1–1 ı C
Temperature dependence of thermovoltage
0:1–1 ı C
Heat radiation
2–10 ı C
The Celsius Scale
The astronomer Anders Celsius (1701–1744) proposed 1742 to
use the expansion of a mercury column for the measurement
of temperatures (mercury thermometer). Two fix points were
defined for the temperature scale: The melting point of ice
(TC D 0 ı C) and the boiling point of water (TC D 100 ı C at
a pressure of 1 atm D 1013:25 hPa). The range between these
two fix points is divided into 100 equal units, where each unit
corresponds to 1 ı C.
Note: We will label the Celsius temperature with TC in order to
distinguish it from the Fahrenheit temperature (TF ) and Kelvin
temperature T.
10.1.1.2
Figure 10.2 Comparison of Celsius and Fahrenheit scale. For the Kelvin scale
there is only one fixpoint (triple point of water at 0:01 ı C)
The Fahrenheit Scale
The Fahrenheit scale, which was proposed by Daniel Gabriel
Fahrenheit (1686–1736) is still used in the USA. It defines as
fix points the melting point of a defined ice-water-ammonia
chloride solution at TF D 0 ı F ( 17:8 ı C) and the normal
body temperature at TF D 100 ı F (C37:7 ı C). The range between the two fix points is equally divided into 100 units where
1 unit corresponds to 1 ı F. From this definition it follows that
0 ı C D 32 ı F and 100 ı C D 212 ı F. The conversion between
the two scales is as follows (Fig. 10.2):
TC = ı C D
TF = ı F D
10.1.1.3
5
9
9
5
.TF = ı F
32/
.TC = ı C C 32/ :
(10.2)
The Absolute Temperature Scale
The absolute temperature scale needs only one fix point, which
is the triple point of water (see below). It is measured with gas
thermometers.
Its definition is: The Kelvin is the unit of the thermodynamic
temperature scale. 1 K is the 273.16th part of the temperature
Tp of the triple point of water. The zero point of the Kelvin
scale is the lower absolute limit of possible temperatures and is
defined by general laws of thermodynamics (see Sect. 10.3).
10.1.1.4 Accuracy of Thermometers
The temperature scale of liquid thermometers depends on the
choice of the liquid and also of the glass of the thermometer capillary, because not only the liquid but also the glass expands with
rising temperature. The thermal expansion of liquids and solids
is generally not constant over the temperature range measured
by thermometers and is not necessarily linear (see next Section).
For mercury, the deviation from linearity is small. The comparison with an alcohol thermometer shows that its scale differs
from that for the mercury thermometer and is not equidistant
(Fig. 10.3).
Chapter 10
10.1.1.1
Temperature range = ı C
256
10 Thermodynamics
Table 10.2 Thermal expansion of solids and liquids at T D 293 K D 20 ı C
Solids
Aluminium
Linear
expansion
coefficient
˛=.10 6 K 1 /
23.8
Liquids
Water
Volume
expansion
coefficient
=.10 4 K 1 /
2.07
Iron
12
Ethanol
11
V2A Steel
16
Acetone
14.3
Copper
16.8
Benzene
10.6
Sodium
71
Mercury
1.8
Tungsten
4.3
Glycerin
Invar
1.5
N-Pentane
Cerodur
<0.1
Water at
Hard rubber
75–100
T D 0 ıC
T D 20 ı C
5.0
15
0.7
C2.07
Figure 10.3 Demonstration of non-uniform expansion of liquids by comparing
mercury and alcohol themometers
Chapter 10
If higher accuracy is demanded, other thermometers have to be
found which have a better linear temperature scale. A possible solution is the thermal expansion of gas volumes at constant
pressure or the pressure increase at constant volume that are utilized in the gas thermometer (see Sect. 10.1.3). They are used
for the definition of the absolute temperature scale (Kelvin scale,
Sect. 10.1.4).
10.1.2
Thermal Expansion of Liquids and
Solids
The length L of a rod changes with temperature. Experiments
show that the relative length change L=L within a restricted
temperature range is approximately proportional to the temperature change T:
(10.3)
L.TC / D L.0/ .1 C ˛TC / :
The expansion coefficient
(10.4)
˛ D .dL=dT/=L
gives the relative length change for a temperature change T D
1 ı C.
the other end B where a turnable tongue is connected to the rod
that shows the length change on a calibrated scale. When hot
water vapour streams through the tube, it expands and turns the
tongue. The scale is set to zero at TC D 20 ı C. The temperature
of the tube is measured with a calibrated thermo-element. Metal
tubes can be also heated by an electric current through the tube.
The reason for the thermal expansion is the asymmetric potential
of the interaction between neighbouring atoms (Fig. 10.5). The
atoms of a solid are not fixed at a constant value r0 of the distance between neighbouring atoms but oscillate around r0 (see
Vol. 3). The length L of a rod is determined by the mean distance
hri of this oscillation. Increasing the temperature causes an increase of the vibrational energy and of the oscillation amplitude
r.t/. Because of the asymmetric potential the mean value hri increases with increasing amplitude r.t/ thus causing an increase
of the length L.
More detailed measurements prove that the thermal expansion
is not strictly linear. Expansion of (10.4a) gives
Integration of (10.4) gives
L.T/ D L.0/ e˛T
Figure 10.4 Demonstration of thermal expansion of metal tubes
with
T D T
T0 :
(10.4a)
Table 10.2 compiles numerical values of ˛ for some materials.
One can see that for most materials ˛ is positive, i. e. the length
L increases with T. The coefficients ˛ can be measured with
the device shown in Fig. 10.4. A tube made of the material to
be inspected, is clamped on one end A, but can freely slide on
L.T/ D L.0/.1 C ˛ T C 21 .˛T/2 C : : :/ :
(10.4b)
This nonlinear expansion can be also expressed by a
temperature-dependent expansion coefficient ˛
˛.TC / D ˛.TC D 0/ C ˇ TC D ˛0 C ˇ TC :
(10.4c)
10.1 Temperature and Amount of Heat
Figure 10.5 Atomic model of thermal expansion due to the anharmonic interaction potential
Figure 10.6 Bimetal thermometer. a Principle, b technical design
Instead of (10.3) the more accurate equation is then
For homogenous and isotropic bodies applies
(10.5)
However, within the temperature range between 0 ı C and 100 ı C
the deviation from linearity is very small, i. e. ˇ TC ˛0 . For
small temperature intervals T the length L.TC / can be approximated by a straight line with a slope dL=dT D ˛0 C ˇ TC that
slightly depends on TC .
At room temperature the expansion coefficients of aluminium are ˛.TC D 20 ı C/ D 23:8 10 6 = ı C, ˇ D
1:8 10 8 =. ı C/2 .
For some alloys the expansion coefficient is very small. Examples are INVAR (64% iron and 36% nickel) or the glass ceramics
CERODUR (see Tab. 10.2 and 10.3) [10.1].
Since all length dimensions of a three-dimensional body vary
with the temperature, also the volume of the body must change.
T=K
50
100
150
200
250
300
350
400
500
Al
3.5
12.0
17.1
20.2
22.4
23.8
24.1
24.9
26.5
Cu
3.8
10.5
13.6
15.2
16.1
16.8
17.3
17.6
18.3
Fe
1.3
5.7
8.4
10.1
11.1
12.0
12.6
13.2
14.3
Al2 O3
0.0
0.2
1.0
2.8
4.0
5.0
6.0
6.4
7.2
SiO2
0:86
0:80
0:45
0:1
C0:2
C0:4
C0:5
C0:55
C0:58
V0 D V.TC D 0 ı C/
˛TC 1
D 3˛ :
(10.6)
For non-isotropic bodies the expansion may differ for the different directions and one obtains, instead of (10.6), the equation
D V0 .1 C 3˛TC /
The relative proportion of the nonlinear expansion is
therefore ˇ TC =˛ D 7:5 10 4 . This implies that the
coefficient ˛ changes within a temperature range T D
J
100 ı C only by 7:5%.
6
with
for
with
V.TC / D V0 .1 C ˛1 TC / .1 C ˛2 TC / .1 C ˛3 TC /
V0 Œ1 C .˛1 C ˛2 C ˛3 /TC
Example
Table 10.3 Dependence of mean thermal expansion coefficient ˛=10
on temperature (given in K)
V.TC / D V0 .1 C ˛TC /3
V0 .1 C 3˛TC /
D V0 .1 C TC /
K
1
with
˛ D 31 .˛1 C ˛2 C ˛3 / :
(10.6a)
The difference of expansion coefficients of different metals
is utilized for bimetal thermometers (Fig. 10.6). When two
metal strips of different materials are bonded (e. g. by welding
or soldering) the double strip will bend when the temperature
changes. A special device converts the bending, which is proportional to the temperature change T, into the turn of a hand
with a scale (Fig. 10.6b) where after calibration the temperature
can be read on the scale.
If the thermal expansion should be prevented by an external
force very large forces are necessary, as the following experiment demonstrates (Fig. 10.7). A thick rod S made of wrought
iron, is clamped between two stable mountings L1 and L2 . On
one end a bolt B with 5 mm diameter fixes the rod S to the
mounting L1 . Now the rod is heated with a Bunsen burner until
it is red glowing. The resulting thermal expansion loosens the
screw M on the right side at L2 , which is tightened again at the
highest temperature of the rod. Now the rod cools down and
contracts. The contraction force is so large that the bold at the
left side cracks.
A quantitative calculation of the forces necessary to prevent
thermal expansion or contraction proceeds as follows:
The force necessary to achieve an elongation of a rod with
length L and cross section A L2 and with an elastic modulus E is according to (6.2) and (10.4)
F D E A L=L D E A ˛ T :
(10.7)
Chapter 10
L.TC / D L0 1 C ˛0 TC C ˇTC2 :
257
258
10 Thermodynamics
Figure 10.7 Bolt cracker. Demonstration of large forces when the thermal
expansion is hindered. (B D bolt, S D hot rod, L1 , L2 D mounts, M D screw
nut)
If the thermal expansion should be prevented by application of
external pressure, we obtain from (6.7) and (10.7) the required
pressure
˛E
pD
T ;
1 2
(10.8)
where is the transverse contraction ratio (Poisson number).
Examples
Chapter 10
1. A steel rod (E D 120 GN=m2 , ˛ D 1610 6= ı C) with
the cross section A D 100 cm2 suffers a temperature
change T D 30 ı C. In order to prevent its expansion
a force F D 5:76 105 N is necessary.
2. A section of a railroad track of steel with L D 20 m
and ˛ D 16 10 6 = ı C expands for a temperature difference T D 40 ı C by L D ˛ L T D 1:3 cm.
For modern railroad tracks all sections are welded together at T D 20 ı C without gap. Without strong
mountings each section between the welding spots
would bend in such a way, that the length expansion L D ˛ LT could be realized. This would
give for T D 60 ı C a maximum deviation from the
straight line of about 30 cm. This bending is prevented
by strong supports where at every meter the rails are
mounted. The force on the welding surfaces with a
cross section A D 0:02 m2 (d D 10 cm, b D 20 cm)
is then (see Probl. 10.2) F D 1:5 106 N. When the
rail track cools down to T D 20 ı C a tensile force
of the same magnitude acts onto the welding surfaces
which corresponds to a tensile stress of 8 107 N=m2 .
This is still sufficiently far below the break stress of
J
7 108 N=m2 .
For the thermal expansion of liquids only the volume expansion
can be given. When measuring this volume expansion one has to
take into account that the solid container also expands. For the
measurement of thermal expansion of liquids a device proposed
by Dulong and Petit (Fig. 10.8) has been developed. The liquid
is contained in a U-shaped tube where one side is encased in a
jacket containing melting ice, while the other side is heated to
100 ı C by water vapour. Of course the inspected liquid should
not boil at 100 ı C and should not freeze at 0 ı C. Since the total
Water-ice
mixture 0°C
Boiling
water
100°C
Figure 10.8 Design of Dulong–Petit for the measurement of thermal expansion of liquids
mass M D % V of the liquid is constant, independent of the
temperature, the density
%0
%.TC / D
1 C TC
does depend on T: It is smaller in the hot side and larger in the
cold side. The height h0 of the liquid surface will be therefore
lower in the cold side by h. Rearrangement of the equation
yields
1
%0
%
1
D
1 D
:
TC %.TC /
TC %.TC /
From the equilibrium condition
gives
h0 %0 g D h.TC / %.TC / g
%0 =% D h=h0 :
The thermal volume expansion coefficient
D
is then
1 h
:
TC h0
(10.9)
Table 10.2 compiles some values of .
Note, that they are much larger than the volume expansion coefficients 3˛ of solids. This justifies the neglect of the glass tube
expansion for liquid thermometers.
10.1.3
Thermal Expansion of Gases;
Gas Thermometer
Experiments show that the volume of ideal gases (see Sect. 7.3)
increases at constant pressure proportional to the temperature.
V.TC / D V0 .1 C
V
TC / ;
(10.10)
where the temperature is measured in ı C and V0 D V.TC D 0 ı C/.
10.1 Temperature and Amount of Heat
is therefore determined by a pressure measurement. At the temperature TC D 0 which is realized by immersing the gas volume
into a melting ice-water mixture, the height difference is adjusted to h D 0. The pressure in the gas volume V is then
p D p0 .
Table 10.4 Thermal expansion coefficient of some gases
3
=.10 = K/
3.661
3.660
3.671
3.674
3.726
10.1.4
The expansion coefficient
D
V
V.TC / V0
V0 TC
(10.11)
gives the relative change V=V0 per 1 ı C. The experimentally
obtained numerical values of V are compiled in Tab. 10.4.
For Helium, which comes closest to an ideal gas, one finds
V
D
1 ı
C
273:15
1
D 3:661 10
3ı
C
1
with
p
p
TC /
D
V
D
D
1 ı
C
273:15
1
:
(10.12)
The gas thermometer (Fig. 10.9) utilizes this pressure dependence for the measurement of temperatures. The volume V is
connected with a U-shaped tube filled with mercury. The height
of Hg in the left side of the U-tube can be changed by up- and
down lifting of the right side, which is connected with the left
side by a flexible tube. When the gas volume is heated, the pressure rises. In order to keep the gas volume constant, the level
of the Hg in the left side is always kept at the same height. The
pressure is then indicated by the difference h between the left
and the right side. It is p D %Hg g h. The temperature,
obtained from (10.12)
1 p
p ı
p0
D 273:15
C
p0
p0
(10.14a)
relating pressure p, particle number N in the gas volume V with
T. The absolute temperature T was defined by Eq. 10.1, which is
based on the results of the kinetic gas theory (see Sect. 7.3). The
general gas equation (10.14a) states that for constant pressure
and temperature the gas volume has a definite value, which is
the same for all ideal gases independent of the specific kind.
At a temperature T0 (at 0 ı C) and a pressure p0 D 1 bar D
103 hPa (normal conditions) Eq. 10.14a becomes
(Law of Gay-Lussac).
TC D
We will now discuss the relation between the Celsius scale and
the absolute temperature scale, which is also called (after its
creator Lord Kelvin) Kelvin scale or thermo-dynamical temperature scale. In Sect. 7.3 the absolute temperature T was
used in the general gas equation
pV D NkT ;
:
Accurate experiments performed at a constant gas volume give
for the temperature dependence of the pressure the completely
analogue relation
p D p0 .1 C
Absolute Temperature Scale
(10.13)
(10.14b)
p0 V0 D N k T0 :
From (10.14a) and (10.14b) we can conclude
p D p0
V0 T
:
V T0
(10.14c)
In the gas thermometer the volume V D V0 is kept constant.
The comparison of (10.14c) with (10.12) yields
p D p0
T
D p0 .1 C
T0
This gives with the experimental value
lation
T D T0 .1 C TC / D T0 C
(10.15)
TC / :
D .273:15/
T0
TC
273:15
1
the re-
(10.16)
between the absolute temperature T and the Celsius scale TC .
Note: The unit oft the absolute temperature scale is the Kelvin.
It is the 273.16th part of the thermodynamic temperature of
the triple point of water.
Figure 10.9 Gas thermometer
A definition of the absolute temperature that is independent
of the specific substance, can be given with their help of the
Carnot-Cycle (see Sect. 10.3.5)
Chapter 10
Gas
Ideal Gas
He
Ar
O2
CO2
259
260
10 Thermodynamics
10.1.5
Amount of Heat and Specific Heat
Capacity
Handle
When a defined energy W is transferred to a body, its temperature rises by T W. A simple demonstration (Fig. 10.10)
uses an immersion heater which is immersed into water in a
thermally isolated Dewar flask and heated for a time t. The
electric energy W D I U t (I D electric current, U D
voltage, see Vol. 2 Chap. 2) causes a temperature rise T that
depends on the mass of the water. The increase Q of the heat
Q (often also called the amount of heat) is given by
Q D W D c M T :
In former times the unit was the large calorie (1 kcal). This is the
amount of heat that increases the temperature of 1 kg water from
14:5 to 15:5 ı C. Nowadays the unit is 1 Joule (1 J D 1 W s D
1 N m). It has the great advantage, that for the conversion of
heat into electrical or mechanical energy the same units are used
and therefore the conversion factor is 1. This is not the case, if
the unit calorie is used. Here measurements give the electrical
heat equivalent
Chapter 10
Q Œcal
D 0:23885 Œcal=W s :
Wel ŒW s
Thermometer
(10.17)
The proportional constant c is the specific heat. It depends on
the specific material of the heated body. It gives the amount of
heat that increases the temperature of a body with M D 1 kg by
T D 1 K. The product C D c M is the heat capacity of a body
with mass M.
WEel D
Ball
bearing
(10.18)
This equation means: If Wel is measured in Joule, but Q in
calories, the ratio Q=Wel has the numerical value 0:23885
i. e. 1 W s D 0:2389 cal or 1 cal D 4:1868 W s.
The temperature rise in the experiment shown in Fig. 10.10 does
not occur abruptly but continuously over the time interval t of
the heating (Fig. 10.10b). During this time interval, a steady
Figure 10.11 Measurement of mechanical heat equivalent
heat flux takes place between the hot water and its surrounding which decreases the temperature difference T. In order to
consider this, a temperature progression for a sudden change
of T is simulated, indicated by the vertical dashed line in
Fig. 10.10b. The time t1 is chosen such that the areas A1 and
RA2 are equal
R in order to maintain the same value for the integral
Tdt dQ D Q.
Instead of heat generation by electrical power, heat can be also
produced by mechanical work due to friction. This is demonstrated by the arrangement in Fig. 10.11. Here a metal tape is
wrapped around a copper cylinder, filled with water. A weight
with mass m presses the tape tightly onto the cylinder. Now the
cylinder with radius r is turned by a handle during a time interval t with such a frequency f that the weight G D m g is
just compensated by the friction force between tape and cylinder. The work performed against the friction force by turning
the cylinder N-times during the time interval t is
W D m g 2r N
D .cW MW C cCo MCo /T1 ;
(10.19a)
where MW is the mass of the water and MCo that of the copper cylinder. Repeating the experiment without water filling a
larger temperature difference T2 is measured. From these two
measurements we obtain from the relation
.cW mW C cCo mCo /T1 D cCo mCo T2 D W
for the heat, put into the water the relation
Q D cW MW T1 D 1
T1
T2
Wmech :
(10.19b)
The mechanical heat equivalent, determined with such experiments is
Figure 10.10 Measurement of the electric heat equivalent with immersion
heater and Dewar flask. a Experimental setup; b time progression of electric
power and temperature
WEmech D
Q=cal
D 4:186 ;
Wmech =Nm
(10.19c)
10.1 Temperature and Amount of Heat
261
as 12 g carbon 12 C. The molar mass of a substance y X with the
atomic mass number y is then equal to .y=12/ m.12 C/g.
The molar volume VM contains 1 mol of the gas.
Examples
A2
1 mol helium gas He are 4 g He,
1 mol hydrogen gas H2 are 2 g H2 ,
1 mol nitrogen N2 are 28 g N2 .
A1 = A2
J
A1
The number of atoms or molecules per mole is the Avogadro
constant NA . This number is independent of the specific substance.
should of course have the same mechanical value as the electrical heat equivalent, because of the definition 1 W s D 1 N m
in the SI system.
The specific heat cK of a body can be measured with the mixture
calorimeter shown in Fig. 10.12. In a well-isolated Dewar container is water with the mass MW at the temperature T1 . Now
a solid body with the mass MK that had been heated up to the
temperature T2 > T1 is inserted into the water. The temperature
TM .t/ of the mixed system (body + Dewar + water) is measured
as a function of time (Fig. 10.12b).
It can be measured with different methods (see Vol. 3, Chap. 2).
The average value of many measurements is
NA D 6:022 1023=mol :
One mole of atoms or molecules always fills the same volume
under equal external conditions, independent of their specific
kind. One finds under normal conditions
VM .p D 1 atm D 101:3 kPa; TC D 0 ı C/ D 22:4 dm3
VM .p D 1 bar D 100 kPa; TC D 0 ı C/ D 22:7 dm3 :
The general gas equation (10.14a) can be written for 1 mol with
V D VM and N D NA
The heat (cK MK .T1 TM / transferred from the body to water
plus Dewar is equal to the heat change cW MW C cD MD of
water plus Dewar. This gives the specific heat of the body
cK D
.MW cW C CD /.TM
MK .T2 TM /
T1 /
(10.21)
where the general gas constant
;
(10.20)
where CD D cD MD is the heat capacity of the dewar. The temperature TM of the mixture is determined in the same way as in
Fig. 10.10. The measured curve T.t/ in Fig. 10.12b is replaced
by the simulated red curve where the vertical line is placed at
the time t1 where the areas A1 D A2 . This takes into account the
heat loss during the heat transfer form body to water. The intersection points of the vertical line with the two horizontal red
curves give the correct temperatures T1 and TM .
The heat capacity CD D cD MD of the Dewar can be measured when two portions of water with masses M1 and M2 at
temperatures T1 and T2 are mixed in the Dewar and the mixing
temperature TM is measured [10.1].
10.1.6
p VM D NA kT D R T ;
Molar Volume and Avogadro Constant
One mole is according to the definition given in Sect. 1.6 the
amount of a substance that contains as many atoms or molecules
R D NA k D 8:31 J=.K mol/
(10.22)
is the product of Avogadro number NA and Boltzmann constant
k. All gases that obey this equation are called ideal gases.
For an arbitrary volume V D VM Eq. 10.14a can be written
as
pV DRT ;
(10.21a)
where the number quotes how many moles are contained in V.
10.1.7
Internal Energy and Molar Heat
Capacity of Ideal Gases
The amount of heat Q supplied to one mole of a gas with molar
mass M (kg=mol) leads to a temperature rise T:
Q D c MM T D C T :
Chapter 10
Figure 10.12 Measurement of specific heat cK of a solid body using a mixture
calorimeter. a Experimental setup; b measurement of time-dependent temperature (t1 D immersion of solid body, T1 D initial temperature, TM D temperature
of mixture)
262
10 Thermodynamics
The product C D c MM of specific heat and molar mass is
the molar heat capacity with the unit ŒC D ŒJ=.mol K/. It
is the heat energy that increases the temperature of 1 Mole by
T D 1 K. For an arbitrary mass M D MM is
Q D C T :
The quotient Q=T D C ŒJ=K is the heat capacity of the
body with mass M.
The molar specific heat of a gas depends on whether the gas is
heated at constant volume or at constant pressure. We will at
first discuss the situation for a constant volume.
We define the internal energy of a gas with volume V as the
total energy of its N molecules. It is composed of translational
energy plus possible rotational and vibrational energy. For nonideal gases also the potential energy of their mutual interaction
has to be taken into account (Fig. 10.5). The internal energy
of a gas depends on the number f of degrees of freedom of the
molecules. In Sect. 7.3 it was shown that the mean energy of a
molecule is hEi D f 21 kT. The internal energy of a gas volume
with N molecules is then
(10.23)
Chapter 10
Under thermal equilibrium the energy U is uniformly distributed among all degrees of freedom.
This equipartition is accomplished by collisions between the
molecules (see Sect. 4.2 and Vol. 3, Chap. 8).
When the heat Q is supplied, the internal energy U increases
by U D Q, if the volume V of the gas stays constant. We
therefore obtain the equation
Q D CV T C p V :
(10.27)
The general gas equation for 1 mol of the gas before and after
the expansion is
p V D R T;
p.V C V/ D R .T C T/ :
Subtraction yields
p V D R T :
Inserting this into (10.27) we obtain
Q D .CV C R/T D Cp T :
(10.24)
and with U D 12 f R T the molar heat capacity at
constant volume
CV D 21 f R :
(10.26)
This energy W must be supplied additionally. The heat Q is
therefore
and for 1 mol with N D NA it is
Q D U D CV T ;
realized when the piston with area A moves by the distance x
against the external pressure p. This requires the work
W D F x D p A x D p V :
U D 21 f N kT
U.VM / D 21 f NA kT D 12 f R T :
Figure 10.13 Determination of Cp . Heating of gas a at constant volume, b at
constant pressure
(10.28)
The factor Cp is the molar specific heat at constant pressure
Cp D CV C R :
(10.25)
(10.29a)
With
CV D 12 f R ! Cp D 21 .f C 2/R ;
10.1.8
Specific Heat of a Gas
at Constant Pressure
When a gas is heated at constant volume the pressure increases
according to the general gas equation (10.14a). In order to
achieve a temperature increase at constant pressure, the gas volume must expand (Fig. 10.13b). Such an expansion can be
(10.29b)
the quotient Cp /CV is called adiabatic index or specific heat
ratio
D
Cp
f C2
D
:
CV
f
(10.29c)
10.1 Temperature and Amount of Heat
10.1.9
Molecular Explanation of the Specific
Heat
263
a
Since atoms or molecules can move into three directions, they
have three degrees of freedom of translation. Their mean translational energy is therefore
Etrans D 3 12 kT ;
b
the molar specific heat of atomic gases is therefore
For molecules the supplied energy can be also converted into
rotational or vibrational energy. Nonlinear molecules can rotate
around three orthogonal axis. They have therefore three degrees
of freedom for the rotation. Linear molecules have only 2 rotational degrees of freedom, because of the following reason:
The rotational energy
Erot D L2 =2I
is determined by the angular momentum L and the moment
of inertia I (see Sect. 5.5). As shown in Quantum physics
(see Vol. 3 Chap. 4) the angular momentum has the amount
L D .l .l C 1//.1=2/ „ with p
l D 1I 2I 3I : : :. The smallest angular
momentum is then Lmin D 2 „, where „ D h=2 is Planck’s
quantum constant, divided by 2. The moment of inertia for
a rotation around the axis of a linear molecule is very small
because the heavy nuclei are located on the axis and the light
electrons do not contribute much to I. Therefore the rotational
energy is very large, generally much larger than the translational
energy at accessible temperatures. Collisions cannot excited this
rotation and it therefore cannot contribute to the accessible energy.
The vibration of diatomic molecules is one-dimensional and
has therefore only one degree of freedom. However, the vibrational energy has two contributions: The kinetic and the
potential energy (see Sect. 11.6). The mean value of both contributions is equal to 21 kT and the thermal energy of the vibration
is kT. Therefore two degrees of freedom .f D 2/ are formally attributed to the vibration. A diatomic molecule has then
f D 3 C 2 C 2 D 7 degrees of freedom, if the temperature is
sufficiently high to excite the vibrations.
Note: Quantum Theory shows (see Vol. 3) that the classical
model of a vibrating oscillator with regard to the total energy
E D Ekin C Epot is correct, but that the energy can be only absorbed in discrete quanta h . This does, however, not influence
our argumentation above.
Figure 10.14 Excitation of rotational degrees of freedom of a diatomic
molecule induced by collisions. The collision with an atom A causes a rotation of the molecule about an axis perpendicular to the drawing plane (a), or
causes the excitation of a molecular vibration (b)
Since collisions can transfer translational energy to rotations or
vibrations only if the thermal energy of the collision partners
is sufficiently high, at low temperatures only translational degrees of freedom are available and feff D 3. With increasing
temperature at first the rotation can be excited (feff D 6 resp.
