Classical Mechanics: An Introduction 9783540736158, 9783540736165, 3540736158

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Classical Mechanics: An Introduction
 9783540736158, 9783540736165, 3540736158

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Dieter Strauch

Classical Mechanics An Introduction

123

Classical Mechanics

Dieter Strauch

Classical Mechanics An Introduction

With 149 Figures

ABC

Professor Dr. Dieter Strauch Universit¨at Regensburg Institut Theoretische Physik Universit¨atsstr. 31, 93053 Regensburg, Germany E-mail: [email protected]

ISBN 978-3-540-73615-8

e-ISBN 978-3-540-73616-5

DOI 10.1007/978-3-540-73616-5

Library of Congress Control Number: 2009929359 © 2009 Springer-Verlag Berlin Heidelberg This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilm or in any other way, and storage in data banks. Duplication of this publication or parts thereof is permitted only under the provisions of the German Copyright Law of September 9, 1965, in its current version, and permission for use must always be obtained from Springer. Violations are liable to prosecution under the German Copyright Law. The use of general descriptive names, registered names, trademarks, etc. in this publication does not imply, 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. Typesetting: Data prepared by the Author and by SPi Technologies Cover design: WMXDesign GmbH, Heidelberg SPIN 11780212 Printed on acid-free paper 987654321 springer.com

Preface

On this Textbook This book has evolved from the series of lecture notes that I had handed out to the physics students at the University of Regensburg in Bavaria, Germany. Over the years, various bits and pieces had been added to the contents of these lecture notes, and others had to be left out for reasons of time limitations. These notes differed from the common textbooks, and as the students seemed to like them, I have collected all those pieces in this book. The Scope of this Book The scope of this book is twofold. The reader can learn that alternative sets of the very few principles of Classical Mechanics carry on very far. Thus, the book contains an amount of applications of varying degrees of sophistication. Also, different physical problems require different methods for their solutions with varying degrees of mathematical sophistication. The Organization of this Book In order not to blur the physics with mathematical intricacies, the necessary mathematical techniques are transferred to appendices. The largest difference of this textbook from other books on Classical Mechanics may be that I have tried to make a particularly strong separation between axioms and fundamental experiences, on the one hand, and between claims, their proofs, various comments, on the other, rather than telling a more or less continuous story. Also, frequent references are made to other parts of the book or to other physical disciplines. If needed, the reader can skip proofs, comments, applications, and footnotes and thus follow only the main ideas. An asterisk has been attached to the more advanced topics of this book, which might also be skipped upon first reading.

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Preface

The Readership On the one hand, the book aims at the beginners by being very explicit on going through the proofs of statements being made and of the examples being given. On the other hand, the applications of the principles are extended to a number of topics which might be of interest to the more advanced students. The Contents of this Book For the beginners I have often shown different paths to solutions of a problem, so they might estimate the (dis)advantage of one method over another. Thus, the reader should appreciate the employment of symmetry at an early stage. It is equally important to realise that physics, unlike mathematics, is not an exact science, but an approximation to the mathematical description of processes in nature, usually relying on a number of simplifying assumptions. In selecting the various assumptions the physicist is trained to choose the important over the unimportant effects and to set up a sensibly simple model. In fact, classical mechanics has its limitations, as astronomers found out a long time ago, and nowadays these limitations have been overcome by taking account of relativistic effects.1 Also, the mechanics of the microscopic world is governed by quantum theory.2 Thus, I have tried to concentrate on the practical applications of the principles and on the calculational techniques rather than on the construction of a theory as such. Even though the structure and method of theoretical physics is an exciting topic (in particular for the advanced physicist), I believe that it is indispensable to have a hands-on understanding of the theory with its basic results and of the techniques if one wants to handle and apply them in practice. Therefore, this book is an introduction to theoretical mechanics rather than an introduction to formal theoretical physics. By no means have I tried to be complete. Thus, some topics are missing which a physicist may need to know, like non-linear dynamics and (deterministic) chaos, and possibly others. Relativistic mechanics is only touched upon.3 Also the statistical methods of the many-body systems is a subject of its own.4 Even though skipped in most textbooks on theoretical mechanics, I have included a short chapter on continuum mechanics, since the mechanics of deformable bodies presents a first example of a (classical) field theory. Here,

1 2 3

4

See the Course on Special Relativity. See the Course on Quantum Mechanics. More details can be found in the separate Course on Special Relativity where the principle of relativity is applied to problems in mechanics, optics, electromagnetism, and quantum theory. See the Course on Thermodynamics and Statistics.

Preface

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numerous concepts are introduced which will be found again in electromagnetism,5 as, for example, the energy-momentum tensor, a lesser liked detail from electromagnetism, and other (classical and quantum) field theories. Material for Electromagnetism and Optics, Quantum Mechanics, Thermodynamics, Statistics, and Special Relativity are planned to be published in the near future.

What is Theoretical Mechanics? Physics is the mathematical description of events occurring in nature. (Comment: Nature offers numerous additional facets which are recognised by man: esthetic, emotional, economical, etc. Let us approach all these with wonder and responsibility!) Theoretical physics is the classification of these physical events (respectively of the corresponding mathematical ways of description) under a minimum number of axiomatic basic assumptions. (A detailed exposition on the nature of a physical theory can be found in the book by Ludwig [4].) These physical axioms cannot be proved (but can be falsified by just one experiment); they can only turn out to be useful. For the application of a physical theory to natural phenomena one presumes in general an idealization of nature, in that the considered aspect of nature is approximated by a model (point masses, neglect of the friction by the air in the case of free motion or by the support in the case of rolling cylinders, etc.). The topic of classical mechanics is the description of the motion of massive macroscopic bodies, namely the kinematics: the geometrical form of the trajectories (e.g. the elliptical trajectories in the Kepler problem) and the dynamics: the time development of the particle position and/or orientation (e.g. the orientation of the children’s top). One can treat only the motion of just one or two particles (e.g. the pendulum or a central star and a satellite, respectively) in an exact way. The motion of three particles cannot be treated analytically. For the treatment of the properties of very many particles (like fluids and gases) statistical methods are needed.6 A special case, however, which can be treated in the context of this Course, are extended rigid or (slightly) deformable solid bodies. Regensburg, January 2009 5 6

Dieter Strauch

Electromagnetism is likewise a field theory, see the Course on Electromagnetism. See the Course on Thermodynamics and Statistics.

Contents

Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

The Newtonian Mechanics of Point-Mass Systems: General Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Point Masses, Open and Closed Systems . . . . . . . . . . . . . . . . . . . . 1.1.1 The Point Mass as an Idealization . . . . . . . . . . . . . . . . . . . 1.1.2 Open and Closed Systems, Internal and External Forces 1.2 Newton’s Axioms (1687): A New Era . . . . . . . . . . . . . . . . . . . . . . 1.3 Einstein’s Equivalence Principle (1916): Another New Era . . . . 1.3.1 Mass, Weight, Force, and the Like . . . . . . . . . . . . . . . . . . . 1.3.2 Einstein’s Equivalence Principle . . . . . . . . . . . . . . . . . . . . . 1.4 Types of Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.1 Central Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.2 Conservative Forces (I) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.3 Central Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.4 Potential and Potential Energy . . . . . . . . . . . . . . . . . . . . . . 1.4.5 The Lorentz Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.6 Frictional Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5.2 Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6 Conservation Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6.1 Conservation of Momentum . . . . . . . . . . . . . . . . . . . . . . . . . 1.6.2 Angular-Momentum Conservation and Central Forces . . 1.7 Energy-Conservation Theorem and Conservative Forces . . . . . . 1.7.1 Conservative Forces (II) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.7.2 Potential Energy, Work, and Power . . . . . . . . . . . . . . . . . . 1.7.3 Energy Conservation Theorem . . . . . . . . . . . . . . . . . . . . . . 1.8 Invariances and Conservation Laws . . . . . . . . . . . . . . . . . . . . . . . . 1.8.1 Translations and Rotations . . . . . . . . . . . . . . . . . . . . . . . . .

v 1 1 1 2 4 8 8 9 9 9 11 12 12 12 13 14 14 15 17 17 18 18 18 19 20 20 20

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1.8.2 Invariances and Conservation Laws: Noether’s Theorem (I) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.8.3 The Usefulness of the Conservation Laws . . . . . . . . . . . . . 1.8.4 The Usefulness of Symmetry Analysis . . . . . . . . . . . . . . . . 1.9 * Virial Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.10 The Basic Problem of the Mechanics of a Point Mass . . . . . . . . Summary: Newtonian Mechanics – General Properties . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

21 23 23 23 25 25 27

2

Newtonian Mechanics: First Applications . . . . . . . . . . . . . . . . . . 2.1 One-Dimensional Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.1 Constant Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.2 Time-Dependent Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.3 Velocity-Dependent Force . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.4 Coordinate-Dependent Force . . . . . . . . . . . . . . . . . . . . . . . . 2.1.5 Example: Plane Pendulum (with Small Amplitude) . . . . 2.1.6 * Example: Plane Pendulum with Large Amplitude . . . . 2.2 Motion in a Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.1 Fixed and Moving Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.2 Time Dependence of the Moving Basis . . . . . . . . . . . . . . . 2.2.3 Velocity, Acceleration, etc. in Plane Polar Coordinates . 2.2.4 Example: Plane Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Two-Particle Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.1 Center-of-Mass and Relative Coordinates . . . . . . . . . . . . . 2.3.2 Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.3 Closed Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Summary: Newtonian Mechanics – First Applications . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

29 29 29 30 31 32 34 38 41 41 42 44 46 48 48 49 50 51 52

3

Lagrangian Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Motion with Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Constraints and Generalized Coordinates . . . . . . . . . . . . . . . . . . . 3.2.1 Examples of Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.2 Classification of Constraints . . . . . . . . . . . . . . . . . . . . . . . . 3.2.3 Degrees of Freedom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.4 Generalized Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.5 Example: Plane Double Pendulum . . . . . . . . . . . . . . . . . . . 3.3 Forces of Constraint . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 The Principle of d’Alembert . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.1 Real and Virtual Displacements . . . . . . . . . . . . . . . . . . . . . 3.4.2 The Principle of Virtual Work . . . . . . . . . . . . . . . . . . . . . . 3.4.3 The Principle of d’Alembert . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Lagrangian Equations of the Second Kind . . . . . . . . . . . . . . . . . . 3.5.1 External and Internal Forces . . . . . . . . . . . . . . . . . . . . . . . . 3.5.2 Conservative Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

55 56 57 57 58 61 62 62 63 64 64 66 67 67 68 68

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3.5.3 3.5.4 3.5.5 3.5.6

Inertial Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 Lagrangian Equations of Motion . . . . . . . . . . . . . . . . . . . . 70 Example: Plane Motion in a Central Potential . . . . . . . . 72 Example: Bead Sliding on a Uniformly Rotating Rod . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 3.5.7 Example: Plane Double Pendulum . . . . . . . . . . . . . . . . . . . 74 3.5.8 Separable Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 3.6 Generalized Momentum, Force, etc. . . . . . . . . . . . . . . . . . . . . . . . . 78 3.6.1 Generalized Force and Generalized Momentum . . . . . . . . 78 3.7 Velocity-Dependent Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 3.7.1 Generalized Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 3.7.2 Lorentz Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 3.8 Gauge Invariance and Form Invariance . . . . . . . . . . . . . . . . . . . . . 81 3.8.1 (Un)Ambiguity of the Lagrangian Function: Gauge Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 3.8.2 * Form Invariance of the Lagrangian Equations under Point Transformations . . . . . . . . . . . . . . . . . . . . . . . . 83 3.9 * Lagrangian Equations of the First Kind . . . . . . . . . . . . . . . . . . 84 3.9.1 Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 3.9.2 Lagrangian Equations of the First Kind . . . . . . . . . . . . . . 85 3.9.3 Example: Atwood Machine . . . . . . . . . . . . . . . . . . . . . . . . . 86 3.10 Hamilton’s Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92 3.10.1 The Action . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92 3.10.2 Hamilton’s Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 3.11 Symmetries and Conservation Laws . . . . . . . . . . . . . . . . . . . . . . . . 95 3.11.1 Noether’s Theorem (II) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 3.11.2 Cyclic Coordinate and Conservation of the Conjugate Momentum . . . . . . . . . . . . . . . . . . . . . . . . 96 3.11.3 Homogeneity in Time and Energy Conservation . . . . . . . 97 3.11.4 Space Homogeneity and Conservation of Momentum . . . 100 3.11.5 Isotropy and Angular-Momentum Conservation . . . . . . . 100 Summary: Lagrangian Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 4

Harmonic Vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 4.1 The Simple Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117 4.1.1 Eigen Solutions of the Homogeneous Differential Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117 4.2 The Free Oscillator with Damping . . . . . . . . . . . . . . . . . . . . . . . . . 120 4.2.1 Eigen Solutions of the Homogeneous Differential Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120 4.2.2 Reality of the Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 4.2.3 Initial Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

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4.3 The Forced Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 4.3.1 Equation of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 4.3.2 Solution of the Inhomogeneous Linear Differential Equation by Fourier Transformation . . . . . . . . . . . . . . . . . 126 4.3.3 Green Function of the Damped Oscillator in Frequency Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 4.3.4 Time-Dependent Green Function . . . . . . . . . . . . . . . . . . . . 131 4.3.5 Energy Considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 4.4 Coupled Oscillators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 4.4.1 Introductory Example: Stretching Vibrations in the CO2 Molecule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 4.4.2 General Coupled Vibrations . . . . . . . . . . . . . . . . . . . . . . . . 145 4.4.3 Normal Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 Summary: Harmonic Vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 5

Central Potentials and the Kepler Problem . . . . . . . . . . . . . . . . 157 5.1 Central Force and Motion in a Plane . . . . . . . . . . . . . . . . . . . . . . . 157 5.1.1 Central Potential, Central Force, and Angular-Momentum Conservation . . . . . . . . . . . . . . . . . . . 157 5.1.2 Central Potential and Effective Radial Potential . . . . . . . 159 5.1.3 Central Potential and Trajectory . . . . . . . . . . . . . . . . . . . . 160 5.2 Kinematics of the Kepler Motion . . . . . . . . . . . . . . . . . . . . . . . . . . 163 5.2.1 Kepler’s Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163 5.2.2 Polar Representation of the Conics . . . . . . . . . . . . . . . . . . 165 5.2.3 Determination of the Force and Potential from the Trajectory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166 5.2.4 Determination of the Trajectory from the Potential . . . . 168 5.2.5 Trajectory and Rotation Periods . . . . . . . . . . . . . . . . . . . . 171 5.2.6 The Laplace–Runge–Lenz Vector . . . . . . . . . . . . . . . . . . . . 173 5.2.7 Perihelion Rotation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175 Summary: Central Potentials and the Kepler Problem . . . . . . . . . . . . 177 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178

6

Collision and Scattering Problems . . . . . . . . . . . . . . . . . . . . . . . . . 183 6.1 Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 6.1.1 Scattering and Collision . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184 6.1.2 Momentum Change and Impulsive Force . . . . . . . . . . . . . 184 6.1.3 Laboratory System and Center-of-Mass System . . . . . . . 185 6.1.4 Consequences of the Conservation of Momentum . . . . . . 186 6.1.5 Elastic and Inelastic Scattering . . . . . . . . . . . . . . . . . . . . . . 187 6.1.6 Consequences of the Angular-Momentum Conservation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189

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6.2 Collision of Hard Spheres . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 6.2.1 Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 6.2.2 Elastic Collision of Smooth Spheres (Laboratory System) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 6.3 Scattering by a Central Potential . . . . . . . . . . . . . . . . . . . . . . . . . . 194 6.3.1 Central Potential and Scattering Angle . . . . . . . . . . . . . . . 194 6.3.2 Scattering by the Gravitational Potential . . . . . . . . . . . . . 195 6.4 * The Cross-Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196 6.4.1 The Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196 6.4.2 Scattering of Many Probe Particles by Many Target Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 6.4.3 Scattering from a Particle at Rest . . . . . . . . . . . . . . . . . . . 197 6.4.4 The Differential Cross-Section . . . . . . . . . . . . . . . . . . . . . . . 198 6.4.5 The Total Cross-Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200 6.4.6 The Rutherford Cross-Section . . . . . . . . . . . . . . . . . . . . . . . 200 6.4.7 Scattering Cross-Section for a Collision of Hard Spheres . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201 Summary: Collision and Scattering Problems . . . . . . . . . . . . . . . . . . . . 202 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 7

Moving Reference Frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207 7.1 Translations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208 7.1.1 The Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208 7.1.2 Inertial Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209 7.2 Rotation Around a Fixed Point . . . . . . . . . . . . . . . . . . . . . . . . . . . 209 7.2.1 Active and Passive Rotation . . . . . . . . . . . . . . . . . . . . . . . . 209 7.2.2 Infinitesimal Rotations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210 7.2.3 Representation in Different Coordinate Systems . . . . . . . 211 7.2.4 Uniformly Rotating System: Centrifugal and Coriolis Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214 7.2.5 The Foucault Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214 7.3 Galilean and Lorentz Transformation . . . . . . . . . . . . . . . . . . . . . . 217 7.3.1 The Relativity Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218 7.3.2 General and Special Transformation . . . . . . . . . . . . . . . . . 218 7.3.3 The Galilean Transformation . . . . . . . . . . . . . . . . . . . . . . . 219 7.3.4 Galilean Invariance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220 7.3.5 The Lorentz Transformation . . . . . . . . . . . . . . . . . . . . . . . . 221 Summary: Moving Reference Frames . . . . . . . . . . . . . . . . . . . . . . . . . . . 222 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223

8

Dynamics of a Rigid Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227 8.1 The Rigid Body as a System of N Point Masses . . . . . . . . . . . . . 227 8.2 Translational and Rotational Energy: Inertia Tensor . . . . . . . . . 228 8.2.1 The Center of Mass as the Specific Point . . . . . . . . . . . . . 228 8.2.2 The Instantaneously Fixed Point as the Specific Point . . 231

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8.3 Transition to the Continuum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232 8.3.1 Example: Center of Mass of a Homogeneous Hemisphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233 8.3.2 Examples of Inertia Tensors . . . . . . . . . . . . . . . . . . . . . . . . 234 8.4 Change of the Reference System: Steiner’s Theorem . . . . . . . . . . 238 8.4.1 Steiner’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238 8.4.2 Angular Momentum and Torque . . . . . . . . . . . . . . . . . . . . 239 8.4.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241 8.5 Principal Moments of Inertia and Principal Axes . . . . . . . . . . . . 242 8.5.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242 8.5.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244 8.6 Rotation Around a Fixed Axis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247 8.6.1 Inertia Tensor and Moment of Inertia . . . . . . . . . . . . . . . . 248 8.6.2 The Equation of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249 8.6.3 Example: Motion in the Homogeneous Gravitational Field: The Physical Pendulum . . . . . . . . . . . . . . . . . . . . . . 249 8.6.4 Example: A Cylinder Rolling on an Inclined Plane . . . . . 251 8.7 Rotation Around a Fixed Point: The Top . . . . . . . . . . . . . . . . . . . 253 8.7.1 Space-Fixed (Inertial) and Body-Fixed Coordinate System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253 8.7.2 * Euler Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254 8.7.3 * The Euler Equations of Motion . . . . . . . . . . . . . . . . . . . . 257 8.8 The Force-Free Symmetrical Top . . . . . . . . . . . . . . . . . . . . . . . . . . 258 8.8.1 The Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . 258 8.8.2 * Stability of the Rotational Motion of the Top . . . . . . . 260 Summary: Dynamics of a Rigid Body . . . . . . . . . . . . . . . . . . . . . . . . . . . 262 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263 9

Hamiltonian Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269 9.1 Hamiltonian Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . 269 9.1.1 The Hamiltonian Function . . . . . . . . . . . . . . . . . . . . . . . . . . 269 9.1.2 Canonical (Hamiltonian) Equations of Motion . . . . . . . . . 271 9.1.3 * (Un)Ambiguity of the Hamiltonian Function: Gauge Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274 9.2 * Poisson Brackets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275 9.2.1 Definition 29. (Poisson Brackets) . . . . . . . . . . . . . . . . . . . . 275 9.2.2 Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276 9.2.3 Fundamental Poisson Brackets . . . . . . . . . . . . . . . . . . . . . . 277 9.2.4 Properties of the Poisson Brackets . . . . . . . . . . . . . . . . . . . 277 9.2.5 Example: Harmonic oscillator . . . . . . . . . . . . . . . . . . . . . . . 277 9.2.6 Transition to Quantum Mechanics . . . . . . . . . . . . . . . . . . . 278 9.3 Configuration Space and Phase Space . . . . . . . . . . . . . . . . . . . . . . 278 9.3.1 Configuration Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278 9.3.2 Phase Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279

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9.4 * The Modified Hamilton Principle . . . . . . . . . . . . . . . . . . . . . . . . 281 9.5 * Canonical Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282 9.5.1 Point Transformation in Configuration Space . . . . . . . . . 282 9.5.2 Point Transformation in Phase Space . . . . . . . . . . . . . . . . 283 9.5.3 Canonical Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . 284 9.5.4 Time Development as a Canonical Transformation . . . . . 287 9.5.5 Canonical Invariants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288 9.5.6 Generating Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289 9.5.7 The Theorem of Liouville . . . . . . . . . . . . . . . . . . . . . . . . . . . 294 9.6 * Hamilton–Jacobi Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296 9.6.1 Action Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296 9.6.2 The Characteristic Function . . . . . . . . . . . . . . . . . . . . . . . . 298 9.6.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299 Summary: Hamiltonian Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304 10 * Introduction to the Mechanics of Continua . . . . . . . . . . . . . . 309 10.1 Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 309 10.2 Lagrangian Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311 10.2.1 Kinetic Energy Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311 10.2.2 Potential Energy Density . . . . . . . . . . . . . . . . . . . . . . . . . . . 312 10.2.3 Lagrangian Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314 10.3 Hamilton’s Principle and Lagrangian Equations . . . . . . . . . . . . . 314 10.4 The Energy–Momentum Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . 317 10.4.1 The Conservation Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 317 10.4.2 Energy Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318 10.4.3 Energy–Current Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319 10.4.4 Energy Conservation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 320 10.4.5 The Strain–Momentum Density . . . . . . . . . . . . . . . . . . . . . 320 10.4.6 The Stress Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322 10.4.7 Conservation of the Strain Momentum . . . . . . . . . . . . . . . 323 Summary: Mechanics of Continua . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324 A

Physical Constants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325

B

Scalars, Vectors, Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329 B.1 Definitions and Simple Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329 B.1.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329 B.1.2 Behavior Under Inversion . . . . . . . . . . . . . . . . . . . . . . . . . . . 330 B.2 Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330 B.2.1 Rules for Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330 B.2.2 Dyadic Product (Tensor Product) of Vectors . . . . . . . . . . 331 B.2.3 Vector Product (Cross Product) of Vectors . . . . . . . . . . . 331 B.2.4 Triple Scalar Product of Vectors . . . . . . . . . . . . . . . . . . . . . 332 B.2.5 Multiple Products of Vectors . . . . . . . . . . . . . . . . . . . . . . . . 332

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C

Rectangular Coordinate Systems . . . . . . . . . . . . . . . . . . . . . . . . . . 333 C.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333 C.2 Cartesian Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333 C.3 Spherical Polar Coordinates (Spherical Coordinates) . . . . . . . . . 334 C.4 Cylindrical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335 C.5 Plane Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335 C.6 Inverse Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336

D

Nabla (Del) Operator and Laplace Operator . . . . . . . . . . . . . . . 339 D.1 Representations of the Nabla and Laplace Operators . . . . . . . . . 339 D.2 Standard Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 340 D.3 Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341 D.4 Integral Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342 D.4.1 The Theorem of Gauß . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342 D.4.2 Stokes’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343 D.4.3 The Theorem of Green . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344

E

Variational Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347 E.1 Functions and Functionals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347 E.2 Variational Problem and Euler Equation . . . . . . . . . . . . . . . . . . . 348 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 350

F

Linear Differential Equations with Constant Coefficients . . . 351 F.1 Homogeneous Linear Differential Equations . . . . . . . . . . . . . . . . . 351 F.2 Inhomogeneous Linear Differential Equations . . . . . . . . . . . . . . . 352 F.3 Stability of Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353

G

Quadratic Matrices and Their Eigen Solutions . . . . . . . . . . . . . 355 G.1 The Eigen Value Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355 G.2 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355 G.3 Properties of the Eigen Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357 G.4 Properties of the Eigen Vectors of Hermitian Matrices . . . . . . . . 358

H

Dirac δ-Function and Heaviside Step Function . . . . . . . . . . . . . 361 H.1 Properties of the Dirac δ-Function and of the Heaviside Step Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361 H.2 Representation of the δ-Function by Functional Sequences . . . . 362 H.3 Integral Representation of the δ-Function . . . . . . . . . . . . . . . . . . . 363 H.4 Periodic δ-Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363 H.5 The δ-Function in R3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363 H.6 The δ-Function as an Inhomogeneity of the Poisson Equation . 364

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I

Fourier Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365 I.1 The Transformation: Fourier Integral . . . . . . . . . . . . . . . . . . . . . . 365 I.1.1 Examples and Applications . . . . . . . . . . . . . . . . . . . . . . . . . 366 I.1.2 Convolution Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 370 I.1.3 Parseval’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371 I.1.4 Uncertainty Relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372 I.2 Fourier Transformation in R4 : Plane Waves . . . . . . . . . . . . . . . . . 374 I.2.1 The Whole R3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374 I.2.2 Normalization Volume V . . . . . . . . . . . . . . . . . . . . . . . . . . . 375 I.3 Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 376 I.3.1 The Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 376 I.3.2 Examples and Applications . . . . . . . . . . . . . . . . . . . . . . . . . 378 I.3.3 Convolution Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380 I.3.4 Parseval’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382 I.3.5 Fourier Series in R3 : Lattices . . . . . . . . . . . . . . . . . . . . . . . 382 I.3.6 Functions with Lattice Periodicity . . . . . . . . . . . . . . . . . . . 383 Summary: Fourier Integral and Fourier Series . . . . . . . . . . . . . . . . . . . . 384 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385

J

Change of Variables: Legendre Transformation . . . . . . . . . . . . . 387

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 389 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 391

Symbols

a da A A b b B c C d d D e e e E f f F g g G h H H i I j k k K

Acceleration, general vector Surface (“area”) element (oriented outwards) Work Electromagnetic vector-potential field Impact parameter General vector Magnetic-induction field General vector Curve Differential General vector General matrix eccentricity Unit vector, eigenvector Exponential function Electrical field Force-density field Force (density) Generating function, area, gauge function (Sect. 3.8) Green matrix Gravitational acceleration, general vector field Reciprocal-lattice vector Lattice vector Hamiltonian function Hamiltonian density Imaginary unit Moment of inertia Energy current density Force constant, wave vector in R1 Wave vector Hamiltonian-type function

xx

Symbols

l L L L m M n N N p p P P q q Q r R R s s ds S t T T T u U v V V Z

Angular momentum Lagrangian function, canoid lengths Lagrangian density Total angular momentum (Point) Mass Mass matrix, general matrix Vector normal to a surface element Total number of particles Torque Canonical momentum Momentum, electrical dipole Power, transformed canonical momentum Total momentum Canonical coordinate, electrical charge Vector within Brillouin zone Generalized force, transformed canonical coordinate, electrical charge Position vector Radius of sphere Center of mass, translation vector Impact parameter Small deviation from ω (Sect. 8.8.2) Line element Tension Time Kinetic energy Energy-momentum tensor Stress tensor Displacement field, general vector Transformation matrix, unitary matrix Velocity, general vector Potential energy Center-of-mass velocity Force of constraint

α β γ δ Δ  ε η θ

Stokes frictional force constant (1.18), angle, cartesian index Velocity relative to velocity of light (Sect. 7.3), cartesian index Damping constant, constant in SR Dirac delta function Difference Levi-Civita tensor, small quantity Reduced energy Viscosity coefficient Heaviside theta function, (Euler) angle, scattering angle

Symbols

Θ Θ κ λ λ Λ μ ν ξ π Π ρ  σ τ φ ϕ ϕ χ ψ ω ω dΩ

Moment of inertia Moment-of-inertia tensor Kinetic-energy expansion coefficient, counter for function zeroes Lagrangian multiplier, kinetic-energy expansion coefficient, wavelength, eigenvalue Runke-Lenz vector Lagrangian-type function Sliding-friction coefficient, reduced mass, kinetic-energy expansion coefficient, four-index in SR Frequency, four-index in SR Complex coordinate, curvilinear coordinate Number pi, canonical momentum density Transformation of canonical coordinates Mass density, charge density, particle density Damping constant, cylindrical radial coordinate Cross section Time increment Electromagnetic scalar potential Rotation angle Angular vector of rotation Electromagnetic gauge function Euler angle Vector of angular velocity Angular velocity Space angle element

xxi

1 The Newtonian Mechanics of Point-Mass Systems: General Properties

Starting from the Newtonian axioms general conclusions concerning the motion of masses shall be drawn in this chapter, and we will concentrate on theorems which one obtains without explicit calculation, i.e., without explicit solutions of the equations of motion. The integration of the equations of motion will then be carried out for various examples in Chap. 2. The primary quantity of this chapter is the force, while in Chap. 3 the primary quantity is the (kinetic and potential) energy. Even though the force on the one hand and the energy on the other as the cause of the motion should be known from the elementary lecture on phenomenological (experimental) mechanics, the force as agens movendi shall be put in the foreground in this introduction, since the Newtonian mechanics is more intuitive than the more abstract Lagrangian theory, even though in many cases the latter method is more useful and comfortable in setting up the equations of motion.

1.1 Point Masses, Open and Closed Systems 1.1.1 The Point Mass as an Idealization Of course, all bodies are extended and not point-like, but in most of those cases, which shall be investigated here, the extension of the various bodies is small compared to their mutual distances, and in this case it is a good approximation to consider these bodies as point-like.1 Then one talks of point masses. Examples: The extension of moon and earth is small compared to their mutual distance. – The extension of the earth–moon system is small compared to the distance to the sun. – The extent of the system of the sun with its planets is small compared to the distance to the nearest star. – The extension of the molecules of a dilute gas is small compared to the average distance 1

See also the comment to the center of mass at the end of Sect. 1.5.

2

1 Newtonian Mechanics: General Properties

Fig. 1.1. The internal and external forces in a system of particles

between the molecules; but for dense gases at high pressures one may approach the limit of the validity of the approximation by a point-like “extent”.2 – Idealizations similar to the massive particles one has likewise for the extended electrically charged particles. 1.1.2 Open and Closed Systems, Internal and External Forces Definition 1. (Internal and External Forces): If one combines a chosen number of particles to a system then the forces between these selected particles of the system are inner forces, and the (“externally” acting) forces of the other (external) particles on these selected (“inner”) particles are external forces. The force F i (r i ) on a particle (index i) at position ri thus is composed of the external force F ext i (r i ) and of the internal forces F ij , which are caused by the particles j at the positions r j ,3  F i (r i ) = F ij (r 1 , r 2 , . . .) + F ext (1.1) i (r i ), j

see Fig. 1.1. Definition 2. (Closed System): A closed system is a system without external forces. Definition 3. (Open System): An open system is a system with external forces. 2 3

Extended solid bodies are treated in Chap. 8 and extended elastic bodies in Chap. 10. The other way around, Scheck denotes F ij as the force of the particle i on the particle j.

1.1 Point Masses, Open and Closed Systems

3

Fig. 1.2. Left: Earth and moon with negligible further external forces as a closed system. Right: Earth and moon as an open system with external forces exerted by the sun

Example: One can consider the motion of the moon around the earth and the motion these two around the sun (neglecting the influence of the other planets) in (at least) two ways: Firstly, one can combine moon and earth to an open system; then the force between moon and earth is an internal force, and the forces by the sun on the system of earth and moon are external forces. Alternatively, one can combine all three bodies to a (closed) system; then all forces between these three bodies are internal forces. Finally, if one would take into account the forces of the other planets and stars acting on the three-body system of sun, moon, and earth, then those forces would be external forces, and the three-body system would be an open system. Further Examples: • • • •

Planets in the gravitational field of the sun (open) Planets and sun (closed), see Fig. 1.2 Charge in a condenser (open) Charge and condenser (and possibly the source of the current) (closed), see Fig. 1.3

Comment: Whether one considers an open or a closed system, often is a question of expediency. For example, in an approximate way one can often investigate the motion of the planets in the field of a central star assumed as fixed; the only change for the planetary motion taking account of the moving

Fig. 1.3. A charge in a condenser as an (left) open system and (right) closed system

4

1 Newtonian Mechanics: General Properties

Fig. 1.4. The internal forces Fij and external forces Fiext acting on particle i

central star would be the replacement of the mass of the planet by the reduced mass (of planet and central star).4 In the (academic) limit of an infinitely heavy central star the two systems are identical.

1.2 Newton’s Axioms (1687): A New Era The original formulation of the Newtonian axioms have been different, of course. In modern formulation they read: (I) Law of Inertia: There are reference systems, namely so-called inertial systems, in which each particle (index i) remains in rectilinear uniform motion (in particular at rest), if there is no resulting force acting on it, Fi = 0



p˙ i = 0



pi = const.

(1.2)

(II) Law of Motion: In inertial systems the time change of the momentum is equal (in modulus and direction) to the acting force, F i = p˙ i .

(1.3)

(III) Reaction Law: The force F ji of a particle i acting on a particle j is oppositely equal to the force F ij of particle j acting on particle i, F ji = −F ij .

(1.4)

(IV) Superposition Principle: The resulting force F i of all forces F ij acting on a particle i are superposed according to the rule of vector addition, (1.1)  Fi = F ij + F ext (1.5) i . j

Cf. Fig. 1.4. 4

See the two-body problem in Sect. 2.3.

1.2 Newton’s Axioms (1687): A New Era

5

Agreement 1 (time derivatives): A time derivative of a quantity will be denoted by a dot over the symbol. Comments: • These laws presume that space and time exists, that one can thus measure distances |Δr| and time differences Δt, and that the length and time scales are constant in space and time. The time is homogeneous, and the space is homogeneous and isotropic. Newton assumed an absolute space and an absolute time which is not quite correct, as one knows since Einstein’s formulation of the Special Relativity theory at the beginning of the twentieth century.5 • Starting from these laws, the motion of particles under the influence of forces shall be described in this chapter. (In the following we will investigate primarily point masses.) Thus the main object is to describe the form of the trajectories of all particles, for example the Kepler ellipses in the case of the planetary motion. This is the topic of the kinematics. Or, and this is the more frequent case, one is rather interested in how, i.e., with which time dependence, the ith particle moves along the trajectory. Then one is interested in the time dependence of the coordinate ri (t). This is the topic of the dynamics. From the latter the form of the trajectory is given in one of many other possible forms of parametric representations, namely that with the time t as parameter. • Furthermore, the differentiability etc. is assumed, from which we will always start in the following, if not stated otherwise, such that the following quantities (of a particle) are defined: velocity acceleration

˙ v(t) = r(t) ≡ a = v˙ = r¨.

dr dt

= limΔt→ 0

r(t+Δt)−r(t) Δt

In classical mechanics one also uses, e.g., the mass density dm/dV , which presents difficulties in the microscopic regime because of the corpuscular nature of matter. • The systems postulated in the first Newtonian law are the so-called inertial systems. In contrast, an isolated particle in all accelerated systems is thus subject to forces, even if there are no external forces acting.6 Since the forces decrease at large distances between the particles7 and since forces originate only from other particles, one can define an inertial system equivalently by the requirement that such a system is sufficiently far away from 5

6 7

Even though one can measure distances Δl = |r 1 − r 2 | and time differences Δt = t1 − t2 (Euklidian structure of the Rt × R3 ), these may depend upon the reference system: There are the effects of length contraction and time dilation, see Sect. 7.3. These are so-called inertial forces, see, e.g., Sects. 7.1.1 and 7.2.4. This holds for the gravitational force and for the Coulomb force and, as far as one knows, also for all other forces, like e.g., nuclear forces – with the exception of the confinement forces between the quarks.

6

• • • •

1 Newtonian Mechanics: General Properties

all other particles and does not rotate against the system of the far-away particles or is in any other way accelerated.8 Newton’s first law may seem as a special case of the second law. This is not the case, since it postulates the existence of the inertial reference systems and defines them. Useful (approximate) inertial systems are – for many problems: earth – for Coriolis effects on the earth: planetary system, galaxy With his formulation of the second axiom in the form of F = p˙ (and not ˙ Newton was (unknowingly) ahead of his time.9 F = mv) In general one has p˙ = m v˙ + m ˙ v,

and the equation F = mv˙ is a special case. – The case m ˙ = 0 appears, e.g., in the treatment of the rocket, the mass of which decreases with time due to burning of the fuel. • Implicitly two-particle interactions are assumed in the third law.10 This applies to the gravitational interaction of macroscopic physics as well as to the Coulomb interaction of macroscopic and microscopic physics. It does not apply to the phenomenological frictional force,11 which one would have to derive from microscopic interactions, as well as to the phenomenological angular forces in molecules and solids. • The sum of the internal forces vanishes because of Newton’s third axiom,  F ij = 0, (1.6) ij

even if they are not central forces.12 8 9

In accelerated reference systems one has the appearance of the so-called inertial forces, see Sect. 7.2.4. In the Special Relativity theory one has p= 

m0 v(t) 1 − (v(t)/c)2

with the rest mass m0 . This is sometimes written in the form m0 p = mv with m =  . 1 − (v(t)/c)2

10 11 12

But properly speaking, in Special Relativity it is not the mass, which appears as velocity dependent, but it actually is the derivative with respect to time which is ˙ See the Course on Special Relativity. modified, e.g., in v = r. Cf. Fig. 1.5. See Sect. 1.4.6. See Fig. 1.5 and Sect. 1.4.

1.2 Newton’s Axioms (1687): A New Era

7

Fig. 1.5. The oppositely equal forces; left: central forces; right: noncentral forces

• In inertial systems the Newtonian laws are form-invariant under Galilean transformations,13 i.e., the Newtonian laws have the same mathematical form, independent of the special inertial system. • Newton’s first and second laws are fundamental insofar as they united the celestial and earthly mechanics. Before Newton the idea was that processes on earth depended on action by man or nature and were unforeseeable. In contrast, the motion of the stars and planets have been extremely steady, thus they were not only considered as extraterrestrial but as supernatural; in fact, in some ages before that a “soul” has been assigned to them because of their seemingly fore-free motion. Actually, Galileo Galilei has had similar ideas. • Newton’s first and second laws have been also a remarkable revolution of physical thought, insofar as from empirical experience it needed a permanent force to keep a body in steady motion. • The case of vanishing friction was not realized before Newton’s time, except possibly by Galileo Galilei. He was among the first who, among many other accomplishments, started from notions, assumptions, and models to draw conclusions in a way which we consider today as scientific. However, instead of the rectilinear motion he assumed the circular motion to be the force-free motion like that of the celestial bodies. Agreement 2 (constant mass): If not stated otherwise, the mass m of a particle will always be assumed as constant in the following m ˙ = 0. Agreement 3 (reference system): If not stated otherwise, an inertial system will always be assumed in the following. Definition 4. (Configuration space): The trajectory14 r(t) : Rt → R3 of a single particle is a mapping of the one-dimensional space Rt of the time onto 13 14

See Sect. 7.3.3. In Special Relativity theory this is the so-called world line, see Sect. 7.3.

8

1 Newtonian Mechanics: General Properties

the R3 , the so-called configuration space. If one has N particles with their trajectories r i (t) (with i = 1, . . . , N ) the configuration space is analogously the N -fold product R3 × R3 × · · · × R3 = R3N .

1.3 Einstein’s Equivalence Principle (1916): Another New Era An implicit additional principle is Einstein’s equivalence principle of the equality of gravitational and inertial mass: 1.3.1 Mass, Weight, Force, and the Like With length and time as measurable quantities one can easily define velocity, acceleration, volume, etc. But there remains the problem of measuring mass, weight, force, momentum, energy or the like without a new unit. Even if well known to the physicist from the introductory lectures on phenomenological mechanics one should be aware that the latter of the abovementioned quantities have been difficult to define some ages ago. While a spring can be stretched by a muscle independent of whether the spring is aligned horizontally or vertically, the extension of the spring by a mass in the gravitational field depends on the orientation of the spring relative to the vertical direction; there seems to be a difference between “mass” and “weight” (of course, as we know today). Anyway, another unit had to be introduced. In Newton’s second law either the mass or alternatively the force or the momentum is defined. With force (and momentum) given, the relation p = mi v defines the inertial mass mi .15 The gravitational mass mg is defined by the gravitational force F = mg g in a (homogeneous) gravitational field. By equating these forces, F = mg g

˙ and F = mi v,

(1.7)

one obtains v˙ =

mg g mi

for all bodies, thus a proportionality of inertial and gravitational mass. 15

This is the easy route: Of course, if the mass were given, the force (and momentum) would be derived quantities.

1.4 Types of Forces

9

1.3.2 Einstein’s Equivalence Principle The equality mg = mi

(1.8)

is known as the Einstein equivalence principle. Comments: Even though the mass equivalence is usually referred to as being stated in the 1905 paper of Einstein, this latter paper contains the energymass equivalence of Special Relativity, while the mass equivalence is the basic assumption of the General Relativity theory of the 1916 paper. Originally postulated, the equality has been found from experiments to hold to within 10−12 . In fact, already in 1916 the mass equality was experimentally confirmed to within some parts of 10−6 by E¨ otv¨ os (1828–1919). Agreement 4 (mass): Henceforth we will not distinguish between inertial and gravitational mass.

1.4 Types of Forces 1.4.1 Central Forces Definition 5. (Fields): A scalar field or a vector field shall be understood as a scalar or vectorial quantity, respectively, depending on a coordinate, i.e., a mapping of configuration space to a scalar or vectorial space. Comment: The notion of a field may be more familiar from electromagnetism: The connection of the electrical field E and of the force F on a charge q at the position r is F (r) = qE(r). A scalar field is thus a mapping of the R3 onto an R1 , and a vector field is a mapping of the R3 onto an R3 . Definition 6. (Central force field): A central force field is a radially oriented vector field (for F < 0 or F > 0 oriented to or from the center, respectively) F (r) = F (r) er ,

er =

r . r

(1.9)

er is the unit vector (with modulus 1) pointing in the radial direction, see the left part of Fig. 1.5. Agreement 5 (unit vectors): In general we will denote unit vectors by e.

10

1 Newtonian Mechanics: General Properties

Basic experience for masses (gravitational force): The gravitational force between two (point) masses m1 and m2 at the positions r1 respectively r2 is oriented radially and proportional to the squared inverse of their distance,16 mi mj F Grav = −G eij (1.10) ij |ri − r j |2 with the unit vector eij =

r i − rj . |r i − rj |

(1.11)

Comments: • Like the axioms the basic experiences cannot be proven on mathematical grounds. They can only be verified by experiment to within a certain approximation; presently the error is less than 10−10 . • With the mass and the force defined in (1.7) and with the symmetry of the interaction, the force in (1.10) must necessarily contain a proportionality constant, the gravitational constant G. • Kepler’s laws17 laid the basis for the gravitational potential. They could be taken as a set of alternative basic experiences. However, from the one gravitational potential all three of Kepler’s laws can be concluded. Further Examples: The Coulomb force18 between two point charges q1 and q2 at the positions r1 respectively r 2 is qualitatively similar to the gravitational force, F Coul = ij

qi qj 1 eij , 4πε0 |r i − rj |2

(1.12)

see Fig. 1.6. An example for a non-central force is the force, which an electrical point dipole p at r = 0 exerts on a charge Q at r (or also reversely), F (r) = −

16 17 18

1 p − 3 er p · er , 4πε0 r3

The gravitational constant is G = 6.67 × 10−11 m3 kg−1 s−2 ; a more precise value can be found in Appendix A. See Sect. 5.2.1. Unfortunately still two unit systems are customary in electromagnetism; both systems can be used here simultaneously if one defines  x in the SI system [x] = 1 in the cgs system. In particular one has 1 = 4πc2 × 10−7 4πε0 with the velocity of light c = 2.997 924 58 × 108 m s−1 .

1.4 Types of Forces

11

Fig. 1.6. The force F on an electrically (left) positive and (right) negative charge q in an electrical field E

Fig. 1.7. The forces of an electrical dipole p on positive charges at equal distances but different positions, namely parallel, respectively, perpendicular to the dipole orientation

see Fig. 1.7 (The force field is, however, a superposition of central (Coulomb) forces originating from the two charges forming the dipole, see the Course on Electricity.) Additional examples for noncentral forces one finds for the interactions between elementary particles. 1.4.2 Conservative Forces (I) Definition 7. (Conservative force field): A force field F (r) is called conservative,19,20 if it can be written as a gradient of a scalar field V (r), F (r) = −∇V (r).

(1.13)

Notation: This scalar field V (r) is called potential energy. Agreement 6 (del operator): Instead of the nabla operator (del operator) ∇ we will often formally write ∂/∂r, F (r) = −∇V (r) ≡ −

∂V (r) , ∂r

well knowing that one cannot divide by a vector. However, this will turn out to be useful in the context of the chain rule of differentiation. Agreement 7 (conservative forces): If not stated otherwise, we will treat the inner forces as conservative forces. In particular for the gravitational and for the Coulomb force one has 19 20

Alternative definitions will be found in Sect. 1.7. The denotation stems from the fact that the energy is conserved in a system with conservative forces; see Sect. 1.7.

12

1 Newtonian Mechanics: General Properties

mi mj |ri − rj | qi qj 1 . = 4πε0 |ri − r j |

VijGrav = −G

(1.14)

VijCoul

(1.15)

with ∇i

1 d |ri −r 1 1 j| = ∇i |r i − r j | = − eij |r i − r j | d|r i − r j | |r i − rj |2

as in (1.10) and (1.12). 1.4.3 Central Potentials Definition 8. (Central potential): A central potential is radially symmetric and has the form V (r) = V (|r|) = V (r). (1.16) Comments A central potential leads to a conservative and central force, see Problem 1.4. 1.4.4 Potential and Potential Energy In mechanics the notions potential and potential energy are mostly used with same meaning. Properly speaking, e.g., the gravitational potential U (r) at the position r, caused by a mass M at the position r M , is given by U (r) = −G

M , |r − rM |

and the potential energy of a mass m at the position rm in this potential is V (r m ) = mU (rm ) = −G

mM . |r m − r M |

Comment: For reasons of lingual simplification the notion “potential” is often incorrectly used in the sense of “potential energy.” In electromagnetism one is more careful in differentiating between these two notions. 1.4.5 The Lorentz Force Basic experience for charges in electromagnetic fields: The force of electromagnetic fields on a point charge q does not only depend upon its coordinate r but also upon its velocity and is21 21

See footnote 1.18.

1.4 Types of Forces

13

Fig. 1.8. The vectors of the magnetic induction B, of the velocity v, and of the Lorentz force F = qv × B

 [c] v(r, t) × B(r, t) , F (r, v, t) = q E(r, t) + c 

(1.17)

see Fig. 1.8. Here the first term is the Coulomb force, and the second term is the Lorentz force, the latter depicted in Fig. 1.8. In general, an external force on a particle can thus depend upon its coordinate r as well as upon its velocity r˙ and explicitly upon the time t, ˙ t). F = F (r, r, 1.4.6 Frictional Forces Frictional forces are primarily velocity dependent. Examples for velocity-dependent forces are • Stokes frictional force (fast ballistics) F = −αvev .

(1.18)

(For a sphere with radius R, which moves in a medium with the viscosity coefficient η, one has α = 6πRη.) • Newtonian frictional force (slow ballistics) F = −βv 2 ev . • Coulomb friction (sliding and static friction)  −μG F⊥ ev for v =  0 F = −F  for v = 0

(1.19)

(1.20)

with F < μH F⊥ . Here, F  is the component (parallel to the supporting plane section) of the force on the body, and F⊥ is the component (perpendicular to the supporting plane section) of the force between the body and the support (see Fig. 1.9), and μG and μH are the coefficients for sliding and static friction, respectively. Agreement 8 (velocity-independent forces): If not stated otherwise, we will restrict ourselves to velocity-independent forces in the following.

14

1 Newtonian Mechanics: General Properties

Fig. 1.9. The components F ⊥ and F  perpendicular and parallel to the supporting surface

1.5 Equations of Motion In this paragraph some consequences of Newton’s axioms shall be investigated and applied. The whole system of (interacting) point-like particles at the positions ri in an external force field F (r i ) shall be considered (with (1.6) the internal forces cancel). As a special case one obtains the corresponding results for a system, which consists of only a single particle. 1.5.1 Definitions Definition 9. (for a single particle, characterized by the index i): mi ri pi = mi r˙ i l i = r i × pi Fi Ti = 12 mi r˙ 2i

the mass of the particle; the coordinate (the position) of the particle; the momentum of the particle; the angular momentum of the particle with respect to the origin (at r = 0) the force acting on the particle; kinetic (translational22 ) energy23 of the particle.

Definition 10. (for a system of N particles):  N The sums are understood as ≡ i=1  total mass M = mi 1  center of mass R= M mi r i   ˙ total momentum P = pi = mi r˙ i = M R   total angular momentum L = li = ri × pi  1 2 1  internal total energy E= ij Vij 2mi pi + 2 (conservative systems) 22 23

Extended particles have also rotational energy in addition to the translational energy, see Chap. 8. This is generally not equal to 12 mr˙ 2 , see, e.g., the motion in a plane, Sect. 2.2.

1.5 Equations of Motion

15

Fig. 1.10. The vectors of the position r, of the momentum p, and of the angular momentum l; l points into the plane of the drawing

as well as F ext =

(external) total force (external) total torque work of the external field power of the external field



F ext (r i )  N ext = r × F ext i (r i )  ext ext F (r i ) · dr i A = ext ext P = A˙ .

Definition 11. (Vectors and Pseudovectors): 1. Vectors, which transform into their negative under space inversion, are also called polar vectors. 2. Vectors, which are invariant under space inversion, are also called axial vectors or pseudo vectors. Examples for (polar) vectors: The position vector r and its derivatives ˙ r¨ with respect to time, the momentum vector p, the force F , etc. r, Examples for pseudovectors: The angular momentum vector l = r × p, see Fig. 1.10, or all cross products of polar vectors.

1.5.2 Equations of Motion The equations of motion for the total system are: (i)

Equation of motion for the (total) momentum: The time change of the momentum of the total system is equal to the external force (axiom II), P˙ = F ext .

(1.21)

(ii) Equation of motion for the (total) angular momentum: The time change of the angular momentum of the total system is equal to the external torque, ˙ = N ext , L (1.22) where the internal forces are assumed as central forces.

16

1 Newtonian Mechanics: General Properties

(iii) Equation of motion for the (total) energy: The time change of the energy of the system is equal to the external power, E˙ = P ext .

(1.23)

Proof: (i)

According to the second axiom one has ⎡ ⎤   (1.3)  ⎣ ⎦ (1.6) P˙ = p˙ i = F ij + F ext = 0 + F ext . i i

i

j

(ii)    ˙ = d L r i × pi = r i × p˙ i r˙ i × pi + dt i i i   = mi r˙ i × r˙ i + ri × F i (pi = mi r˙ i , i

i

= 0+



ri × F i =





i

ri × ⎝

i



p˙ i = F i )

⎞ ⎠. F ij + F ext i

j

The first term in the brackets can be rewritten after interchange of the indices and with the third axiom,    r i × F ij = 12 ri × F ij + 12 r j × F ji ij

ij

=

1 2



ij

(ri − rj ) × F ij .

ij

According to the assumption the internal forces are central forces, such that one has F ij  r i − rj and thus ˙ = 0+ L



r i × F ext = N ext . i

i

(iii) With the chain rule one has ⎡ ⎤   d ⎣ 1 1 ˙ ˙ E˙ = Vij (r i , rj )⎦ 2 mr i · r i + 2 dt i ij    r˙ i · ∇i Vij + 12 r˙ j · ∇j Vij = m¨ ri · r˙ i + 12 i

ij

ij

1.6 Conservation Laws (1.13)

=



m¨ ri · r˙ i −

1 2

i

(1.4)

=

 i

=





r˙ i · F ij −

1 2

ij

m¨ ri · r˙ i −

1 2

 ij

⎛ ⎝F i −

i





17

r˙ j · F ji

ij

r˙ i · F ij −

1 2



r˙ i · F ij

ij



F ij ⎠ · r˙ i =

(1.5)

j



F ext · r˙ i i

i



with the axioms II and IV. Comments:

• The preceding equations of motion are valid in particular for a single particle. • Only the external, but not the internal forces contribute to the equations of motion of the total quantities P , L, and E. • The equation of motion for the total momentum implies that the center of mass moves in such a way as if the total external force acts on a point mass, consisting of the total mass and located at the center of mass. This is the ground on which for the idealization by point masses in the mechanics of massive bodies rests.

1.6 Conservation Laws The next to last comment also implies that in a closed system (i.e., with = 0) the total of the momentum, angular momentum and energy are F ext i conserved quantities. Denotation: The conserved quantities are also called constants of the motion or integrals of the motion. Comment: The last name originates from the fact that the conserved constant can be given as a result of a (first) integration of the equations of motion; then the (first) integrals of the motion contain only the first derivative of the coordinate with respect to time, while the equations of motion themselves contain the second  derivative. For example, the (conserved) total angular momentum L = i mi r i ×r˙i only contains the first derivative, while the equation of motion (1.22) contains the second derivative. 1.6.1 Conservation of Momentum Conservation of Momentum: The momentum is a conserved quantity, if there are no external forces, F ext = 0



P˙ = 0



P = const.

(1.24)

Of course, this is clear from the second axiom, and it also follows from (1.21).

18

1 Newtonian Mechanics: General Properties

1.6.2 Angular-Momentum Conservation and Central Forces According to (1.22) the angular momentum is a conserved quantity, if the external torque vanishes. This is trivially fulfilled for vanishing external forces. But this is also fulfilled for central forces: Angular-momentum Conservation Theorem: If the force is a central force, then there is no torque, and the angular momentum is conserved, F i = Fi (r) er,i



N =0



˙ =0 L



L = const. (1.25)

Proof: (1.9)

N = r × F = r × F er =

F r×r=0. r 

1.7 Energy-Conservation Theorem and Conservative Forces 1.7.1 Conservative Forces (II) Definition 12. Alternative to Definition 7 (conservative force): A force F is called conservative, if alternatively one has: (a) The conservative force field F (r) can be written as the gradient of a potential energy V (r), (1.13)

F (r) = −∇V (r).

(1.26)

(b) The conservative force field F is rotation-free, ∇× F (r) = 0.

(1.27)

(c) The line integral over the conservative force F is independent of the path C (dependent only upon the initial and final point of the path C),  F (r) · dr independent of C. (1.28) C

(d) The closed line integral over the conservative force F vanishes,  F (r) · dr = 0 for all closed paths.

(1.29)

1.7 Energy-Conservation Theorem and Conservative Forces

19

Proofs (extracts): (a) ⇒ (b) ∇× F (b) ⇒ (d)  F · dr

(D.36)

= −∇×∇V



(D.31)

=

0

(1.27)

∇ × F · da = 0 for area F arbitrary

=

∂F

(a) ⇒ (c)

(1.26)

F



(1.26)

F · dr = −



 ∇V · dr = −

dV

C

with the last integral depending only upon boundary points.



Comment: Possibly, the condition (b) is not defined everywhere; for example, the force field of a point mass or of a point charge at their respective positions is not defined. (This is one of the difficulties with the idealizations; in reality the masses or charges are extended, and then the force field at least does not diverge, even though the direction still is not defined everywhere.) 1.7.2 Potential Energy, Work, and Power The reverse of (1.13)=(1.26) is: The potential energy V (r) belonging to a conservative force F (r) is the indefinite line integral  V (r) = − F (r) · dr + V0 (1.30) and is defined up to a constant V0 . The definite integral is  r V (r) − V (r 1 ) = − F (r) · dr. r1

Definition 13. (Work): The work of the external force fields on a particle (not necessarily conservative forces) is   A = dA = + F · dr. (1.31) Definition 14. (Power): The power of the external force fields on a particle (not of the particle on the field) is P = A˙ = F (r) · r˙

(1.32)

dr (from dA = F · dr ⇒ dA dt = F · dt ). Comment: The line integral over a force exists for non-“pathological” force fields, even if this force field is not conservative; however, the integral then does depend upon the path. This is true in particular for frictional forces.

20

1 Newtonian Mechanics: General Properties

Also in thermodynamics it turns out that in general the result of the integral dA depends upon the path. In this case an (infinitesimal) work element is not denoted by dA but by δA or −A. d 1.7.3 Energy Conservation Theorem In a conservative force field the total energy is a conserved quantity, F is conservative





E˙ = 0

E = const.

(1.33)

Proof:  d m d ¨ · r˙ + ∇V · r˙ (T + V ) = r˙ · r˙ + V (r) = m r E˙ = dt dt 2 = (m r¨ + ∇V ) · r˙

(1.3,1.26)

=

(F − F ) · r˙ = 0. 

1.8 Invariances and Conservation Laws 1.8.1 Translations and Rotations In the following section we will investigate the invariance of a system under infinitesimal coordinate transformations. To this end we start with some preparatory remarks: Translation in Time An infinitesimally small time translation from t to t by a time interval dt is described by t = t + dt. Translation in Space The translation of the coordinate from r to r  by an infinitesimally small space-translation vector dr is described by r  = r + dr.