5 for linear molecules) and at still higher temperature also the
vibrations contribute to the specific heat, because their energy is
higher than that of the rotations (feff D 3 C 3 C 2 .3j 6/ resp.
3 C 2 C 2.3j 5/ for linear molecules). This gives for diatomic
molecules feff D 3 C 2 C 2 D 7.
The molar specific heat is
CV D
@U
@T
V
D
1
feff R :
2
(10.31)
Here the partial derivative is used, because U can depend on
several variables (p, V, T). The index V indicates that the energy
supply occurs at constant volume.
Examples
1. For the atomic gas Helium is f D 3. Since the translational energy is not quantized all three degrees of
freedom are excited even at low temperatures. Therefore the specific heat of Helium is independent of the
temperature (Fig. 10.15).
For polyatomic molecules with j atoms each atom has three degrees of freedom. If we subtract 3 degrees of freedom for the
translational motion of the whole molecule and 3 degrees of
freedom for the rotation (2 degrees for a linear molecule) we
end up with fvib D 3j 6 (3j 5 for linear molecules) vibrational degrees of freedom.
The total internal energy U of a molecule with j atoms is then
UD
1
2
f NA kT
with
f D ftrans C frot C fvib :
(10.30)
Figure 10.15 Temperature dependence of molar heat capacity of helium, nitrogen N2 and nitrogen dioxyde NO2 (see also Sect. 10.1.10)
Chapter 10
CV D .3=2/R :
264
10 Thermodynamics
2. For nitrogen gas consisting of the diatomic molecules
N2 kT is for very low temperatures smaller than the
energy of the lowest rotational level (see Probl. 10.3).
The rotational energy cannot be excited. Therefore
feff D 3. With increasing temperature when kT
Erot feff approaches the value feff D 5. For still higher
temperatures kT Evib feff becomes feff D 7 because
the vibrational degrees of freedom are counted twice
(see Sect. 10.1.9). The specific heat of a molecular gas
is therefore dependent on the temperature and reaches
its maximum value only if kT is sufficiently high to
excite all degrees of freedom.
3. Polyatomic gas (e. g. NO2 at T > 200 K where NO2
has a sufficiently high gas pressure). At this temperature already all three rotational modes can be excited.
We then obtain f D 3 C 3 D 6.
Above T D 300 K the bending vibration can be excited, rising f to f D 8. Only above T D 800 K
all three vibrational mods can be excited and we have
f D 12. The molar specific heat is then CV D 6R. J
10.1.10
Specific Heat Capacity of Solids
Chapter 10
With decreasing temperature all gases become liquids and pass
finally into the solid state (except helium, which solidifies only
under high pressures). Considerations about the specific heat of
solids require a knowledge about the possible degrees of freedom for atoms and molecules in solids. Since the atoms in a
solid can only oscillate in three directions around their fixed
equilibrium positions but have no degrees of freedom for translation or rotation we would expect f D 2 3 D 6. However,
it turns out that the oscillation frequencies of all atoms are not
the same but spread over a large frequency range. In order to
get an idea about the frequency distribution, we regard a onedimensional arrangement of atoms in an ideal crystal where
all atoms are regularly placed at a distance d (Fig. 10.16a).
When an atom oscillates around its equilibrium positions, the
oscillation can be transferred to the neighbouring atoms, due
to the coupling force between the atoms. This results in elastic waves travelling through the crystal (see Sect. 11.8). The
waves are reflected at the end faces of the crystal, superimpose
with the incoming waves and form stationary standing waves.
Longitudinal as well as transversal standing waves can develop,
depending on whether the oscillation occurs in the direction of
wave propagation or perpendicular to it. The standing wave with
the smallest possible wavelength (i. e. the highest frequency
D c=) is realized, when the neighbouring atoms oscillates
against each other (Fig. 10.16a, b lowest line). The oscillation
with the largest possible wavelength ( D L with L D length of
the crystal) has the lowest energy h. At low temperatures only
those vibrations with the lowest energy can be excited. With increasing temperature more and more vibrations can be excited.
The number of possible vibrations Z N 3 is proportional to
Figure 10.16 Stationary vibrational modes of a linear chain. a Transversal;
b longitudinal standing waves. c Number of vibrational modes per energy interval dE of a solid body as function of temperature T
the third power of the number N of atoms in the crystal. This
means that the specific heat rises continuously with the tempermax
all vibrations are excited
ature (Fig. 10.16c) until at kT Evib
and the specific heat takes its maximum value. Since the interaction between neighbouring atoms depends on the specific
kind of atoms the progression C.T/ differs for the different materials (Fig. 10.17). However, all curves C.T/ approach for high
temperatures the same value of the molar specific heat
CV D 6 12 NA k D 3R
(Dulong–Petit law) :
(10.32)
Measurements of the temperature-dependent progression of
CV .T/ gives information about the distribution of the vibrational
frequencies and therefore about the coupling forces between the
atoms of the solid. They are furthermore a convincing experimental proof of quantum theory (see Vol. 3).
Table 10.5 gives numerical values of CV for some materials.
Figure 10.17 Qualitative temperature dependence of molar heat capacity of
different solids
10.1 Temperature and Amount of Heat
265
Table 10.5 Specific heat c of some materials at 20 ı C and 1013:25 hPa, specific heat of fusion f and heat of evaporation e
10.1.11
c=kJ kg 1 K
4.182
2.43
0.14
0.896
0.45
0.13
0.383
2.1
1
f =kJ kg
333.5
105
12.4
397
277
65
205
332.8
1
e =kJ kg
2256
840
285
10,900
6340
16,500
4790
–
1
Fusion Heat and Heat of Evaporation
When a constant heat power dQ=dt is supplied to a container
with 1 kg ice (specific heat ci ) at a temperature TC < 0 ı C the
temperature
T.t/ D Ti C a t
with
a D .dQ=dt/=ci
Figure 10.19 Energy distribution N .E / and mean total energy E of molecules
just below the melting temperature, illustrated by the interaction potential V .r /
Therefore the supplied energy .dQ=dt/ .t2 t1 / must have increased the potential energy of the atoms or molecules. This can
be explained by the molecular model as follows:
(10.33)
rises linearly with the slope a D .dQ=dt/ ci , (ci D specific heat
of ice) up to Tm D 0 ı C at t D t1 . Here the temperature stays
constant until t2 , when the ice is completely molten, in spite of
a constant power supply dQ=dt (Fig. 10.18). Then the temperature rises again but with a different slope b D .dQ=dt/=cW
(cW D specific heat of water) up to T D 100 ı C at t D t3 , where
the water starts to boil (at p D bar 1). Again the temperature remains constant until t D t4 when the whole water is evaporated.
Then the temperature rises further with the slope .dQ=dt/=cvap.
The energy dQ=dt/ .t2 t1 / supplied during the melting process
is called fusion heat, the energy dQ=dt/ .t4 t3 / is the heat of
evaporation.
The energy f D .dQ=dt/=m ŒJ=kg necessary to melt 1 kg of a
substance is the specific fusion heat while the molar fusion heat
is labelled by f ŒJ=mol. Analogue label e and e the specific
and the molar heat of evaporation.
Since the temperature has not changed during the melting
process, also the kinetic energy must have stayed constant.
Figure 10.18 Temperature T .t / of water under constant energy supply within
the temperature range from below the melting temperature up to above the
evaporation temperature (ice–water–water vapor)
The molecules in a solid body are bound to the equilibrium
positions by attractive forces. This means that all molecules
vibrate
Pin a potential V.r/ that is determined by the vector sum
FD
Fi D rV.r/ of all forces. At the melting temperature
the mean total energy E D Ekin C Epot is illustrated in Fig. 10.19
by the horizontal line close to the dissociation energy ED of
the interaction potential between the atoms or molecules. The
energy distribution N.E/ of the molecules follows a Maxwell–
Boltzmann distribution, as depicted in Fig. 10.19. Those
molecules with E > ED can leave their fixed equilibrium position without changing their kinetic energy.
Continuous energy supply increases the number of molecules
that leave their fixed position until all molecules can freely
move: The solid body has dissolved and has become a liquid.
The energy supplied during the melting process keeps the
kinetic energy constant but increases the potential energy.
An analogous process occurs during the evaporation process.
The molecules from the higher energy part of the Maxwell–
Boltzmann-distribution have sufficient energy to leave the liquid
against the attractive forces and enter the vapour phase. Since
the density of the vapour at atmospheric pressure is about 3
orders of magnitude smaller than that of the liquid, the mean
distance between the molecules if about 10 times larger. The
negative potential energy of the mutual attraction is therefore in
the gas phase negligible against their kinetic energy. Similar to
the melting process the supplied energy increases the potential
energy but not the kinetic energy because the temperature remains constant. The potential energy increases from a negative
value (work function, surface tension, see Sect. 6.4) to nearly
zero.
The numerical values of fusion energy and evaporation energy
depend on the substance. In Tab. 10.5 the values for some materials are listed.
Chapter 10
Substance
Water
Ethanol
Mercury
Aluminium
Iron
Gold
Copper
Ice at 0 ı C
266
10 Thermodynamics
10.2
Heat Transport
Always when a temperature difference exists between two different locations, heat is transported from the warmer to the
colder region (see Sect. 7.5.3) Such a heat transport is very
important for many technical problems and also for different
measuring methods. In many cases one tries to maximize heat
transport (for example for cooling heat generating systems) in
other cases it is minimized (for heat isolating devices such as
Dewars or refrigerators).
There are essentially three mechanisms of heat transport: Convection, heat conduction and thermal radiation.
10.2.1
Convection
When the bottom of a container with water is heated,
(Fig. 10.20) the lowest liquid layer is heated first. Its temperature increases and its density therefore decreases. This causes
a rise of this lower layer through the layers across, which sink
down. This process is called convection. It results in a heat
transport from the warmer to the colder region. This convection of liquids can be demonstrated by colouring the lower layer
and observing how this coloured layer moves upwards when the
bottom of the container is heated.
Chapter 10
Convection occurs also for gases. It plays an essential role in
the earth atmosphere and is responsible for the generation and
the equalization of pressure differences (Fig. 10.21). Heated air
rises from the bottom just above the earth surface, creating a local low pressure region. Air from the surrounding with higher
pressure streams into this region. The wind transports not only
mass but also heat [10.2a, 10.2b]. This mass- and heat transport
depends on the wind velocity and the temperature difference between high and low pressure region. The wind flow can be either
laminar or turbulent, depending on the boundary conditions.
Although the total energy received by the earth is due to radiation from the sun, the local distribution of this energy is
essentially determined by convection. This is illustrated by sudden local temperature changes when the wind direction changes,
although the intensity of the sun radiation has not changed.
Figure 10.20 Convection in a liquid. a Lamination of dyed and pure water at
equal temperatures; b mixing of the layers by convection due to heating at the
bottom
Figure 10.21 Convection in the earth atmosphere. a Onshore wind during the
daytime, when the ocean surface is colder than the land surface; b wind flow
into a low pressure region which is created by uprising air (thermal lift)
Also the temperature distribution in the oceans is mainly determined by convection. Examples are the gulf-stream, which
influences the climate in the northern part of Europe, or the
Humboldt current along the west coast of south America, causing the Atacama desert, because the cold water induces the
moisture of the west wind to rain down before it reaches the
dry areas.
When the temperature gradient of a liquid, heated at the bottom, exceeds a certain value that depends on the viscosity of the
liquid, ordered macroscopic structures of the velocity field can
develop. Current roles are created and the liquid moves along
cylindrical stream lines (Fig. 10.22b). This sudden start of a
Figure 10.22 a Linear vertical temperature gradient; b Bénard instability
10.2 Heat Transport
10.2.2
Heat Conduction
Contrary to the situation for convection, for heat conduction
only energy transport takes place, but generally no mass transport. Heat conduction can only occur in matter, i. e. in vacuum
no heat conduction is possible while thermal radiation also takes
place in vacuum (otherwise we would not receive the sun radiation).
We will at first discuss heat conduction in solids, where the
atoms or molecules are bound to fixed equilibrium positions and
no convection can happen.
10.2.2.1
Heat Conduction in Solids
A rod with length L and cross section A is connected at both
ends with thermal reservoirs that keep the two ends always at
the fixed temperatures T1 and T2 with T1 < T2 (Fig. 10.23).
After a sufficient long time a stationary state appears, where a
temperature gradient dT=dx is established that depends on the
temperature difference T D T1 T2 , and on the length L. If
we neglect heat losses through the side wall of the rod, a constant heat energy
dQ
dT
D A
(10.34a)
dt
dx
Table 10.6 Heat conduction coefficient of some materials at 20 ı C
Substance
Aluminium
Iron
Gold
Copper
Zinc
Lead
Normal concrete
Foamed concrete
Glas
Glas wool
Wood
Ice
Water
Air (p D 1 atm)
CO2 (p D 1 atm)
Helium (p D 1 atm)
=.W m 1 K 1 /
221
67
314
393
112
35
2.1
0.22
0.8
0.04
0.13
2.2
0.6
0.026
0.015
0.14
The integration constant C is determined by the boundary condition T.x D 0/ D T1 D C. The energy supply necessary
to maintain the given temperature gradient dT=dx, is obtained
from
dQ=dt D A .T1 T2 /=L :
For the general nonstationary heat conduction through inhomogeneous bodies with variable cross section the temperature
function T.x; t/ is more complicated. For its derivation we regard a volume element dV between the planes x D x1 and x2
(Fig. 10.24).
flows per sec through the cross section A of the rod . The constant .Œ D ŒW m 1 K 1 / depends on the substance of the
rod and is called heat conductivity. In Tab. 10.6 the heat conductivities of some substances are listed.
For a homogeneous rod with constant cross section A, the stationary temperature T(x) is a linear function of x, as can be seen
by integrating (10.34a), which yields
T.x/ D
dQ=dt
xCC :
A
Figure 10.23 Heat conduction in a stab
(10.34b)
Figure 10.24 Derivation of the heat conduction equation
Chapter 10
“self-organization” is called Bénard-instability. When the liquid is further heated, the rolls begin a wave-like motion along
the cylinder axis. Such organized motions that develop from
random conditions play an important role for the creation of organized structures from disordered systems. These processes
are investigated in the rapidly developing field of Synergetics,
which represents a frontier area between Physics, Chemistry,
Biology and Computer Science [10.3, 10.4].
267
268
10 Thermodynamics
For the one-dimensional case (for instance a thin homogeneous
rod) the temperature changes only in one direction and we obtain for the heat power, transported through the cross section A
at the position x1
dQ1
@T
D A
:
dt
@x
(10.35)
The partial derivative is used here, because the temperature
T.x; t/ depends on the two variables x and t. At the plane
x2 D x1 C dx the temperature has changed to
T.x2 / D T.x1 / C .@T=@x/ dx :
The heat passing per sec through the plane at x D x2 is
@T
@
dQ2
TC
D A
dx :
(10.36)
dt
@x
@x
When the temperature is higher at x1 than at x2 , the heat dQ1 =dt
flows per sec from the left side in Fig. 10.24 into the volume
dV D A dx, and the heat dQ2 =dt leaves it per sec to the right
side. The change dQ=dt of the heat per second in the volume dV
is then
dQ
dQ1 dQ2
@2 T
D
D 2 A dx
dt
dt
dt
@x
(10.37)
@2 T
D 2 dV :
@x
Because dQ D c m dT and m D % dV this net supply of heat
power dQ changes the temperature T according to (10.37) by
Chapter 10
@T
@2 T
:
D
@t
% c @x2
(10.38a)
If the rod has heat losses H D dQV =dt through the side walls
(for example through cooling by the surrounding air) a loss term
H D h.T T0 / has to be added to (10.37) which is proportional
to the temperature difference between the rod temperature at the
position x and the surrounding temperature T0 . The factor h has
the unit ŒW K 1 . Equation 10.38a can then be generalized with
h D h=.c m/ as
2
@T
@ T
D
@t
% c @x2
h .T
T0 / :
(10.38b)
If T depends also on y and z all net heat power contributions
supplied from all directions to the volume element dV add to
the total energy increase of dV. One obtains for this threedimensional case the general equation for the heat conduction
2
@T
@T
@2 T
@2 T
C 2 C 2
D
@t
c % @x2
@y
@z
D
T D T T ;
c%
(10.39)
with the Laplace operator (see Sect. 13.1.6). The factor T D
.=c %/ is the thermal diffusitivity.
Figure 10.25 Demonstration of the different heat conduction of some metals
The heat conduction in solids is accomplished by the coupling
between adjacent atoms, which causes the transport of the vibrational energy of atoms at the plane x to the neighbouring position
x C dx without a transport of the atoms themselves.
In metals the freely moving electrons contribute essentially to
heat conduction by collisions with each other and with the
atoms. Because of their small mass, their thermal velocities
and in particular their Fermi-velocities (which can be only explained by quantum theory, see Vol. 3) are very high. They can
therefore transfer their large kinetic energy much faster by collisions. The heat conductivity in metals is therefore mainly due
to the electrons. Experiments confirm in deed that for metals the
thermal conductivity () is proportional to the electrical conductivity (), which is solely caused by electron transport.
This is expressed by the Wiedemann–Franz law
= D a T
with a D 2 k2 =3e2
D 2:45 10 8 V2 =K2 ;
(10.40)
where the constant a is determined by the Boltzmann constant k
and the elementary charge e.
This can be readily demonstrated by a simple experiment
(Fig. 10.25). The red centre plate of a cross with four arms of
different metals is heated by a small burner. At the ends of the
arms four matches are placed. After the heating starts it takes
different times ti until the ends of the arms reach the ignition
temperature. The matches are ignited at times t1 < t2 < t3 < t4 .
This time sequence reflects the electrical conductivity of the
metals, where the arm 1 (Cu) has the highest electrical and thermal conductivities.
In solids (even in non-metals) the thermal conductivity is much
larger than in gases, because of the much larger density and the
resulting larger coupling strength between neighbouring atoms
(see Tab. 10.6). However, the coefficient of heat conductivity
T D .c %/, which gives the time constant of reaching a
stationary temperature, is for solids and gases nearly the same
because of the much smaller density % of gases.
10.2 Heat Transport
269
In gases temperature differences are equated in times comparable to those in solids.
One of the reasons is the much smaller heat energy to reach a
temperature rise T for a given volume of a gas than for the
same volume of a solid.
For the measurement of heat conductivities stationary as well as
time resolving techniques have been developed [10.5].
For the stationary methods a constant heat power dQ=dt is supplied to one end of the body (for instance a rod), which is
extracted on the other side by cooling. According to (10.35)
this results for a rod with constant cross section A in a constant
temperature gradient
(10.41)
which can be determined by measuring the temperatures T1 and
T2 and the length L.
The dynamical methods for the measurement of the heat conduction under non-stationary conditions are based on a timedependent supply of the heat power. The heat power dQ=dt is
either periodically modulated or supplied in short pulses. If for
example the heat power supplied at x D 0 is
dQ=dt D dQ0 =dt C a cos.!t/ ;
the temperature at x D 0 is
and one obtains from the heat conduction equation (10.38b) for
a thin cylindrical rod with heat losses h .T T0 / through the
side walls (Fig. 10.26) the solution
T.x; t/ D T0 C .T1 T0 /e ˛1 x
C Te ˛2 x cos.!t kx/ :
(10.42)
Inserting this into (10.38b) yields for the coefficients
˛1 D
˛2 D
kD
%ch
D
the frequency ! and the wavelength T of the temperature wave.
The phase velocity of the wave
vPh
1=2
2! 2 T
!
D
D
k
.! 2 C h2 /1=2 C h
(10.42a)
depends on the frequency !.
Temperature waves show dispersion!
T.0; t/ D T1 C T cos.!t/ ;
r
Figure 10.26 Damped temperature waves in a stab under periodic energy supply
s
h
;
T
Without heat losses (h D 0) (10.42) reduces to
T.x; t/ D T1 C Te
with
˛x
cos.!t
kx/
1=2
˛ D k D .!=2T /
:
(10.42b)
10.2.2.2 Heat Conduction in Liquids
.h2 C ! 2 /1=2 C h
2T
.h2 C ! 2 /1=2
2T
When amplitude and phase of temperature waves are measured
at selected points x for different frequencies !, the quantities h
and T D =% c can be obtained.
h
1=2
1=2
I
:
The temperature T.x/ along the rod is a superposition of a constant time-independent contribution that decays exponentially
with x due to the heat losses through the sidewalls, and a damped
temperature wave with an exponentially decreasing amplitude.
The phase of this wave is determined by the loss coefficient h,
In liquids, there are no shear forces (see Sect. 6.2). Therefore,
the coupling between neighbouring atoms is much weaker than
in solids and the heat transport is slower. The heat conduction in liquids that have no electrical conductivity, is therefore
smaller than in solids (see Tab. 10.6). However, in liquids the
freely moving molecules can transfer energy by collisions. The
effective energy transfer depends on the mean velocity of the
molecules, the time between two collisions and the cross section for energy transferring collisions.
In electrically conducting liquids (for example mercury or
melted metals) the free electrons make the major contribute to
Chapter 10
@T
T1 T2
1
dQ
D const D
D
;
@x
L
A dt
270
10 Thermodynamics
Tungsten wire
Water vapor
Grid
Figure 10.28 Demonstration of heat conduction in gases and its dependence
on the molecular mass
Ice
Figure 10.27 Demonstration of the small heat conductivity of water
the energy transfer, similar to the situation in solids. Their heat
conductivity is therefore much larger than for non-conductive
liquids, because of their much larger mass, the ions give only a
minor contribution.
The influence of the molecular mass m on the heat conduction
can be demonstrated by the device shown in Fig. 10.28. A tungsten wire runs coaxially through two separated parts of a glass
tube, which are filled with hydrogen gas in the left part and nitrogen gas in the right part. When the wire is heated by an electric
current the right part has a higher temperature and is glowing
red while the left part remains much colder due to the different heat conduction of the two gases. When removing the two
gases both parts of the wire glow equally strong. This effect is
intensified by two causes:
Because of the free mobility of the molecules in liquids, generally convection occurs besides heat conduction. This can be 1. The electrical resistance R of tungsten decreases with deprevented when the liquid is heated from above, because then
creasing temperature T. Therefore the electrical power
the hot liquid layer has a smaller density and therefore remains
dW=dt D I 2 R supplied to the wire is smaller in the cold
at the top.
part.
2.
The
visible radiation power of the glowing wire is proporThe small heat conductivity of water can be demonstrated by
tional
to T 4 . Even a small change of the temperature T
the experiment shown in Fig. 10.27. At the bottom of a glass
results in a large change of the radiation power.
tube filled with water, are small ice cubes which are prevented
Chapter 10
from uprising by a mesh. One can heat the upper part of the
water until it boils and emits water vapour. Nonetheless, the
ice cubes do not melt in spite of the temperature difference of
T D 100 ı C between the lower and upper part of the tube due
to the poor heat conductivity of glass and water and the absence
of convection.
10.2.2.3
Heat Conduction in Gases
In Sect. 7.5 it was shown, that heat conduction in gases is caused
by collisional energy transfer between the molecules which
move with thermal velocities. According to Eq. 7.49 the heat
energy transferred per m2 between two parallel walls at temperatures T1 and T2 is
Jw D .T1
A modification of the demonstration experiment (Fig. 10.29)
uses a vertical glass tube with the coaxial wire, which is filled
with a gas mixture of H2 and N2 . At first the heated wire glows
equally bright along the whole tube. After some minutes, the
lighter H2 -gas diffuses to the upper part while the heavier N2 gas sinks to the bottom (see Sect. 7.6 and Fig. 7.34). This effect
is even amplified by convection where the hot gas around the
wire rises up while the colder gas close to the inner wall of the
glass tube sinks down. Now the lower part of the wire is brighter
than the upper part.
T2 / :
According to (7.49a) the heat conduction coefficient is
˛ D n v k f =8 n
p
T=m :
Uniformly mixed
Demixed
Because of the much smaller density n of gases compared to
liquids the heat conduction in gases is generally much smaller,
except for ionized gases where the electrons contribute essentially to heat conduction. For neutral gases it is maximum for
hydrogen because of the small mass m of hydrogen molecules.
When the mean free path is larger than the dimensions of the
gas container, the heat conduction becomes independent of the
gas pressure.
Initial
Later
Figure 10.29 Demixing of a gas mixture by convection and diffusion in the
gravitation field of the earth
10.2 Heat Transport
10.2.3
The Heat Pipe
Often the problem arises that heat produced in a volume V
should be extracted as effectively as possible, in order to reach
a sufficient cooling power. For the solution of this problem a
special device was developed, which uses evaporation of a liquid on the hot side and condensation of the vapour on the cold
side. The heat transport occurs by convection. This heat pipe
allows a heat transport through the unit area that is larger by two
orders of magnitude than can be achieved with metals. Its basic
principle is illustrated in Fig. 10.30.
A tube of metal or another material is connected at the hot
side (left) with the volume at the temperature T1 that should be
cooled and on the cold side with a cooling bath at T D T2 < T1 .
The evacuated tube is filled with a substance that has an evaporation temperature Te < T1 and a melting temperature Tm < T2 .
For instance, if water is used, the temperatures should be T1 >
100 ı C and T2 > 0 ı C.
At the hot side the substance boils which extracts the evaporation heat from the volume to be cooled. The vapour streams to
the cold end where it condenses and delivers its heat of fusion to
the cooling bath. Along the tube a gradient of the vapour density
develops and an opposite gradient of the liquid density. An essential part of the heat pipe is a mesh that is wrapped around the
inner part of the tube close to the wall. For the correct choice of
the materials for tube and mesh the liquid substance wets both
the mesh and the inner wall of the tube. Due to capillary action
the liquid then flows between mesh and wall from the fusion
zone back to the evaporation zone where it can be again evaporated and extract heat. The heat transport of this cyclic process
depends on the vapour density and its flow velocity from the hot
to the cold zone, but mainly on the magnitude of evaporation
and fusion energy. For cooling media with a large evaporation
energy (for example water) and a large convection velocity a
very large heat transport per sec can be achieved.
Figure 10.30 Heat pipe
With the mass dm=dt evaporated per sec the energy extracted
per sec from the hot volume is
dW=dt D e dm=dt ;
where e is the specific evaporation energy. This is much larger
than the heat cp .dm=dt/T transported through the heat pipe
with a temperature difference T D T1 T2 (see Tab. 10.5). At
the cold end the heat
dW2 =dt D .f Ccp T/ dm=dt
has to be transferred to the cooling water.
More details about the technical design and the applications of
heat pipes can be found in [10.6].
10.2.4
Methods of Thermal Insulation
While in Sect. 10.2.3 the realization of devices with a maximum
heat transport was discussed, in this section we will treat methods to prevent heat transport out of a volume or to make it at
least as small as possible. In order to reach this goal one has
to take into account the contributions of all three heat transport
mechanisms and minimize them. We will illustrate this by consider the thermal isolation of a residential house.
The heat transport between the inside and outside is mainly governed by heat conduction through walls and windows and to a
minor part by air convection through leaky joints and during airing of a room. Depending on the size and the technical features
of the windows also heat radiation can be important for heat exchange.
The heat flux through the area A of walls or windows with thickness d and a temperature difference T D Ti To between
inside and outside is
dQ=dt D
.=d/ A T ;
(10.43)
where ŒW=.m K/ is the heat conductivity, which depends
on the material. It is generally characterized by the constant
k D =d, which gives the energy flux through the unit area
A D 1 m2 at a temperature difference T D 1 K. For most estimations of the heat isolation of houses the k value ŒW=.m2 K/
of walls and windows is given. For good heat insulation it
should be as small as possible. In Fig. 10.31 the k-values and
the temperature rise from an outside temperature T0 D 15 ı C
to the room temperature Ti D C20 ı C are depicted for different wall compositions. These figures illustrate, that even a
thin layer of Styrofoam considerably improves the thermal insulation. The largest heat losses are caused by the windows,
where the heat transport process is more complex. We regard at first a single-layer window (Fig. 10.32). In spite of the
small heat conductivity of glass ( D 0:9 W=.mK/) the k-value
k D 200 W=.m2 K/ is much larger than that of the thick walls,
due to the small thickness (d D 4 mm) of the window.
Because of the temperature gradient in the air layers close to the
inside and outside of the glass a convective air current develops,
Chapter 10
For most situations the convection in gases at atmospheric pressure gives a larger contribution to the energy transport than heat
conduction.
271
272
10 Thermodynamics
which is for To < Ti downwards at the inside and upwards at
the outside (Fig. 10.32). Due to friction between glass and air
a thin boundary layer of air adheres on both sides of the glass
(see Sect. 8.4). The heat passes through these layers to the convective air layers. Since the heat conduction is smaller for gas
at atmospheric pressure than for glass, the k-value is smaller for
these adhered air layers then for the pane of glass.
From Eq. 8.24 one obtains a thickness of 5 mm for the boundary
air layer with a k-value k D 3:4 W=.m2 K/ for heat conduction,
compared to k D 200 W=.m2 K/ for the pane of glass.
Another heat transport mechanism is heat radiation (see
Sect. 10.2.5). The room temperature in the inner part of the
house causes infrared radiation, which can escape through the
windows. The heat loss can be estimated as 4:6 W=.m2 K/.
This gives a total k-value of the inner air layer k D
8 W=.m2 K/.
For the outer convective air layer the k-value is different because
the air flows upwards against the gravitation. Detailed calculations give a value k D 20 W=.m2 K/ including radiation losses.
For successive layers the reciprocal k-values add (analogous to
electrostatics where the reciprocal electric conductivities add)
and we obtain from
Figure 10.31 Thermal insulation. a Temperature behaviour across a plastered
claybrick wall; b plastered wall of pumice stone; c pumice stone wall with styrofoam layer. The arrows give the direction of the convection current
1
1
1
1
C
D C
k
ki
kg
ko
(10.44)
the total k-value k D 5:5 W=.m2 K/. The comparison with the
k-value of the walls k < 1 W=.m2 K/ shows that windows with
a single pane of glass constitute a major heat loss.
Chapter 10
A much better heat insulation can be achieved with windows of
two panes of glass and an inert gas enclosed between the panes
(Fig. 10.33a).
The k-value of the gas depends on the thickness d of the gas
between the glass panes. For d 1 cm the heat conduction is
dominant, while for larger values of d convection undertakes the
major part of heat transfer. Fig. 10.33b shows, that for d D 1 cm
the minimum k-value is reached because the boundary layers
that adhere to the glass walls, prevent convection.
Figure 10.32 Heat transport across a single pane window
For such a double glass window, the k-value for heat conduction
is substantially smaller than for a single pane window. In order
Figure 10.33 Double pane glas window. a Composition and temperature change across the window; b k -values as a function of the thickness d of the gas layer
between the two glas panes; c decrease of k -values through technical progress
10.2 Heat Transport
to minimize also the radiation losses, the glass panes are covered by a thin dielectric layer [10.8], which reflects the infrared
radiation from the inside of the house (see Vol. 2, Sect. 10.4).
Without a reflecting layer, k-values of k 3 W=.m2 K/ can be
realized, while with reflecting layer the k-value decreases down
to k 0:6 W=.m2 K/. The k-values are then comparable to
those of the walls [10.8, 10.9].
Since the intensity and the spectral distribution of the radiation
emitted by a body depends essentially on the temperature of
the body, this radiation is called heat radiation or thermal radiation. In this section we will discuss the characteristics of
thermal radiation.
The considerations above illustrate that all three heat-transfer
processes as heat conduction, convection and radiation have to
be taken into account in order to optimize the heat insulation of
a house. In Fig. 10.33c the technical progress of minimizing the
total k-value is illustrated.
At first we will experimentally study, how the intensity of thermal radiation depends on the surface conditions of the body. We
use a metal hollow cube filled with hot water, where the four side
walls have a different surface structure (black, white, shiny and
rough). All side walls have the same temperature. Four equal radiation detectors, which measure the total radiation (integrated
over all wavelengths) are placed at the same distance d from the
four walls (Fig. 10.35). They all show different radiation powers. When the cube is turned by n 90ı (n D 1, 2, 3, : : :) about
a vertical axis, it can be proved that the difference is not due to
differences of the detectors but that the different sidewalls really
emit different radiation powers. The experiment shows surprisingly that the black side wall emits the maximum power and the
shiny white surface the minimum power. The radiation power
emitted from the surface area dA into the solid angle d˝ can be
quantitatively described by
10.2.5
Thermal Radiation
10.2.5.1 Emissivity and Absorptivity of a Body
Every body at a temperature TK exchanges energy with its
surrounding. If TK is higher than the temperature TS of the
surrounding, the energy emitted by the body is larger than the
energy received from the surrounding. If no energy is supplied
to the system body +surrounding, the system approaches thermal equilibrium, where the temperature of the body is equal to
that of the surrounding (Fig. 10.34). This energy balance can be
reached by heat conduction, convection or radiation. If the body
is kept in vacuum, (for instance our earth) radiation is the only
way to exchange energy with the surrounding, because both heat
conduction and convection need matter for the transport of energy.
dW
D E dA d˝ :
dt
The constant E is the emissivity of the surface. It gives the
radiation power dW=dt, integrated over all wavelengths that is
emitted from a surface element dA D 1 m2 into the solid angle
d˝ D 1 sr around the surface normal (Fig. 10.36). According
to the experiment the emissivity E of a black surface is larger
than that of a white surface at the same temperature.
Extensive experiments have proved, that radiation emitted by
hot bodies represents electromagnetic waves, which can transport energy through matter and also through vacuum.
The ratio
Figure 10.34 Energy exchange by thermal radiation between a body and its
surroundings. At thermal equilibrium is dW1 =dt D dW2 =dt and TK D T2
The integral absorptivity A is defined as the mean value of the
quotient A D absorbed radiation power = incident radiation
power, averaged over all wavelengths.
K.T/ D
E .T/
A .T/
Figure 10.35 Experimental setup for the measurement of emission
(10.45)
Chapter 10
A more quantitative representation of heat insulation can be
found in [10.7] and the references gives there and also in many
books on energy saving new house construction [10.9].
273
274
10 Thermodynamics
0
A0
1 > A2 (a black surface has a higher absorptivity than a
shiny one) it follows that dW10 =dt > dW20 =dt ! T1 > T2 .
Now the cube is turned about a vertical axis by 180ı and
the surface A1 now faces A02 and A2 faces A01 (Fig. 10.37b).
The absorbed powers are
Figure 10.36 Illustration of the emissivity E of a surface element dA
depends solely on the temperature T and not on the material of
the body, as can be demonstrated by the following experiment:
dW1 =dt E2 A0
1
and dW2 =dt E1 A0
2 :
The experimental result is now T1 D T2 ! dW1 =dt D
dW2 =dt.
)
E2 .T/
E1 .T/
D
:
A1
A2
(10.46)
J
Experiment
We place in Fig. 10.37a in front of the black surface A1
of the hot cube an equivalent surface A01 of the detector at
a distance d and in front of the shiny surface A2 a shiny
detector surface A02 at the same distance d. Measuring the
temperatures T1 of A01 and T2 of A02 one finds that T1 > T2 .
Since the surface structure of A01 is equal to that of A1 , and
that of A02 is equal to that of A2 , the absorptivity A1 must
0
be equal to A0
1 and A2 D A2 .
A separate experiment proves that the absorptivity of the surfaces does not depend on the temperature at least within the
temperature range from 0–100 ı C, which is covered in the experiment above. Therefore it follows from (10.46) for an arbitrary
body
E .T/
E1 .T/
D 2 D K.T/ :
(10.47)
A1
A2
The ratio of emissivity to absorptivity can be described
for any body by a function K.T/ that depends solely on
the temperature T.
Chapter 10
A body with A D 1 is called a black body.
It completely absorbs any incident radiation. According to
(10.47) a black body must also have the maximum emissivity
compared to all other bodies with equal temperature.
Figure 10.37 Derivation of (10.47)
The power absorbed by the two detector surfaces is
dW10 =dt E1 A0
1
and
dW20 =dt E2 A0
2 :
Since E1 > E2 (according to the foregoing experiment,
a black surface emits a larger power than a shiny one) and
Note: Bodies with a large absorption coefficient ˛ but a sudden increase of ˛ at the glossy surface are not a black body,
because their reflectivity also increases (Fig. 10.38a). Therefore the major part of the incident radiation is reflected and only
the minor part, that penetrates into the body is absorbed (see
Vol. 2, Chap. 8). In order to realize a black body, the absorption coefficient should not increase suddenly at the surface but
must continuously increase over a distance z > ( D wavelength of the incident radiation) from zero to its maximum value
(Fig. 10.38b). This can be for instance realized by a roughened
surface (black velvet, soot or graphite with a rough surface)
where the optical density rises slowly from the outside to the
inner part of the body. The sun is an example of a nearly perfect
black body, because the gas density and with it the absorptivity
increases slowly from the diffuse outer edge of the photosphere
to the interior.
Often the problem arises to keep a body at a constant temperature TK , that differs from the temperature TS of its surrounding
by supporting (TK > TS ) or extracting (TK < TS ) energy. This
energy can be minimized when heat conduction, convection and
radiation are minimized. The experimental realization uses materials with low heat conductance and radiation shields.
10.2 Heat Transport
Figure 10.38 a For bodies with a large gradient d˛=dz at the surface incident
radiation is mainly reflected and only partly absorbed inspite of the large absorption coefficient ˛. b Most of the incident radiation is absoorbed, if d˛=dz
is small, i. e. ˛.z / rises slowly from ˛ D 0 to ˛ D ˛max
275
1. A thermos bottle (Fig. 10.39a) consists of a double
wall glass flask. The space between the two walls
is evacuated and the two inner sides of the walls are
mirrored. The vacuum prevents heat conduction and
convection. The reflective walls minimize the escape
of thermal radiation to the outside. Therefore, the heat
losses from the inner volume are very small and the
coffee stays hot for a long time or cold drinks remain
cold.
2. For the storage of liquid nitrogen a Dewar is used
(Fig. 10.39b), which is based on the same principle as
the thermos bottle. Here the heat transfer from the outside is minimized in order to keep the evaporation of
the cold liquid nitrogen (T D 77 K) as low as possible.
The small portion of evaporating nitrogen extracts the
heat of evaporation and keeps the temperature in the
Dewar low.
If liquid air is used (78% N2 and 21% O2 /, the nitrogen
evaporates faster because of its higher vapour pressure
and the concentration of the reactive oxygen increases
until an explosive concentration is reached. Therefore,
generally liquid air is dangerous and is only used for
J
special purposes.
Examples
Reflective coating
Vacuum
Glass wall
10.2.5.2 Characteristic Features of Thermal Radiation
The energy that is emitted by the surface element dA into the
solid angle d˝ around the direction against the surface normal can be measured with a radiation detector (for example a
thermo-couple connected to a black surface). The detector area
dA2 at a distance r from the radiation source receives the radiation within the solid angel
d˝ D
Hot liquid
Heat insulation and
shock absorption
a)
dA2
:
r2
Experiments prove that for many radiation sources the angular
distribution of the measured radiation power is
dW./=dt D S cos dA d˝ :
Vaporization
Glass or steel wall
(10.48)
The quantity S is the emittance or radiation density of the
source. It describes the radiation power per m2 of the radiation source, emitted into the solid angle d˝ D 1 sr around the
surface normal (Fig. 10.40a).
The radiant intensity
Liquid air
J./ D
Z
S cos dA ;
ŒJ D 1
W
sr
(10.49)
F
Vacuum
b)
Figure 10.39 a Thermos bottle; b Dewar gasket
is the total radiation power emitted by the radiation source into
the solid angle d˝ D 1 sr around the direction against the
surface normal.
Note: The relation between the radiation density S and the
emissivity E is outlined in Sect. 10.2.5.3.
Chapter 10
Shive with low
heat conductance
276
10 Thermodynamics
Figure 10.41 The surface element A2 receives from A1 the radiation
power dW =dt D .S A1 A2 cos 1 cos 2 /=r 2
The ratio
Figure 10.40 a Illustration of radiant intensity J . /. b The length of the arrow
is proportional to the radiant intensity J . /
dW2 =dt
D
A2
Z
A1
dA1 S1 cos 1 cos 2 =r2
(10.54)
is the irradiance or intensity at the detector ŒW=m2 .
The emitted radiation power generally depends on the wavelength resp. the frequency D c= of the electromagnetic
wave. The spectral radiation intensity S is defined by the equation
Z1
S D
S d :
(10.50)
D0
Chapter 10
The radiation of the source results in an electromagnetic field
with the energy density w ŒJ=m3 and the intensity I ŒW=m2 .
The spectral energy density w is the energy per m3 within the
spectral frequency interval D 1 s 1 . It is related to the total
energy density w by
Z
w D w d :
(10.51)
For a radiation source with isotropic radiation (for instance the
sun) the relation between I D jSj and w is
I D .c=4/ w ;
(10.52a)
and similar for the spectral quantities
I D .c=4/ w ;
(10.52b)
where c is the velocity of light. For plane waves the relations
are I D c w and I D c w .
The detector element A2 at a distance r from the isotropic
source receives from the source element A1 the radiation
power
dW1
D S1 cos 1 A1 ˝
dt
D S1 cos 1 A1 A2 cos 2 =r2 ;
Note: The radiation power, absorbed by the detector with the
normalized absorptivity A , the reflectivity R and the transmission T (A C R C T D 1) is
dWabs
dW1
D A
D .1
dt
dt
R
T/
dW1
;
dt
because the fraction (R C T) of the incident radiation is reflected
and transmitted.
10.2.5.3 Black Body Radiation
A black body with the absorptivity A can be experimentally realized by a cavity with absorbing walls and a small hole with
an area A that is small compared to the total inner wall area A
of the cavity (Fig. 10.42). Radiation that penetrates through the
hole into the cavity, suffers many reflections at the absorbing
walls before it can eventually escape with a very small probability through the hole. The absorptivity of the hole area A is
therefore A 1.
When the cavity is heated up to a temperature T, the hole area
A acts as radiation source with an emissivity E that is, according to (10.47) larger than that of all other bodies with A < 1
at the same temperature T (the black body radiation is therefore
also called cavity radiation). This can be demonstrated by the
following experiment (Fig. 10.43):
The letter H is milled deeply into a graphite cube. At room
temperature, the letter H appears darker than the surface of the
cube (left picture of Fig. 10.43). When the cube is heated, up to
(10.53)
where ˝ D A2 cos 2 =r2 is the solid angle under which the
tilted surface element A2 appears from the source (Fig. 10.41).
The Equation 10.53 is symmetric. Replacing S1 by the radiation
intensity of the surface element A2 the equation describes the
radiation power dW2 =dt received by A1 from A2 .
Figure 10.42 A cavity with a small hole A absorbs nearly all of the radiation
incident onto A
10.2 Heat Transport
Low temperature
277
High temperature
Figure 10.44 A body inside a closed cavity at thermal equilibrium with the
radiation field
Figure 10.43 The letter H milled into a graphite block appears darker as its
surrounding at low temperatures but brighter at high temperatures
Some simple considerations allow one to postulate some basic
laws of the black body radiation:
Under stationary conditions (T D const) emission and absorption of the cavity walls must be balanced. This implies
for all frequencies of the radiation that the absorbed power
of an arbitrary surface element A of the walls must be equal
to the emitted power:
dWa ./=dt D dWe ./=dt :
At this equilibrium we define the temperature T of the black
body radiation as the temperature of the walls.
The black body radiation is isotropic. The spectral irradiance
I ŒW=.m2 s 1 sr is for any point in the cavity independent of the direction in the cavity and also of the material
or structure of the walls. If the radiation were not isotropic,
one could place a black disc into the cavity and orientate it
in such a way, that its surface normal points into the direction of maximum radiation intensity S . The disc would then
absorb more radiation power and would heat up to higher
temperatures than the walls. This contradicts the second law
of thermodynamics (see Sect. 10.3).
The black body radiation is homogeneous, i. e. its energy
density is independent of the specific location inside the cavity. Otherwise one could construct a perpetuum mobile of
the second kind (see Sect. 10.3).
When we place a body in the radiation field of the cavity, the
spectral radiation power S d dA d˝, falls within the solid
angle d˝ onto the body. The spectral power absorbed by the
surface element dA is
dWa
D A S dA d˝ d ;
dt
(10.55a)
while the emitted power is
dWe
D E dA d˝ d :
dt
(10.55b)
At thermal equilibrium the absorbed power must be equal to the
emitted power. Since the cavity radiation is isotropic, this must
E =A D S .T/ :
(10.56)
For all bodies in the radiation field of the cavity the ratio of
spectral emissivity and absorptivity equals the spectral radiation
density S of the radiation field.
For a black body is A D 1 for all frequencies . We can therefore conclude:
The spectral emissivity E of a black body is equal to the
spectral radiation density S of the cavity radiation.
10.2.5.4 The Emitted Radiation Power of a Hot Body
The surface S of a black body at the temperature T emits, according to the Stefan–Boltzmann Law, (see Vol. 2, Sect. 12.3)
the radiation power
dW
D S T4 :
dt
(10.57)
For a black surface with A D 1 the Stefan–Boltzmann constant
has the numerical value D 5:67051 10 8 W=.m2 K/. For
bodies with A < 1 the emissivity is smaller and therefore also
the emitted radiation power at the same temperature is smaller
than for a black body. The Stefan–Boltzmann Law can be derived from Planck’s radiation law (see Vol. 3, Chap. 3). The
deviation of the experimental results for small wavelengths from
those predicted by the Stefan–Boltzmann law,. gave the impetus
for the development of quantum theory.
Note:
The radiation power of a hot body is proportional to the
fourth power of the surface temperature. With increasing
temperature it therefore represents an increasing fraction of
the total energy loss of a body.
The thermal radiation is an electromagnetic wave and therefore propagates also through vacuum. The energy transport
by radiation is not bound to matter. We own our existence to
the heat radiation from the sun because this is the only energy transport mechanism from the sun to the earth (except
the negligible contribution of particles such as electrons and
protons emitted by the sun).
Chapter 10
T D 1000 K the letter H appears much brighter than the other
surface elements (right figure).
be valid for all directions. Therefore it follows from (10.55a,b)
the Kirchhoff-Law
278
10 Thermodynamics
A more detailed and quantitative treatment of heat radiation will
be postponed to Vol. 3, because it demands some basic knowledge of quantum theory.
10.2.5.5
Practical Use of Solar Energy
The radiation energy of the sun, received on earth, can be either directly converted to heat by solar energy collectors or
transformed into electrical power by photovoltaic semiconductor elements. While the second technique is treated in Vol. 3,
the first will be shortly discussed here [10.10, 10.11].
The radiation power of the sun, incident on 1 m2 of a surface
black surface element outside of the earth atmosphere has an
annual average Pˇ D 1:4 kW=m2 (solar spectral irradiance).
However, even at a clear day without clouds only a smaller
power PE reaches the earth surface because of absorption and
light scattering in the atmosphere. For geographical latitudes
' D 40ı –50ı one measures PE 0:5Pˇ . For an inclination angle ˛ of the incident radiation to the surface normal the received
power at a clear sky is PE 730 cos ˛W=m2 .
With the absorptivity A of the surface, the power absorbed
within the time interval t by a plane surface with area A is
Figure 10.45 Variation of the sun radiation incident onto a sun radiation collector at the lattitude ' D 49ı as a function of daytime for three different times
of the year
The area under the curves gives the integrated energy (J=m2 sun
hours) received during a whole day between sunrise t1 and sun
set t2 .
Zt2
WE D PE cos ˛ dt
(10.60)
t1
Pa D A A PE cos ˛ t :
This results in a temperature increase T of a sun collector with
mass m and specific heat c
Chapter 10
T D A A PE cos ˛ t=.c m/ ;
(10.58)
if no heat losses occur.
The temperature increases with irradiation time if the heat is
not dissipated. This dissipation can be achieved, when on the
backside of the sun collector tubes are welded with a good heat
contact to the sun collector and a liquid is pumped through the
tubes, which takes away the heat. In order to keep the temperature of the sun collector constant, the pumping speed is chosen
such that the heat transport just balances the received radiation
power.
With a mass flow dml =dt of the heat transporting liquid with
the specific heat cl and the temperature increase T the energy
balance is given by the equation
A PE A cos ˛ D .dml =dt/ cl T C .dW=dt/v :
The average radiation power per day is then hPE cos ˛i D
WE =.t2 t1 /.
Example
A D 0:8; hPE cos ˛i D 250 W=m2 during a clear day
in August at ' D 45ı ; A D 8 m2 . With water as the
heat transporting liquid (cW D 4186 Ws=.kg K/) which
is heated from 20 to 60 ı C. With a good heat insulation
the heat losses dWl =dt can be kept down to 50 W=m2 for
a temperature difference of T D 40 ı C. The amount
of water heated per sec is then given by dmW =dt D
.A hPE cos ˛i dWl =dt/ A=.cW T/ D 0:0072 kg=s.
Within one hour 26 l water are heated from 20 to 60 ı C.
J
Figure 10.46 shows a possible realization of a sun power collector for the heating of houses. It consists of a blackened
(10.59)
The angle ˛ depends on the inclination of the energy collecting
plane, on the latitude ' and on the daytime. In Fig. 10.45 the
daytime dependence of the sun energy received by a collector
with ˛ D 45ı in Kaiserslautern (' D 49ı ) is illustrated for
three different dates. Two effects cause this variation with the
daytime: 1) The variation of the angle ˛ due to the apparent
motion of the sun and 2) The variation of the path length of
the sun radiation through the atmosphere during the day, where
absorption and scattering attenuates the radiation energy.
Solar radiation
Heat
radiation
Glas cover
Air
Absorbing sheet
Water pipes
Insulation
Figure 10.46 Cross section of a flat solar radiation collector which is mounted
on house roofs
10.3 The Three Laws of Thermodynamics
10.3
Hot water
Additional
heating
Heat
exchanger
Cold water
Figure 10.47 Thermal solar radiation collector for heating water with heat
exchanger
absorber plate with pipes connected to the backside with good
heat contact. Through the pipes a water-glycol mixture (to avoid
freezing at low temperatures) is pumped. In cases where temperatures above 100 ı C are reached, low viscosity oil is used. The
absorber plate is placed inside a heat insulating housing with a
glass plate in front. Heat losses are due to reflection of the sun
radiation by the glass plate, by heat radiation of the black absorber plate and by heat conduction from the hot part of the sun
radiation collector (including the pipes for the transport of the
hot liquid) and convective cooling by the wind that blows along
the device.
The heated liquid transfers its heat through a heat exchanger to
a thermal storage system inside the house that generates hot service water (Fig. 10.47). A temperature sensor and a feedback
system controls the temperature of the service water and takes
care that it always has the wanted temperature. In case the sun
energy is not sufficient, a conventional heating system is connected which only operates if the temperature sinks below the
wanted value. When the hot water is used for room heating,
a floor heating system is advantageous, because here the water
temperature can be lower than that for radiator heating [10.9–
10.11].
In large facilities for thermal solar energy conversion, it is more
effective to heat the liquid above its boiling point. The generated vapour drives turbines which can produce electric current
through electric generators. The technical realization uses large
spherical mirrors that focus the sun radiation onto a black surface connected to a pipe system that transports the hot vapour.
Temperatures above 1000 ı C can be achieved and an electric
output power of many kW has been demonstrated. The installation costs for such systems are up to now very high and therefore
only a few pilot plants have been built. One example is the system in Almeria in Spain.
The Three Laws of
Thermodynamics
We will define a thermodynamic system as a system of atoms
or molecules that interacts with its surroundings by exchange of
energy in form of heat or mechanical work. The system can be
described by physical quantities such as temperature, pressure,
volume, particle density etc. In this section we will discuss,
how the state of such a system changes by the exchange of
energy with its surroundings. The results of all investigations
can be condensed in three laws of thermodynamics, which have
a comparable importance for Physics as the conservation laws
of mechanics for momentum, angular momentum and energy.
These three laws are solely based on experimental data and cannot be derived mathematically from first principles contrary to
a widespread false opinion.
At first we must discuss, which quantities are necessary to describe the state of a thermodynamic system.
10.3.1
Thermodynamic Variables
The state of a system is defined by all characteristic properties,
which are determined by the external conditions. A thermodynamic system is completely determined if the chemical
composition is known and the quantities pressure p, volume V
and temperature T are given. If these quantities do not change
with time, the system is in an equilibrium state and it is called
a stationary system. Most of the thermodynamic considerations
deal with stationary systems. Often a system changes so slowly,
that it can be described by a succession of equilibrium states.
Systems far away from equilibrium play an important role for
all chemical and biological reactions and they are intensively
discussed in modern physics. They are therefore shortly treated
at the end of this chapter. In this section, we will restrict the
discussion to ideal gases. The thermodynamics of real bodies
will be discussed later.
An equilibrium state of a system is unambiguously determined,
if the three quantities pressure p, volume V and temperature T
are fixed. These quantities are therefore called thermodynamic
variables.
Definition
A thermodynamic variable is a variable in the equation of
state of a thermodynamic system. It describes the momentary state of the system and is independent of the way on
which the system has reached its momentary state. Besides V, p and T also the total energy, the entropy and the
enthalpy are thermodynamic variables.
Chapter 10
Pump and
control system
279
280
10 Thermodynamics
The isothermal compressibility
T
DD
1
V
@V
@p
(10.66)
T
gives the relative volume change V=V for a pressure change
p at a constant temperature T.
As recollection keep in mind:
isothermal:
isobaric:
isochoric:
T D const
p D const
V D const :
Figure 10.48 a Heating at constant pressure; b heating at constant volume;
c no heat supply
The thermodynamic variables are related to each other by the
equation of state for a gas volume V of an ideal gas with N
molecules
pV DRT ;
(10.61)
where D N=NA is the number of moles and NA the Avogadro
number. Also for real gases a corresponding equation can be
derived (see Sect. 10.4). For a given volume V and a pressure p
the temperature T determines the internal energy
UD
1
2
f RT
(10.62)
Chapter 10
of molecules with f degrees of freedom. For ideal gases (for
instance helium) is f D 3. If the volume decreases (dV < 0) at
a constant pressure p the necessary work is
dW D
p dV :
The total change dV of the volume V.p; T/, when both quantities
p and T are changing is
@V
@V
dV D
dp C
dT
@p T
@T p
(10.67)
D V dp C
p
V dT :
For isochoric processes the volume V stays constant, i. e. dV D
0. Then (10.67) reduces to
0D
V .dp/V C
) dp D
p
p
V .dT/V
dT :
Division by dT yields with .dp=dT/V D
p
D
V
V
p
(10.68)
p the relation
(10.69)
(10.63)
The sign is chosen in such a way that the applied work
is positive, if the energy of the system increases. Work
performed by the system means a decrease of its internal
energy and is therefore defined as negative.
When a gas is heated at constant pressure p its volume increases
(Fig. 10.48a). The quantity
1
@V
;
(10.64)
p D
V
@T p
that describes the relative volume change per Kelvin temperature rise, is the isobaric expansion coefficient.
In an analogous way the heating of a gas at a constant volume
(Fig. 10.48b, where the pressure increases, is described by the
isochoric pressure coefficient
@p
1
;
(10.65)
D
V
p
@T V
which describes the relative pressure increase p=p for a temperature rise of 1 K.
between isobaric expansion coefficient p , isothermic compressibility , isochoric expansion coefficient V and pressure p.
10.3.2
The First Law of Thermodynamics
The heat Q applied to a system can be either used for rising the
temperature T at a constant volume V, or for the expansion of
the volume V against the external pressure p where the system
has to perform the work W. Energy conservation demands
Q D U
W ;
(10.70a)
where, as defined before, W < 0 if the system performs
work (which decreases its own energy). This sign definition
is in agreement with the definition (2.35) for the work. If the
system, for instance, performs work against an external force
F D p A when a piston with area A is moved along the distance x against the external pressure p, the work is
W D F x D p A x D p V
with V > 0 :
10.3 The Three Laws of Thermodynamics
Equation 10.70 is the first law of thermodynamics. It is a special case of the general law of energy conservation. It can be
formulated as:
The sum of the external heat Q, applied to a thermodynamic system, and the supplied mechanical energy W is
equal to the increase U of the total internal energy U.