(1.34)

Rotation in Space Let the vector ϕ be directed along the direction of the axis of rotation, and let the modulus have the value of the rotational angle. The transformation of

1.8 Invariances and Conservation Laws

21

Fig. 1.11. Axis of rotation Δϕ (as a vector) and angle of rotation Δϕ (as its modulus)

the angle ϕ in an infinitesimal rotation by an angle dϕ along the direction eϕ is described by24 ri = ri + dr = ri + dϕ × r i (1.35) with the rotation vector dϕ = ϕeϕ ,

(1.36)

see Fig. 1.11. Comment: For a time-independent axis one has e˙ ϕ = 0. 1.8.2 Invariances and Conservation Laws: Noether’s Theorem (I) The invariance of a system under any one of the continuous transformations leads to a conservation law (Noether’s theorem).25 This theorem plays an important role in the modern physics, in particular in elementary-particle physics. In classical mechanics one deals primarily with three possible invariances of the potential energy and thus with conserved quantities of the system: (i)

Space homogeneity of the system, space translational symmetry: The invariance under continuous space translations leads to the conservation of the total momentum. (ii) Space isotropy of the system, rotational symmetry: The invariance under continuous space rotations leads to the conservation of the total angular momentum. (iii) Time homogeneity of the system: The invariance under continuous time translations leads to the conservation of the total energy.26 24 25 26

See also Sect. 7.2.2 further below. This is considered in more detail in the context with the Lagrangian formulation of mechanics, Sect. 3.11. This one has only in the case of the motion without rheonomous constraints, see Sect. 3.11.3.

22

1 Newtonian Mechanics: General Properties

Proof: (i) Invariance under continuous space translations means V (r 1 , . . . , rN ) = V (r 1 + r, . . . , rN + r)

for arbitrary r.

If one expands the right side into a Taylor series and compares with the left side, then one sees that the terms of all orders in r vanish except for that of zeroth order, since the different powers of r are linearly independent. In particular one finds for the first-order term 0 = V (r 1 + r, . . . , rN + r) − V (r 1 , . . . , r N ) =

N 

(r · ∇i )V (r 1 , . . . , r N ) = −

i=1



r · F i = −r · F = −r · P˙ . (1.37)

i

Here ∇i is the gradient with respect to the coordinate ri , F i the force on the ith particle and F the total force, which is responsible for the motion of the center of mass. Since r is arbitrary with respect to modulus and direction, the total force and the time change of the total momentum must vanish, F =0



P = const.

For the proofs to (ii) and (iii) see problem 1.8. Comments:



• If the invariance applies only to one or two components in (i) or (ii), then only the respective components of the conserved quantities are conserved. • If a closed system has certain invariance properties, then these are generally destroyed by additional external fields. For example, a homogeneous gravitational field destroys the invariance of a free point mass under an (arbitrary) translation in vertical direction. For the potential energy of a point mass in a gravitational field one namely has V (x, y, z) = mgz = V (x, y, z + h) = mg (z + h) . In contrast, the translation invariance is conserved in the horizontal directions, V (x, y, z) = mgz = V (x, y + y0 , z) = V (x + x0 , y, z). In contrast to the momentum components in horizontal direction the vertical component of the momentum is not a conserved quantity. • Even though the translation of a crystal by a lattice vector, which maps every point of the crystal onto an equivalent one (neglecting surface effects), is an invariance property, this translation is discrete and not continuous, since it is not infinitesimally small and cannot be chosen arbitrarily.27 27

However, in solid-state physics this lattice translation invariance is connected to the conservation of the so-called quasi-momentum, which leads, among others, to the Laue spots of X-ray structure determination. This, in turn is related to the “quantized” structure of the Fourier coefficients of periodic structures.

1.9 * Virial Theorem

23

1.8.3 The Usefulness of the Conservation Laws The energy is a first so-called integral of the motion. Therefore, if one knows that the force is conservative and that thus the energy is conserved, one can immediately start from an equation like 1 ˙2 2 mr

+ V (r) = E

rather than the equation of motion mr˙ = −∇V (r). For a given system, it is therefore of practical importance to know the constants of the motion, since with their help a first integration of the equations of motion is performed more or less automatically. 1.8.4 The Usefulness of Symmetry Analysis The various conservation laws are interconnected with different invariances, i.e., with different symmetries of the systems. It is, therefore of great practical advantage to investigate the symmetry of the system. Obeying this advice reduces the labor by an amount which is the more extensive the more complex the system is. (None of the problems in modern physics is as simple as the harmonic oscillator or the Kepler problem.)

1.9 * Virial Theorem In a closed conservative system the total energy is a conserved quantity; the energy is distributed over kinetic and potential energy in a time dependent way. In the virial theorem a statement is made on the time-average values of kinetic (and potential) energy for finite motions (|ri | = ∞ and |pi | = ∞) and holds in particular for one-particle systems. Time-average Value Definition 15. (Time Average): The time-average value A t of a timedependent quantity A(t) is defined as 1 A t = lim τ →∞ τ



τ

dt A(t + t ).

(1.38)

0

This makes sense only if this mean value is independent of the arbitrarily chosen time t.

24

1 Newtonian Mechanics: General Properties

Comment: For a periodic quantity A with the period28 T it suffices to average over one single period,  1 T  dt A(t + t ). A t = T 0 The Virial Theorem Virial Theorem: Let the positions r i and momenta pi be finite, and let F i be the force acting on the particle i; then one has for the time average value T t of the kinetic energy29 T 2 T t = −

 r i · F i t .

(1.39)

i

For conservative systems one has 2 T t =



r i · ∇i V t .

i

For a single particle one has in particular (1.39)

2 T t = − r · F t

generally

(1.3)

= r · ∇V t for a conservative force dV (r) = r

t for a central potential dr = n V t for V (r) = arn .

Proof (for a one-particle system): One obtains by partial integration  1 τ  (1.38) − r · F t = − lim dt r(t + t ) · F (t + t ) τ →∞ τ 0  1 τ  (1.3) = − lim dt r(t + t ) · m¨ r (t + t ) τ →∞ τ 0  1 τ  ˙ + t ) · r(t ˙ + t ) dt mr(t = lim τ →∞ τ 0 1 ˙ + τ ) − r(t) · r(t)] ˙ − lim m [r(t + τ ) · r(t τ →∞ τ = mr˙ 2 (t) + 0 = 2 T t . Since the quantities r and r˙ are finite by supposition, also the difference of the scalar products remains finite, and the second (integrated) term vanishes in the limit τ → ∞. The first term is just the average value of twice the kinetic energy.  28 29

Both, the period as well as the kinetic energy are denoted by T . See footnote 28.

Summary: Newtonian Mechanics – General Properties

25

1.10 The Basic Problem of the Mechanics of a Point Mass The basic problem of the mechanics of a point mass thus consists of the ˙ t); to be found is the trajectory r(t) of a point following: Given a force F (r, r, mass under the influence of the force F , i.e., the solution r(t) of the differential equation d ˙ = F (r, r, ˙ t). (m r) dt Sometimes also reversely: Given a trajectory; to be determined is the force and/or the potential. This is the object of the dynamics. In the kinematics one determines only the form of the trajectory from the knowledge of the forces or of the potential and not the time dependence with which the trajectory is traced out (the time dependence would be but one of various possible parametric representations of the trajectory). Reversely one can determine the potential from the form of the trajectory, see, e.g., the consideration of the planet trajectories in Sect. 5.1.3.

Summary: Newtonian Mechanics – General Properties Newton’s Axioms (I) Law of inertia: postulate of inertial systems: (1.2)

Fi = 0



p˙ i = 0



pi = const.

(II) Law of motion: dynamics in inertial systems (1.3)

F i = p˙ i . (III) Reaction law: (1.4)

F ji = −F ij . (IV) Superposition principle: (1.5)

Fi =



F ij .

j

Einstein’s Equivalence Principle (1.8)

mi = mg

26

1 Newtonian Mechanics: General Properties

Equations of Motion Momentum: (1.21)



= F ext

Angular momentum: ˙ (1.22) L = N ext Energy: (1.23) E˙ = P ext

Special Symmetries Lead to Conservation Laws (and vice versa) Conservation of momentum: (1.24)



P˙ = 0

F =0

Angular-momentum conservation: ˙ =0 L

(1.25)



F = F (r) er

Energy conservation: E˙ = 0

(1.33)

⇔ ⇔

F conservative.  (1.27) (1.29) ∇ × F = 0, F · dr = 0, F

(1.26)

= −∇V

Work, Potential Energy

(1.31)

dA = +F · dr,

dV

(1.30)

= −F · dr

Power

(1.32)

P = A˙ = F · r˙ Virial Theorem

(1.39)

2 T t = − r · F t generally = r · ∇V t for conservative force   dV (r) for central potential = r dr t = n V t for V (r) = arn .

Problems

27

Problems 1.1. Mechanics of the Many-Body System. Given be a system of n point masses, {m1 , m2 , . . . , mn }. Acting between these masses are internal forces F ik of the form r ik F ik = Fik (rik ) , rik = |r ik | . rik In addition, the system shall be subject to external forces F i . (a) Which restriction for the internal forces follows from the ansatz for F ik ? (b) Set up the equation of motion for the particle i in such a system. (c) Prove under these conditions (i) the center-of-mass theorem, ¨ = MR

n 

Ki

with

M=

i=1

n 

mi

R=

i=1

n 1  mi r i ; M i=1

(ii) the angular momentum theorem, ˙ = L

n 

ri × K i

with L =

n 

i=1

li ;

i=1

(iii) the energy-conservation theorem, n  d (T + U ) = (v i · K i ) with dt i=1

T =

n  i=1

Ti ,

U=

n n  

Uik (rik ).

i=1 k=i+1

(d) Consider now a closed system. Which conservation theorems apply? 1.2. Torque on a Spool. A nearly empty spool rests on a horizontal plane. A horizontal force F pulls on the free end of the thread, see Fig. 1.12. Determine the direction of the torque on the spool by the force with respect to the spool axis and to the axis through the point of contact (A) of the spool with the plane. 1.3. Conservative Forces. Examine (by investigating ∇ × F ) whether the following forces are conservative or not, and, if so, determine the potential energy V ; then calculate −∇V , and convince yourself that the original force is obtained.   F (α) (x, y, z) = α 2xyzex + x2 zey + x2 yez   F (β) (x, y, z) = β y e−γr ex + z e−γr ey + xyez with r2 = x2 + y 2 + z 2 and the constants α, β, and γ.

28

1 Newtonian Mechanics: General Properties

F A

Fig. 1.12. The spool and the force F due to the thread

1.4. Central Potentials. Prove that energy and angular momentum is conserved for a particle in a central potential V (r) = V (r). 1.5. Central Force and Central Potential. In problem 1.4 it is concluded that the force corresponding to a central potential is a central force. Determine the condition for the reverse case. 1.6. Rotational Motion. Prove that for a purely rotational motion the power P = N · ω is compatible with P the angular velocity ω = dϕ/dt.)

(1.32)

=

F · v. (ω is the vector of

1.7. Particle in a Static Electric and Magnetic Field. (a) Prove that the static electric field E = −∇φ represents a conservative force field (F = qE for a particle with the charge q). Determine the potential energy. (b) Prove that the kinetic energy of a particle in a magnetic induction field B is a conserved quantity. 1.8. Invariances and Conservation Laws. Prove that the invariances of a system under time translations and under space rotations lead to the conservation of the energy and of the angular momentum, respectively. 1.9. The Virial Theorem. Prove the virial theorem 1  dV  r T = 2 dr for a central potential V (r). Discuss the case V (r) = a · rn+1 , in particular for n = 2 and n = −1. (Give examples for these latter cases.)

2 Newtonian Mechanics: First Applications

In this chapter we shall treat some examples, useful in the following, for the motion of one and two particles. Thereby the separation into center-of-mass and relative coordinates shall be performed and some techniques shall be learned, mostly for the solution of typical equations of motion.

2.1 One-Dimensional Motion The equation of motion m¨ x = F (x, x, ˙ t)

(2.1)

is a differential equation of second order. The general solution contains two integration constants. These integration constants are to be fixed by two initial conditions (boundary conditions). Even though there are no general solution methods, the solution in special cases can be given. 2.1.1 Constant Force Let the force be F (x, x, ˙ t) = F0 . Then one can integrate once immediately,  t  t  m x ¨ dt = F0 dt = m [x(t) ˙ − x(t ˙ 0 )] = F0 (t − t0 ) t0

t0

or x(t) ˙ = v0 +

F0 (t − t0 ) m

˙ 0 ). A second integration leads to with v0 = x(t

30

2 Newtonian Mechanics: First Applications

x(t) = x0 + v0 (t − t0 ) +

F0 2 (t − t0 ) 2m

with x0 = x(t0 ). The two integration constants are v0 and x0 . Example: Particle in the homogeneous gravitational field. A particularly simple example for an equation of motion, which can be integrated directly, is the motion in a homogeneous gravitational field. In this case the force is constant, m¨ z = −mg. Details will be treated in problem 2.2. 2.1.2 Time-Dependent Force Let the force be F (x, x, ˙ t) = F (t). Then a first integration (over t ) yields  t  t  t dt x ¨(t ) = dt F (t ) = m [x(t) ˙ − v0 ] = dt F (t ) m t0

t0

t0

and from this x(t) ˙ = v0 +

1 m



t

dt F (t ).

t0

A second integration (over t ) leads to x(t) = x0 + v0 (t − t0 ) +

1 m



t

t0

dt



t

dt F (t )

t0

with x0 = x(t0 ) and v0 = x(t ˙ 0 ). Example: Motion of a particle with charge q in an alternating electric field E(t) = E0 sin(ωt) of a condenser. The force is F (t) = qE(t) = qE0 sin(ωt). The first integration yields  qE0 t  qE0 [cos(ωt) − cos(ωt0 )] dt sin(ωt ) = v0 − x(t) ˙ = x(t ˙ 0) + m t0 mω with v0 = x(t ˙ 0 ). If one chooses t0 = 0 for simplicity one obtains x(t) ˙ = v0 −

qE0 [cos(ωt) − 1] . mω

2.1 One-Dimensional Motion

31

An additional integration leads with x0 = x(0) to  qE0 t  qE0 t dt cos(ωt ) + mω 0 mω qE0 = x0 + v0 t − [ωt − sin(ωt)] . mω 2

x(t) = x0 + v0 t −

2.1.3 Velocity-Dependent Force Let the force be F (x, x, ˙ t) = F (x). ˙ This can be integrated directly once with the method of the separation of variables. With x(t) ˙ = v(t) one obtains mv˙ = m

dv = F (v) dt

or m

dv = dt F (v)

and from this with v = v(t) and v0 = v(t0 ) 

v

m v0

dv  = F (v  )



t

dt = t − t0 .

t0

This has the form t = t0 + f (v, v0 ) and in most cases can be solved for dx = v = g(t, t0 , v0 ), dt from which the solution of the equation of motion is obtained by a second integration,  t dt g(t , t0 , v0 ). x(t) − x(t0 ) = t0

Example: and 2.7.

For a motion under frictional forces see, e.g., problems 2.6

32

2 Newtonian Mechanics: First Applications

2.1.4 Coordinate-Dependent Force Let the force be F (x, x, ˙ t) = F (x). Every one-dimensional force F (x) is in general a conservative force, since the potential V (x) can be calculated from the integral  V (x) = − dx F (x). (2.2) With the so-called “integration of the energy conservation theorem” one obtains

⇒ ⇒ ⇒ or

m¨ x = F (x) dx m dx˙ 2 = F (x)x˙ = F (x) m¨ xx˙ = 2 dt dt m 2 d(x˙ ) = F (x) dx 2  x(t) m 2 m x˙ (t) − x˙ 2 (t0 ) = dx F (x) = −V (x) + V (x(t0 )) 2 2 x(t0 ) m 2 m x˙ + V (x) = x˙ 2 (t0 ) + V (x(t0 )) = E. 2 2

(2.3)

Comments: • This procedure, namely the multiplication with the velocity, leads in many cases to a first integral, which is easy to obtain. The next integration can be performed again via the separation of variables: From (2.3) one obtains  dx 2  = x˙ = ± E − V (x) (2.4) dt m or  dx m  dt = ± . 2 E − V (x) If the particle was at the position x0 = x(t0 ) a time t0 , it will be at the position x at a time t, given by   x  t m dx  dt = ± . (2.5) t − t0 = 2 x0 E − V (x ) t0 This equation yields – in principle – the time t(x) as function of position x and, if one can invert this relation, also reversely the position x(t) is given as a function of time t.

2.1 One-Dimensional Motion

33

V

E

x

forbidden

allowed x1

forbidden x2

x3

Fig. 2.1. Allowed and forbidden regions of motion in a one-dimensional potential V (x) at a given energy E

With a given energy E those regions of x are allowed, in which the radicand (namely x˙ 2 ) in (2.5) is positive, E − V (x) ≥ 0. If one assumes for example a potential as in Fig. 2.1, i.e., a potential of the form V (x) > V (x1 ) x < x1 V (x) < V (x1 ) x1 < x < x2 V (x) > V (x3 ) x2 < x < x3 V (x) < V (x3 ) x > x3 with V (x1 ) = V (x2 ) = V (x3 ) = E, then there are three types of motion: (i) Bound Motion: The energy lies between the local minimum and the local maximum of the potential energy. The motion consists of an (anharmonic) oscillation with turning points x1 and x2 , at which the kinetic energy vanishes and at which the total energy is thus equal to the potential energy. (ii) Free (Unbound) Motion: The energy lies above of the local maximum. The motion consists of an approach towards the center (x = 0) up to the point x1 and/or towards infinitely large values of x. This is the process of scattering or of a collision. (iii) The energy is equal to the potential energy of the local maximum. The motion ends at the x-value of the local maximum.

34

2 Newtonian Mechanics: First Applications

(iv) A motion with an energy below that of the local minimum is again a scattering or a collision, however with an approach up to x3 . In the case (i) of the oscillation the regime x1 ≤ x ≤ x2 is traced once in half a period, and one finds from (2.5) for the period of the oscillations   x2 m dx  . (2.6) T =2 2 x1 E − V (x) Notice that the turning points x1 and x2 depend upon the energy E! Comments: • Also in the case that the kinetic energy has the form T (x, x) ˙ = a(x) x˙ 2 + b(x) x˙ + c(x), there is a solution x(E, ˙ V (x)) from the energy conservation theorem, E = T (x, x) ˙ + V (x)

⇒ ⇒

x˙ = x(E, ˙ V (x))   dx dt = x(E, ˙ V (x))

(2.7)

in analogy to (2.4), see for example the pendulum in the following section or the Kepler motion in Sect. 5. • With the energy as sketched in Fig. 2.1 the regions x < x1 and x2 < x < x3 are forbidden for the motion. This is not so in quantum mechanics of the microscopic particles, in which the so-called tunneling effect does allow the particle to be found in these regions. 2.1.5 Example: Plane Pendulum (with Small Amplitude) In this example1 one deals with a motion under constraints2 namely under the condition of the motion to be along the circular arc. Even though the problem can be treated more elegantly with the Lagrangian formalism,3 it shall be presented here as an example for the motion in a one-dimensional configuration space. The experienced physicist will notice that the system is conservative and that the energy conservation theorem can be exploited. Here, we want to trod along the path which one would take if one did not know the symmetry of the 1 2

3

See also Sect. 2.2.4. Here one has the constraint r = l = const. and a force of constraint S. The tension S = −S er (with S > 0) is that (oriented towards the center, the hinge) force of constraint, which (in addition to the external forces) lets the pendulum execute its motion in such a way that the constraint is fulfilled. Thus the constraint consists of the fact that the motion is restricted (constrained) to a circular arc. For the general treatment of constraints see Chap. 3.

2.1 One-Dimensional Motion

35

ϕ

Fig. 2.2. The definition of the (mathematically) positive angle ϕ and of the positive force Fϕ (in the direction of increasing angle!) as well as of the positive force Fr (in the direction of increasing distance r)

ϕ

l S m mg

mg cos ϕ mg sin ϕ

Fig. 2.3. The decomposition of the forces acting on the plane pendulum

system. Also, so far we have not presented a criterion according to which one could decide whether or not the system is conservative. The motion can be described by the coordinate ϕ (Fig. 2.2). The forces acting on the pendulum mass are the (known) gravitational force F = mg and the (so far unknown) tension S = −Ser (in the direction towards the hinge) (Fig. 2.3). There is no motion in radial direction, r = l = const, such that the tension S = −Ser , the fugal force mrϕ˙ 2 er , and the radial component of the gravitational force add to zero,4 0 = m¨ r = Fr = −S + mrϕ˙ 2 + mg cos ϕ.

4

See also Sect. 2.2.4.

36

2 Newtonian Mechanics: First Applications

The motion along the circular arc is forced by the tangential component Fϕ of the gravitational force, Fϕ = −m g sin ϕ. (Fϕ points into the direction of decreasing angle ϕ!) The equation of motion is then (with the line element ds = l dϕ of the circular arc) m (¨ r )ϕ = m¨ s = m l ϕ¨ = Fϕ = −mg sin ϕ or ϕ¨ = −

g sin ϕ. l

(2.8)

First Integration, First Alternative The integration of this equation of motion is done implicitly by the use of the energy conservation theorem (for conservative forces). It is always the same trick: One multiplies the equation of motion with ϕ˙ (in the general case with a first derivative with respect to time), g ϕ˙ ϕ¨ = − ϕ˙ sin ϕ, l what one writes in the form 1 d 2 g d ϕ˙ = cos ϕ, 2 dt l dt and can integrate directly once,  g 1 2 ϕ˙ (t) − ϕ˙ 2 (t0 ) = [cos ϕ(t) − cos ϕ(t0 )]. 2 l

(2.9)

The additional integration is not elementary;5 the methodological procedure has been presented in Sect. 2.1.4. First Integration, Second Alternative If one multiplies (2.9) with ml2 and reorders, one obtains the energy conservation theorem, E(t) = 12 ml2 ϕ˙ 2 (t) − mgl cos ϕ(t) = 12 ml2 ϕ˙ 2 (t0 ) − mgl cos ϕ(t0 ) = E(t0 ), from which one could have started right away. 5

See (2.6) in Sect. 2.1.6.

2.1 One-Dimensional Motion

37

w

f

1

0

E

φ0 / π −1 −1

0

0 −2

1

φ/π

−1

0

1

2

φ/π

Fig. 2.4. The (reduced) force f = Fϕ /mg acting on the mathematical pendulum (left) and the (reduced) potential energy w = V /mgl (right) as a function of the angle ϕ ≡ φ; the approximation of small amplitudes is shown by the broken lines

Second Integration in the Case of Small Amplitudes To begin with, small displacements6 shall be considered in the following. Here, the one-dimensional path is the path x = lϕ on the circular arc. One has (Fig. 2.4) line element ds = l dϕ m 2 m 2 2 kinetic energy T = s˙ = l ϕ˙ 2 2 potential energy V = mgl (1 − cos ϕ) ≈

m glϕ2 2

(2.10)

total energy E = T + V. The (total) energy and the turning points are interconnected; at the turning points ±ϕ0 the kinetic energy vanishes, ϕ˙ = 0,

E = V (ϕ0 ) = mgl (1 − cos ϕ0 ) ≈

m glϕ20 . 2

One obtains7 from (2.5) by choosing the initial condition8 as ϕ(t) = 0 for t = t0 = 0  t= 0

6 7 8

ϕ



l dϕ 2 m 2 m 2 gl (ϕ0



ϕ2 )

=

l g

 0

ϕ

dϕ  ϕ20 − ϕ2

GR 2.261

=

1 ϕ arcsin ω0 ϕ0

Large displacements are treated in Sect. 2.1.6. With the integral # 2.261 with c = −1 and Δ = −4ϕ20 from the extremely voluminous tables in [11]. In order to perform one integration one needs one initial condition.

38

2 Newtonian Mechanics: First Applications K(z)

p/2

0

0

z

1

Fig. 2.5. Graphical representation of K(z); see also [10], p. 592

with ω02 = l/g. From this one obtains by solving for ϕ ϕ = ϕ0 sin(ω0 t). Comment: The solution of the linear differential equation ϕ¨ = − gl ϕ is simple within the standard procedure; here, however, the method of the integration of the energy conservation theorem according to (2.7) was to be demonstrated. 2.1.6 * Example: Plane Pendulum with Large Amplitude As an example, the plane pendulum shall be considered again, but now for displacements which are not necessarily small; the pendulum is assumed to oscillate, but not to rotate. The kinetic energy T is thus smaller than the potential energy for the upright standing pendulum, T ≤ 2mgl. The onedimensional path x is as in Sect. 2.1.5 again the path on the circular arc. The evaluation of the integral in (2.5) for arbitrary times t and angles ϕ poses difficulties, but the period can be given. The (total) energy and the turning points are related to each other; at the turning points ±ϕ0 one has, cf. Fig. 2.4, (2.11) ϕ˙ = 0 , E = V (±ϕ0 ) = mgl (1 − cos ϕ0 ) . The regime 0 < ϕ < ϕ0 is traced in a quarter period, and one obtains from (2.5) and (2.10) for the period   ϕ0 m l dϕ (2.5)  T = 4 2 0 E − V (ϕ)   ϕ0 m l dϕ (2.11)  = 4 2 0 mgl (cos ϕ − cos ϕ0 )

2.1 One-Dimensional Motion

=

8l g

= 4



ϕ0



0

dϕ cos ϕ − cos ϕ0

39

(2.12)

ϕ0 " l ! K sin g 2

with the so-called complete elliptic integral of the first kind9  π/2 dα  K(z) = , 0 1 − z 2 sin2 α see Fig. 2.5. Proof: With cos ϕ = 1 − 2 sin2

ϕ 2

(2.13)

one obtains for the radicand

! ϕ0 ϕ" cos ϕ − cos ϕ0 = 2 sin2 − sin2 . 2 2

One defines sin α :=

sin ϕ2 sin ϕ20

(2.14)

and has at the limits of integration  0 for ϕ = 0 α= π/2 for ϕ = ϕ0 . The differential of sin α is d(sin α) = cos α dα



 2 ϕ 1 1 − sin 1 cos 2 dϕ = = 2 sin ϕ20 2 sin ϕ20 sin ϕ20 cos α dα . dϕ = 2  1 − sin2 ϕ20 sin2 α

ϕ 2



(For the sign of the square root one should observe that one has cos ϕ2 > 0 for 0 ≤ ϕ ≤ ϕ0 ≤ π.) Then, with (2.13) and (2.14) one has for the square root in the denominator of (2.12)  √ √ ϕ0 ϕ − sin2 cos ϕ − cos ϕ0 = 2 sin2 2 2 √ ϕ0  2 1 − sin α = 2 sin 2 √ ϕ0 cos α, = 2 sin 2 9

An extensive collection of special functions, their properties, tabulated values, and graphical representations can be found in [10].

40

2 Newtonian Mechanics: First Applications

and for the period one obtains  8l ϕ0 1 dϕ √ T = g 0 cos ϕ − cos ϕ0  ϕ 8l π/2 2 sin 20 cos α dα 1  = ·√ g 0 2 sin ϕ20 cos α 1 − sin2 ϕ20 sin2 α l g

=4



π/2 0

dα  . 1 − sin2 ϕ20 sin2 α 

Small Amplitudes For small amplitudes one obtains from K(z) →  g , ω= l

π 2

for z → 0 the known result

and the period diverges with the amplitude ϕ0 approaching π. Proof: One expands for small amplitudes in (2.12), cos ϕ − cos ϕ0 ≈

 1 2 ϕ0 − ϕ2 2

and obtains for the period  ϕ0 l dϕ  T ≈4 =4 g 0 ϕ20 − ϕ2

l g

 0

ϕ0

dϕ/ϕ0  . 1 − (ϕ/ϕ0 )2

With the substitution ϕ = ϕ0 sin β one obtains T =4 ⇒

ω=

l g

2π = T



π/2

0



cos β dβ =4 cos β

lπ g2

g . l

Who ever looks for exceptional integrals should consult alternatively the integral tables in [11], see footnote 7. There one finds ϕ   ϕ0 l dϕ l l π ϕ 0 GR 2.261  T ≈4 . = 4 =4 arcsin 2 2 g 0 g ϕ0 0 g 2 ϕ0 − ϕ 

2.2 Motion in a Plane

41

Intermediate Amplitudes One can determine corrections of the harmonic result by expansion of the elliptic integral K in terms of small ϕ0 . With 1 ϕ0 1  sin2 α + . . . = 1 + sin2 2 2 2 ϕ0 2 1 − sin 2 sin α one obtains !

 1 2 ϕ0 2 sin α + . . . dα 1 + sin 2 2 0   π 1 2 ϕ0 1 = · + ... 1 + sin 2 2 2 2   π ϕ2 ≈ 1 + 0 + ... , 2 16

ϕ0 " K sin = 2



π/2



and from this one obtains for the period T = 2π

  ϕ20 l 1+ + ... . g 16

Thus, the period increases with increasing amplitude. In the case ϕ0 = 12 π the period diverges: The pendulum remains at rest in the indifferent equilibrium.

2.2 Motion in a Plane In this paragraph we continue with the treatment of a single particle, but we leave the one-dimensional motion and consider the motion in a plane. Examples are the plane pendulum or, as one will see in Chap. 5, the motion of a particle in a central force field. In many cases there is a force center, and the motion in the plane suggests the use of plane polar coordinates10 (with the origin in the force center). 2.2.1 Fixed and Moving Basis The transformation between plane polar coordinates and cartesian coordinates is x = r cos ϕ y = r sin ϕ. 10

For the extension to spherical polar coordinates and to cylindrical coordinates in the three-dimensional case see Appendix C.

42

2 Newtonian Mechanics: First Applications

y

eϕ er

ey



ey

r

r

ϕ

er

ϕ

ex

X ex

Fig. 2.6. The fixed (ex , ey ) and moving (er , eϕ ) basis. The trajectory r(t) is indicated by the broken line

In addition, it is practical to introduce a fixed (cartesian) and a moving basis (with radial and tangential unit vectors er and eϕ , respectively). (The basis vectors point into the direction of increasing coordinates!) The position vector can be written in the alternative forms r = x ex + y ey r = r er as shown in the left part of Fig. 2.6. The relation between the basis vectors is $# $ # $ # cos ϕ sin ϕ ex er = , (2.15) − sin ϕ cos ϕ eϕ ey see the left part of Fig. 2.6, with the inverse relation # $ # $−1 # $ # $# $ cos ϕ sin ϕ er cos ϕ − sin ϕ ex er = = . ey eϕ eϕ − sin ϕ cos ϕ sin ϕ cos ϕ The matrix of the inverse transformation is thus the transpose of the original transformation matrix, since the transformation is a unitary transformation (since the determinant of the transformation matrix is equal to unity). Representation of a Vector in the Fixed and Moving Basis Analogously to the vector r of the position other vectors can be represented in the fixed and in the moving basis. An example of the vector F of the force is given in Fig. 2.7. 2.2.2 Time Dependence of the Moving Basis The basis vectors of the moving basis are varying with time (except for a purely radial motion); for the derivatives with respect to time11 (Fig. 2.8) one obtains 11

For the space derivatives see Appendix D.

2.2 Motion in a Plane

43

Fig. 2.7. The decomposition of the force F (r) in the (left) fixed and (right) moving basis

e˙ r = ϕ˙ eϕ e˙ ϕ = −ϕ˙ er .

(2.16)

Proof: (1) Intuitive derivation: The derivative of the radial unit vector with respect to time is Δer , Δer = er (t + Δt) − er (t). e˙ r = lim Δt→ 0 Δt The arc length is |Δer | = Δϕ |er | = Δϕ, and the direction for |Δer |  1 (Δϕ  π) is Δer  eϕ , thus der eϕ dϕ Δer = = = ϕ˙ eϕ . Δt→ 0 Δt dt dt

e˙ r = lim

Analogously one finds e˙ ϕ .

er (t+Dt) = er¢

De er (t)

eϕ¢

ϕ

eϕ D er

er¢



er Dϕ

Fig. 2.8. The time dependence of the moving basis: In the left panel the trajectory of the particle is indicated by the broken line; also shown are the basis vectors at two different times. In the right panel the same basis vectors are shown on an expanded scale

44

2 Newtonian Mechanics: First Applications

¤ v

¤ vϕ

¤ vr

¤ r ϕ Fig. 2.9. The decomposition of the velocity v in its radial and angular part. The trajectory r(t) is indicated by the solid broken line

(2) Formal way: From #

one finds #

e˙ r e˙ ϕ

$

er eϕ

# = ϕ˙ #

$

# =

cos ϕ − sin ϕ

− sin ϕ cos ϕ − cos ϕ − sin ϕ

sin ϕ cos ϕ $#

ex ey

$#

ex ey

$

$

$# $# $ − sin ϕ cos ϕ cos ϕ − sin ϕ er = ϕ˙ − cos ϕ − sin ϕ sin ϕ cos ϕ eϕ # $# $ # $ 0 1 er eϕ = ϕ˙ = ϕ˙ . −1 0 eϕ −er



2.2.3 Velocity, Acceleration, etc. in Plane Polar Coordinates Position, Velocity, Acceleration In addition to the position, also the velocity, acceleration, etc. can now be decomposed into radial and tangential components. The position vector has, of course, only a radial component, r = r er . For the velocity one obtains r˙ = r˙ er + r e˙ r (2.16)

= r˙ er + r ϕ˙ eϕ = vr er + vϕ eϕ

(2.17)

2.2 Motion in a Plane

45

(the first term of the result is the component of the velocity in the direction of the position vector, and the second term is the component perpendicular to it), see Fig. 2.9. From this one obtains with er · eϕ = 0 the kinetic energy     T = 12 mr˙ 2 = 12 m vr2 + vϕ2 = 12 m r˙ 2 + r2 ϕ˙ 2 . (2.18) For the acceleration one obtains analogously with (2.16) and (2.17) r¨ = (¨ r er + r˙ e˙ r ) + (r˙ ϕ˙ eϕ + r ϕ¨ eϕ + r ϕ˙ e˙ ϕ ) = r¨ er + r˙ ϕ˙ eϕ + r˙ ϕ˙ eϕ + r ϕ¨ eϕ − r ϕ˙ ϕ˙ er   = r¨ − r ϕ˙ 2 er + (r ϕ¨ + 2 r˙ ϕ) ˙ eϕ

(2.19)

= b r er + b ϕ eϕ . The four terms in (2.19) are the radial, centripetal, angular and (negative) Coriolis acceleration (of the particle). Inertial Forces The centrifugal force mrϕ˙ 2 and the Coriolis force −2mr˙ϕ˙ are also called inertial forces,12 which appear in a moving (here rotating) system, even if the particle is at rest in this system and even if there are no external forces acting.

Momentum and Angular Momentum From the velocity one obtains the momentum p = m r˙ = m (r˙ er + r ϕ˙ eϕ ) and the angular momentum (with respect to the coordinate origin) l = r × p = m (r × v) = m r er × (r˙ er + rϕ˙ eϕ )   = m rr˙ er × er + r2 ϕ˙ er × eϕ = 0 + m r2 ϕ˙ eω = m r2 ω = θ ω with the angular velocity ϕ˙ = ω and the vector ω = ω eω of the angular velocity13 as well as the moment of inertia θ = m r2 of a point mass m at a distance r from the axis of rotation.14

12 13 14

More on inertial forces further below in Sect. 7 on moving coordinate systems. Compare the infinitesimal vector in (1.36). More on moments of inertia and time dependent axes of rotation in Chap. 8 on rigid bodies.

46

2 Newtonian Mechanics: First Applications

Radial and Circular Motion Special Case: Radial motion: ϕ = const, ϕ˙ = 0. One obtains r = r er v = r˙ = r˙ er a = r¨ = r¨ er p = m r˙ = mr˙ er l =r×p = 0 T = 12 mr˙ 2 = 12 mr˙ 2 . Special Case: Circular motion r = const, r˙ = 0. One obtains r = r er v = r˙ = rϕ˙ eϕ a = r¨ = −rϕ˙ 2 er + rϕ¨ eϕ p = mr˙ = mrϕ˙ eϕ = m vϕ eϕ l = r × p = mr2 ϕ˙ eω = mr2 ω T = 12 mr˙ 2 = 12 mr2 ϕ˙ 2 . 2.2.4 Example: Plane Pendulum Again this example:15 In the homogeneous gravitational field the pendulum mass is subject to the gravitational force mg and to the tension S by the pendulum rod (assumed as massless) (Fig. 2.10), F = mg + S. The equation of motion ¨ F = mr can be decomposed into the radial and tangential components,   Fr = F · er = mg cos ϕ − S = m r¨ − r ϕ˙ 2 = m 0 − lϕ˙ 2 = −ml ϕ˙ 2 Fϕ = F · eϕ = −mg sin ϕ = m (rϕ¨ + 2r˙ ϕ) ˙ = m (lϕ ¨ + 0) = mlϕ¨ with the radial and tangential component of the acceleration from (2.19). (Because of the appearance of the tangential component Fϕ the force is not a central force.) The second equation is a differential equation for ϕ, 15

See also Sect. 2.1.5.

2.2 Motion in a Plane

ϕ

47

l S m mg

mg cos ϕ mg sin ϕ

Fig. 2.10. The plane pendulum, cf. Fig. 2.3

g ϕ¨ = − sin ϕ, l

(2.20)

cf. Sect. 2.1.5. The first equation is an algebraic equation for the stress S,   S = m g cos ϕ + l ϕ˙ 2 , (2.21) which is compensated by the sum of the radial component of the fugal force and of the gravitational force (both oriented outward). In this form one obtains the stress as a function of ϕ and ϕ. ˙ One can also express ϕ˙ by other quantities, e.g., by the energy or by the amplitude, see problem 2.8. Small Displacements For small displacements, sin ϕ ≈ ϕ, the differential (2.20) for ϕ takes the form ϕ¨ +

g ϕ = 0. l

The eigen solutions are ϕ± (t) ∝ e±iωt with ω2 =

g . l

The general solution is the superposition of the eigen solutions, ϕ(t) = a+ eiωt + a+ e−iωt

48

2 Newtonian Mechanics: First Applications

With the (initial) conditions (by choice) ϕ(t = 0) = 0,

ϕmax = ϕ0

the solution is ϕ = ϕ0 sin(ωt). The stress S as a function of time t can now be obtained from the known solution for ϕ(t),   (2.21) S = m g cos ϕ + lϕ˙ 2  # $  1 ≈ m g 1 − ϕ2 + lϕ˙ 2 2   l 1 2 2 = mg 1 − ϕ0 sin (ωt) + ω 2 ϕ20 cos2 (ωt) 2 g   3 2 2 2 = mg 1 + ϕ0 − ϕ0 sin (ωt) 2 with ω 2 = g/l and cos2 (ωt) = 1 − sin2 (ωt).

2.3 Two-Particle Systems Now we leave the one-particle systems. For the following it is essential that the interaction energy between the two particles is assumed to depend only upon the relative coordinate, F (r 1 , r2 ) = F (r 1 − r 2 ), (and not in a more complex way upon the individual coordinates of the two particles). Because of the homogeneity of space this condition is practically always fulfilled, and with additional isotropy one even has F (r1 , r2 ) = F (r 1 − r 2 ) = F (|r1 − r 2 |) as in the case of the gravitational or Coulomb interaction. Then the twoparticle problem can be reduced to two one-particle problems. 2.3.1 Center-of-Mass and Relative Coordinates Definition 16. (see Fig. 2.11) m1 r 1 + m2 r 2 m1 + m2 r = r1 − r2

Center-of-mass coordinate R = Relative coordinate Total mass Reduced mass μ

M = m1 + m2 1 1 1 = + . μ m1 m2

2.3 Two-Particle Systems

49

r

r2

R

r1 Fig. 2.11. The center-of-mass and relative coordinates R and r, respectively

The inverse relations are μ r m1 μ r. r2 = R − m2

r1 = R +

2.3.2 Equations of Motion The equations of motion, expressed in terms of the coordinates of the two particles, are ¨1 = F 1 (r 1 ) + F 12 (r 1 − r2 ) m1 r

(2.22)

¨2 = F 2 (r 2 ) + F 21 (r 1 − r2 ) m2 r

(2.23)

with the external forces F i and the internal forces F ij . If one introduces center-of-mass and relative coordinates, one obtains the equations of motion,16 ¨ = F 1 (r, R) + F 2 (r, R) MR (2.24) # $ μ μ ¨ = F 12 (r) + F 1 (r, R) − F 2 (r, R) . (2.25) μr m1 m2 Proof: For (2.24): Adding the two equations (2.22) and (2.23) yields (2.22) + (2.23) d2 d2 ¨ (m r + m r ) = MR=MR 1 1 2 2 dt2 dt2 = F 1 + F 2 + F 12 + F 21 = F 1 + F 2 .

¨1 + m2 r¨2 = = m1 r

For (2.25): The (weighted) difference of the two equations (2.22) and (2.23) yields μ μ ¨ 2 ) = μ¨ (2.22) − (2.23) = μ (¨ r1 − r r m1 m2 μ μ μ μ = F1 − F2 + F 12 − F 21 m1 m2 m1 m2 # $ 1 μ μ 1 = F1 − F2 + μ + F 12 m1 m2 m1 m2 μ μ = F1 − F 2 + F 12 .  m1 m2 16

Equation (2.24) is identical to (1.21).

50

2 Newtonian Mechanics: First Applications

Comment: The equations of motion (2.22) and (2.23) for r1 and r2 are coupled through the inner forces F ij , and the equations of motion (2.24) and (2.25) for r and R are coupled only through the external forces F i ; in a closed system (i.e., without external forces) or in a system with constant external forces the latter equations are thus decoupled; this is the reason for the transformation from single-particle to center-of-mass and relative coordinates. 2.3.3 Closed Systems In a closed system (F 1 = 0, F 2 = 0) (2.24) and (2.25) are simplified to the form ¨ =0 MR ¨ = F 12 (r). μr The center-of-mass and relative motions are decoupled: The center-of-mass motion is force-free; the relative motion is determined by the relative force F 12 . The relative motion is the same as that of one particle in the field of the other (assumed as fixed); only the mass (m1 or m2 ) is to be substituted by the reduced mass (μ). Comments: • Except for the influence of the mutual gravitational force the motion of a planet around the sun is also subject to the forces of moons and other planets in this example, but these latter forces are comparatively weak, such that the planet and the sun can be considered as a closed system in good approximation. On the other hand, the system of moon and earth is not a closed system; because of nearly the same distance from the sun one has F 1 /m1 ≈ F 2 /m2 , and even though the external forces in (2.25) for the relative motion approximately vanish, the center-of-mass motion, however, is subject to the influence of the force F , which the sun exerts on the total mass of earth and moon. • For the center-of-mass motion in the closed system, of course, the centerof-mass energy, momentum, and angular momentum are conserved.17 Of interest is thus the relative motion. • In the limit m1  m2 one obtains (in the case of vanishing external forces, i.e., in a closed system) R = r2 M = m2 μ = m1

17

See Sect. 1.6.

Summary: Newtonian Mechanics – First Applications

51

and ¨ =0 m2 R ¨1 = F 12 . m1 r The light particle moves in a field of the practically resting (or uniformly moving) heavy mass. If one chooses an inertial system, in which one has R = r2 = 0, then the second equation m1 r¨1 = F 12 has the form like the equation of motion (of particle 1) in an open system. This is the basics of the dynamics of one-particle-systems in external fields, like they are caused by sufficiently heavy systems, on which the dynamics of the light mass practically does not act back on the heavy mass.

Summary: Newtonian Mechanics – First Applications One-Dimensional Motion:

(2.1)

m¨ x = F (x, x, ˙ t) F (x, x, ˙ t) = F (x) is always conservative:  (2.2) V (x) = − F (x) dx with E = T + V . Motion in a Plane (central force F (r) = F (r) er ): #

er eϕ

$

(2.15)

#

=

d dt

#

er eϕ

1 ∇V m

$

(2.16)

= ϕ˙

#

$#

eϕ −er

ex ey

$ .

$

 r¨ − rϕ˙ 2 er + (rϕ¨ + 2r˙ ϕ) ˙ eϕ $ # 1 ∂V 1 ∂V = − er + eϕ m ∂r r ∂ϕ

(2.19)

r¨ = =−

cos ϕ sin ϕ − sin ϕ cos ϕ



52

2 Newtonian Mechanics: First Applications

Two-Body Problem:

(2.22)

m1 r¨1 = F 1 (r 1 ) + F 12 (r 1 , r2 ) (2.23)

m2 r¨2 = F 2 (r 2 ) + F 21 (r 1 , r2 ) ¨ (2.24) MR = F 1 (r, R) + F 2 (r, R) ! " (2.25) μ r¨ = F 12 (r) + mμ1 F 1 (r, R) − mμ2 F 2 (r, R) ⇒ free center-of-mass motion F i (r i ) = 0 ⇒ relative motion in plane: μ r¨ = F 12 (r).

Problems 2.1. Free Particle. Solving the equation of motion describe the motion of a point mass without forces. Write down the solution for the initial conditions r (t) = r 0 and v (t) = v 0 for t = 0, and illustrate the trajectory of the point mass with a sketch. Determine the angular momentum. Write down the conserved quantities. Hint: Avoid a special choice of the reference point (namely that on the trajectory). 2.2. Motion in a Homogeneous Gravitational Field. Determine the solution of the equation of motion m¨ r = mg ˙ = v 0 . Employ two different with the initial conditions r(0) = r0 and r(0) methods: (1) Decompose the equation of motion in cartesian coordinates. (2) Use a coordinate-free procedure. 2.3. Freely Falling Body from Large Height. A body falls from large height h perpendicularly to the earth’s surface (radius R of the earth) with the force F = −γ

mM . r2

Determine the velocity upon hitting the earth. Compare the result with that one which results for small height h  R. Up to which height is the difference between these two calculations less than 1%? 2.4. Vertical Throw, Gravitational Field. Discuss the vertical throw of a mass m in the gravitational field of the earth

Problems

53

(a) Let the initial velocity in the throw of the mass from the earth’s surface be v0 . Determine the velocity v of mass as a function of the distance z from the center of the earth. (b) What is the minimum value of v0 such that the mass leaves the gravitational range of the earth? (G = 6.67 × 10−11 N m2 kg−2 , M = 5.98 × 1024 kg, R = 6.37 × 106 m) 2.5. The Inclined Throw. For an inclined throw, determine duration, height, and range of the throw. Let the initial velocity be v0 ; the initial angle be α. 2.6. Freely Falling Body in a Homogeneous Gravitational Field with Friction. Investigate the motion of a body initially at rest under the influence of the homogeneous gravitational field and of the Stokes and alternatively of the Newton friction, and sketch the results. Discuss the result for large times, and determine the correction to v(t) by the friction by the air if this is small. 2.7. Motion with Friction on an Inclined Plane. Investigate the motion of an initially resting body on an inclined plane as a function of the inclination angle α (against the horizontal direction) under the influence of the homogeneous gravitational fields and of the Coulomb friction (Fig. 2.12). What changes, if one introduces an additional force on the body, acting in the direction parallel or perpendicular to the inclined plane?

α mg

Fig. 2.12. Motion on an inclined plane

2.8. The Plane Pendulum of Sect. 2.2.4. Prove with the help of the energy conservation theorem that the energy can be expressed by the amplitude ϕ0 of the vibration. Prove then that the stress in the hinge is given by S = mg (3 cos ϕ − 2 cos ϕ0 ) . 2.9. Motion in a Central Force Field. A point mass moves in a central force field a F = −m 3 er r with a > 0. Investigate the constants of the motion as well as the form of the motion for large and small values of a.

54

2 Newtonian Mechanics: First Applications

2.10. Circular Trajectory and Potential. Consider a circular orbit r = const. of a particle. Under which condition for the r-dependence of the potential energy V (r, ϕ) is this orbit a solution of the equation of motion? 2.11. Rocket in the Homogeneous Gravitational Field. A rocket starts from ground vertically. The initial mass of the rocket is m0 , the rate of mass change is constant in time (m ˙ = −α), and the exhaust velocity u0 of the material from the engine relative to the rocket is kept constant in time. (a) At which time tE does the engine stop burning fuel, if the final mass of the rocket is mE ? (b) Which velocity relative to the earth does the exhaust matter have at leaving the rocket, if the rocket moves with velocity v relative to the earth? (c) Set up the equation of motion of the rocket in an appropriate coordinate system. Hint: Formulation of the momentum conservation for the rocket and the ejected mass. (d) Which condition has to hold for the rocket to lift off at t = 0? (e) Write down the velocity v(t) as well as its maximum value vmax , and sketch v(t) in a graph for t ≤ tE . (f) Write down the distance s(t) traveled by the rocket, and sketch its time development for t ≤ tE . 2.12. Dynamics of the Universe. From the red shift of the light from far-away galaxies one knows the expansion of the universe (Hubble law: the change of the distance between two galaxies with time is proportional to this distance) which – if it always existed – which has started with the Big Bang. The expansion of the universe counteracts the gravitation. (a) Give an expression for the equation of motion for a galaxy of mass m at the distance R(t) from the origin. Rationalize that only the mass of the galaxies inside of the sphere of radius R(t) act on m. Assume an average mass density ρ(t). Write down the equation of motion for R(t) with the assumption of the mass conservation in the universe. (b) Reformulate the equation of motion such that one can notice a constant of the motion. What is the meaning of this quantity, and by which quantity is it determined? (c) Solve the differential equation for R(t), where the integration constant from b) occurs as a parameter (distinguish different cases; use an integral table), and discuss the different cases of the solution by means of the graphs R(t).

3 Lagrangian Mechanics

In this chapter an alternative formulation of the mechanics of point masses will be treated, which will turn out to be of particular advantage, if the motion is restricted by constraints. A first example for a motion with constraints has been encountered in the context of the plane pendulum in Sect. 2.1.5: The constraints restrict the configuration space. (In the example of the plane pendulum the configuration space is no longer the R3 but a circle, an R1 ; the configuration space of the spherical pendulum is the surface of a sphere, an R2 .) Another example is the force which keeps a train on the rails. In other cases it may be more difficult to determine the forces, which restrict the motion of one or several particles to the reduced configuration space. In this chapter we will thus present the Lagrangian formulation of the equations of motion with implicit1 and explicit2 consideration of the constraints. We will first consider the various forms of the constraint3 and then the forces of constraint,4 which would be needed for the Newtonian dynamics. But we will see that the need of these forces, which may be very difficult to obtain,5 can be circumvented.6 Like many other phenomena of physics the Lagrangian equations of motion can be derived from an extremum principle.7 In the following we will continue to assume inertial reference frames as postulated by Newton’s first law.

1 2 3 4 5 6 7

See Sect. 3.5. See Sect. 3.9. See Sect. 3.3. See Sect. 3.4. The stress in the pendulum rod has been determined only after the equation of motion has been solved, see Sect. 2.2.4. See Sect. 3.5. See Sect. 3.10

56

3 Lagrangian Mechanics

3.1 Motion with Constraints According to the second Newtonian axiom the motion of point masses is subject to the influence of forces,  m¨ r i = F ext + F ij + Z i = F i + Z i . (3.1) i j

Even though in many cases the (internal and external) forces F i are known, the so-called forces of constraint Z i , which force (“constrain”) the motion of the ith particle to a subspace of the R3 , are not known a priory. On the other hand, the constraints themselves are in most cases (as in the example of the pendulum) very apparent and easy to formulate. In addition, it may be of an advantage to employ coordinates other than cartesian coordinates, e.g., polar coordinates as known from the treatment of the pendulum. Often it is also of an advantage to use the (still to be developed) Lagrangian equations of motion instead of the Newtonian equations of motion, and there are very general prescriptions for setting up these Lagrangian equations of motion; this makes the Lagrangian method often very much easier to handle than the Newtonian method. Three Good Reasons Thus the motivation for the Lagrangian formulation will be firstly the easy consideration of the constraints. Secondly, it will be of an advantage that one will have to deal with scalars, while the Newtonian equations of motion are vectorial relations. Finally, the form invariance8 of the Lagrangian equations of motion under coordinate transformation, i.e., the independence of the equations of motion in various (also curvilinear) coordinate systems, is an advantage. Comment: The Newtonian equation of motion m¨ r=F is not form-invariant under coordinate transformation: In cartesian representation one has for the motion in a plane m¨ x = Fx m¨ y = Fy

8

See also Sect. 3.8.2 further below.

3.2 Constraints and Generalized Coordinates

57

and in plane polar coordinates   m r¨ − rϕ˙ 2 = Fr m (rϕ¨ + 2r˙ ϕ) ˙ = Fϕ with the notation as in Fig. 2.7. Even though the Newtonian equations of motion in abstract representation are form-invariant under Galilean transformations (from one inertial system to another), they have a different form in the different representations. This will be not so in the Lagrangian equations of motion.

3.2 Constraints and Generalized Coordinates The example of the plane pendulum has shown that it may be more practical to work with coordinates other than cartesian coordinates. 3.2.1 Examples of Constraints If the motion of a particle is restricted to a sheet in R3 , then the coordinates of the particle must obey the equation G(r) = 0 describing this sheet. As an example, the spherical pendulum moves on the surface of a sphere; G(r) = x2 + y 2 + z 2 − l2 = 0 G(r) = r2 − l2

(cartesian coordinates) (spherical coordinates).

This sheet may also be varying in time (e.g., a sheet fixed in the laboratory but moving along with the laboratory on the rotating earth), G(r, t) = 0. The motion of a particle may also be restricted to a curve in space; the curve is an intersection of two sheets, and in this case one has two constraints, G1 = 0 G2 = 0, e.g., for a plane pendulum (in the (x, z)-plane) 0 = G1 = y 0 = G2 = x2 + y 2 + z 2 − l2 = x2 + z 2 − l2 .

58

3 Lagrangian Mechanics

Other constraints cannot be expressed in the form of equations. If in the case of the spherical pendulum the mass is not fixed to a (rigid) rod, but to a (flexible) string, then the mass can reach every point in the interior of a sphere, (3.2) x2 + y 2 + z 2 − l2 ≤ 0. Analogous relations hold for more particles, e.g., one has for a rigid dumb-bell with two masses at a distance l G(r 1 , r 2 ) = (r 1 − r 2 )2 − l2 = (x1 − x2 )2 + (y1 − y2 )2 + (z1 − z2 )2 − l2 = 0.

3.2.2 Classification of Constraints Consider a system of N particles with 3N coordinates and k independent constraints of the form 3N 

Gji dxi + Gj0 dt = 0

(j = 1, . . . , k)

(3.3)

i=1

then the jth constraint is called (a) Rheonomous (“running,” “flowing”), if it is varying in time, Gj0 = 0 (b) Scleronomous (“rigid,” “solid”), if it is independent of time, Gj0 = 0 (c) Holonomic (“total”), if it can be written in integrated (not only in differential) form Gj = cj = const, i.e., if one has Gji =

∂Gj , ∂xi

Gj0 =

∂Gj . ∂t

The integrability condition (x0 ≡ t) ∂ 2 Gj ∂ 2 Gj = ∂xi ∂xl ∂xl ∂xi

(i, l = 0, 1, . . . , 3N and j = 1, . . . , k)

thus reads ∂Gji ∂Gjl = . ∂xi ∂xl (d) Nonholonomic, if the constraint cannot be written in the form as in case (c). Inequalities like, e.g., (3.2) are nonholonomic constraints.

3.2 Constraints and Generalized Coordinates

59

Fig. 3.1. The plane pendulum

Fig. 3.2. The bead on the rotating rod

Examples: For (b) and (c): Plane pendulum, see Fig. 3.1. The constraint is scleronomous and holonomic, in cartesian coordinates G = x2 + z 2 − l2 = 0 ⇒ 2x dx + 2z dz = 0 or, in plane polar coordinates, r−l =0



dr = 0.

For (a) and (c): Bead on a rod rotating uniformly (with the angular velocity ω), see Fig. 3.2. In plane polar coordinates the angular dependence is given by ϕ = ωt



dϕ − ω dt = 0

(in plane polar coordinates). The constraint is rheonomous and holonomic,

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3 Lagrangian Mechanics

y

j

q x

Fig. 3.3. For the constraints of a skate



y − x tan(ωt) = 0 xω dt = 0 dy − tan(ωt) dx − cos2 (ωt)

(in cartesian coordinates). For (d): Skate (ideally touching the ice at one point), see Fig. 3.3.9 The system is not a point particle and has three degrees of freedom, the position (x, y) and the orientation θ. The constraint is10 tan θ =

dy dx

or tan θ dx − dy = 0 or 0 = tan θ dx − 1d y + 0 dθ = Gx dx + Gy dy + Gθ dθ. This is not of the form dG(x, y, θ) = 0 and, therefore, cannot be integrated to yield a form like G(x, y, θ) = const. Namely, one has 9 10

A similar example is the wheel, see Sect. 3.9.3 below. A rigid body has six degrees of freedom, three of the position and three of the orientation; the trivial constraints, namely the restriction to the positions in the plane and the restriction to one degree of freedom of the orientation are not mentioned here explicitly.

3.2 Constraints and Generalized Coordinates

61

∂2G d ∂Gx = = tan θ = 0 ∂θ ∂θ ∂x dθ ∂Gθ ∂2G ∂ = = 0=0 ∂x ∂x ∂θ ∂x and thus ∂2G ∂2G ∂Gθ ∂Gx = = = . ∂θ ∂θ ∂x ∂x ∂θ ∂x The constraint is nonholonomic. 3.2.3 Degrees of Freedom A system of N (point-like) particles without constraints has 3N degrees of freedom, and the motion is described in a 3N -dimensional configuration space and can thus be represented by 3N independent coordinates. Through k independent constraints of the form (3.3) given above the number of the independent degrees of freedom (the dimension of the configuration space) is reduced by k and only is f = 3N − k. Example: Plane pendulum. A particle has 3 coordinates; the plane pendulum has 2 constraints (motion in a plane, motion on a sphere) and, therefore, has 3 − 2 = 1 degree of freedom. Comments: • The nonholonomic constraint (3.2), x2 + y 2 + z 2 < l2 , for the pendulum on a string does not reduce the number of the degrees of freedom in comparison to that of a free particle (but this condition restricts the domain of definition of the coordinates). • Of the 12 N (N − 1) (pairwise) constraints |ri − r j | = cij for all interparticle distances of the N particles of a rigid body only 3N − 6 are independent of each other: Given three distances, the position of a particle in a rigid body is (nearly unambiguously) fixed. The number n = 3N − 12 N (N − 1) would be negative for sufficiently large N : N 2 3 4 5 6 7 8 n 5 6 6 5 3 0 −4 Therefore one has indeed f = 6 degrees of freedom, three each for the positions and for the orientation.