10.3.3
281
Special Processes as Examples of the
First Law of Thermodynamics
Note: We will discuss the following processes for one mole of
a gas where the number of moles is D V=VM D 1.
10.3.3.1 Isochoric Processes (V = const)
With dV D 0 it follows from (10.71)
U D Q C W
(10.70b)
When the system performs work against an external force, is
W < 0 and therefore U < 0. Many inventors have tried
to construct machines that deliver more energy than they consume. Such a machine could use part of the delivered energy for
its own operation. It could run continuously delivering energy
without external energy input. Therefore this hypothetical machine is called a perpetuum mobile. Because it contradicts the
first law of thermodynamics it is also called perpetuum mobile
of the first kind.
Equation 10.70 can be also formulated in a more floppy way as:
dQ D dU D CV dT :
(10.72)
The heat supplied to the system is used solely for the increase of
the internal energy U. We can therefore relate the specific heat
to the internal energy U by
CV D
@U
@T
:
(10.73)
V
10.3.3.2 Isobaric Processes (p = const)
The first law of thermodynamics has now the form
Note: This statement cannot be proved mathematically. It is
solely based on empirical knowledge.
For ideal gases the work performed during the expansion of the
volume V by infinitesimal amount dV against the external pressure p is
dW D p dV :
The first law of thermodynamics for ideal gases can therefore be
written in a differential form as
dQ D dU C p dV D Cp dT ;
where we have used (10.28). When we introduce the enthalpy
H DUCpV
p dV :
(10.75)
as new thermodynamic variable with
dH D dU C p dV C V dp D dQ C V dp ;
(10.76)
we can write the first law of thermodynamics as
dH D dU C p dV D dQ :
dU D dQ
(10.74)
(10.77)
(10.71)
For dV > 0 the system releases energy and according to (10.71)
dU < dQ, i. e. the loss of internal energy cannot be compensated
by the supplied heat dQ. For dV < 0 the volume is compressed
and the system gains the energy p dV. Now dU > dQ, the gain
of internal energy is larger than the supplied heat.
The relation between the thermodynamic variables p, V, T can
be derived from (10.71) for special processes where in each case
one of the variables p, V, T or the quantity Q is kept constant.
Note, that the quantity Q is not a thermodynamic variable! The
state of a system does change with the supply of heat dQ, but
one cannot unambiguously determine the final state of the system, because either U or V or both variables can change. In a
mathematical language this means: dQ is not a complete differential.
For isobaric processes the increase dH of the enthalpy H
is equal to the supplied heat dQ.
The specific heat at constant pressure is then
Cp D
@H
@T
:
(10.78)
p
The variable H is often used for phase changes, chemical reactions or other processes that take place at constant pressure, but
where the volume can change. A further example is the expansion of a gas from a reservoir with constant pressure into the
vacuum where the pressure p D 0 is maintained by pumps.
Chapter 10
A perpetuum mobile of the first kind is impossible.
282
10 Thermodynamics
oscillation period T D 1= nearly no energy exchange between maxima and minima of the wave can occur.
The first law of thermodynamics (10.71) can be written with
(10.73) for adiabatic processes
dU D CV dT D p dV :
(10.82)
From the equation of state (10.21) p V D R T we obtain
p D R T=V. Inserting this into (10.82) yields
CV dT=T D R dV=V :
Integration gives
Figure 10.49 Isothermal and adiabatic curves in a p -V -diagram
10.3.3.3
Isothermal Processes (T = const)
Since the internal energy per mole of a gas depends solely on
the temperature T but not on the pressure p or the volume V, for
isothermal processes U must be constant, i. e. dU D 0. From
(10.71) it follows
dQ D p dV :
(10.79)
The external heat energy dQ supplied to the system is completely transferred to the work p dV that the system releases
to the outside. Its internal energy does not change.
With R D Cp
Chapter 10
which are called isotherms (black curves in Fig. 10.49).
We will now discuss how large the work is that a system has
to perform for an isothermal expansion from a volume V1 to
V2 > V1 at constant temperature T.
WD
ZV2
p dV D R T
V1
D R T ln
10.3.3.4
ZV2
dV
V
V1
(10.81)
V1
V2
D R T ln
:
V1
V2
CV /
D const :
(10.83a)
The 1=CV -th power of (10.83a) yields with the adiabatic index
D Cp =CV the equation
T V
1
D const ;
(10.83b)
because T D p V=R, this can be also written as
p V D const :
(10.80)
The state p.V/ of the system can be plotted in a p-V-diagram
(Fig. 10.49) for isothermal and adiabatic processes at different
temperatures T. This gives for isothermal processes the hyperbolas
R TK
const
pD
D
;
V
V
CV this can be written as
T CV V .Cp
The equation of state p V D R T then reduces to the Boyle–
Marriott law (see Sect. 7.1)
p V D const :
CV ln T D R ln V C const
) ln T CV V R D const :
(10.83c)
The Eq. 10.83a–c describe the relations between the thermodynamic variables T, p, V for adiabatic processes. They are called
Poisson-adiabatic equations.
In a p–V-diagram (Fig. 10.49) the red adiabatic curves p.V/ /
1=V ( > 1) are steeper than the isothermal curves p.V/ /
1=V.
For an ideal gas is f D 3 and D .f C 2/=f D 5=3. For
molecular nitrogen N2 is f D 5 ! D 7=5.
Example
In the pneumatic cigarette lighter, the volume V filled
with an air–benzene-mixture is suddenly compressed to
0:1 V. According to (10.83b) the temperature T rises from
room temperature (T1 D 293 K) to T2 D T1 .V1 =V2 / 1 .
For air is D 7=5 which gives T2 D 736 K D 463 ı C.
This is above the ignition temperature of the air–benzenemixture.
J
Adiabatic Processes
During adiabatic processes no heat is exchanged between the
system and its surroundings. Adiabatic processes occur in nature, when changes of volume or pressure are so fast, that the
energy exchange during this short time period can be neglected.
An example is the propagation of acoustic waves at high frequencies through a medium (see Sect. 11.9). During one
10.3.4
The Second Law of Thermodynamics
While the first law of thermodynamics represents the energy
conservation when thermal energy is converted into mechanical
10.3 The Three Laws of Thermodynamics
283
energy, the second law of thermodynamics gives the maximum
fraction of thermal energy that can be really transferred into mechanical energy.
As we will see, this is connected with the question, into which
direction the transfer of one form of energy into another form
proceeds by its own, i. e. without external action. All of our
experience tells us, that heat flows by its own only from the
hotter region into the colder one, not vice versa. Furthermore all
experiments show, that mechanical energy can be completely
converted into heat, but that for the opposite process only part
of the heat can be converted into mechanical energy.
Figure 10.51 Carnot’s cycle
Heat flows by its own only from the warmer body to the
colder one, never into the opposite direction.
real world of many-particle systems only idealized “Gedankenexperiments”, which represent limiting cases of real processes
that are always irreversible.
We will now discuss more quantitatively the transformation of
heat into mechanical work. This will be illustrated by considering thermodynamic cyclic processes, which leads us to a
quantitative formulation of the second law of thermodynamics.
All periodically operating machines, such as steam engines or
car motors traverse irreversible cyclic processes. Although they
arrive at the end of the cycle again at the initial state, if they are
regarded as isolated systems, but they have lost energy during
the cycle (for instance friction losses) which has to be replaced
for each cycle.
10.3.5
The Carnot Cycle
A thermodynamic cycle is a series of processes where a thermodynamic system passes through several different states until
it finally reaches again its initial state. At the end of this cycle,
the system shows again the same thermodynamic variables as in
the initial state, although it has passed during the cycle through
different states with different variables. A simple example is a
system that is heated and then cooled again until it has reached
the initial temperature.
If the cyclic process can traverse into both directions the cycle is
called reversible (Fig. 10.50) otherwise it is called irreversible.
Although such reversible processes can occur in micro-physics
if only a few particles are involved, they represent in the
Figure 10.50 Thermodynamic cycle from the state 1 .T1 ; p1 ; V1 / via the state
2 .T2 ; p2 ; V2 / back to the state 1. a In a p –V -diagram; b in the temperaturetime diagram. Note: The cycle shown here, can only proceed, if energy is fed
into the system during the first step and energy is taken away from the system
during the second step
The most famous reversible cyclic process is the Carnot-cycle
that represents an idealized loss-free cycle. It was published
in 1824 by Nicolas Leonard Sadi Carnot. This cyclic process
will enable us to calculate the maximum fraction of heat that
can be transformed into mechanical energy and therefore allows
the quantitative formulation of the second law of thermodynamics. Furthermore, it illustrates nicely the difference between
reversible and irreversible processes.
The Carnot Cycle is a “Gedanken-Experiment”, where a thermodynamic system passes through two isothermal and two adiabatic processes during two expansion and compression events,
until it finally reaches its initial state again (Fig. 10.51).
Note: The following considerations are valid for 1 mol of an
ideal gas, where in Eq. 10.61 V D VM and D 1.
The state of the system at the starting point 1 is defined by the
thermodynamic variables .V1 ; p1 ; T1 /. The isothermal expansion brings the system to the state 2 D .V2 ; p2 ; T1 /. During this
process, the heat Q1 has to be supplied to the system in order
to keep the temperature constant. Now an adiabatic expansion
follows and the system gets to the state 3 D .V3 ; p3 ; T2 < T1 /.
In the next step the system is isothermally compressed and
reaches the state 4 with the conditions (V4 , p4 , T2 ). Here
the heat Q2 has to be removed from the system. Finally
an adiabatic compression brings the system back to its initial
state 1 D .V1 ; p1 ; T1 /. Such a virtual thermodynamic system
that passes through a Carnot cycle is called a Carnot Machine.
We will now calculate the heat energies Q1 and Q2 which
are exchanged between the system and a heat reservoir during
the isothermal processes.
1st process: Isothermal expansion from the state 1 to the
state 2. According to the first law of thermodynamics we ob-
Chapter 10
This fact, that is based solely on experimental experience, is
formulated in the second law of thermodynamics:
284
10 Thermodynamics
tain for an isothermal expansion
dQ D p dV :
The heat supplied to the system is equal to the mechanical
work the system performs during the expansion.
With (10.81) it follows:
Q1 D W12 D
ZV2
p dV
V1
Figure 10.52 Heat exchange and net mechanical energy W D W34
W12 of Carnot’s cycle
(10.84a)
D R T1 ln.V2 =V1 / :
The net work is then
2nd process: Adiabatic expansion from state 2 to state 3.
For adiabatic processes the heat exchange is zero. We therefore
obtain:
dQ D 0 ! dU D
p dV D W23 :
(10.84b)
The work performed during the expansion is negative, because
it is delivered from the system to the surrounding. This results
in a decrease U D U.T2 / U.T1 / of the internal energy U
because T2 < T1 .
Chapter 10
3rd process: Isothermal compression from state 3 to state 4.
Similar to step 1 is the heat Q2 delivered at the lower temperature T2 to the heat reservoir equal to the work W34 necessary
to compress the volume V
W34 D R T2 ln.V3 =V4 / D Q2 > 0 :
(10.84c)
4th process: Adiabatic compression from state 4 to the starting conditions in state 1. Similar to step 2 is here no heat
exchange with the surrounding and the work performed during
the compression is converted to the increase U of the internal
energy
U D U.T1 /
(10.84d)
U.T2 / :
Total energy balance: The work delivered to the surrounding
during the 2nd process is equal to the work supplied to the
system during the 4th process. Therefore, only during the
isothermal processes a net energy is transferred. The net mechanical work during the Carnot cycle (Fig. 10.52) is
W D W12 C W34
D R T1 ln.V1 =V2 / C R T2 ln.V3 =V4 / :
For the adiabatic processes 2 ! 3 and 4 ! 1 the relations hold
T1 V2
T1
1
V1 1
D T2 V3
D T2
1
V4 1
and
:
T2 / ln.V1 =V2 / :
W D R .T1
The Carnot Engine has received the heat Q1 and has supplied
the mechanical work W < 0 to the outside.
Such a machine that transfers heat into mechanical energy is
called heat engine.
The heat Q2 supplied to the surrounding, is generally lost.
Therefore the efficiency of the engine is defined as the mechanical work supplied by the engine divided by the heat Q1 put
into the engine.
The efficiency of the Carnot Engine is then
ˇ
ˇ
ˇ W ˇ
ˇ D R.T1 T2 / ln.V2 =V1 / D T1 T2
D ˇˇ
Q1 ˇ
R T1 ln.V2 =V1 /
T1
D
T2
T1
T1
:
ln.V1 =V2 / :
(10.86)
This is a remarkable result: During the cycle the total received
heat cannot be transformed into mechanical work, but only the
fraction D .T1 T2 /=T1 < 1. This fraction is called exergy. The remaining part (1 ) of the input energy (Anergy) is
exchanged as heat Q2 to the surrounding at the lower temperature T2 . The conservation of total energy can be written as
Energy D Exergy C Anergy :
The efficiency of the Carnot Engine increases with increasing
temperature difference T1 T2 . It is therefore advantageous to
choose T1 as high as possible and T2 as low as possible. We will
see in Sect. 10.3.10 that it is impossible to reach T2 D 0 K.
Division of the two equations yields
V2 =V1 D V3 =V4 ) ln.V3 =V4 / D
(10.85)
This implies that is always smaller than 1.
10.3 The Three Laws of Thermodynamics
When the Carnot cycle is traversed into the opposite direction,
heat is transported from the lower temperature T2 to the higher
temperature T1 . This requires the work
W D R .T2
T1 / ln.V1 =V2 / :
This represents the ideal limiting case of a heat pump, which
is also used as refrigerating machine (see Sect. 10.3.14). Its
coefficient of performance (also called figure of merit) is defined
as the ratio of delivered heat Q to the input work W.
"hp D
285
than the Carnot Engine, it needs less heat from the reservoir at the temperature T1 for its operation than the Carnot
Engine transports to this reservoir. It furthermore delivers
less heat to the cold reservoir at T2 than the Carnot engine
needs for its operation as heat pump.
Q1
1
T1
D :
D
W
T1 T2
Note, that "hp > 1.
Example
J
Note:
1. The heat pump does not contradict the second law of thermodynamics, because here the heat does not flow by its own
from the colder to the hotter place but needs mechanical
work to drive this heat transport.
2. The Carnot Engine works with an ideal gas and all energy
losses are neglected. The Carnot Cycle is reversible. Real
engines have always losses that cannot be avoided. They are
due to friction of the moving pistons, friction in the non-ideal
gas, heat conduction from the system to the surroundings etc.
These losses decrease the efficiency of the engine. We will
now indeed prove, that:
There is no periodically working machine with
a higher efficiency than that of the Carnot engine.
Proof
Assume, there is a machine Mx with a higher efficiency
than the Carnot Engine. This “magic machine” needs
for a given mechanical energy output a smaller heat input than the Carnot Engine, i. e. Qx < Q1 . We now
combine Mx with a Carnot engine that passes the cycle in opposite direction, i. e. it works as a heat pump
(Fig. 10.53). We adapt the size of Mx in such a way that it
delivers just the mechanical work W, which the Carnot
engine needs as heat pump. The Carnot engine then transports the heat
Figure 10.53 Proof of the impossibility of the perpetuum mobile of
the second kind
The combined system therefore transports heat from the
colder to the hotter reservoir without mechanical energy
input. This contradicts the second law of thermodynamics
which has been proved by numerous experiments. Therefore a heat engine with a higher efficiency than that of the
Carnot engine is not possible!
J
Remark. These considerations can be also applied to a heat
pump, where the cycle is traversed into the opposite direction. We replace the Carnot Engine in Fig. 10.53 by a “magic
heat pump” and the magic machine Mx by the Carnot Engine
(Fig. 10.54) and assume that the coefficient of performance "x
is larger than that of a Carnot heat pump. An analogous consideration shows that "x < "C D 1=C . This can be seen as
follows:
The Carnot engine in Fig. 10.54 now runs as heat engine that
extracts the heat Q1 from the hot reservoir at the temperature
T1 and delivers the heat Q2 D Q2 W to the cold reservoir
at T2 < T1 . The output energy W is transferred to the magic
heat pump, which takes the heat Q4 from the cold reservoir
|Q3|
Carnot
engine
|Q2|
|
|
Magic
heat pump
|Q4|
jQ1 j D jQ2 j C jWj
from the colder to the warmer reservoir. Since we have assumed that the magic machine Mx has a higher efficiency
Figure 10.54 Proof, that the energy efficiency ratio of a heat pump is always
smaller than that of a Carnot-engine that operates as a heat pump
Chapter 10
T1 D 30 ı C D 303 K,
T2 D 10 ı C D 283 K ) "hp D 15:2.
286
10 Thermodynamics
and transports the heat Q3 D Q C W to the hot reservoir.
Assume that the coefficient of performance "x D Q3 =W of
the magic heat pump is larger than "C D 1=C D Q1 =W of
the Carnot engine. Then jQ3 j > jQ1 j and jQ4 j D jQ3 j
W > jQ2 j D jQ1 j This implies that the combined system
Carnot engine plus magic heat pump can pump heat from the
cold to the hot reservoir without mechanical energy input. This
again contradicts the second law of thermodynamics.
The coefficient of performance "x of an arbitrary heat
pump cannot be larger than "C D 1=C where C is the
efficiency of the Carnot engine.
With other words: The coefficient of performance "x of any heat
pump cannot be larger than that of a Carnot heat pump "C D
T1 =.T1 T2 /.
From the considerations above it follows: All reversible cycles
have the same efficiency
D W=Q1 D .T1
T2 /=T1 ;
independent of the working material, which can be different
from an ideal gas.
10.3.6
Example
A perpetuum mobile of the second kind could be a ship
with engines that receive their energy solely from the heat
of the sea. Such a ship could move without additional
energy and would not need oil or coal.
J
A perpetuum mobile of the second kind does not contradict the
first law of thermodynamics, because it does not violate the energy conservation. Therefore numerous inventors have tried to
construct such machines, however unsuccessful!
The Carnot cycle allows a method to measure the Kelvin temperature, which is independent of the thermometer substance and
works down to very low temperatures where gas thermometers
are no longer useful, because all gases condense at such low
temperatures. From (10.84) we can deduce the ratio of the heat
energies Q1 and Q2 supplied from and released to the heat
reservoirs
Q1
T1
D
:
Q2
T2
The temperatures of the two heat reservoirs can be compared,
when the heat energies, exchanged between the system and the
reservoirs, are measured. For instance, if one of the reservoirs
is kept at the temperature T1 D 273:16 K of the triple point of
water the temperature T2 is obtained from
Chapter 10
Equivalent Formulations of the Second
Law
The considerations above allow the following statements:
The efficiency D W=Q1 < 1 of any heat engine is
always smaller than 100%. This means that heat cannot completely converted into mechanical work.
The Carnot engine has the maximum possible efficiency:
D .T1 T2 /=T1 .
The value D 1 would be only possible for T2 D 0. However,
we will see that this is excluded by the third law of thermodynamics (see Sect. 10.3.13)
The first and second laws of thermodynamics are purely rules of
thumb, based on numerous experimental facts. They cannot be
proved mathematically without additional assumptions.
The second law can be formulated in different ways:
Heat flows by its own only from the hot to the cold region,
never into the opposite direction.
There is no periodically acting machine that can convert heat
completely into mechanical work without additional energy
supply.
Such a machine is called a perpetuum mobile of the second kind.
The second law can then be formulated similar to the first law:
The realization of a perpetuum mobile of the second kind is impossible.
T2 D 273:16
jQ2 j
D 273:16 .1
jQ1 j
/ :
The heat reservoir, kept at the temperature T1 , can be electrically
heated, which allows the determination of Q1 . The efficiency
can then be measured as the ratio of mechanical work W D
p A x when a piston with area A moves by the distance x
against the external pressure p and the supplied heat Q1 .
The temperature scale obtained by this way is called the thermodynamic temperature scale.
One Kelvin (1 K) is 1/273.16 times the temperature of the
triple point of water.
10.3.7
Entropy
By introducing the entropy as new thermodynamic variable, the
second law of thermodynamics can be mathematically formulated in an elegant way. When the heat dQ is supplied to a
system at the temperature T we define as reduced heat the ratio dQ=T.
For the Carnot cycle in Fig. 10.51 we can bring the system from
the point 1 to the point 3 on two different ways: 1 ! 2 ! 3
or 1 ! 4 ! 3. Only during the isothermal processes, heat is
10.3 The Three Laws of Thermodynamics
We introduce the thermodynamic variable S called the entropy
with the dimension ŒS D ŒJ=K, in the following way. We define
the change dS of the entropy as the reduced heat exchanged on
an infinitesimal part of a reversible process
Sisobar D CV ln
In the Carnot Cycle the reduced heat energies change only during the isothermal processes. According to (10.84) the entropy
then changes by
Q
V2
S D
:
(10.87)
D ˙R ln
T
V1
For the complete reversible cycle we have
Q1
D
T1
Q2
;
T2
and therefore
S D 0 :
T2
T2
D .Cp R/ ln
T1
T1
T2
T2
D Cp ln
R ln
T1
T1
Sisochor D CV ln
Processes where S D const are called isentropic processes. For
these processes is S D 0 and therefore Q D 0 and T D
const. During isentropic processes the system must be kept at
a constant temperature. This distinguishes isentropic processes
from adiabatic processes where also Q D 0 but where the
temperature changes.
With the first law of thermodynamics (10.71) the entropy change
dS during reversible processes of an ideal gas can be calculated
as
dQrev
dU C p dV
dS D
D
:
(10.88)
T
T
For 1 mol of the gas is dU D CV dT and p VM D R T. This
converts (10.88) to
dT
dV
CR
:
T
V
Sisochor D Cp ln
T2
T1
R ln
p2
:
p1
(10.91)
Since the entropy S is a thermodynamic variable, its change S
does not depend on the kind of process but only on initial and
final state of the process. We can therefore determine S also
for irreversible processes. This can be seen as follows:
We consider a substance at the temperature T1 (e. g. a solid
body) in a gas volume V. The body should be in thermal contact
with the gas. Now the gas is slowly expanded in an adiabatic
process, which results in a slow decrease of the temperature. If
this proceeds sufficiently slowly, the temperature of the body
is always equal to that of the gas, because sufficient time is
available for reaching temperature equilibrium. Finally the temperature has decreased to T2 . This process is reversible because
the initial state can be retrieved by slow adiabatic compression.
When the solid body is regarded as isolated body without the
gas, the cooling process is irreversible, because heat is transferred to the surrounding. The entropy change of the body is
For a reversible cycle the entropy S is constant.
dS D CV
(10.90)
In a similar way one obtains for isochoric processes (V D const)
with CV D Cp R and p1 =T1 D p2 =T2
dS D dQ=T :
Since the change S for a system that is brought from a defined
initial state into a defined final state is independent of the way
between these two states, and depends solely on initial and final
states of the system, the quantity S is a thermodynamic variable which describes together with pressure p, temperature T
and volume V the state of a thermodynamic system.
V2
T2
C R ln
:
T1
V1
(10.89)
Integration over the temperature range from T1 to T2 where the
molar heat capacity can be assumed as constant, yields for isobaric processes where V and T can change but p D const.
Qirr
T
ZT2
dT
D CV
T
S D
(10.92)
T1
D CV ln
T2
<0
T1
for
T2 < T1 ;
as in the reversible process. However, since heat has been
transferred to the surroundings the entropy of the surroundings
increases. For the total change of the system body + surroundings Sirrev > 0, i. e. the entropy increases!
Example
1. We regard in Fig. 10.55 two equal bodies with mass
m and specific heat c which have been brought by different energy supply to different temperatures T1 and
T2 < T1 . Their heat energies are then Q1 D m c T1
and Q2 D m c T2 . When they are brought into
Chapter 10
exchanged with the surroundings. The absolute values of the
reduced heat energies jQ1 j=T1 and jQ2 j=T2 on the two ways
are equal, as can be seen from (10.84a–c). This means: The
reduced energies do not depend on the way but only on starting
and final state of the system. This is not only valid for the Carnot
cycle but for all reversible processes.
287
288
10 Thermodynamics
thermal contact heat flows from the hot body 1 to
the colder body 2 until the temperatures are equal to
the average temperature Tm . If no heat is transferred
to the surroundings the body 1 has lost the energy
Q1 D m c.Tm T1 / and the body 2 has received
the energy Q2 D m c .T2
Tm /. Because
Q1 D Q2 we obtain the average temperature
Tm D
T1 C T2
:
2
sufficiently long time t > 0 they are uniformly distributed over the whole volume V D V1 C V2 . The
gas temperature remains constant during this isothermal expansion (experiment of Gay-Lussac) because
no work is needed for the expansion into the vacuum
(p dV D 0).
(10.93)
The entropy change S of body 1 is
S1 D
ZTm
dQ
D mc
T
ZTm
dT
T
Figure 10.56 Diffusion of molecules from a small volume V1 through
a hole into the large volume V2 . After a sufficiently long time the
molecules are uniformly distributed over the total volume V1 C V2
T1
T1
D mc ln.Tm =T1 / :
Since Tm < T1 ! S1 < 0.
The change of S2 is accordingly
S2 D mc ln.Tm =T2 / ;
where S2 > 0. The total change of the entropy of the
system of bodies is therefore
S D S1 C S2
Chapter 10
D mc ln
Tm2
:
T1 T2
(10.94)
Since Tm D 21 .T1 C T2 / we get Tm2 =.T1 T2 / > 1
because the arithmetic mean is always geometric
mean. This gives S > 0. The entropy increases
during the irreversible process. The combination of
the two bodies at different temperatures to a combined system is an irreversible process, because the
cooled body cannot heat up again by cooling the other
body without the supply of energy from outside (second law).
Figure 10.55 Increase of entropy during the equalization of temperatures at the contact of two bodies with different temperatures
2. The second example, which will give us a deeper insight into the meaning of entropy, is related to the
diffusion of an ideal gas from a small volume V1
through a hole into a larger volume V2 . Initially (for
times t 0) the gas is confined to the small volume V1 . At t D 0 the hole in the barrier separating
V1 from V2 is opened and the gas molecules diffuse
into the evacuated volume V2 (Fig. 10.56). After a
The diffusion is irreversible, because it is highly improbable that all gas molecules diffuse back through
the hole into the small volume V1 (see below). It is,
however, nevertheless possible to calculate the entropy
change by using as reversible substitute process the
isothermal expansion (against an external pressure)
with the same initial and final states as the diffusion.
For this process, the supply of heat Q is necessary in
order to keep the temperature constant (Sect. 10.3.5).
Since the reduced heat does not depend on the way
during the expansion but solely on initial and final
states the entropy change S
S D R ln
V
V1
(10.95)
for the adiabatic expansions must be the same as for
the diffusion.
This can be also understood, when we substitute the
diffusion by a Gedanken-experiment, where the diffusion is separated into two steps (Fig. 10.57). The
gas drives during the isothermal expansion a piston
and extracts from a heat reservoir the heat Q1 as in
the Carnot cycle. The work W D Q1 performed
during the expansion drives a stirrer that releases the
heat Q1 again to the heat reservoir due to frictional
losses. For this Gedankenexperiment initial and final
states are identical to those of the diffusion. Therefore,
the entropy change must be the same. Since in (10.95)
V V1 ! S > 0.
Based on this diffusion process a statistical explanation of the entropy can be derived. We regard a
molecule in volume V1 . Before the hole in the barrier is opened, the probability of finding the molecule
in V1 is P1 D 1, because it must be for sure in V1 . After opening the hole the probability has decreased to
P1 D V1 =.V1 C V2 / D V1 =V.