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3 Lagrangian Mechanics

x r

y

ϕ

Fig. 3.4. Two different sets of coordinates, (x, y) and (r, ϕ), respectively, for the description of the motion in a plane

3.2.4 Generalized Coordinates In principle, instead of the 3N cartesian coordinates x1 , . . . , x3N one could employ any other coordinates q1 , . . . , q3N for the description of the particle positions xi (t) = xi (q1 , . . . , q3N , t)

(i = 1, . . . , 3N ),

in the example of the motion in the (x, y)-plane, e.g., the plane polar coordinates (r, ϕ) instead of the cartesian coordinates (x, y), see Fig. 3.4. Comment: For practical reasons one chooses (if possible at all) such coordinates, in which the constraints %j (q1 , . . . , q3N , t) = 0 Gj (r 1 , . . . , rN , t) = Gj (r1 (q1 , . . . , q3N , t), . . . , t) = G each describe a whole coordinate sheet in R3N (i.e., a sheet qi = const.). Then it is often very easy to find the 3N − k = f independent coordinates. Example: The spherical pendulum. Transition from cartesian coordinates x, y, and z to spherical polar coordinates r, ϑ, and ϕ leads to the constraint 0 = G(x, y, z) = x2 + y 2 + z 2 − R2 % ϑ, ϕ). = r2 − R2 = G(r, The mentioned coordinate sheet is the surface of the sphere r = R = const. The independent coordinates are (q1 , q2 ) = (ϑ, ϕ). Definition 13. (Generalized coordinates): The (arbitrarily chosen) independent coordinates q1 , . . . , qf are denoted as the generalized11 coordinates. 3.2.5 Example: Plane Double Pendulum The motion shall take place in the (y, z)-plane. The constraints are (in cartesian coordinates) 11

Very rarely the q1 , . . . , q3N are denoted as generalized coordinates.

3.3 Forces of Constraint

63

z

ϑ1

y l1 m1 ϑ2 ϑ1

l2

m2

Fig. 3.5. The notations for the plane double pendulum

G1 = x1 = 0 G2 = x2 = 0 G3 = y12 + z12 = l12 2

2

G4 = (y2 − y1 ) + (z2 − z1 ) = l22 . The first two conditions describe the motion in the plane x = 0, and the last two describe the fixed pendulum lengths. For practical reasons one chooses ϑ1 , ϑ2 , as the two remaining independent variables (the generalized coordinates), see Fig. 3.5. Introduction of the generalized coordinates: x2 = 0 x1 = 0 y1 = l1 sin ϑ1 y2 = y1 + l2 sin ϑ2 = l1 sin ϑ1 + l2 sin ϑ2 z1 = −l1 cos ϑ1 z2 = z1 − l2 cos ϑ2 = −l1 cos ϑ1 − l2 cos ϑ2 The equations of motion and their solution are treated in Sect. 3.5.7.

3.3 Forces of Constraint If a particle moves on a predetermined trajectory or sheet, then except for the external and internal forces also further forces, the so-called forces of constraint, must act on the particle, which force (“constrain”) it on this curve or sheet. These forces of constraint themselves do not accelerate the particle(s) along a curve or sheet. Also, the forces of constraint do not have a component parallel to the respective curves or sheets and thus can act only perpendicular to the respective curves or sheets. The constraint G = 0 describes a sheet in coordination space, and the force of constraint has thus the direction of ∇G(r, t)

64

3 Lagrangian Mechanics

(for a single force of constraint). The force of constraint for a motion on a sheet can then be written as Z = λ∇G with a still to be determined function λ, the so-called Lagrangian multiplier, or, for a motion along the intersecting line of two sheets with constraints G1 = 0, G2 = 0, as Z = λ1 ∇G1 + λ2 ∇G2 . In general one has for several particles Gj (r 1 , . . . , rN , t) = 0

(j = 1, . . . , k),

and the force of constraint on the ith particle is Zi =

k  j=1

λj ∇i Gj =

k 

λj Gji

(i = 1, . . . , 3N )

(3.4)

j=1

with Gij from (3.3). Comment: Frictional forces may be connected with the motion along a sheet. Frictional forces are not forces of constraint. Like all other forces they influence the trajectory of the particle (or of the particles), but they do not force the particle(s) onto a fixed sheet or trajectory.

3.4 The Principle of d’Alembert Even though the forces of constraint are not known because of the unknown Lagrangian parameters, in this section we will derive some properties of the forces of constraint which are essential for what follows. With the principle of d’Alembert considered in this section or with the equivalent principle of virtual work statements are made concerning the forces of constraint. Comment: Later in this chapter we will derive the equations of motion from the rather general Hamilton principle.12 In contrast to this latter principle, the former principles are tailored for applications in mechanics. 3.4.1 Real and Virtual Displacements Consider the motion of particles with constraints. In a time interval dt the particle (index i) moves by the real “displacements” dr i . Definition 18. (Virtual Displacement): A virtual displacement δr i is a displacement with fixed time and consistent with the constraints. 12

See Sect. 3.10.

3.4 The Principle of d’Alembert

65

t2 δr

dr

t = const t1

Fig. 3.6. Real trajectory (thick line) and varied trajectories (thin line) in an R2 with a real displacement dr and a virtual displacement δr q(t)+δq(t)

q

δq

q(t)

t1

t2

t

Fig. 3.7. Time dependence of a “real” coordinate q and of the virtually displaced coordinate q + δq

Comments: • The virtual displacement δr i with fixed time is in contrast to the real displacement dri along the real trajectory. • The virtual displacements are thus tangential vectors in configuration space. The vectors δr i thus point at positions on different, geometrically possible trajectories of the ith particle at a given time. For example, a given trajectory can be realized by a given initial conditions, but δri may also point at other imaginative trajectories, see Figs. 3.6 and 3.7. Example: Bead on a Rotating Rod At every instance, the virtual displacements δr are all oriented along the rod, which has a given orientation at a given time. In general, the real displacements dr are not oriented along the rod, since the rod rotates during the time interval dt, see Fig. 3.8. If one considers a particle and its different trajectories r(t, ), “varied” by a (small) parameter , with r(t, 0) = r(t), then δr(t) = r(t, ) − r(t, 0) = 

dr && + O(2 ) & d =0

is the virtual displacement, and dr(t) = r(t + dt) − r(t) = dt is the real displacement.

dr + O(dt2 ) dt

66

3 Lagrangian Mechanics

dr dr

Fig. 3.8. The real and the virtual displacements of a bead on the rotating rod with the axis perpendicular to the paper plane

If one expresses the displacements by generalized coordinates ql , then one has ri = ri (q1 , . . . , qf , t)

with

f = 3N − k,

and the infinitesimally small real and virtual displacements are dr i =

f  ∂ri l=1

δri =

∂ql

f  ∂ri l=1

∂ql

dql +

∂ri dt ∂t

δql + 0

δt ≡ 0 (!).

3.4.2 The Principle of Virtual Work Consider the motion of a particle with a constraint G = 0. The virtual displacement δr is oriented tangential to the sheet described by the constraint, the force of constraint Z ∝ ∇G is oriented perpendicular to the sheet G = 0; thus one has Z ⊥ δr



Z · δr = 0.

(In Fig. 3.8 the vector of the force of constraint is oriented in the plane of the figure and perpendicular to the rod.) Thus the so-called virtual work of the force of constraint vanishes. In generalization of the preceding argument to N particles, the principle of virtual work states that the forces of constraint do not perform any virtual work: N  Z i · δr i = 0 (3.5) i=1

with the forces of constraint Z j of Sect. 3.3.

3.5 Lagrangian Equations of the Second Kind

67

Comments: • Virtual work and real work:  i Z i · δri is the virtual work of the forces of constraint i F i · δri is the virtual work of the internal and external forces − i p˙ i · δri is the virtual work of the inertial forces i Z i · dr i is the real work of the forces of constraint • If the constraints are rheonomous, Z i = Z i (t), the forces of constraint can perform real work,  Z i · dri = 0; i

for example, the bead sliding on the rotating rod of Fig. 3.8 is accelerated in the direction of the rod. • For scleronomous constraints (for one particle) one has dr ⊥ Z, because not only δr but also dr is oriented tangentially to the sheet G = 0 fixed in time. 3.4.3 The Principle of d’Alembert If one combines the Newtonian law (3.1)

p˙ i = F i + Z i with the principle of the virtual work N 

(3.5)

Z i · δr i = 0,

i=1

one obtains the equivalent principle of d’Alembert N 

(F i − p˙ i ) · δri = 0.

(3.6)

i=1

3.5 Lagrangian Equations of the Second Kind In this section finally the equations of motion with implicit13 consideration of the constraints are set up. 13

Explicit consideration will be dealt with in Sect. 3.9.

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3 Lagrangian Mechanics

3.5.1 External and Internal Forces Starting from the principle of d’Alembert one can write the first term in (3.6) as N 

F i · δr i =

i=1

f 

Ql δql

l=1

with the number f = 3N − k of the (independent) degrees of freedom and the generalized forces (type I)14 Ql =

N 

Fi ·

i=1

∂r i ∂ql

(l = 1, . . . , f ).

(3.7)

Proof: via δr i =

f  ∂ri l=1

∂ql

δql .

(3.8)

3.5.2 Conservative Forces For conservative forces one has F i = −∇i V, and one obtains the generalized forces again as the (negative) derivative of the potential, here, however, with respect to generalized coordinate, Ql = −

∂V . ∂ql

(3.9)

Proof: Ql =

N  i=1

Fi ·

N  ∂r i ∂ri ∂V =− ∇i V · =− . ∂ql ∂ql ∂ql i=1



14

There is no standard definition of the generalized force; further customary definitions are given in Sects. 3.6 and 3.7 below. To differentiate between the differently defined generalized forces they are characterized by an addition in this book. While reading other text books one should look up the definitions.

3.5 Lagrangian Equations of the Second Kind

69

3.5.3 Inertial Forces The second part in (3.6) of d’Alembert’s principle can be written as  d # ∂T $ ∂T   − δql p˙ i · δri = dt ∂ q˙l ∂ql i

(3.10)

l

with the kinetic energy15 1 T = mi v 2i 2 i

with

vi =

d r i (q1 , . . . , qf , t). dt

Proof: Let ri = ri ({ql }, t). Then one has r˙ i =

 ∂ri d ∂r i ri = q˙l + dt ∂ql ∂t

(3.11)

l

and from this

∂ r˙ i ∂r i = . ∂ q˙l ∂ql

From T =



(3.12)

1 ˙ i )2 2 m (r

i

follows

' (  d # ∂  mi $   d ∂T  2 r˙ δql = δql dt ∂ q˙l dt ∂ q˙l i 2 i l l  # $  d ∂ r˙ i r˙ i · = mi δql dt ∂ q˙l i l  # $ ∂r i d (3.12)   r˙ i · = mi δql dt ∂ql i l $  # ∂ri ∂ r˙ i ¨i · = mi r + r˙ i · δql ∂ql ∂ql i l (3.8)  = mi (¨ ri · δri + r˙ i · δ r˙ i ) i

=



(p˙ i · δri + pi · δ r˙ i )

i 15

T = has

1 2

 i

mi v 2i only holds in non-relativistic mechanics. In Special Relativity one T = 

m0 c2 − m0 c2 . 1 − v 2 /c2

See the Course on Special Relativity theory.

70

3 Lagrangian Mechanics

and

' (  ∂T  ∂ # mi $ 2 r˙ δql = δql ∂ql ∂ql i 2 i l l   ∂ r˙ i = mi r˙ i · δql = pi · δ r˙ i ∂ql i i l

as well as finally    d # ∂T $ ∂T  p˙ i · δri . − δql = dt ∂ q˙l ∂ql i l

 Comments: • The two terms on the right side of (3.11) lead to the consequence that the kinetic energy T can be expressed by the generalized coordinates and their derivatives with respect to time, T =

 1 μll q˙l q˙l + λl q˙l + κ 2  ll

(3.13)

l

with μ (q1 , . . . , qf , t) =

3N 

ll

mi

i=1



∂xi ∂xi ∂ql ∂ql

∂xi ∂xi ∂ql ∂t i # $2 ∂xi 1 κ(q1 , . . . , qf , t) = mi . 2 i ∂t

λl (q1 , . . . , qf , t) =

mi

(3.14) (3.15) (3.16)

The kinetic energy T is thus of quadratic form in the velocities (in the q˙i ), and since the xi may depend nonlinearly upon the ql , the coefficients λ, μ and κ in general depend upon the coordinates ql . For example the kinetic energy in plane polar coordinates reads T = 12 m(r˙ 2 + r2 ϕ˙ 2 ). • One has λl = 0 and κ = 0 for scleronomous constraints. • In contrast, for the bead on the uniformly rotating rod with rheonomous constraint one has ϕ˙ = ω = const. and T = 12 m(r˙ 2 + r2 ω 2 ) (i.e., μ(r) = m, λ(r) = 0, κ(r) = 12 mr2 ω 2 ). 3.5.4 Lagrangian Equations of Motion Substituting in the principle of d’Alembert, one finds

3.5 Lagrangian Equations of the Second Kind (3.6)

0 =

N 

(3.10)

(p˙ i − F i ) · δr i =

i=1

 f   d ∂T ∂T − − Ql δql dt ∂ q˙l ∂ql

71

(3.17)

l=1

with f = 3N − k. While in the case of existing constraints the r i and thus the δri are not linearly independent, the δql are so (if the constraints are holonomic). From the vanishing of expression (3.17) for arbitrary δql one concludes that the expression in the square brackets must vanish, $ # d ∂T ∂T = Ql l = 1, . . . , f. (3.18) − dt ∂ q˙l ∂ql This equation describes the equality of the inertial forces to the internal and external forces. Newtonian Equation of Motion as a Special Case Only if constraints are missing, Z i = 0, all the δr i are linearly independent. Then the Newtonian equation of motion p˙ i = F i of axiom II follows from the left side of (3.17). (Alternatively one can use for example cartesian coordinates xi , yi , and zi as generalized coordinates and combine the three components of the coordinates of a particle to form the vector r i .) Conservative Force as a Frequent Special Case • For a conservative force there is a potential energy V (q1 , . . . , qf ) with (3.9)

Ql = −

∂V , ∂ql

∂V = 0. ∂ q˙l

With the Lagrangian function L L=T −V ,

T = T ({ql }, {q˙l }, t) V = V ({ql })

(3.19)

one finds from (3.9) and (3.18) d ∂L ∂L − = 0. dt ∂ q˙l ∂ql

(3.20)

The equations (3.20), occasionally also (3.18), are denoted as the Lagrangian equations of motion of the second kind and sometimes as the Euler–Lagrangian equations.16 16

In general, the Euler–Lagrangian equations are the result of a variational method, see, e.g., Hamilton’s principle treated further below in Sect. 3.10.

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3 Lagrangian Mechanics

• If one can separate the forces into conservative forces Ql or F i (which R can be derived from a potential V ) and non-conservative forces QR l or F i (e.g., frictional forces), then one has d ∂T ∂T − = Ql + QR l dt ∂ q˙l ∂ql d ∂L ∂L − = QR l dt ∂ q˙l ∂ql

(3.21)

L =T −V ∂V Ql = − ∂q  l ∂ri R Ql = FR . i · ∂ql i

(3.22)

with

Form Invariance The form of the Lagrangian equation of motion (3.18) or (3.20) is independent of the special choice of the coordinate transformation xi (q1 , . . . qf ), and thus the form of (3.18) is independent of the choice of the generalized coordinates, in contrast to the Newtonian equation of motion.17 3.5.5 Example: Plane Motion in a Central Potential For practical reasons one chooses plane polar coordinates (in the plane z = 0) as generalized coordinates. Then the kinetic and potential energies as well as the Lagrangian function are   T = 12 mr˙ 2 = 12 m r˙ 2 + r2 ϕ˙ 2 V = V (r)   L = T − V = 12 m r˙ 2 + r2 ϕ˙ 2 − V (r). For the equations of motion one needs



17

∂V ∂L = mrϕ˙ 2 − ∂r ∂r ∂L = mr˙ ∂ r˙   ∂V d ∂L ∂L − = m r¨ − rϕ˙ 2 + 0= dt ∂ r˙ ∂r ∂r

See also the comment at the end of Sect. 3.1 and in particular Sect. 3.8.2.

3.5 Lagrangian Equations of the Second Kind

73

Fig. 3.9. The bead sliding on a uniformly rotating rod, cf. Fig. 3.2



∂L =0 ∂ϕ ∂L = mr2 ϕ˙ ∂ ϕ˙ d ∂L ∂L d  2  d 0= − = mr ϕ˙ + 0 = l dt ∂ ϕ˙ ∂ϕ dt dt

with the angular momentum l as in Sect. 2.2.3. 3.5.6 Example: Bead Sliding on a Uniformly Rotating Rod We repeat the example of Sect. 3.2.2 (Fig. 3.9). One has to deal with the motion in a plane (z = 0). The (additional) constraint can be given most easily in plane polar coordinates, ϕ = ωt. The (only) generalized coordinate is r. With this one obtains V =0

  L = T = 12 mr˙ 2 = 12 m r˙ 2 + r2 ϕ˙ 2   = 12 m r˙ 2 + r2 ω 2 and ∂L = mrω 2 ∂r ∂L = mr˙ ∂ r˙   d ∂L ∂L − = m r¨ − rω 2 . ⇒0= dt ∂ r˙ ∂r The equation of motion is thus r¨ = ω 2 r. For the solution of this equation see problem 3.10.

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3 Lagrangian Mechanics

3.5.7 Example: Plane Double Pendulum Here we repeat the double pendulum of Sect. 3.2.5. With the coordinates x1 = 0 y1 = l1 sin ϑ1 z1 = −l1 cos ϑ1

x2 = 0 y2 = y1 + l2 sin ϑ2 = l1 sin ϑ1 + l2 sin ϑ2 z2 = z1 − l2 cos ϑ2 = −l1 cos ϑ1 − l2 cos ϑ2

one obtains the velocities (which one needs for the kinetic energy) y˙ 1 = l1 ϑ˙ 1 cos ϑ1 z˙1 = l1 ϑ˙ 1 sin ϑ1 y˙ 2 = l1 ϑ˙ 1 cos ϑ1 + l2 ϑ˙ 2 cos ϑ2 z˙2 = l1 ϑ˙ 1 sin ϑ1 + l2 ϑ˙ 2 sin ϑ2 and thus r˙ 21 = l12 ϑ˙ 21 r˙ 22 = l12 ϑ˙ 21 + l22 ϑ˙ 22 + 2l1 l2 ϑ˙ 1 ϑ˙ 2 (cos ϑ1 cos ϑ2 + sin ϑ1 sin ϑ2 ) = l2 ϑ˙ 2 + l2 ϑ˙ 2 + 2l1 l2 ϑ˙ 1 ϑ˙ 2 cos(ϑ1 − ϑ2 ). 1 1

2 2

The energies are V = g {m1 l1 (1 − cos ϑ1 ) + m2 [l1 (1 − cos ϑ1 ) + l2 (1 − cos ϑ2 )]} T = 12 mr˙ 21 + 12 mr˙ 22 = 1 m1 l2 ϑ˙ 2 + 1 m2 [l2 ϑ˙ 2 + l2 ϑ˙ 2 + 2l1 l2 cos(ϑ1 − ϑ2 )ϑ˙ 1 ϑ˙ 2 ]. 2

1 1

2

1 1

2 2

The Lagrangian equations are obtained with ∂L ∂T = = m1 l12 ϑ˙ 1 + m2 [l12 ϑ˙ 1 + l1 l2 cos(ϑ1 − ϑ2 )ϑ˙ 2 ] ˙ ˙ ∂ ϑ1 ∂ ϑ1 ∂L = m2 [l22 ϑ˙ 2 + l1 l2 cos(ϑ1 − ϑ2 )ϑ˙ 1 ] ∂ ϑ˙ 2 ∂L = −m2 l1 l2 sin(ϑ1 − ϑ2 )ϑ˙ 1 ϑ˙ 2 − g[m1 l1 sin ϑ1 + m2 l1 sin ϑ1 ] ∂ϑ1 ∂L = +m2 l1 l2 sin(ϑ1 − ϑ2 )ϑ˙ 1 ϑ˙ 2 − gm2 l2 sin ϑ2 ∂ϑ2 d ∂L ∂L − dt ∂ ϑ˙ 1 ∂ϑ1 = m1 l12 ϑ¨1 + m2 [l12 ϑ¨1 − l1 l2 sin(ϑ1 − ϑ2 )(ϑ˙ 1 − ϑ˙ 2 )ϑ˙ 2 +l1 l2 cos(ϑ1 − ϑ2 )ϑ¨2 ] +m2 l1 l2 sin(ϑ1 − ϑ2 )ϑ˙ 1 ϑ˙ 2 +g[m1 l1 sin ϑ1 + m2 l1 sin ϑ1 ]

3.5 Lagrangian Equations of the Second Kind

75

= (m1 + m2 )l12 ϑ¨1 + m2 l1 l2 cos(ϑ1 − ϑ2 )ϑ¨2 + m2 l1 l2 sin(ϑ1 − ϑ2 )ϑ˙ 22 +g(m1 + m2 )l1 sin ϑ1 d ∂L ∂L − ˙ dt ∂ ϑ2 ∂ϑ2 = m2 [l2 ϑ¨2 − l1 l2 sin(ϑ1 − ϑ2 )(ϑ˙ 1 − ϑ˙ 2 )ϑ˙ 1 + l1 l2 cos(ϑ1 − ϑ2 )ϑ¨1 ] 2

−m2 l1 l2 sin(ϑ1 − ϑ2 )ϑ˙ 1 ϑ˙ 2 + gm2 l2 sin ϑ2 = m2 l22 ϑ¨2 − m2 l1 l2 sin(ϑ1 − ϑ2 )ϑ˙ 21 + m2 l1 l2 cos(ϑ1 − ϑ2 )ϑ¨1 +gm2 l2 sin ϑ2 . Case of Small Displacements For small displacements one obtains 0 = (m1 + m2 )l12 ϑ¨1 + m2 l1 l2 ϑ¨2 + g(m1 + m2 )l1 ϑ1 0 = m2 l22 ϑ¨2 + m2 l1 l2 ϑ¨1 + gm2 l2 ϑ2 . One divides the first equation by (m1 + m2 )l12 and the second by m2 l1 l2 , and with the abbreviations μ=

m2 m1 + m2

and λ =

l2 l1

one can write the system of equations in the form # $# $ # $# $ g 1 0 1 μλ ϑ¨1 ϑ1 + = 0. 1 λ ϑ2 ϑ¨2 l1 0 1 With the ansatz

one obtains

#

#

− ω2 −ω 2

g l1

ϑ1 ϑ2

$

∝ e−iωt

−μλω 2 g 2 l1 − λω

$#

ϑ1 ϑ2

$ = 0.

The solubility condition for this homogeneous system of equations is the vanishing of the secular determinant of the coefficient matrix, $# $ # g g − ω2 − λω 2 − ω 4 μλ = 0 l1 l1 # $2 g g 4 2 ⇒ ω (λ − μλ) − ω (λ + 1) + =0 l1 l1 # $2 g λ+1 1 g 1 ⇒ ω4 − ω2 =0 + λ 1 − μ l1 l1 λ(1 − μ)

76

3 Lagrangian Mechanics

with the solution 2 ω±

1λ+1 1 g = 2 λ 1 − μ l1

) 1±

λ 1 − 4(1 − μ) (λ + 1)2

* .

Case of Equal Pendulum Lengths (and Small Displacements) For the case of equal pendulum lengths, l1 = l2 = l, λ = 1, one obtains # $2 g 2 −ω = ω4μ l



ω2 −

√ g = ±ω 2 μ l



2 ω± =

1 g √ . l 1∓ μ

With the knowledge of the eigen values the eigen vectors can be determined: one inserts the eigen values into the eigen value equation, # $ g 2 2 − ω± ϑ1,± = μω± ϑ2,± l 2√ 2 = ∓ω± μ ϑ1,± = μω± ϑ2,± ϑ1,± √ = ∓ μ. ϑ2,± The general solution is the superposition of the eigen solutions, ϑi (t) = a+ ϑi,+ e−iω+ t + a− ϑi,− e−iω− t + c.c. Comments: • Vibrations are treated in more detail in Chap. 4. • At this point it shall be stressed that in an eigen vibration the two oscillators vibrate with a fixed phase. • The term abbreviated as “c.c.” denotes the complex-conjugate of the preceding expression(s). 3.5.8 Separable Systems In general, the Lagrangian equations of motion are coupled differential equations for the coordinates q1 , . . . qf . However, if there are two (or more) mutually independent systems as sketched in Fig. 3.10, i.e., if the Lagrangian function decomposes into two (or more) parts such that the coordinates and their corresponding velocities are contained in separate parts, L(q1 , q˙1 , . . . , qf , q˙f , t) = L1 (q1 , q˙1 , . . . , qμ , q˙μ , t) + L2 (qμ+1 , q˙μ+1 , . . . , qf , q˙f , t),

3.5 Lagrangian Equations of the Second Kind

77

Fig. 3.10. The system of the particles i = 1, . . . , μ is separated from the system of the particles i = μ + 1, . . . , N

i.e., if one can separate the coordinates (here the coordinates q1 , . . . , qμ from the coordinates qμ+1 , . . . , qf ), then the time dependence of the coordinates q1 , . . . , qμ is independent of that of the coordinates qμ+1 , . . . , qf . Then there are thus two (or more) decoupled sets of (coupled) differential equations. This follows immediately from 0=

∂L d ∂L1 ∂L1 d ∂L − = − dt ∂ q˙l ∂ql dt ∂ q˙l ∂ql

(l = 1, . . . , μ)

and correspondingly for the other coordinates qj (with j = μ+1, . . . , f ). Thus, the equation of motion for q1 contains also the coordinates q2 , . . . , qμ , but not the coordinates qμ+1 , . . . , qf . Example: Relative and Center-of-Mass Coordinates For two particles without external forces the relative and center-of-mass motion can be separated from each other: While one has, e.g., in the picture of the individual particles the (nonseparable) Lagrangian function L(r1 , r 2 ) =

r˙ 21 r˙ 2 + 2 − V (r 1 − r 2 ), 2m1 2m2

(3.23)

one has in the picture of the relative and center-of-mass motion the (separable) Lagrangian function L(R, r) =

˙2 r˙ 2 R + − V (r) = L1 (R) + L2 (r). 2M 2μ

(3.24)

While the equations of motion for the coordinates r1 and r 2 following from the Lagrangian function (3.23) are coupled, the equations of motion for R and r following from the Lagrangian function (3.24) are decoupled (separated) from each other.

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3 Lagrangian Mechanics

3.6 Generalized Momentum, Force, etc. The conjugate coordinates ql and momenta pl introduced in the following play a central role in the further development of theoretical physics, in particular in the Hamiltonian mechanics and in the commutation rules of quantum mechanics. 3.6.1 Generalized Force and Generalized Momentum So far, the generalized quantities ql generalized coordinates q˙l generalized velocities have been introduced. In the following, additional generalized quantities will be defined. Definition 19. (Canonical momentum): The generalized (also: canonical or conjugate18 ) momentum is pl =

∂L . ∂ q˙l

(3.25)

Definition 20. (Generalized force II): The generalized force (type II)19 is Fl =

∂L . ∂ql

Comments: (3.9)

• The generalized force Fl of type II is distinguished from the force Ql = −∂V /∂ql of type I by Fl = Ql +

∂T . ∂ql

• With these generalized quantities one can write the Lagrangian equations d ∂L ∂L = + QR l dt ∂ q˙l ∂ql

(3.26)

in a form similar to the Newtonian equations as p˙ l = Fl + QR l . 18 19

Conjugate to the generalized coordinate ql See footnote 14 in Sect. 3.5.1: This notation is not customary in general.

(3.27)

3.6 Generalized Momentum, Force, etc.

79

• For velocity independent potentials20 one has ∂V =0 ∂ q˙l



∂L ∂T = , ∂ q˙l ∂ q˙l

and in systems without constraints one can use the cartesian coordinates as the generalized coordinates, ql = xl , and from (3.26) or (3.27) one immediately obtains the Newtonian law of motion in cartesian representation. • Of the inertial forces21 appearing in rotating coordinate systems the centrifugal force is contained in the generalized force (type II), the Coriolis force in p˙ ϕ .

Example: Plane Motion in a Central Potential The potential V (r) depends only upon the coordinate r, but not upon the velocities r, ˙ ϕ˙ (and also not upon ϕ). The generalized momenta are pr =

∂T ∂L = = mr˙ ∂ r˙ ∂ r˙

pϕ =

∂T ∂L = = mr2 ϕ˙ ∂ ϕ˙ ∂ ϕ˙

and represent the component of the momentum in the radial direction and the angular momentum, respectively. The generalized forces of type I are Qr = −

∂V = −V  ∂r

Qϕ = −

∂V = 0. ∂ϕ

With the kinetic energy   T = 12 m r˙ 2 + r2 ϕ˙ 2 the difference between the generalized forces of type I and type II is

20 21

Fr − Qr =

∂T = mrϕ˙ 2 ∂r

Fϕ − Qϕ =

∂T = 0. ∂ϕ

For velocity dependent potentials see Sect. 3.7 further below. For the inertial forces and the Coriolis force see Sect. 7.2.4 further below.

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3 Lagrangian Mechanics

3.7 Velocity-Dependent Forces 3.7.1 Generalized Potential The equation of motion QR l =

d ∂L ∂L − dt ∂ q˙l ∂ql

with

L = T − V ({ql })

for a particle under the influence of conservative and other forces, −∂V /∂ql and QR l , respectively, suggest the question whether some of these other forces QR can be described by a generalized potential l V R ({ql }, {q˙l }, t); the generalized forces (type III)22 are then QR l =

d ∂V R ∂V R − . dt ∂ q˙l ∂ql

(3.28)

Comment: For a velocity independent potential one obtains from (3.28) the generalized force of type I. 3.7.2 Lorentz Force The definition (3.28) would be superfluous, if there were no application, namely, e.g., that to the Lorentz force: A particle with the mass m and the charge23 q in an electromagnetic field is subject to the (velocity dependent) Lorentz force24 $ # [c] F =q E+ v×B . c The Lorentz force is obtained as a generalized force from the generalized potential25 [c] V = qφ − q A · v (3.29) c with the scalar potential φ(r, t) and the vector potential A(r, t) and the fields26 22 23 24 25

26

See footnote 14 in Sect. 3.5.1. Not to be confused with the generalized coordinate ql . [x] = x in SI-units, [x] = 1 in Gauß units, cf. footnote 1.18. The potential energy of a charge in the electromagnetic field is Epot = qφ+q [c] A· c v. Notice the different sign of the term with the vector potential A in the energy and in (3.29). See the Course on Electrodynamics.

3.8 Gauge Invariance and Form Invariance

E = −∇φ − B = ∇ × A.

81

[c] ˙ A c

˙ Proof: The force (type III) is given by (v = r) F

d ∇r˙ V − ∇r V dt # $ # $ d [c] [c] = −q A − q∇ φ − (A · v) dt c c $ #  [c]  [c] ∂A = −q + r˙ · ∇A − q∇φ + q v × (∇ × A) + v · ∇A c ∂t c [c] = qE + q v × B c

(3.28)

=

(D.26)

with ∇(A · v) = v × (∇ × A) + v · ∇A.  For the particle (without constraints) in an electromagnetic field one obtains the canonical (generalized) momentum p, p=

[c] ∂L = mv + q A. ∂v c

(3.30)

Comments: • For Lorentz (and other velocity dependent) forces one has thus to distinguish between – The canonical momentum p = ∂L/∂v and – The dynamical (kinetic) momentum π = mv = mr˙ = p − q [c] c A. • The Lagrangian function L=

m 2 [c] v + q A · v − qφ 2 c

(3.31)

here has not the simple form L = Ekin − Epot . • The Lagrangian function is not unique but subject to gauge transformations, see the following section. In the form given here the form is known as the “minimal coupling.”

3.8 Gauge Invariance and Form Invariance 3.8.1 (Un)Ambiguity of the Lagrangian Function: Gauge Transformation As known from the Newtonian formalism the potential energy is determined except for an additive constant and is thus not unique; likewise the Lagrangian function L is not unique:

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3 Lagrangian Mechanics

Proposition: Gauge invariance of the Lagrangian equations of motion (of the second kind): Given two Lagrangian functions L and L , which are connected to each other by a so-called gauge transformation L = L +

dF dt

with

∂F =0 ∂ q˙l

(3.32)

with the gauge function of the form F = F ({ql }, t) (notice ∂F/∂ q˙l = 0), thus with ∂F  ∂F dF = + q˙l . (3.33) dt ∂t ∂ql l



Then, the functions L and L of (3.32) lead to the same equations of motion, $ $ # # d ∂L d ∂L ∂L ∂L = =0 − − dt ∂ q˙l ∂ql dt ∂ q˙l ∂ql Proof: (for one coordinate q): # $ # $ d ∂L d ∂L ∂L ∂L = +X − − dt ∂ q˙ ∂q dt ∂ q˙ ∂q with the additional term X=

# $ d ∂ dF ∂ dF , − dt ∂ q˙ dt ∂q dt

which is to be shown to vanish. With dF ∂F ∂F = q˙ + dt ∂q ∂t one obtains (∂F/∂ q˙ = 0) ∂ dF d ∂ dF − dt ∂ q˙ dt ∂q dt   ∂F ∂ dF d ∂ ∂F + q˙ − = dt ∂ q˙ ∂t ∂q ∂q dt   d ∂F ∂ dF d ∂F ∂ dF = − = 0. = 0+ − dt ∂q ∂q dt dt ∂q ∂q dt

X=

 Comments: • With the gauge transformation (3.32) the canonical momentum transforms according to ∂F pl = pl + , (3.34) ∂ql see problem 3.23.

3.8 Gauge Invariance and Form Invariance

83

3.8.2 * Form Invariance of the Lagrangian Equations under Point Transformations Let qi = qi (qj , t) be a point transformation. In the form given here transformations to moving coordinates are also included. With the equation of motion d ∂L ∂L − =0 dt ∂ q˙i ∂qi one has d ∂L ∂L −  = 0,  dt ∂ q˙j ∂qj where L (qj , q˙j , t) = L(qi (qj , t), q˙i (qj , q˙j , t), t) arises simply by inserting.27 Notice that with q˙i =

∂qi dqi ∂qi =  q˙j + dt ∂qj ∂t

the (generalized) velocity q˙i depends upon qβ , q˙β , and t. The Lagrangian equations of motion are thus form-invariant under point transformations. In contrast, the Newtonian equations are not form-invariant under transformations to curvilinear coordinates (and to noninertial reference systems): Form invariance of the Newtonian equations would mean that m¨ xi = Fi (xi ) in the system S and m¨ xi = Fi (xi ) in the system S would be true; but, in fact, this is true only for a Galilean transformation of one inertial system S to another inertial system S in the standard configuration.28 In addition, for the motion in a plane29 in a central potential V = α/r with F = F (r) er , F (r) = −α/r2 , one has m¨ x=α m¨ y=α 27 28 29

x (x2

3

+ y2) 2 y 3

(x2 + y 2 ) 2

,

Without proof. For details on coordinate transformations see Chap. 7. See, for example, Sects. 2.2.4 or 5.1.

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3 Lagrangian Mechanics

but α + mrϕ˙ 2 r2 r˙ ϕ˙ mϕ¨ = −2m . r m¨ r=

3.9 * Lagrangian Equations of the First Kind In the foregoing, the dependent coordinates (because of the constraints) and along with it the forces of constraint have been eliminated from the equations of motion. As a result, the Lagrangian equations of the second kind have been obtained. Now these considerations shall be extended to the explicit treatment of the dependent coordinates and of the forces of constraint. One starts from (1) The Newtonian equations of motion, (3.1)

mi x ¨α = Fα + Zα

α = 1, . . . , 3N.

(3.35)

(2) The constraints (Gjα = ∂Gj /∂xα for holonomic constraints), (3.3)

dGj =

3N 

Gjα dxα + Gj0 dt = 0

(3.36)

Gjα δxα = 0 j = 1, . . . , k.

(3.37)

α=1

δGj =

3N  α=1

(3) The principle of the virtual work, 3N 

(3.5)

Zα δxα = 0.

(3.38)

α=1

3.9.1 Lagrange Multipliers Because of the k constraints only f = 3N − k of the coordinates xα of the N particles are independent, and thus only f of the virtual displacements δxα are linearly independent. The dependent displacements δxα can now be eliminated. To this end one takes the difference of the equations (3.37) and (3.38), (3.38) −

k  j=1

λj (3.37) =

$ 3N # k   λj Gjα δxα = 0. Zα − α=1

j=1

(3.39)

3.9 * Lagrangian Equations of the First Kind

85

Comment: The λj are called Lagrange multipliers. They depend upon the time, λj = λj (t), but are independent of the coordinates, ∂λj /∂xα = 0. One divides the sum over α in (3.39) into a sum over the independent coordinates (α = 1, . . . , f with f = 3N − k) and a second sum over the k dependent coordinates (α = f + 1, . . . , 3N ), $ $ f # 3N # k k     λj Gjα δxα = λj Gjα δxα Zα − Zα − α=1

α=1

j=1

+

j=1

$ 3N # k   λj Gjα δxα . (3.40) Zα − j=1

α=f +1

One chooses the Lagrange multipliers λj (j = 1, . . . , k) such that in the second k sum in (3.40) the summands (Zα − j=1 λj Gjα ) vanish. These are k linear algebraic equations for the k different Lagrange multipliers λj . In the remaining first sum in (3.40) the virtual displacements δxα for α = 1, . . . , f are linearly independent. From the linear independence of these δxα it follows then that all summands of the first sum vanish. Thus one has for all α Zα =

k 

λj Gjα

α = 1, . . . , 3N,

(3.41)

j=1

one part because of the choice of the λj , the other part because of the linear independence of the virtual displacements δxα . 3.9.2 Lagrangian Equations of the First Kind Instead of (3.35) and (3.36) one now has because of (3.41) the equations  Gjα dxα + Gj0 dt = 0 j = 1, . . . , k (3.42) α

mα x¨α = Fα +

k 

λj Gjα

α = 1, . . . , 3N.

(3.43)

j=1

These are 3N + k (differential) equations for the 3N + k unknown quantities xα and λj . If one rewrites (3.42) and (3.43) in terms of arbitrary, not necessarily independent and not necessarily cartesian coordinates,30 one obtains, analogously to the results of Sect. 3.5.4,  % jl dql + G %j0 dt = 0 j = 1, . . . , k G (3.44) l

30

See, e.g., Goldstein [1] Sects. 2–4.

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3 Lagrangian Mechanics

 d ∂L ∂L % Jl − = QR λj G l + dt ∂ q˙l ∂ql j

l = 1, . . . , 3N

(3.45)

with xα = xα ({ql }, t)  ∂xα ∂xα dt dxα = dql + ∂ql ∂t l  ∂xα %jl = G Gjα ∂ql α  ∂xα %j0 = Gj0 + G . Gjα ∂t α Comments: • For scleronomous constraints one has ∂xα = 0. ∂t

Gj0 = 0, • For holonomic constraints one has Gj ({xα }, t) = 0



dGj =

 ∂Gj α

∂xα

dxα +

∂Gj dt, ∂t

(3.46)

and (3.42) can be written in the form dGj = 0; thus one has in (3.43) Gjα = ∂Gj /∂xα and Gj0 = ∂Gj /∂t, thus mα x¨α = Fα +

k 

λj

j=1

∂Gj ∂xα

α = 1, . . . , 3N

(3.47)

or, instead of (3.45), %j d ∂L ∂L  ∂ G − = λj + QR l dt ∂ q˙l ∂ql ∂q l j

l = 1, . . . , 3N

%j ({ql }, t) = Gj ({xα ({ql }, t)}, t). with G 3.9.3 Example: Atwood Machine Two masses m1 and m2 are connected to each other by a string of length l strung over a (massless) pulley and can move along the vertical (z) direction, see Fig. 3.11. The (holonomic) constraint reads G = z1 + z2 − l = 0.

(3.48)

3.9 * Lagrangian Equations of the First Kind

87

m2 m1

Fig. 3.11. The Atwood machine

The equations of motion (of the first kind) read mα x ¨α = Fα + λGα , or, with ∂G/∂zi = 1, m1 z¨1 = −m1 g + λ

∂G = −m1 g + λ ∂z1

(3.49)

m2 z¨2 = −m2 g + λ

∂G = −m2 g + λ. ∂z2

(3.50)

In (3.49) λ is the force of constraint on m1 , in (3.50) it is the force of constraint on m2 , i.e., in both cases the tension in the string. For the solution of (3.48)–(3.50) one eliminates λ, (3.49) − (3.50) = m1 z¨1 − m2 z¨2 = − (m1 − m2 ) g d2 m1 2 (3.48) = m1 z¨1 + m1 z¨2 = 0. dt Subtraction of the first from the second equation leads to (m1 + m2 ) z¨2 = (m1 − m2 ) g or z¨2 =

m1 − m2 g. m1 + m2

This can be integrated trivially; with z2 (t0 ) = z2,0 and z˙2 (t0 ) = v2,0 one obtains z2 (t) =

1 m1 − m2 2 g (t − t0 ) + v2,0 (t − t0 ) + z2,0 . 2 m1 + m2

The solution for z1 is z1 = l − z2 .

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3 Lagrangian Mechanics

z

z–h α h

x x

Fig. 3.12. For the moving plane the angle α is fixed, and the height h(t) (at x = 0) is a function of time

For the force of constraint one obtains from the equations (3.49) and (3.50), respectively, z1 + g) = m2 (z¨2 + g) λ = m1 (¨   m1 − m2 +1 g = m2 m1 + m2 m1 m2 g. =2 m1 + m2 The force of constraint has a positive z (upward) component. Example: Point Mass on a Moving Inclined Plane With the choice of the coordinate system as in Fig. 3.12 the constraint is G = z − h(t) − x tan α = 0.

(3.51)

The constraint is holonomic and rheonomous.31 With the Lagrangian function   L = 12 m x˙ 2 + y˙ 2 + z˙ 2 − mgz the Lagrangian equations of the first kind are ∂G = −λ tan α ∂x ∂G =0 m¨ y−0=λ ∂y

m¨ x−0=λ

m¨ z + mg = λ

∂G =λ ∂z

(3.52)

(3.53)

or m¨ r =F +Z 31

For the treatment with the Lagrangian equations of the second kind see problem 3.15.

3.9 * Lagrangian Equations of the First Kind

89

with the components ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ x ¨ 0 −λ tan α ⎠. m ⎝ y¨ ⎠ = ⎝ 0 ⎠ + ⎝ 0 z¨ −mg λ The vector of the force of constraint is thus ⎛ ⎞ − tan α ⎠. Z = λ⎝ 0 1 The equation y¨ = 0 can be integrated trivially and leads to uniform motion in the y direction. There remain the three coupled equations for x, z, and λ to be solved. For the elimination of λ from the equations of motion one has (3.52) (3.53)

⇒ ⇒

λ = −m¨ x cot α λ = m (¨ z + g)

or x ¨ cot α + z¨ + g = 0 and from (3.51) ¨=0 ¨=x −G ¨ tan α − z¨ + h These are two equations for x and z. Elimination of z¨ by adding these two equations leads to ¨=0 x ¨ (cot α + tan α) + g + h or x ¨=−

! " ¨ g+h ¨ sin α cos α. =− g+h cot α + tan α

Now this can be integrated for a given time dependence of h(t). Now one can insert successively. One obtains z(t) from the constraint, z = h + x tan α, and finally ! " λ = m (¨ z + g) = −m¨ x cot α = +m g + ¨h cos2 α. See also problem 3.15.

90

3 Lagrangian Mechanics z

ϕ

y

ϑ x

v

Fig. 3.13. The definition of the angles ϕ and ϑ for the wheel

Example: The Vertical Wheel In this section the motion of a wheel shall be investigated. The wheel (with the radius R) shall roll on a plane (the (x, y)-plane); an inclination of the wheel shall be neglected. As in the example of the skate of Sect. 3.2.2 one has to deal not with a system of point masses, but with an extended body, whose properties will be treated in a more general context of rigid bodies further below.32 With the notation as in Fig. 3.13 the generalized coordinates are x, y, ϑ, and ϕ. The constraint (the condition for rolling) is ds = R dϕ or, if one decomposes the path into its components, 0 = G1x dx + G1ϕ dϕ = dx − R cos ϑ dϕ

(3.54)

0 = G2y dy + G2ϕ dϕ = dy − R sin ϑ dϕ.

(3.55)

These are nonholonomic constraints as in the example of the skate in Sect. 3.2.2. The Lagrangian function is   L = T = 12 m x˙ 2 + y˙ 2 + 12 Iϕ ϕ˙ 2 + 12 Iϑ ϑ˙ 2 with the moments of inertia Iϕ and Iϑ = Iz . The Lagrangian equations of the first kind are ∂L  d ∂L − − λj Gjl 0= dt ∂ q˙l ∂ql j q1 = x : q2 = y :

0 = m¨ x − 0 − λ1 0 = m¨ y − 0 − λ2

q3 = ϕ :

0 = Iϕ ϕ¨ − λ1 (−R cos ϑ) − λ2 (−R sin ϑ) 0 = Iϑ ϑ¨ − 0 − 0.

q4 = ϑ : 32

See Chap. 8 further below.

3.9 * Lagrangian Equations of the First Kind

91

From the last equation (for q4 = ϑ) one obtains ⇒

ϑ¨ = 0 ϑ˙ = ϑ˙ 0 = Ω = const.



ϑ = Ωt + ϑ0

with ϑ(t) = ϑ0 for t = 0. The forces of constraint can be expressed by ϑ and ϕ: With the time derivative of the constraints (3.54) and (3.55) x˙ = Rϕ˙ cos ϑ y˙ = Rϕ˙ sin ϑ one obtains

x ¨ = Rϕ¨ cos ϑ − Rϕ˙ ϑ˙ sin ϑ y¨ = Rϕ¨ sin ϑ + Rϕ˙ ϑ˙ cos ϑ



! " x = mR ϕ¨ cos ϑ − ϕ˙ ϑ˙ sin ϑ λ1 = m¨ ! " λ2 = m¨ y = mR ϕ¨ sin ϑ + ϕ˙ ϑ˙ cos ϑ .

(3.56)

This one can insert into the equation for q3 = ϕ and obtains 0 = Iϕ ϕ¨ + R (λ1 cos ϑ + λ2 sin ϑ)   ¨ = Iϕ + mR2 ϕ. Because of Iϕ + mR2 = 0 one obtains ϕ¨ = 0



ϕ˙ = ϕ˙ 0 = ω = const.

and with this from (3.56) λ1 = −mRϕ˙ 0 ϑ˙ 0 sin ϑ λ2 = mRϕ˙ 0 ϑ˙ 0 cos ϑ. If one inserts this into (3.54) and (3.55) for the constraints, one finally obtains # $ # $ x˙ = Rϕ˙ 0 cos ϑ x˙ cos ϑ ⇒ v= = Rϕ˙ 0 y˙ = Rϕ˙ 0 sin ϑ y˙ sin ϑ with ϑ(t) = Ωt + ϑ0 , and

#

# $ λ1 = λ2 # $ − sin ϑ = mRϕ˙ 0 ϑ˙ 0 . cos ϑ

Z=

Zx Zy

$

ϕ˙ 0 = ω

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3 Lagrangian Mechanics

z ϑ0 > 0 Fig. 3.14. The direction of the force of constraint for the wheel

The direction of the force of constraint is perpendicular to the velocity, see also Fig. 3.14. The force of constraint points into the direction of the wheel axis, i.e., the torque for the rotation around the wheel axis vanishes. (⇒ ϕ˙ = const.!) The motion of the wheel is such that the contact point runs around a circular orbit. However, for this motion a torque arises, which would lead to an inclination of the wheel. Thus, in this example also a torque of constraint must act in addition, which compensates this inclination.

3.10 Hamilton’s Principle In this paragraph it shall be shown that the Lagrangian equations of motion, which had been derived above with physical arguments from d’Alembert’s principle, also can be derived as the Euler equations of an extremum principle, in this case of Hamilton’s principle. Comment: In an axiomatic approach to classical mechanics, Hamilton’s principle could be a starting point, alternative to the Newtonian axioms. 3.10.1 The Action Definition 20. (Action): The action W is33 

t2

W =

L dt t1

with the Lagrangian function L = L({ql (t)}, {q˙l (t)}, t) . 33

The action is often denoted by S.

(3.57)

3.10 Hamilton’s Principle

93

q

t1

t2

t

Fig. 3.15. Two varied trajectories with the same initial and final points

Comments: • The notation “action” is an unfortunate choice. The physical action has nothing to do with performance, but rather with effectiveness. • The action W is a functional34 of the trajectories ql (t), W = W [ql (t)] . 3.10.2 Hamilton’s Principle The action for the real trajectories is “stationary,” i.e., extremal, 

t2

δW = δ

L dt = 0,

(3.58)

t1

with fixed initial and final points, δql (t1 ) = 0 = δql (t2 ). Comments: • There are thus several varied trajectories ql (t) + δql (t), which all have the same initial and final point, compared with the real trajectory ql (t) of a single particle, see Fig. 3.15. • It may look as if, according to Hamilton’s principle, for the prediction of the trajectory the total trajectory should be known (because otherwise one could not evaluate the integral), and as if this principle would contradict the principle of the causality. However, one could argue against this that only a piece of the trajectory (namely that between the arbitrarily chosen times t1 and t2 ) must be known. Thus, the Newtonian laws enable the prediction of the trajectory, starting from any given time,35 i.e., if one knows the coordinates ql as well as, e.g., the velocities q˙l at a given time.

34 35

See Appendix E.1. See, for example, the discussion in Sommerfeld [6].

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3 Lagrangian Mechanics

• The result of the variational method36 (with an integral over time) is a set of Euler–Lagrangian equations (of the second kind) (3.20) (at an arbitrary time), see the following. • One finds that the inanimate nature apparently behaves such that some quantities become extremal. For example, a ray of light, emitted from a point, does not “know”, of course, where it will eventually arrive. In fact, the “trajectory” of a ray of light is such that the arrival at this final point is in the shortest time possible. (Additional examples for extreme quantities: minimum energy, maximum entropy, etc.) • Hamilton’s principle has applications not only in mechanics, but also in other fields of physics and technology. The variational problem (3.58) implies37 the Lagrangian function (3.20) of the second kind,38 δW = 0



d ∂L ∂L − = 0. dt ∂ q˙l ∂ql

(3.59)

Proof: 



t2

δW = δ

L dt = t1

⎞  ∂L  ∂L ∂L ⎝ δt⎠ dt. δql + δ q˙l + ∂q ∂ q ˙ ∂t l l j j ⎛

t2

t1

(3.60)

The third term vanishes because of δt = 0; partial integration of the second term leads to  t2  t2 ∂L d ∂L δ q˙l dt = − δql dt, (3.61) dt ∂ q˙l t1 ∂ q˙l t1 where the integrated term vanishes because of vanishing variations δq(t1 ) = 0 = δq(t2 ) at the limits, and one obtains $  t2  # ∂L d ∂L 0 = δW = − (3.62) δql dt. ∂ql dt ∂ q˙l t1 j Since the limits of integration are arbitrary, the integrand (the sum) vanishes, and since the δql are linearly independent, the summands (in the brackets) vanish.  Comment: The gauge transformation (3.32) L = L + 36 37 38

dF dt

See Appendix E. See Appendix E.2. For the derivation of the Lagrangian equations of the first kind from Hamilton’s principle see, e.g., Goldstein [1], Sects. 2–4.

3.11 Symmetries and Conservation Laws

95

leaves the variation of the action unchanged, δW  = δW. Proof: # $ dF δW = δ L+ dt dt t1  t2  t2 dF dt = δW + δ dF = δW + δ t1 dt t1 &t2 &t2 & & = δW + δF & = δW + [F (q + δq) − F (q)]& = δW + 0 



t2

t1

t1

because of δq(t1 ) = 0 = δq(t2 ).



3.11 Symmetries and Conservation Laws39 Further above,40 the conservation laws have been inferred from the form of the Newtonian forces. In this paragraph we shall analyze the Lagrangian function for conservation laws. 3.11.1 Noether’s Theorem (II)

Noether’s theorem: [15] For each invariance of the Lagrangian function (or of the action integral) under a continuous transformation there is a conservation theorem.

In the following we will start from the Lagrangian equations of the second kind, d ∂L ∂L − = 0, dt ∂ q˙l ∂ql here for reasons of simplicity without constraints, d ∂L ∂L − = 0. ˙ dt ∂ r i ∂ri 39 40

According to Hill [13] See Sect. 1.6.

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3 Lagrangian Mechanics

Examples: invariance of the system under . . . . . . (cyclic) variable, in particular : . . . time translation (homogeneity of time) . . . space translation (homogeneity of space) . . . space rotation (isotropy of space) . . . space inversion . . . Galilean transformation Comments:

conservation theorem for . . . . . . canonical momentum . . . Hamiltonian function . . . momentum . . . angular momentum . . . parity . . . center of mass

• It is important to observe the fact that the transformation is continuous. For example, the symmetry operation, which maps a crystal onto itself (by a shift of a lattice vector, by a mirror operation, etc.), is not a continuous transformation, since it cannot be described by a continuous parameter (an arbitrary angle of rotation, etc.). • Additional invariances are connected with additional conserved quantities like the charge, the number of baryons, etc.; the invariance under Lorentz transformations of Special Relativity theory41 is connected with a maximum velocity (the velocity of light). • The Lagrangian densities (for the continuous quantum-mechanical fields of the elementary particles instead of the discrete coordinates of the point mechanics) play a role not only in classical field theories like the continuum theory of deformable bodies42 and in classical electrodynamics43 but in particular also in the modern physics of the elementary particles with correspondingly conserved quantities. 3.11.2 Cyclic Coordinate and Conservation of the Conjugate Momentum Definition 21. (Cyclic Coordinate): A coordinate ql is called cyclic, if the Lagrangian function L is independent of this coordinate, ∂L =0 ∂ql



ql is cyclic.

(3.63)

Proposition: The conjugate momentum pl corresponding to a cyclic coordinate ql is a conserved quantity (if the system can be described by a Lagrangian function, i.e., if all forces are conservative), ∂L =0 ∂ql 41 42 43



pl = const.

See the Course on Special Relativity. See Chap. 10. See the Course on Classical Electrodynamics.

(3.64)

3.11 Symmetries and Conservation Laws

97

Proof: For the derivative of the generalized momentum with respect to time one obtains with the use of the Lagrangian equation of motion (3.63)

0 =

∂L ∂ql

(3.20)

=

d ∂L dt ∂ q˙l

(3.25)

= p˙ l .

(3.65) 

Comments: • The existence of conserved quantities very obviously facilitates the integration of the equations of motion. Thus the aim is to make a clever choice of the coordinate system (i.e., the system of the generalized coordinates ql ) such that as many coordinates as possible are cyclic. Example: Motion in a Central Potential In cartesian coordinates the Lagrangian function is    L = L(x, y, z, x, ˙ y, ˙ z) ˙ = 12 m x˙ 2 + y˙ 2 + z˙ 2 − V ( x2 + y 2 + z 2 ), and none of these coordinates is cyclic. In spherical polar coordinates it is " ! L = 12 m r˙ 2 + r2 ϕ˙ 2 sin2 ϑ + r2 ϑ˙ 2 − V (r), and ϕ is a cyclic coordinate, ∂L =0 ∂ϕ



pϕ =

∂L = mr2 ϕ˙ sin2 ϑ = const. ∂ ϕ˙

In this example, the z component of the angular momentum is thus conserved. 3.11.3 Homogeneity in Time and Energy Conservation The invariance of the Lagrangian function L of a system under a (continuous) time translation τ is equivalent to writing L(ql , q˙l , t) = L(ql , q˙l , t + τ ) for arbitrary τ , or equivalent to the fact that the Lagrangian function L has no explicit time dependence, ∂L = 0, ∂t but in general an implicit dependence, dL/dt = 0, because of the time dependence of the coordinates ql (t). One obtains from the homogeneity of time the so-called Hamiltonian function

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3 Lagrangian Mechanics

H=

f 

q˙l pl − L

(3.66)

l=1

as a conserved quantity, ∂L =0 ∂t

dH = 0. dt



(3.67)

Proof: One obtains dL dt

=

l

(3.20)

=



# ∂L

0 =

#

∂ q˙l

q¨l +

∂L q˙l ∂ql

$ +

∂L ∂t

$  d # ∂L $ ∂L d ∂L q¨l + q˙l q˙l +0= ∂ q˙l dt ∂ q˙l dt ∂ q˙l l l $ $ # # d  dH d  ∂L . q˙l − L = pl q˙l − L = dt ∂ q˙l dt dt l

l

 Comments: • Equation (3.66) is a so-called Legendre transformation,44 by which the function L(q, q) ˙ is transformed into a function H(q, p), L(q, q) ˙ → H(q, p). • Legendre transformations play a vital role in thermodynamics in the context of the different thermodynamic potentials. With the additional assumptions T =

1 μll ({ql })q˙l q˙l 2 

(3.68)

ll

V = V ({ql }),

∂V = 0 (for all l) ∂ q˙l

(3.69)

then follows H = T + V = E = const. If the system can be described by a Lagrangian function L = T − V with the given restrictions (i.e., if there are no velocity dependent forces), the energy E is thus a conserved quantity (a “constant of the motion”). 44

See the Appendix J.