For two molecules the probability of finding both
molecules in V1 is equal to the product P2 D P1 P1 D
10.3 The Three Laws of Thermodynamics
P21
of the probabilities for each molecule. For N
molecules we therefore obtain
N
V1
PN D
:
(10.96)
V
For 1 mol is N D NA D R=k, where k is the Boltzmann
constant and NA the Avogadro number. We then get
PNA D
V1
V
R=k
(10.97)
:
Piston
Stirer
Heat bath
Figure 10.57 Separation of the diffusion process in Fig. 10.56 into two
steps: Isothermal expansion and conversion of the mechanical work into
heat at the heat reservoir
J
289
S during the transition from state 1 (all molecules are in volume V1 ) to state 2 (all molecules are distributes over the volume
V D V1 C V2 the result
S D S.V/
D k ln
S.V1 / D k .ln P.V/
Pfinal
:
Pinitial
ln P.V1 //
(10.99a)
This illustrates that the entropy change S during the diffusion
from state 1 with the probability Pinitial to the state 2 with the
probability Pfinal
Pfinal
(10.99b)
S D k ln
Pinitial
is a measure for the probability that a system undergoes a transition from the initial to the final state. This can be formulated
in a more general way:
The probability that a system occupies a state i is proportional to the number Z of possible ways that lead to this
state.
For V1 D .1=2/V, NA D 6 1023 =mol the probability that
all molecules are found in V1 is
PN D 2
61023
23
D 10 1:810
0 (Fig. 10.58) :
When N particles, each with the energy ni E0 that are integer multiples of a minimum
P energy E0 occupy a state
with the total energy E D ni E0 , the number Z of possible realizations of this state equals the number of possible
combinations of the integers ni that fulfil the condition
P
ni D E=E0 .
J
The entropy S of a thermodynamic system state that can be realized by Z D P possible ways is
S D k ln P :
(10.99c)
The entropy S of a thermodynamic state is proportional to
the number of possible realizations for this state.
Figure 10.58 The probability P that all N molecules are simultaneously in the volume V1 D .1=2/V is P D .1=2/N
J
Because of the large numbers in the exponent it is more convenient to use the logarithm of the probability P. From (10.97) we
obtain
V1
V
k ln P D R ln
:
(10.98)
D R ln
V
V1
According to (10.95) the right side of (10.98) is equal to the
change S of the entropy. We therefore obtain for the change
As a third example, we will discuss the increase of entropy
for the mixing of two different kinds of molecules X and Y.
Initially all NX molecules X should be in volume V1 and all
NY molecules Y in volume V2 . We assume that pressure p
and temperature T are equal in both volumes, which demands
NX =V1 D NY =V2 . When we open a hole in the barrier between
the two volumes the molecules will diffuse through the hole until a uniform distribution of both kinds of molecules is reached.
This process is irreversible and the entropy increases because
the molecules NX as well as the molecules NY fill now a larger
volume V D V1 C V2 , and the number of possible realizations of
this situation is larger than that of the initial state. The increase
Chapter 10
Example
Example
290
10 Thermodynamics
of entropy for the NX molecules is
V
V NX
SX D k ln
D k NX ln
V1
V1
NX C NY
;
D k NX ln
NX
and for the NY molecules
SY D k NY ln
NX C NY
:
NY
The total change of entropy (called mixing entropy) is then
Sm D SX C SY
NX C NY
NX C NY
:
C NY ln
D k NX ln
NX
NY
(10.100)
These examples illustrate that only changes of entropy can be
measured. The absolute value of the entropy S.V; p; T/ of a thermodynamic state
S D S0 C S
(10.101)
Figure 10.59 Reversible collision process. Reversing the time course of the
process inverts the direction of all momentum vectors
is only defined, if the constant term S0 is known. We will show
in Sect. 10.3.10, how S0 can be determined.
10.3.8
Reversible and Irreversible Processes
Chapter 10
For a completely elastic collision between two particles, energy and momentum of the two-body system are conserved
(Sect. 4.2). A movie of such a collision process could run
backwards and the observer would not notice this, because the
reverse process is equally probable (Fig. 10.59). The collision
process is reversible. One can also say that the process is timeinvariant, i. e. one con exchange t with t without violating any
physical law.
Contrary to this situation, the collision process in Fig. 10.60,
where a bullet hits a glass ball that shatters in numerous pieces,
is irreversible. The inverse process, where the pieces come again
together to form a glass ball, which then emits the bullet, is
highly improbable. The following question now arises: Since
the glass ball consists of atoms and for each atom a reversible
collision process should occur, why is the macroscopic process
irreversible?
The answer to this question leads us again to the number of possible realizations of a macroscopic state that depends on a huge
number of atoms or molecules. While before the collision the
glass ball was at rest and the bullet had a well-defined energy
and momentum the final state could have a very large number of
possible realizations, because energy and momentum conservation still allows many different flight paths of the fragments as
long as the sum of all individual pieces fulfils the conservation
laws. The distribution of the fragments, observed for one experiment, represents only one of many possible distributions. At
Figure 10.60 Collision of a steel ball with a glas hollow sphere, which bursts
into many fragments, as example of an irreversible process
a second experiment under identical conditions, another distribution will be observed although both distributions obey energy
and momentum conservation. The entropy increases during this
collision process because the process starts from a state with a
well-defined realization and ends at a large number of possible
realizations. This is the signature of an irreversible process.
We can define an irreversible process as follows:
The change of a thermodynamic state of a closed system is irreversible, if the reverse process that leads to the initial state,
does not proceed by its own but only with additional energy
supply from outside.
The change of a thermodynamic state is irreversible, if the
entropy increases during this process.
The transition from an ordered state to a non-ordered state
(for instance the melting of a crystal) always increases the
entropy.
Note, that these statements are only valid for closed systems,
which do not interact with their surroundings. For a macroscopic subsystem the entropy can indeed decrease if that of the
10.3 The Three Laws of Thermodynamics
dF D dQirr C dW
Examples
1. When single crystals are formed out of the molten
bath, a non-ordered state (the liquid) is transferred into
an ordered state (the single crystal). The entropy S of
the crystal is lower than that of the liquid, however, the
decrease of S for a subsystem is over-compensated by
the increase of S for the surroundings of the crystal.
2. All living beings (plants, animals, humans) decrease
their entropy S by building up ordered structures, but
at the expense of an increase of S of their surroundings
(for example the digestion of food increases S).
J
In all of these cases the entropy of the total system increases!
Ordered structures therefore cannot be formed in closed systems. Their formation needs open systems far away from
thermal equilibrium. This nonequilibrium allows the exchange
of energy between the open system and its surroundings, which
can induce the decrease of the entropy of the open system.
For all macroscopic closed systems strictly speaking reversible
processes cannot occur, because always part of the kinetic energy (even if it is very small) is converted into heat by the
unavoidable friction. One of many examples is a swinging
pendulum, where the amplitude continuously decreases due to
friction by the air. The oscillation to the right will not be exactly
reproduced by the following oscillation to the left, because of
this amplitude decrease.
The interesting question, why the time has only one direction,
can be related to the increase of entropy. The time-derivative
dS=dt can define a time arrow that allows us to distinguish
between past and future [10.12]. For completely reversible processes time reversal would not change the validity of physical
phenomena.
10.3.9
With dQrev D TdS and inserting the free energy F D U TS )
dF D dU T dS S dT, this becomes
Free Energy and Enthalpy
The first and second law of thermodynamics contain the essential statements of thermodynamics. For their application to
special problems it is useful to introduce as a new parameter the
free energy
FDU
T S :
(10.102)
With the entropy S we can formulate the first law in a more
specific form:
dU D dQrev C dQirr C dW :
(10.103a)
S dT dW
SdT ;
(10.103b)
where the equal sign holds for reversible and the <-sign for irreversible processes.
For isothermal processes is dT D 0. This reduces (10.103) to
dF dW ) dW
dF :
(10.104a)
This means:
For isothermal processes, the maximum increase of the
free energy is equal to the mechanical work supplied to
the system.
The maximum work that a system can deliver during an
isothermal process, is equal to the decrease of its free
energy.
The difference
F DTS
U
is called bound energy. From the relations
dU D dQ C dW
and dF dW
follows by subtraction
d.U
F/ dQ :
For isothermal processes the bound energy U F is completely
converted into heat and is therefore not available for mechanical
work. This explains the label “bound energy”.
The second law of thermodynamics makes the following statement:
For isothermal processes, the change of the bound energy
is at least equal to the supplied heat. The increase of the
free energy is at most equal to the supplied mechanical
energy.
If the isothermal process occurs at a constant volume (dV D 0)
no mechanical energy is exchanged, i. e. dW D 0. Then
dF dW D 0 ;
(10.104b)
which means that the free energy decreases.
A spontaneous isothermal process without exchange of work always proceeds in the direction where the free energy decreases.
The entropy S then increases because of
TSDU
F
and
U D const :
Chapter 10
other subsystems which, interact with the selected one, does increase.
291
292
10 Thermodynamics
Since most of the processes occurring in nature are irreversible,
the free energy of the universe decreases and therefore also the
capability to perform mechanical work. All irreversible processes always tend to decrease existing temperature differences,
because then the entropy increases (see the examples in the previous section).
Pessimists say: “The universe strives towards its heat death.
This means, that all temperature differences approach zero,
where no longer any chemical and biological processes are
possible. However, it will take quite a while until this might
happen and furthermore it is not clear, which fate the universe
after many billion years will suffer, because it is still an open
question, whether the universe represents a closed or an open
system.”
As the last thermodynamic parameter we will introduce besides
the ethalpiy H D U C p V the free enthalpy G (also called
Gibb’s chemical potential) defined by the relation
G D U C pV
TS D H
TS :
(10.105)
The total differential of G is
dG D dU C p dV C V dp
T dS
S dT :
(10.106a)
Chapter 10
dU C p dV D T dS ;
(10.106b)
dG D V dp
(10.106c)
this converts to
10.3.10
Chemical Reactions
The mixing of the different components increases the entropy
(see (10.100)) by the amount
where xi D i =
P
X
i ln xi ;
iD1
i Ai !
p
X
j Bj :
(10.108)
jDkC1
The number of moles can change by the reaction. For instance
for the reaction
2H2 C O2 ! 2H2 O
P
P
is 1 D 2; 2 D 1 and 3 D 2 ! i ¤ j .
If the number i of moles for the i-th component changes by i ,
the free enthalpy G changes for processes with p D T D 0
according to (10.106a) by
G D
D
X
X
i i
i i
where x0i D .i C i /=
T Sm
X
RT
.i C i / ln x0i ;
(10.109a)
P
.i C i /.
When a reaction proceeds by its own (without external energy
supply), its free enthalpy must decrease; i. e. G < 0. Equilibrium is reached if G becomes minimal.
K D exp
hX
i
i Gi =RT :
The change of the free enthalpy can then be written as
h
i
X
G D RT ln K C
i ln i ;
(10.109b)
where i is the fraction of the component i that reacts.
Chemical reactions represent the basis of all living processes.
The utilization of food or the decomposition of waste products
proceed by chemical reactions. It is therefore of essential interest, under which conditions chemical reactions proceed by
themselves and when they need external energy supply for their
start. For all reactions that proceed at constant pressure and constant temperature the Gibbs’ potential is constant. That is the
reason why G is called chemical potential. Often several components react with each other. If i moles of the i-th component
P
exist before the reaction, the total free enthalpy is G D
i i
(i D Gibbs potential for one mole of the i-th component).
Sm D R
k
X
If the number of moles does not change during the reaction, one
can define a chemical equilibrium constant K by
With the first law of thermodynamics
S dT :
A chemical reaction between the molecules A which results in
the formation of molecules B is then described by
(10.107)
i
i is the mole fraction of the i-th component.
When the quantity of all components is one mole (i D 1)
Eq. 10.109b can be reduced to
G.1 mol/ D RT ln K :
(10.109c)
The equilibrium constant K is therefore directly related to the
change G of the chemical potential G.
10.3.11
Thermodynamic Potentials; Relations
Between Thermodynamic Variables
The thermodynamic variables: internal energy U, free energy
F, Gibbs’ potential G and enthalpy H are also called thermodynamic potentials. The advantage of their introduction is based
on the fact that all thermodynamic variables can be written as
partial derivatives of these potentials. The total differentials of
10.3 The Three Laws of Thermodynamics
the potentials are
Examples
@F
@V
T
@F
@T
ˇ
@U ˇˇ
DTI
@S ˇV
ˇ
@F ˇˇ
D pI
@V ˇ
dT
V
@U
dS
@S V
S
@G
@G
dG D
dp C
dT
@p T
@T p
@H
@H
dH D
dS C
dp ;
@S p
@p S
dU D
@U
@V
dV C
dV C
T
SdT ;
(10.103)
dU D pdV C TdS ;
(10.88)
dG D Vdp
(10.106c)
SdT ;
dH D dU C pdV C Vdp
(10.76)
D dQ C Vdp
gives the following relations between the thermodynamic variables and the potentials.
For the entropy we obtain
p
T
@G
@T
D
@F
@T
D
@U
@V
;
(10.110a)
;
(10.110b)
V
and for the pressure
pD
@F
@V
S
while the relation for the volume V is
@G
@H
VD
D
:
@p T
@p S
J
V
An example for the application of thermodynamic potentials is
given in Sect. 10.4.2.
where the lower index at the brackets denotes the quantity that
is kept constant. The comparison with the equations derived in
the previous sections
dF D pdV
ˇ
@G ˇˇ
D SI
@T ˇp
ˇ
@H ˇˇ
DV :
@p ˇ
(10.110c)
In the “Guggenheim Square”, each thermodynamic potential
can be placed in such a way, that the results of their derivatives can be immediately seen by the following procedure: One
goes in the scheme from the potential symbol to the derivative
variable and from there to the opposite corner on the diagonal.
If this way on the diagonal is in the direction of the arrow, the
result is positive, if it is opposite to the arrow, it is negative.
10.3.12
Equilibrium States
The thermodynamic potentials play a comparable role in thermodynamics as the mechanical potential Ep that determines the
forces F D grad Ep which governs the motion of particles.
In a similar way the gradient of the thermodynamic potentials
keeps the chemical processes running until the minimum of the
potentials is reached.
A system is at equilibrium, if without the action of external influences the state of the system does not change. If the state of a
system changes due to external action, but returns to its original
conditions after the external action ends, the equilibrium is stable. If, however, the system further removes from equilibrium,
even after the termination of the external influence, the equilibrium is unstable. A mechanical example is a mass m which is
fixed to a rigid rod that can rotate around a horizontal axis. At
the minimum of the potential energy, where the mass is just below the horizontal axis the mass is in a stable equilibrium. When
the mass is at its maximum height vertically above the horizontal axis, the equilibrium is unstable. Every slight perturbation
brings m downwards. In a thermodynamic system, the thermodynamic potentials take the role of the potential energy in our
example. We will illustrate this for several specific processes.
We assume a thermodynamic system with the internal energy U
and the volume V at the temperature T and the pressure p. An
arbitrary change of the conditions of the system is described by
the differentials dU, dV, dT and dp. If the change is reversible,
the work dW D p dV performed by the system during an
adiabatic expansion, causes a decrease dU D dW of the internal
energy.
For irreversible processes, the system loses heat which causes a
decrease of the total energy. Equilibrium is reached, if no further irreversible process is possible. Since for all irreversible
processes with constant volume the entropy increases, the equilibrium condition can be formulated as
dS 0 :
(10.111)
For all possible processes, which bring a thermodynamic
system away from equilibrium the entropy must decrease.
Chapter 10
dF D
SD
293
294
10 Thermodynamics
With other words:
10.3.13
A closed system with constant volume is in an equilibrium
state if its entropy is maximal.
The thermodynamic potentials of an equilibrium state have their
minimum value. This can be seen as follows:
With the change of entropy dS D dQ=T we obtain from the first
law of thermodynamics (10.71)
dU C p dV
T dS D 0 :
TdS/ 0 ! dF 0 :
(10.113)
Under isothermal and isochoric conditions a system has
reached its equilibrium state, if the free energy F has its
minimum value.
Under isothermal-isobaric conditions ( dT D 0 and dp D 0)
equilibrium is reached if
Chapter 10
dU C pdV
TdS D d.U C p V
We have seen in Sect. 10.3.7 that the entropy S is only determined apart from an additive constant S0 . We will now show,
that for T ! 0
lim S.T/ D 0 :
This fixes the constant S0 D S.T D 0/ D 0.
For the proof we start with the free energy F D U
Because of (10.110a) this can be also written as
F DUCT
(10.112)
For isothermal-isochoric processes is dV D 0 and T D const.
From (10.111) it follows: dS 0. Inserting this into (10.112)
gives
d.U
The Third Law of Thermodynamics
T S/ D dG D 0 ;
because for all processes that drive the system away from equilibrium dG > 0, as can be seen in an analogous way as the
arguments above for dF > 0.
Under isothermal-isobaric conditions a system is in an
equilibrium state, if the Gibbs’ potential is minimal.
In a similar way it can be proved, that for adiabatic-isobaric processes (dQ D 0 and dp D 0) the system is in an equilibrium state
if the enthalpy H D U C p V is minimal.
@F
@T
:
T S.
(10.114a)
V
We regard an isothermal chemical reaction where the system
starts with the free energy F1 and ends at F2 . The change F D
F1 F2 is
@
:
(10.114b)
F D U C T
F
@T
V
For T > 0 the changes F and U differ, but for T ! 0 the
difference approaches zero
lim.F
U/ D 0 :
(10.114c)
Nernst observed that with decreasing temperature the derivatives d.F/=dT and d.U=dT/ decreased and that they approached zero for T ! 0. This means that the curves F.T/
and U.T/ come towards each other with horizontal slopes
(Fig. 10.61).
Nernst therefore postulated that also for the general case
@F
D 0 and
T!0
@T V
@U
lim
D0:
T!0
@T V
lim
(10.115a)
(10.115b)
because of (10.114a) it follows then
lim
T!0
@U
@T
@F
@T
D0:
(10.115c)
For adiabatic-isochoric processes (dQ D 0 and dV D 0) the
internal energy < U must be minimal at equilibrium.
All reactions that are possible without external interaction
must start from states far away from equilibrium.
Therefore the thermodynamic treatment of chemical reactions
and biological processes is based on the description of systems
that are not at thermodynamic equilibrium.
Figure 10.61 Nernst’s theorem: Temperature course of U .T /; F .T / near
the absolute zero temperature
10.3 The Three Laws of Thermodynamics
With
@
F
@T
V
It is impossible to reach the absolute zero T D 0 of the
thermodynamic D absolute temperature scale.
D S
) lim S.T/ D 0 :
This can be seen as follows by an experimental argument.
T!0
This means that at sufficiently low temperatures reactions in
pure condensed substances proceed always without changes of
entropy i. e. S D 0. These reactions therefore proceed reversible.
If one tries to reach experimentally the absolute zero T D 0 this
could be only realized by an adiabatic process, because every
cooling process where heat is exchanged, requires a system that
is colder than the system to be cooled.
Investigating the dependence S.T/ of condensed substances
(liquified or solid gases) at very low temperatures, one finds
indeed that the entropy does not depend on the crystal modification or on the specific substance as long as it is a pure substance,
i. e. not a mixture of different substances. This suggests that the
entropy of all pure substances approaches for T ! 0 the same
value. Quantum theoretical considerations show (see Vol. 3) that
for all pure substances the entropy S approaches zero for T ! 0.
During an adiabatic process no entropy change occurs because
dQ D S dT D 0. For an adiabatic isobaric process is
dS D
This gives
dT D
(10.116)
The relations (10.115)–(10.116) are named the 3rd law of thermodynamics or Nernst’s Theorem.
Regarding the statistical interpretation of the entropy S D kln W
the 3rd law can be also formulated as:
The thermodynamic equilibrium state at T D 0 is a state
with maximum order, which has only one possible realization with P D 1. The entropy is then S D 0.
Note: The statement S.T D 0/ D 0 is only valid for pure substances. Mixed substances (for example mixed crystals) have
even for T D 0 an entropy S > 0, called the mixing entropy (see
Sect. 10.3.7).
S.T/ D
0
ZT
0
(10.118a)
and with (10.110a) and (10.114a) it follows
@S
@ @F
:
D
@V T
@V @T
For T ! 0 is with (10.115c)
@F
@U
!
D CV ;
@T
@T
and we obtain from (10.118a)
dT D T
CV
CV
D T
:
Cp
CV C R
(10.118b)
This shows that for T ! 0 also dT ! 0. The absolute zero
T D 0 for the temperature can be therefore not reached.
For one mole one obtains
dQrev
D
T0
.@S=@V /T
dV :
.@S=@T/p
For the partial derivative applies
Cp
@S
S
1 Q
D lim
;
D lim
D
T!0 T p
T!0 T T p
@T
T
The definition S0 D 0 for the zero point of the entropy allows
the determination of the absolute value of S.T > 0/.
ZT
@S
@S
dV C
dT D 0 :
@V
@T
Chapter 10
lim S.T/ D 0 :
C.T 0 / 0
dT :
T0
295
(10.117)
In order to fulfil the condition lim S.T ! 0/ D 0 the specific
heat C.T/ must converge sufficiently fast towards zero for T !
0 This is indeed observed experimentally (see Sect. 10.1.10).
More detailed measurements show that for solids at very low
temperatures C.T/ / T 3 (see Vol. 3). This is indeed observed
experimentally. The entropy S.T/ is then, according to (10.117),
also proportional to T 3 .
Remark. The first and second law of thermodynamics could
be formulated as the impossibility to realize a perpetuum mobile
of the first resp. the second kind. Also the third law can be
formulated as an impossibility statement:
10.3.14
Thermodynamic Engines
When the Carnot cycle in Fig. 10.51 is traversed into the opposite direction, i. e. counterclockwise, the corresponding engine
uses mechanical work to transport heat from the cold to the
warmer part of a system (Fig. 10.62). This has technical applications in refrigerators and heat pumps.
10.3.14.1 Refrigerators
In refrigerators the heat Q2 is extracted at a temperature T2 from
the volume V2 that should be cooled and a larger heat energy
Q1 D Q2 C W is transported to a warmer environment. This
296
10 Thermodynamics
Figure 10.62 Principle of a refrigeration machine and heat pump based on the
inverse Carnot’s cycle
Figure 10.63 Technical realization of a refrigerator
demands the supply of mechanical or electrical work W to the
system. We have neglected all energy losses by friction or heat
conduction. The coefficient of performance
Kref D
Q2
dQ2 =dt
D
W
dW=dt
(10.119a)
gives the ratio of cooling rate dQ2 =dt and power input dW=dt.
From the efficiency of the Carnot engine we obtain for the
inverse Carnot cycle the coefficient of performance for the refrigerator the relation
Kref D 1= D
T2
T1
T2
:
(10.119b)
Chapter 10
This shows that a refrigerator works more efficiently for small
temperature differences (T1 T2 ) between the cooled volume
and the warmer environment.
10.3.14.2
Heat Pumps
Heat pumps use the heat reservoir of the environment (air,
ground) for heating water for floor heating of rooms or for
swimming pools. The basic principle is the same as that of refrigerators. Heat is transported from a cold to a warmer volume.
This demands the supply of mechanical or electrical energy.
The useful energy is the heat transported to the warmer volume. Therefore the coefficient of performance is defined as in
(10.119b)
Khp D
dQ1 =dt
T1
Q1
:
D
D
W
dW=dt
T1 T2
(10.120)
Contrary to the efficiency < 1 of the Carnot engine the coefficient Khp D 1= is larger than 1! It increases with decreasing
temperature difference T D T1 T2 .
Example
A heat pump used for heating a swimming pool takes the
heat from a river with a water temperature of 10 ı C D
283 K and heats the swimming pool to a temperature of
T D 27 ı C D 300 K. The maximum coefficient of performance is then Khp D 17:6. One therefore saves a
factor of 17:6 of heating costs compared with the direct
heating of the swimming pool. In this idealized example
all other losses of the heat pump system have been neglected. Realistic values, taking into account all losses,
are Khp D 5–10.
J
For practical applications heat pumps and refrigerators operate
with special cooling liquids, which are not permanent gases but
evaporate and condense during one cycle. This is illustrated
schematically in Fig. 10.63. The heat Q2 is transported from
the room to be cooled to the liquid cooling agent at the low temperature T2 in the evaporator. The resulting temperature increase
of the cooling liquid results in the evaporation of the liquid . In
the condenser the heat Q1 is extracted by a heat exchanger from
the vapour at high pressure. This causes the condensation of the
vapour. The liquid under high pressure expands through a throttle valve, which decreases its temperature and is again used for
heat extraction from the volume to be cooled.
10.3.14.3 Stirling Engine (Hot Air Engine)
The Stirling engine uses air as working agent, which is periodically expanded and compressed in a cycle of two isotherms and
two isochors (Fig. 10.64a). The red arrows indicate the heat exchange between the environment (white) and the system (red).
During the isothermal expansion 1 ! 2 the heat Q1 is supplied
at the temperature T1 to the system. During the isochoric cooling 2 ! 3 the temperature drops to T2 < T1 . Now isothermal
compression 3 ! 4 occurs where the heat Q2 is transported
to the environment. Finally the isochoric compression 4 ! 1
with heat supply Q4 brings the system back to its original state
1. The heat Q4 is necessary to increase the temperature from T2
to T1 > T2 . Since no work is performed during the isochoric
processes, the energy balance demands
Q2 D
Q4 D Cv T :
10.3 The Three Laws of Thermodynamics
297
Figure 10.67 Stirling motor with two pistons in one cylinder
Figure 10.65 Stirling engine with two pistons and two cylinders
This can be technically realized, at least approximately, by using
two pistons, the working piston and the displacer piston in two
different cylinders: a hot cylinder and a cold one (Fig. 10.65).
The two pistons are driven by the same crankshaft with a 90ı
phase shift against each other. The two cylinders are connected
by a pipe filled with an energy storage material (regenerator).
When the piston compresses the gas in the hot cylinder the hot
gas flows from the hot to the cold cylinder through the connecting pipe and heats up the storage material. In the next step the
cold cylinder is compressed and the cold gas flowing through
the pipe is heated by the storage material. About 80% of the
energy exchanged during one cycle can be stored in the regenerator. In the diagrams of Fig. 10.66 the time sequence of the
total volume, the hot and the cold volume are depicted for the
Stirling engine, used as heat engine and as heat-pump.
Figure 10.66 Volume-diagram V .t / of a Stirling engine: Left diagram: used as heat engine, right diagram: used as heat pump
Chapter 10
Figure 10.64 Cycle of a Stirling engine, b Otto engine (gasoline engine),
c Diesel engine, d steam engine. The red curve gives the vapor pressure p .V / of
water vapor
When the extracted heat Q2 can be stored and resubmitted to
the system during the process 4 ! 1, the system does not loose
energy during a cycle and the efficiency of the Stirling process
would be comparable to that of the Carnot cycle.
298
10 Thermodynamics
Heated wall: T1
Slave piston
Heat storage
Working piston
Cold wall: T2
Isothermal
expansion
Isochoric
cooling
Isothermal
compression
Isochoric
heating
Figure 10.68 Positions of working piston and slave piston during the four sections of a Stirling cycle. The energy necessary for the operation of the engine is
supplied by heating the upper wall
Another version of the Stirling engine uses only one cylinder
but still two pistons. The displacer piston presses the air periodically into the upper hot volume and in the next step into
the lower cold volume. During these processes the air streams
through a hole in the piston which is filled with metal cuttings.
They are heated during the passage 2 ! 3 of the hot air and
they transfer their heat during the passage 4 ! 1 to the cold
air.
Chapter 10
In Fig. 10.67 and 10.68 the different steps durng a working cycle
are illustrated for a Stirling engine with one cylinder and two
pistons.
10.3.14.4
The Otto-Engine
The Otto-engine is used in many cars as effective drive. In the pV-diagram it passes a cycle consisting of two isentropic and two
isochoric processes (Fig. 10.64b). In the state 1 the gasoline-air
mixture is sucked in and compressed. At point 2 the ignition
oocurs where the mixture explodes so fast that the volume does
not change essentially. The heat Q1 released at the explosion
is fed into the system and increases the pressure very fast up
to point 3. Now an isentropic expansion (no further heat supply) follows until point 4 is reached. Here the exhaust valve is
opened and the exhaust gas streams into the exhaust pipe. This
causes a decrease of the pressure, a release of the heat Q2 and
the restitution of state 1.