3.11 Symmetries and Conservation Laws

99

Fig. 3.16. The bead sliding on a rotating rod, cf. Fig. 3.2

Proof: With 

q˙l pl =

l

one obtains d dH = 0= dt dt



q˙l

l

#

∂L  ∂T = q˙l = 2T ∂ q˙l ∂ q˙l l

$ q˙l pl − L

l

=

d d (2T − (T − V )) = (T + V ) dt dt

d = E. dt  Example: Bead Sliding on the Rotating Rod For the example of the bead sliding on the rod rotating (uniformly) with the angular velocity ω (Fig. 3.16) from Sect. 3.5.6, the Lagrangian function   L = T = 12 m r˙ 2 + r2 ω 2 is time independent. With the canonical momentum p=

∂L = mr˙ ∂ r˙

the Hamiltonian function   m p2 − ω 2 r2 = H(r, p) H = rp ˙ − L = 12 m r˙ 2 − ω 2 r2 = 2m 2 is a conserved quantity. Since here the kinetic energy is not of the form (3.68) (because of the term 12 mr2 ω 2 originating from the constraint), the energy E = H − mω 2 r2 is not conserved. In fact, here the force of constraint does not do virtual, but real work.

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3 Lagrangian Mechanics

3.11.4 Space Homogeneity and Conservation of Momentum Space homogeneity of the system is equivalent to the invariance of the system under continuous45 translations, L(ri , r˙ i , t) = L(ri + a, r˙ i , t) with an arbitrary vector a or   ∂L ≡ ∇i L = 0. ∂ri i i This implies the conservation of the (canonical) total momentum P = 

∇i L = 0

i

Proof:



dP = 0. dt

 i

pi ,

(3.70)

46

(i) From 0 = L(r i + a, r˙ i , t) − L(r i , r˙ i , t) =



a · ∇i L + O(a2 )

i

and for an arbitrary translation a one has  ∇i L = 0. i

(ii) From the Lagrangian equation(3.20) with ∂L/∂ r˙ i = ∂ mi r˙ i = pi one has 0=

 ∂L ∂ri i

(3.20)

=



1 ˙ 2j /∂ r˙ i j 2 mj r

d  ∂L d  d = pi = P = 0. dt i ∂ r˙ i dt i dt

=



3.11.5 Isotropy and Angular-Momentum Conservation The isotropy of a system is equivalent to the invariance under continuous rotations. Rotations are described by r i → ri = ri + Δϕ × ri 45 46

The importance of the continuity of an invariance has already pointed out in Sect. 1.8. The proof is exactly along the lines of that in Sect. 1.8.

Summary: Lagrangian Mechanics

101

with the rotation vector47 Δϕ, thus L(r i , r˙ i , t) = L(ri + Δϕ × r i , r˙ i + Δϕ × r˙ i , t). This implies the conservation of the angular momentum, d d  L= ri × pi = 0. dt dt i

(3.71)

Proof: Analogously to the proof of (3.70) one obtains 0 = L(r i + Δϕ × ri , r˙ i + Δϕ × r˙ i , t) − L(ri , r˙ i , t)   ∂L ∂L = + (Δϕ × r˙ i ) · (Δϕ × r i ) · + O(Δϕ2 ) ˙i ∂r ∂ r i i   d ∂L ∂L (3.20) = Δϕ · + r˙ i × ri × + O(Δϕ2 ) ˙i ˙i dt ∂ r ∂ r i d  ∂L = Δϕ · ri × + O(Δϕ2 ). dt ∂ r˙ i i

Since Δϕ is arbitrary, one concludes 0=

d  ∂L d  d  d ri × = r i × pi = li = L. dt i ∂ r˙ i dt i dt i dt

 Comment: Here, the canonical momentum corresponding to the “coordinate” ϕ is the total angular momentum L.

Summary: Lagrangian Mechanics Alternative Principles: Principle of virtual work Principle of d’Alembert Principle of Hamilton

(3.5)

Z i · δr i = 0 (3.6)

(p˙ i − F i ) · δr i = 0 (3.58) δW = δ L dt = 0

Euler–Lagrangian Equations (Lagrangian equations of the second kind): d ∂L ∂L − dt ∂ q˙l ∂ql d ∂T ∂T − dt ∂ q˙l ∂ql 47

See also Sect. 1.8.1.

(3.21)

= QR l

(3.21)

= Ql + QR l

(l = 1, . . . , f ; f = 3N − k)

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3 Lagrangian Mechanics

Gauge Transformation leaves equation of motion invariant: (3.32)

L = L +

dF dt

∂F =0 ∂ q˙l

with

Lagrangian Equations of the first kind: d ∂L ∂L − dt ∂ q˙l ∂ql

(3.45)

= QR l +

k 

% jl λj G

j=1

(l = 1, . . . , 3N ) and

% jl dql + G %j0 dt (3.44) G = 0

(j = 1, . . . , k)

In most cases: L = T ({ql }, {q˙l }, t) − V ({ql }, t) ≡ L(q, q, ˙ t) Canonical Momentum: (3.25)

pl =

∂L ∂ q˙l

Hamiltonian Function: (Legendre transformation): (3.66)  H = pl q˙l − L = H({ql }, {pl }, t) l

Conservation Laws due to symmetries/invariances (Noether’s theorem): ql cyclic



∂L =0 ∂ql ∂L =0 ∂t

(3.64)



(3.67)



pl = const. H = const.

Charge q in an Electromagnetic Field: [c] r˙ · A(r, t) c $2 # 1 [c] p − q A(r, t) + qφ(r, t) 2m c

(3.31) 1 ˙2 2 mr

˙ t) = L(r, r,

(3.72)

H(r, p, t) =

− qφ(r, t) + q

Problems 3.1. Brachistochrone in the Homogeneous Gravitational Field I. Two points P1 (x1 , y1 ) and P2 (x2 , y2 ) shall be connected by a curve y(x), on which a point mass sliding without friction moves from P1 to P2 under the influence of the homogeneous gravitational field (in the y-direction) in minimal time. The velocity of the point mass is zero at the point P1 .

Problems

103

(a) Determine the time T which the point mass needs to get from P1 to P2 , as a function of the velocity and of the path traveled. Formulate the problem in a variational form  2 f (y, y  , x) dx = 0. δT = δ 1

(b) Using the Euler equation for f (y, y  , x) write down the differential equation for y(x), and solve the differential equation. Comment: With the substitution y  = u(y) the resulting differential equation 2yy  + y 2 + 1 = 0 can be reduced to a differential equation of first order for u(y). Then the differential equation y  = u(y) can be solved with the substitution y = a sin2 ϕ. With x1 = y1 = 0 the solution then has the form of a cycloid, a a x = (2ϕ − sin 2ϕ), y = (1 − cos 2ϕ). 2 2 (c) Which parameter regime ϕ of the cycloid is traced by the point mass for a a 3πa a a P1 = (0, 0) and P2 = ( πa 4 − 2 , 2 ), P2 = ( 4 + 2 , 2 ), or P2 = (πa, 0)? 3.2. Brachistochrone in the Homogeneous Gravitational Field II. A particle of mass m may move without friction under the influence of gravity from a point (0, y0 ) to (x0 , 0) with the initial velocity v = 0. What is the form of the trajectory of the particle, if the point (x0 , 0) shall be reached in the shortest possible time? (a) With the help of the energy conservation determine the total time T as a functional of the cartesian position coordinate. (b) What is the variational problem (boundary conditions)? (c) What is the Euler equation? To this end show that one has $ #  x0  x0 ∂F  d ∂F δy dx = − δy dx. ∂y  dx ∂y  0 0 (d) The functional F does not depend explicitly upon x. Show that F − y

∂F = const. ∂y 

is a first integral of the Euler–Lagrangian equation. (e) With the help of the transformation ∂y/∂x = − cot(θ/2) determine the parameter representation of the trajectory B = B(x(θ), y(θ)). 3.3. Catenary Curve. A string of length l is suspended in the homogeneous gravitational field between two points the distance of which is smaller than the string length. (a) Give an expression for the extremal condition. (b) Write down the constraint and the boundary conditions.

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3 Lagrangian Mechanics

(c) Integrate the resulting differential equation  k dy = (y + μ)2 − k 2 dx (Lagrange multiplier μ; k = const.). With the transformation y + μ = k cosh z a direct integration is possible. (d) Determine the constants with the help of the constraint and of the boundary conditions. 3.4. Atwood Machine. Two masses m1 and m2 are connected by a string of length l strung without friction across a massless wheel (see Fig. 3.17). (a) Give an expression for the equation of motion for the total mass m1 + m2 . (b) What is the constraint, and of which kind is it? (c) Give an expression for the d’Alembert principle, and derive from this the equations of motion. (d) Introduce the coordinate of a mass as the generalized coordinate q, and express the coordinate of the other mass by q, by the radius of the wheel, and by the string length . (e) Write down the Lagrangian function, and set up the Lagrangian equation of the second kind. (f) Solve the equation of motion. How does the acceleration change in the equation of motion, if the mass of the wheel has to be taken into account (qualitative arguments)?

R

y m1 m2

α1

α2

x

Fig. 3.17. Left: Atwood machine (for problem 3.4). Right: A modified Atwood machine (for problem 3.5)

Problems

105

3.5. A Modified Atwood Machine. Two masses m1 and m2 are positioned on inclined planes with the inclination angles α1 and α2 , respectively. They are connected by a massless rope of constant length, see Fig. 3.17. (a) How many constraints apply to the two masses? Write the constraints in the form fi (x1 , y1 , x2 , y2 ) = 0. (b) With the help of Lagrange multipliers set up the equations of motion for the two masses in cartesian coordinates. xi , y¨i ) of each (c) Determine the acceleration y¨1 and the modulus of bi = (¨ particle. Hint: First, eliminate the Lagrange multipliers and then, with the help of the constraints, all coordinates except for y1 . (d) Determine the constraining forces for the case α1 = α2 = 90◦ . 3.6. The Wheel and Axis. The sketched wheel and axis of mass m2 is fixed by a rope to the ceiling (Fig. 3.18); the rope is tied around the smaller disk (radius r). Tied around the larger disk (radius R > r) runs a rope at which the mass m1 pulls under the influence of gravity.

r

R

m2 m1

Fig. 3.18. The wheel and axis (for problem 3.6)

(a) Was happens without the mass m1 ? (b) Give an expression (now with m1 ) for the equilibrium condition in the static case. What are the constraints? How is the mass m1 to be chosen for static equilibrium with r, R, and m2 given. (c) Now consider the wheel and axis with the attached mass m1 as a dynamical system. Write down the kinetic and the potential energies, and formulate the Lagrangian function in an appropriately chosen generalized coordinate. (d) What is the Lagrangian equation of the second kind for the system? Interpret this equation, in that you identify cause and effect of the motion. Set up the relation to the static case.

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3 Lagrangian Mechanics

(e) Now write down the Hamiltonian function and the canonical equations. 3.7. Motion on a Spiral. A particle moves under the influence of the gravitational force without friction on a spiral of radius R and of threading 2π/α which rotates with the angular velocity ω. The equation for the geometrical position of all points on the spiral thus is x = R cos (αz − ωt) ; y = R sin (αz − ωt) . (a) Set up the equations of motion for the particle’s coordinates. (b) Solve the equation for the motion in the z-direction. (c) Determine the constraining forces acting on the particle, and discuss their origin. 3.8. Particle on Logarithmic Spiral. Let the trajectory of a point mass be r(φ) = a ebφ with a, b > 0 (logarithmic spiral). Which central force acts on the point mass? Hint: φ˙ can be eliminated using conserved quantities. 3.9. Motion Along a Parabola. A point mass moves in the gravitational 1 x2 . field g = (0, −g) in the (x, y)-plane on the parabola y = 2a (a) From the information given above, are there conserved quantities? If so, which? Of which kind is the constraint? (b) Write down the kinetic and the potential energies for the point mass. What is the Lagrangian function, if one chooses x as the generalized coordinate? (c) Give an expression for the Lagrangian equation of the second kind for x(t), and write down the canonical momentum p conjugate to x. (d) Determine the Hamiltonian function, and approximate this with x  a such that one obtains a quadratic form in x and p. (e) Write down the canonical equations for this approximate Hamiltonian function, and solve this with the initial conditions x(0) = 0, p(0) = p0 . 3.10. Bead Sliding on a Uniformly Rotating Rod. A bead slides without friction on a horizontal wire rotating about a vertical axis with the angular velocity ω (Fig. 3.19).

Fig. 3.19. The bead sliding on a rotating rod (like Fig. 3.9) (for problems 3.10 and 3.11)

Problems

107

(a) What are the constraints, and of which kind are they? (b) The equation of motion is r¨ = ω 2 r. Determine the eigensolutions and the general solution. (c) Determine the solution for the initial conditions r(0) = r0 and r(0) ˙ = v0 . Explain the possible forms of motion. (d) What are the conserved quantities of the system? Compare in particular the Hamiltonian function with the total energy. 3.11. Motion of a Bead on a Rotating Rod. A bead of mass m moves in the gravitational field on a rod which rotates horizontally with constant angular velocity ω. The frictional force on the bead is of the form F r = −γAv/|v|, where v is the velocity along the rod; A is the force pressing the bead to the rod. (a) Set up the Lagrangian equations of the first kind. Which physical quantity is hidden in the Lagrange multiplier? Write down the constraining force, and determine the force by which the mass is pressed against the rod. (b) Solve the equations of motion (neglecting gravity) with the initial conditions r(t) = r0 and v(t) = 0 for t = 0. 3.12. A Bead on a Rotating Wire. A bead of mass m slides along a straight wire which is fixed to a vertical axis at an angle α; the vertical axis rotates with the angular velocity ω. Solve the equation of motion for the initial conditions r(0) = 0, r(0) ˙ = v. 3.13. The Rotating Plane Pendulum. A plane pendulum with its hinge in the coordinate origin swings in the (x, z) plane (Fig. 3.20). This plane rotates with constant angular velocity ω around the z-axis (antiparallel to the homogeneous gravitational field). Employ spherical polar coordinates (r, ϑ, ϕ). (a) Set up the constraints. (b) Determine the Lagrangian function. (c) Determine the equation of motion. (d) A stationary solution is given by a vanishing acceleration (ϑ¨ = 0). Show that there are one or more stationary solutions, and determine the corresponding equilibrium positions ϑ0 as solutions of the stationary equation. (e) Investigate the (time dependent) motions in the neighborhood of the stationary solutions. To this end, make the ansatz ϑ = ϑ0 + δ, and expand the equation of motion to lowest order in δ; this will result in a linear differential equation. Determine the solutions δ(t). An equilibrium position ϑ0 is called stable, if the corresponding solution δ is restricted to a (finite) neighborhood of ϑ0 ; otherwise the equilibrium position is called unstable. Which equilibrium positions are thus stable or unstable?

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3 Lagrangian Mechanics

Fig. 3.20. Rotating plane pendulum (for problem 3.13)

3.14. Point Mass on Inclined Plane. A point mass m slides on an inclined plane with inclination angle α. What is a useful generalized coordinate to describe the motion of the point mass? Set up the Lagrangian function, and from this derive the equation of motion. 3.15. Point Mass on a Moving Inclined Plane. A point mass moves without friction under the influence of the homogeneous gravitational field (in negative z-direction) on an inclined straight line (Fig. 3.21). The straight line moves in the x direction, such that the height h(t) (along the z axis) is a function of time t. z

z–h α h

x x

Fig. 3.21. Point mass on a moving inclined plane (for problem 3.15)

(a) Set up the constraint. How many degrees of freedom does the system have? (b) Determine the Lagrangian function of the mass. (c) Determine the equation of motion of the mass. (d) Determine x(t) and z(t) for the cases h(t) = const., h(t) = vt, and h(t) = 12 at2 . (e) Determine the force of constraint.

Problems

109

3.16. Particle Sliding on a Cone. A point mass m moves without friction under the influence of gravity on the surface of a circular cone. (a) Write down the kinetic and the potential energies in cartesian coordinates. Give an expression for the constraint. (b) Introduce the coordinates x = r cos ϕ and y = r sin ϕ, and employ the relation z = r cot α from the constraint. (c) What are the Lagrangian equations of the second kind for the generalized coordinates r and ϕ? Which conserved quantity can be derived directly from these equations? What is the physical meaning of this quantity? (d) Employ this conserved quantity to eliminate one of the variables. Interpret the (one-dimensional) equation of motion for the radial motion. (e) Write down the condition for the point mass not to accelerate radially. 3.17. Crank Mechanism. Consider a piston engine (e.g., a steam engine, see Fig. 3.22), which runs slowly enough such that inertial forces can be neglected. Via the crank mechanism the piston force D is transformed into the crank force K which is tangential to the radius r. K

r

α

D β

x Fig. 3.22. Crank mechanism (for problem 3.17)

(a) Give an expression for the condition for static equilibrium (principle of virtual work). Write down the constraint relating the virtual displacements δx of the piston to the rotation angle δα. Hint: First employ the angle β in addition to x and α, and eliminate β by geometrical considerations. (b) How can K be represented as a function of D and α? 3.18. Kinetic Energy as a Homogeneous Function. Show that for a mechanical system with holonomous, skleronomous constraints the kinetic energy is a homogeneous function of the generalized velocities of order n = 2, i.e., T (αq˙1 , . . . , αq˙m ) = α2 T (q˙1 , . . . , q˙m ). 3.19. A Particle in an Electric and Magnetic Field. The Lagrangian function for a particle of mass m and of charge q, which moves in an electrical field E and a magnetic field B, has the following form:

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3 Lagrangian Mechanics

L=

1 mr˙ 2 + q · Ar˙ − qφ. 2

The fields E and B are related to the scalar potential φ(r) and the vector potential A(r) via the relations B = ∇ × A and E = −∇φ − ∂A/∂t. (a) Using the Lagrangian equations of the second kind show that m¨ r = q(E + r˙ × B). (b) Determine the canonical momentum conjugate to r, and determine the Hamiltonian function. (c) Is the total energy of the system conserved? Try a physical interpretation of this matter. (d) At time t = 0 the particle is placed into an electrical field E = (E, 0, 0) and a magnetic field B = (0, 0, B) at the point r 0 = 0, where it has the initial velocity v 0 . • Solve the equations of motion for the particle under the given initial conditions, and describe the trajectory of the particle. Hint: To solve for x(t) and y(t) introduce the complex quantity u(t) = x(t) + iy(t). • Discuss the trajectory for the case E = 0. • How does the motion of the particle change, if the electrical field also has a z-component? 3.20. A Point Mass on a Sphere. A point mass may move under the influence of gravity on the surface of a sphere (Fig. 3.23).

m φ mg Fig. 3.23. A point mass on the surface of a sphere (for problem 3.20)

(a) Write down the constraints and the constraining force in polar coordinates. (b) Give an expression for the equation of motion, and decompose this into the radial and azimuth components. (c) Integrate the azimuth equation of motion to obtain the energy conserva˙ = 0) = φ˙ 0 . tion for the initial conditions φ(t = 0) = 0, φ(t (d) Via the radial equation of motion determine the Lagrange multiplier and the constraining force.

Problems

111

(e) At which angle does the point mass detach from the surface of the sphere? 3.21. Particle on a Rotation Paraboloid. A particle of mass m moves under the influence of gravity on the internal area of a rotation paraboloid x2 + y 2 = az without friction. (a) Set up the equation of motion for the particle in Cartesian coordinates. Employ the Lagrangian equations of the first kind. (b) How large is the angular velocity ω of the particle, if it moves on the circle which results from the cut of the rotation paraboloid with the plane z = h? How large is the radius ρ0 of the circle? 3.22. Dissipation Function and Free Case with Friction. The friction as a force depending upon the velocity can be accounted for via the dissipation function D=

n 1  βlm q˙l q˙m 2 l,m=1

(R)

in the Lagrangian formalism by a generalized force Qj Lagrangian equation of the second kind then reads

= −∂D/∂ q˙j . The

d ∂L ∂L ∂D − + = 0. dt ∂ q˙j ∂qj ∂ q˙j (a) Give an expression for the differential energy change connected with the (R) frictional forces Qj , and express the time rate of change of this energy by the dissipation function D. (b) A point mass falls under the influence of gravity and of friction, which is described by the dissipation function D = α2 v 2 . Starting from the Lagrangian function and the dissipation function give an expression for the equation of motion. Determine the velocity for the point mass to be at rest at t = 0 (as the initial condition). Discuss the result with the help of the graph for v(t). 3.23. Gauge Transformation of the Canonical Momentum. Given be the gauge transformation (3.32)

L = L +

dF dt

with

∂F = 0. ∂ q˙l

Show that the canonical momenta transform according to pl = pl +

∂F . ∂ql

112

3 Lagrangian Mechanics

3.24. Hamiltonian Function of a Charged Particle. Given a particle with charge q with the Lagrangian function (3.31) 1 ˙2 2 mr

˙ t) = L(r, r,

− qφ(r, t) + q

[c] r˙ · A(r, t). c

Prove that the Hamiltonian function is 1 H(r, p, t) = 2m

$2 # [c] p − q A(r, t) + qφ(r, t). c

(3.72)

4 Harmonic Vibrations

In this chapter vibrations shall be investigated, and in particular an oscillator • Without and with damping • Without and with coupling to other oscillators, as well as • Without and with external force In the context of the external force we shall encounter a very simple case of a Green function, which plays an important role in statistical mechanics. In this context we shall study the tools to treat • (Un)coupled, (in)homogeneous, linear differential equations • Eigen values and eigen vectors of matrices, and finally • The Fourier transformation The prototype of the equation of motion is the equation of motion of a simple harmonic oscillator, for which the restoring force is proportional to the displacement (Hooke’s law, ω02 ), damped by Stokes friction1 (γ), and forced by an external field (f (t)), x ¨ + 2γ x˙ + ω02 x = f (t).

(4.1)

In the case of the oscillator without damping and without external force one obtains the simpler equation x ¨ + ω02 x = 0

(ω02 = k/m).

The restoring force −kx = −mω02 x of a spring with the spring constant k can be derived from the potential (Fig. 4.1) V (x) = 12 kx2 . The potential is symmetrical with respect to the origin, V (x) = V (−x); thus the turning points are located symmetrically. For a given energy E (as a 1

See Sect. 1.4.

114

4 Harmonic Vibrations 5 4

V/f

3 2

K/ f

1 0 3 2 1 0 −1 −2 −3 −3

−2

−1

0

x

1

2

3

Fig. 4.1. The potential V (top) and the force F (bottom) of the simple harmonic oscillator in units of the force constant k as a function of the displacement x

constant of the motion!) the form of V (x) is arbitrary for all x with V (x) > E. (This is wrong in the quantum-mechanical case because of the tunneling effect.) The harmonic oscillator is a model for many systems: Spring For the spring one has a force proportional to the change from the equilibrium length l0 of the spring, and for a displacement in the direction of the axis of a spiral spring (chosen as the z direction) one has m¨ z = −k (z − z0 ) .

(4.2)

With the substitution ζ = z − z0 one obtains mζ¨ = −kζ. Comments: • For a real spring Hooke’s law is only approximately valid and only for small elongations. For large elongations one observes the phenomenon of flowage (see Fig. 4.2), for which the spring does not return to its original form after the tension is released. • The restoring force for a small displacement perpendicular to the axis of an unextended spring is proportional to the third power of the displacement.2 2

See problem 4.3.

4 Harmonic Vibrations

115

V

x Fig. 4.2. The phenomenon of flowing of a material at larger stretching

Three-Dimensional Oscillator In the case of the three-dimensional oscillator one has in generalization of (4.2) m¨ r = −k (r − r0 ) . This equation can be decomposed into the three orthogonal cartesian components, and for each single component one has a one-dimensional equation (4.2) as in the case of the pendulum, m¨ xi = −k (xi − xi,0 ) . Molecule The (potential) energy of the diatomic molecule as a function of the distance of the cores has a minimum and is likewise approximately harmonic near the minimum (the broken curve in Fig. 4.3). Crystals In crystals the potential between two atoms as a function of their distance qualitatively has the form as in the case of the molecule. For small displacements out of the equilibrium positions the restoring forces are again harmonic.

v

r

Fig. 4.3. The potential energy V as a function of the interatomic distance r for the diatomic molecule (schematic)

116

4 Harmonic Vibrations

RLC Circuit ˙ The equation of motion for the electrical charge Q (with the current I = Q) as a function of the voltage U ¨ + RQ˙ + 1 Q = U LQ C of the RLC circuit, see Fig. 4.4, is the equation of a damped forced oscillator.

L

R

C

U Fig. 4.4. The RLC circuit

Mathematical Pendulum The potential energy is a periodic function of the angle, see Fig. 4.5, V (ϕ) = mgl (1 − cos ϕ) (with V (0) = 0). Differently from the other mentioned examples, here there are bound and free motions,3 depending on the energy (0 ≤ E < mgl and E > mgl, respectively). For small displacements from the minimum position the potential can be expanded into a Taylor series, V (ϕ) = 12 mglϕ2

(ϕ  π).

The potential is thus approximately harmonic in the neighborhood of the minimum (and only there) and can be approximated by a parabola (broken curve in Fig. 4.5). In this approximation one obtains the equation of motion ml2 ϕ¨ = −mglϕ (ϕ  π).

3

See the reasoning in the context of Fig. 2.1 on p. 33.

4.1 The Simple Oscillator

117

V

2p

0

j

Fig. 4.5. The potential energy V as a function of the displacement angle ϕ for the mathematical pendulum

Molecular Crystals The rotational potential of a molecule in a lattice (assumed as rigid) of a molecular crystal is likewise periodic in most cases, even though not of the simple form as in the case of the mathematical pendulum.

4.1 The Simple Oscillator Here the “simple” oscillator shall be investigated, by which we mean the onedimensional, free (i.e., without external force), harmonic oscillator without damping. 4.1.1 Eigen Solutions of the Homogeneous Differential Equation The equation of motion of the simple oscillator reads (4.1)

m¨ x + kx = 0

(4.3)

x¨ + ω02 x = 0

(4.4)

or with ω02 = k/m. Eigen Solutions For the solution of the linear differential (4.4) one makes an exponential ansatz eλt or, since one expects oscillations, the ansatz4 x(t) = x0 e−iωt .

4

The sign of the phase is chosen here as in nonrelativistic quantum mechanics. In classical mechanics one could also choose the opposite sign.

118

4 Harmonic Vibrations

Inserting this into the differential (4.4) one obtains  2  −ω + ω02 x0 e−iωt = 0. With this equation fulfilled for all times t, the factor in front of the exponential function must vanish, and one obtains an algebraic equation  2  −ω + ω02 x0 = 0 for the amplitude x0 instead of the differential (4.4) for the displacement x(t). For nontrivial solutions x0 = 0 the secular equation −ω 2 + ω02 = 0 must be fulfilled with the solutions ω± = ±ω0 = ω1,2 .

(4.5)

Thus, one obtains two eigen solutions (for the differential equation of second order) x1 (t) = e−iω1 t

and x2 (t) = e−iω2 t .

General Solution The general solution is the superposition of the eigen solutions, x(t) = a1 x1 (t) + a2 x2 (t) = a1 e−iω1 t + a2 e−iω2 t

(4.6)

with two (superposition) constants a1 and a2 . Comment: Instead of the solutions e±ω0 t one also could have obtained the eigen solutions sin(ω0 t)

and

cos(ω0 t),

(as a combination of the exponential functions) and the general solution would be x(t) = c sin(ω0 t) + d cos(ω0 t). Reality of the Solution The factors e±iωt in (4.6) are complex, e±iz = cos z ± i sin z. But since the displacement x(t) is real, the superposition coefficients ai must be complex. For example, one obtains from (4.6) the general solution x(t) = a cos(ω0 t) + b sin(ω0 t)

4.1 The Simple Oscillator

119

with real coefficients a and b, ib = a1 − a2 .

a = a1 + a2 ,

Proof: From the reality of the solutions follows x(t) = a1 e−iω0 t + a2 e+iω0 t = x∗ (t) = a∗1 e+iω0 t + a∗2 e−iω0 t . This must hold for all times t, and from the linear independence of the functions e±iω0 t one has a1 = a∗2

(a∗1 = a2 ).

If one sets a1 =

1 2

(a + ib) ,

a2 =

1 2

(a − ib) .

then one gets

With this one obtains for the solution x(t) =

1 2

(a + ib) [cos(ω0 t) − i sin(ω0 t)] +

1 2

(a − ib) [cos(ω0 t) + i sin(ω0 t)]

= a cos(ω0 t) + b sin(ω0 t).  Comments: • Of the four constants, which determine the real and imaginary part of a1 and a2 of the general solution (4.6), only two are independent of each other. • Thus one can write the solutions in the alternative forms x(t) = a cos(ω0 t) + b sin(ω0 t) = a1 e−iω0 t + a∗1 eiω0 t = a1 e−iω0 t + c.c.

(4.7) (4.8) (4.9)

with a1 = 12 (a + ib) and where “c.c.” stands for the complex conjugate of the preceding expression.

120

4 Harmonic Vibrations

Comment: (Concerning the use of complex quantities) Because of the convenience offered by the exponential function one mostly works with the complex quantities e±iωt , without saying explicitly that one means the real part of the corresponding expression. This works fine for all linear operations (addition, differentiation, etc.), e.g.,  d  Re (α e−iωt ) + Re (β eiωt ) dt   d = Re α e−iωt + β eiωt dt  d  −iωt = Re + β eiωt . αe dt Taking the square and other nonlinear operations one has first to take the real part and then to square because of Re (z1 z2 ) = Re z1 · Re z2 .

4.2 The Free Oscillator with Damping Here now the one-dimensional, free (i.e., without external force), harmonic oscillator with damping shall be investigated. The development follows very closely that one of the previous section. 4.2.1 Eigen Solutions of the Homogeneous Differential Equation For Stokes friction5 the homogeneous equation of motion of a linear oscillator reads (4.1)

m¨ x + x˙ + kx = 0

(4.10)

x ¨ + 2 γ x˙ + ω02 x = 0

(4.11)

or with 2γ = /m and ω02 = k/m. Eigen Solutions For the solution of the linear differential (4.11) one makes an exponential ansatz eλt or, since one expects oscillations, the ansatz6 x(t) = x0 e−iωt . 5 6

See Sect. 1.4. The sign of the phase is chosen here as in nonrelativistic quantum mechanics. In classical mechanics one could also choose the opposite sign.

4.2 The Free Oscillator with Damping

121

If one inserts this into the differential (4.11), one obtains  2  −ω − 2iωγ + ω02 x0 e−iωt = 0. With this equation fulfilled for all times t, the factor in front of the exponential function must vanish, and one obtains an algebraic equation  2  −ω − 2iωγ + ω02 x0 = 0 for the “amplitude” x0 instead of the differential (4.11) for the displacement x(t). For nontrivial solutions x0 = 0 the secular equation −ω 2 − 2iωγ + ω02 = 0 must be fulfilled with the solutions   for γ 2 < ω02 −iγ ± ω02 − γ 2 = −iγ ± Ω ω± = ω1,2 = −iγ ± i γ 2 − ω02 = −i(γ ∓ μ) for γ 2 > ω02 with Ω=

 ω02 − γ 2

and μ =

(4.12)

 γ 2 − ω02 .

Thus, one obtains two eigen solutions (for the differential equation of second order) x1 (t) = e−iω1 t

and x2 (t) = e−iω2 t .

General Solution The general solution is the superposition of the eigen solutions, x(t) = a1 x1 (t) + a2 x2 (t) = a1 e−iω1 t + a2 e−iω2 t + c.c.

(4.13)

with two (superposition) constants a1 and a2 . The Case γ = ω0 In the case γ = ω0 (i.e., ω1 = ω2 = −iγ) it looks as if the two eigen solutions would be the same. However, for a twofold zero of the secular equation one has the eigen solutions e−γt

and t e−γt

and the general solution x(t) = (c1 + c2 t) e−γt with two constants c1 and c2 .

(4.14)

122

4 Harmonic Vibrations

4.2.2 Reality of the Solution In the case γ 2 = ω02 the two eigen solutions (4.12) for the frequency ω are complex; also e−iωt is complex, ez = ex+iy = [cosh x + sinh x] [cos y + i sin y] with x and y real. But since the displacement x(t) is real, the superposition coefficients ai must be complex in addition to the exponential functions. Case γ < ω0 (Underdamped Case) For example, in the case ω02 > γ 2 , one obtains from (4.12) the general solution  x(t) = e−γt [a cos(Ωt) + b sin(Ωt)] , Ω = ω02 − γ 2 real with real coefficients a and b, ib = a1 − a2 .

a = a1 + a2 ,

Proof: From the reality of the solutions follows   x(t) = e−γt a1 e−iΩt + a2 e+iΩt   = x∗ (t) = e−γt a∗1 e+iΩt + a∗2 e−iΩt . Since this holds for all times t, from the linear independence of the functions e±iΩt one has a1 = a∗2

(a∗1 = a2 ).

If one sets a1 =

1 2

(a + ib) ,

a2 =

1 2

(a − ib) .

then one gets

With this one obtains for the solution x(t) = e−γt 12 {(a + ib) [cos(Ωt) − i sin(Ωt)] + (a − ib) [cos(Ωt) + i sin(Ωt)]} = e−γt [a cos(Ωt) + b sin(Ωt)] . 

4.2 The Free Oscillator with Damping

123

Comment: Of the four constants, which determine the real and imaginary part of a1 and a2 of the general solution (4.13), only two are independent of each other. Thus one can write the solutions in the alternative forms x(t) = e−γt [a cos(Ωt) + b sin(Ωt)]   = e−γt a1 e−iΩt + a∗1 eiΩt =e

−γt

a1 e

−iΩt

(ω02 > γ 2 )

(4.15)

+ c.c.

with a1 = 12 (a + ib)  Ω = ω02 − γ 2 . Case γ > ω0 (Overdamped Case) In the case ω02 < γ 2 one has x(t) = c e−(γ+μ) t + d e−(γ−μ) t = e−γt [C cosh(μt) + D sinh(μt)]

(ω02 < γ 2 )

(4.16)

with C =d+c

D =d−c

d = 12 (C + D)

c = 12 (C − D).

Case γ = ω0 (Critically Damped Case) Finally, in the case ω02 = γ 2 the solution is x(t) = a e−γt (1 + b t)

(ω02 = γ 2 ).

4.2.3 Initial Conditions The so far still undetermined constants of the general solutions (4.15) or (4.16) have to be fixed by conditions, typically by initial conditions, e.g., by x(0) = x0 x(0) ˙ = v(0) = v0 . With the eigen values (4.12) in the solution (4.13) one obtains x(0) = x0 = a1 + a2 x(0) ˙ = v0 = −iω1 a1 − iω2 a2

124

4 Harmonic Vibrations

or, in matrix notation,7 #

1 −iω1

1 −iω2

$#

a1 a2

$

# =

x0 v0

$

with the solution # $ # a1 1 = a2 −iω1

$−1 # $ 1 x0 −iω2 v0 # $# $ 1 −iω2 −1 x0 = iω1 1 v0 −iω2 + iω1 # $ # $ v0 x0 ω2 1 + = . ω2 − ω1 −ω1 i (ω2 − ω1 ) −1

With this one obtains   1 (x0 ω2 − iv0 ) e−iω1 t − (x0 ω1 + iv0 ) e−iω2 t . x(t) = ω2 − ω1

(4.17)

Comment: With ω2 − ω1 real or purely imaginary, as well as with x0 and v0 real one recovers the relation a2 = a∗1 . Case γ < ω0 (Underdamped Case) With ω1,2 = −iγ ± one obtains from (4.17) x(t) = e

−γt



 ω02 − γ 2 = −iγ ± Ω

 x0 γ + v0 sin(Ωt) + x0 cos(Ωt) . Ω

Case γ > ω0 (Overdamped Case) With −iω1,2 = −γ ±

 γ 2 − ω02 = −γ ± μ

one obtains

, 1 + [x0 (−γμ ) − v0 ] e−γt+μt − [x0 (−γ + μ) − v0 ] e−γt−μt −2μ   x0 γ + v0 sinh(μt) + x0 cosh(μt) . = e−γt μ

x(t) =

7

The multiplication of matrices with each other and with vectors is denoted here and in the following without a dot (product sign).

4.3 The Forced Harmonic Oscillator

1

γ/ω0 = 0.0 γ/ω0 = 0.2 γ/ω0 = 1.0 γ/ω0 = 2.0

0,5

(ω 0/v 0) x

125

0

−0,5

−1

0

10

20

ω0 t

30

Fig. 4.6. The displacement x of the oscillator hit by an impulsive force at time t = 0 for various damping constants

Case γ = ω0 (Critically Damped Case) From γ → ω0 , one obtains x(t) = lim e

−γt

Ω→0



Ω→0

 sin(Ωt) + x0 cos(Ωt) (x0 γ + v0 ) Ω

= e−γt [x0 + (x0 γ + v0 ) t] . Comment: The time dependence of x(t) for the case x0 = 0 is shown in Fig. 4.6 and is the same as for the Green function g(t) considered further below in Sect. 4.3.4 and represented in Fig. 4.12.

4.3 The Forced Harmonic Oscillator 4.3.1 Equation of Motion In the following a simple oscillator under the influence of a driving force f (t) shall be considered. The equation of motion is (4.1)

mDt x ≡ m¨ x + x˙ + kx = f (t)

(4.18)

126

4 Harmonic Vibrations

with the differential operator Dt given by d2 d +  + k. 2 dt dt

mDt = m

This is an inhomogeneous linear differential equation; the inhomogeneity here is the driving force f (t). The general solution x of the inhomogeneous linear differential (4.18) mDt x(t) = f (t) is a superposition of the general solution xhom of the homogeneous equation,8 mDt xhom (t) = 0, with any particular solution xpart of the inhomogeneous equation (4.18), x(t) = xhom (t) + xpart (t). In the case of the damped oscillator the solution (4.13) or (4.14) of the homogeneous equation (4.10) decays in time and after a sufficiently long time is small in comparison to the particular solution. This section thus deals with the construction of a particular solution. 4.3.2 Solution of the Inhomogeneous Linear Differential Equation by Fourier Transformation For linear differential equations there is, in principle, a general procedure, namely that of the Fourier transformation.9 One performs a Fourier decomposition of the driving force,  ∞ dω −iωt f (t) = e F (ω) (4.19) −∞ 2π with 



F (ω) =

dt eiωt f (t).

−∞

For the particular solution one makes thus the ansatz analogous to (4.19)  dω −iωt e x(t) = X(ω) (4.20) 2π

8 9

See Sect. 4.2 above. See Appendix I.

4.3 The Forced Harmonic Oscillator

127

(under omission of the suffix “part”), and by inserting the ansatz into the equation of motion Dt x = x¨ + 2γ x˙ + ω02 x − one obtains 

f (t) =0 m

   dω −iωt  2 1 2 −ω − 2iωγ + ω0 X(ω) − F (ω) = 0. e 2π m

Since this must hold for all times, it follows that the contents of the square brackets must vanish,   1 D(ω)X(ω) := −ω 2 − 2iωγ + ω02 X(ω) = F (ω). m Now one can solve for X(ω) and obtains for the particular solution X(ω) =

1 1 1 F (ω) = G(ω)F (ω) F (ω) = m ω02 − 2iωγ − ω 2 mD(ω)

(4.21)

with the Green function (response function), see Fig. 4.7, G(ω) =

1 1 1 = . mD(ω) m ω02 − 2iωγ − ω 2

(4.22)

Comments:

γ/ω0 = 0.2 γ/ω0 = 1.0 γ/ω0 = 2.0

1

2

Im G(ω) mω02

Re G(ω) mω0

2

• One obtains x(t) from (4.20) with X(ω) from (4.21). • By the use of the Fourier transformation  ∞ dω −iωt f (t) = e F (ω) −∞ 2π

0

−1 −3

1

γ/ω0 = 0.2 γ/ω0 = 1.0 γ/ω0 = 2.0

0

−1 −2

−2

−1

0

ω/ω0

1

2

3

−3

−2

−1

0

ω/ω0

1

2

3

Fig. 4.7. Real part (left panel ) and imaginary part (right panel ) of the Green function (in units of 1/mω02 ) for three different damping constants

128

4 Harmonic Vibrations

– because of the property of the exponential function – the Fourier transform of the time derivatives is particularly simple; with h(t) = df /dt one obtains   d dω −iωt d dω −iωt e e h(t) = H(ω) = f (t) = F (ω) 2π dt dt 2π  dω −iωt e = (−iω)F (ω) 2π and thus h(t) = df /dt



H(ω) = −iωF (ω).

(4.23)

This is one of the reasons why one often prefers to work in frequency space instead of working in time space. • In particular, by the Fourier transformation the differential operator Dt has turned into a multiplicative factor D(ω). • The property G(ω) = G∗ (−ω)

(4.24)

is a very general property of any response function (Green function), which describes the reaction (here the displacement) of a system (here of the oscillator) to an external perturbation (here to the driving force). • The Green function is determined only by the properties of the system (here of the oscillator) and is independent of the externally applied field (here of the force). With the introduction of the Green function the problem has not become more simple, but more general, since with the knowledge of the Green function one can find the particular solution to a differential equation with an arbitrary inhomogeneity. The larger difficulty generally lies in the determination of the Green function and less in the subsequent application. Periodic External Force For a periodic external force with the period T one can expand the force (like any other periodic function) in terms of a Fourier series, f (t) =

∞ 

Fn e−iωn t

n=−∞

with ωn = n and with the expansion coefficients10 10

See Appendix I.3.

2π T

4.3 The Forced Harmonic Oscillator

129

Fig. 4.8. The vectors of the force F (ω) and of the displacement X(ω) (in arbitrary units) in the complex plane. The arrows indicate the evolution with time in clockwise direction

1 Fn = T



T

dt f (t) eiωn t . 0

For the corresponding expansion of the displacement of the oscillator then one obtains analogously Xn = G(ωn ) Fn . Harmonic External Force See problem 4.13. 4.3.3 Green Function of the Damped Oscillator in Frequency Space Since the force f (t) is real, the Fourier transform F (ω) is complex; likewise X(ω) is complex. The Green function (4.22) in frequency space is G(ω) =

1 1 = |G(ω)| eiϕ(ω) . m ω02 − 2iωγ − ω 2

With X(ω) = G(ω) F (ω) the displacement X(ω) has a phase shift ϕ(ω) relative to the driving force F (ω), see Figs. 4.8 and 4.9. The Green function can be written in the form 1 1 m (ω − ω1 )(ω − ω2 ) # $ 1 1 1 1 − =− m ω1 − ω2 ω − ω1 ω − ω2

G(ω) = −

with ω1,2

  −iγ ± Ω (γ < ω0 ) 2 2 = −iγ ± ω0 − γ = −i(γ ± μ) (γ > ω0 ).

(4.12)

130

4 Harmonic Vibrations 0,5

2

1

0 −3

γ/ω0 = 0.2 γ/ω0 = 1.0 γ/ω0 = 2.0

−2

−1

φ/π

|G(ω)| mω02

3

0

ω/ω0

1

2

3

0

−0,5 −3

γ/ω0 = 0.2 γ/ω0 = 1.0 γ/ω0 = 2.0

−2

−1

0

ω/ω 0

1

2

3

Fig. 4.9. The modulus (left panel ) and the phase ϕ ≡ φ (right panel ) of the Green function for various strengths of the damping constant

Fig. 4.10. The position of the poles of the Green function in the complex frequency plane for increasing strength of the damping (from left to right, see also Fig. 4.11) 1

Ω/ω0, −(γ + μ)/ω0

0 −1 −2 −3 −4 0

Ω/ω0 −(γ+μ)/ω0

0, 5

1

γ/ω0

1, 5

2

Fig. 4.11. Position of the real and imaginary part of the poles of the Green function as a function of the damping strength in the complex frequency plane, see also Fig. 4.10

Comments: • The poles of the Green function lie at ω = ω1,2 in the lower complex frequency half-plane and are given by the eigen values ω1,2 of the secular equation, see Figs. 4.9–4.11.

4.3 The Forced Harmonic Oscillator

131

• In the underdamped case (ω0 > γ), with  Ω = ω02 − γ 2 , one can write the Green function in the form 1 1 m Ω 2 − (ω + iγ)2   1 1 1 + = . 2Ωm Ω − (ω + iγ) Ω + (ω + iγ)

G(ω) =

The Green function has poles at the positions ω = ±Ω − iγ. In particular, the real part of the pole is equal to the eigen frequency, and the imaginary part of the pole is equal to the damping constant. • In the overdamped case one has to set μ = iΩ as in Sect. 4.2, and the poles are located at ω = i(−γ ± μ) on the negative imaginary frequency axis. 4.3.4 Time-Dependent Green Function From X(ω) = G(ω)F (ω) (in frequency space) from (4.21) one obtains according to the convolution theorem11 ∞ x(t) =

dt g(t − t ) f (t )

−∞

(in time space) with    dω i 1 g(t) = G(ω) e−iωt = (4.25) θ(t) e−iω1 t − e−iω2 t 2π m ω1 − ω2 ⎧  sin(Ωt) ⎪ 2 2 ⎪ for ω0 > γ ⎪ ⎨ Ωt with Ω = ω0 − γ e−γt  = θ(t) t × sinh(μt) (4.26) with μ = γ 2 − ω02 for ω0 < γ ⎪ m ⎪ ⎪ μt ⎩ 1 for ω0 = γ  with the Heaviside step function12 θ and ω1,2 = −iγ ± ω02 − γ 2 and shown in Fig. 4.12. Proof: The integral  dω g(t) = G(ω) e−iωt 2π 11 12

See Appendix I.1.2 See Appendix H.1.

132

4 Harmonic Vibrations 1

1 γ/ω0 = 0.5

γ/ω0 = 0.2

γ/ω0 = 1.0 γ/ω0 = 2.0

mg(t)/T

0,5 0 −0,5 −1

0

1

2

t/T

3

4

5 0

1

t/T

2

0

0

t/T

1

Fig. 4.12. Green function g(t) (in units of T m−1 ) as a function of time t (in units of the period T = 2π/ω0 ) for different damping constants. The dotted lines are the functions ±e−γt . Note the different time scales. Compare with the displacements in Fig. 4.6

can be evaluated most comfortably with the calculus of residues, usually not known in the early years of a physics student.13 Alternatively the proof can be given by the back transformation: If one inserts the proposed form (4.25) of g(t), one obtains  ∞ G(ω) = dt eiωt g(t) −∞  ∞   i 1 (4.25) = dt eiωt θ(t) e−iω1 t − e−iω2 t m ω1 − ω2 −∞  ∞ ! " i 1 = dt ei(ω−ω1 )t − ei(ω−ω2 )t m ω1 − ω2 0 # $ 0−1 1 i 0−1 = − m ω1 − ω2 i(ω − ω1 ) i(ω − ω2 ) # $ 1 1 1 1 − = − m ω1 − ω2 ω − ω1 ω − ω2 with e−iωj t ∝ e−γt → 0 for t → ∞ (since ωj is complex!).



Comments: • The Green function fulfills the inhomogeneous differential equation   2 d d 2   Dt g(t − t ) ≡ + 2γ (4.27) + ω 0 g(t − t ) = δ(t − t ), dt2 dt 13

For this one needs the knowledge of the theory of complex functions, which is taught generally at a later stage. However, since the proof is a beautiful example for the application of the Cauchy integral theorem, it is given as example (v) in Appendix I.1.1, so it can be read at a later stage.

4.3 The Forced Harmonic Oscillator

133

where the inhomogeneity is given by the Dirac δ-function. (Without proof.) The inhomogeneity of the equation of motion (the force) can be looked at as a superposition of δ functions,  f (t) = dt δ(t − t ) f (t ), such that the particular solution of the linear differential equation can be looked at as a superposition of particular solutions to δ-function-like inhomogeneities,  2   d d 2 + ω + 2γ dt g(t − t ) f (t ) Dt x(t) ≡ x(t) = D t 0 dt2 dt   = dt Dt g(t − t ) f (t ) = dt δ(t − t ) f (t ) = f (t). • For the complete solution for the Green function one still needs initial (and/or boundary) conditions. • The Heaviside step function14 θ in the retarded Green function g(t) of (4.25) expresses the causality (as one of the boundary conditions): Only those times t contribute to the integral for x(t) which are before the considered time t, g(t) = 0

for t < 0

and thus also #

dg(t) dt

$n = 0 for t < 0.

• Since the Green function g(t) describes the reaction of the oscillator, i.e., its displacement u(t), and since this displacement as a physical observable is continuous, also the Green function g(t) must be continuous,15 g(−0) = 0 = g(+0), which can be taken as a first initial condition. The second initial condition is g(+0) ˙ = 1. Proof: From 

14 15

See Appendix H.1. g(−0) ≡ limt0 g(t).

 d2 d (4.27) 2 + 2γ + ω0 g(t) − δ(t) = 0 2 dt dt

134

4 Harmonic Vibrations

one finds



τ

0 = lim

τ →0

−τ

  dt g¨(t) + 2γ g(t) ˙ + ω02 g(t) − δ(t)

  2 = lim [g(τ ˙ ) − g(−τ ˙ )] + 2γ [g(τ ) − g(−τ )] + ω0 τ →0

1

τ

−τ

dt g(t) − 1

= [g(+0) ˙ − g(−0)] ˙ + 2γ [g(+0) − g(−0)] + ω02 × 0 − 1. With g(−0) ˙ = 0, g(−0) = 0, and g(+0) = 0 one obtains g(+0) ˙ = 1.



• The initial conditions of the Green function are the same as those for the oscillator initially at rest, and thus the form (4.26) of the solution for g(t) is the same as that for the displacement of the kicked oscillator. • The derivatives of the Green function of linear differential equations of higher order all vanish at t = +0 except the one of first order. • The differential equation (4.27) for the Green function is of second order and thus has a second solution, the so-called advanced Green function, which has a factor of θ(−t) instead of the factor θ(t) in (4.25). This mathematical solution is unphysical, since it violates the principle of the causality. • Many of the quantities related to each other by causality are interconnected via a convolution. Transformation to the dual space then leads to a problem which can be treated more comfortably there. 4.3.5 Energy Considerations Energy Dissipation In order to arrive at an expression for the energy, one multiplies the equation of motion x ¨ + 2γ x˙ + ω02 x =

f m

with mx˙ (this is typical for the energy conservation theorem) and obtains d ! m 2 m 2 2" x˙ + ω0 x + 2γmx˙ 2 f x˙ = dt 2 2 d = (T + V ) + 2mγ x˙ 2 . dt (F R = −2mγ x˙ is the frictional force.) For the free (not forced) oscillator (f = 0) thus one finds dE = −2mγ x˙ 2 < 0. dt The energy of the oscillator decreases with time (is converted into frictional heat). This process is also called dissipation of energy.

4.3 The Forced Harmonic Oscillator

135

The Stationary State For the forced oscillator in the stationary state (after the transient oscillation have decayed) the energy E of the oscillator averaged over a period remains constant in time, d E = 0, dt and thus 2mγ x˙ 2 = f x . ˙ The energy converted into frictional heat per unit time P R = F R x ˙ = 2mγ x˙ 2 becomes substituted by the power f x ˙ of the driving force. Power The power p(t) of the force f (t) is thus p(t) = v(t) f (t) with the Fourier transform (from the convolution theorem16 )  dω  V (ω  ) F (ω − ω  ) P (ω) = 2π for a force with a general time dependence or, for a force with a periodic time dependence,  Pn = Vm Fn−m , m

respectively; the mean power is (in the periodic case) 1 p t = T



T

p(t) dt = P0 =



0

Vn F−n =

n



Vn Fn∗ .

n

With Vn = −iωn Xn = −iωn G(ωn ) Fn one obtains Re P0 = Re

 n

16

See Appendix I.1.2.

2

−iωn G(ωn ) |Fn | =

 n

2

ωn Im G(ωn ) |Fn | .

136

4 Harmonic Vibrations 2,5

ωω0 Im G(ω)

2 γ/ω0 = 0.2 γ/ω0 = 1.0

1,5 1 0,5 0 −3

−2

−1

0

1

2

3

ω/ω0

Fig. 4.13. Absorbed power of the damped oscillator as a function of frequency

Harmonic External Force For a purely harmonic external force f (t) = F cos(ωt) =

 F  −iωt + eiωt e 2

with the period T = 2π/ω the coefficients of the Fourier series are + Fn = F/2 n = −1, 1 0 otherwise with F real. The mean supplied (respectively dissipated) power is then  Re P0 = ω Im G(ω) |Fn |2 = 12 ω Im G(ω)F 2 . n=±1

Absorption The imaginary part of the Green function thus is connected to the energy dissipation. This is a general result for generalized susceptibilities (as relations between the reactions of systems and the forces causing these reactions). For example, the absorption of light is described by the imaginary part of the dielectric susceptibility, which in the case of the absorption of infra-red light by lattice vibrations has a form which is very similar to the form of the Green function of the simple oscillator. With the Green function of the simple oscillator of Sect. 4.3.3 one obtains the absorption (except for constants thus the absorbed power) as a function of the frequency given by 1

(4.22)

a(ω) = mω Im G(ω) = ω Im =

− − 2iωγ 2ω 2 γ . (ω02 − ω 2 )2 + (2ωγ)2 ω02

ω2

4.3 The Forced Harmonic Oscillator

137

This function in the form of a Lorentz curve is shown in Fig. 4.13. The maximum of the absorption lies at the (undamped) eigen frequency, ωmax = ±ω0 with a(ωmax ) =

1 2γ

with a half width (FWHM)17 of Δω ≈ 2γ

(γ  ω0 ),

defined via a(ωmax ± 12 Δω) = 12 a(ωmax ). Proof: (i) One obtains the maximum from da(ω) da(ω) 2ω 2 γ d = 2ω = 2ω 2 2 2 dω dω dω (ω0 − ω 2 )2 + (2ωγ)2 ⎞ ⎛   2 2 2 2 ω − ω0 + (2γ) ⎟ 1 ⎜ = 4ωγ ⎝ − ω2  2 ⎠ . 2 2 2 2 2 2 (ω0 − ω ) + (2ωγ) (ω02 − ω 2 ) + (2ωγ)

0=

If one introduces the common denominator, from the vanishing of the numerator one has 2    2 2 0 = ω02 − ω 2 + (2ωγ) + 2ω 2 ω02 − ω 2 − (2ωγ)     2 = ω0 − ω 2 ω02 + ω 2 . For real frequencies one obtains thus the maximum at ω = ωmax = ±ω0 . (ii) If one inserts this into the expression for the absorption, one obtain amax = a(ωmax ) =

2 γ 2ωmax 1 . = 0 + (2ωmax γ)2 2γ

(iii) The half width (FWHM) Δω = 2γ one obtains from a(ω) = 12 amax 17

Full Width at Half Maximum; some authors denote the width by γ.

138

4 Harmonic Vibrations

or (ω02

2ω 2 γ 1 = − ω 2 )2 + (2ωγ)2 4γ

or (2ωγ)2 =

1 2

  2 2 ω02 − ω 2 + (2ωγ)

or (2ωγ)2 = (ω02 − ω 2 )2 with the solution ω 2 = ω02 + 2γ 2 ± 2γ

 ω02 + γ 2 ≈ ω02 ± 2γω0

(γ  ω0 )

and ω ≈ ω0 ± γ.



4.4 Coupled Oscillators We shall start with a simple example in this section and then continue with the general case in the following section. 4.4.1 Introductory Example: Stretching Vibrations in the CO2 Molecule In this section we will treat the linear molecule. We will disregard the bending vibrations and concentrate on the stretching vibrations only; we will also neglect the effect of damping. Assuming a Hooke-like behavior the restoring forces are proportional to the change of the binding length. For a stretching vibration the change in length is equal to the difference of the displacements. Either from the consideration, which displacements lead to which forces and along with it to which accelerations, or (more easily) from the Lagrangian function       2 2 L = 12 M u˙ 21 + u˙ 23 + mu˙ 22 − 12 k (u1 − u2 ) + (u3 − u2 ) (for the numbering see Fig. 4.14) one obtains the equations of motion Mu ¨1 = k (u2 − u1 ) m¨ u2 = k [(u1 − u2 ) + (u3 − u2 )] Mu ¨3 = k (u2 − u3 )

4.4 Coupled Oscillators f

139

f

O

C

O

1

2

3

Fig. 4.14. The numbering of the atoms of the CO2 molecule

or, in matrix notation, ⎛ ⎞⎛ ⎞ ⎛ M u¨1 −k ⎝ ⎠ ⎝ u¨2 ⎠ = ⎝ k m u¨3 M 0

k −2k k

⎞⎛ ⎞ 0 u1 k ⎠ ⎝ u2 ⎠ , u3 −k

in short M¨ u = Ku,

(4.28)

This is a system of three coupled linear homogeneous differential equations of second order (with constant coefficients), which can be solved with an exponential ansatz. There are different procedures for the solution of a system of linear homogeneous differential equations: (I) Direct diagonalization (II) Decoupling by clever linear combinations (III) Transformation each of which will be considered in the following. (I) Direct Diagonalization With the ansatz, ⎛ ⎞ ⎛ ⎞ u1 (t) w1 ⎝ u2 (t) ⎠ = ⎝ w2 ⎠ e−iωt , u3 (t) w3

in short u(t) = w e−iωt

(4.29)

with the phase being the same for all “particles” in the differential (4.28) one obtains ⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ M w1 −k k 0 w1 ⎠ ⎝ w2 ⎠ e−iωt = ⎝ k −2k k ⎠ ⎝ w2 ⎠ e−iωt . m −ω 2 ⎝ w3 w3 M 0 k −k Because of the still to be determined values of ω this system represents an eigen value problem. Since this must hold for all times t,

140

4 Harmonic Vibrations



k − ω2M ⎝ −k 0 in short

⎞⎛ ⎞ 0 w1 ⎠ ⎝ w2 ⎠ = 0, −k k − ω2M w3

−k 2k − ω 2 m −k

  K − ω2M w = 0

(4.30)

has to hold. By the ansatz (4.29) for the displacements ui (t) the system of linear differential equations has turned into a system of linear algebraic equations for the amplitudes wi . These equations are homogeneous (like the differential equations). The condition for the solubility of any system of linear homogeneous equations is the vanishing of the secular determinant, ⎛ 2 ⎞ k 0 ω M −k   2 ⎠ k ω 2 m − 2k k 0 = det ω M − K = det ⎝ 2 0 k ω M −k  2 2  2    2 2 ω m − 2k − 2k ω M − k = ω M −k  2    2   = ω M −k ω M − k ω 2 m − 2k − 2k 2    = ω 2 M − k ω 4 M m − ω 2 k (2M + m) # $  2  2 2M + m 2 = ω M − k ω Mm ω − k Mm with the three solutions ω12 = 0 ω22 = k ω32 = k

1 M #

2 1 + m M

$ .

The first part of the eigensolutions thus consists of the eigen values ±ωj (j = 1, 2, 3). Comments: • The eigen values ωj2 are the eigen values of the differential equation (4.28) or of the algebraic equation (4.30). One obtains the eigen value problem of the theory of matrices in the standard form, if one multiplies (4.30) with M−1/2 (which exists, since M is diagonal), ! "! " ! " % − ω21 w % = 0. M−1/2 KM−1/2 − ω 2 1 M1/2 w ≡ K % is symmetrical because of Newton’s reaction • The coefficient matrix K or K law and thus has real eigen values ω 2 .18 18

See Appendix G.3.

4.4 Coupled Oscillators

141

• The stability of the system under displacements from the equilibrium configuration (like for the minimum of the potential of the single oscillator) has the consequence that the eigen values ωj2 not only are real but that also ωj2 ≥ 0 is true. With the knowledge of the eigen frequencies one can determine the eigen vectors as the second part of the eigen solution. For the calculation of the eigen vector belonging to an eigen value one substitutes the eigen value in the eigen value equation (4.30). Case ω 2 = ω12 = 0: The eigen value (4.30) takes the form ⎛ ⎞ ⎛ (1) ⎞ w1 −k k 0 (1) ⎟ ⎝ k −2k k ⎠ ⎜ ⎝ w2 ⎠ = 0. (1) 0 k −k w3 If one writes explicitly for the first row, one obtains " ! (1) (1) = 0, −k w1 − w2 from which one obtains (1)

(1)

w1 = w2 and analogously from the third row (1)

(1)

w3 = w2 . Of course, the eigen vector19 is determined up to a normalization factor N , ⎛ ⎞ 1 w(1) = N (1) ⎝ 1 ⎠ . 1 Case ω 2 = ω22 = k/M : The eigen value equation (4.30) takes the form ⎛ ⎞ ⎛ (2) ⎞ w1 0 k 0 (2) ⎟ ⎝ k k( m − 2) k ⎠ ⎜ ⎝ w2 ⎠ = 0. M (2) 0 k 0 w3 From the first row one obtains (2)

kw2 = 0 19



(2)

w2 = 0.

The three components of the eigen vector are the displacements of the three atoms in the binding direction.

142

4 Harmonic Vibrations j = 1: Translation of the molecule as a rigid unit; all eigen vector components are equal; there is no restoring force, for which reason one has ω = 0 j = 2: Vibration of even parity; upon inversion the vibrational pattern remains unchanged j = 3: Vibration of odd parity; upon inversion all displacements change their sign

Fig. 4.15. The eigen vectors (these are the displacement amplitudes of the atoms) of the three stretching eigen vibrations with j = 1, 2, 3 (from top to bottom) of the CO2 molecule of Fig. 4.14 (2)

The same is obtained from the third row. The second row is, with w2 = 0, ! " (2) (2) (2) (2) k w1 + w3 = 0 ⇒ w1 = −w3 . The eigen vector belonging to the second eigen value ω22 is (up to a normalization factor) ⎛ ⎞ 1 w(2) = N (2) ⎝ 0 ⎠ . −1 2 Case ω32 = k( m +

1 M ):

With the analogous procedure one obtains ⎛ M ⎞ ⎛ (3) ⎞ w1 0 2m 1 ⎜ (3) ⎟ m ⎠ 1 k⎝ 1 ⎝ w2 ⎠ = 0 M M (3) 0 1 2m w3

and from the first and from the third row (3)

w2 = −2

M (3) M (3) w = −2 w3 . m 1 m

The eigen vector is (up to a normalization factor) ⎛ ⎞ m w(3) = N (3) ⎝ −2M ⎠ . m Comments: • One obtains here three so-called normal vibrations (eigen vibrations), which are defined as those vibrations in which all particles vibrate with the

4.4 Coupled Oscillators

143

same frequency and fixed phase relation. In systems with n degrees of freedom one obtains correspondingly n normal vibrations. The corresponding frequencies are the eigen frequencies. • The interpretation of the normal vibrations (eigen vibrations) with the frequencies ωj and eigen vectors w(j) can be taken from Fig. 4.15. • The “vibration” with j = 1 describes the motion of the center of mass. For the vibrations with j = 2 and j = 3 the center of mass is at rest. (The eigenvectors are orthogonal to each other.20 ) The eigen solutions are thus u1,+ (t) = w(1) ,

u1,− (t) = w(1) t,

u2,± (t) = w(2) e±iω2 t , u3,± (t) = w(3) e±iω3 t (ω1 = 0 is a twofold zero!21 ). These are six eigen solutions, from the three equations of second order. The general solution is a superposition of the eigen solutions, u(t) = w(1) (a + bt) +

3 ! 

" aj w(j) e−iωj t + c.c. .

j=2

where c.c. stands for the complex-conjugate expression. One can compare the nonzero frequencies by eliminating the force constant k. The theoretical value for the frequency ratio from this simple model is   2 1 ω3 M 16 m + M + 1 = 2 + 1 = 1.915. = = 2 1 ω2 m 12 M From the experimentally determined frequencies22 of the stretching vibrations CO2 one finds23 1/λ3 = 2, 396 cm−1 and 1/λ2 = 1, 351 cm−1 and thus ω3 1/λ3 2396 = 1.77. = = ω2 1/λ2 1351 The ratio of the frequencies is thus obtained in the correct order of magnitude with an error of approximately 10%. Actually, there is also a coupling between the two oxygen atoms, which has been neglected in the model and which is responsible for the deviation of the theoretical from the experimental value. The force constant for the stretch vibration can thus be determined with an error of approximately 20%. 20 21 22 23

See Appendix G.4. See (F.5). The energy unit customary in infra-red spectroscopy is the wave number λ = 1/λ = ν/c = ω/(2πc). From a very old compilation of spectroscopic data of molecules [12].

144

4 Harmonic Vibrations

(II) “Clever” Linear Combination In systems with inversion symmetry, i.e., in systems, which are transformed onto themselves under space inversion, one can classify the vibrations as those with either even or odd parity, i.e., displacements, which under space inversion are transformed either into themselves or into their negative. If the positions of the particles with the indices i and −i are related to each other by inversion symmetry, one can form the combinations ui ± u−i . One obtains for the CO2 molecule ⎞ ⎛ ⎛ ⎞⎛ M −k u ¨1 − u¨3 ⎝ ⎠⎝u ¨1 + u¨3 ⎠ = ⎝ 0 M u¨2 m 0

0 −k k

⎞ ⎞⎛ 0 u1 − u3 2k ⎠ ⎝ u1 + u3 ⎠ . u2 −2k

From here on the exponential ansatz must be made again for the solution. From the block-diagonal form of the force-constant matrix one can see immediately that there is one vibrations of even and two vibrations of odd parity, (+) (+) 2 = k/M and w1 = −w3 for the vibration with of even and the solution ω+ parity (with j = 2 from above, see also Fig. 4.15) can be read off immediately from the block-diagonal form of the force-constant matrix. For the subsequent evaluation the exponential ansatz can be used again. Comment on Practical Implications: Usefulness of the Symmetry Analysis At this place the symmetry of the system is employed in order to reduce the expense. In the example of the CO2 molecule one has, besides the identity only one further symmetry operation, namely the inversion (identical with a mirror operation). In the case of other molecules as well as of solids often there are essentially more symmetry operations, and the consequence for the symmetry properties of the eigensolutions is the topic of the theory of groups and representations. In most cases, the form of the eigen vectors is not apparent, except possibly for the parity in systems with inversion symmetry. However, for a closed system there are always the rigid translations in the three orthogonal directions with eigen frequencies ω = 0 because of the conservation of the center of mass; in addition the other eigen vectors have to be orthogonal to these eigen vectors of the rigid translation, i.e., keeping the center of mass at rest. (III) Transformation The decomposition in vibrations of even and of odd parity one can also perform by the (orthogonal) transformation24 24

The transformation matrix T is orthogonal, T−1 = TT , see Appendix G.2.

4.4 Coupled Oscillators



b T = ⎝b 0 One obtains ⎛ M T⎝

0 −b 0 b ⎠ 1 0

with

1 b= √ 2

⎞ u¨1 ⎠ ⎝ u¨2 ⎠ M u¨3 ⎛ ⎞ ⎞⎛

m

⎞ ⎛ u¨1 M ⎠ T−1 T ⎝ u¨2 ⎠ = ⎝ = T⎝ m u¨3 M ⎛ ⎞ ⎛ ⎞ u1 −k k 0 = T ⎝ k −2k k ⎠ T−1 T ⎝ u2 ⎠ u3 0 k −k ⎛ ⎞⎛ ⎞ −k 0 0 b (u1 − u3 ) = ⎝ 0 −k 2bk ⎠ ⎝ b (u1 + u3 ) ⎠ . 0 2bk −2k u2 ⎛

145



M

⎞ ¨3 ) b (¨ u1 − u ⎠ ⎝ b (¨ ¨3 ) ⎠ u1 + u u ¨2 m ⎞⎛

M

For the consecutive evaluation the exponential ansatz can be used again. 4.4.2 General Coupled Vibrations The equation of in the form ⎛ m1 ⎝ 0 .. .

motion for a system of N coupled oscillators can be written 0 m2 .. .

⎞⎛ ⎞ ⎛ ... u¨1 a11 . . . ⎠ ⎝ u¨2 ⎠ = − ⎝ a21 .. .. .. . . .

a12 a22 .. .

⎞⎛ ⎞ ... u1 . . . ⎠ ⎝ u2 ⎠ , .. .. . .

in short M¨ u = −Ku.

(4.31)

The displacement vector u contains the 3N components of the N oscillators, and the matrices K and M are (3N × 3N )-matrices. With the ansatz u(t) = w e−iωt one obtains from the system of (homogeneous linear) differential equation (4.31) −ω 2 Mw e−iωt = −Kw e−iωt a system of (homogeneous linear) algebraic equations   K − Mω 2 w = 0.

(4.32)

146

4 Harmonic Vibrations

The solubility condition for the linear homogeneous system of equations (4.32) is the vanishing of the secular determinant,   det K − Mω 2 = 0, from which one obtains the eigen values ωj2 and with these the eigen vectors25 wj ; eigen values and eigen vectors fulfill the equation   K − Mωj2 wj = 0. (4.33) and in combination form the eigen solutions,26 wj e±iωj t

(j = 1, . . . , 3N ).

The eigen vectors fulfill the orthogonality relation27 wi† · M · wj = δi,j and the completeness relation    wj wj† · M = M · wj wj† = M1/2 · wj wj† · M1/2 = 1, j

j

(4.34)

(4.35)

j

i.e., the vectors M1/2 wj are orthonormal. Here, in the completeness relation the tensor product28 wj wj† of the eigen vectors is to be taken. The completeness relation says that each displacement u can be described as a superposition of eigen displacements wj , "  ! † wj wj · M · u . u=1·u= j

Proof: (i) Orthogonality: Since K and M are symmetrical matrices, K† = K



aij = aji ,

one has †  † wj† · K = (K · wj ) = ωj2 M · wj = ωj2 wj† · M, 25 26 27

28

For simplicity we use a lower index here. For multiple zeroes see (F.5). The notation w† stands for the complex-conjugate and transpose of the quantity w; the transpose of a row vector is a column vector and vice versa, see Appendix G.2. The tensor product ab of two vectors a and b (with the elements ai and bj , respectively) is a tensor with the elements (ab)ij = ai bj . Sometimes the tensor product is also written as a ⊗ b. To distinguish the tensor product from the scalar product it is thus important that the scalar product (dot product, inner product) a · b is written with a multiplication dot.

4.4 Coupled Oscillators

147

and thus one obtains from the eigen value equation 0 = wj† · (K − ωi2 M) · w(i) = wj† · M · wi (ωj2 − ωi2 ). This expression vanishes for different frequencies only if the orthogonality relation is fulfilled. For equal eigen values the eigen vectors can be chosen orthogonal. The normalization factor in front of the Kronecker-symbol in (4.34) is chosen to be unity. (ii) Completeness: See problem 4.4.  Comment: For the case of the CO2 molecule see problem 4.4. The general solution is a superposition of the eigen solutions29  u(t) = aj wj e−iωj t + c.c. j

Here, u(t) and wj are vectors, and aj are superposition coefficients. 4.4.3 Normal Coordinates In Sect. 4.4.1 use has been made of the symmetry of the CO2 molecule in order to bring the matrix of the force constant to a block-diagonal form; even though these blocks describe the vibrations of even parity decoupled from those of odd parity, the oscillators within a given parity are still coupled to each other. The diagonalization of the force-constant matrix described by (4.33) has effected a transformation from coupled harmonic oscillators to completely decoupled so-called normal vibrations. One writes the displacements u(t) (the “normal” coordinates with 3N components for N particles) as a (time dependent) superposition of the eigen displacements wj ,  u(t) = wj Qj (t). (j = 1, . . . , 3N ) (4.36) j

The superposition coefficients Qj (t) are the so-called normal coordinates. The reverse of (4.36) is Qj = wj† · M · u.

(4.37)

Proof: wj† · M · u = wj† · M · (4.36)

 j

wj† Qj 

(4.34)

=



δj,j  Qj  = Qj .

j

 29

For multiple zeroes see (F.5).

148

4 Harmonic Vibrations

The equation of motion for a normal coordinate is ¨ j = −ω 2 Qj . Q j

(4.38)

Proof: With (4.37) one obtains ¨ j = w† · M · u ¨ = −wj† · K · u Q j  †  † =− wj · K · wj  Qj  = − wj · M · wj  ωj2 Qj  j

=−



j

δj,j  ωj2 Qj  = −ωj2 Qj .

j

 Comments: • The eigen solutions Qj (t) ∝ e±iωj t of (4.38) are obvious. • The normal coordinates are the starting point for the quantization of vibrations in quantum mechanics.

Summary: Harmonic Vibrations Equation of Motion of Harmonically Coupled Damped Vibrations: (A, B, . . . are (N × N )-matrices, u, w are (N × 1) vectors):   ¨ ≡ A¨ u + Bu˙ + Cu ≡ Ad2t + Bdt + C u = f(t) (with dt ≡ d/dt) Dt u with

 f(t) =

dω −iωt e Fω . 2π

The General Solution is the superposition of the general solution of the homogeneous equation and of a particular solution of the inhomogeneous equation, u(t) = uhom (t) + upart (t) with Dt uhom (t) = 0 With the ansatz uhom (t) = w e−iωt

Problems

149

one obtains   D(ω) w ≡ −ω 2 A − iωB + C w with the solution condition for the N values of ω 2 (secular equation)   det −ω 2 A − iωB + C = 0 A Particular Solution is



upart (t) =

dω −iωt part e Uω 2π

with Upart = [D(ω)] ω

−1

Fω .

Problems 4.1. Normal Coordinates. Prove that the energy of a system of coupled oscillators, expressed by normal coordinates Qj , can be written in the form " ! Q˙ j Q˙ †j + ωj2 Qj Q†j . E = 12 j

4.2. The Harmonic Oscillator. For the harmonic oscillator the force F acting on a particle is proportional to its displacement r from the equilibrium position and oriented oppositely, F = −kr. (a) The One-dimensional Harmonic Oscillator: Set up the equation of motion for a one-dimensional harmonic oscillator, and solve this. From the integration of the energy conservation theorem one obtains  x dx  t − t0 = 2  x0 m [E − U (x )] with x0 = x(t0 ). From this derive x(t). (b) The n-Dimensional Harmonic Oscillator: Given be an n-dimensional harmonic oscillator. By integration of the energy conservation theorem derive an equation equivalent to the case (a). Rationalize your procedure.

150

4 Harmonic Vibrations

4.3. Transverse Elongation of a Spring. Consider the sketched system of a mass m and two equal springs (spring constant k, equilibrium length l0 ) shown in Fig. 4.16. One end of each spring is attached to the mass, and the other end is fixed. The distance of the two fixed ends is 2l. The potential energy of a spring of the length l obeying Hooke’s law is 2

V = 12 k (l − l0 ) . Calculate the restoring forces for a (small) displacement z of the mass in the direction perpendicular to the springs for the cases l = l0 and l = l0 to leading order of z.

Fig. 4.16. Transverse displacements of springs (for problem 4.3)

4.4. Eigenvectors of Coupled Oscillators. Determine the normalization constants N (j) (1 ≤ j ≤ 3) of the eigen vectors w(j) of the CO2 molecule of Sect. 4.4.1, and prove the orthogonality and completeness of the eigen vectors. 4.5. Coupled Pendulums. Two equal pendulums (small amplitudes) are coupled to each other by a harmonic spring. (a) Write down the normal vibrations of the system. (b) Solve the eigen value problem, if at time t = 0 these two pendulums are in the equilibrium positions and the left pendulum has the velocity v. 4.6. Diatomic Molecule. Two atoms of masses m1 and m2 , respectively, (for example Na and Cl or C and O,but also H and H) shall be coupled to each other by a harmonic potential (spring constant f ) see Fig. 4.17. Determine the eigen vibrations in the direction of the connecting line (valence vibrations) by the solution of the coupled equations of motion. Identify the eigen vectors.

M1

f

M2

Fig. 4.17. The model of the diatomic molecule (for problem 4.6)

Problems

151

4.7. Eigen Modes of a Linear Array of Atoms. Consider a linear chain of three equal masses. The masses are coupled to each other and to the walls by springs of equal strength. The springs shall be stretched even in the equilibrium state of the system by the force F . Let the equilibrium distance of the masses be l0 (see Fig. 4.18). (a) Set up the system of the equations of motion for transverse vibrations with small displacements. Here, consider only the interaction of a mass with its nearest neighbors. (b) Solve the equations of motion, and determine the eigen frequencies and eigen vectors for transverse vibrations. Represent the calculated eigen vectors in a graph.

l 1

2

3

l0 Fig. 4.18. The chain of three particles (for problem 4.7)

Hint: It is useful to interpret the system of equations to be solved as an eigen value problem. 4.8. Valence vibrations of the CH4 Molecule. In the equilibrium state of the CH4 molecule the H atoms are positioned at the corners of a symmetrical tetrahedron around the C atom, see Fig. 4.19. To describe the valance vibrations, during which the H atoms move along the lines connecting the edges of the tetrahedron to the central C atom, assume that forces act only between C and H which are described by a potential of the form V (xH i )=

2 1  H f xi − xC 2

C (i = 1, . . . , 4). Here, xH i and x denote the positions of the H atoms and of the C atom, respectively.

(a) Write down the position vector for the center of mass of the molecule, and set up the equations of motion of the single atoms with respect to an arbitrary coordinate origin. Do the equations of motion change upon the transition to the center-of-mass system? (b) Which simplifications of the equations of motion are possible, if only the valence vibrations are to be determined? Give an expression for the equations of motion for the valence vibrations.

152

4 Harmonic Vibrations

H2 H1 C H4

→H

x1

→C

x

H3 Fig. 4.19. The CH4 molecule (for problem 4.8)

(c) Solve the coupled equations of motion, and write down the eigen frequencies and eigen vectors. What influence does the special configuration of the H atoms have on the eigen vibrations? Illustrate the eigen vectors by sketches. 4.9. A Pendulum with Mobile Suspension. A mass m2 bound to a string of length  oscillates in a plane under the influence of gravity. The suspension is formed by a mass m1 which is freely mobile along a line in horizontal direction (like a trolley on a (fixed) crane’s horizontal extension arm and a bucket of concrete hanging from it), see Fig. 4.20.

Fig. 4.20. The pendulum with mobile suspension (for problem 4.9)

(a) Set up the Lagrangian function in appropriate generalized coordinates, and from this derive the equations of motion. Which quantities are conserved? (b) Solve the equations of motion for small angular displacements of the mass m2 , and determine the vibrational frequency. Hint: It is useful first to approximate the equations of motion for small angles and then to separate the independent variables. (c) Discuss the solution, in particular in the limits m1  m2 and m2  m1 . What kind of motion is performed by the mass m1 ? 4.10. The Cycloid Pendulum After Huygens (Patent 1657). Two blocks of cycloidic form constrain the space of motion of a pendulum. Let A be the end point of the line touching the blocks, see the left part of Fig. 4.21.

Problems

153

The parameter representation of a cycloid results from the motion of a point on a circle, if the circle rolls along a straight line, ξ = R (ϕ − sin ϕ), ζ = R (1 − cos ϕ). (a) Introduce the angle ϕ as the generalized coordinate, and determine the coordinates of the point P . (b) Choose the length  of the pendulum such that a most simple coordinate representation results (trigonometry!) (c) Write down the kinetic and potential energy of the point mass as a function of ϕ. Choose the potential energy to vanish at z = −. (d) Introduce a new generalized coordinate q which replaces the old coordinate ϕ, and determine the Lagrangian function. (e) Write down the equation of motion for q. Is there something special with the vibrational period?

z 0 x

z

A

j P

x

Fig. 4.21. Left: The cycloid pendulum. Right: Geometrical construction of the cycloid (for problem 4.10)

4.11. Diatomic Linear Chain. Consider a diatomic linear chain of atoms, sketched in Fig. 4.22, with masses m and M ; a is the lattice constant (i.e., the periodicity length). Between neighboring atoms there is a harmonic force with the spring constant k. In equilibrium the distance of the equilibrium positions x0n and yn0 , respectively, of neighboring atoms is equal to 12 a, see Fig. 4.22.

f

m

M

Fig. 4.22. The linear chain (for problem 4.11)

(a) Set up the Lagrangian function for the system. Use the displacements un = xn − x0n and vn = yn − yn0 out of the equilibrium position x0n and yn0 ,

154

4 Harmonic Vibrations

respectively, as coordinates. Then derive the equations of motion, and compare with the Newtonian equations of motion. (b) Solve the system of differential equations with the ansatz (“Bloch ansatz”) un = Aei(qna−ωt)

and vn = Bei(qna−ωt) ,

(4.39)

respectively, for the two different kinds of atoms, i.e., determine the eigen frequencies and eigen vectors as well as the general solution. (c) The system has infinitely many degrees of freedom, and therefore one obtains a continuum of possible eigen frequencies. Derive the relation between the frequency ω and the wave number q, the so-called “dispersion relation.” (d) Assume at first M > m. Sketch the dispersion relation, and comment on the periodicity of ω(q). What are the frequencies of the two dispersion branches at the points q = 0 and q = π/a? Calculate and sketch the displacement pattern of the atoms at the special points q = 0 and q = π/a. The two branches of the dispersion relation are called “optical” and “acoustic” branch, respectively, why? (e) Calculate the “sound velocity” in the chain. (f) The frequencies of the two branches fill the so-called allowed bands. Between the bands there is a band gap. What kind of wave vector belongs to a (so-called forbidden) frequency outside of the allowed bands? What is the consequence for the position dependence of the wave, if one forces the system with a frequency outside of the allowed bands? Sketch the dispersion relation including complex wave vectors. (g) Now set M = m. How does the dispersion relation change? What would one have obtained, if one had started the calculation with a monatomic chain? 4.12. Monatomic Linear Chain – Transverse Vibrations. Consider a monatomic linear chain at rest at the ends of which a force F acts along the chain outwards (mass m, lattice constant a of the stretched chain, spring constant k). (a) The masses of the chain are now displaced transversely (perpendicular to the chain). Show that in linear approximation the forces along the chain are independent of the displacement. (b) Set up the equations of motion for the transverse displacements, and solve them with the ansatz un (q, t) = ei(qan−ωt) . How does ω depend upon q? (c) Perform the limit to the string (a → 0). How does one have to scale the masses and the force constant k with a?

Problems

155

4.13. Harmonically Driven Oscillator with Friction. The external forces acting on a vibrating body are the frictional force and a periodic force F0 cos(ωt), such that one has the following equation of motion for the body, m¨ x + k x˙ + Dx = F0 sin(ωt). (a) Show that the displacement due to this force is x(t) = x0 sin(ωt + ϕ). (b) Sketch the vibrational amplitude x0 as a function of ω, and discuss the limiting case of vanishing damping (k → 0). (c) Sketch and discuss the phase shift ϕ between the periodic force and the vibrational displacement as a function of ω. (d) How does the result x(t) = x0 sin(ωt + ϕ) correlate to the expression x(t) = x1 cos(ωt) + x2 sin(ωt)? (e) Determine the most general solution of the equation of motion.

5 Central Potentials and the Kepler Problem

In this chapter the investigations of the Sects. 2.3 and 2.2 are applied to the bound1 motion of a particle (of mass m) in a central potential V (r) = V (|r|) = V (r).

(5.1)

Included in this chapter is the case of the relative motion of two particles of a closed system (i.e., without external forces), the interaction energy of which is described by a central potential; this case can be reduced to the case of a single particle, if only the mass m of the single particle is substituted by the reduced mass μ of the pair.2 The specialization to the gravitational potential (i.e., to the Kepler problem) will be made in Sect. 5.2.

5.1 Central Force and Motion in a Plane 5.1.1 Central Potential, Central Force, and Angular-Momentum Conservation Central Force The isotropy of the potential (5.1) leads to a conservative central force F (r) = −∇V (r) = −

dV (r) er = F (r) er dr

(5.2)

and thus the angular momentum3 l (with respect to the center) and the energy4 E are conserved quantities. 1 2 3 4

The free (unbound) motion is treated in the following Chap. 6 on the scattering of particles. See Sect. 2.3. See Sect. 1.6.2. See Sect. 1.7.

158

5 Central Potentials and the Kepler Problem

Alternatively: The system of a particle in a time independent potential is invariant under translations in time, and with this (because of the absence of any, in particular rheonomous, constraints) the energy is a constant of the motion.5 In addition, the system of a particle in a central potential V (r) is invariant under (space) rotations, and with this the angular momentum is a constant of the motion.6 Plane Motion Because of l = r × p = const.



r(t) ⊥ l

and p(t) ⊥ l

the position r and the momentum p always lie in a plane perpendicular to the angular momentum l, i.e., the motion takes place in a plane perpendicular to l. Well-known examples are the motion of planets in the gravitational field of the sun (neglecting the influence of other planets) or that of a satellite in the field of the earth (neglecting the influence of sun, moon, and other planets). The investigation of the generally three-dimensional motions under the action of central forces can thus be reduced to two dimensions.7 (Here the use of plane polar coordinates is then of an advantage, and one obtains, e.g., l = mr2 ϕ˙ for the modulus of the angular momentum.) Conservation of Areal Velocity In the context of the planetary motion in former years one has talked of the conservation of the areal velocity F˙ , which states nothing else than the conservation of the angular momentum l = |l|, l = 2mF˙ = const.

(5.3)

The conservation of areal velocity F˙ says that the area F covered by the radius per unit time is a constant. Even though mentioned in general in the context of the gravitational potential, one has the conservation of areal velocity for any central potentials. Proof: The intuitive derivation of the conservation of areal velocity goes as follows: In a small time interval dt the position vector changes from r to r + dr, see Fig. 5.1. The area covered during this time interval is dF = 5 6 7

1 2

|r × dr|

See Sect. 3.11.3. See Sect. 3.11.5. The vanishing of the coordinate perpendicular to the plane, in which the motion takes place, is then not a constraint but a consequence of the form of the potential.

5.1 Central Force and Motion in a Plane

159

dF dr r

Fig. 5.1. Notations in the context of the conservation of areal velocity

and dF dt

=

1 2

(2.17) 1 = 2 = 12 = 12

˙ |r × r| |r er × (r˙ er + rϕ˙ eϕ )| |r r˙ er × er + r2 ϕ˙ er × eϕ | |0 + r2 ϕ˙ ez | =

1 2

r2 ϕ˙ = const.

Except for the factor 2m this is just the angular momentum.



5.1.2 Central Potential and Effective Radial Potential As obtained in Sect. 5.1.1 the energy and angular momentum are conserved quantities for a central potential,  m 2 m 2 r˙ + V (r) = r˙ + r2 ϕ˙ 2 + V (r) = const. 2 2 l = m r2 ϕ˙ = const.

E = T +V =

(5.4) (5.5)

If one expresses the angular velocity ϕ˙ in (5.4) by the angular momentum l of (5.5), one obtains m 2 l2 r˙ + E= + V (r). (5.6) 2 2mr2 The second term of the right-hand side is the so-called centrifugal potential. If one combines this centrifugal potential with the external potential V (r), one obtains an effective potential Veff (r) = V (r) +

l2 , 2mr2

(5.7)

valid for the radial motion. Instead of (5.6) the energy can thus be written also in the form m 2 r˙ + Veff (r). E= (5.8) 2 An example is plotted in Fig. 5.2.

160

5 Central Potentials and the Kepler Problem 5

0

−1/x 0.12/x2

−5

−1/x + 0.12/x2

0

1

x

Fig. 5.2. The form of the effective potential (full line) l2 1 + 2 x x as the superposition of the true potential (dash-dotted line; here the gravitational (or Coulomb) potential) −1/x and the centrifugal potential (dashed line) l2 /x2 for the case l = 0.1. See also Fig. 5.4 below −

Comments: • Introducing the effective potential one has to remember that it is for just the radial motion. The angular motion has been eliminated by the use of the conservation theorem for the angular momentum. • For the (effectively) one-dimensional radial motion one can use all results of Sect. 2.1 for the one-dimensional motion with the replacement x(t) → r(t). 5.1.3 Central Potential and Trajectory Dynamics of a Particle in a Central Potential Equation (5.8) for the energy can be integrated as in the case of the onedimensional motion in Sect. 2.1. From (5.8) one obtains 1 dt 1 = = ± r˙ dr 2 m [E − Veff (r)] dt = ± 

dr 2 m

[E − Veff (r)]

(5.9)

(5.10)

5.1 Central Force and Motion in a Plane

161

and from there8 

r

t(r) − t0 = ±



r0

dr 2 m

(5.11)

(E − Veff (r ))

with r0 = r(t0 ). From this one obtains the time t(r) as a function of the distance r and from this the distance r(t) as a function of time t, if the mapping r → t is injective. Now (5.5) for the angular momentum – in principle – can also be integrated,  t dt l ϕ(t) − ϕ0 = (5.12) 2 m t0 r (t ) with r(t ) from the inversion of (5.11). Proof: With the angular momentum l = m r2 ϕ˙ one obtains dϕ l = dt mr2 (t) ⇒

ϕ(t) − ϕ0 =

l m



t

t0

dt r2 (t ) 

with ϕ0 = ϕ(t0 ). Kinematics of a Particle in a Central Potential

The functions r(t) and ϕ(t) together serve as a parameter description of the trajectory (with the time t as the parameter). Eliminating t one obtains the trajectory in the form r(ϕ). The form r(ϕ) of the trajectory one can also be deduced directly from the conservation laws, 

r

ϕ(r) − ϕ0 =

dr 

r0

8

l/mr2 2 m

[E − Veff (r)]

.

According to Goldstein [1], Sect. 3.5, the integral dt = ± 

dr 2 m

[E − V (r)]

can be expressed in term of trigonometric functions for elliptical functions for

with

V (r) = a r n

n = −2, −1, 2 , n = −6, −4, −3, 1, 4, 6 .

(5.13)

162

5 Central Potentials and the Kepler Problem

From this one obtains the angle ϕ(r) as a function of the distance r and from this the distance r(ϕ) as a function of the angle, if the mapping r → ϕ is injective. Proof: Eliminating dt from the energy and angular-momentum conservation theorems one obtains (5.10)

dt =

dr 2 m

[E − Veff (r)]

l dt mr2 dr l (5.10)  = 2 mr 2 m [E − Veff (r)]   r l2 dr  . ϕ − ϕ0 = 2m r0 r2 E − Veff (r)







(5.5)

dϕ =

(5.14)

 Determination of the Potential from the Trajectory Reversely, one can determine the potential from the form r(ϕ) of the trajectory with the help of (5.6), # $2 dr l2 l2 − 2mr2 2mr4 dϕ '# $2 ( d 1r l2 1 =E− + 2 . 2m dϕ r

V (r) = E −

(5.15)

For the force one obtains dV er = −F (r) er ( 'dr # $2 l2 dV 2 dr d2 r = F (r) = − − −r dr mr4 dϕ2 r dϕ   l2 1 d2 1r 1 =− + . m r2 dϕ2 r F = −∇V (r) = −

Proof: With d dϕ d (5.5) l d = = dt dt dϕ mr2 dϕ ˙ one obtains for the radial component of the force (from l = mr2 ϕ)

(5.16)

5.2 Kinematics of the Kepler Motion

Fr

'#

$2

l d r−r mr2 dϕ ( # $2 ' 2 # $2 l d r 2 dr = m − −r mr2 dϕ2 r dϕ # $2  2 1  d r l 1 + = −m . mr dϕ2 r

(2.19)

= m¨ r − mrϕ˙ 2 = m

#

l mr2

163

$2 (

 Comment: The final forms in (5.15) and (5.16) will turn out to be practical in the context of the trajectories (conics), along which a particle moves under the influence of the gravitational potential, see the following section.

5.2 Kinematics of the Kepler Motion In this section the investigations of Sect. 5.1 on the motion in a central force field will be specialized to the gravitational field. Again, in this section only the relative motion in the gravitational potential will be investigated. In this section the reduced mass will be denoted again by m. Since, in the case of the motion of the earth around the sun, the mass of the sun (m = 1.99 × 1030 kg) is very much larger than the mass of the earth (m = 5.98 × 1024 kg), thus the reduced mass is nearly equal to the mass of the earth. One has similar conclusions for the motion of the moon (m = 7.34 × 1022 kg) around the earth. One could start from the known gravitational potential m1 m2 V (r) = −G |r1 − r 2 | and deduce the planetary motion. However, in the following we want to start from Kepler’s laws and from those draw conclusions about the existence and the form of the potential. 5.2.1 Kepler’s Laws Kepler’s First Law (1609) The planet trajectories are ellipses, the focus of which is the sun. Comment: In fact, there is an additional effect, the perihelion rotation, which is caused by the influence of the other planets and by relativistic corrections.9 From the fact of the periodic motion one concludes that one has energy conservation for the motion, properly speaking actually only on the time average; but if one assumes in the following that the energy is conserved also locally in time (which would have to be proven), then the force must also be conservative, and thus a potential energy V must exist. 9

See Sect. 5.2.7 further below.

164

5 Central Potentials and the Kepler Problem

Kepler’s Second Law (1609) Conservation of areal velocity: The radius vector (from the sun to the planets) traces equal areas Δa in equal times Δt, a˙ = 0, compare Fig. 5.1. Together with the plane of the motion claimed in the first law this theorem is identical to the angular-momentum conservation theorem,10 (5.3)

˙ l = 2ma. The force must thus be a central force. Together with the first Kepler law the potential energy thus must depend upon the distance only, V (r) = V (r)

(5.17)

Proof: See (1.13) in Sect. 1.4. In plane polar coordinates one has11 F = −∇V (r, ϕ) = −er

∂V 1 ∂V − eϕ . ∂r r ∂ϕ

From F  er follows ∂V = 0. ∂ϕ  Comments: • From the elliptic form of the trajectory one obtains V (r) = −

k r

with

k=

l2 . mp

(5.18)

For the proof of this relation and for the determination of the constant k see (5.24) of Sect. 5.2.3 further below. • The first two laws already suffice to fix the distance dependence of the gravitational potential. Kepler’s Third Law (1619) The cubes of the large half-axes of the different planets are proportional to the squares of the rotation periods, # $2 l k 1 a3 = = x = const. (5.19) = T2 2πm p 4π 2 m For the determination of the proportionality constant see Sect. 5.2.5 further below. 10 11

See Sect. 5.1.1. See the representation of the nabla operator in cylindrical coordinates in Appendix D.

5.2 Kinematics of the Kepler Motion

165

Comments: • Actually this law is true only in the limit of the mass of the planet being small compared to the mass of the sun, i.e., of the reduced mass thus being approximately equal to the mass of the planet, see Sect. 5.2.5. This is equivalent to the approximate assumption that the sun is fixed (in the galaxy or in the universe). • For the constant x in (5.19) to be independent of the mass m of the planets one has to conclude that the constant k in the potential (5.18) is proportional to the mass m of the planet, and because of the equality of action and reaction (actio and reactio) thus proportional to the mass M of the central star, mM (5.20) V (r) = −G r with the gravitational constant12 G = 6.67 × 10−11 m3 kg−1 s−2 .

(5.21)

• With Kepler’s third law then the gravitational potential is fixed (except for the constant G). In fact, Kepler’s laws have been the experimental basis for the derivation of the gravitational potential.13 (Kepler himself had assumed a 1/r force rather than the actual 1/r2 force.) 5.2.2 Polar Representation of the Conics As is well known the trajectories of the particles in the gravitational potential are conics, the focus of which coincides with the center of the force. If one chooses the focus as the origin of a system of plane polar coordinates, then the conics are given by r=

p , 1 + e cos ϕ

Here e is the so-called eccentricity. The form of the conics is (Fig. 5.3) e=0 0 < e1

circle ellipse parabola hyperbola.

Comments: • If one replaces e by −e, the conics are rotated by π. 12 13

A more precise value can be found in Appendix A. See Sect. 1.4

(5.22)

166

5 Central Potentials and the Kepler Problem

b a

r

r

ϕ

ea

j

p

p

b

a

Fig. 5.3. The polar representation of the ellipse (left) and of the hyperbola (right)

• For the bound motion,14 the points of minimum and maximum distance from the force center (perihelion and aphelion, respectively) is given by ϕ = 0 and ϕ = π, respectively, p p and rmax = (|e| ≤ 1), rmin = 1+e 1−e and in the perpendicularly oriented direction (ϕ = π/2) the distance of the ellipse from the focus is equal to the half parameter p (the latus rectum). The large and small half-axes of the ellipse are p p a= , b= √ , (5.23) 1 − e2 1 − e2 respectively. 5.2.3 Determination of the Force and Potential from the Trajectory Determination of the Potential from the Trajectory One obtains V (r) = −

k r

with

k=

l2 mp

as stated in (5.18). Proof: With 1 1 = (1 + e cos ϕ) r p one obtains l2 2mr2 l2 (5.22) = E− 2mp2 l2 = E− 2mp2

(5.15)

V (r) = E −

14



l2 2m

#

d 1r dϕ

$2

l2 2 (−e sin ϕ) 2mp2   1 + e2 + 2e cos ϕ , 2

(1 + e cos ϕ) −

For the unbound motion see Sect. 6.3.2.

(5.24)

5.2 Kinematics of the Kepler Motion

and with (5.22)

2e cos ϕ =

167

2p −2 r

one obtains from this V (r) = E +

 l2  l2 1 . 1 − e2 − 2 2mp mp r

With V (r) → 0 for r → ∞ one obtains not only V (r) = − but also E=−

l2 1 mp r

 l2  1 − e2 . 2mp2

(5.25) 

Comments: • The potential V (r) is determined solely by the parameter p of the trajectory. (The parameter e describes only the particular type and form of the conic.) • Kepler’s third law is not needed here. Determination of the Force from the Trajectory From the form of the trajectory one can infer the form of the (gravitational) force: According to Kepler’s second law the force is a central force, F = Fr (r) er , and as a conservative force it must then have the form F = Fr (r) er . The radial component of the equation of motion is   Fr = m r¨ − rϕ˙ 2 . Looking for the r-dependence of Fr one obtains Fr = −

k r2

with k =

l2 . mp

Proof: One obtains for the force F

(5.16c)

=

# −m

l mr

$2 

 d2 1r 1 + . dϕ2 r

With the polar representation 1 r

(5.16c)

=

1 + e cos ϕ p

(5.26)

168

5 Central Potentials and the Kepler Problem

of the trajectory’s ellipse one obtains d2 1r 1 1 1 = − e cos ϕ = − dϕ2 p p r and thus F (r) = −

k l2 1 = − 2. mp r2 r

Alternatively the force can be obtained from F = −∇V with V from (5.24).  Comment: Analogous to the potential V (r) thus the force F (r) is determined solely by the parameter p of the trajectory. (The parameter e describes only the particular form of the trajectory.) Parameters of the Trajectory and Conserved Quantities In reverse of the foregoing one can relate the parameter of the polar representation of the ellipse to the constants of the motion, i.e., to the energy E and to the angular momentum l, r=

p 1 + e cos ϕ

with l2 p= mk

 and e =

1+

(5.27)

2El2 . mk 2

(5.28)

Proof: One determines p from (5.24) or (5.26) as well as e from p and (5.25).  5.2.4 Determination of the Trajectory from the Potential Kepler’s laws (reversely) can be deduced from the potential V (r) = −k/r: The trajectories are conic curves (5.22) with the parameters from equation (5.28). Proof: With (5.13) for the trajectory of arbitrary central potentials and with V (r) = −k/r one obtains ϕ(r) = √

l 2m



dr  r2 E + kr −

l2 2mr 2

.

5.2 Kinematics of the Kepler Motion

169

Here it is practical to substitute u = 1/r (and du = −dr/r2 ),  du = −  ϕ(u) − ϕ0 2mE 2mk 2 l2 + l2 u − u  du = − √ a + bu + cu2 1 2cu + b GR 2.261 √ arcsin √ (c < 0, 4ac − b2 < 0) = −c b2 − 4ac −2u + b = arcsin √ b2 + 4a with an integration constant ϕ0 and the constants a = mE/l2 , b = 2mk/l2 and c = −1 of the table in [11]. Comment: The minimum of the effective potential is V0 = Veff (r0 ) = −

mk 2 2l2

with r0 =

l2 , mk

such that with E ≥ V0 ⇒ −E + V0 ≤ 0 one has # $2 2mk 2m 8m ac − b2 = −4E 2 − = 2 (−E + V0 ) ≤ 0. l l2 l The reverse yields b 1 2 1 b + 4a sin(ϕ(u) − ϕ0 ) =u= − r 2) 2 *  4a b 1 − 1 + 2 sin(ϕ − ϕ0 ) . = 2 b Choosing the integration constant ϕ0 such that one obtains r = rmin for ϕ = 0 (thus ϕ0 = 12 π), one has ) *  b 4a 1 = 1 + 1 + 2 cos ϕ . r 2 b With this one can identify 2 l2 = b mk  4a e = 1+ 2 = b

p=

8mE 1+ 2 l

#

l2 2mk



$2 =

1+

2El2 . mk 2 

170

5 Central Potentials and the Kepler Problem

One can now classify the trajectories by the energy instead of the eccentricity,15 e = 0 ⇔ E = −mk 2 /2l2 circle 0 < e0 hyperbola

−1/x + l 2/x 2

0

l = 0.0 l = 0.2 l = 0.3 l = 0.4

−5 0

x

1

Fig. 5.4. The form of the effective potential in the case of the gravitational (or Coulomb) potential for different angular momenta.

If one considers the effective potential (5.7)

Veff (r) = −

l2 k + r 2mr2

for the radial motion (plotted in Fig. 5.4) one finds,16 that there is a minimum distance for l = 0. In the case E < 0 there is thus a maximum distance, and one denotes this by a bound state. In the case E > 0 there is no maximum distance; the particle arrives from infinitely large distances and then vanishes there; one denotes this by a scattering or unbound state. For a bound state the ratio of the half-axes is   2|E|l2 b 2 = 1−e = . a mk 2 Comments: • With increasing angular momentum (at fixed energy) or with decreasing energy (increasing modulus of the energy) (at fixed angular momentum) 15 16

See also p. 165. See Sect. 2.1.4.

5.2 Kinematics of the Kepler Motion

171

the ratio b/a approaches unity. In contrast, in quantum mechanics one has spherically-symmetrical situations for vanishing angular momentum. • In the general case V (r) = krn−1 (except for n = 0 and n = 3) the trajectories of the bound motion are not closed, but form rosette-like curves with minimum and maximum distances. The point of nearest approach to the center of the force is different from one revolution to the next (perihelion rotation). One would obtain closed trajectories only if the angle for a full revolution (from one maximum value of the distance r to the next) would be a rational multiple of 2π. It seems to be an exceptional accident that the time variation of the angle and of the distances in the Kepler problem have the same period.17 5.2.5 Trajectory and Rotation Periods First Alternative: Conservation of Areal Velocity Kepler’s third law can be verified with the quantities e and p of the parameter description (5.22). From the conservation of areal velocity one obtains for the rotation period T of the periodic motion πab F l (5.3) ˙ = = F = 2m T T √ (5.23) (5.23) and from this with the parameters a = p/(1 − e2 ) and b = p/ 1 − e2 of the ellipse # # $2 # $2 $2 l l l p 1 − e2 1 a3 3 . = a = = T2 2πabm 2πm 1 − e 2 p2 2πm p One can now relate the trajectory parameter p to the constant k of the potential, see (5.24) or (5.26) # # $2 $2 a3 l l 1 mk k = . = = T2 2πm p 2πm l2 4π 2 m This result no longer depends upon the parameters p and e of the trajectory, thus not upon the energy E or the angular momentum l of the planet, but upon its mass m. If one substitutes the reduced mass, which was named m so far, by the actual expression mM/(m + M ) of the two masses involved, and the constant k by GmM , then G(m + M ) GM a3 = ≈ 2 2 T 4π 4π 2 17

(5.29)

Actually the perihelion rotation does not vanish because of the influence of the other planets and because of effects of General Relativity theory. For the perihelion rotation by a perturbation of the (1/r)-potential see Sect. 5.2.7.

172

5 Central Potentials and the Kepler Problem

is approximately independent of the mass of the planets. Thus one finds that Kepler’s third law is valid only approximately (for a comparably large mass M of the central star). Second Alternative: Integration of the Motion Along the Trajectory Kepler’s third law makes two statements, namely firstly the proportionality between the square of the period and third power of the half-axis and secondly the independence of the proportionality factor from the mass of the planets. The first statement is a consequence of the first and of the second law: From the angular momentum conservation theorem (conservation of areal velocity) l = mr2 ϕ˙



dt =

mr2 dϕ l

(5.30)

one has for a rotation period T

(5.30)

= =

GR 2.554.3

=

GR 2.553.3

=

=

m l



2π 0

2mp l

2



(5.22)

r2 dϕ =

mp2 l

 0



dϕ (1 + e cos ϕ)2

π

dϕ (1 + e cos ϕ)2 0 &π  π   2mp2 1 e sin ϕ && dϕ − − l 1 − e2 1 + e cos ϕ &0 0 1 + e cos ϕ &π ( ' √ 1 − e2 tan(ϕ/2) && 2mp2 1 2 arctan 0+ √ & & l 1 − e2 1+e 1 − e2 0 1 2mp2 1 2 π 4mp2 √ . arctan ∞ = 2 l 1−e l (1 − e2 )3/2 2 1 − e2 (5.23)

(5.24)

If one now inserts also 1 − e2 = p/a and p = l2 /mk, one obtains # T2 =

2πmp2 l

$2 # $3 # $2 2 2πm a l 3 m a = 4π 2 a3 = p l mk k

with k ∝ m. Third Alternative: Mechanical Similarity Reversely, the power of the central potential can be obtained from Kepler’s third law as follows:18 One starts from a potential V (r) = krn−1 , and the Newtonian equation of motion is 18

Compare Landau and Lifshitz [3] Sect. 10.

5.2 Kinematics of the Kepler Motion

m

173

d2 r = −∇V (r) = ∇krn−1 = (n − 1) krn−2 er . dt2

If one scales the time with α and the length with β, t = αt one obtains

r = βr  ,

d2 r  β m 2 = β n−2 ∇ krn−1 2 α dt

or m

d2 r  = α2 β n−3 ∇ krn−1 . dt2

One obtains for the scaled and for the unscaled quantities identical equations of motion, if one chooses α2 = β 3−n . One obtains thus

Thus, Kepler’s third law

(T /T  )2 α2 = = β −n . (a/a )3 β3 (T /T  )2 =1 (a/a )3

leads again to n = 0.

5.2.6 The Laplace–Runge–Lenz Vector In the case of the motion in a potential V (r) = −k/r there is a further conserved quantity besides that of the energy and the angular momentum, namely the Runge–Lenz vector19

Λ=

p×l r − = e eperihelion mk r

˙ =0 Λ

|Λ| = e.

(5.31)

(p is the momentum, l is the angular momentum, eperihelion is the vector from the focus to the perihelion, and e is the eccentricity of the trajectory.) The Runge–Lenz vector thus stands for the constancy of the perihelion direction. (In the more general case V (r) = f (r) with f (r) = r−1 the direction of the perihelion is time dependent.)

19

Often kΛ and sometimes mkΛ (in Goldstein [1], for example) is denoted as the Runge–Lenz vector.

174

5 Central Potentials and the Kepler Problem

−er l Λ r

−er p Λ p x l /mk

Fig. 5.5. On the Runge–Lenz vector

Proof: For the notation see Fig. 5.5. (i) For the derivative with respect to time one obtains ! " ˙ = 1 p˙ × l + p × l˙ − 1 r˙ + r 1 r · r. ˙ Λ mk r r3 With constant angular momentum l˙ = 0 and with the equation of motion p˙ = −

k r r3

one obtains ˙ = − 1 r × l + 0 − 1 r˙ + r 1 r · r˙ Λ mr3 r r3 1 1 1 (B.3) ˙ + 0 − r˙ + r 3 r · r˙ = 0. = − 3 r × (r × r) r r r (ii) Since the Runge–Lenz vector is a constant, it can be evaluated at an arbitrary point of the trajectory, in particular at the perihelion (index 0, r = r0 , etc.). At the perihelion one has ϕ0 = 0 and r˙0 = 0 and thus v 0 = r0 ϕ˙ 0 e0ϕ . With l = r × p = mr2 ϕ˙ ez one obtains at the perihelion p × l = mr0 ϕ˙ 0 e0ϕ × mr02 ϕ˙ 0 ez = m2 r03 ϕ˙ 20 e0r

5.2 Kinematics of the Kepler Motion

and

# Λ=

175

$ mr03 ϕ˙ 20 − 1 e0r . k

The energy is an additional constant of the motion and can be evaluated at the perihelion, where the radial part of the kinetic energy vanishes. With E=

l2 k − 2mr02 r0

l = mr02 ϕ˙ 0 and 2El2 mk 2

(5.28)

= e2 − 1

one obtains k (e − 1) 2E 2E e + 1 . ϕ˙ 0 = l e−1 r0 =

If one inserts for r0 and ϕ˙ 0 in Λ, one obtains the results.



5.2.7 Perihelion Rotation While the nonrelativistic two-body theory yields a fixed perihelion (constant Runge–Lenz vector), a rotation of the perihelion direction of the planet trajectories is actually observed. The origins of it are firstly the influence of the other planets, which have been neglected so far, and secondly the effects of the General Relativity theory. For the planet Merkurium the first effect leads to a rotation by 531 per century and the second by 43 . The qualitative form of the trajectory is shown in Fig. 5.6 in a much exaggerated form. For a small perturbation δV of the gravitational potential one obtains an open trajectory; the position of the perihelion changes from rotation to rotation. One obtains the rotation angle δϕ of the perihelion caused by the perturbation δV  d 2m π dϕ r2 (ϕ) δV (r(ϕ)) + O(δV 2 ). δϕ = dl l 0 Proof: From dϕ dr

(5.14)

=

r2

l 1 √  2m E − V (r) −

l2 2mr 2

176

5 Central Potentials and the Kepler Problem

one obtains the polar angle ϕ as a function of the distance r replacing V (r) → V (r) + δV (r),  r dr l  ϕ(r) − ϕ0 = √ 2m r0 r2 E − V (r) − δV (r) − l2 2mr 2   √ d r l2 = − 2m dr E − V (r) − δV (r) − . dl r0 2mr2

Fig. 5.6. A rosette-type trajectory; the two arrows mark the positions of the perihelion at two consecutive revolutions

In particular one obtains half a rotation (from perihelion to aphelion), if one integrates from rmin to rmax ,   √ d rmax l2 ϕ = −2 2m dr E − V (r) − δV (r) − dl rmin 2mr2 ⎡  √ d rmax ⎣ l2 = −2 2m dr E − V (r) − dl rmin 2mr2 ⎤ 1 2 δV

(r)

− E − V (r) −

l2 2mr 2

+ O(δV 2 )⎦ .

The first term yields just the full rotation angle of 2π for the unperturbed gravitational potential. The second term yields the perihelion rotation angle δϕ. If one substitutes in this term the square root by dϕ/dr, see above, one obtains  √ d rmax 1 1 δϕ = 2 2m dr 2 δV (r)  dl rmin l2 E − V (r) − 2mr 2

Summary: Central Potentials and the Kepler Problem (5.14)

=

=

d 2m dl l d 2m dl l



rmax

dr rmin



177

dϕ 2 r δV (r) dr

π

dϕ r2 (ϕ) δV (r(ϕ)). 0

 In particular one obtains for a perturbation potential of the form δV (r) =

w r2

a perihelion rotation of   d 2m π d 2m π 2πmw 2 δϕ = dϕ r (ϕ) δV (r(ϕ)) = dϕ w = − 2 . dl l 0 dl l 0 l

Summary: Central Potentials and the Kepler Problem ⇒

V (r) = V (r)

F (r) = F (r) er

As conserved quantities one has the angular momentum (5.4)

l = μr2 ϕ˙ and the energy E=

μ 2 l2 μ 2 r˙ + V (r) = r˙ + + V (r) 2 2 2μr2 (5.7) 1 = 2 μr˙ 2

+ Veff (r)

From the energy conservation one finds r˙ =

dr dt



(5.10)

dt =

dr  (2/μ)[E − Veff (r)]

(5.11)



t(r)



From the angular-momentum conservation one finds (5.10)

dϕ =

(5.14)

=

l dt ⇒ ϕ(t) μr2 (t) dr  2 r 2μ[E − Veff (r)]

(5.13)



ϕ(r)

In particular for the gravitational potential (5.18)

V (r) = −

k r

(5.20)

= −G

Mm r



r(ϕ)

r(t)

178

5 Central Potentials and the Kepler Problem

one obtains Kepler’s laws p 1 + e cos ϕ (5.3) l = const. (5.22)

r =

conics conservation of areal velocity

a2 T3

rotation periods/axes

(5.29)

= const.

with (5.28)

p =

l2 kμ

(5.28)

e =

1+

2El2 μk 2

as well as the conserved Runge–Lenz vector (5.31)

Λ =

p×l r − = e eperihelion μk r

Problems 5.1. Kepler’s Laws. Why does a month last for about four weeks? Determine the length of the month from the following quantities: G = 6.68 × 10−11 m3moon kg−1 s−2 , mmoon = 7.34 × 1022 kg, mearth = 5.98 × 1024 kg and earth–moon distance R = 3.84 × 108 m. Which of these quantities is not needed (rationalize)? 5.2. Kepler’s Laws and Circular Orbits. (a) Assume that Kepler’s first law describes the planetary orbits as circles with the sun in the center. Which consequences result for Kepler’s second and third laws? (b) Using Kepler’s modified laws from (a) determine the force underlying the motion of the planets. 5.3. Central Potential and Circular Orbit. Under the influence of a central potential a mass m moves on a circular orbit which passes through the potential center. Determine the potential as a function of the distance from the potential center. 5.4. Trajectories for Motion with Central Force. The energy-conservation theorem (as a first integral of the equation of motion) reads E=

μ 2 L2 r˙ + + U (r). 2 2μr2

To obtain the expression for the trajectory r(ϕ) (in plane polar coordinates) perform the second integration.

Problems

179

5.5. Motion in the Central Field: Determination of the potential from the knowledge of the motion. (a) Determine the radial and azimuthal components of the position vector r, ˙ and acceleration vector r¨ of a moving particle. velocity vector r, Hint: Use plane polar coordinates: Let er = (cos ϕ, sin ϕ) and eϕ = (− sin ϕ, cos ϕ) be the unit vectors in radial and azimuthal direction, respectively. Show that e˙ r = ϕ˙ eϕ and e˙ ϕ = −ϕ˙ er . (b) Which central force has to act on a point mass for the trajectory to be (i) a logarithmic spiral r = eaϕ , (ii) a reciprocal Archimedes spiral r−1 = bϕ (a > 0, b > 0)? Sketch! 5.6. Double Stars. Two bodies (masses M1 and M2 ) rotate around each other at constant distance. (a) Determine the position of the center of mass for a given distance R . What is the position of the center of mass of the moon–earth system, what is it in the earth–sun system? Mmoon = 7.34 × 1022 kg, Mearth = 5.98 × 1024 kg, Msun = 1.99 × 1030 kg Rmoon−earth = 3.84 × 108 m, Rsun−earth = 1.496 × 1011 m. For comparison: radius of the earth rearth = 0.637 × 107 m , radius of the sun rsun = 0.696 × 109 m) (b) Which forces occur in a system with the origin in the center of mass and with one of the coordinate axes along the connecting line between the two bodies? How can one conclude the period of the double-star system from the constancy of the distance? Determine the period for the two systems treated in (a). (c) Which period would an artificial satellite have which is in an orbit close to the earth? (d) Can one imagine a double-star system with relativistic velocities? 5.7. Applications in Astronomy. (a) The parallaxe of the sun yields the average distance between sun and earth as aE = 1.49×108 km. The period of Jupiter around the sun is 11.86 years. How large is the average distance between sun and Jupiter? (b) When sun and Jupiter are in opposition the trajectory radius of Jupiter’s satellite Io appears under an angle of 2.3 . How large is the radius of the trajectory of Io? (c) The period of Io’s orbit around Jupiter is 1.77 days. What is the mass of Jupiter? (d) In the telescope Jupiter is resolved as a disk with an apparent equatorial radius of 38 , if the sun and Jupiter are in quadrature to each other. How

180

5 Central Potentials and the Kepler Problem

large is the radius RJ of Jupiter (oblateness neglected), and thus how large is its average density? How large is the gravitational acceleration at the surface of Jupiter? Astronomical notations: Opposition: A constellation in which, observed from the earth, the elongation (the length difference) between sun and star is 180◦ . Quadrature: A constellation in which, observed from the earth, the elongation (the length difference) between sun and star is 90◦ . 5.8. A particle in the Central Force Field. Determine the orbit of a point mass m in the central force field V (r) = −k/r2 , (k > 0). (a) First, investigate qualitatively the possible forms of the trajectories for the radial motion with the help of an effective potential Veff (r) = V (r) +

l2 . 2mr2

Discriminate the cases l2 /2m > k, l2 /2m < k, E > 0, and E < 0 (angular momentum l, total energy E). (b) Determine the possible trajectories r = r(ϕ) in plane polar coordinates. Discuss and sketch the different trajectories. Hint: 

x dx √ = arc sin 2 2 a a −x

 √ 

x dx = Ar Sinh a x2 + a2

dx x √ = Ar Cosh 2 2 a x −a

5.9. Free Fall into the Center. (a) Under which conditions can a mass reach the center of the field V (r) = −α r−n with nonvanishing angular momentum? Is the number of the revolutions for the free fall into center finite or infinite? (b) Show that the time needed for the free fall into the center is always finite and that the velocity and angular velocity always tend to infinity.