The efficiency depends on the compression ratio V1 =V2 . One
obtains (see Probl. 10.12)
Example
V1 =V2 D 9 and D 9=7 ! D 0:44. Note, that the real
efficiency is only about 0:3–0:35 due to energy losses by
J
friction and heat conduction.
10.3.14.5 Diesel Engine
For the Diesel engine the cycle in the p-V-diagram (Fig. 10.64c)
consists of two isentropic, one isobaric and one isochoric process. In the state 1 air is sucked in and the volume is compressed
until point 2 is reached. The compression ratio is much larger
(up to 1:20) as in the Otto-engine. During this compression the
temperature rises to 700–900 ı C, which is above the ignition
termperature of Diesel-fuel. Now Diesel fuel is injected, which
does not explode as in the Otto-engine but burns more slowly
(there is no electrical ignition). This causes the air-fuel-mixture
to expand isobaric until point 3 is reached, where the combustion stops. The volume now further expands isentropically to
the point 4, where the exhaust valve opens and the pressure suddenly drops to the atmospheric pressure outside. Here the intial
point 1 is reached again.
The efficiency of the Diesel engine is higher thatn that of the
Otto-engine because of the higher compression ratio. Its theoretical value is about 0:55 but due to unavoidable losses the real
enegines only reach about 0:45. The disadvantage of the Diesel
engine is the higher output of NOx gases and soot particles.
10.3.14.6 Steam Engine
D1
1
.V1 =V2 /
1
;
where D Cp =Cv is the specific heat ratio.
(10.121)
In a Steam Engine the cyclic process (Clausius–Rankine
Process) consists of two isentropic and two isobaric parts
(Fig. 10.64d). In the initial state 1 the system contains water.
A pump increases the pressure at a constant volume isentropic
10.4 Thermodynamics of Real Gases and Liquids
The red curve in Fig. 10.64d gives part of the Van der Waals
curve p.V/ for water vapour (see Sect. 10.4.1). Inside this curve
water and vapour can exist simultaneously, in the region left of
the curve only the liquid phase exists, to the right hand of the
curve only the vapour phase.
10.3.14.7
Thermal Power Plants
In thermal power plants, heat is produced by burning fossil fuels,
such as coal, oil, wood or gas, or by fission of atomic nuclei.
For fossil fuels the heat comes from the reaction heat that is
released during the oxidation of atoms or molecules and is due
to the different chemical binding energies of reaction partners
and reaction products. The essential part of this energy stems
from the oxidation of carbon atoms C to CO2 . The produced
heat is 8 kcal D 33 kJ for 1 g C. The fission of 1 g Uranium
produces an energy of 2:5 107 kJ. This is 7:5 105 times more!
The heat produced in thermal power plants is converted into
the generation of hot water vapour under high pressure, driving turbines that propel electric generators for the production of
electric energy. The maximum efficiency depends, according to
the second law, on the initial temperature T1 and the final temperature T2 .
The initial temperature is limited by technical conditions (heat
and pressure resistance of the hot vapour tank. Typical values
are between 600 and 700 ı C. Only for the high temperature reactors, temperatures above 800 ı C are realized.
For the choice of the final temperature T2 two options exist:
1. One chooses T2 D 100 ı C (condensation temperature of water) and uses the rest energy of the hot water for heating of
houses. The efficiency for the conversion of heat into mechanical (or electric) energy is then for an initial temperature
of T1 D 600 ı C D 873 K: D 500=873 D 0:57. In addition the heat Q of the hot water can be delivered to houses
nearby the power station.
2. The final temperature of the water vapour is chosen as
T2 D 30 ı C. In order to avoid condensation one has to
lower the pressure below the atmospheric pressure by pumping the expanding volume. This increases the efficiency to
D 570=873 D 0:65. The work needed for evacuating the
expanding volume against the external pressure is smaller
than the additional energy gain due to the lower final temperature.
In case 1 one does not win the total energy of the hot water Q
compared to case 2 because here one could use the extra energy
due to the higher efficiency 2 D .600 30/=873 to transport
electric energy for heating. The increase of the efficiency for
case 2 compared to case 1 is
2
1 D .70=873 "/Q
D .0:19 "/Q ;
where " Q is the mechanical work of the pump, necessary
to evacuate the volume down to a pressure that is equal to the
vapour pressure of water at T D 30 ı C.
10.4
Thermodynamics of Real Gases
and Liquids
Up to now, we have discussed the thermodynamics of ideal
gases, where the interaction between the atoms of the gas has
been neglected.
We will now discuss, which rules have to be generalized and
which are still valid without restrictions, when we treat the thermodynamics of real atoms and molecules including their size
and their mutual interactions.
While ideal gases remain gaseous at any temperature, real gases
condense below their boiling temperature and they can even become solids below the melting temperature. In this section, we
will investigate what are the conditions for transitions between
the different phases solid, liquid and gaseous and what are the
equilibrium conditions of the different phases.
10.4.1
Van der Waals Equation of State
At very high pressures, the density of atoms or molecules becomes so high, that the internal volume of the molecules (also
called covolume) cannot be neglected compared with the free
volume V that is available for the molecules.
When we describe the atoms as rigid balls with radius r, two
atoms cannot come closer to each other than at a minimum distance d D 2r. If one atom is in the volume V the other atoms
cannot penetrate into the volume Vforbidden D .4=3/d 3 D 8Va ,
where Va is the volume of one atom in the model of rigid balls
(Fig. 10.69a). Furthermore the centres of all balls must have the
minimum distance d D r from the walls of the container.
Assume there were only two atoms in the cubic volume V D l3 .
The volume available for the second atom is then
V2 D .L
2r/3
8Va
(Fig. 10.69b) :
A third atom could only be found in the volume
V3 D .L
2r/3
2 8Va ;
and the rest volume available for the n-th atom is then
Vn D .L
2r/3
.n
1/ 8Va :
Chapter 10
from p1 to p2 . From point 2 to 3 heat is supplied at constant
pressure, which causes the volume to expand and increases the
temperature above the boiling point of water. The hot vapour
drives a piston during the isentropic expansion and the system reaches point 4 where the temperature is cooled down, the
vapour condenses and the heat Q2 is transferred to the surrounding. Now the initial point 1 is reached again. Mechanical work
is performed on the part 3 ! 4.
299
300
10 Thermodynamics
Figure 10.70 Illustration of the internal pressure. Forces on an atom A a inside
a gas, b at the boundary between wall and gas
Figure 10.69 Illustration of the co-volume. a The center of B cannot be in the
bright red circle. b Forbidden volume (bright red ) of molecule B in the volume
L 3 with one atom A
Chapter 10
An estimation of the real sizes shows, however, that, for example, for N D 1020 atoms in a volume with L D 0:1 m the
forbidden volume at the wall is with r 10 10 m completely
negligible compared to the internal volume N Va of the N atoms
in the volume V. The average over all N atoms gives then for
the mean volume available to each atom
V D .L
D L3
L3
2r/3
6r L2
4NVa
N
1X
.n
N nD1
1/ 8Va
6r2 L C 8r3
4NVa
(10.122)
for N 1 :
We therefore have to replace in the general gas-equation (10.21)
the volume V by the reduced volume
V
bDL
4 N Va
with
pb D a %2 / a=V 2 ;
which acts onto the atoms in addition to the external pressure p.
because the 2nd, 3rd and 4th term are neglible compared to the
1st and last term.
3
cancels for atoms inside a liquid or a gas volume, because the
interaction forces are in the average uniformly distributed over
all directions (Fig. 10.70a) (see the similar discussion about the
surface tension in Sect. 6.4.1). At the boundaries between liquid
and gas or between gas and wall, the interaction forces are no
longer uniformly distributed but are directed only into the half
space of the medium. They do not cancel and the total force Fa
on one atom is not zero but is proportional to the number density na of atoms in the half-sphere shown in Fig. 10.70b, which
means to the density % D M=V, where M is the total mass of the
gas in the volume V.
P
The amount of the total force F D j Fi j / na Fa onto all na
2
atoms is therefore proportional to na / %2 . The force is directed
towards the interior of the gas and causes an intrinsic pressure
b D 4 N Va :
For the situation N 1 the volume Vavailable available to
the N atoms in a volume V is reduced by 4 times the total
atomic volume N Va D N .4=3/ r3 .
The next question concerns possible corrections for the pressure
p due to the attractive or repulsive forces between the atoms. At
low temperatures or for high densities the interaction between
the atoms can be no longer neglected. The total force on a selected atom resulting from the interaction with all other atoms
Taking into account this intrinsic pressure and the co-volume
b D 4N Va the general gas equation for one mole of an ideal
gas
p VM D R T
has to be modified to the van der Waals-equation of real gases
pC
a
V2
.VM
b/ D R T ;
(10.123)
where the constant b D 4 Na Va gives 4-times the internal
volume of the NA atoms in the mole volume VM .
The progression of the function p.V/ at constant T for a real
gas, described by (10.123), depends on the constants a and b. In
Fig. 10.71 the isotherms of CO2 are shown for different temperatures. They confirm, that for high temperatures (Ekin jEpot j)
the curves are similar to those of an ideal gas, but for low
temperatures closely above the condensation temperature they
deviate strongly.
10.4 Thermodynamics of Real Gases and Liquids
301
Figure 10.72 Measurement of vapor pressure curve ps .T /
Solving (10.123) for p at constant T gives a polynomial p.V/
of third order, which shows for low temperatures a maximum
and a minimum (Fig. 10.71). How looks the comparison of this
theoretical curve with experimental results? Let’s see this for
the example of CO2 .
If one mole of CO2 is compressed at the temperature T D 0 ı C
starting at low pressures one finds in deed that the curve p.V/
follows quite nicely the theoretical curve until the point A in
Fig. 10.71. Further compression does not increase the pressure
p, which stays constant until the point C is reached where the
pressure shows a steep increase and follows again the van der
Waals curve.
The reason for this strange behaviour is the condensation of the
CO2 vapour that starts at the point A. On the way from A to
C the fraction of the liquid phase continuously increases until
in C the vapour is completely liquefied. On the way from C to
smaller volumes, the pressure increases steeply because of the
small compressibility of the liquid. Between A and C gas and
liquid can both exist (co-existence range).
For a quantitative description of the condensation process we
must discuss the different phases (aggregation states) in more
detail.
10.4.2
Matter in Different Aggregation States
The different aggregation states of matter (solid, liquid gas) are
called its phases. In this section we will discuss, under which
conditions a phase transition solid ! liquid, liquid ! gas or
solid ! gas can occur and when two or three phases can exist
side by side.
10.4.2.1
Vapour Pressure and Liquid–Gas Equilibrium
When a liquid is enclosed in a container which it fills only partly,
one finds that part of the liquid is vaporized and in the volume
above the liquid surface a vapour phase has established at a
vapour pressure ps .T/ that acts upon the walls and the liquid
surface. The dependence of the vapour pressure ps .T/ on the
temperature can be measured with the pressure tank shown in
Fig. 10.72 that is equipped with a thermometer and a manometer.
At a constant temperature T a constant saturation vapour pressure ps .T/ is present where the liquid and the gaseous phase can
exist simultaneously under stable conditions.
The explanation given by molecular physics is based on the kinetic gas theory (Sect. 7.4). Similar to the situation in a gas also
the molecules in a liquid show a velocity distribution with kinetic energies that follow the Maxwell–Boltzmann distribution.
The fastest molecules in the high energy tail of the velocity distribution can leave the liquid, if their energy is larger than the
surface tension of the liquid (See Sect. 6.4). On the other hand,
when molecules in the gas phase hit the liquid surface, they can
enter into the liquid.
At the saturation vapour pressure ps .T/ the liquid and the gasphase are at equilibrium, which means that the rate of molecules
leaving the liquid is equal to the rate of molecules that reenter
the liquid from the gas phase.
The higher the temperature the more molecules have sufficient
energy to leave the liquid, i. e. the vapour pressure rises with increasing temperature (Fig. 10.72). The quantitative form of the
vapour pressure curve ps .T/ can be calculated in the following
way:
In Fig. 10.73 we regard for 1 mol of the evaporating liquid a
cyclic process in the p, V-diagram of Fig. 10.71. In the state
C0 .T C dT; ps C dps / the vapour should be completely condensed and the liquid occupies the volume Vl . Now the volume
is isothermally expanded at the temperature T C dT, while the
pressure is kept constant. Here the heat dQ1 D , which is
equal to the evaporation energy of 1 mol, has to be supplied
in order to keep the temperature constant. At A0 the total liquid is evaporated. During the next step, the adiabatic expansion
A0 ! A, pressure and temperature are lowered by an infinitesimal small amount. The system remains in the vapour phase
and reaches the point A.ps ; T/. Now the vapour is isothermally
compressed while the pressure remains constant, because condensation progresses during the path from A to the point C. The
Chapter 10
Figure 10.71 Van-der-Waals-isotherms of CO2 for different temperatures
302
10 Thermodynamics
contribution is by far the largest one. It is therefore nearly equal
to the heat of evaporation.
Example
Figure 10.73 Carnot cycle C0 A0 ACC0 in a p -V diagram of Fig. 10.71 illustrating
the derivation of the Clausius–Clapeyron-equation
The volume of 1 kg water expands during the evaporation
from Vl D 1 dm3 to Vv D 1700 dm3 . The work performed
during the expansion against the external pressure of 1 bar
is W D p dV D 105 Nm 1:7 m3 D 170 kJ. The measured
specific evaporation heat is D 2080 kJ=kg. Therefore
J
the first contribution only amounts to 8%.
condensation heat dQ2 is released to the surrounding. The step
A ! C corresponds to the curve ABC in Fig. 10.71. The liquid
state in point C is then transferred by an infinitesimal step to the
initial point C0 .p C dp; T C dT/.
As one of many applications of the thermodynamic potentials we will derive the Clausius–Clapeyron equation (10.124)
with the help of the thermodynamic potentials, where here the
Gibbs’-potential G.p; T/ of (10.105) is used.
The temperature of the system changes only on the short paths
A0 ! A and C ! C0 . During the isothermal expansion C0 ! A0
the system has delivered the work dW1 D .ps C dps / .Vl Vv /,
while during the compression A ! C the work dW2 D ps .Vv
Vl / has to be supplied to the system. The net work is therefore
dW D dW1 C dW2 D .Vl Vv / dps .
In Sect. 10.3.5 it was shown that the efficiency of the Carnot
engine for an arbitrary working material is
jWj
.Vv Vl /dps
D
D
Q1
Chapter 10
because here is dT T. This gives for the evaporation energy
for 1 mol evaporated liquid the Clausius–Clapeyron equation
dps
.Vv
dT
Vl / :
dG D
ˇ
ˇ
@G ˇˇ
@G ˇˇ
dp
C
dT :
@p ˇT
@T ˇp
The compilation scheme of the potentials in Sect. 10.3.11 shows
ˇ
@G ˇˇ
DV
@p ˇT
ˇ
@G ˇˇ
D S:
@T ˇp
and
At the phase equilibrium is dG1 D dG2
T C dT T
dT
D
;
T C dT
T
DT
Differentiation of (10.105) gives
(10.124)
The evaporation heat is proportional to the difference of the
mole-volumes of the liquid and gaseous phases and to the slope
dps =dT of the vapour pressure curve.
.S2
! dG1 D V1 dp
S1 dT D V2 dp
S1 /dT D .V2 V1 /dp
dp
S2 S1
D
:
dT
V2 V1
S2 dT D dG2
From the definition of the entropy we conclude
S2
S1 D
Z2
1
dQrev
D
;
T
T
which finally gives
Note: Often the specific evaporation energy ŒkJ=kg is given
instead of the molar evaporation energy ŒkJ=mol. The conversion factor is
dp
D
:
dT
T.V2 V1 /
1 kJ=mol D .10 3 M/ kJ=kg ;
The heat supply does not increase the kinetic energy of the
molecules, (because the temperature stays constant), but only
the potential energy. Therefore in Fig. 10.18 the long horizontal
line T(t) appears during the evaporation process.
where M is the molar mass in g=mol.
The evaporation energy pro molecule is w D =NA with NA D
Avogadro number.
The heat of evaporation has two causes: The first cause is the
energy necessary to enlarge the volume Vl of the liquid to the
larger volume Vv of the vapour against the external pressure p.
Since in (10.124) Vv Vl we can neglect Vl in (10.124) and we
can also approximate in the general gas equation p V D RT the
volume V Vv ! Vv D R T=ps . Inserting this into (10.124)
we get
The second cause is the energy spend to enlarge the distance between the molecules against their mutual attraction. The second
1 dps
D
:
ps dT
RT 2
10.4 Thermodynamics of Real Gases and Liquids
Integration yields
ln ps D
CC
RT
with the integration constant C. This gives with the boundary
condition ps .T0 / D p0
ps D p0 A e
=.RT/
with
A D e=RT0 :
(10.125)
This van’t-Hoff equation shows that the vapour pressure rises
proportional to exp. 1=T/.
Along the vapour pressure curve ps .T/ the vapour phase and the
liquid phase are at equilibrium, i. e. at each temperature there
exists a vapour pressure ps .T/ where the two phases exist simultaneously and are both stable.
Figure 10.75 Behaviour of Van-der-Waals isotherms p .V / around the critical
point .pc ; Tc /
When the volume V is compressed, the real pressure curve
(Fig. 10.75) shows a kink at V D V2 and follows until V1 not
the van der Waals curve but the horizontal line p D const because here condensation takes place. The dashed black curve in
Fig. 10.75 gives the volume V2 where condensation starts and
V1 where the whole gas is liquefied. At the critical temperature
Tc the curve p.V/ has no longer minima and maxima but only an
inflection point, which indicates that there are no longer phase
transitions but only a unique phase is present, which is called the
supercritical phase. The tangent to the curve p(V) in the critical
point p.Tc ; Vc / is horizontal. The critical point can be calculated
from the van der Waals equation (10.123) with the conditions
@p
@V
Tc ;Vc
D0
and
@2 p
@V 2
Tc ;Vc
The vapour pressure curve divides the area in the p-T-diagram
into two sections (Fig. 10.74). For p.T/ < ps .T/ only the vapour
phase exists under equilibrium conditions, for p.T/ > ps .T/
only the liquid phase.
This gives for pc and Tc the results
The vapour pressure curve terminates at the critical temperature
T D Tc . The corresponding vapour pressure pc D ps .Tc / is the
critical pressure. Above the critical temperature Tc no distinction between liquid and vapour phase is possible. The densities
of both phases become equal. The slope of the vapour pressure
curve is there
dps
pc
:
(10.126)
D
dT Tc
RTc2
and for the van der Waals constants a and b
The evaporation heat decreases with increasing temperature and
becomes zero at the critical temperature Tc . Just below Tc part of
the liquid changes statistically into the vapour phase and back.
This causes striations in the optical density which can be seen
in the transmitted light.
10.4.2.2 Boiling and Condensation
The critical temperature is related to the interaction potential between the molecules. Above Tc the mean kinetic energy of the
molecules is larger than the amount of the mean potential energy. In the p-V-diagram of Fig. 10.71 the isotherms have for Tc
three intersection points with the horizontal line p D const < pc .
pc D
1 a
I
27 b2
Vc D 3b I
a D 3pc Vc2 I
bD
Tc D
D0:
8 a
;
27 R b
1
Vc :
3
(10.127a)
(10.127b)
It is therefore possible to gain information about the attractive interaction between the molecules and their internal volume
from measurements of the critical parameters pc and Tc .
If the vapour pressure ps becomes larger than the external pressure p acting on the liquid surface, vapour bubbles can form in
the inside of a liquid. They rise, due to buoyancy, to the liquid
surface: The liquid boils. The boiling temperature Tb depends
on the external pressure p. From (10.125) one obtains
Tb .p/ D Tb .p0 /
1
1
RTb .p0 /
ln.p=p0/
:
(10.128)
Chapter 10
Figure 10.74 Phase diagram with vapor pressure curve ps .T / representing the
separating line between liquid and vapor phase from the triple point up to the
critical point P .pc ; Tc / and melting curve Psl .T / as separation line betweeen solid
and liquid phase
303
304
10 Thermodynamics
Example
10.4.2.3 Liquefaction of Gases;
Joule–Thomson Effect
Water boils under a pressure p D 1 bar at Tb D 373 K D
100 ı C. For p D 400 mbar Tb D 77 ı C. Since the cooking
time of food strongly depends on the temperature, cooking at high altitudes becomes tedious. Therefore one uses
a pressure cooker, which operates at about 1:5–2 bar and
J
reduces the cooking time considerably.
In order to liquefy gases at the pressure p one has to lower their
temperature below the pressure-dependent boiling temperature
Ts .p/. There are several experimental realizations:
If the vapour pressure becomes smaller than the external pressure the vapour starts to condensate.
In our atmosphere the air mixed with water vapour generally
does not reach an equilibrium state (p, T), because the conditions in the atmosphere change faster than the time necessary to
establish an equilibrium. The water vapour pressure is therefore
in general lower than the saturation pressure.
The concentration of water vapour in our atmosphere, measured
in g=m3 , is called the absolute humidity 'a . The maximum possible concentration of water vapour is reached, when the water
vapour pressure pw is equal to the saturation pressure ps . The
humidity ' at this pressure is the saturation humidity 's .
The relative humidity is the quotient
'rel D
'a
pw
D
:
's
ps
(10.129)
Adiabatic Cooling with Energy Output
Here the internal energy U of the gas at the pressure p1 decreases because the expanding gas delivers the work dW D pdV
against the lower external pressure p, while no heat is exchanged
(dQ D 0). From the first law (10.82) we obtain for 1 mol
dU D Cv dT D p dV :
This yields the temperature decrease
dT D
pp
dV :
Cv
Example
10 mol of a gas at room temperature T D 300 are expanded against an external pressure of 10 bar D 106 Pa
by V D 10 2 m3 (this corresponds to 5 mol volumes.
With Cv D 20:7 J=.mol K/ we obtain T D 4:8 K.
J
Example
Chapter 10
A relative humidity of 40% is reached, when the vapour
pressure of water is pw D 0:4ps (H2 O).
J
For a given absolute humidity the relative humidity increases
with decreasing temperature because the vapour pressure of water decreases with T. (Fig. 10.76). When 'rel D 1 it starts to
rain. The temperature Td where 'rel D 1 is the dew point or
saturation temperature.
For the operation of air conditioning systems, this has to be
taken into account. If the air is cooled below the dew point,
the water vapour will condense and increase the humidity in the
cooled room. The air has therefore to be dried before it is cooled
down.
This adiabatic cooling can be realizes for ideal and also for real
gases. It comes from the decrease U of the internal energy due
to the partial transfer into mechanical work.
Joule–Thomson Effect
For real gases cooling can be also achieved without the transfer into mechanical work. The expansion of the volume V
increases the mean distance between the molecules. This requires work against the attractive forces between the molecules,
which means that the potential energy of the system increases at
the expense of the kinetic energy and the temperature decreases.
When a real gas expands adiabatically through a nozzle at a
pressure p1 that is kept constant, from the volume V1 into the
volume V2 (Fig. 10.77) with the pressure p2 < p1 there is no
heat exchange with the surrounding (dQ D 0) and the enthalpy
H D U C p V, is constant because the cooling is due to the
work against the attractive forces between the molecules during
the expansion.
The internal energy U of a real gas is the sum of the kinetic
energy Ekin D .f =2/ R T and a the potential energy
Ep D
Figure 10.76 Illustration of relative and absolute humidity of air and of dew
point
ZV1
1
a
dV D
V2
a
;
V1
which is due to the attractive forces between the molecules and
causes the internal pressure (cohesion pressure).
10.4 Thermodynamics of Real Gases and Liquids
Solving the van der Waals equation (10.123) for p, we obtain
RT
pD
V b
a
:
V2
The enthalpy then becomes
f
H D U C p V D RT
2
f
V
D RT
C
2
V b
a
C
V
2a
:
V
RT
V b
a
V2
V
(10.130)
Since H is constant during the adiabatic expansion through the
nozzle, we get
dH D
) dT D
@H
@H
dV C
dT D 0
@V
@T
bT
2a
@H
dV
2
2
.V
b/
RV
@V
D
dV
f
V
@H
C
2
V b
@T
(10.131)
bRT 2a
dV :
. 12 f C 1/RV 2
For temperatures below the inversion temperature
Ti D
2a
;
bR
(10.132)
we get dT < 0. The gas cools down although no heat exchange with the surrounding takes place. The value of the
inversion temperature depends on the ratio of the amount of
the attractive forces (described by the constant a) and the covolume b D 4N Va of the molecules. For ideal gases is
a D b D 0 ) dT D 0 and no cooling occurs. The cooling of
real gases through adiabatic expansion through a nozzle is the
Joule–Thomson effect. It is only realized for real gases, not for
ideal gases. This can be seen as follows: When the gas flows
Figure 10.78 Schematic illustration of the Linde-process for liquefaction of air
through the nozzle, driven by the pressure p1 , the energy p1 V1
is released. The gas, streaming into V2 , builds up the pressure p
which requires the energy p2 V2 . Energy conservation demands
U1 C p1 V1 D U2 C p2 V2 ) H1 D H2 (10.76). The expansion
therefore proceeds at constant enthalpy.
For temperatures above the inversion temperature is dT > 0, i. e.
the gas heats up. In order to use the Joule–Thomson effect for
cooling, the gas has at first to be precooled below the inversion
temperature. For higher pressures the density of molecules increases and with it the relative share of the covolume b D 4N Va
and the inversion temperature Ti becomes pressure dependent.
More detailed information on the curves Ti .p/ can be found
in [10.14].
In Tab. 10.7 the maximum values of Ti are compiled for some
gases. The numbers show that for air the inversion temperature lies above room temperature. Therefore precooling is not
necessary. The gases N2 and O2 can be cooled below their condensation temperatures solely with the Joule–Thomson effect.
This is realized with the Linde-gas liquefying system, which
uses the counter-current principle (Fig. 10.78). The gas is compressed by the piston K and streams through the valve Vl1 into
the volume V2 where it is dehumidified. It then passes through a
cooling system where it is precooled, before it streams through
the counter current cooler and finally through a nozzle into the
container D at low pressure. During this last step, it further cools
down. The cooling rate is for air T=p D 0:25 K=bar. For a
pressure difference p D 100 bar one reaches a cooling rate of
25 K per step. The cold vapour is guided through the countercurrent cooler and helps to precool the incoming gas. Finally, it
is sucked in through the valve Vl2 into the initial chamber during the expansion phase of the piston. The next step starts then
already with a colder gas and reaches therefore a lower final
temperature. After several steps the cooling during the expansion through the nozzle reaches the condensation temperature
and the gas is liquefied.
During the cooling of air, which is composed of N2 and O2 , at
first the higher condensation temperature of oxygen is reached
Chapter 10
Figure 10.77 Comparison between adiabatic cooling and Joule–Thomson effect. a Adiabatic expansion with work delivery W D pa V ; b adiabatic
expansion through a nozzle without work output
305
306
10 Thermodynamics
Table 10.7 Critical temperatures Tc , critical pressure pc , maximum inversion temperature Ti and boiling temperature for some gases
Gas
Helium
Hydrogen
Nitrogen
Oxygen
Air
CO2
NH3
Water vapor
Tc =K
5.19
33.2
126
154.6
132.5
304.2
405.5
647.15
Pc =bar
2.26
13
35
50.8
37.2
72.9
108.9
217.0
a=N m4 =mol2
0.0033
0.025
0.136
0.137
–
0.365
0.424
–
before N2 liquefies. Therefore the two gases can be readily separated.
Nowadays liquid nitrogen rather than liquid air is used for
many applications, because liquid oxygen contains the explosive ozone O3 . Liquid air that is kept in a Dewar increases its
O2 and O3 concentration in the course of time since N2 evaporates faster due to its higher vapour pressure and therefore after
some time liquid air reaches a critical concentration of O3 which
explodes above a critical temperature.
The gases H2 , He of Ne can be liquefied by precooling them
with liquid nitrogen below the inversion temperature before they
can be further cooled by the Joule–Thomson effect.
10.4.2.4
Equilibrium Between Solid and Liquid
Phase; Melting Curve
Chapter 10
If the temperature of a solid material is increased above a certain
temperature that depends on the material, the solid phase starts
to convert into the liquid phase. Only at the melting temperature Tm , both phases can coexist under equilibrium conditions.