Problems

181

5.10. Geosynchronous Satellite. A (geosynchronous) satellite (without propulsion) is steadily positioned above the same point of the earth’s surface. (a) What is the relation between the angular velocities of the rotation of the earth and the motion of the satellite around the axis of the earth? (b) Write down the angular momentum of the geosynchronous satellite (relative to the center of the earth) using spherical polar coordinates. Show that the satellite can move only in the equatorial plane. (c) Show that the satellite describes a circular orbit, and from the equation of motion determine its radius and the distance of the satellite from the earth’s surface. 5.11. Perihelion Rotation in the Kepler Problem. (a) Starting from the total energy of the unperturbed Kepler problem derive the differential equation for ϕ (r) with V (r) = − λr and (λ > 0) in the form dϕ L 1 =± 2  dr mr 2 λ m E+ r −

L2 2mr 2



(5.32)

(b) Determine the aphelion and perihelion from (a) as an expression in terms of the conserved quantities. Write down corresponding formulations for the parameters e and p of the conic. (c) Using r+ for aphelion and r− for perihelion show that ϕ(r) can be expressed as  dϕ 1 r+ r− =± (5.33) dr r (r − r− ) (r+ − r) (d) For a (1/r) potential aphelion, perihelion, and center of force always lie on a straight line. Prove this claim by integrating the differential equation (5.33). Hint: Use ϕ(r = r− ) = 0 for the integration. (e) Now determine ϕ(r) for the case V (r ) = −

B λ + 2, r r

and discuss the angular difference of successive perihelion positions and the angular velocity of the perihelion rotation as a function of B. (f) What may be the origin of such an additional term in the potential? 5.12. Ballistic Trajectory. Neglecting the friction by the air and the acceleration phases, the trajectory of a ballistic rocket describes a section of an ellipse limited by the earth’s surface (Radius RE ), see the sketch in Fig. 5.7.

182

5 Central Potentials and the Kepler Problem

→ vo h

RE

α

β θ

Fig. 5.7. The ballistic trajectory

(a) Using the Runge-Lenz vector show that the range βRE of the rocket is determined by sin α cos α , tan(β/2) = gRE 2 − sin α v02 where α denotes the initial angle from the vertical direction and v0 the initial velocity. (g is the gravitational acceleration.) (b) Determine the initial angle which leads to the maximum range for a given initial velocity v0 . (Here, 0 < β < π is assumed.) Discuss the cases v0 → 0 and RE → ∞.

6 Collision and Scattering Problems

6.1 Kinematics In this chapter the unbound motion of two interacting particles shall be investigated. Here, we will restrict ourselves to the kinematical aspects, as they are of primary interest in the microscopic regime of physics, in which one cannot observe the dynamics (i.e., the time evolution of the motion) of individual particles. In contrast, the dynamics would be of practical interest for the prediction of the trajectories of planets and moons as well as in particular of artificial satellites. In the following it shall be assumed (if not stated otherwise) • That no external forces act on the two-particle system and • That the interaction between the collision or scattering partners has a range which is small in comparison to the lengths of the paths traveled as sketched in Fig. 6.1. An exception is the r−1 interaction potential, see Sect. 6.3.2 and in particular Sect. 6.4.6.

Fig. 6.1. Sketch of the trajectories of two particles in a scattering event within an interaction range indicated by the circle

184

6 Collision and Scattering Problems

Comments: • With the first assumption the center-of-mass and relative motions are decoupled, and of interest here is only the relative motion (as in the Kepler problem). • With the second assumption the potentials are weak at times long before and long after the scattering process, and in good approximation the initial and final states are straight trajectories (as asymptotes of the hyperbola trajectories in the case of the r−1 -potential), and the trajectories differ notably from a straight line only in regions of small distances between the two partners. • An example for a short-range interaction is that between a (practically point-like) electron and a neutral atom (the latter having a nonvanishing electrical potential practically only in its interior). • In contrast, the Coulomb potential acting between two charges or the gravitational potential acting between two masses is of long range.1 Even though the existence of the interaction potential shall be assumed in this context, its particular form shall not be essential at the moment. Even without the knowledge of the form of the interaction potential one can determine the kinematics of the unbound relative motion to a large extent using the conservation laws for energy, momentum, and angular momentum. 6.1.1 Scattering and Collision In the mechanics of unbound microscopic particles one generally talks of scattering. In the mechanics of unbound macroscopic particles the typical case is that of a collision, e.g., of billiard balls. For the collision of two hard (macroscopic) spheres (with radii R1 and R2 , respectively) one has an ideal so-called hard-core potential  ∞ r < R1 + R2 V (r) = 0 r > R1 + R2 . 6.1.2 Momentum Change and Impulsive Force During a scattering process the momentum p of the particle is changed due to the force F acting on it, p˙ = F . The equation can be integrated over the (short) time interval t1 ≤ t ≤ t2 of the interaction with the result  t2 dt F . (6.1) p(t2 ) − p(t1 ) = t1

The integral on the right side is called the impulsive force. 1

For the corresponding treatment of the scattering process see the Rutherford scattering in the Kepler problem, Sect. 6.4.6.

6.1 Kinematics

185

E2¢,p2¢

E2,p2

Fig. 6.2. The notation in the collision or scattering process. (A more precise sketch one finds in Fig. 6.3)

E1¢,p1¢

E1,p1 Comments:

• In the example of the collision of hard spheres the time interval, during which the force acts, is infinitesimally small. In practically all cases, however, the integral (6.1) is not evaluated; rather, for the analysis of a collision of macroscopic spheres only the conservation of energy, momentum, and angular momentum is employed.2 • In the case of scattering by a potential, the form of the potential will enter the scattering angle3 and the cross-section.4 In reverse, the analysis of the latter can be used to determine the potential. 6.1.3 Laboratory System and Center-of-Mass System In the following the quantities long before and long after the collision or scattering process shall be characterized by unprimed and primed symbols, respectively, see Fig. 6.2. In principle, the kinematics of the scattering or collision process can be described in any inertial system. Particularly useful, however, are either the laboratory system, in which one of the scattering partners (the so-called target) is at rest before the scattering process, or the center-of-mass system, in which the center of mass is at rest. In the laboratory system the velocities v 1 and v 2 of the two collision or scattering partners and the velocity V of the center of mass before the scattering process are v L1 = 0,

v L2 = 0,

VL =

m1 v L1 ; m1 + m2

in the center-of-mass system one has before the scattering process v CM = 0, 1 2 3 4

v CM = 0, 2

See Sect. 6.2 further below. See Sect. 6.3.1 further below. See Sect. 6.4.4.

V CM =

m1 v CM + m2 v CM 1 2 = 0. m1 + m2

186

6 Collision and Scattering Problems

1

1

θ1

θs

b 2

θ2

2

Fig. 6.3. The motion of two particles with repulsive interaction in the laboratory system (left panel ) and in the center-of-mass system (right panel ) (schematic)

The relation between the quantities in both systems is m2 vL m1 + m2 1 m1 =− vL m1 + m2 1

= v L1 − V L = v CM 1 v CM = v L2 − V L 2 or

μv L1 = m1 v CM = −m2 v CM 1 2

(6.2)

with the reduced mass μ. In the following the velocities after the collision or scattering process are to be determined. 6.1.4 Consequences of the Conservation of Momentum Since there are no external forces assumed to act, the total momentum is a conserved quantity, P = p1 + p2 = p1 + p2 . In the center-of-mass system the center-of-mass momentum vanishes, P = 0, and one has before the scattering process p1 = −p2

(6.3)

p1 = −p2 .

(6.4)

and after the scattering process

In the center-of-mass system, the two scattering partners have thus oppositely equal momenta before as well as after the scattering process.

6.1 Kinematics

187

In the laboratory system one has P = p1 + 0 = p1 + p2 .

(6.5)

For the decomposition of the momenta (after the scattering process) into components parallel and perpendicular to the direction of the incoming momentum one obtains p1 + p2 = p1 p1⊥ + p2⊥ = 0. The two angles (θ1 and θ2 in the left part of Fig. 6.3) denote the directions in which the particles travel after scattering process and which are unknown so far. The momentum conservation theorem yields three conditions for the originally six unknown quantities (the three components of the two momenta p1 and p2 ). An additional condition is obtained from the energy conservation theorem.5 Central potentials lead to a motion in a plane and with this to an additional condition.6 Then one last quantity remains unknown, which can be determined from the special form of the potential.7 6.1.5 Elastic and Inelastic Scattering Definition 22. (Elastic and inelastic scattering): An elastic collision or scattering process is a process, in which the (mechanical) energy (of the two collision or scattering partners) is a conserved quantity; in an inelastic collision or scattering process a part of the mechanical energy is converted into other forms of the energy, e.g., into deformation energy, excitation energy of the atomic nucleus or of the electron cloud, or any other type of energy. Comments: • In a completely inelastic collision or scattering process the two scattering partners remain attached to each other after the collision or scattering process. • An example for a completely inelastic collision or scattering process is the capture reaction of a particle hitting an atomic nucleus, where the projectile remains stuck in the nucleus; the reverse process is that of the decay (into two particles).8 If one denotes the kinetic energies (long before and long after the scattering) by p2 p2 and Ei = i , Ei = i 2mi 2mi 5 6 7 8

See the following Sect. 6.1.5. See the Sect. 6.1.6. See the Sect. 6.3.1. Here, one must take account of effects of Special Relativity.

188

6 Collision and Scattering Problems

respectively, then the energy conservation theorem (because of negligible potential energies) reads E1 + E2 = E1 + E2 + Q, where Q is the energy converted into deformation, heat, or other excitation energy. One obtains Q = E1 − E1 + E2 − E2 =

1 1 (p2 − p2 (p2 − p2 1)+ 2 ). 2m1 1 2m2 2

(6.6)

Scattering in the Center-of-Mass System (6.3)

In the center-of-mass system9 one has p1 = −p2 , thus p1 = p2 and analogously p1 = p2 . With this one obtains from (6.6) $ #   1 1 1  2 1  2 p1 − p2 Q= p − p2 + 1 = 1 2 m1 m2 2μ 1 and thus



p1 = p1 p1 = p1

for Q = 0 (elastic scattering process) for Q =  0 (inelastic scattering process).

(6.7)

Comments: • The maximum energy, which can be converted, is Qmax =

p21 2μ

for p2 1 = 0 (completely inelastic scattering process); this is of interest for nuclear reactions, which one induces by collisions (in a linear accelerator or in a storage ring). Notice that this maximum energy is given by the momentum in the center-of-mass system; since there the total momentum vanishes, the total energy can be converted into excitation energy; this is not so in the laboratory system, in which the total momentum does not vanish. Elastic Process in the Laboratory System In the laboratory system10 the target particle is at rest initially, p2 = 0. The energy conservation theorem reads p21 p2 p2 = 1 + 2 . 2m1 2m1 2m2 9 10

Compare Fig. 6.3 (right). Compare Fig. 6.3 (left).

(6.8)

6.1 Kinematics

189

Fig. 6.4. The three cases of scattering: central scattering, no scattering, and oblique scattering (from left to right)

From the momentum conservation theorem p1 = p1 + p2 one obtains taking squares  1 2 1  2   p1 + p2 p1 = 2 + 2p1 · p2 . 2m1 2m1

(6.9)

If one subtracts the two equations (6.8) and (6.9), one obtains  1  2 1 2 p p + 2p1 · p2 = 2m1 2 2m2 2



p1 · p2 =

m1 − m2 2 p2 . m2

Case m1 = m2 : In the case of equal masses of the two scattering partners one obtains p1 · p2 = 0 and from this, cf. Fig. 6.4, either p2 = 0 ⇒ p1 = p1 (no scattering/collision) or p1 = 0 ⇒ p2 = p1 (central collision with billiards) or p1 ⊥ p2 (oblique collision with billiards). Comment: For the oblique collision of equal masses the scattering partners in the laboratory system thus travel at right angles after the collision.11 6.1.6 Consequences of the Angular-Momentum Conservation Central potentials lead to central forces and thus to the conservation of angular momentum; thus the scattering process takes place in a plane. Definition 23. (Impact parameter): The distance of one of the scattering partners from the asymptote of the trajectory of the other scattering partner is the impact parameter.12 11 12

This is modified, if one takes into account the effects of Special Relativity. This is denoted by b in Fig. 6.3.

190

6 Collision and Scattering Problems

The angular momentum is related to the collision parameter. In an elastic scattering process this distance (i.e., the impact parameter) is the same before and after the scattering process because of the angular-momentum conservation; this one can see most easily in the center-of-mass system, see Fig. 6.3; the angular momentum of the system in an elastic scattering process is = b  p i l = bpCM i

CM

(center-of-mass system),

(6.10)

and since the moduli of the momenta are conserved because of the conservation of the kinetic energy (in an elastic scattering process in the center-of-mass system), one has thus b = b .

6.2 Collision of Hard Spheres 6.2.1 Notations In an collision process the contact plane is the instantaneous common tangential plane. The collision normal is the normal of the tangential plane at the point of contact. For a collinear collision the velocity vectors v 1 and v 2 of the two collision partners lie parallel to each other; otherwise the collision is called oblique. For a central collision the centers of the masses of the two collision partners lie on the collision normal before the collision takes place; otherwise the collision is eccentric. Ideally, for a smooth collision the impulsive force has the direction of the collision normal; otherwise it is rough. in the smooth collision of spheres no angular momentum is exchanged. Comment: In a rough collision of macroscopic spheres thus a component of the impulsive force is acting in the tangential plane, and angular momentum is exchanged; in the (ideally) rough collision and at the moment of contact the two collision partners roll against each other without sliding (like tooth wheels). 6.2.2 Elastic Collision of Smooth Spheres (Laboratory System) In this case the scattering angle is fixed by the scattering geometry (and is not calculated from the hard-core potential): In a collision of smooth spheres the impulsive force acts only along the collision normal, and the angular momenta of both spheres are unchanged. Since the impulsive force has only a component along the collision normal, only the momentum component of the two particles along the collision normal is changed, while the perpendicular components remain unchanged. The two particles may have the masses m1 and m2 , respectively; their radii shall be R1 and R2 , respectively. In the laboratory system the particle at rest before the collision thus does not have any momentum component perpendicular to the collision normal

6.2 Collision of Hard Spheres V’1 ❘ V1 ❘

V1❘❘

V’1❘❘

191

V’1 θ1

V1 b

θ2 ➛ V’2

Fig. 6.5. The collision of hard spheres

after the collision; after the collision it thus travels in the direction of the collision normal. Let the angles of the directions after the collision against the incoming direction be θ1 and θ2 (here for simplicity without a prime; the two angles are oriented oppositely, but are both taken as positive), see Fig. 6.5. The impact parameter is b. The momentum and velocity components in the direction of the collision normal (index ) and perpendicular (index ⊥) to it are  = v1 cos(θ1 + θ2 ) v1 v1 = v1 cos θ2  v1⊥ = v1 sin θ2 v1⊥ = v1 sin(θ1 + θ2 )   v2 =0 v2 = v2   v2⊥ =0 v2⊥ =0 The angle θ2 is given by the collision parameter b, sin θ2 =

b R1 + R2

(b ≤ R1 + R2 ).

(6.11)

The conservation of momentum, p1 + p2 = p1 + p2 , leads to

m1 v1 cos θ2 + 0 = m1 v1 cos(θ1 + θ2 ) + m2 v2

for the components parallel to the collision normal and to m1 v1 sin θ2 + 0 = m1 v1 sin(θ1 + θ2 ) + 0 for the components perpendicular to the collision normal. This is supplemented by the energy conservation m1 2 m1 2 m2 2 v = v + v . 2 1 2 1 2 2

192

6 Collision and Scattering Problems

cotθ cotθ2

θ1+ θ2

0

θ2

m1 − m2 cotθ2 m1 + m2

θ

π

π 2

Fig. 6.6. For the solution of (6.12): Determination of θ1 for a given θ2 . θ1 + θ2 is indicated for the case m1 > m2

These are three equations for v1 , v2 , and θ1 depending upon θ2 , b, and v1 . One obtains v2 = v1 v1 = v1 cot(θ1 + θ2 ) =

2m1 cos θ2 m1 + m2 #

m1 − m2 m1 + m2

$2 +

4m1 m2 sin2 θ2 (m1 + m2 )2

m1 − m2 cot θ2 . m1 + m2

(6.12)

Since one has |(m1 − m2 )/(m1 + m2 )| < 1, one always finds 0 < θ1 < π − 2θ2 . For a graphical solution of this equation see Fig. 6.6. Proof: If the terms with (θ1 + θ2 ) of the momentum-conservation theorem are brought to one side, m1 v1 cos(θ1 + θ2 ) = m1 v1 cos θ2 − m2 v2 m1 v1 sin(θ1 + θ2 ) = m1 v1 sin θ2 , taking squares of all sides, and taking the sum of both equations, one obtains m21 v12 = m21 v12 − 2m1 m2 v1 v2 cos θ2 + m22 v22 . If one subtracts from this the energy-conservation equation in the form m21 v12 = m21 v12 − m1 m2 v22 ,

6.2 Collision of Hard Spheres

one obtains

193

0 = m2 v2 (m1 v2 − 2m1 v1 cos θ2 + m2 v2 ).

From this one obtains either the trivial solution v2 = 0 (no collision) or v2 =

2m1 v1 cos θ2 . m1 + m2

From this and from the energy-conservation equation one obtains v12 = v12 − ' =

v12

m2 2 v m1 2

m2 1− m1

#

2m1 m1 + m2

(

$2 2

cos θ2

 4m1 m2 2 (1 − sin θ ) 2 (m1 + m2 )2 '# ( $2 m1 − m2 4m1 m2 2 2 = v1 + sin θ2 . m1 + m2 (m1 + m2 )2  = v12 1 −

If one finally divides the two sides of the conservation of momentum theorem, one obtains m2 v2 1 m1 v1 sin θ2 m2 2m1 = cot θ2 − cot θ2 . m1 m1 + m2 m1 − m2 = cot θ2 . m1 + m2

cot(θ1 + θ2 ) = cot θ2 −

 Case of Equal Masses In the case of equal masses (m1 = m2 , R1 = R2 = R) one obtains v2 = v1 cos θ2 v1 = v1 sin θ2 = v1 θ1 + θ2 =

π . 2

b 2R

After the collision the spheres travel at right angle, see the case m1 = m2 at the end of Sect. 6.1.6.

194

6 Collision and Scattering Problems

Case of a Fixed Scattering Center In the limit of a fixed scattering center (m2  m1 ) one obtains v2 = 0

v1 = v1 θ1 = π − 2θ2 .

(6.13)

(The angles θ1 and θ2 are oppositely oriented. θ1 is the scattering angle.)

q s

y Fig. 6.7. The definition of the scattering angle θ in the center-of-mass system

6.3 Scattering by a Central Potential 6.3.1 Central Potential and Scattering Angle A relation between the polar angle of the trajectory of a particle and the (central) potential one had already obtained in (5.13). For the unbound motion one had found a minimum distance rmin (perihelion) (from the vanishing of the radial velocity r). ˙ Definition 24. (Scattering angle): The scattering angle θ is the angle between the asymptotes of the incoming and outgoing particle, see Fig. 6.7. Comment: For multiple rotation around the scattering center the angle is larger than 2π, but this does not occur for the special case of the r−1 potential. In the center-of-mass system the angle θ is as shown in Fig. 6.7. From the figure one obtains π − θ = 2(π − ψ) and from this the scattering angle θ = 2ψ − π.

(6.14)

Here ψ is the angle between the direction to the perihelion and the direction of the asymptote of the incoming particle,

6.3 Scattering by a Central Potential



(5.13)



ψ =

dr rmin

r2

195

l  2m [E − V (r)] − l2 /r2

With (6.10) and E = p21 /(2m) one obtains this angle as a function of the impact parameter b, 



ψ=

b rmin

r2

dr  . 1 − 2mV (r)/p21 − b2 /r2

(6.15)

Comments: • With (6.14) the scattering angle θ is a functional of the potential V (r), the functional being quite complicated because rmin depends on the potential V . From the dependence of the scattering angle upon the impact parameter the form of the potential V (r) can thus be determined. • But since the impact parameter in experiments with microscopic particles cannot be fixed (there is a whole distribution of them), one investigates the cross-section, in which an average is taken over the distribution of the impact parameter, see Sect. 6.4. 6.3.2 Scattering by the Gravitational Potential For the gravitational potential the scattering angle θ, see Fig. 6.7, is given by tan

k θ = 2 2Es

with k = GmM.

(6.16)

Proof: The kinetic energy of the mass13 m is larger than the potential energy (E > 0), and the mass moves along a hyperbola. For very large distances (r → ∞) the trajectory approaches the asymptote of this hyperbola, see Fig. 6.7. The corresponding polar angle ψ is given by (5.22)

cos ψ = −

1 e

(r → ∞).

(6.17)

The energy and the angular momentum are conserved quantities, and if one evaluates the two quantities for r → ∞, one obtains the relations14 l = msv∞ m l2 E = v 2∞ = . 2 2ms2 13 14

(6.18)

Properly speaking one has to take the kinetic energy of the relative motion with a reduced mass μ. Differently from Sect. 6.1.6, the impact parameter is denoted here by s (and not by b) in order to avoid a confusion with the notation for the small half-axis.

196

6 Collision and Scattering Problems

The scattering angle is the angle θ between the asymptotes. The impact parameter is the distance s of the focus from the asymptote. With (6.14) (6.17) 1 sin( 12 θ) = sin(ψ − 12 π) = − cos ψ = e one obtains sin( 12 θ) 1 1/e = √ =  2 2 cos( 12 θ) e −1 1 − (1/e)  mk 2 (6.18) k (5.28) . = = 2El2 2Es

tan( 12 θ) =



Fig. 6.8. Scattering of a projectile (probe S) from the attractive potential (V < 0, left panel ) and repulsive potential (V > 0, right panel ) (schematic)

6.4 * The Cross-Section 6.4.1 The Problem The motion of the planets or comets (i.e., of macroscopic particles) along their trajectories can be observed, and from this one can infer the form of the potential.15 In contrast, one can neither observe the trajectories of microscopic projectiles, nor can one fix (or even observe) a certain impact parameter experimentally. The problem thus lies in the fact that in general one does not know the impact parameter. Rather, in general a beam of probe particles (projectiles) (with a given distribution of impact parameters) hits on a more or less extended distribution of target particles. The distribution of impact parameters leads to a distribution of scattering angles. One is thus interested in the answer to the question of how many particles are deflected by an angle θ (and 15

See (5.15).

6.4 * The Cross-Section

197

arrive on a detector with the opening angle dΩ). The distribution of the scattered particles over different scattering angles is described by the (differential) cross-section.16 A beam of (“probe”) particles strikes on a substance (“target”). The beam consists of many particles (e.g., of electrons, ions, photons, etc.), which are scattered by the many particles of the target (electrons, nuclei, etc.). The scattering is caused by the interaction between the probe and target particles, which is described by an interaction potential V . If one picks one probe and one target particle, then the probe particle is deflected towards the target for an attractive potential V and away from the target for a repulsive potential, see Fig. 6.8. The deflection is the larger the stronger the potential is; as a rule (for a potential decreasing with increasing distance), the probe particle is thus the less affected the larger the impact parameter is. 6.4.2 Scattering of Many Probe Particles by Many Target Particles If radiation penetrates matter a given projectile can undergo multiple scattering processes. For the following a number of assumptions shall be made: 1. The beam of the incoming projectiles consists of particles of a single kind. 2. The distribution of the target atoms is sufficiently dilute and/or the interaction potential is of sufficiently short range, such that a single scattering process is a good approximation (i.e., a particle scattered once is not scattered again).17 3. The incoming beam is collimated, i.e., the velocities of the incoming particles are parallel. 4. The incoming beam is monochromatic, i.e., the incoming particles all have the same energy (and thus the same velocity). 5. The density of the incoming probe particles is homogeneous. (The leads to an axially symmetric distribution of scattering angles.) 6.4.3 Scattering from a Particle at Rest In the following we shall start in particular from assumption (2), i.e., from the assumption of a single scattering process. For further simplification we shall make the following assumptions in addition: 6. We will use the frame of the target particles initially at rest. 7. The target particle is so heavy or fixed to its surrounding that it is subject to a negligibly small recoil. (Then the laboratory and the center-of-mass systems are identical, and one deals with the motion of a projectile around a fixed force center.) 16 17

See Sect. 6.4.4 further below. In quantum mechanics this approximation is known as the Born approximation.

198

6 Collision and Scattering Problems

8. For sufficiently large distances of the probe particle from the target the interaction energy is so weak that it is a good approximation to consider the motion of the probe particle as free long before the scattering process (t → −∞) and long after the scattering process (t → +∞). 9. The target has no internal degrees of freedom, to which energy of the probe particle can be transferred. This is thus an elastic scattering process with E CM (t → −∞) = E CM (t → +∞) and thus vi = vs . 10. A central potential V (r) = V (|r|) = V (r) shall be assumed. (The above-mentioned hard-core potential is an example.) Then the angular momentum conservation theorem applies: The trajectory of the probe particle lies in a plane (the so-called scattering plane),

y

y f q

z

x

Fig. 6.9. Axial symmetry and notation of the angles in the scattering process

and with the angular-momentum conservation the distance of the target from the asymptote of the trajectory before the scattering process (i.e., the impact parameter) is equal to the one after the scattering process. Namely, the angular momentum (perpendicular to the plane of motion) is bvi = bvs . The problem has axial symmetry in addition, i.e., for a given impact parameter b the scattering angle θ is independent of the azimuth angle φ, see Fig. 6.9. 11. The interaction potential V (r) is a monotonic18 function of the distance r (like the gravitational potential). 6.4.4 The Differential Cross-Section Particles with different impact parameters b are scattered with different scattering angles θ. A measure for the number of the particles, which are scattered into a certain space angle element dΩ, is the differential cross-section 18

For nonmonotonic interaction potentials see Goldstein [1] Sects. 3–10.

6.4 * The Cross-Section

d2 N/(dt dΩ) df dσ = 2 = . dΩ d N/(dt df ) dΩ

199

(6.19)

d2 N/(dt dΩ) is the number of particles, which are scattered into the space angle element dΩ in the time unit dt, and d2 N/(dt df ) is the particle flux density of the incoming particles (i.e., the number of the particles, which pass through an areal element df per time unit dt.) All particles which pass through an areal element df = b db dϕ per unit time are scattered into a space angle element dΩ = sin θ dθ dϕ. This can be inverted, since θ is monotonic in the impact parameter b because of the assumption (11): All particles, which are scattered into a space angle element, have passed through an areal element df .19 For the notation see Fig. 6.10.

b

b

Θ db



Fig. 6.10. Notation for the differential cross-section with one trajectory drawn

In order to arrive at an unambiguous20 relation, one makes the additional assumption: 12. The maximum scattering angle is π. Now one can write the differential scattering cross-section (6.19) as & & & & & & & df & & b db dϕ & & & dσ &=& & = b & db & (6.20) = && & & & dΩ dΩ sin θ dθ dϕ sin θ & dθ & with dθ/db from (6.14) and (6.15). (Typically the impact parameter b decreases with increasing impact parameter b; for the case db/dθ < 0 one has thus to take the modulus of it.) 19

20

For nonmonotonic interaction potentials there are more impact parameters bi , which lead to the same scattering angle θ; the scattering cross-section is then a sum over the different contributions; see Goldstein [1] Sects. 3–10. Sufficiently strong attractive potentials can lead to the fact that the trajectory of the probe particle winds (possibly multiply) around the target particle; then there are again more impact parameter bj to a given angle θ; see Goldstein [1] Sects. 3–10.

200

6 Collision and Scattering Problems

Comment: The dependence θ(b) upon b is not directly apparent; but since this dependence is a functional of the potential energy V (r), the differential scattering cross-section is thus a functional of V (r), and from the dependence of the differential cross-section upon the scattering angle θ one can infer the form of V (r). 6.4.5 The Total Cross-Section The total scattering cross-section is defined as  dσ . σtot = dΩ dΩ With the differential scattering cross-section of Sect. 6.4.4, with the assumptions (1) to (11), and with the space angle element dΩ = sin θ dθdϕ (in spherical coordinates) one obtains & &  bmax  2π  π b && db && = 2π dϕ sin θ dθ b db = πb2max . σtot = & dθ & sin θ 0 0 0 Comment: If the interaction potential does not vanish outside of a maximum distance, the total cross-section is divergent.21 This is true in particular for the gravitational potential. 6.4.6 The Rutherford Cross-Section With the relation (6.16) between the impact parameter b and the scattering angle θ in the case of the gravitational potential one obtains for the differential scattering cross-section of (6.20) dσ = dΩ

#

k 4E

$2

1 . sin4 12 θ

(6.21)

This scattering cross-section was originally derived by Rutherford for the scattering of (charged) α-particles by the Coulomb potential of (charged) atomic nuclei (with the nuclear charge Ze and the charge 2e of the α-particle, the constant of the potential is k = 2Ze2 /4πε0 ). Proof: The relation between the impact parameter b and the scattering angle θ is k 2E tan θ2 k = − . 4E sin2 θ2

(6.16)

b = ⇒ 21

db dθ

The calculated quantum-mechanical scattering cross-section is finite also for potentials, which do not have a cut-off radius, but decrease sufficiently fast (faster than r −2 ).

6.4 * The Cross-Section

With

θ θ cos 2 2

sin θ = 2 sin one obtains dσ dΩ

201

& & & 1 ds & k 1 k &= = &&b sin θ dθ & 2E tan θ2 2 sin θ2 cos θ2 4E sin2 # $2 1 k = . 4E sin4 θ2

(6.20)

θ 2

 For the total scattering cross-section one obtains  σtot =

dσ = dΩ dΩ

#

k 4E

$2 







π

sin θ dθ

0

0

1 sin4

θ 2

As an exception, one does not substitute t = cos θ here, but # # $ $ θ θ θ θ sin θ dθ = 2 sin cos dθ = 4 sin d sin . 2 2 2 2 With the substitution x = sin θ2 one obtains # σtot = 8π

k 4E

$2 

1 0

dx = −4π x3

#

k 4E

$2

&1 1 && . x2 &0

The total scattering cross-section diverges; the reason is that the scattering angle even for arbitrarily large impact parameters (small θ) does not vanish. The gravitational potential is of “long range.” 6.4.7 Scattering Cross-Section for a Collision of Hard Spheres The differential scattering cross-section is dσ = dΩ

1 4

2

(R1 + R2 ) ,

independent of the scattering angle. Proof: With (6.11)

b = (R1 + R2 ) sin θ2 in the laboratory system and (6.13) 1 2π

θ2 =

− 12 θ1

202

6 Collision and Scattering Problems

for a fixed collision center (m2  m1 ) one finds b = (R1 + R2 ) sin( 12 π − 12 θ1 ) = (R1 + R2 ) cos( 12 θ1 ) & & & db & 1 1 & & & dθ1 & = 2 (R1 + R2 ) sin( 2 θ1 ) and thus dσ dΩ

& & & b db & & = && sin θ1 dθ1 &

(6.20)

=

(R1 + R2 ) cos 12 θ1 1 (R1 + R2 ) sin( 12 θ1 ) 2 sin( 12 θ1 ) cos( 12 θ1 ) 2

=

1 4 (R1

+ R2 )2 . 

R1

R2

Fig. 6.11. For the total scattering cross-section for a collision of hard spheres

The total scattering cross-section is then  dσ σtot = dΩ = π(R1 + R2 )2 . dΩ This result expresses the fact that all those particles are scattered whose impact parameter lies within the area π(R1 + R2 )2 , see Fig. 6.11.

Summary: Collision and Scattering Problems Laboratory and center-of-mass system: (6.2)

μv L1 = m1 v CM = −m2 v CM 1 2 Scattering off a central potential (impact parameter b): scattering angle  ∞ dr (6.15)  θ = |2ψ − π| with ψ = b 2 1 − 2mV (r)/p21 − b2 /r2 rmin r

Problems

203

scattering angle (V (r) = −k/r) tan

θ 2

(6.16)

=

k 2Es

differential cross-section dσ dΩ

(6.19)

=

df d2 N/dt dΩ = 2 d N/dt df dΩ

(6.20)

=

b sin θ

& & & db & & & & dθ &

Rutherford scattering cross-section (V (r) = −k/r) # $2 dσ (6.21) k 1 = 4 1 dΩ 4E sin 2 θ

Problems 6.1. Energy Transfer in the Laboratory System. A target of mass m2 at rest is hit by a particle of mass m1 and kinetic energy E. Which fraction of this energy is at disposal for the physically interesting interaction energy? Discuss the limits m1  m2 and m1 ≈ m2 . 6.2. Scattering of an α-Particle under Nuclear Excitation. An α-particle of mass mα and velocity v α collides with a nucleus of mass mK at rest; it excites the nucleus and is deflected by an angle β from its original direction. The excitation energy is Q. (Given are mα , v α , mK , β, and Q.) How large is the velocity v α of the α-particle after the collision in the laboratory system? 6.3. Particle Decay. A particle with the energy E (internal energy) decays spontaneously (i.e., without external influence) into two fragments “1” and “2.” (a) Give an expression for the energy conservation for the decay in the centerof-mass system. Write down the “decay energy.” This is that energy which is converted into kinetic energy in the decay process. (b) Let v be the velocity of the decaying particle in the laboratory system, u1 the velocity of the fragment “1” in the center-of-mass system, and v 1 that in the laboratory system. Discuss v 1 as a function of v and u1 . First, assume v > u1 , and then investigate the case v < u1 . Display your result in a graph. (c) The energy of fragment “1” in the laboratory system depends upon the angle δ between v and v 1 . Discuss the maximum and minimum kinetic energy Tmax and Tmin of particle “1” as a function of the mass ratio x = m2 /m1 . Let the reaction energy ε and the kinetic energy E = 12 M V 2 of the compound particle be constant. Consider in particular the cases x → 0 and x → ∞. (d) What is the main difference a decay into two or into more particles?

204

6 Collision and Scattering Problems

6.4. Two-Body Collision in the Laboratory and in the Center-ofMass System. Two particles of mass m1 and m2 collide elastically, where the particle 2 shall be at rest before the collision (laboratory system). After the collision particle 2 moves at an angle φ with respect to the direction of the incoming particle 1 in the laboratory system (see left and right part of Fig. 6.12). Show that φ can be expressed by the scattering angle θ∗ in the center-of-mass system according to φ=

π − θ∗ . 2

Hint: Use a vector relation between the momentum in the center-of-mass and in the laboratory system. s

m1

m1

r

Θ s

Θ*

Θ*

Φ m2

m2

Fig. 6.12. Collision in the laboratory system (left) and in the center-of-mass system (right) (for Problem 6.4)

g s

g

Q R

Fig. 6.13. Collision with the sphere (for Problem 6.5)

6.5. Scattering from an Ideally Reflecting Sphere. Discuss the elastic scattering of point masses from a fixed sphere of radius R (see Fig. 6.13). (a) Derive the relation between impact parameter s and scattering angle θ. The reflection law “incoming = outgoing angle” shall be assumed.

Problems

205

(b) Determine the differential scattering cross-section σ(θ). (c) Determine the total scattering cross-section σtot (θ). 6.6. Rutherford Scattering. A particle of charge q1 is scattered from a heavy center of force of charge q2 (Rutherford scattering). ˙2 (a) Find the relation of the energy E = m 2 r + V (r) to the angular momentum 2 l = mr ϕ˙ of the particle and the impact parameter s. This latter is defined as the perpendicular distance between the center of force and the incoming velocity. (b) Starting from the energy and angular-momentum conservation theorems, find a differential equation for the trajectory r(ϕ) of the particle. (c) Solve this differential equation (the result must be a conic), and determine the parameter of this solution from the q1 , q2 , E, and the impact parameter s. (d) Discuss the form of the possible trajectories as a function of the eccentricity ε=

# $1/2 2El2 1+ . mq12 q22

(e) Determine the scattering angle. (f) Discuss the difference between the trajectories for q1 q2 > 0 and q1 q2 < 0. 6.7. A Scattering Problem. Discuss the scattering process of point-like particles from a spherically symmetric potential of the form  −V0 for r < a V (r) = 0 for r > a with V0 > 0. (a) Discuss the scattering process qualitatively, and make a sketch for the trajectory of the particle. Include the scattering angle θ and the impact parameter s in the sketch. (b) Where on the trajectory does a force act on the particle? Determine the relation between the directions of the momentum for r > a and r < a using the energy E of the particle and the potential V (r). (Which momentum component changes?) (c) Write down the scattering angle using the results of part (b). Determine s(θ) and the differential scattering cross-section. Discuss the result. How large is the total scattering cross-section? (d) To which extent is there an analogy between this mechanical problem and the scattering of light from a glass sphere?

7 Moving Reference Frames

In this chapter we will treat reference systems which move relative to each other as well as the transformation of the laws of motion of one of these systems into the other. In the context of systems in motion relative to each other we have to distinguish between two inertial systems on the one hand, which are defined in the first Newtonian axiom and which move with constant relative velocity, and those systems on the other, which are accelerated relative to an inertial system and in which also so-called inertial forces appear in addition to the inner and/or external forces. Let us consider two coordinate systems S and S . The system S is assumed to be an inertial system; for simplicity one could imagine this system as the system of the observer (i.e., as that of the reader and as at rest). The system S moves relative to the system S. This motion of the system S is a superposition of a translation of the origin and of a rotation with respect to the origin as indicated in Fig. 7.1. The two systems S and S are thus distinguished by a generally time dependent rotation and translation. In the following first the time dependent

z S

y y



x z

x

Fig. 7.1. The motion of a system S relative to a system S. The path of the origin of the system S in the system S is represented by the broken line

208

7 Moving Reference Frames



S¢ R

P

r

S Fig. 7.2. Coordinates of a point P in two different coordinate systems S and S

translation1 and then the rotation2 will be treated as well as the transformation of the equations of motion from the frame S to the form valid in the frame S . As an important special case the translation with constant relative velocity will be treated finally, and in particular from the point of view of Special Relativity theory.3

7.1 Translations 7.1.1 The Transformation The position of the origin of the system S (relative to the origin of the system S) is given by the (generally time dependent) vector R. A point P is described by the position vector r in the inertial system S and by the position vector r in the system S , see Fig. 7.2. Then one has r(t) = r  (t) + R(t).

(7.1)

Comments: • The vectors r and r describe the same point in different coordinate systems. ¨ = 0 are special: If S is an inertial • Among the translations those with R  system, then also S is an inertial system.4 • If one takes account of the fact that the force on a particle (at position r) is caused by other particles (at the positions r i ), then (because of the homogeneity and isotropy of space) this force is dependent only upon the relative positions r − r i of the particles,  F i (r − r i ). F (r) = i 1 2 3 4

See Sect. 7.1. See Sect. 7.2. See Sect. 7.3. This case is treated in Sect. 7.3.3.

7.2 Rotation Around a Fixed Point

The relative positions

209

r − ri = r − ri

are invariant under the translation (7.1), and thus the force F is invariant under a translation, (7.2) F (r) = F  (r  ). • Successively performed translations commute, R = R1 + R2 = R2 + R1 , i.e., the succession of the translations is arbitrary. 7.1.2 Inertial Forces The equation of motion in the system S is m¨ r=F ¨ ¨ = r¨  + R) and in the system S (with r ¨ m¨ r = F − mR.

(7.3)

Comments: • An acceleration of the system S leads thus (in S ) to an additional force, ¨ this force does not oriented oppositely to the relative acceleration (−R); appear in the inertial system S. Such additional forces in noninertial systems are called inertial forces.5 (The force, with which one is pressed against the back in an accelerated vehicle, is such an inertial force.) • Even though the force is invariant under a translation according to (7.2), the Newtonian equations of motion are in general not invariant under a translation, as (7.3) shows. ¨ = 0 (Galilean transformation) will be taken up in • The special case of R Sect. 7.3.3.

7.2 Rotation Around a Fixed Point 7.2.1 Active and Passive Rotation The active rotation is the rotation of a vector in a fixed coordinate system. The passive rotation is the rotation of a coordinate system, i.e., the representation of the same vector in different coordinate systems. Here we will treat the passive rotation. 5

Inertial forces appear not only in some translated systems but in particular also in all rotating systems, see Sect. 7.2.5.

210

7 Moving Reference Frames

ey ey r

y

ex x′

y′ x

ex

Fig. 7.3. The representation of a vector r in two different coordinate systems (in R2 ) rotated against each other by an angle ϕ

The transformation of the coordinates of a point from one coordinate system to that of a rotated coordinate system is described by an orthogonal transformation U,6 r = U · r. (7.4) Comments: • If the rotation angles are time independent and if the vectors r 0 and v in the translation vector R = r0 + vt of Sect. 7.1.1 are time independent, then one speaks of a general Galilean transformation, r = U0 · r  + r 0 + vt

with U˙ 0 = 0 and r˙ 0 = 0.

(7.5)

• The specialty of the special Galilean transformation is that the coordinate systems are not rotated against each other, U0 = 1 (as well as r 0 = 0). • For a rotation in R2 one has # $ # $# $ x cos ϕ sin ϕ x = y − sin ϕ cos ϕ y as in Fig. 7.3, and thus the transformation matrix for this rotation is # $ cos ϕ sin ϕ U= . − sin ϕ cos ϕ

7.2.2 Infinitesimal Rotations Consider two systems S and S. The system S shall have the same origin as the inertial system S, but it rotates with the (generally time dependent) angular ˙ relative to the system S. Consider a fixed position vector r in velocity ω = ϕ the rotating system S . In an infinitesimally small time interval dt the system S rotates by an angle dϕ, and the vector r in the system S changes by 6

More details are given in Sect. 8.7.2.

7.2 Rotation Around a Fixed Point

211

Fig. 7.4. On the coordinate transformation for infinitesimal rotations; see also Fig. 1.11

(1.35)

dr = dϕ × r 

(7.6)

(infinitesimal rotation), see Fig. 7.4. Comments: • From (7.6) one has

˙ × r . r˙ = ϕ

• Infinitesimal rotations commute. • Finite rotations generally do not commute. Proof: Consider two consecutively performed infinitesimal rotations dϕ1 and dϕ2 , which lead to changes dr 1 = dϕ1 × r 1 dr 2 = dϕ2 × r 2

with r 2 = r1 + dr 1 .

The total change of r is dr = dr1 + dr 2 = dϕ1 × r1 + dϕ2 × r2 = dϕ1 × r1 + dϕ2 × (r 1 + dϕ1 × r 1 ) = (dϕ1 + dϕ2 ) × r 1 + dϕ2 × (dϕ1 × r 1 ). In the first term the order of the rotations is arbitrary. The last term shows that the order of the rotations is not arbitrary, but for infinitesimally small rotations is small of second order and in this case can be neglected.  7.2.3 Representation in Different Coordinate Systems The quantities position, momentum, etc. are represented differently in different systems. Consider two coordinate systems S and S with the same origin. The inertial system S with the axes ei rotates with the angular velocity ω relative

212

7 Moving Reference Frames

to the system S with the axes ei . With (7.6) one has thus for the axes in the coordinate system S (S)

e˙ i

(S) e˙ i (S ) e˙ i

=0 = ω × ei

(S )

= 0.

The same position vector r is represented differently in the two systems, depicted in Fig. 7.3,  xi ei = r (S) r= i

=





xi ei = r (S ) .

i

Comments: 

• Notice that the vectors r(S) and r(S ) with different upper indices S and S , respectively, are the same quantity and that the index only denotes the different representation of the same quantity in different coordinate  systems, see also Fig. 7.2, where r ≡ r (S ) . • The vector ω describes an instantaneous axis of rotation. For the velocity one obtains with (7.6) v = r˙ (S)   = r˙ (S ) + ω × r (S ) .

(7.7)

Proof: In the two representations with e˙ i = 0 one obtains v = r˙ (S) =

 d  xi ei = x˙ i ei dt i i

and, with e˙ i = ω × ei , v=

    d       x˙ i ei + xi e˙ i = xi ei = (x˙ i ei + xi ω × ei ) dt i i i 



= r˙ (S ) + ω × r (S ) .  Comments: • The second term in (7.7) is the velocity (in system S) of a point resting in the rotating system S , see (7.6); the first term is the additional velocity, with which a point moves in the system S .

7.2 Rotation Around a Fixed Point

213

• One can abbreviate the relation between the time derivatives in the two systems by the operator identity #

$(S) # $(S ) d d ... ... + ω × ... = . dt dt

(7.8)

For the acceleration one obtains analogously 







r¨(S) = r¨(S ) + ω × (ω × r (S ) ) + 2ω × r˙ (S ) + ω˙ × r (S )

(7.9)

or alternatively '#

d dt

$(S) (2

'# ... =

d + ω× dt

$(S ) (2 .... 

Proof: With the result for the velocity one obtains (with r  = r(S ) )  d   r˙ + ω × r dt ' (  d      = x˙ e + ω × xi ei dt i i i i

r¨ =

=

      x ¨i ei + x˙ i e˙ i + ω × x˙ i ei + xi e˙ i + ω˙ × xi ei i

=



[(¨ xi ei + x˙ i ω × ei ) + ω × (x˙ i ei + xi ω × ei ) + ω˙ × xi ei ]

i

     ¨ + ω × r˙  + ω × r˙  + ω × r  + ω˙ × r = r ¨  + ω × (ω × r ) + 2ω × r˙  + ω˙ × r  . =r Alternatively # r¨ =

d + ω× dt

$(S )



r˙  + ω × r 



     ¨ + ω˙ × r  + ω × r˙  + ω × r˙  + ω × r = r = r¨ + ω × (ω × r  ) + 2ω × r˙  + ω˙ × r .  Comment: Notice ω˙ = 0.

214

7 Moving Reference Frames

7.2.4 Uniformly Rotating System: Centrifugal and Coriolis Force If one uses (7.9) in the Newtonian equation of motion, one obtains ω

ωxr r

–ω ´ ( ω ´ r )

Fig. 7.5. On the centrifugal force

m¨ r(S) = F   m¨ r (S ) = F − mω × (ω × r(S ) ) 

− 2mω × r˙ (S )

(7.10) 

with ω˙ = 0. This shows that in the rotating system S inertial forces appear, see Fig. 7.5, namely the centrifugal force, 





−mω × (ω × r (S ) ) = mω 2 r(S ) − mω(ω · r (S ) ) and the Coriolis force

(7.11)



−2mω × r˙ (S ) as well as an additional (unnamed) force −mω˙ × r , which vanishes in a uniformly rotating system. Example: On the northern hemisphere of the earth the Coriolis force leads to a clockwise deflection, cf. Fig. 7.6. 7.2.5 The Foucault Pendulum Without consideration of the rotation of the earth one deals with a spherical pendulum or, with appropriate initial conditions, a plane pendulum. For small displacements from the equilibrium position the vertical displacements of the pendulum are small in comparison with the horizontal displacements  ), such that the problem can be reduced (Δz   Δxmax and Δz   Δymax approximately to a two-dimensional (horizontal) problem. For small displacements one obtains the equation of motion r¨ = −ω02 r 

7.2 Rotation Around a Fixed Point

ω

215

ω

v

–ω ´ v v

r

r

–ω ´ v

Fig. 7.6. The Coriolis force for a motion along a latitude (left) and along a longitude (right) on the earth

Ω

y′ x′ R β

Fig. 7.7. For the Foucault pendulum

with ω02 = g/l as well as r  = (x , y  ) and with the origin on the surface on the earth, see Fig. 7.7. The equation of motion with consideration of the rotation of the earth (with constant angular velocity Ω) reads (7.10)

¨ = −ω02 r  − Ω × [Ω × (R + r )] − 2Ω × r˙  r with Ω = Ω (0, 0, 1) Ω  = (Ωx , Ωy , Ωz ) = Ω (0, cos β, sin β)

216

7 Moving Reference Frames

in coordinate systems which are fixed in space and fixed on the earth with unprimed and primed, respectively, notation (at R as in Fig. 7.7), where β is the geographic latitude. If one neglects the centrifugal term which is quadratic in Ω (i.e., small), the equation of motion is r¨ + ω02 r + 2Ω × r˙  = 0 or, in components, x ¨ + ω02 x − 2Ωz y˙  = 0 y¨ + ω02 y  + 2Ωz x˙  = 0. Here the motion in the vertical (z  -)direction is neglected (z˙   x, ˙ y) ˙ again. Thus, two coupled linear differential equations evolve. For the solution of this system of equations one can look at the components x and y  as components of a complex quantity ξ = x + iy 

(7.12)

and obtains by multiplication of the second equation with i and subsequent addition ξ¨ + ω02 ξ + 2iΩz ξ˙ = 0. This is a linear differential equation. With the exponential ansatz ξ ∝ eiαt one obtains

 2  −α + ω02 − 2Ωz α ξ = 0.

The vanishing of the secular equation yields the solutions for α,  α = −Ωz ± Ωz2 + ω02 ≈ −Ωz ± ω0 because of Ωz  ω0 . Without consideration of the rotation of the earth (Ω = 0, α = ±ω0 ) the solution, known from Sect. 4.2, is reproduced, ξ0 (t) = A eiω0 t + B e−iω0 t . W ith consideration of the rotation of the earth one can write the solution in the form (7.13) ξ = ξ0 (t) e−iΩz t . On the northern hemisphere one has Ωz = Ω sin β > 0

7.3 Galilean and Lorentz Transformation

217

because of β > 0. The initial conditions shall be chosen such that, neglecting the rotation of the earth, the pendulum would move in a plane. Now consider the rotation of the earth with the same initial conditions. Because of the sign in the exponent in (7.13) this pendulum plane rotates clockwise (in mathematically negative sense) with the frequency Ωz . Comments: • If one chooses ξ0 (t) real one has x = Re ξ = ξ0 cos(Ωt) y = Im ξ = −ξ0 sin(Ωt). • For Vancouver one has β ≈ 49◦ , sin β ≈ 0.75, and the period is T =

2π 1 1 2π = ≈ 24 h = 32 h. Ω sin β Ω sin β 0.75

• A second solution would be obtained by ξ˜ = x − iy  . and α ˜ = Ωz ±

 Ωz2 + ω02

≈ Ωz ± ω0 If one chooses ξ0 as real for simplicity one has ! " ! " ! " ˜ −Im ξ(t) ˜ x (t), y  (t) = Re ξ(t), Im ξ(t) = Re ξ(t), ! " = ξ0 (t) cos(Ωz t), − sin(Ωz t) , thus in both cases a clockwise rotation. • The two solutions one could have obtained in a standard way with the ansatz # $ # $ x0 x ∝ eiαt y y0 much more laboriously. The comfortable way for problems involving rotations (e.g., motion of a charged particle in a magnetic field) thus is often a complex superposition of the coordinates like in (7.12).

7.3 Galilean and Lorentz Transformation The Special Relativity theory is treated in detail in a separate Course.7 Here the (nonrelativistic) Galilean transformation is contrasted to the (relativistic) Lorentz transformation, the latter being valid in the relativistic and nonrelativistic regime. 7

See the course Special Relativity.

218

7 Moving Reference Frames

7.3.1 The Relativity Principle In the theory of Special Relativity considerations are made concerning the laws of nature in inertial systems moving relative to each other. Inertial systems are those systems, which are defined in the first Newtonian axiom8 Since a force-free particle moves at constant speed in any inertial system, different inertial systems must move with constant velocity relative to each other. In principle it is not possible to ascertain which of the different inertial systems is at rest. One can only determine the relative velocity between two (or more) different inertial systems. The relativity principle reads: Relativity principle: The laws of nature have the same form in all inertial systems.

(7.14)

Comments: • In the case of the Newtonian relativity the transformation from one inertial system to another is performed by the Galilean transformation and in the case of the Einstein relativity by the Lorentz transformation. • It turns out that the Newtonian relativity is a limit of the Einstein relativity (i.e., if the relative velocity is small compared to the velocity of light) and that the laws of mechanics and of electrodynamics (the Newtonian and the Maxwellian theory) have a common transformation behavior under the Lorentz transformation. • The influence of forces due to external masses and thus of gravitational potentials is described by the General Relativity theory; in Special Relativity one assumes that those masses, if there are any, are infinitely far away (and thus without any effect). 7.3.2 General and Special Transformation General Transformation Consider an inertial system S , which moves with constant velocity v relative to an inertial system S. In the general transformation the two coordinate systems can be rotated against each other by a fixed angle (matrix U) from (7.4) and translated by a fixed vector (translation vector r0 ), see Fig. 7.8, (7.5)

r = U · r − vt + r0 ,

U = U−1 0

but U and r 0 shall not be of interest in the following. 8

See Sect. 1.2.

7.3 Galilean and Lorentz Transformation

219

r r’

S

v

r0

S’

0

Fig. 7.8. The (general) transformation between different inertial systems

vt

Special Transformation Choose the x-directions of the coordinate systems S and S along v. At time t = 0 the two inertial systems shall coincide. This is the special transformation, depicted in Fig. 7.9. We will also refer to these systems as them being in the standard configuration.

Fig. 7.9. Inertial systems moving relative to each other in the standard configuration

Agreement 9 In the following always the standard configuration shall be chosen, if not stated otherwise. 7.3.3 The Galilean Transformation In the standard configuration the (special) Galilean transformation is x y z t

= x − vt =y =z = t.

(7.15)

Comments: • The last of these equations seems obvious, but will be modified in Special Relativity. In the light of the latter an absolute time is (wrongly) assumed and can be looked at as a principle of Newtonian mechanics.

220

7 Moving Reference Frames

• The Galilean transformation turns out to be the nonrelativistic limit of the Lorentz transformation (in which the considered velocities v are small compared to the velocity of light).9 • To facilitate the comparison with the Lorentz transformation below one can write the coordinate transformation in the form10 ⎛

⎞ ⎛ t 1  ⎜ x ⎟ ⎜ −v ⎝ ⎠=⎝ y z

1 1

⎞⎛ ⎞ t ⎟⎜x⎟ ⎠⎝ ⎠ y 1 z

.

(7.16)

7.3.4 Galilean Invariance Galilean Invariants Definition 25. (Galilean invariants): Galilean invariants are quantities, which are invariant under Galilean transformations. The relative positions are invariant under (special) Galilean transformations, r1 − r2 = r 1 − r2 , and in particular the distances are invariant under (general) Galilean transformations. Only because one assumes in Newtonian mechanics that the internal forces typically depend upon the instantaneous11 relative positions of two particles, these forces are then invariant under Galilean transformations. Invariant under Galilean transformations are many scalars, besides the distance among others the mass m. Form Invariance Under Galilean Transformation Claim: The Newtonian equations of motion are invariant under a Galilean transformation, m¨ r = F (r) ⇔ m¨ r  = F  (r  ).

(7.17)

Proof: According to (7.2) the force F is invariant under a translation, F = F  , and thus in particular under a Galilean transformation. In addition, the acceleration is invariant under a Galilean transformation, r¨ = r¨  . Finally, the mass is an invariant scalar, m = m .  9 10 11

See the Course on Special Relativity theory. Missing elements in the transformation matrix stand for a zero. Actually, this is an approximation.

7.3 Galilean and Lorentz Transformation

221

Fig. 7.10. Inertial systems moving relative to each other in the standard configuration, cf. Fig. 7.9

Comments: • Equation (7.17) exhibits the equations of motion of two different inertial systems in the same form; this is the form invariance under Galilean transformation. • The Newtonian equations of motion are form-invariant only in their coordinate-free notation. 7.3.5 The Lorentz Transformation In addition to the relativity principle (7.14) the Special Relativity theory is subject to a second principle: The maximum velocity is the velocity of light.

(7.18)

The corresponding (special) Lorentz transformation can be derived under the following assumptions: 1. The transformation of space and time is linear. 2. The relativity principle holds. 3. The maximum velocity is the velocity of light, c. Under these assumptions the Lorentz transformation in the standard configuration (Fig. 7.10) turns out to be ⎛

⎞ ⎛ ct γ  ⎜ x ⎟ ⎜ −βγ ⎝ ⎠=⎝ y z with β=

v , c

⎞⎛

−βγ γ 1

⎞ ct ⎟⎜ x ⎟ ⎠⎝ ⎠ y 1 z

(7.19)

1 γ=  . 1 − β2

Comments: • With the first two assumptions one finds that there is a maximum velocity,12 which then is postulated as being the velocity of light. 12

See the Course on Special Relativity.