The pressure dependence dTm =dp of the melting temperature
is much smaller than that of the evaporation temperature, i. e.
the slope of the curve p.T/ in the p-T-diagram of Fig. 10.79
is much larger than that of the evaporation curve. One of the
reasons is the much smaller change of the volume during the
melting process, compared with the much larger change during
the evaporation process. A similar consideration as that resulting in Eq. 10.124 for the heat of evaporation gives the heat of
fusion
dp
m D T
(10.133)
.Vliquid Vsolid / :
dT
b=106 m3 =mol
24
27
38.5
31.6
–
42.5
37.2
–
Ti =K
30
200
620
765
650
>1000
>1000
–
Ts =K at p0 D 1:013 bar
4.2
20.4
77.4
90.2
80.2
194.7
–
373.2
H2 O molecules but also contains multimers (H2 O)n in a concentration that depends on the temperature and on the distance
from the surface of water. In the multimers the different H2 O
molecules are connected by hydrogen bonds. At higher temperatures theses weak bonds break and a structural change results
in a change of the mean distance and therefore also a change of
the density. In the solid phase the H2 O-molecules form a regular
lattice with empty space between the molecules. Therefore the
density of the solid is smaller than that of the liquid phase.
Example
At T D 0 ı C the density of solid ice is % D 0:917 kg=dm3,
that of sea water is % D 1:04 kg=dm3. Therefore, only
about 12% of an iceberg are above the seawater surface,
but 88% are below.
J
Application of external pressure decreases the mean distance between the molecules and therefore the ice can melt, according to
the principle of minimum constraint. This is utilized by skaters,
For most materials the density decreases during the melting
process, i. e. Vliquid > Vsolid . This gives dp=dT > 0, because m > 0. There are some substances (e. g. water) where
Vliquid < Vsolid . For these substances is dp=dT < 0, the melting
curve has a negative slope (anomaly of water) (Fig. 10.79b).
Note: The fact that for water Vliquid < Vsolid ! %liquid > %solid
is essential for many processes in nature. Lakes freeze up from
the top to the bottom. Since the heat conductivity of ice is small,
this gives an isolating layer at the top, preventing the complete
freezing of the water, thus protecting fishes and other sensitive
creatures.
The fact, that water has its maximum density at T D 4 ı C is
called its anomaly. It is due to the temperature dependent molecular structure of water. Liquid water does not solely consist of
Figure 10.79 Melting curve, vapor pressure curve and triple point for a a positive, b a negative slope of the melting curve. a represents the phase diagram
of CO2 , b that of water
10.4 Thermodynamics of Real Gases and Liquids
Wire
Ice block
307
that none of the two variables p and T can be changed without
leaving the triple point.
This can be generalized by Gibb’s phase rule, which relates the
number f of the degrees of freedom in the choice of the variables
p and T with the number q of coexisting phases. It states:
f D3
mg
Figure 10.80 Apparently convincing demonstration of the lowering of the
melting temperature by pressure
At the triple point is q D 3 ! f D 0, i.e. no degree of freedom
in choosing the variables p and T. If only one phase is present
.q D 1/ we obtain f D 2. The pressure p as well as the temperature T can be chosen independently (within certain limits). On
the vapor curve is q D 2 and therefore f D 1. We can choose
one variable and the other is then fixed.
For a mixture of different chemical components, which can be
present in different phases the generalized Gibbs phase rule
states:
f DkC2
q
where the high pressure below the sharp ice skates forms a thin
liquid layer with low friction (see, however, the remarks below).
where k is the number of components.
The lowering of the melting temperature Tm D .dTm =dp/ p
is often demonstrated by a wire that is pulled through an ice
block by a heavy weight (Fig. 10.80).
10.4.3
Remark. More detailed calculations show, however, that the
major part of the necessary melting energy comes from heat
conduction from the higher temperature of the wire to the ice
surface (see Probl. 10.11).
Even without external pressure a thin liquid layer is formed at
the surface of ice above T D 33 ı C. The necessary energy
for melting this layer is provided by the gain in surface energy. The boundary ice-air needs more energy for evaporating
molecules than the boundary ice-liquid. A liquid layer therefore
has a lower potential energy and the loss of potential energy is
larger than the melting energy.
10.4.2.5
Coexistence of different phases; Triple Point
Since the melting curve in the p-V-diagram of Fig. 10.76 has a
larger slope than the vapor curve the two curves must intersect
in a point .ptr ; Ttr /, called the triple point. Here the three phases
solid, liquid and gase can coexist.
For T < Ttr there is one boundary curve (sublimation curve)
that separates the solid and the gaseous phases. It has in the
.p; V/-diagram generally a positive slope. Solid materials can
directly pass into the gaseous phase without becoming liquid.
This process is called sublimation. Because of the small vapor
pressure of solids this process is, however, very slow.
If there are more than one phase of a material in a container,
pressure p and temperature T are no longer independent of each
other. For example, the coexistence of the liquid and the vapor
phase is only possible on the vapor curve pS .T/. This implies
that p and T are related by the evaporation coefficient in
(10.125). It is possible to change T but then pS .T/ is fixed.
At the triple point .ptr ; Ttr / p and T are connected by two conditions: the vapor curve and the sublimation curve. This means
(10.135)
Solutions and Mixed States
Up to now we have discussed only pure substances, which are
composed of only one component and do not contain any impurities. We have explained the different phases of solid, liquid and
gaseous states and possible transitions between these phases.
In nature, however, often mixed substances are present where
molecules of different species are mixed together. Examples are
NaCl-molecules or sugar molecules, which are dissolved in water and dissociate into their atomic components. Other examples
are metal alloys
For the complete characterization of such mixed states pressure
and temperature are not sufficient, but also the concentration of
the different components have to be defined.
The concentration of a substance dissolved in a liquid is generally given in g/litre or in mole/litre. Often not the complete
substance has dissolved but a rest remains as solid sediment
(if %solid > %liquid ) or as layer on the liquid surface (if %solid <
%liquid ).
The solution of substances can alter the characteristic features of
the liquid considerably. In this section we will shortly discuss
the most important features of solutions.
10.4.3.1 Osmosis and Osmotic Pressure
Assume a container with a semipermeable membrane including
a solution with the concentration c of the dissolved substance
is submerged into a reservoir with the pure liquid (Fig. 10.81).
One observes that the level of the solution in a standing pipe
rises above the level of the pure solution, if the molecules of the
solvent can penetrate through the semipermeable membrane but
not the molecules of the dissolved substance. Such permeable
Chapter 10
Weight
(10.134)
q
308
10 Thermodynamics
h
Semipermeable
membrane
Solution
Pure solvent
Figure 10.81 Demonstration of osmosis in a Pfeffer cell
membranes with substance-specific transmission play an important role in biological cells.
In the example of Fig. 10.81 the concentration difference of the
dissolved substance between outside and inside of the container
results in a diffusion of the solvent molecules into the solution
through the permeable membrane. This builds up a pressure
difference, indicated by the height h in the standing pipe.
Figure 10.82 a Decrease of vapor pressure and increase of boiling temperature Tb of a solution compared to that of a pure solvent; b decrease of melting
temperature Tm
p D % g h ;
which stops the net diffusion, because now an equal number of
molecules diffuses into and out of the container.
Chapter 10
The net diffusion caused by the concentration difference is
called Osmosis and the pressure difference p is the osmotic
pressure.
The osmotic pressure posm is proportional to the concentration
of the dissolved molecules and to the temperature.
Experiments show that
The vapour pressure reduction p is proportional to the concentration of the dissolved molecules (if their vapour pressure is
negligible).
Francois Marie Raoult formulated in 1882 the law
ps
D
ps0
posm V D R T ;
(10.136)
where is the number of moles dissolved in the volume V of
the solvent.
This van’t Hoff’s Law is the analogue to the general gas equation
pV D RT :
The osmotic pressure of a solution exerted onto the walls
of the container equals the pressure that would be present,
if the dissolved molecules were in the gas phase at the
temperature T.
10.4.3.2
solvent molecules increases. This means that at identical temperatures less molecules evaporate than in pure liquids. The
vapour pressure is therefore lower than in a pure solvent.
Reduction of Vapour Pressure
Due to the additional attractive forces between the dissolved
molecules and the solvent molecules the work function of the
1
;
0 C 1
(10.137a)
here ps0 is the vapour pressure of the pure solvent, 0 is the
number of moles of the solvent and 1 that of the dissolved substance. For diluted solutions is 1 0 and (10.137a) reduces
to
ps D ps0
1
:
0
(10.137b)
The lowering of the vapour pressure causes an increase Tb of
the boiling temperature as shown in Fig. 10.82a. The vapour
pressure has to rise by ps to reach the external pressure pa .
From the vapour pressure curve ps .T/ in Eq. 10.125 we can
derive the relation between ps and Tb . Differentiation of
(10.125) gives
dps
RT 2 ps
p
)
T
D
:
D
s
dT
RT 2
ps
(10.137c)
10.5 Comparison of the Different Changes of State
Together with (10.137) this yields the Raoult’ Law
RT 2 1
:
0
(10.138a)
When several substances with the molar concentrations i are
dissolved, this generalizes to
RT 2 X
Tb D
i :
0 i
Comparison of the Different
Changes of State
Here we will summarize all possible changes of thermodynamic
states and the corresponding equations.
1. Isochoric processes: V D const
dQ D CV dT
(10.140a)
(10.138b) 2. Isobaric processes: p D const
dQ D Cp dT D dU C p dV
Since Tb depends on the molar evaporation heat , it is dependent not only on the dissolved substances but also on the specific
3. Isothermal processes: T D const
solvent.
(10.140b)
For dissolved substances that partly dissociate (for instance disdU D 0; dQ D p dV; p V D const
(10.140c)
sociates NaCl into NaC + Cl ) the sum in (10.138b) extends
over all dissociated and non-dissociated components dissolved 4. Adiabatic processes: dQ D 0
in the solvent.
p V D constI D Cp =CV
(10.140d)
The lowering of the vapour pressure also results in a lowering of
the melting temperature Tm (Fig. 10.82b). Similar to (10.138a)
5. Isentropic processes: S D const
one gets
RT 2 1
dS D CV dT=T C R dV=V D 0
Tm D
;
(10.139)
(10.140e)
m 0
) T V 1 D const
where m is the molar melting heat.
A reversible adiabatic process is always isentropic, but
not every isentropic process is also adiabatic.
Example
6. Isoenthalpic processes
For water with the concentration of 1 moles of a disH D U C p V D const
solved substance is the lowering of the melting tempera(10.140f)
ture
dH D .@H=@p/TDconst C .@H=@T/pDconst
Tm D
1:85 K 1 :
When 50 g NaCl are dissolved in 1 litre water, (1 mol
NaCl are
P58 g), the lowering of the melting temperature
is with
i D 2 50=58 D 1:72 mol: Tm D 3:2 K.
J
Seawater has a melting temperature that lies several degrees below 0 ı C depending on the salt concentration.
The lowering of the melting temperature is used to clear icy
roads from ice and snow by salting the roads.
The zero point of the Fahrenheit temperature scale is defined by
the melting temperature of a specific salt-water solution. From
(10.2) and (10.139) the zero point can be obtained as
0 ı F D 17:8 ı C :
Solutions with dissolved substances have generally a
larger temperature range of the liquid phase than pure solvents, because the boiling point rises and the melting point
is lowered.
10.6
Energy Sources and Energy
Conversion
The supply of sufficient energy that can replace to a large extent
manual work, has changed our life considerably. It is fair to say
that only the provision of sufficient and affordable energy has
essentially improved our standard of life. This is the reason why
in developing countries the desire for more energy will cause a
drastic increase of worldwide energy consumption.
The first law of thermodynamics teaches us, however, that energy can be neither generated nor annihilated. The phrase
“energy generation” (for example in power stations) means correctly speaking the conversion of energy from a specific form
into another (for instance from thermal energy into electric energy).
In fossil power stations the potential energy of CO- and
CO2 -molecules is transferred into heat (kinetic energy of the
molecules and atoms), which is further converted via turbines
into mechanic energy of the rotating turbine, which drives an
electric generator that produces electric energy.
Chapter 10
Tb D
10.5
309
310
10 Thermodynamics
Chapter 10
Figure 10.83 Cooling towers of the coal power plant Staudinger. The plant delivers 500 MW electric power and 300 MW heat power. It reaches an efficiency of
42.5% (With kind permission of Preußen Elektra AG, Hannover)
In car engines this molecular potential energy is converted into
mechanical energy that drives the car. In nuclear power stations
the potential energy of uranium nuclei (which exceeds that of
molecular bindings by 6 orders of magnitude) is converted by
nuclear fission into kinetic energy of the fission products and
then into heat of circulating cooling water.
Wind energy converters convert the kinetic energy of airflow
into rotation energy of the converter rotor blades, which drive
an electric generator. The wind energy has its origin in the solar
radiation energy, which in turn stems from nuclear fusion energy
in the interior of the sun.
In order to realize an energy conversion efficiency as high as
possible, one has to understand the basic physical processes
of the different conversion processes. We have learned in
Sect. 10.3.3 that the maximum possible conversion factor for
the conversion of heat into mechanical energy is given by the
10.6 Energy Sources and Energy Conversion
The increasing concern about the warming of our atmosphere
(global warming) by man-made emission of molecular gases
such as CO2 , CH4 , NO2 etc., which absorb the infrared emission of the earth surface thus heating up the atmosphere, has led
to the proposal and partly realization of several different “energy
sources”, i. e. energy conversion processes. In particular regenerative energy sources, where the working material is available
in unlimited quantities, or where the consumption of the working material is replaced by nature over time intervals of many
centuries, are favourable candidates. Such energy conversion
processes should not contribute to global warming. Examples
are nuclear energy conversion, wind energy, solar energy and
energy conversion based on the tides of the ocean. The most
important renewable energy conversion processes include:
Hydro-electric power plants (based on the potential or kinetic
energy of water)
Wind-energy converters
Geothermic plants
Solar-thermal power plants
Solar-electric conversion (photo-voltaic devices)
Bio-energy (burning of regrowing biological material such as
wood, plants)
quadrillion Btu
250
Table 10.8 Worldwide total energy consumption (in 103 TWh) and electric energy
Year
1990
2000
2012
Total energy
71
117
155
Electric energy
6
15
23
liquids
coal
natural gas
150
renewables
100
nuclear
50
0
1990
2000
2007
2015
2025
2035
Source: U.S. Energy Information Administration
(Report #.DOE/EIA-0484(2010))
Figure 10.84 Worldwide energy consumption of different energy sources in
units of 1015 Btu 1018 J 300 TWh
contributions of the different energy sources to the total worldwide energy consumption are illustrated and Tab. 10.9 lists some
countries with the highest energy consumption. It illustrates the
enormous increase during the last 40 years.
The units for energy and their abbreviations are given below:
1 Kilojoule
1 Megajoule
1 Gigajoule
1 Terrajoule
1 Petajoule
1 Exajoule
D 1 kJ
D 103 J
D 1 GJ
D 109 J
D 1 PJ
D 1015 J
D 1 MJ D 106 J
D 1 TJ
D 1 EJ
D 1012 J
D 1018 J
1 Kilowatt hour D 1 kWh D 3:6 MJ
At first we will clarify some often used definitions.
In Tab. 10.8 the increase of the worldwide primary energy consumption is summarized from 1990 to 2012. Note the large
increase of the electric power consumption. In Fig. 10.84 the
projections
200
Some examples shall illustrate these different “energy sources”.
The primary energy is the energy directly obtained from the
different sources (coal, oil, gas, water, wind, sun radiation, nuclear fission) while the secondary energy is won by conversion
of the primary energy into other energy forms (mechanical energy, electric energy, etc.). The conversion of primary into
secondary energy has an efficiency < 1. This means a fraction (1 ) is lost and is delivered as heat into the surrounding.
If the consumption of primary energy in a country is larger than
the production of energy sources, the country has to import coal,
oil or gas.
history
It is interesting to compare the total energy consumption of Germany (13 400 PJ per year) with the energy that it receives per
year from the sun. The energy of solar radiation per sec and m2
outside the atmosphere (solar constant) is 1:367 kW=.m2 s/.
During its transit through the atmosphere, the radiation power
decreases through backscattering (30%) and absorption (20%)
Table 10.9 The countries with the highest consumption of primary energy (in
Megatons Oil-Units) [10.31]
Pos.
1
2
3
4
5
6
7
8
9
10
Country
China
USA
Russia
India
Japan
Canada
Germany
Brasilia
South Korea
France
1970
202.1
1627.7
483.0
64.8
279.9
156.4
309.6
36.8
14.3
155.8
2000
980.3
2313.7
619.4
295.8
518.0
303.0
333.0
185.8
193.9
258.7
2010
2339.6
2284.9
674.1
510.2
506.8
315.6
322.5
257.4
254.6
253.3
2013
2852.4
2265.8
699.0
595.0
474.0
332.9
325.0
284.0
271.3
248.4
%
22.4
17.8
5.5
4.7
3.7
2.6
2.6
2.2
2.1
2.0
Chapter 10
efficiency of the Carnot engine which depends on initial and final temperature during the conversion. The maximum initial
temperature is generally limited by the material of the container
walls which enclose the working gas. The lowest final temperature is often limited by the temperature of the surrounding. By
using the rest energy of the cooled gas for heating (combined
heat and power) the energy efficiency can be improved. This
reduces the waste of energy which would otherwise heat up the
environment. The non-usable rest heat energy is taken away by
cooling towers (Fig. 10.83).
311
312
10 Thermodynamics
and therefore only about 50% reach the earth surface. Since
the sun radiation generally does not incide vertically onto a
surface element but under an angle ˛ that depends on the daytime, the latitude and the yearly season, the annual average
P D P0 cos ˛ of the incident radiation intensity is for a latitude
of 45ı about 300 W=.m2 s/. With an annual sunshine duration
of 1000 h=year (3:6 106 s=year) we receive the annual average
of solar radiation energy of about 1 GJ per m2 and year. In order
to match the energy consumption one needs in Germany an area
of 3 104 km2 for solar energy collectors with an efficiency of
50%.
For the whole earth (the hemi-sphere with the area of 2:55
108 km2 ) the total incident sun radiation power is 1015 2:55
108 D 2:55 1023 J=year. The total energy consumption in the
year 2011 was, however, only 51020 J=year which is about 0.2%
of the incident sun energy.
10.6.1
Hydro-Electric Power Plants
Most of the hydro-electric power plants use water reservoirs
where the water outlet streams through pipes and drives turbines
that generate electric power. Here the potential energy of the
damned up water is converted into kinetic energy of the water
flowing through the pipe.
For a storage height h an area A of the reservoir and a density %
of the water the total potential energy is
Chapter 10
Epot D % g h A h ;
when the storage height is lowered by h h.
Example
A D 1 km2 D 106 m2 , h D 30 m, h D 5 m ! Epot D
1:5 1012 J D 1:5 TJ.
J
10.6.2
Tidal Power Stations
Tidal power stations use the tidal range between low and high
tide for power generation. This range is in particular large in
the mouth of rivers, where it can reach up to 16 m. The water
passes through turbines built into logs in the river. At low tides,
the water streams seawards and at high tides against the river.
This streaming water drives the turbines at low tides as well as
at high tides (Fig. 10.85), which activates generators for producing electric energy. At a water level difference h between the
dammed river and the sea level the energy that can be converted
is
Z
W D .dM=dt/ g h.t/dt ;
where dM=dt is the mass of water passing pro second through
the turbines, h is the time dependent level difference and T
(about 5 h) the time duration of low resp. high tide.
Here the gravitational energy of earth-moon attraction and the
decrease of the rotation energy of the earth (due to friction by the
tides) are the primary energy sources. During the time intervals
where h D 0 the tidal power station cannot deliver energy.
The first tidal power station was built in France in the mouth
of the river Rance (Fig. 10.86) where a tidal range of 16 m is
obtained. The river dam is 750 m long and has 24 passages
where the turbines are located. The total power station delivers an electric power of 240 MW and per year an electric energy
of 600 GWh. This equals the energy delivered by 240 wind converters with 1 MW power each and 3000 hours of full operation
per year.
The disadvantage of such tidal power stations is the separation
of the bay at the mouth of the river. This can change the biological conditions for plants and fishes and it can furthermore
influence the tidal range in neighbouring bays with the danger
of flooding.
Some hydro-electric power stations use the flow energy of
rivers, where in most cases, however, the river has several barrages where again the potential energy of the dammed river is
used to drive turbines. This method was often used in earlier
times to drive corn mills and hammer mills which only need
moderate powers.
Example
When a channel with a width of 5 m and a depth of
3 m is branched off a river the water with a velocity of
v D 6 km=h D 1:67 m=s drives a turbine, the maximum
available power is
P D 12 Mv 2 D % B h v 2 =2 :
With the numerical values given above this yields P D
21 kW.
J
Figure 10.85 Schematic illustration of a tidal power station
10.6 Energy Sources and Energy Conversion
313
10.6.3
Wave Power Stations
Wave power stations use the kinetic and potential energy of sea
waves for the generation of electric energy. Their basic principle is illustrated in Fig. 10.87. A pneumatic chamber is filled
with air in its upper part while the lower part has a connection
to the sea. The incoming waves induce a periodic change of
the water level in the lower part of the chamber. This causes
a periodic change of the air pressure in the upper part and an
air flow through the pipe at the top of the chamber that periodically changes its direction. In the upper part of the pipe a
Generator
Turbine
Air flow wells
Water
Ground
Figure 10.87 Concept of a wave-power station
Wave power stations do not use the tide difference between high
and low tide but the wave energy, which is in turn driven mainly
by the wind energy and only to a minor part by the tides [10.32,
10.33].
10.6.4
Incomming
wave
Periodic
lift
Wells-turbine is installed, that always rotates in the same direction independent of the direction of the air flow. This turbine
has symmetric blade profiles in contrast to normal turbines that
have asymmetric blade profiles, optimized for one direction of
the airflow. The efficiency of the Wells-turbine is smaller than
that of normal turbines. It has, however, the advantage that it
rotates continuously for both directions of the air flow.
Geothermal Power Plants
The temperature of the earth’s interior increases with increasing depth by about 3–5 ı C=100 m, because heat flows from the
hot kernel to the outer parts of the earth. The heat in the kernel
was mainly generated in the formation period of the earth (about
4 billion years ago) where heavier elements dropped down to the
kernel due to gravitational forces. This increased the temperature of the kernel. Another cause for the production of heat is
the radioactive decay of elements such as Uranium, Thorium
and Potassium that are contained in the kernel as well as in the
earth mantle.
Chapter 10
Figure 10.86 Tidal Power Station St.Malo. Aerial view of the power station in the estuary of the river Rance in France. Here the tidal amplitude is about 8 m (see
Sect. 6.6). With kind permission © Foto dpa
314
10 Thermodynamics
Heat exchanger
Sun light
Turbine
Absorber
pipe
Pump
Hot water
Water
input
Parabolic
reflector
Pump
Hot rock
Figure 10.88 Schematic illustration of a geothermic power station
In regions with volcanic activities, water rising from the interior
to the surface of the earth has a sufficiently high temperature to
be useful as energy source. For example in Iceland residences,
green houses and swimming pools are heated by hot water from
the earth interior. This energy streaming out of the earth interior
is called geo-thermal energy.
Iceland can cover about 80% of its primary energy consumption
(40 PJ D 40 1015 J) by this geothermal sources.
Chapter 10
In regions without such active volcanism, one can drill deep
bore holes (3000–5000 m) in order to utilize geothermal energy.
The temperature at a depth of 5000 m is about 200–300 ı C. For
energy production water is pumped into the bore holes which
interacts with the hot rock and is heated up. The hot water is
pumped back to the surface and can be used for heating purposes
or, if its temperature is above 100 ı C it can produce through heat
exchangers hot steam that drives turbines (Fig. 10.88).
The limitations of the geothermal energy usage in non-volcanic
regions is the slow transport of heat from the surrounding of a
bore hole, which is mainly due to heat conduction. When the
heat extraction becomes larger than the supply of energy from
the surroundings the temperature drops and the efficiency of the
plant decreases accordingly [10.24]. A much more serious problem are possible geological dislocations. The water pumped
under high pressure into the bore hole can modify the rock in
the surroundings of the bore hole and can increase the volume
of such chemically altered porous rocks. This will cause local
uplifts at the earth surface which can damage buildings. Such
geothermal plants should be therefore operated far away from
inhabited areas.
10.6.5
Solar-Thermal Power Stations
These power stations use the heating of material that absorbs the
sun radiation and transfers the heat to a liquid transport medium,
such as water or oil. In order to reach sufficiently high temperatures the sun radiation is focused by parabolic or spherical
mirrors onto the heated devices.
Pipes heated by
sunradiation
Figure 10.89 Thermal solar power station using parabolic reflectors
Stirling engine
Tracking
parabolic mirror
Figure 10.90 Solar power station with parabolic mirrors, that follow up the
sun position and focus the sun radiation onto a Stirling motor [10.26]
In the parabolic gullies construction the water or oil is pumped
through pipes that are located in the focal line of cylindrical
mirrors with parabolic profile (Fig. 10.89), which concentrates
the sun radiation onto the pipes [10.25].
Another modification consists of several hundred parabolic mirrors (heliostats) that follow up the changing sun position during
the day (Fig. 10.90) and concentrate the sun radiation, nearly
independent of the position of the sun, onto a small volume at
the top of a high tower (Fig. 10.91).
The achievable radiation density of this device is much higher
than in the parabolic gully construction and temperatures of
about 1000 ı C can be reached. This increases the efficiency for
the conversion into electric energy. The generated hot steam
drives turbines as in fossil power stations.
An example of such a solar-thermal power station is the plant
“Plataforma solar de Almeria” in Spain (Fig. 10.92). Here
300 heliostats with 40 m2 parabolic mirror surface each concentrate the sun radiation onto the radiation collector at the top of an
10.6 Energy Sources and Energy Conversion
Sun radiation
Solartower
with absorber
315
Upwind
Chimney
Turbine
Glas roof
Heliostats
Air
Generator
Figure 10.91 Solar tower power station
Figure 10.93 Upwind solar power station
80 m high tower. This plant produces an average electric power
of 40 MW.
10.6.6
Here another type of solar power stations, the upwind plant,
is favourable (Fig. 10.93). The air above a large area under a
sloped glass roof that is formed like a cone, is heated by direct or diffuse sun radiation. The hot air streams to the centre
of the area where it rises up into a chimney, driving a turbine.
Although the efficiency of such a plant is only about 5% it still
pays off because of the low construction and operation costs.
An example is the plant “Fuente el Fresno” in Spain, where an
area of 2:5 km2 is covered by the glass roof. The chimney is
750 m high and the delivered electric power amounts to 40 MW.
The large area is not lost for agriculture but can be used for
growing plants and fruits even during colder periods, since it
operates like a green house.
Here the sun radiation energy is directly converted into electric
energy by photovoltaic semiconductors. The efficiency amounts
to 5–20% depending on the semiconductor material. The basic
physics of these devices is explained in Vol. 3.
The price per kWh was very high in the beginning but decreases
now rapidly due to mass production of solar cells or thin film
photovoltaic devices. In view of the rising prices for fossil energy sources it will soon be able to compete with conventional
power plants.
The large disadvantage of all solar power plants is the dependence on the unreliable sunshine duration. It is therefore
necessary to realize energy storage devices which can bridge
the time periods where the sun does not shine.
Figure 10.92 Gemasolar power station close to Sevilla. 2650 mirrors reflect the sun light onto a tower where a salt solution is heated. It will deliver electric
energy of 110 GWh per year. (Torresol Energy Investment S.A.)
Chapter 10
The solar-thermal power plants discussed so far only work efficiently for direct sun radiation, i. e. under a clear sky, because
the concentration by the heliostats onto a small volume does not
work efficiently for the diffuse radiation at cloudy days.