222

7 Moving Reference Frames

• The inverse transformation is13 ⎛ ⎞ ⎛ ct γ ⎜ x ⎟ ⎜ βγ ⎝ ⎠=⎝ y ct

βγ γ

⎞ ct ⎟ ⎜ x ⎟ ⎠⎝  ⎠ y 1 z 1 ⎞⎛

(7.20)

• In the nonrelativistic case v  c one obtains the Galilean transformation (7.16) from the Lorentz transformation (7.19).14 • The concept (or postulate) of an absolute time must be given up in Special Relativity, where one finds the effect of the time dilation. • The distance is a Galilean invariant,15 but not a Lorentz invariant; instead, one finds the effect of the length contraction. • With the validity of the Lorentz transformation (rather than the Galilean transformation) the mechanics (Einstein) and the electrodynamics (Maxwell) have a unified transformation behavior. Until the beginning of the twentieth century the different behavior of the two theories (Newtonian and Maxwellian, respectively) presented a serious puzzle, since both theories were interconnected by the Lorentz force. • The invariance of the Maxwell equations in electrodynamics against Lorentz transformations was originally considered merely as a form invariance, and as such one could have transposed the Lorentz transformation to mechanics. Einstein’s prime achievement was the interpretation of the Lorentz transformation as the transformation between inertial systems. For more details see the first chapters in the Course on Special Relativity, where the effects of length contraction and time dilation are discussed and where subsequently also the transformation of velocity, momentum, energy, force, etc. are treated.

Summary: Moving Reference Frames General transformation (rotation and translation) r  (t) = U(t) · r(t) − R(t) rotating systems (S rotates with ω relative to S) #

13 14 15

d dt

$(S)

(7.8)

... =

For the proof see problem 7.5. For the proof see problem 7.6. See Sect. 7.3.4.

#

d + ω× dt

$(S ) ...

Problems

223

external, centripetal, Coriolis, additional force: m¨ r(S) = F 





(7.10)



m¨ r (S ) = F − mω × (ω × r (S ) ) − 2mω × r˙ (S ) − mω˙ × r (S ) General Galilean transformation (between inertial systems) (7.5)

r = U0 · r − r0 − vt

with

U˙ 0 = 0 and r˙ 0 = 0

Special Galilean transformation (r 0 = 0, U0 = 1, v = vex ) x y z t

(7.15)

= = = =

x − vt y z t

⎞ ⎛ 1 ct  x −β ⎜ ⎟ ⎜ ⎝ ⎠=⎝ y z

⎞⎛

⎛ ⇔

1

⎞ ct ⎟⎜ x ⎟ ⎠⎝ ⎠. 1 y 1 z

Relativity Principle: The laws of nature are the same in all inertial systems. This is a result of the Galilean transformation in the context of classical mechanics, and this is a first postulate of Special Relativity. Second postulate of Special Relativity: The maximum velocity is the speed of light. Special Lorentz transformation ⎛ ⎞ ⎛ ⎞⎛ ⎞ ct γ −βγ ct  (7.19) x −βγ γ ⎜ ⎟ ⎜ ⎟⎜ x ⎟ ⎝ ⎠ = ⎝ ⎠⎝ ⎠ y 1 y z 1 z with β=

v , c

1 γ=  . 1 − β2

Problems 7.1. Earth as an Accelerated Reference System. Even though the equations of motion are more simple in inertial systems, one usually describes a motion on the earth in a reference system (laboratory system) rotating along with the earth. Strictly speaking, this laboratory system is not an inertial system because of the rotation of the earth. At a point on the earth with the geographic latitude ϕ a Cartesian coordinate system S  is defined, see Fig. 7.11: x ¯3 -axis vertically upwards x ¯2 -axis towards north x ¯1 -axis towards east.

224

7 Moving Reference Frames

For the angular velocity of the earth one has |ω| =

2π −1 h = 7.27 × 10−5 s−1 . 24

(a) What is the equation of motion of a point mass in this coordinate system close to the earth’s surface (neglect the terms in ω 2 )? (b) How does the Coriolis force depend upon the geographic latitude? ¯3 -axis is perpendicular to (c) Orient the coordinate system S  such that the x the earth’s surface. Which equations of motion are then to be solved for a point mass close to the earth’s surface? (The Coriolis force can be taken from (b).) (d) A body initially at rest is dropped from the height H. Solve the equation ¯˙ 2 remain negligibly of motion in (c) under the supposition that x ¯˙ 1 and x small during the time of the free fall. Determine the deflection towards east originating from the earths’s rotation. Determine the value of the deflection towards east for a height H = 100 m at ϕ = 45◦ .

x2

w

w j

r0

x3

r0 j R

Fig. 7.11. For the deflection towards east in the free fall (for problem 7.1)

7.2. Coriolis Force. In which direction does the Coriolis force point (on earth), if the velocity has radial direction? 7.3. Centrifugal and Coriolis Force on the Southern Hemisphere. Compare the direction of the centrifugal and of the Coriolis force on the southern and northern hemispheres of the earth. 7.4. Coriolis Force and Meteorology. (a) Explain into which directions the winds from north, east, south, and west are deflected on the northern and southern hemisphere (see left part of Fig. 7.12). Neglect the centrifugal acceleration, and assume that the air moves at constant height.

Problems

225

Z ω1

ω e3 Q

e2 r

R

E3 λ E1

0

ρ

z ω2

60⬚

y P

T

0⬚

e1

H

x

E2

Y

X

Fig. 7.12. Left: For the Coriolis-forces on the earth. Right: High- and low-pressure regions on the meteorological chart (for problem 7.4)

(b) Explain the origin of cyclones (see the right part of Fig. 7.12). Which sense of rotation do they have on the northern and southern hemisphere? 7.5. Inverse Lorentz Transformation. Prove that the transformation matrix of the inverse Lorentz transformation (in the standard configuration) is given by ⎛ ⎞ γ βγ ⎜ βγ γ ⎟ ⎝ ⎠. 1 1 7.6. The Galilean Transformation as the Nonrelativistic Limit of the Lorentz Transformation. Prove that the Galilean transformation is the nonrelativistic limit of the Lorentz transformation. 7.7. Lorentz Transformation of the Velocity. Prove that the Lorentz transformation of the velocity (in the standard configuration) is given by ⎛ ⎞ ⎛  ⎞ ux − v ux 1 ⎝ uy /γ ⎠ . ⎝ uy ⎠ = 2 1 − vu x /c ux /γ uz Hint: Write ux etc. as dx/dt.

8 Dynamics of a Rigid Body

In the previous chapters the motion of one or two particles has been investigated, whose point-like extension is a meaningful ansatz, if one deals, e.g., with the motion of the center of mass or if the extension of the two particles under consideration is small compared to their distance. In condensed matter the latter approximation is not valid: The extension of the atoms is comparable to the distance between the neighboring atoms. In this chapter extended particles shall thus be investigated. We have in mind, however, the ideal case of a rigid body, and therefore we exclude elastic matter1 as well as fluids in particular.

8.1 The Rigid Body as a System of N Point Masses One can think of an extended body as to be composed of point masses (atoms). If the body is rigid, the positions of the various point masses are connected to each other by constraints such that the distances of pairs (i, k) of atoms are fixed (time independent). These constraints are G(i k) = ri2k − c2i k = 0

(i k) = 1, . . . , N (N2 −1) ;

they are not independent of each other, because for a sufficiently large N one would have more constraints (≈ 12 N 2 ) as coordinates (3N ).2 The forces of constraint are Z i k = λ(i k) ∇i G(i k) = λ(i k) r i k Z k i = λ(i k) ∇k G(i k) = −λ(i k) r i k = −Z i k (from the independent constraints). The forces of constraint cancel pairwise. 1 2

For elastic materials, see Chap. 10. This was noted already in Sect. 3.2.3.

228

8 Dynamics of a Rigid Body

Comments: • One is not interested in the trajectory of every single particle at all. How ever will one survey 1023 to 1024 trajectories (per mole), and what could one do with them (even if one had them)? • The large number of constraints makes their explicit consideration unpractical. • In addition, the pairwise oppositely oriented forces of constraint do not enter the conservation of total momentum, angular momentum, and energy:  P˙ = F ext F ext = F ext i i

˙ = N ext L

N

ext

=



r i × F ext i

i

In the following thus the translation and rotation (three degrees of freedom each, possibly reduced due to additional constraints) of the rigid body will be investigated. While a point mass has only translational energy, an extended body has thus rotational energy in addition. The latter is the primary object of this chapter.

8.2 Translational and Rotational Energy: Inertia Tensor For an extended rotating body there are two specific points: the center of mass3 and the point at instantaneous rest.4 The instantaneously fixed point may lie in the interior of the body (as in the case of the wheel rotating around a fixed axis) or outside (as in the case of the pendulum) or also on the surface (as in the case of the rolling body). 8.2.1 The Center of Mass as the Specific Point If one chooses the center of mass as the specific point, one can divide the kinetic energy into translational and rotational energy. To this end one decomposes the coordinates ri of the point masses into the center-of-mass coordinate R and the coordinates ri relative to the center of mass, see Fig. 8.1, ri = R + r i with R= 3 4

See Sect. 8.2.1. See Sect. 8.2.2.

1  mi r i , M i

 i

mi r i = 0.

8.2 Translational and Rotational Energy: Inertia Tensor

S

229

ri′

R Fig. 8.1. The notation of the coordinates with respect to an arbitrary reference point O and with respect to the center of mass S

ri 0

Kinetic Energy and Inertia Tensor One obtains for the kinetic energy T =

1 2



˙ 2 + 1 ω · Θ(S) · ω. mi r˙ 2i = 12 M R 2

(8.1)

i

Here the inertia tensor (relative to the center of mass S) is Θ(S) =



! " 2 mi ri 1 − r i ⊗ r i

(8.2)

i

with



⎞ 0 0⎠ 1

1 0 1 = ⎝0 1 0 0



xx and r ⊗ r = ⎝ yx zx

xy yy zy

⎞ xz yz ⎠ zz

and with the components (S)

Θij =



! " 2 mi ri δi,j − xi,i xi,j .

(8.3)

i

Proof: (i) Inserting one obtains T =

1 2

=

1 2

 i

=



mi r˙ 2i =

1 2



! "2 ˙ + r˙  mi R i

i

! 2 " ˙ + 2r˙  · R ˙ + r˙  2 mi R i i

i 1 ˙2 2MR

˙ · +R



mi r˙ i +

i

1 2



mi r˙ i 2 .

i



The term in the middle vanishes, since i mi r i gives the (fixed) position of the center of mass in the center-of-mass system, and this is the origin for the primed coordinates.

230

8 Dynamics of a Rigid Body

(ii) For the rotation of a rigid body around the center of mass with the (instantaneous) angular velocity ω the velocity of a point mass i is (7.7)

r˙ i = ω × r i . With this one obtains for the last term (with ω 2 = ω · 1 · ω)      2  2 (B.4) 1 2 2  2 1 1 ˙ r r m = m (ω × r ) = m ω − (ω · r ) i i i i i i i 2 2 2 i

i

=

1 2ω

·



i

  2 mi ri 1 − r i ⊗ ri · ω = 12 ω · Θ(S) · ω.

i

 Comments: • According to (8.1) the kinetic energy can thus be decomposed into a part, which contains the translational energy of the total mass, thought of as concentrated at the center of mass, and into a part, which contains the rotational energy relative to the center of mass. • Compare this result for the center of mass with the result for the instantaneously fixed point in Sect. 8.2.2 further below. Angular Momentum and Inertia Tensor In a similar way one obtains for the angular momentum of a body rotating with the angular velocity ω ˙ + L(S) L = MR × R

(8.4)

and for the angular momentum with respect to the center of mass L(S) = Θ(S) · ω with the components (S)

Li

=

3 

(8.5)

(S)

Θij ωj .

j=1

Proof: L(S) =



mi ri × r˙ i =

i

(B.3)

=



mi ri × (ω × ri )

i 2 mi [ri ω

i (S)

= Θ





r i (r i

· ω)] =



mi [ri 1 − r i ⊗ ri ] · ω 2

i

· ω. 

8.2 Translational and Rotational Energy: Inertia Tensor

ri′′

P ρ

231

ri Fig. 8.2. The notation of the coordinates relative to an (instantaneous) point P

0

Comment: The angular momentum L(S) and the rotation vector ω are not parallel in general; they are parallel only if the inertia tensor is proportional to the unit matrix, i.e., if the rotation vector is parallel to one of the three principal axes.5 8.2.2 The Instantaneously Fixed Point as the Specific Point The instantaneously fixed point with the position vector ρ shall be denoted by P. One introduces the coordinates ri of the point masses relative to the point P, see Fig. 8.2, r i = ρ + r i . As an alternative to (8.1) the kinetic energy can be written as T =

1 2



mi r˙ 2i = 12 ω · Θ(P) · ω

(8.6)

i

with the inertia tensor Θ(P) with respect to the instantaneously fixed point P ! "  2 Θ(P) = mi ri 1 − ri ⊗ r i . i

Proof: As above with R → ρ, but with ρ˙ = 0. Comment: Θ(P) and Θ(S) are different from each other in general. The relation between the inertia tensors with respect to the center of mass and with respect to other points, respectively, is given by Steiner’s theorem.6 Summarizing one can treat the motion of a rigid body either as a pure rotation around an (instantaneously) fixed point or as a translation of the center of mass and a rotation around the center of mass.

5 6

See Sect. 8.5 further below. See Sect. 8.4 further below.

232

8 Dynamics of a Rigid Body

8.3 Transition to the Continuum Definition 26. (Mass density): The mass density is ρ(r) = =

dm(r) Δm(r) = lim dV (r) ΔV (r)→0 ΔV (r) 1 ΔV (r)→0 ΔV (r)

r i ∈ΔV (r)

lim



mi .

(8.7)

i

Comments: • The limit (8.7) to the continuum has its limitations because of the corpuscular nature of matter. Thus, the limits are always to be understood in the way that the considered volumes ΔV are small in comparison to macroscopic dimensions but always still large in comparison to atomic dimensions (such that the average is taken over a sufficiently large number of atoms to create a continuous function of the position). See Fig. 8.3.

Fig. 8.3. Macroscopic and microscopic view of solid matter

• Exactly the same consideration is made in electrodynamics with the definition, e.g., of the electrical charge density. • In fact, the Heisenberg uncertainty relation of quantum mechanics tells that one cannot locate the position of a mass or of a charge (simultaneously with the momentum), but that there is a probability distribution for the position (and momentum). If one considers the point masses in the limit of infinitesimally small (and correspondingly many) masses and the following sum as a Riemann sum, one obtains for any function f (r) 

 mi f (ri ) →

 dm(r)f (r) =

d3r ρ(r) f (r).

i

These and the following relations one formally obtains with the density

(8.8)

8.3 Transition to the Continuum

ρ(r) =



233

mi δ(r − r i )

(8.9)

i

of a point-mass distribution. In particular one has the total mass   M= mi = d3rρ(r);

(8.10)

i

the position vector of the center of mass R=

 1 1  mi r i = d3rρ(r) r; M i M

the total momentum 

P =

(8.11)

 mi r˙ i =

˙ d3rρ(r) r;

(8.12)

˙ d3rρ(r) r × r;

(8.13)

i

the total angular momentum L=



 mi r i × r˙ i =

i

and the components of the inertia tensor      (8.3)  Θij = mi ri2 δi,j − xi,i xi,j = d3rρ(r) r2 δi,j − xi xj .

(8.14)

i

8.3.1 Example: Center of Mass of a Homogeneous Hemisphere Consider a hemisphere with the mass M and the radius R. For reasons of symmetry the center of mass lies on the rotational axis, which is chosen as the z axis. Then the z component of the center of mass is  (8.11) 1 Z = d3r ρ(r) z M in the coordinate system of Fig. 8.4. With spherical coordinates as the coordinates adapted to the problem one obtains z = r cos ϑ d r = r2 dr sin ϑ dϑ dϕ 3

and the mass density ρ(r) =

M θ(R − r) θ( 12 π − ϑ) V

with V =

1 4π 3 R 2 3

234

8 Dynamics of a Rigid Body z

y

Fig. 8.4. The coordinate system for the hemisphere

x

as well as Z=

1 2π 3 3 R





R



π/2

r2 dr



sin ϑ dϑ

0

0

dϕ r cos ϑ. 0

With (t = cos ϑ) 

R

r3 dr = 0



R4 4 

π/2

cos ϑ sin ϑ dϑ = 0





1

cos ϑ d(cos ϑ) = 0

1

t dt = 0

1 2



dϕ = 2π 0

one obtains Z=

2π 3

3 1 R4 1 · · 2π = R. · 3 4 2 8 R

8.3.2 Examples of Inertia Tensors (i) Inertia Tensor of a Homogeneous Sphere with Respect to the Center of Mass With the mass M and the radius R one has the (homogeneous) mass density ρ(r) =

M θ(R − r), V

V =

and the moment of inertia Θij =

2 M R2 δi,j . 5

4π 3 R 3

8.3 Transition to the Continuum

235

Fig. 8.5. Coordinate system for the sphere

Proof: Using the coordinates of Fig. 8.5 the moment of inertia is    (8.14) Θij = d3 r ρ(r) r2 δi,j − ri rj . In cartesian coordinates one obtains    Θzz = d3 r ρ(r) r2 − z 2 = Θxx = Θyy  Θxy = d3 r ρ(r) (0 − xy) = Θyx = Θxz = . . . . As in Sect. 8.3.1 the integrations is most comfortably performed with spherical coordinates, x = r sin ϑ cos ϕ y = r sin ϑ sin ϕ z = r cos ϑ d3r = r2 dr sin ϑ dϑ dϕ. With this one obtains for the zz-element of the inertia tensor  π  2π    M R 2 Θzz = r dr sin ϑ dϑ dϕ r2 − r2 cos2 ϑ V 0 0 0  π  2π    M R 4 2 = r dr dϕ. 1 − cos ϑ sin ϑ dϑ V 0 0 0 With the substitution t = cos ϑ one obtains  π  1     4 2 2 1 − cos ϑ sin ϑ dϑ = 1 − t2 dt = 2 − = 3 3 0 −1 and thus Θzz =

2 M R5 4 2 4π 3 5 · 3 · 2π = 5 M R . 3 R

236

8 Dynamics of a Rigid Body

z

y x

Fig. 8.6. The coordinate system for the cuboid

Alternatively one employs symmetry and obtains 1 (Θxx + Θyy + Θzz ) 3   1 M R 4 2 2r2 ρ(r) d3r = 3 = 2r dr = M R2 . 3 R 0 5

Θzz =

Furthermore, Θxy = − =−

M V M V





R



0



π

r2 dr 0



R

r4 dr 0



sin ϑ dϑ

dϕ (r sin ϑ cos ϕ) (r sin ϑ sin ϕ) 0



π

sin3 ϑ dϑ 0



dϕ cos ϕ sin ϕ = 0 , 0

since the integral over ϕ vanishes. (Or: The integrand in in xi and/or xj with respect to the integration regime).



ρ(r)xi xj d3r is odd 

(ii) Inertia Tensor of a Homogeneous Cuboid with Respect to the Center of Mass Cartesian coordinates are adapted to the problem, see Fig. 8.6. One chooses the coordinate system with the origin in the center of mass and the axes parallel to the edges (with the edge lengths Li ) of the cuboid. With the mass density ρ=

M 1 θ( Lx − |x|) θ( 12 Ly − |y|) θ( 12 Lz − |z|) V 2

with V = Lx Ly Lz one obtains the inertia tensor (S)

Θij = δi,j

M 2 (L + L2y + L2z − L2i ). 12 x

8.3 Transition to the Continuum

237

Proof: The moment of inertia is    Θij = ρ(r) r2 δi,j − xi xj d3r . With r2 = x2 + y 2 + z 2 d3 r = dx dy dz one obtains Θxx

M = V M = V





Lx /2



Ly /2

dx −Lx /2

'

Θxy = −

M V

=−

M V

−Ly /2



Lx /2

'

 

Lz /2

y dy #

 dx

 2

−Ly /2

2 Lx 3

dz (y 2 + z 2 )

−Lz /2

Ly /2

dx −Lx /2

M = Lx Ly Lz

Lz /2

dy

Ly 2

−Lz /2

$3

( dz + (y ↔ z) (

Lz + (y ↔ z) =

M 2 (L + L2z ) 12 y

 dy

dz xy 

Lx /2

x dx

 y dy

dz = 0,

−Lx /2

since the integrand in the integral over x (and in that over y) is odd in the (symmetrical) integration regime, and the integral(s) thus vanish(es).  (iii) Inertia Tensor of a Homogeneous Cuboid with Respect to a Corner One chooses the coordinate system as in the previous example, but with the origin in a corner, see Fig. 8.7. Then the mass density is ρ(x, y, z) =

M θ(Lx − x) θ(Ly − y) θ(Lz − z) V ×θ(x) θ(y) θ(z)

with V = Lx Ly Lz . The moment of inertia is 1 M [(L2x + L2y + L2z ) − L2i ] i = j (0) Θij = 3 1 i=  j. − 4 M Li Lj

238

8 Dynamics of a Rigid Body

z

Lz

Ly

Lx

y Fig. 8.7. The coordinate system for the cuboid

x

Proof: The moment of inertia is    Θij = ρ(r) r2 δij − xi xj d3r  Ly  Lz    M Lx dx dy dz y 2 + z 2 Θxx = V 0 0 0 ' (  Ly  Lz  Lx  Ly  Lz Lx M 2 2 = dx y dy dz + dx dy dz z V 0 0 0 0 0 0 ' (  L3y M 2 L3z M Lx Lz + Lx Ly = Ly + L2z = V 3 3 3 Θxy

M = V



M =− V =−



Lx



Ly

dx 0



0 Lx



dx x 0 L2x

Lz

dy 0 Ly

dz (0 − xy) 

Lz

dy y 0

dz 0

L2y M M Lz = − Lx Ly . V 2 2 4

(The integrand is an odd function of x and y, but not with respect to the integration regime; thus the integral does not vanish.) 

8.4 Change of the Reference System: Steiner’s Theorem 8.4.1 Steiner’s Theorem Steiner’s theorem gives a relation between the inertia tensor relative to the center of mass (S) and that with respect to an arbitrary point (Q). Let R be the vector,7 which points from the point Q to the center of mass S (or the 7

Equation (8.15) is invariant under the replacement R → −R.

8.4 Change of the Reference System: Steiner’s Theorem

239

Fig. 8.8. Center of mass S and an arbitrary point Q, which may also be outside of the body (The vector R may equivalently point from Q to S)

reverse), see Fig. 8.8 analogous to Fig. 8.2 (if one substitutes there the point P by the point Q). Then one has   Θ(Q) = Θ(S) + M R2 1 − R ⊗ R (8.15) with the total mass M

(8.10)

=

 ρ(r) d3r.

Proof: One chooses Q as the coordinate origin and r = R + r . Then the inertia tensor with respect to the point Q is    (Q) = ρ(r) r2 1 − r ⊗ r d3r Θ    2 = ρ(r ) (R + r  ) 1 − (R + r  ) ⊗ (R + r  ) d3r    = ρ(r ) R2 + 2r · R + r2 1  − (R ⊗ R + r  ⊗ R + R ⊗ r + r  ⊗ r  ) d3r . Now one has



ρ(r ) r d3r = 0,

since the vectors r  have their origin at the center of mass. Of the seven terms thus the second, fifth, and sixth term vanish, and one obtains       ρ(r  ) d3r Θ(Q) = ρ(r  ) r2 1 − r  ⊗ r  d3r + R2 1 − R ⊗ R   = Θ(S) + M R2 1 − R ⊗ R .  8.4.2 Angular Momentum and Torque In a way analogous to the inertia tensor one finds a decomposition of the other quantities connected with the rotation, of the angular momentum, and of the torque L(Q) = L(S) + R × P

(8.16)

240

8 Dynamics of a Rigid Body

and N (Q) = N (S) + R × F .

(8.17)

Proof: (i) The angular momentum with respect to the point Q is  ˙ d3r L(Q) = ρ(r) (r × r)  ˙ + r˙  ) = d3r ρ(r  ) (R + r ) × (R  ˙ + r × R ˙ + R × r˙  + r  × r˙  ). = d3r ρ(r  ) [R × R Since the primed coordinate system has its origin at the center of mass,  ρ(r ) r d3r = 0, the second term vanishes; with r˙  = ω × r  thus the third vanishes, and one obtains   ˙ L(Q) = d3r ρ(r ) r × r˙  + d3r ρ(r  ) R × R = L(S) + R × P 

with ˙ P = M · R,

M=

ρ(r  ) d3r .

Alternatively (and much simpler):   L(Q) = Θ(Q) · ω = Θ(S) · ω + M R2 ω − R R · ω = L(S) + M R × (ω × R) = L(S) + R × P . (ii) The proof for the torque is quite analogous: ˙ (Q) = L ˙ (S) + R ˙ × P + R × P˙ = N (S) + 0 + R × F . N (Q) = L  Comments: • A quantity connected with the rotation (torque, angular momentum) with respect to an arbitrary point Q is thus equal to that with respect to the center of mass plus a term which describes the rotation of the total mass thought of as assembled at the center of mass. In particular, the total angular momentum is equal to the sum of orbital angular momentum R × P and eigen angular momentum L(S) (spin, as called in quantum mechanics).

8.4 Change of the Reference System: Steiner’s Theorem

241

Fig. 8.9. Center of mass S and a point Q at a corner of a cuboid

8.4.3 Examples (i) Homogeneous Cuboid The homogeneous cuboid has been treated further above in the examples (ii) and (iii) of Sect. 8.3.2. The result was  1 M (L2x + L2y + L2z − L2i ) i = j (S) Θij = 12 0 i = j 4 1 2 2 2 2 M (Lx + Ly + Lz − Li ) i = j (Q) Θij = 3 1 i = j − 4 M Li Lj with the notation as in Fig. 8.9. Here one can verify the Steiner theorem: On the one hand, one obtains 41 M (L2x + L2y + L2z − L2i ) i = j (Q) (S) Θij − Θij = 4 1 i = j. − 4 M Li Lj With the center-of mass-coordinate (with respect to the selected corner) R=

1 (Lx , Ly , Lz ) 2

one obtains on the other hand (Q)

Θij

(S)

− Θij

= M (R2 1 − R ⊗ R) ⎛ ⎞ ⎛ 2 ⎡ 1 0 0 Lx  M ⎣ 2 Lx + L2y + L2z ⎝ 0 1 0 ⎠ − ⎝ Ly Lx = 4 0 0 1 Lz Lx ⎛ 2 ⎞ L + L2z −Lx Ly −Lx Lz M⎝ y = −Ly Lx L2z + L2x −Ly Lz ⎠ . 4 −Lz Lx −Lz Ly L2x + L2y

Lx Ly L2y Lz Ly

⎞⎤ Lx Lz Ly Lz ⎠⎦ L2z

242

8 Dynamics of a Rigid Body

(ii) Inertia Tensor of a Homogeneous Full Sphere with Respect to a Point on the Surface The coordinate system shall be chosen such that the specific point lies in the origin; in addition the z axis shall coincide with the symmetry axis as in Fig. 8.10. Then the vector to the center of mass is R = (0, 0, R) and the inertia tensor Θ(0) = Θ(S) + M (R2 1 − R ⊗ R) with

2 M R2 1 5 of example (i) of Sect. 8.3.2. With this one obtains ⎛ ⎞ ⎛ ⎡ $ 1 0 0 # 0 0 2 Θ(0) = M R2 ⎣ + 1 ⎝0 1 0⎠ − ⎝0 0 5 0 0 1 0 0 Θ(S) =

⎞⎤ 0 0 ⎠⎦ , 1

thus 7 M R2 5 2 = M R2 5 =0 (i = j).

(0) (0) = Θyy = Θxx (0) Θzz (0)

Θij

Here one deals with a symmetrical top with the principal axes8 ez and, e.g., ex and ey .

8.5 Principal Moments of Inertia and Principal Axes 8.5.1 Definitions Since the inertia tensor is not a scalar, the vectors L of the angular momentum and ω of the angular velocity are not parallel in general because of (8.5)

L = Θ · ω. But one can ask whether there are directions in which L and ω are parallel, L = Θ · ω = I ω. 8

See the following Sect. 8.5.

8.5 Principal Moments of Inertia and Principal Axes

243

z

S

0

y

x

Fig. 8.10. The coordinate system for the sphere

This poses an eigen value problem.9 Since Θ is a real and symmetrical tensor, one finds indeed always (three) real eigen values I with orthogonal eigen vectors e, i = 1, 2, 3. (8.18) Θ · ei = Ii ei Notations: • The eigen values Ii are called principal moments of inertia, and the eigen vectors ei are called principal axes. • If two of the principal moments of inertia are equal, e.g., I1 = I2 = I3 , one speaks of a symmetrical body. • If all three principal moments of inertia are equal, I1 = I2 = I3 , one speaks of a spherical body. • If one point is fixed,10 one speaks of a top (instead of a body). For the case in which the vector of the angular velocity is oriented along one of the principal axes of inertia, then also the angular momentum is oriented along this direction. For a symmetrical body with I1 = I2 one has this parallelity for all vectors which lie in the plane spanned by e1 and e2 : Let ω = a e1 + b e2 . Then one has L = Θ · ω = Θ · (ae1 + be2 ) = aI1 e1 + bI2 e2 = I1 (ae1 + be2 ) . For the spherical body finally one has the parallelity for the vectors in arbitrary direction. In the general case one can use the property (9) of 9 10

See Appendix G. See Sect. 8.7 further below.

244

8 Dynamics of a Rigid Body

z

y x

Fig. 8.11. The coordinate system for the cube

Appendix G.4 and obtains the angular-momentum components in the basis of the principal axes (normalized to unity),11 Θ=



ei Ii eT i with |ei | = 1



L=Θ·ω =

i



ei Ii eT i ·ω =



i

ei L i .

i

Comment: The reference point P is arbitrary; however, the inertia tensor Θ(P) and the moments of inertia I (P) depend upon the reference system. 8.5.2 Examples (i) Inertia Tensor of a Homogeneous Cube with Respect to the Center of Mass Let the edge length be L (Fig. 8.11). The inertia tensor is according to example (ii) of Sect. 8.3.2 1 Θij = δi,j M L2 . 6 The inertia tensor is already diagonal. The principal moments of inertia are all equal, 1 Ii = M L2 i = 1, 2, 3. 6 this is a spherical top. The principal axes of inertia, are for example, e1 = ex , e2 = ey , e3 = ez .

11

If e is a (column) vector, then eT denotes the transposed (row) vector.

8.5 Principal Moments of Inertia and Principal Axes

245

(ii) Inertia Tensor of a Homogeneous Cube with Respect to a Corner According to example (iii) of Sect. 8.3.2 the inertia tensor is, in the notation given in Fig. 8.12,

z

Lz Lx

Ly y Fig. 8.12. The coordinate system for the cube (Lx = Ly = Lz = L)

x

4 Θij =

2 2 3ML 1 − 4 M L2

i=j i = j

⎛ ⎞ +8 −3 −3 M L2 ⎝ Θ= −3 +8 −3 ⎠ . 12 −3 −3 +8 The principal moments of inertia are Ii = M L2 γi with γ1 =

1 , 6

γ2 = γ3 =

11 12

with the principal axes of inertia (normalized to 1) ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 1 0 −2 1 ⎝ ⎠ 1 ⎝ 1 e1 = √ 1 , e2 = √ 1 ⎠ , e3 = √ ⎝ 1 ⎠ . 3 1 2 −1 6 1 Proof: The eigen values I = M L2 γ are obtained from the secular equation ⎛2 ⎞⎤ ⎡ − 14 − 14 3 −γ ⎜ ⎟⎥ ⎢ 2 det ⎣ M L2 ⎝ − 14 − 14 ⎠ ⎦ = 0. 3 −γ 2 − 14 − 14 3 −γ

246

8 Dynamics of a Rigid Body

This is an equation of third order for γ, # $3 $ # $2 # $3 # 2 1 1 2 0= −γ −2 −γ −3 3 4 3 4 $ $2 $ # $2 ( #  '# # 2 1 2 2 2 2 −γ − −γ + −γ + = 3 4 3 4 3 4 # $ $  # 2 2 2 2 1 = −γ − −γ + . 3 4 3 4 From this one can determine the eigen values easily, 4−3 1 2 2 − =2 = 3 4 12 6 8+3 11 2 1 γ2 = γ3 = + = = . 3 4 12 12 For the eigen vectors one obtains by inserting the eigen values into the eigen value equation (as solution of the homogeneous equation) ⎛ ⎞ ⎛ ⎛ ⎞ ⎞ 1 0 −2 1 ⎝ ⎠ 1 ⎝ 1 e1 = √ 1 , e2 = √ 1 ⎠ , e3 = √ ⎝ 1 ⎠ . 3 1 2 −1 6 1 γ1 =

Comment: If one can guess the eigen vectors, find. By comparison of the right and left side of ⎛ a M e = λ e with M = ⎝ b b one obtains ⎛ ⎞ 1 M e1 = (a + 2b)e1 e1 = ⎝ 1 ⎠ , 1 ⎛ ⎞ 0 e2 = ⎝ 1 ⎠ , M e2 = (a − b)e2 −1 ⎛ ⎞ 2 e3 = ⎝ −1 ⎠ , M e3 = (a − b)e3 −1

 the eigen values are easy to b a b

⎞ b b⎠ a

# $ 1 2 1 λ1 = a + 2 b = + 2 − = 3 4 6 λ2 = a − b =

11 2 1 + = 3 4 12

λ3 = (a − b) =

11 12

The third of the eigen vectors is less obvious than the other two. However, if one has determined two eigen vectors e1 and e2 , one obtains the third from e3 = ±e1 × e2

8.6 Rotation Around a Fixed Axis

247

(for right- or left-handed systems) because of the orthogonality of the eigen vectors. The eigen vectors are fixed except for a factor (the so-called normalization factor), which in general is determined such that the eigen vectors are unit vectors.

8.6 Rotation Around a Fixed Axis The motion of the rigid body around a fixed axis is a pure rotation around this axis. The motion has one degree of freedom: the rotation angle ϕ. If one denotes the unit vector in the direction of the (fixed) axis of rotation by eω , then one can assign a rotation vector ϕ to the rotation, (7.6)

ϕ = ϕeω as in Fig. 8.13.

Fig. 8.13. On the infinitesimal rotations (previously shown in Figs. 1.11 and 7.4)

Let a point be described by the position vector r i in the body-fixed system (with the origin on any point on the axis of rotation). Then the change dr i of a position r i in the space fixed system is described by (1.35)

dri = dϕ × r i . The velocity of a point with the position vector ri is then r˙ i = ω × r i

(8.19)

with ˙ and e˙ ω = 0. ˙ ω=ϕ ω = ωeω = ϕe Comment: In contrast to the single degree of freedom here, the motion of a top (i.e., the motion around an instantaneous axis with one fixed point), considered in the following paragraph 8.7, has three degrees of freedom: the rotational angle (one degree of freedom) and the orientation of ω (two further degrees of freedom). This motion is likewise described by (8.19), but with e˙ ω = 0. In this paragraph the more simple case of the fixed axis shall be treated.

248

8 Dynamics of a Rigid Body

8.6.1 Inertia Tensor and Moment of Inertia The part of the energy connected with the rotation around the fixed axis is Trot = 12 ω · Θ · ω = 12 ω eω · Θ · eω ω = 12 I ω 2 with the moment of inertia (rotational inertia) I = eω · Θ · eω .

(8.20)

For the representation of the body by point masses one obtains  2 ri⊥ mi I=

(8.21)

i

and for the representation by a mass distribution  I=

 2 r⊥ dm =

2 3 ρ(r) r⊥ d r,

(8.22)

where r⊥ is the distance of the point with the position vector r from the axis of rotation, cf. Fig. 8.14.

Fig. 8.14. The definition of r⊥

Proof: I = eω · Θ · e ω = =

 i



mi eω · (ri2 1 − ri ⊗ ri ) · eω

i

    2 mi ri2 − (eω · r i )2 = mi (eω × ri )2 = mi ri⊥ . i

i

 In cartesian (x, y, z) or sometimes better in cylindrical coordinates 2 = x2 + y 2 one chooses customarily (r⊥ , ϕ, z) with r⊥ ez = eω .

8.6 Rotation Around a Fixed Axis

Then one has I = Θzz =



249

mi (x2i + yi2 )

i



and I=

 dx

 dy

dz ρ(x, y, z)(x2 + y 2 ).

Also here Steiner’s theorem (8.15) can be of use, if one knows the moment (S) of inertia I (S) = Θzz with respect to the center of mass or if one can calculate it easily, then (S) 2 I = Θzz = Θzz + M R⊥ , where R⊥ is the distance of the center of mass from the axis of rotation. 8.6.2 The Equation of Motion The equation of motion reads in general ˙ N = L,

L = Θ · ω,

˙ = 0, since the moment of inertia changes with and in general12 one has Θ the change of the orientation with time. In this paragraph, however, one deals with the motion around a fixed axis of rotation, ω = ω eω N = N eω I = eω · Θ · e ω , and then one has e˙ ω = 0



I˙ = 0,

thus N = I ω˙

(fixed axis of rotation).

(8.23)

Comment: With eω = ez fixed also Θzz is fixed, but for example Θxx (in the space coordinate system) changes with time. However, the change of Θxx and Θyy and possibly Θxy is of no interest for the motion with the fixed axis ez , unless one wants to calculate the torque of constraint. 8.6.3 Example: Motion in the Homogeneous Gravitational Field: The Physical Pendulum The plane physical pendulum, shown in Fig. 8.15, is an example for the abovetreated rotation around a fixed axis of rotation. 12

See Sect. 8.7 further below.

250

8 Dynamics of a Rigid Body





0 →

ϕ

R S





K = Mg

Fig. 8.15. The notations for the physical pendulum

In the homogeneous gravitational field the torque is N = R × F = M R × g, where R is the vector from a point on the axis of rotation to the center of mass and F = M g the force on the total mass M thought of as assembled in the center of mass. Proof: For a system of point masses the torque is   ri × F i = r i × mi g N = i

=



i

mi r i × g = M R × g = R × F .

i

 The component in direction of the (horizontal) axis of rotation eω is N = eω · N = eω · R × F = eω · (−RM g sin ϕ eω ) = −RM g sin ϕ, where ϕ is the angle between R and F . The angular momentum is Lz = Θzz ϕ˙ = I ϕ. ˙ For a pure rotational motion the equation of motion is thus I ϕ¨ = −RM g sin ϕ. The equation of motion is the pendulum equation ϕ¨ = −

RM g sin ϕ. I

For comparison: The equation of motion of the mathematical pendulum is ϕ¨ = −

g sin ϕ. l

8.6 Rotation Around a Fixed Axis

251

In order to make the similarity more apparent, one introduces an effective length l∗ = I/RM and obtains

g sin ϕ. l∗ For small amplitudes one has then the frequency   g RM g ω= . = ∗ l I ϕ¨ = −

8.6.4 Example: A Cylinder Rolling on an Inclined Plane In this example the direction of the axis is fixed; however, the body can execute a translation in addition. In the (x, y)-system of Fig. 8.16 the forces on the body are (1) the gravitational force M g = M g (sin α, − cos α); it causes an acceleration in the x direction and a torque around the point of contact A, but not, however, around the center of mass; (2) the force N 0 of the support (as a force of constraint); it does not cause an acceleration or a torque, neither around the point of contact nor around the center of mass; (3) the frictional force R0 (as a force of constraint); it prevents sliding and can be substituted by the condition of rolling; it causes an acceleration in negative x direction and a torque with respect to the center of mass (but not with respect to the point of contact). The rolling condition is (with the angle ϕ as in Fig. 8.16, which is not the customary angle ϕ of the plane polar coordinates)

Fig. 8.16. The coordinate system for the body rolling on the inclined plane

252

8 Dynamics of a Rigid Body



dx = r dϕ

x = rϕ + x0 ,

and as an additional constraint one can use y = const = 0. As treated in Sect. 8.2 one can choose the position of the axis of rotation. Either the axis through the center of mass or the one through the (instantaneously fixed) point of contact are obvious. Method I: Axis of Rotation Through the Center of Mass If one chooses a coordinate system as in Fig. 8.16, the equations of motion are Mx ¨ = M g sin α − R0 M y¨ = −M g cos α + N0 = 0 I (S) ϕ¨ = r R0 . The gravitational force M g here does not cause any torque (since the lever vanishes). With ϕ¨ = x ¨/r one obtains on the one hand R0 = I (S)

ϕ¨ x ¨ = I (S) 2 r r

from the angular momentum conservation and on the other hand ¨ R0 = M g sin α − M x from the momentum conservation and from this # (S) $ I x ¨ + M = M g sin α r2 or x¨ = g

sin α 1+

I (S) M r2

,

i.e., a constant acceleration. Alternatively with the Lagrangian instead of the Newtonian equations of motion: The Lagrangian function is # $2 1 x˙ 1 L = T − V = M x˙ 2 + I (S) + M gx sin α. 2 2 r The equation of motion is d ∂L ∂L 0= − = dt ∂ x˙ ∂x with the same result x ¨=

#

I (S) M+ 2 r

$

g sin α . 1 + I (S) /M r2

x¨ − M g sin α

8.7 Rotation Around a Fixed Point: The Top

253

Method II: Instantaneous Axis of Rotation The torque is now caused by the gravitational force (and not by the frictional force), N (A) = r × M g = rM g sin(π − α) e = rM g sin α e ˙ (A) = I (A) ω˙ = L with e = −ez and with the moment of inertia (with respect to the point of contact, according to the theorem of Steiner) I (A) = I (S) + M r2 . With ω˙ = ω˙ e, one obtains

ω˙ = ϕ¨ = x ¨/r

" ! rM g sin α = I (A) ϕ¨ = I (S) + M r2 x¨/r

or x¨ =

r2 M g sin α g sin α = I (S) I (S) + M r2 1+ M r2

as above.

8.7 Rotation Around a Fixed Point: The Top 8.7.1 Space-Fixed (Inertial) and Body-Fixed Coordinate System In the previous Sect. 8.6 the motion of the rigid body around a fixed axis has been investigated. In this paragraph the motion of the top shall be treated, i.e., the motion around an instantaneous axis with one fixed point.13 The velocity of a point mass is (8.19)

r˙ i = ω × ri . In contrast to the previous Sect. 8.6 here the direction of the rotational vector ω of the angular velocity is not fixed. This motion has three degrees of freedom: the rotational angle (one degree of freedom) and the orientation of ω (two additional degrees of freedom). The inertia tensor Θ of a body is a fixed quantity in the body system; however, in the space coordinate system it depends upon the instantaneous orientation of the body. To investigate the free14 motion of a body it would 13 14

The fixed point can be the (possibly force-free) center of mass. For the motion in an external torque field see the heavy top in Sect. 8.7.3 or Goldstein [1], Sects. 5–7.

254

8 Dynamics of a Rigid Body

thus be very disadvantageous, to work in an inertial system (S) (in which the inertia tensor Θ is time dependent); rather it is advantageous to work in the body system (S ) (in which Θ is independent of time). For the construction of a Lagrangian function one needs, among others, the kinetic energy of the rotation Trot = 12 ω · Θ · ω, and it would thus be advantageous to express the angular velocity ω (and the inertia tensor Θ) in the body system (S ). z z⬘

. ψ

ϕ

. ϕ

x⬘ Fig. 8.17. Exemplary description of the rotation in the body system

As a special example15 for a body-fixed system a body is considered, which rotates with the angular velocity ϕ˙ around its body x axis as shown in Fig. 8.17. The x axis of the body rotates around the (perpendicularly ori˙ Then the vector of the angular ented) space z axis with the angular velocity ψ. velocity in the space system is given by ˙ 0, 1) ˙ ψ, sin ψ, 0) + ψ(0, ω (S) = ϕ(cos and in the body system by 

˙ sin ϕ, cos ϕ). ω (S ) = (ϕ, ˙ 0, 0) + ψ(0, 8.7.2 * Euler Angles The Transformation The transformation of the coordinates of a point from one coordinate system to a second coordinate system rotated relative to the first is described by an orthogonal transformation U, r = U · r. 15

The general motion is described by the Euler angles and their derivatives with respect to time. See Sect. 8.7.2.

8.7 Rotation Around a Fixed Point: The Top

255

Fig. 8.18. The Euler angles

A general rotation is given by three angles16 and can be described by three consecutive rotations. Definition 27. (Euler angles): The orientation of a body can be described by the Euler angles ϕ, θ, and ψ, which are defined as follows:17 The primed and unprimed coordinates are fixed in the body and space system, respectively. The first rotation is around the z axis with the angle ϕ, compare Fig. 8.18, ⎞⎛ ⎞ ⎛ ⎞ ⎛ cos ϕ sin ϕ 0 x x1 ⎝ y1 ⎠ = ⎝ − sin ϕ cos ϕ 0 ⎠ ⎝ y ⎠ ⇔ r 1 = U1 · r. z1 0 0 1 z The second rotation is around the x1 axis with the angle θ, ⎛ ⎞ ⎛ ⎞⎛ ⎞ 1 0 0 x1 x2 ⎝ y2 ⎠ = ⎝ 0 cos θ sin θ ⎠ ⎝ y1 ⎠ ⇔ r 2 = U2 · r1 . z2 z1 0 − sin θ cos θ The third rotation is around the z2 axis with the angle ψ, ⎛ ⎞ ⎛ ⎞⎛ ⎞ x cos ψ sin ψ 0 x2 ⎝ y  ⎠ = ⎝ − sin ψ cos ψ 0 ⎠ ⎝ y2 ⎠ ⇔ r = U3 · r2 . z z2 0 0 1 16

17

For example the orientation of a point of the body is determined by two angles; the third angle describes then the rotation of the body around the axis, which goes through this point and the origin. See Goldstein [1]; Sommerfeld [6] has interchanged the notation of ϕ and ψ.

256

8 Dynamics of a Rigid Body

With r1 = U1 · r, one obtains

r = U · r

r2 = U2 · r 1 , with

r  = U3 · r2

U = U3 · U2 · U1 .

(8.24)

Comments: • In contrast to translations consecutively performed rotations do not commute, U1 · U2 = U2 · U1 , i.e., the succession of the rotations is not arbitrary. • Of this one can convince oneself easily with an example. To this end one considers, e.g., the rotations Ry and Rz of a position vector r = (x, y, z) around the y and around the z axis by 90◦ each with ⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ x 0 0 1 x z Ry ⎝ y ⎠ = ⎝ 0 1 0 ⎠ ⎝ y ⎠ = ⎝ y ⎠ z −1 0 0 z −x ⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ x 0 −1 0 x −y Rz ⎝ y ⎠ = ⎝ 1 0 0 ⎠ ⎝ y ⎠ = ⎝ x ⎠ . z 0 0 1 z z If one interchanges the order of the two rotations, one obtains a different result, ⎛ ⎞⎛ ⎞ ⎛ ⎞ 0 −1 0 0 0 1 0 −1 0 0 1⎠ Rz Ry = ⎝ 1 0 0 ⎠ ⎝ 0 1 0 ⎠ = ⎝ 0 0 0 1 −1 0 0 −1 0 0 ⎛ ⎞⎛ ⎞ ⎛ ⎞ 0 0 1 0 −1 0 0 0 1 Ry Rz = ⎝ 0 1 0 ⎠ ⎝ 1 0 0 ⎠ = ⎝ 1 0 0 ⎠ . −1 0 0 0 0 1 0 1 0 • The angles θ and ϕ describe the orientation of the body axis; the angle ψ describes the (eigen) rotation of the body around the body axis. ˙ • The angular velocity of the precession is ϕ, ˙ the one of the nutation is θ, ˙ and the one of the eigen angular momentum (spin) is ψ. The Angular Velocity If one has the kinetic energy T = 12 ω · Θ · ω in the space system, then not only the angular velocity ω but also the inertia tensor Θ is time dependent; thus one wants to know these quantities in the body system, in which the inertia tensor is constant. The angular velocity one be described by the Euler angles and their time dependence. In the space system, the angular velocity ωϕ of the rotation with ϕ˙ around the body axis (z3 axis) is

8.7 Rotation Around a Fixed Point: The Top



257



sin ψ sin θ ω ϕ = U3 · U2 · U1 · ez ϕ˙ = ϕ˙ ⎝ cos ψ sin θ ⎠ cos θ The angular velocity ω θ of the rotation with θ˙ around the x1 axis is ⎛ ⎞ cos ψ ω θ = U3 · U2 · ex θ˙ = θ˙ ⎝ − sin ψ ⎠ 0 The angular velocity ω ψ of the rotation with ψ˙ around the z axis is ⎛ ⎞ 0 ω ψ = ez ψ˙ = ψ˙ ⎝ 0 ⎠ 1 The total angular velocity is ω = ωϕ + ω θ + ω ψ . 8.7.3 * The Euler Equations of Motion In the space (inertial) system S the equations of motion for the top (pure rotations, no translations) is ˙ (S) = N (S) L with L(S) = Θ(S) · ω (S) . The angular velocity ω as well as the inertia tensor Θ in the space system (S) are varying in time. The transformation from the space to the body system (S ) is  ˙ (S) (7.8) ˙ (S ) + ω × L(S ) . = L N (S) = L One has thus



˙ (S ) = N (S) − ω × L(S ) , L

(8.25)

(S)

where N and ω are in the space system S. Agreement: For the rest of this paragraph the angular momentum L and the angular velocity ω shall be considered as given in the body-fixed system, and the indices S shall be omitted. ˙ ≡ 0 (in the system S ) and Θij = Ii δi,j in the With L = Θ · ω, Θ body principal-axis system one then finds the so-called Euler equations as the component-wise representation of (8.25) I1 ω˙ 1 = N1 + (I2 − I3 ) ω2 ω3 I2 ω˙ 2 = N2 + (I3 − I1 ) ω3 ω1 I3 ω˙ 3 = N3 + (I1 − I2 ) ω1 ω2 .

(8.26)

258

8 Dynamics of a Rigid Body

Proof: In the principal-axis body system one has on the one hand ⇒

Li = ωi Ii

L˙ i = ω˙ i Ii ;

on the other hand, one has ˙ (8.25) L = N − ω × L,



L˙ i = Ni −



ijk ωj Lk

jk

and from these two relations ω˙ i Ii = Ni −



ijk ωj ωk Ik

jk

(and cyclic) with the Levi-Civita tensor with the elements 4 +1 if (ijk) is an even permutation of (123) ijk = −1 if (ijk) is an odd permutation of (123) 0 otherwise.  Comment: Notice that the Euler equations (8.26) are nonlinear coupled differential equations (of first order) (for ω1 , ω2 and ω3 ). Since there is thus in general18 no “harmonic” approximation, the treatment of the dynamics of the angular momenta is much more difficult than that of oscillations.

8.8 The Force-Free Symmetrical Top 8.8.1 The Equations of Motion In the case of the symmetrical top the nonlinear system of (8.26) can be treated analytically. Let the principal moments of inertia be I1 = I2 = I3 . The force-free top is (by definition) without external torque, N = 0, thus, either the external field vanishes, or the top is placed in the homogeneous gravitational field and is suspended at the center of mass. Then the Euler equations of motion (8.8) reduce to I1 ω˙ 1 = −(I3 − I1 ) ω3 ω2 I1 ω˙ 2 = (I3 − I1 ) ω3 ω1 I3 ω˙ 3 = 0. 18

See but the following Sect. 8.8.

8.8 The Force-Free Symmetrical Top

259

From this one finds immediately a first solution ω3 = const.

(8.27)

With this a system of two coupled linear equations for ω1 and ω2 is obtained, which one can solve with an exponential ansatz ωi ∝ eλt One obtains

#

I1 λ −(I3 − I1 ) ω3

(i = 1, 2).

(I3 − I1 ) ω3 I1 λ

$#

ω1 ω2

$ = 0.

The solubility condition is the vanishing of the secular determinant, (I1 λ)2 + [(I3 − I1 ) ω3 ]2 = 0. One obtains the eigen values λ = ±i ω3 and the eigen vectors

I3 − I1 = ±i Ω I1

#

1 ∓i

(8.28)

$

(except for a phase and except for a normalization factor). The general solution is # $ # $ # $ 1 1 ω1 iΩt e−iΩt =a e +b i −i ω2

e3

ω

Ωt+ϕ

e2 e1

Fig. 8.19. The motion of the force-free symmetrical top (I3 > I1 ) in the body system

260

8 Dynamics of a Rigid Body

or ω1 = A cos(Ωt + ϕ) ω2 = A sin(Ωt + ϕ) ω3 = C with (8.28)

Ω = ω3

I3 − I1 . I1

The body does not rotate about a fixed symmetry axis. Rather, in the body system one has thus the following picture: The vector ω of the angular velocity of the rotating body precesses with constant angular velocity Ω (for I3 > I1 in the same direction as ω3 ) around the body symmetry axis e3 with ω3 = const, see Fig. 8.19. Example: The motion of the poles of the earth:19 The earth is flattened at the poles, such that one finds20 for the moments of inertia 1 I3 − I1 . ≈ I1 300 From this one obtains Ω≈

ω3 , 300

T3 =

2π = 1d ω3



T =

2π ≈ 300 d, Ω

i.e., the instantaneous axis of rotation of the earth should precess with the so-called Euler period of about 300 days. However, the yearly period (of 365 days) causes atmospheric redistributions of the mass, which changes the Euler period of 300 days. In addition, the earth is not rigid but elastic and primarily liquid in the interior; this reduces the effect; the resulting so-called Chandler period is T = 415 to 433 days [14, 16]. The radius of the pole motion is A ≈ 10 m, but varying in time/“Chandler wobble”). The rotation of the inclination with time against the ecliptic is another effect, which originates from the forces of the sun and the moon; the corresponding period is T ≈ 26,000 years. The geometrical north pole is the cut of the body symmetry axis through the surface of the earth; the kinematical north pole is that of ω. 8.8.2 * Stability of the Rotational Motion of the Top Consider a free (asymmetrical) top with three different principal axes Ii . The uniform rotations 19 20

The center of mass of the earth is considered as a fixed point; the rotation around the sun is thus neglected. (I3 − I1 )/I1 = 1/306 according to Goldstein [1], Sect. 5.6.

8.8 The Force-Free Symmetrical Top

ω i = ω i ei ,

(j = i)

ωj = 0

261

(8.29)

around each of the three principal axes of inertia ei are (stationary) solutions of the force-free equations of motion (8.26) (with N = 0) I1 ω˙ 1 − (I2 − I3 ) ω2 ω3 = 0 I2 ω˙ 2 − (I3 − I1 ) ω3 ω1 = 0 I3 ω˙ 3 − (I1 − I2 ) ω1 ω2 = 0.

(8.30)

Not all stationary solutions (8.29) are stable:21 Now let ω = (0, 0, ω3 ) be the solution (8.27), the stability of which is to be determined. To this end one investigates the behavior of a neighboring solution ω + s (0, 0, ω3 ) + (s1 , s2 , s3 ) with

si  ω 3 .

If one inserts this into the equation of motion (8.30) and, because of the smallness of the deviations si , takes into account only the terms linear in si and neglects the terms of higher order, one obtains a system of coupled linear equations, I1 s˙ 1 − (I2 − I3 ) ω3 s2 = 0 I2 s˙ 2 − (I3 − I1 ) ω3 s1 = 0 (8.31) I3 s˙ 3 = 0. One finds from the last of the three equations s3 = const. For the two other components one makes the ansatz si ∝ eλt and obtains

#

λI1 −(I3 − I1 )ω3

i = 1, 2

−(I2 − I3 )ω3 λI2

$#

(8.32) s1 s2

$ = 0.

The vanishing of the secular determinant yields λ2 =

(I2 − I3 )(I3 − I1 ) 2 ω3 . I1 I2

(8.33)

Case λ2 < 0: If I3 is the largest or the smallest of the three principal moments of inertia, then λ is imaginary; according to (8.32) the deviation s oscillates with time (in the (x, y)-plane around the stationary solution (along the z direction), whence the solution is stable. (In contrast to the exact oscillation in the case of the symmetrical top of Sect. 8.8 one has the solution here only for small deviations from the stationary solution.) Case λ2 > 0: If I3 is between I1 and I2 , the solution (8.32) has one part decreasing and one part increasing exponentially with time. Because of the increasing term the solution for s(t) moves away from the stationary solution ω = (0, 0, ω3 ), which is thus unstable. 21

For the stability see Appendix F.3.

262

8 Dynamics of a Rigid Body

Case I1 = I2 (λ2 < 0): For the symmetrical top the rotation around the symmetry axis e3 is thus stable, see Sect. 8.8, where this solution has been found already as an exact (and not only approximate) solution. Case I2 = I3 or I1 = I3 : For the rotation around the axis e3 perpendicular to the symmetry axis e1 one would obtain λ = 0 according to (8.33); thus one must find a second solution. With I2 = I3 one obtains from (8.31) the equations of motion I1 s˙ 1 = 0 I2 s˙ 2 − (I3 − I1 ) ω3 s1 = 0 I3 s˙ 3 = 0. With the solutions s1 = const. and s3 = const. one finds a solution linearly increasing or decreasing with time, s2 (t) = s2 (0) +

I3 − I1 s1 ω3 t; I3

in this case the stationary solution (8.29) is thus unstable.