Photovoltaic Power Stations
316
10 Thermodynamics
10.6.7
Bio-Energy
The burning of farming refuse, such as waste wood, stray,
garbage or biogas, which remains in agriculture can produce
useful energy, named bio-energy, because the burned material
is of biological origin. The advantage of this energy source is
that it is renewable as long as the consumption does not exceed
the natural production. Its disadvantage is the emission of CO2
which generally exceeds the consumption of CO2 by the growing plants, although the net emission balance is more favourable
than for conventional fossil power plants. Furthermore other
species such as SO2 , phosphor and heavy metal compounds
are emitted. If only substances as burning material are used,
which cannot be utilized for other purposes, the bio-energy can
be judged positively. However, if food is used for the production of gasoline, this is contra-productive and should be rejected.
Also the burning of wood pellets only makes sense, if they are
produced from wood waste, but this technique is nonsense if the
pellets are pressed from trees that could have been used elsewhere.
10.6.8
Electrolysis
of H2O
2 H2O
Chapter 10
The oldest energy storage systems are pumped hydro storage
plants. Here water is pumped from a lower storage reservoir
into a higher one during times, where sufficient energy is available. During periods where more energy is needed, the water
runs back from the higher into the lower reservoir and drives
turbines, which activate electric generators. The generators are
used during the up-pumping period as electric motors that drive
the pumps. This method is up to now the most efficient, but
it needs sufficient space on the top of mountains for the upper
reservoir. One of many examples is the Walchensee plant in
Bavaria, Germany, where the water is pumped from the lower
Kochelsee into the 200 m higher Walchensee.
For bridging the night periods, where solar plants cannot work,
salt storage systems have been developed. Here the surplus energy produced during daytime is used to heat up and melt a salt
solution. During night time the heat of the hot solution and the
heat of fusion that is released when the solution solidifies, can
be used to bridge the energy gap. With multi-component salt
solutions, there are several melting temperatures and the heat of
fusion is more uniformly delivered during the cooling of the solution. Examples of such salt solutions are Mg.NO3 /2 6H2 O,
or CaCl2 6H2 O.
For small energy demands during night-time compact lithium
batteries have been developed which have a storage capacity of
2 H2 + O2
H2reservoir
O2
Combustion
H2O
Figure 10.94 Hydrogen gas as energy reservior
0:2 kWh per kilogram mass. For a volume of 0:5 m3 of the battery system one can reach a storage energy of 20 kWh. This
is sufficient for most private households, which may have solar collectors on the roof and can provide with such a combined
system their energy demands during day and night.
A promising storage medium is hydrogen gas H2 , which can
be produced by electrolysis during times where surplus solar or
wind energy is available. According to the endothermic reaction
2H2 O ! 2H2 C O2 ;
Energy Storage
The increasing production of energy from renewable resources
that are not continuously available, demands the realization of
sufficient energy storage systems in order to bridge time periods
where these sources cannot deliver sufficient energy. There are
several proposals for such storage systems, where some of them
have been already realized.
H2
(a)
(Fig. 10.94) hydrogen gas is produced by electrolysis of water.
In the reverse exothermic reaction
2H2 C O2 ! 2H2 O
(b)
energy is released. The advantage is, that no environmentally
dangerous gases such as CO2 or NH3 are emitted. The electrolytic systems can be placed directly inside the tower of wind
converters and the produced hydrogen gas can be stored in high
pressure bottles. The systems can be controlled in such a way,
that reaction (a) operates during the time period of wind energy
surplus and reaction (b) during times of wind energy shortage.
For large plants the hydrogen gas is stored in huge underground
caverns, for instance in no longer used salt mines and is transported by underground pipes to special power stations which can
burn hydrogen gas. Meanwhile long-time experience is present
for the storage of hydrogen gas in caverns.
Example
The cavern Clemens Dome, close to Lake Jackson USA,
has a volume of 580 000 m3 . The stored gas at a pressure
of 10 MPa can deliver an energy of 90 GWh. It is operated
since 1986.
J
While for H2 storage the reaction energy of the reaction (b) is
used, for air storage at high pressures the potential energy pV of
the gas volume is utilized. When the gas flows from the storage
tank through a pipe, the potential energy is converted into kinetic
energy, which is used to drive a turbine (Fig. 10.95).
Summary
Low pressure
valve
High pressure
valve
Turbine
Generator
317
Example
p D 100 atm D 10 MPa, V D 500 000 m3 ! Epot D
J
p V D 5 1012 N m D 5 1012 Ws D 1:39 GWh.
Compressor
Electrical
current
Gas reservoir
Figure 10.95 Scheme of a high pressure gas reservior
Summary
T=K D TC = ı C C 273:15;
T=F D .9=5/TC ı C C 32
D .9=5/ŒT=K 273:15/ C 32
D .9=5/T=K 459:67 :
For temperature measurements all quantities can be used,
that depend on the temperature (expansion of a liquid volume, electric resistance, thermo-voltage, conductivity of
semiconductors).
The thermal expansion of bodies is caused by the nonharmonic interaction potential between neighbouring atoms.
The absolute temperature is determined with the gas thermometer, where the increase of the gas pressure with temperature in a constant volume is proportional to the temperature
increase.
The thermal energy of a body is determined by the kinetic
and potential energy of the atoms or molecules. The temperature increase T of the system is proportional to the supplied
heat energy Q D C T.
The molar heat capacity for a constant volume of a gas
CV D R f =2, is equal to the product of gas constant R D
k NA times one half of the number f of degrees of freedom
of the atoms or molecules in the gas.
The molar heat capacity at constant pressure is Cp D CV C R
The transition from the solid to the liquid phase requires the
molar melting energy W D m per mole. During the melting the potential energy of the atoms or molecules increases
while the kinetic energy stays constant. Similar the transition from the liquid to the gaseous phase needs the energy
per mole W D e (heat of evaporation).
Thermal energy can be transported from one area to another
– by heat conduction
– by convection
– by thermal radiation
The amount of heat transported per second by heat conduction in the direction r through the area A is dQ=dt D
A .grad T/r , i. e. the product of heat conductivity ,
area A and temperature gradient in the direction of r.
For metals the heat conductivity is proportional to the electrical conductivity, which indicates that the electrons are
mainly responsible for both conductivities.
The thermodynamic state of a system is unambiguously determined by the state variables pressure p, volume V and
temperature T. For moles of an ideal gas in the volume
V the general gas equation is
pV DRT :
The number of internal state variables in real gases is given
by Gibbs’ phase rule (10.134).
The entropy S of a system is a measure for the number of
possible ways the state of the system can be realized. The
change of the entropy is dS D dQ=T where dQ is the heat
energy supplied to or by the system.
The first law of thermodynamics U D Q C W describes the energy conservation. The change U of internal
energy U D N .f =2/kT of a system with N atoms or
molecules equals the sum of supplied heat Q and mechanical work W performed on or by the system. For real gases
is U D Ekin C Epot , because the interaction energy between
the atoms has to be taken into account.
Special processes in a system of an ideal gas are:
isochoric processes (V D const) ) dU D CV dT,
isobaric processes (p D const) ) dU D dQ pdV,
isothermal processes (T D const) ) p V D constant,
adiabatic processes (dQ D 0) ) dU D dW and p V D
constant with D Cp =CV D adiabatic index.
The second law of thermodynamics states that at the conversion of heat into mechanical energy at most the fraction
D .T1 T2 /=T1 can be converted when the heat reservoir
is cooled from the temperature T1 to T2 .
The entropy S D k ln P is a measure for the number P of realization possibilities for a system with a given temperature
T and total energy E.
Chapter 10
The temperature of a body is given either as absolute temperature T in Kelvin or as Celsius temperature TC = ı C or in the
US as Fahrenheit temperature. The relations are
318
10 Thermodynamics
Reversible processes are ideal processes where a system
passes a cycle of processes and reaches its initial state without any losses. An example is the Carnot Cycle where the
system passes through two isothermal and two adiabatic processes.
For reversible processes the entropy remains constant. For
all irreversible processes the entropy increases and the free
energy F D U T S decreases.
The entropy S approaches zero for T ! 0 (third law of thermodynamics).
For real gases the Eigen-volume of the atoms and the interaction between the atoms cannot be neglected as for ideal
gases. The equation of state p V D R T of ideal gases is
modified to the van der Waals equation .pCa=V 2 /.V b/ D
R T, where a=V 2 denotes the internal pressure and b=4 the
Eigen-volume of the NA molecules per mole.
The heat of evaporation of a liquid D Tdps =dT.Vv Vl / is
due to the mechanical work necessary to enlarge the volume
Vl of the liquid to the much larger volume Vv of the vapour
against the external pressure and against the internal attractive forces between the molecules. The second contribution
is much larger than the first one.
In a p.T/ phase diagram the liquid and gaseous phases are
separated by the vapour phase curve and the liquid and solid
phase by the melting curve. The two curves intersect in the
triple point (Ttr ; ptr ) where all three phases can coexist.
The vapour pressure of a liquid is lowered by addition of
solvable substances, which increases the evaporation temperature. Also the melting temperature can be lowered.
Problems
10.1 Give a physically intuitive explanation, why the thermal
expansion coefficient for liquids is larger than that of solids.
10.2
Prove example 2 in Sect. 10.1.2.
10.8 What is the entropy increase S1 when 1 kg water is
heated from 0 to 50 ı C? Compare S1 with the entropy increase
S2 when 0:5 kg water of 0 ı C is mixed with 0:5 kg of 100 ı C.
Chapter 10
10.9 A power station delivers the mechanical work W1 when
10.3 A container with 1 mol helium and a container of equal water vapour of 600 ı C drives a turbine and cools down to
size with 1 mol nitrogen are heated with the same heat power 100 ı C.
of 10 W. Calculate after which time the temperature of the gas a) What is the Carnot efficiency?
in the containers has risen from 20 to 100 ı C. The heat capac- b) How many % of the output energy can one win, when the
water of 100 ı C is used for heating and cools down to 30 ı C?
ity of the containers is 10 Ws=K. How long does it take, until
ı
T D 1000 C is reached, when we assume that the vibrational
degrees of freedom of N2 can be excited already at T D 500 ı C? 10.10 A hot solid body (m D 1 kg, c D 470 J=.kg K/,
T D 300 ı C) is immersed into 10 kg of water at 20 ı C.
All heat losses should be neglected.
a) What is the final temperature?
10.4 Give a vivid and a mathematical justification for the time b) What is the entropy increase?
dependent temperature function T.t/ during the mixing experi10.11 Calculate the pressure that a wire with 1 mm diameter
ment of Fig. 10.12.
exerts onto an ice block with a width of 10 cm (according to
10.5 A container (m D 0:1 kg) with 10 mol air at room tem- Fig. 10.80) when both ends are connected with a mass m D 5 kg.
perature rests on the ground. What is the probability that it lifts What is the increase of the melting temperature? What is the
by itself 10 cm above ground? Such an event would cause a heat supplied to the ice block by the wire, if the outside tempercooling (conversion of thermal into potential energy). How large ature and the wire temperature are 300 K‹ How much ice can be
is the decrease of the temperature? (Specific heat of the gas is melted per second by the wire?
.5=2/R, that of the container is 1 kJ=.kg K/.)
10.12 Calculate from the diagram of Fig. 10.64b the theoretical
10.6 A volume of 1 dm3 of helium under standard condi- efficiency of the Otto-motor.
tions (p0 D 1 bar, T0 D 0 ı C) is heated up to the temperature
T D 500 K. What is the entropy increase for isochoric and for 10.13 Show that for a periodically supplied heat at x D 0
Eq. 10.42 is a solution of the Eq. 10.38b for one-dimensional
isobaric heating?
heat conduction.
10.7 The critical temperature for CO2 (M D 44 g=mol) is
Tc D 304:2 K and the critical pressure pc D 7:6 106 Pa, its 10.14 What is the maximum power an upwind power plant
density at the critical point is % D 46 kg=m3 . What are the van can deliver (area 5 km2 , temperature below the glass roof T D
der Waals constants a and b?
50 ı C, height of the tower 100 m, outside temperature 20 ı C at
the top of the chimney).
References
319
10.1. F.W.G. Kohlrausch, An Introduction to Physical Measurements, 2nd ed. (Univ. Toronto Libraries, Toronto,
2011)
10.2a. J.V. Iribane, H.R. Cho, Atmospheric Physics. (D. Reidel,
Dordrecht, 1980)
10.2b. D.G. Andrews, An Introduction to Atmospheric Physics.
(Cambridge Univ. Press, Cambridge, 2010)
10.3. H. Haken, Synergetics. An Introduction. (Springer,
Berlin, Heidelberg, 2014)
10.4. H. Haken, Synergetics. Introduction and Advanced Topics. (Springer, Heidelberg, 2004)
10.5. J.E. Parrot, A.D. Stuckes, Thermal Conductivity of
Solids. (Pion Ltd, London, 1975)
10.6. P. Dunn, D.A. Reay, Heatpipes, 2nd ed. (Pergamon, Oxford, 1978)
10.7. N. Rice, Thermal Insulation. A Building Guide. (NY Research Press, New York, 2015)
10.8. R.T. Bynum, Insulation Handbook. (McGrawHill, New
York, 2000)
10.9. J. Fricke and W.L. Borst, Essentials of Energy Technology: Sources, Transport, Storage Conservation. (Wiley
VCH, Weinheim, 2014)
10.10. Ph. Warburg, Harvest the Sun. America’s Quest for a Solar Powered Future. (Beacon Press, Boston, 2015)
10.11. M. Green, Third Generation Photovoltaics. Advanced
Solar Energy Conversion. Springer Series in Photonics,
Vol. 12. (Springer, Berlin, Heidelberg, 2005)
10.12. S.A. Goudsmit, R. Clayborne, Time. (Time-Life Amsterdam, 1970)
10.13. S.C. Colbeck, Am. J. Phys. 63, 888 (1995)
10.14. CRC handbook of Chemistry and Physics, 96th ed. (CRC
Press, Boca Raton, Florida, USA, 2015)
10.15. http://cdn.intechopen.com/pdfs/20377/InTechPractical_application_of_electrical_energy_storage_
system_in_industry.pdf
10.16. http://www2.hesston.edu/physics/201112/
regenerativeenergy_cw/paper.html
10.17. https://en.wikipedia.org/wiki/Climate_change
10.18. http://www.bounceenergy.com/blog/2013/05/windenergy-grid-part-3-future/,
http://www.pasolar.org/
index.asp?Type=B_BASIC&SEC=%7B9D644D34AF8E-475C-A330-4396D09F454B%7D
10.19. https://en.wikipedia.org/wiki/World_energy_
consumption
10.20. https://en.wikipedia.org/wiki/Electric_energy_
consumption
10.21. https://en.wikipedia.org/wiki/Renewable_energy
10.22. https://en.wikipedia.org/wiki/Tidal_power, https://en.
wikipedia.org/wiki/Rance_Tidal_Power_Station, http://
www.darvill.clara.net/altenerg/tidal.htm
10.23. https://en.wikipedia.org/wiki/Wave_power, https://en.
wikipedia.org/wiki/List_of_wave_power_stations
10.24. https://en.wikipedia.org/wiki/Geothermal_energy,
https://en.wikipedia.org/wiki/Geothermal_electricity
10.25. https://en.wikipedia.org/wiki/Solar_thermal_energy
10.26. http://www.volker-quaschning.de/articles/
fundamentals2/index_e.php,
https://en.wikipedia.
org/wiki/List_of_solar_thermal_power_stations
10.27. http://de.total.com/en-us/making-energybetter/worldwide-%20%20%20projects/
sunpower-puts-total-cutting-edge-solar?
gclid=CLKnluvxjckCFVZAGwod6mQFh
10.28. http://energy.gov/eere/fuelcells/hydrogen-storage,
https://en.wikipedia.org/wiki/Hydrogen_storage
10.29. https://en.wikipedia.org/wiki/Energy_storage
10.30. https://en.wikipedia.org/wiki/Renewable_energy
10.31. BP, Workbook of historical data. Microsoft Excel document
10.32. http://www.darvill.clara.net/altenerg/wave.htm
10.33. http://thinkglobalgreen.org/WAVEPOWER.html
Chapter 10
References
Mechanical Oscillations and
Waves
11.1
The Free Undamped Oscillator . . . . . . . . . . . . . . . . . . . . . . 322
11.2
Mathematical Notations of Oscillations . . . . . . . . . . . . . . . . . 323
11.3
Superposition of Oscillations . . . . . . . . . . . . . . . . . . . . . . . 324
11.4
The Free Damped Oscillator . . . . . . . . . . . . . . . . . . . . . . . . 328
11.5
Forced Oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330
11.6
Energy Balance for the Oscillation of a Point Mass . . . . . . . . . . 333
11.7
Parametric Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . 334
11.8
Coupled Oscillators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335
11.9
Mechanical Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 339
11.10
Superposition of Waves; Interference . . . . . . . . . . . . . . . . . . 352
11.11
Diffraction, Reflection and Refraction of Waves . . . . . . . . . . . . 354
11.12
Standing Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359
11.13
Waves Generated by Moving Sources . . . . . . . . . . . . . . . . . . 363
11.14
Acoustics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 366
11.15
Physics of Musical Instruments . . . . . . . . . . . . . . . . . . . . . . 372
11
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 376
Chapter 11
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 378
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379
© Springer International Publishing Switzerland 2017
W. Demtröder, Mechanics and Thermodynamics, Undergraduate Lecture Notes in Physics, DOI 10.1007/978-3-319-27877-3_11
321
322
11 Mechanical Oscillations and Waves
Mechanical oscillations play an important role in basic sciences as well as for technical applications. Their significance
as sources of acoustic waves and for the realization of musical performances, in sensors for hearing is obvious. Often the
prevention of unwanted acoustic resonances of buildings and
bridges represents a technical challenge. All these points justify a more detailed study of the basic physics of oscillations
and waves.
Their mathematical treatment is in many aspects very similar to
that of electric oscillations and waves (see Vol. 2, Chap. 6). The
investigation of common features and differences between mechanical and electro-magnetic oscillations and waves not only
intensifies our knowledge of macroscopic oscillation phenomena but also gives a deeper insight into the microscopic structure
of matter. (atomic and molecular vibrations in solids).
In this chapter we will discuss mechanical oscillations, where
matter is moved, and mechanical waves where this motion is
transported by couplings between neighbouring layers of gases,
liquids or solids. At the end of this chapter some interesting
applications of ultrasonics in medicine and of acoustics in music
are presented.
Figure 11.1 Undamped harmonic oscillator
In Sect. 2.9.7 we had already derived the oscillation equation for
the simple pendulum, where we had found for small elongations
the same equation 'R C .g=L/' D 0 for the angle ' (Eq. 2.79b).
The equation (11.1) has the solution
x D c et ;
(11.2)
where c is an arbitrary constant. Inserting (11.2) into (11.1)
gives the quadratic equation for the parameter :
2 C !02 D 0 ;
which has the two solutions
11.1
The Free Undamped Oscillator
In Chap. 2 the basic equations of motion for the simplified
model of point masses were derived. In a similar way the basic
facts of mechanical oscillations can be best understood when we
start with the idealized model of point masses before we proceed
to oscillations of extended bodies.
A point mass m suspended by a spring has its equilibrium position at x D 0 where the gravity force is just compensated by
the opposite restoring force of the spring. When the mass m is
removed from its equilibrium position by a small displacement
x (Fig. 11.1) a restoring force occurs, which is, according to
Hooke’s Law (Sect. 6.2) proportional to x:
Chapter 11
FD Dx;
where D is the spring constant that depends on the strength of
the spring. This force drives the mass m back to its equilibrium
position x D 0. The one-dimensional equation of motion is then
d2 x
m 2 D Dx :
dt
(11.1a)
With the abbreviation !02 D D=m this becomes
2
d x
C !02 x D 0 :
dt2
(11.1b)
This is the equation for the harmonic oscillator (which is called
“harmonic” because its oscillation generates a “pure” sinusoidal
tone at the frequency !0 . Together with its overtones n !0 it
forms a superposition of tones that are felt by human ears as
harmony).
and
1 D Ci !0
2 D i !0 :
We therefore obtain the two solutions of (11.1)
x1 .t/ D c1 ei!0 t
and
x2 .t/ D c2 e
i!0 t
;
which are linearly independent for !0 ¤ 0. The general solution of the linear differential equation (11.1) is then the linear
combination of the two solutions
x.t/ D c1 ei!0 t C c2 e
i!0 t
:
(11.3)
Since x.t/ must be a real function (not complex) it follows for
the complex constants c1 D c2 D c. The solution for the oscillation amplitude is then
x.t/ D cei!0 t C c e
i!0 t
with
c D a C ib :
(11.4a)
The real constants a and b can be determined from the initial
conditions for the special oscillation problem.
Example
When the mass m in Fig. 11.1 passes at t D 0 with the
velocity v0 through the equilibrium position x D 0, we
obtain from (11.4a): c C c D 0 ) a D 0 and v0 D
i !0 .c c / D i!0 2ib D> b D v0 =2!0 . Therefore is
x.t/ D
v0
sin !0 t :
!0
J
Remark. The oscillating mass on a spring is only one example for a harmonic oscillator. Other examples are a mass that
oscillates on a parabolic air track, or the simple pendulum suspended by a string, or an electron in the lowest energy level of
the hydrogen atom.
11.2 Mathematical Notations of Oscillations
11.2
323
Mathematical Notations of
Oscillations
When we write the complex amplitudes c and c in (11.4) as
polar representation
c D jcj ei' ;
c D jcj e
i'
:
We obtain the representation
x.t/ D jcj ei.!0 tC'/ C e
i.!0 tC'/
that is equivalent to (11.4a).
;
(11.4b)
Figure 11.3 Period T , amplitude A and phase shift ' of a harmonic oscillation
According to Euler’s formula for complex numbers
eix D cos x C i sin x :
We can write (11.4a) also in the form
x.t/ D C1 cos !0 t C C2 sin !0 t
(
)
C1 D c C c
with
:
C2 D i.c c /
(11.4c)
Figure 11.4 Elongation x .t /, velocity xP .t / and acceleration xR .t / of a harmonic
oscillation
A forth equivalent representation is
(11.4d)
The comparison with (11.4c) gives
C1 D A cos ' ; C2 D A sin '
q
C2
) tan ' D
and A D C12 C C22 :
C1
All 4 representations (11.4a–d) for the solution of (11.1) are
equivalent (Fig. 11.2). They represent a harmonic oscillation
with the frequency !0 and the amplitude A D 2jcj (Fig. 11.3).
For our example above with the initial conditions x.0/ D 0 and
.dx=dt/0 D v0 all forms (11.4a–d) give the solution
x.t/ D
v0
sin.!0 t/ ;
!0
as can be immediately proved by inserting x.t/ into (11.4a–d).
Imaginary part
The argument (!0 t C ') in the cosine function (11.4d), which
determines the momentary value of the elongation x.t/, is called
the phase of the oscillation. The time origin x D 0 can be chosen
in such a way that ' D 0. This reduces (11.4d) to
x.t/ D A cos !0 t :
(11.4e)
After a time t D 2=!0 D T always the same value of x.t/ is
reached. This means
x.t C T/ D x.t/ :
The time interval T is called the oscillation period, while the
reciprocal D 1=T is the oscillation frequency and ! D 2
is the circular frequency. The mass that experiences a restoring
force proportional to the displacement (x x0 ) from the equilibrium position x0 is called a harmonic oscillator.
In Fig. 11.4 the elongation x.t/ D A cos.!t/, the velocity dx=dt
and the acceleration d2 x=dt2 are shown. The figure illustrates
that the acceleration always has the opposite phase as the elongation, i. e. x.t/ shows a phase shift of against d2 x=dt2 .
Example
Real part
Figure 11.2 Relations betwenn different equivalent representations of harmonic oscillations
x1 .t/ D A cos.!0 t/ and x2 .t/ D A cos.!0 C '/ are
two harmonic oscillators with the same frequency and
the same amplitude but with a phase shift ' against each
other. The maxima of the two oscillations are shifted
against each other by the time t D '=!0 (Fig. 11.5).
Chapter 11
x.t/ D A cos.!0 t C '/ :
324
11 Mechanical Oscillations and Waves
Figure 11.5 Two harmonic oscillations with equal frequency but relative phase shift '
J
Figure 11.6 One-dimensional superposition of two oscillations with equal frequencies but different phases '1 and '2
with the relations
11.3
A D a cos '1 C b cos '2 ;
Superposition of Oscillations
B D a sin '1 b sin '2 ;
p
B
:
C D A2 C B2 and tan ' D
A
In nature, pure harmonic sine oscillations are rare. Generally
more or less complex forms of oscillations occur. It turns out, The superposition is therefore again a harmonic oscillation with
however, that even complex non-harmonic oscillations can be the same frequency but amplitude and phase differ from that of
represented by a superposition of pure harmonic oscillations the partial oscillations (Fig. 11.6).
with different amplitudes, frequencies and phases. We will discuss in this Section such superpositions. If the elongations of
all pure harmonic oscillations point into the same direction (for Special Cases:
instance into the x-direction), we have a one-dimensional su- 1. a D b and '1 D '2 D '
perposition. In the general case of two- or three-dimensional
superpositions the elongations of the different oscillators can
) x D x1 C x2 D 2a cos.!t C '/
point into arbitrary directions.
Both oscillations add in phase and the resulting oscillation
has twice the amplitude of the two summands.
2. a D b but '1 ¤ '2
11.3.1
One-Dimensional Superposition
The sum of the different oscillations
X
X
an cos.!n t C 'n /
xn .t/ D
x.t/ D
x.t/ D aŒcos.!t C '1 / C cos.!t C '2 /
D aŒcos !t.cos '1 C cos '2 / sin !t.sin '1 C sin '2 /
(11.5)
n
n
Chapter 11
depends on the amplitudes an , the frequencies !n and the phases
'n of the different summands.
Ansatz:
x.t/ D b cos.!t C '/
D b Œcos !t cos '
sin !t sin '
) a.cos '1 C cos '2 / D b cos '
a.sin '1 C sin '2 / D b sin '
11.3.1.1
Two Oscillations of Equal Frequencies
tan ' D
If the two oscillations
x1 .t/ D a cos.!t C '1 /
with equal frequencies !, but different amplitudes and phases
are superimposed, one obtains according to the addition theorem
of trigonometric functions
x.t/ D x1 .t/ C x2 .t/ D A cos !t C B sin !t
D C cos.!t C '/
(11.6)
)
'1 C '2
sin '1 C sin '2
D tan
cos '1 C cos '2
2
)'D
x2 .t/ D b cos.!t C '2 /
)
'1 C '2
2
p
) b D a 2 C 2 cos.'1 '2 /
p
) x.t/ D a 2 C 2 cos.'1 '2 / cos.!t C '/
The resultant amplitude is smaller than 2a and the phase differs from '1 and '2 . For '1 D '2 C the two oscillations
have opposite phases. The two oscillations cancel each other
and x.t/ 0, i. e. the total amplitude is zero.
11.3 Superposition of Oscillations
11.3.1.2
Different Frequencies, Beats
325
Envelope
A different situation arises when two oscillations with different
frequencies are superimposed (Fig. 11.7). For equal amplitudes
a D b the sum of the two oscillations
x1 .t/ D a cos !1 t I
x2 .t/ D a cos !2 t
gives with the trigonometric theorem
cos ˛ C cos ˇ D 2 cos
˛
ˇ
2
cos
Figure 11.8 Beat pattern of the superposition of two oscillations with frequencies !1 and !2 and ı! ! D 12 .!1 C !2 /
˛Cˇ
;
2
the superposition
x.t/ D 2a cos
!
1
!1 C !2
!2
t cos
t :
2
2
(11.7)
If the two frequencies do not differ much, i. e. .!1 !2 /
! D 21 .!1 C !2 / Eq. 11.7 can be interpreted as an oscillation
with the frequency ! and an amplitude A.t/ D 2a cosŒ 21 .!1
!2 /t that oscillates slowly with a period D 2=.!1 !2 /,
which is long compared with the mean oscillation period T D
2=! (Fig. 11.8). This oscillation x.t/ is called a beat and the
period is the beat period or beat cycle.
Acoustic beats can be realized by two vibrating tuning forks,
which are slightly detuned against each other. With a microphone they can be made audible to