Summary: Dynamics of a Rigid Body Angular momentum w.r.t. center of mass: (8.5)

L(S) = Θ(S) · ω Angular momentum w.r.t. point Q: (8.16)

L(Q) = L(S) + R × P Stoner’s theorem:   (8.15) Θ(Q) = Θ(S) + M R2 1 − R ⊗ R Inertia tensor elements: Θij

(8.14)

=



  mi ri2 δi,j − xi,i xi,j =



  d3rρ(r) δi,j r2 − xi xj

i

Principal axes ei and principal moments of inertia Ii : (8.18)

Θ · ei = Ii ei

i = 1, 2, 3

Problems

263

Kinetic energy: (8.1) 1 ˙ 2 + 1 ω · Θ(S) · ω relative to the center of mass = 2 MR 2 (8.6) 1 T = 2 ω · Θ(P) · ω relative to the instantaneously fixed point (S ) (8.25) (S) (S )

T

˙ L ⇔

= N

−ω×L

Euler equations

(8.26)

I3 ω˙ 3 = N3 + (I1 − I2 ) ω1 ω2

and cyclic

Fixed axis of rotation: e˙ ω = 0, I = eω · Θ · eω , L = eω · L = Iω, N = eω · N : ⇒ Euler angles

r = U · r

N

(8.23)

= I ω˙

with U = U3 · U2 · U1



⎞ cos ϕ sin ϕ 0 U1 = ⎝ − sin ϕ cos ϕ 0 ⎠ 0 0 1 ⎛ ⎞ 1 0 0 sin θ ⎠ U2 = ⎝ 0 cos θ 0 − sin θ cos θ ⎛ ⎞ cos θ sin ψ 0 U2 = ⎝ − sin ψ cos ϕ 0 ⎠ 0 0 1

Problems 8.1. Rolling Motion with Fixed Axis. Consider a cylinder on an inclined plane. Verify the separation of the kinetic energy, which one can write either as sum of translational energy and rotational energy (relative to the center of mass S) or as a pure rotational energy (relative to the instantaneously fixed point of contact P ). 8.2. Center of Mass and Inertia Tensor of a Symmetrical Pyramid. Given be a pyramid of constant density ρ which is bordered by the planes x + y + z = a, x = 0, y = 0, and z = 0. (a) First, making a sketch get insight into the position and symmetry of the pyramid. (b) Determine the center of mass. (c) Which components of the inertia tensors are equal? Determine the inertia tensor with respect to the coordinate origin (0, 0, 0) taking into account the symmetry considerations in part (b).

264

8 Dynamics of a Rigid Body

(d) A principal axis can be noticed from the symmetry of the pyramid. Write down this direction (unit vector), and determine the moments of inertia with respect to this direction. 8.3. Principal Moments of Inertia of Various Bodies. Determine the principal moments of inertia of the following geometrical bodies of homogeneous mass distributions: (a) Cube with edge length a, (b) Sphere of radius R, (c) Cylinder of height h and radius R, (d) Circular cone of height h and radius R, (e) Sphere consisting of two hemispheres of different density. 8.4. Principal Moments of Inertia of a Tetrahedral Molecule. Determine the principal moments of inertia of a tetrahedral molecule of the edge length a with four masses m at the corners. 8.5. The Rotating Cuboid (Rectangular Parallel Epiped). A homogeneous cuboid (edge lengths a, b, c; constant mass distribution ρ) rotates with the angular velocity ω around a volume diagonal. (a) Determine the inertia tensor. (Which coordinate system is useful?) (b) Determine the angular momentum and the kinetic energy which are connected with this rotation. 8.6. Inertia Tensor and Moments of Inertia of a Sphere with Homogeneous Mass Distribution. (a) Determine the inertia tensor of a sphere of radius R with homogeneous mass distribution with respect to the center of mass S. (b) With the help of Steiner’s theorem write down the inertia tensors with respect to the points P1 = (0, R, 0) and P2 = √R2 (0, 1, 1). (c) What are the principal axes of the inertia tensor with respect to the points S, P1 and P2 ? Write down the position of the inertia ellipsoid. (d) Determine the moments of inertia with respect to the axes parallel to the x axis, y axis, and z axis through the points S, P1 , and P2 . 8.7. The Spool Pulled by a Thread. At a spool of radius a is attached a cylindrical center piece of radius b < a (Fig. 8.20). On this a force K acts under the angle ϑ with respect to the horizontal direction. What motion results, if there is no sliding?

W

K Fig. 8.20. Thread and spool (for problem 8.7)

Problems

265

8.8. Cylinder Rolling Inside of a Hollow Cylinder. Determine the kinetic energy T of a homogeneous cylinder of radius a which rolls inside of a surface of a cylinder of radius R > a (Fig. 8.21). First, determine this from the rotational motion of the body around the momentary touching line of these two cylinders. Then compare with the result of the determination of T from the translation of the center of mass of the rolling cylinder and of the rotation around its symmetry axis.

R a Fig. 8.21. The cylinder of radius a rolling inside of a hollow cylinder of radius R (for problem 8.8)

8.9. Cylinder Rolling on an Inclined Plane. Consider a cylinder (radius R) which moves on an inclined plane (inclination α) under the influence of the gravity see Fig. 8.22.

R α

Fig. 8.22. Cylinder on an inclined plane (for problem 8.9)

(a) Give an expression for the Lagrangian function under the condition that the cylinder rolls, and write down the total energy (expressed by the coordinate along the inclined plane). Compare the velocities for a hollow cylinder and a massive cylinder of equal total mass and of equal dimensions at equal total energy. Which cylinder rolls faster? Hint: There is no need for an explicit calculation of the moments of inertia. (b) The cause of the rolling motion is a torque. Give an expression for the rate of change of the angular momentum, once with respect to the contact line and once with respect to the central axis of the cylinder.

266

8 Dynamics of a Rigid Body

8.10. The Rolling Pendulum. Using the Lagrangian equations of the second kind formulate the vibrational equation of motion for the rolling pendulum (see Fig. 8.23). The weightless lever of the pendulum of length l and mass m is rigidly connected to the cylinder (radius R, mass M , moment of inertia I). The cylinder rolls without sliding on the horizontal plane.

Fig. 8.23. Rolling pendulum (for problem 8.10)

(a) Choose an appropriate coordinate for the description of the problem. (b) Express the coordinates of m as well as of the center of mass of the cylinder in terms of this generalized coordinate. (c) What is the Lagrangian function? (d) What is the equation of motion? Linearize the equation of motion. Show that one obtains the same result if one takes into account only bilinear expressions in generalized coordinate and velocity in the Lagrangian function. (e) What is the vibrational frequency for small amplitudes? 8.11. Stable and Unstable Rotation. Solving of the Euler equations for an unsymmetrical top show that the rotation around the axis of largest or smallest principal moments of inertias is stable, but is unstable around the axis of the intermediate principal moment of inertia. 8.12. Linear and Rotational Motion with Friction. A soccer ball is kicked such that at the beginning it slides without rotation across the lawn. Due to the friction with the lawn it gradually starts rotating. Finally the ball rolls without sliding. Let the frictional force Fr be proportional to the weight of the ball, |Fr | = M g μ ; here, M is the mass of the ball, g the gravitational acceleration, and μ the friction coefficient. Let R be the radius and IF the moment of inertia of the ball around an axis through the center. Let the initial velocity be v0 . (a) What are the equations of motion for the translational and the rotational motion? (b) After which time te has the motion of the ball changed into a pure rolling motion? How large is then the translational velocity ve ?

Problems

267

(c) Determine the moments of inertia IF . Use the approximation of the thickness of the leather being small as compared to the radius of the ball. For comparison: The moments of inertia IK of a massive sphere around an axis through the center is IK = 25 MK R2 . Can you tell without explicit calculation which of the two moments of inertia has to be the larger one? (d) Which fraction ΔE/E0 of the initial kinetic energy E0 is consumed by the friction? (e) For which quantity of the motion is there a conservation theorem? 8.13. The Stumbling Cube. Given is a homogeneous cube of edge length a and of mass M which moves with an initial velocity v0 without friction on a horizontal support (Fig. 8.24).

B

B

B⬘ A

A = A⬘

A⬘

A

Fig. 8.24. The stumbling cube: For the parts (b) (left), (c) and (d) (middle), and (e) (right); the obstacle is denoted by the point (for problem 8.13)

(a) Determine the moments of inertia I0 of the cube with respect to an arbitrary axis through the cube center and with respect to an edge of the cube. (b) The edge A of the cube hits an obstacle H, the height of which can be neglected. Which conservation theorems hold true (mathematical formulation!) or not, if the collision is completely inelastic? (c) How large has v0 to be for the cube just about to tip? (d) After the end of tipping the cube edge AB has moved to A B  . Which impulse ( K dt) must act in A or B  for the cube just to remain in this position? Assume that horizontal forces can act only in A . Make a drawing. (e) Which conservation theorems hold true, if the collision against the obstacle occurs completely elastic (the obstacle being, e. g. a railway buffer)? What are the formulae of these conservation theorems if the cube center assumes ˙ Which values the velocity components vx , vy and the rotational velocity ϕ? do these quantities have immediately after the initial kick? What is the condition for tipping?

268

8 Dynamics of a Rigid Body

8.14. Stability of a Heavy Symmetrical Top. Formulate the Lagrangian function for the heavy top. With ω1 = ϑ˙ cos ψ + ρ˙ sin ϑ sin ψ, ω2 = ϑ˙ sin ψ + ρ˙ sin ϑ cos ψ, and ω3 = ϑ˙ cos θ + ψ˙ transform to new variables ϑ, ρ, ψ, and write down the Lagrangian equations for these latter generalized coordinates. Write down the conserved quantities for the system. Using the conserved quantities rewrite the total energy as an ˙ The result corresponds to the expression which depends only upon ϑ and ϑ. energy of a particle in an effective potential. Under which condition has this potential a minimum (stability of the motion of the top) at ϑ = 0? What kind of motion does the top execute if this condition is fulfilled?

9 Hamiltonian Dynamics

9.1 Hamiltonian Equations of Motion In Chap. 3 the dynamics of a system has been derived from the Lagrangian function, and the Hamiltonian function and the canonical momenta (as possible constants of the motion) have evolved as important quantities. The Lagrangian function L(q, q) ˙ and the Hamiltonian function H(p, q) are connected to each other by the simple relation, namely by a Legendre transformation1 H(p, q) = pq˙ − L(q, q) ˙ with

p=

∂L . ∂ q˙

In this chapter the equations of motion shall be derived from the Hamiltonian function rather than from the Lagrangian function. 9.1.1 The Hamiltonian Function In the Lagrangian formulation of mechanics, the Lagrangian function and the Hamiltonian function are related to each other by a Legendre transformation,2 where a transition is made from the generalized velocities to the generalized (canonical) momenta. Let the Lagrangian function be L(q, q, ˙ t) with

q = {ql }

with the canonical momenta pl =

and q˙ = {q˙l }

∂L . ∂ q˙l

Definition 28. (Hamiltonian function): The Hamiltonian function is3  H(p, q, t) = pl q˙l (p, q) − L(q, q(p, ˙ q), t) . (9.1) l

1 2 3

See Appendix J. See Appendix J. This is a reiteration of the Definition (3.66).

270

9 Hamiltonian Dynamics

Example: The Pendulum in a Homogeneous Gravitational Field The Lagrangian function is T = 12 ml2 ϕ˙ 2

V = mgl (1 − cos ϕ) L = T − V = 12 ml2 ϕ˙ 2 − mgl (1 − cos ϕ) . The canonical momentum in this example is the angular momentum pϕ =

∂L = ml2 ϕ. ˙ ∂ ϕ˙

The Hamiltonian function is4 H = pϕ ϕ˙ − L = ml2 ϕ˙ 2 −

1

2 ml

 ϕ˙ − mgl (1 − cos ϕ)

2 2

= 12 ml2 ϕ˙ 2 + mgl (1 − cos ϕ) 1 2 = p + mgl (1 − cos ϕ) = T + V. 2ml2 ϕ

(9.2)

Comments: • The canonical momentum is a function the q and q, ˙ pl = pl (q, q, ˙ t). • The reverse relation q˙ = q(p, ˙ q, t) can be obtained if one has det

∂2L = 0. ∂ q˙i ∂ q˙j

• The Hamiltonian function given in (9.1) is the negative of the usual Legendre transform. • The reverse  pl (q, q, ˙ t)q˙l − H(p(q, q), ˙ q, t) L(q, q, ˙ t) = l

is unique if one has det

4

See also Sect. 3.11.3.

∂2H = 0. ∂pi ∂pj

9.1 Hamiltonian Equations of Motion

271

• If the kinetic energy is quadratic in the generalized velocities and if the potential energy is independent of the generalized velocities, the Hamiltonian function is equal to the total energy.5 This one can see from the example of a system of particles in a potential. The Lagrangian function is  1 ˙2 L= 2 mi r i − V ({r i }) = L({r i }, {pi }). i

The canonical momenta are equal to the kinetic momenta, pi =

∂L = mi r˙ i . ∂ r˙ i

The Hamiltonian function then is  1  r˙ i · pi − L = p2i + V ({r i }) = H({ri }, {pi }). H= 2m i i i • For rheonomous constraints the Hamiltonian function in general is not equal to the total energy.6 9.1.2 Canonical (Hamiltonian) Equations of Motion The canonical (Hamiltonian) equations of motion (for conservative systems) are ∂H ∂H q˙l = , p˙l = − . (9.3) ∂pl ∂ql Proof: One has on the one hand ) *  dH = d pl q˙l − L =



l

(pl dq˙l + q˙l dpl ) −

 # ∂L

dq˙l +

∂ q˙l l # $  d ∂L ∂L = dt dql − q˙l dpl − dt ∂ q˙l ∂t l  ∂L = dt. (q˙l dpl − p˙ l dql ) − ∂t l

∂L dql ∂ql

$ −

∂L dt ∂t

l

On the other hand, one has $  # ∂H ∂H ∂H dt. dH = dpl + dql + ∂pl ∂ql ∂t l

By comparison one finds the proposition. 5 6



Compare again the counter-example in Sect. 3.11.3. This would be for example the case the bead sliding on the rotating rod of Sect. 3.5.6.

272

9 Hamiltonian Dynamics

Example: (A) The Pendulum The Hamiltonian function is (9.2)

H =

1 2 p + mgl (1 − cos ϕ) . 2ml2 ϕ

The Hamiltonian equations of motion are (ql = ϕ) pϕ ∂H = ∂pϕ ml2 ∂H = −mgl sin ϕ. p˙ ϕ = − ∂ϕ ϕ˙ =

(9.4) (9.5)

The familiar second-order differential equation of the pendulum is recovered, if one differentiates the first (9.4) with respect to time and uses the second (9.5) to eliminate p˙ϕ , g p˙ ϕ = − sin ϕ. ϕ¨ = ml2 l Comments: • If qi is a cyclic coordinate, then the corresponding canonical momentum is a constant of the motion (a conserved quantity),7 ∂L =0 ∂qi



pi = const;

the conserved canonical momentum is thus fixed by the initial conditions and is not a (time dependent) variable but simply a parameter. Because of the equation of motion 0 = p˙i = −

∂H ∂qi

for the corresponding canonical momentum the Hamiltonian function H then does not depend upon the cyclic variable qi . • With f degrees of freedom and s cyclic coordinates the Lagrangian equations are f coupled differential equations of second order. (These are differential equations for the f generalized coordinates qi ; however, s equations for the cyclic variables can be integrated once trivially). • In contrast to the previous comment, the Hamiltonian equations of motion are thus only 2(f − s) coupled differential equations of first order for the noncyclic coordinates and the corresponding canonical momenta. • If it is possible to choose the generalized coordinates such that they are all cyclic, then one has 7

See Sect. 3.11.2.

9.1 Hamiltonian Equations of Motion

0=

∂H ∂qi

273

⇒ pi = αi = const ⇒ H = H({pj }) = H({αj }) = const ⇒ q˙i =

∂H ∂H({αj }) = = βi = const ⇒ qi = βi t + γi . ∂pi ∂αi

The equations of motion can then be solved nearly trivially. The difficulty lies in the determination of the appropriate generalized coordinates.8 • A partial decoupling of the Lagrangian equations as a consequence of the separable form of the Lagrangian function9 is transferred to a separable form of the Hamiltonian function and thereby to a decoupling of the Hamiltonian equations of motion, L = L1 + L2



H = H1 + H 2 .

• From (9.3) and the proof for it one has ∂H ∂L dH = =− . dt ∂t ∂t

(9.6)

Definition 29. (Canonical system): A system described by the Hamiltonian function H is called canonical, if the canonical equations (9.3) are fulfilled. Example: (B) Charged Particle in a Static Electromagnetic Field For a particle with mass m and charge q in a static electromagnetic field one had found in Sect. 3.7.2 the Lagrangian function (3.31)

L =

m 2 r˙ − qφ(r, t) + qA(r, t) · r˙ 2

and the canonical momentum p=

∂L = mr˙ + qA ∂ r˙

with A = A[c]/c. The kinetic momentum is mr˙ = p − qA. With this one obtains for the Hamiltonian function H=

8 9

1 2 (p − qA) + qφ. 2m

See the Hamilton–Jacobi theory in Sect. 9.6. See Sect. 3.5.8.

(9.7)

274

9 Hamiltonian Dynamics

Proof: One obtains by inserting H = p · r˙ − L =p· =

p − qA 1 − m m 2

#

p − qA m

$2 + qφ − qA ·

p − qA m

1 (p − qA)2 + qφ. 2m 

Comments: • The Hamiltonian function in the form of (9.7) is used in quantum mechanics by taking it as an operator rather than function. (One denotes this form then as the form of the “minimal coupling.”) • In the form (9.7) the Hamiltonian function H = T + V is the sum of kinetic energy 12 mr˙ 2 and potential energy qφ, while the Lagrangian function cannot be represented as the difference of these quantities, except if one introduces the generalized potential energy of Sect. 3.7.2. 9.1.3 * (Un)Ambiguity of the Hamiltonian Function: Gauge Transformation In Sect. 3.8.1 one had found that the gauge transformation of the Lagrangian function L, d % (3.32) L = L + F (q, t) dt (with ∂F/∂ q˙ = 0) leaves the equations of motion unchanged. In addition, one has for the canonical momentum (3.34)

p% = p +

∂F . ∂q

In the same way the Hamiltonian functions H and the gauged Hamiltonian function % = H − ∂ F (q, t) H (9.8) ∂t lead to identical equations of motion.10 Proof: (i) One needs ∂F ∂F dF = q˙ + . dt ∂q ∂t 10

(9.9)

7 = cL leads to identical Lagrangian equations of moAlso the transformation L tion; but then the Hamiltonian function resulting thereof is (under the usual conditions) the energy scaled by a factor of c.

9.2 * Poisson Brackets

275

(ii) Inserting one obtains % H

(3.34),(3.32)

=

(9.9)

=

$ # $ # ∂F d p+ q˙ − L + F ∂q dt   ∂F ∂F ∂F ∂F q˙ − q˙ + . pq˙ − L + =H− ∂q ∂q ∂t ∂t %= p%q˙ − L

(iii) For the derivation of the equation of motion for p% one must pay attention to which variable one keeps fixed in the partial differentiation. One obtains with q% = q ) * * )  # $ % % ∂ ∂H ∂F (q, t) ∂H (9.8) = = H(q, p%, t) − ∂ q%l ∂ql ∂ql ∂t p % p % p % ⎤ ⎡ $ # 2  # ∂H $ # ∂pj $ ∂H (9.9) ⎦− ∂ F . ⎣ = + ∂ql p ∂pj q ∂ql p% ∂ql ∂t j #

With

one obtains * ) % ∂H ∂ q%l

p %

∂pj ∂ql

= −p˙ l −

$

(3.34)

#

=

p %

 j

q˙j

∂ p%j ∂ql

$ p %



∂2F ∂2F =0− ∂ql ∂qj ∂ql ∂qj

∂2F ∂2F − ∂ql ∂qj ∂ql ∂t

(3.34)

= −p˙ l −

d partialF ˙ . = −p% l dt ∂ql

(iv) The equation of motion for q% = q is ) * * )  # # $  # $ $ % % ∂ ∂p ∂ ∂H ∂H ∂F ∂F = = = H− H− ∂ p% ∂ p% ∂ p% ∂t ∂ p% q ∂p ∂t q q q%

q

= 1 [q˙ − 0] , since F depends only upon q (but not upon p).  Comment: One should not confuse the partial time derivative of the gauge function of the Hamiltonian function in (9.8) with the total derivative of the gauge function of the Lagrangian function in (3.32).

9.2 * Poisson Brackets 9.2.1 Definition 29. (Poisson Brackets): Let p = {pl } and q = {ql } be canonical variables. Let two functions f (p, q, t) and g(p, q, t) be given. The Poisson brackets {. . .} are defined as $  # ∂f ∂g ∂f ∂g {f, g} = − (9.10) . ∂pl ∂ql ∂ql ∂pl l

276

9 Hamiltonian Dynamics

Comment: Some authors write for the Poisson brackets in (9.10) explicitly {f, g}p,q , and many authors11 define the negative of the Poisson brackets, {f, g}q,p = − {f, g}p,q . One must thus check the definition, if there are no indices on the brackets! 9.2.2 Equations of Motion Definition 30. (Observable): An observable is a measurable quantity. The equation of motion of an observables G(p, q, t) can be written in the form ∂G dG = + {H, G} . (9.11) dt ∂t Proof: Differentiation yields    ∂G dG ∂G ∂G = p˙ l + q˙l + dt ∂pl ∂ql ∂t l   ∂G ∂G ∂H ∂G ∂H ∂G (9.3)  = {H, G} + . = + − + ∂pl ∂ql ∂ql ∂pl ∂t ∂t l

 In particular the canonical equations of motion (9.3) for the canonical variables take the form12 p˙ l = {H, pl } q˙l = {H, ql }

(9.12)

An observable G(pl , ql ) (with ∂G/∂t = 0) is a conserved quantity, if one has G˙ = 0



{H, G} = 0 (and

∂G = 0). ∂t

(9.13)

Comment: For a system with explicitly time independent Lagrangian function thus the Hamiltonian function is explicitly time independent, and because of {H, H} ≡ 0 then the Hamiltonian function is a conserved quantity.13 11 12 13

In this way Goldstein [1]. The definition given in (9.10) is also used by Scheck [5]. For an example see Sect. 9.2.4. See also Sect. 3.11.

9.2 * Poisson Brackets

277

9.2.3 Fundamental Poisson Brackets The fundamental Poisson brackets (for canonical variables) are {pi , pj } = 0, Proof: {pi , qj } =

 # ∂pi ∂qj l

{pi , pj } =

{qi , qj } = 0,

∂qj ∂pi − ∂pl ∂ql ∂pl ∂ql

 # ∂pi ∂pj l

∂pl ∂ql



{pi , qj } = δi,j .

$

∂pj ∂pi ∂pl ∂ql

=



(9.14)

(δi,l δj,l − 0) = δi,j

l

$ =



(δi,l 0 − δj,l 0) = 0,

l



etc. 9.2.4 Properties of the Poisson Brackets The properties of the Poisson brackets are {f, g} = −{g, f }

Antisymmetry

{c, f } = 0

Zero element

(c = const.)

(9.16)

{f, (ag + bh)} = a{f, g} + b{f, h}

(9.17)

{f, gh} = {f, g}h + g{f, h}

(9.18)

{f, {g, h}} + {g, {h, f }} + {h, {f, g}} = 0

(9.19)

Distribution law Product rule Jacobi identity

(9.15)

with the functions f , g, and h of q, p, and t and the constants a, b, and c. Comment: The properties given in (9.15)–(9.19) are characteristic of a mathematical framework which is realized not only by the Poisson brackets but for example also by the cross product of vectors in R3 , by the so-called commutator AB − BA of matrices A and B or by the commutator [A, B] ≡ AB − BA of operators A and B in quantum mechanics.14 9.2.5 Example: Harmonic oscillator With the Hamiltonian function H = 12 aP 2 + 12 bQ2 (a = 1/m, b = mω 2 ) one obtains (9.18) Q˙ = {H, Q} = { 21 aP 2 , Q} = 12 aP {P, Q} + 12 a{P, Q}P

(9.14)

= aP

(9.18) (9.14) P˙ = {H, P } = { 21 bQ2 , P } = 12 bQ{Q, P } + 12 b{Q, P }Q = −bQ. 14

See the following in Sect. 9.2.6.

278

9 Hamiltonian Dynamics

Comments: ¨ = −abQ = • From this one can recover the familiar equation of motion Q −ω 2 Q if one ever wants. • In this example one has performed the manipulations in an algebraic way, using the product rule (9.18) and the fundamental Poisson brackets (9.14), but not the special realization (9.10) of the Poisson brackets. Analogous procedures will be applied also in the case of quantum mechanics. 9.2.6 Transition to Quantum Mechanics In quantum mechanics the observables, i.e., the measurable quantities, are replaced by operators. Furthermore, the Poisson brackets {f, g} is substituted by the commutator brackets [f, g] = f g − gf with the replacement i [f, g].  In particular one obtains for conjugate quantities the quantization prescription {f, g} →

[pi , qj ] =

 δi,j . i

The quantum-mechanical replacement of the fundamental brackets (9.14) is basic to the “quantization” in the physics of microscopic particles and waves and likewise represents the grounds in that regime of physics. The mathematical structure of the Poisson-brackets algebra of variables is thus transferred to the mathematical structure of the commutator algebra of operators.

9.3 Configuration Space and Phase Space 9.3.1 Configuration Space The system of the equations of motion q¨ = F (q, q, ˙ t) with q = {ql } = q1 , . . . , qf

(9.20)

for the generalized coordinates ql is a system of f coupled differential equations of second order for the generalized coordinates ql . The solutions q(t) are curves in the f -dimensional configuration space. Definition 31. (State): A state of a system is characterized by a minimum number of quantities, which yield a maximum number of information on the system.

9.3 Configuration Space and Phase Space

279

Comments: • A state of the mechanical system of point masses can thus be fixed, e.g., by the set of the independent variables q and q˙ (at a given time t) (analogously, e.g., to the initial conditions needed for the solution of the equations of motion), i.e., by a point in phase space, which is the union of the configuration space and the corresponding tangential space, see the sketch in Fig. 9.1.

Fig. 9.1. An example for a point of a (twodimensional) configuration space and the related tangent space

• With a given set of q and q˙ (at a given time t) all other (mechanical) observables of the system (momentum, angular momentum, kinetic and potential energies, etc. of the single particles as well as of the total system) are fixed. • A part of the information on the system thus is its development (the process; in mechanics this is the time development; in the equilibrium thermodynamics it is the development with pressure, temperature, etc.) • With the unique solution of the equations of motion by fixing the state at one time t0 thus the state at an arbitrary other time t is uniquely fixed as well. The motion is deterministic. • In quantum mechanics coordinate and canonical momentum cannot be given simultaneously, and the characterization of a state must be done differently. The time development of the state is deterministic nevertheless. 9.3.2 Phase Space The system (9.20) of f differential equations of second order can be rewritten in twice as large a system of differential equations of first order y˙ = F (q, y, t)

(9.21)

q˙ = y

(9.22)

for the f variables q (from the configuration space) and the f variables y (from the corresponding tangential space). At a given time t the state is fixed

280

9 Hamiltonian Dynamics

by a point (q(t), y(t)) in phase space.15 Parameterized by t, the (q(t), y(t)) are curves (trajectories, orbits) in the 2f -dimensional phase space.

V

z1 z2

z1

Fig. 9.2. Potential energy of the (undamped) harmonic oscillator with four different values of the total energy (top) and the corresponding phase-space trajectories (bottom)

Example: (A) Harmonic Oscillator The equations of motion are x˙ = p p˙ = −mω 2 x. The trajectories are given by the ellipses p2 mω 2 2 + x =E 2m 2 (from the energy conservation theorem) which are traced out clockwise. Different trajectories are given by different energies; a given energy can be√realized by a multitude of initial conditions. Using the variables z1 = xω m and √ z2 = p/ m one obtains circles, 1 2 2 z1

+ 12 z22 = E,

see Fig. 9.2. 15

A state is described only incompletely by a point in configuration space: One needs position and velocity.

9.4 * The Modified Hamilton Principle

281

V/V 0

1

0

−1

r/r0

1

0

−1

0

5

10

15

Fig. 9.3. Top: The effective potential energy of the Kepler problem (marked are the total energies E/V0 = 0.75, 0.50, 0.25, 0.01 (from bottom to top). Bottom: Phasespace trajectories for the different values of the total energies

Example: (B) Plane Pendulum See problem 9.8 Example: (C) Kepler Problem Writing the potential energy in the form # $ r02 r0 V = V0 −2 + 2 r r with the position r0 and the value V0 of the potential minimum the parameters of the Kepler ellipse can be given in the form  E p = r0 and e = 1 − . V0 Plotted in Fig. 9.3 is  m r 1 vs. = r˙ = e sin ϕ. r0 1 + 2 cos ϕ 2V0

9.4 * The Modified Hamilton Principle In the Lagrangian formalism one has obtained the f Euler–Lagrangian equations d ∂L (3.20) ∂L = dt ∂ q˙l ∂ql

282

9 Hamiltonian Dynamics

for the f variables ql from the Hamilton principle  t2 (3.58) δ L dt = 0 with δql (t1 ) = 0 = δql (t2 ) t1

with the Lagrangian function L(q, q, ˙ t). In the Hamiltonian formalism one has the 2f variables (ql , pl ). One defines now a function analogous to the Lagrangian function  pl q˙l − H(q, p, t) (9.23) Λ(q, p, q, ˙ p, ˙ t) = l

(independent of p). ˙ Claim: The canonical equations of motion (9.3) can be deduced from a modified Hamilton principle  t2

Λ dt = 0

δ

(9.24)

t1

with fixed values of the q and16 p at the times t1 and t2 . Proof: d ∂Λ ∂Λ = dt ∂ q˙l ∂ql d ∂Λ ∂Λ = dt ∂ p˙ l ∂pl

⇒ ⇒

∂H ∂ql ∂H 0 = q˙l − . ∂pl p˙ l = −

(9.25) (9.26) 

9.5 * Canonical Transformation 9.5.1 Point Transformation in Configuration Space In the context of the Lagrangian formalism the aim of the point transformation is, among others, the introduction of cyclic coordinates. In the example of the central potential the Lagrangian function for the motion in the plane in cartesian coordinates is a function of x, y, x, ˙ and y; ˙ in contrast, in polar coordinates it is a function of r, r, ˙ and ϕ, ˙ but not of the cyclic coordinate ϕ. A point transformation in configuration space has the form17 q → Q = Q(q) 16

17

(9.27)

One can give up the condition δp(t1 ) = 0 = δp(t2 ) and leave the variation of the canonical momenta at the borders to be free; then the various steps of the proof are somewhat more involved. See also (9.45) further below.

9.5 * Canonical Transformation

and

7 ˙ t) = L(q(Q), q(Q, ˙ t) L(q, q, ˙ t) → L(Q, Q, ˙ Q),

283

(9.28)

or, for a transition to a moving coordinate system,

and

q → Q = Q(q, t)

(9.29)

7 ˙ t) = L(q(Q, t), q(Q, ˙ t), t). L(q, q, ˙ t) → L(Q, Q, ˙ Q,

(9.30)

7 ˙ t) thus is obtained simThe transformed Lagrangian function L(Q, Q, ply by inserting (invertability of the transformation equations assumed). The Lagrangian equations of motion are form-invariant under a point transformation18 (even though the form of the resulting differential equations ¨ = f7(Q, Q, ˙ t) can be very different for different transq¨ = f (q, q, ˙ t) and Q formations). Also the canonical equations (9.3) are then form-invariant under a point transformation (9.29). Also, one has p7j = pk

∂qk . ∂Qj

So much for the conclusions drawn from the Lagrangian formalism. 9.5.2 Point Transformation in Phase Space Since the Hamiltonian function is a function of the 2f variables q and p, the transformation in phase space is enlarged to the form q → Q = Q(q, p, t) p → P = P (q, p, t)

(9.31)

(enlarged in comparison to the point transformation in configuration space) and H(q, p, t) → K(Q, P, t). (9.32) While the canonical variables are connected via the transformation equations, the relation between the Hamiltonian functions H and K still is to be established. Comments: • The space of the possible transformations (9.31) in phase space is enlarged in comparison to the space of the point transformations (9.29) in configuration space. • The transformation (9.32) evolves only then by inserting into H(q, p, t) (similarly to the Lagrangian function), if the transformation is time independent.19 18 19

See Sect. 3.8.2. For the relation see further below in Sect. 9.5.6.

284

9 Hamiltonian Dynamics

• The canonical equations (9.3) and the Poisson brackets (9.10) are not form-invariant under all of these transformations. • By the transformation (9.31) the meaning of the Q and P as (generalized) coordinates and momenta, respectively, may get lost. • But the possibility opens up to transform the Hamiltonian function with the transformation (9.31) into a particularly simple form. (The determination, however, of such a transformation may be very involved.) 9.5.3 Canonical Transformation For a given problem there are in general various possibilities to choose the canonical variables (for example, the generalized coordinates q and canonical momenta p or alternatively the Q and P ). For the variables q and p as well as for the variables Q and P then the fundamental Poisson brackets are valid. The corresponding transformations are, however, a subset of the transformations of the kind (9.31). Among the sets of the transformations (9.31) are those, under which the canonical equations are form-invariant, as well as those, for which this is not true (in contrast to the point transformations in configuration space). In the following, however, only the subset of such transformations (9.31) shall be considered under which the canonical equations of motion and with it the Poisson brackets are form-invariant. The subset of these transformations forms the set of the canonical transformations. Since the canonical equations, Poisson brackets, etc., have been derived with (9.23) from the modified Hamilton principle (9.24), a canonical transformation (9.31), (9.32) is of a kind in which the Hamiltonian function H(q, p, t) with the variables q and p as well as the Hamiltonian function K(Q, P, t) with the variables Q and P can be derived from a modified Hamilton principle (as a natural starting point). Definition 23. (Canonical Transformation): The transformation (9.31), (9.32) is then a canonical transformation, • If the canonical equations of motion (9.3) or (9.12) can be derived from a modified Hamilton principle; • Or: if the canonical equations of motion (9.3) or (9.12) are form-invariant; • Or: if the fundamental Poisson brackets (9.14) are form-invariant. The canonical transformation shall be illustrated by three examples: Example: (A) Harmonic Oscillator I A simple transformation of the kind (9.31) is p = ±Q ,

q = P.

9.5 * Canonical Transformation

285

With this the coordinate and momentum is exchanged in the Hamiltonian function H = 12 p2 + 12 q 2 = 12 P 2 + 12 Q2 and the Hamiltonian function retains its form. However, the fundamental Poisson brackets and the equations of motion retain their form only for the lower sign, p = −Q , q = P. (9.33) Thus, the transformation is canonical only for the lower sign. Example: (B) Harmonic Oscillator II With the transformation20  q=

2P sin Q, mω

p=

√ 2mωP cos Q

(9.34)

one obtains for the Hamiltonian function by inserting H=

1 2 mω 2 2 p + q 2m 2



K = ωP.

(9.35)

The equations of motion can be integrated very simply, ∂K Q˙ = =ω ∂P ∂K P˙ = − =0 ∂Q



Q(t) = ωt + α



P (t) = β =

K E = . ω ω

With this one obtains  q(t) =

2E sin(ωt + α). mω 2

One can easily show that the fundamental Poisson brackets for P and Q follow from the fundamental Poisson brackets for p and q.21

20

21

The transformation here comes from out of the blue sky. For the corresponding generating function see example (A) in Sect. 9.5.6. For the corresponding characteristic function the Hamilton–Jacobi theory see example (C) of Sect. 9.6.3 further below. See problem 9.6.

286

9 Hamiltonian Dynamics

Example: (C) Particle in a Homogeneous Magnetic Field Here an example of practical use shall be treated. The Hamiltonian function of a particle with the charge q in the homogeneous magnetic induction field B is 1 (p − qA)2 H= 2m (SI system) with A = 12 B × r. If one orients the z axis in the direction of the magnetic induction field and employs cartesian coordinates, the Hamiltonian function has the form  2  2 1  px − 12 qBy + py + 12 qBx + p2z . H= 2m With a canonical transformation the Hamiltonian function can be brought into a simple form: The transformation22 ! " 2 x1 = 12 x + qB py p1 = px − 12 qBy ! " (9.36) 2 p2 = py − 12 qBx px x2 = 12 y + qB with the reverse px = 12 p1 + 12 qBx2

x = x1 −

1 qB p2

py = 12 p2 + 12 qBx1

y = x2 −

1 qB p1

(9.37)

is a canonical transformation, because the fundamental Poisson brackets hold true. Proof: One inserts; the Poisson brackets of two coordinates (x and y) or of two momenta (px and py ) vanish; t he rest yields {p1 , x1 } = 12 {px , x} − 12 {y, py } = 1 {p2 , x2 } = 12 {py , y} − 12 {x, px } = 1

{p1 , x2 } = 12 {px , y} − 12 {y, px } = 0 {p2 , x1 } = 12 {py , x} − 12 {x, py } = 0

{x1 , x2 } = {x, px }/(qB) + {py , y}/(qB) = 0 {p1 , p2 } = − 12 {px , x}qB − 12 {y, py }qB = 0  The transformation (9.36) leads from the Hamiltonian function H by simply inserting to the transformed Hamiltonian function 22

The transformation here falls from out of the blue sky. In fact, one can bring each expression, which is bilinear in the coordinates and/or momenta, into a “diagonal” form by forming a linear combination of these coordinates and/or momenta.

9.5 * Canonical Transformation

287

1 2 mωc2 2 1 2 p + x + p (9.38) 2m 1 2 1 2m z in the form of a one-dimensional oscillator with the cyclotron frequency K(x1 , x2 , p1 , p2 ) =

[c] qB (9.39) c m (SI and cgs system) and a free motion in the z direction. The solutions one can be given immediately (if one ever has gone through the calculation of the motion of the free particle and of the oscillator): ωc =

x1 (t) = a sin(ωc t) + b cos(ωc t)

(9.40)

p1 (t) = mx˙ 1 (t) x2 (t) = x0

(9.41) (9.42)

p2 (t) = p0 z(t) = z0 + vt .

(9.43) (9.44)

The six integration constants (a, b, x0 , p0 , z0 , and v) are fixed by the initial conditions. (The constants x0 and p0 enter the (constant) center of the cyclotron circular orbit.) Comment: This transformation is extremely helpful in the quantummechanical treatment of the charged particle in a magnetic field with the resulting so-called Landau levels. 9.5.4 Time Development as a Canonical Transformation Let the canonical variables q and p of a system be given at a time t0 with the fundamental Poisson brackets {p, q} = 1,

t = t0 :

{p, p} = 0,

{q, q} = 0

and the canonical variables Q and P at another time t0 + Δt. Then the variables Q and P evolve from the variables q and p by a canonical transformation, i.e., one has t = t0 + Δt :

{P, Q} = 1,

{P, P } = 0,

{Q, Q} = 0.

Proof: The Taylor expansion for small, but otherwise arbitrary, time intervals Δt is Q = q(t0 + Δt) = q + Δt q˙ + 12 (Δt)2 q¨ + . . . and analogously for P . Then one has p, q} + 2{p, ˙ q} ˙ + {p, q}) ˙ {P, Q} = {p, q} + Δt ({p, ˙ q} + {p, q}) ˙ + 12 (Δt)2 ({¨ +... = {p, q} + Δt

d d2 {p, q} + 12 (Δt)2 2 {p, q} + . . . dt dt

= 1 + 0. 

288

9 Hamiltonian Dynamics

9.5.5 Canonical Invariants Let the two sets P , Q and p, q of canonical variables (with fundamental Poisson brackets) evolve by a point transformation in phase space, Q = Q(q, p),

P = P (q, p)



q = q(Q, P ),

p = p(Q, P ).

(9.45)

Then the Poisson brackets are invariant under this point transformation, {f, g}p,q = {f, g}P,Q,

(9.46)

because the choice of the canonical variables is arbitrary. Proof: One conceives the functions f and g as functions of P and Q and the functions P and Q as functions of the p and q; then one rewrites the derivatives of f and g with respect to P and Q as derivatives with respect to p and q. With the Einstein summation convention (over doubly appearing indices) one obtains {f, g}P,Q = = − =

+

= + = =

∂f ∂g ∂g ∂f − ∂Pl ∂Ql ∂Pl ∂Ql # $# $ ∂f ∂pk ∂f ∂qk ∂g ∂qm ∂g ∂pm + + ∂pk ∂Pl ∂qk ∂Pl ∂pm ∂Ql ∂qm ∂Ql # $# $ ∂g ∂pm ∂f ∂pk ∂g ∂qm ∂f ∂qk + + ∂pm ∂Pl ∂qm ∂Pl ∂pk ∂Ql ∂qk ∂Ql  # $ ∂g ∂pk ∂pm ∂pm ∂pk ∂f − ∂pk ∂pm ∂Pl ∂Ql ∂Pl ∂Ql # $ ∂pk ∂qm ∂qm ∂pk ∂g − + ∂qm ∂Pl ∂Ql ∂Pl ∂Ql  # $ ∂g ∂qk ∂pm ∂pm ∂qk ∂f − ∂qk ∂pm ∂Pl ∂Ql ∂Pl ∂Ql # $ ∂qk ∂qm ∂qm ∂qk ∂g − + ∂qm ∂Pl ∂Ql ∂Pl ∂Ql   ∂g ∂g ∂f {pk , pm } + {pk , qm } ∂pk ∂pm ∂qm   ∂g ∂f ∂g {qk , pm } + {qk , qm } ∂qk ∂pm ∂qm     ∂g ∂g ∂f ∂g ∂g ∂f 0+ δk,m + (−δk,m ) + 0 ∂pk ∂pm ∂qm ∂qk ∂pm ∂qm ∂f ∂g ∂g ∂f − = {f, g}p,q . ∂pk ∂qk ∂pk ∂qk 

9.5 * Canonical Transformation

289

9.5.6 Generating Function The relation between the Hamiltonian functions H(q, p, t) and K(Q, P, t) of (9.32) is 

pl q˙l − H(q, q, ˙ t) = c

) 

l

l

* d ˙ ˙ Pl Ql − K(Q, Q, t) + F (q, p, Q, P, t). (9.47) dt

Proof: In order for the canonical equations with the variables p and q and the variables P and Q, respectively, to be fulfilled, they have to be derivable from the modified Hamilton principle. Thus, 

t2

)

δ 



t1 t2

)

and δ

* pl q˙l − H(q, q, ˙ t) dt = 0

l



t1

(9.48)

*

˙ t) dt = 0 Pl Q˙ l − K(Q, Q,

(9.49)

l

has to hold. Since both equations describe the same system, the two integrands have to be equal except for a total time derivative dF/dt (and possibly for a constant factor c), 

) pl q˙l − H(q, q, ˙ t) = c

l

 l

* d ˙ ˙ Pl Ql − K(Q, Q, t) + F (q, p, Q, P, t). dt 

The factor c is irrelevant, since one can get rid of the factor by 7 l = cQl Q

7 = cK(Q/c, P, t). and K

7 Q 7 remains unchanged.) In the following we (Then P˙ = −∂K/∂Q = −∂ K/∂ shall thus use c = 1,  l

pl q˙l − H(q, q, ˙ t) =

 l

˙ t) + Pl Q˙ l − K(Q, Q,

d F (q, p, Q, P, t). dt

(9.50)

Of the 4f variables of the q, p, Q, P of the function F in (9.50) now, however, only 2f are linearly independent. Therefore, there are different types of generating functions F , depending upon wether the function F contains the 2n variables (Q, p), (q, P ), (q, Q), (p, P ), (q, p), (Q, P ) or mixtures thereof.

290

9 Hamiltonian Dynamics

However, not every form of the generating function of a given type leads to a canonical form. This shall be demonstrated with an example of a generating function: An Example: (A) Let F = F1 (q, Q, t). With

 ∂F1 dF1 (q, Q, t)  ∂F1 ∂F1 Q˙ l + = q˙l + dt ∂ql ∂Ql ∂t l

l

one obtains from (9.50) $ $  # ∂F1  # ∂F1 ∂F1 K −H = . − pl q˙l + + Pl Q˙ l + ∂ql ∂Ql ∂t l

(9.51)

l

Since the old and new coordinates ql and Ql , respectively, are independent of each other, the canonicity of the transformation is only then guaranteed, if ∂F1 ∂ql ∂F1 Pl = − ∂Ql ∂F1 K −H = ∂t pl =

(9.52) (9.53) (9.54)

holds. The right sides of each of (9.52) to (9.54) are a function of q and Q. Comments: • The strategy towards determining the transformation is the following: From (9.52) one obtains p = p(q, Q, t). This becomes inverted to Q = Q(q, p, t). This is used in the right side of (9.53), and one obtains P = P (Q, q, t) = P (Q(q, p, t), q, t). This is the mapping (q, p) → (Q, P ). Alternatively one obtains P = P (Q, q, t) from equation (9.53). This is inverted to q = q(Q, P, t). This is used in the right side of equation (9.52), and one obtains p = p(q, Q, t) = p(q(Q, P, t), Q, t). This is the mapping (Q, P ) → (q, p). The right side of (9.54) is & ∂ F1 (q, Q, t)&q=q(Q,P,t) . ∂t This differs from ∂F1 (Q, P, t)/∂t by the explicit time dependence, if q(Q, P, t) is explicitly time dependent. • With the function F1 (q, Q, t) the total set of the transformation equations is determined. The function F1 (q, Q, t) thus generates the transformation. Notation: Because of the property just given the function F in (9.50) is called generating function.

9.5 * Canonical Transformation

291

For a function F1 (q, Q, t) with the properties (9.52) and (9.53) to exist, i.e., for these equations to be integrable the integrability conditions # $ $ # ∂Pl ∂pl =− (9.55) ∂Ql q ∂ql Q must be fulfilled. Proof: ∂pl ∂ 2 F1 ∂ 2 F1 ∂Pj = = =− ∂Qj ∂Qj ∂ql ∂ql ∂Qj ∂ql  Comment: Equation (9.55) can be used as a practical criterium for the canonicity of a transformation. Example: (B) Determination of the Transformation from the Generating Function Let the generating function be F1 (q, Q) = 12 mωq 2 cot Q. From this one obtains ∂F1 = mωq cot Q ∂q ∂F1 mωq 2 P =− = ∂Q 2 sin2 Q p=

and from this23 

2P sin Q mω √ p = 2mωP cos Q. q=

Example: (C) Determination of the Generating Function from the Transformation Equations ... ... by integration of (9.52) and (9.53): Let the transformation be Q = ln p P = −qp. 23

This is a canonical transformation for the harmonic oscillator; see (9.34).

(9.56)

292

9 Hamiltonian Dynamics

The function F1 (q, Q, t) can be found, since the integration condition (9.55) is fulfilled, ∂eQ ∂p = = eQ ∂Q ∂Q ∂ ∂P = − (−qp) = p = eQ . − ∂q ∂q From (9.52) one obtains by integration over q (at fixed Q and t) with an integration “constant” g(Q, t)  ∂F1 (q, Q, t) F1 (q, Q, t) = dq + g(Q, t) ∂q   = p(q, Q, t) dq + g(Q, t) = eQ dq + g(Q, t) = qeQ + g(Q, t). From this one obtains P =

∂F1 ∂Q

(9.53)

= qeQ +

∂g(Q, t) , ∂Q

and comparison with the transformation P = −qp = −qeQ leads to an additive function g(t) which depends only upon time, which is irrelevant for the canonical equations of motion. Thus one has F1 (q, Q) = qeQ . Other Generating Functions The details for the other types of the generating functions can be taken from all text books, in which the canonical transformation is treated, and shall not be followed any further here.24 The result is F1 (q, Q, t) F2 (q, P, t) F3 (p, Q, t) F4 (p, P, t)

∂F1 ∂q ∂F2 p= ∂q ∂F3 q=− ∂p ∂F4 q=− ∂p p=

∂F1 ∂Q ∂F2 Q= ∂P ∂F3 P =− ∂Q ∂F4 Q=− ∂P P =−

∂F1 ∂t ∂F2 K=H+ ∂t ∂F3 K=H+ ∂t ∂F4 K=H+ . ∂t K=H+

(9.57)

In all cases the two Hamiltonian functions differ by a partial time derivative of the generating function. 24

The function F2 (q, P, t) is treated in Sect. 9.6.

9.5 * Canonical Transformation

293

Comments: • The further generating functions F5 (q, p, t) and F6 (Q, P, t) lead to a whole class of transformations. • For an explicitly time independent canonical transformation the generating function is thus explicitly time independent, and then one has K = H, (q, p) → (Q, P ) expl. time independent ⇔

∂F = 0 ⇔ K = H. ∂t (9.58)

Explicitly time independent transformations are thus canonical, if one has K = H, even though one has ! l

or



" dF pl q˙l − Pl Q˙ l = dt

(pl dql − Pl dQl ) = dF.

(9.59)

l

The problem thus consists in finding a canonical transformation or the corresponding generating function, which generates a Hamiltonian function, which leads to equations of motion which are as simply as possible to solve. In one case or another this can turn out to be extremely difficult and possibly does not bring with it any advantages in comparison to the similarly difficult solution of the original (untransformed) canonical equations of motion. In addition, the transformed equations of motion are little intuitive, in contrast to the equations of motion which one obtains from the Newtonian or Lagrangian formalism. In contrast to the point transformations25 in configuration space, the canonical transformations thus cannot be found by intuition. On the other hand, the canonical transformations form a basis not only in classical mechanics but also in quantum mechanics and in statistical physics (and thus in thermodynamics). The idea would be to make all coordinates cyclic by the transformation; then all canonical momenta are conserved quantities, and because of ∂K 0 = P˙l = ∂Ql the Hamiltonian function is then independent of the coordinates, and since all momenta are constants, the Hamiltonian function is a constant, which one could set equal to zero by choice of the energy zero. A way to the determination of the generating function is described by the Hamilton–Jacobi theory.26

25 26

Remember the transformation from cartesian to polar coordinates in systems with a central potential. See Sect. 9.6.

294

9 Hamiltonian Dynamics

9.5.7 The Theorem of Liouville Transformation of the Phase-Space Volume Theorem: The phase-space volume is invariant under a canonical transformation. Proof: Let a transformation Π be given, which transforms the canonical coordinates (q, p) into the coordinates (Q, P ). Given are the phase-space volumes     γ(u) = dq1 . . . dqf dp1 . . . dpf     Γ (U ) = dQ1 . . . dQf dP1 . . . dPf , integrated over a neighborhood u around a point (q0 , p0 ) in the 2f -dimensional phase space and over a neighborhood U = Π(u) around the transformed point (Q0 , P0 ), respectively, where the points of the neighborhood U arise from the original points by a canonical transformation Π. The relation is   Γ (U ) = . . . D dq1 . . . dqf dp1 . . . dpf , integrated over a neighborhood u, with the functional determinant D=

∂(Q1 , . . . , Qf , P1 , . . . , Pf ) . ∂(q1 , . . . , qf , p1 , . . . , pf )

The functional determinant can be rewritten, ∂(Q1 , . . . , Qf , P1 , . . . , Pf ) ∂(q1 , . . . , qf , p1 , . . . , pf ) ∂(Q1 , . . . , Qf , P1 , . . . , Pf ) ∂(q1 , . . . , qf , Q1 , . . . , Qf ) = ∂(q1 , . . . , qf , Q1 , . . . , Qf ) ∂(q1 , . . . , qf , p1 , . . . , pf ) # $−1 ∂(Q1 , . . . , Qf , P1 , . . . , Pf ) ∂(q1 , . . . , qf , p1 , . . . , pf ) = (−1)f ∂(Q1 , . . . , Qf , q1 , . . . , qf ) ∂(q1 , . . . , qf , Q1 , . . . , Qf ) # $−1 ∂(p1 , . . . , pf ) f ∂(P1 , . . . , Pf ) = (−1) . (9.60) ∂(q1 , . . . , qf ) ∂(Q1 , . . . , Qf )

D=

The i, j-element of the two matrices is ∂Pi ∂ 2 F1 =− ∂qj ∂qj ∂Qi

and

∂pi ∂ 2 F1 = ∂Qj ∂Qj ∂qi

with (9.59). Except for the sign the two matrix elements thus belong to transposed matrices, and the two corresponding determinants of the last line in  (9.60) are equal except for a sign (−1)f . Thus one has Γ = γ.

9.5 * Canonical Transformation 3

295

p

2

1

–3

–2

1

–1

2

3

q

–1 –2 –3

Fig. 9.4. Plane mathematical pendulum. Circular area of the initial configurations and its time development. The initial configurations lie all below the orbit of the overdamped case. The four times are (clockwise) t/T0 = 0, 14 , 12 , 1 with the period T0 of the pendulum at small amplitudes. The open points mark the trace of the center of the original area (From Scheck [5], Fig. 3)

The Theorem of Liouville The theorem of Liouville has important applications in statistical mechanics: Theorem of Liouville: The phase-space volume of a canonical system is independent of time. Proof: According to Sect. 9.5.4 the time development of the canonical variables is a canonical transformation, and according to the previous theorem the phase-space volume is invariant under a canonical transformation. Thus the phase-space volume is invariant under a translation in time.  Comments: • For example, in preparing a system of identical particles, e.g., a beam of particles, the states are distributed in general over a certain regime of trajectories (i.e., positions at a given time) and of momenta, i.e., over a certain area in phase space. If one then brings the momenta to a uniform value (e.g., by acceleration of the slow and/or decelaration of the fast particles), then the values of the positions are distributed over a correspondingly larger region such that the phase space volume remains constant. This phenomenon is known from optics: The more precisely one fixes the wave length (or the wave vector), the less precise is the localization in space. This is a consequence of the Fourier transformation. • In quantum mechanics specific statements are made on the minimum possible quantities of the phase space volumes. • An impression of the deformation of the constant phase-space volumes for the example of the damped harmonic oscillator one can obtain from the Figs. 9.4–9.6 (taken from in Scheck’s [5] Figs. 2.13–2.15.)

296

9 Hamiltonian Dynamics 3 p 2

1 –3

–2

1

–1

2

3 pq

–1 –2 –3

Fig. 9.5. Same as Fig. 9.4. The area of the initial configurations touches the orbit of the overdamped case. The four times are t/T0 = 0, 15 , 25 , 34 . The arrows mark the trace of the lower point of the original area (From Scheck [5]) 3 p 2 1 –3

–2

–1

1

2 3

q

–1 –2 –3

Fig. 9.6. Same as Fig. 9.5. The center of the initial configurations lies on the orbit 1 , 14 , 12 (From Scheck [5]) of the overdamped case. The four times are t/T0 = 0, 10

9.6 * Hamilton–Jacobi Equation 9.6.1 Action Function Now we will persue the aim set up at the end of Sect. 9.5.6: We search for that transformation of the canonical variables (q, p) to other canonical variables (Q, P ), which lead to transformed coordinates which are all cyclic, such that all canonical momenta are constant. Then one has

9.6 * Hamilton–Jacobi Equation

0=−

∂K = P˙ l ∂Ql



297

Pl = P l = const.

(9.61)



(9.62)

In addition, one has then ∂K Q˙ l = = V l = const. ∂Pl

Ql = V l t + Ql .

Hereby, the solution of the equations of motion is reduced to most simple integrations. Thus the problem consists in the determination of the transformation. Also the Hamiltonian function as a function only of the (constant) momenta is a constant itself. This constant shall be set equal to zero, and one obtains ∂F (9.54) . (9.63) 0 = K(Q, P, t) = H(q, p, t) + ∂t Using in addition the equation pl =

∂F2 (q, P, t) , ∂ql

(9.64)

which is analogous to (9.52), then (9.63) takes the form of the Hamilton–Jacobi differential equation ! ∂W " ∂W ,t + =0 K = H q, ∂q ∂t

(9.65)

with the generating function of type 2 now named Hamilton’s action function W . Comments: • The Hamilton–Jacobi differential equation is a nonlinear differential equation of first order for the so-called action function W with the f + 1 variables qi and t. • Since only the derivatives of the function W enters (9.65), W is determined except for a constant W0 , which does not play any role in the following. • The strategy towards the solution is the following: One sets up the Hamiltonian function H(q, p, t). One replaces pl = ∂W/∂ql and with this sets up the Hamilton–Jacobi differential equation. One solves the equation with f additional integration constants. One identifies the integration constants with the canonical momenta P l (as the required constants). • With (9.65) the generating function thus can be determined – in principle – and with it the form of the transformation. With W = F2 (q, P, t) one obtains pl =

∂W = pl (q, P, t) ∂ql

and Ql =

∂W = Ql (q, P, t). ∂Pl

The transformation is obtained either by inversion of the first equation to Pl = Pl (q, p, t) or by inversion of the second equation to ql = ql (Q, P, t).

298

9 Hamiltonian Dynamics

• The generating function F1 (q, Q, t) can be generated from the solution of (9.65): With pl (q, P , t) from (9.64) one obtains by integration of (9.52)  (9.66) pl (q, Q, t)dql + g(Q, t). F1 (q, Q, t) = l

On the other hand, (9.53)

Pl =

∂F1 ∂Ql

(9.66)

=

  ∂pl (q, Q, t) ∂g(Q, t) dql + = Pl ∂Ql ∂Ql l

must hold, from which one can determine g(Q, t). • The action function W is known from the Hamilton principle,27  W = L dt + const. Proof: One has to show that dW/dt = L is true: ∂W dW ∂W ˙ ∂W Pl + = . q˙l + dt ∂ql ∂Pl ∂t With

one obtains

∂W = pl , ∂ql

P˙l = 0,

∂W =K −H =0−H ∂t

dW = pl q˙l + 0 − H = L. dt 

9.6.2 The Characteristic Function In the following it is assumed that the Hamiltonian function is not explicitly time dependent. (Then one has K = H.) Then the time dependence can be separated: One writes (9.65) in the form ! ∂W " ∂W H q, =− . ∂q ∂t

(9.67)

The left side is independent of time; then also the right side must be independent of time; since this has to hold for all times, the two sides must thus be a constant E. E is a first integration constant. From − 27

See Sect. 3.10.1.

∂W =E ∂t

(9.68)

9.6 * Hamilton–Jacobi Equation

299

one obtains by integration W (q, P , t) = S(q, P ) − E(P )t

(9.69)

with an integration “constant” S(q, P ), and one obtains instead of (9.67) ! ∂S " H q, = E, ∂q

(9.70)

what I will call the time-independent Hamilton–Jacobi equation. The function S is called characteristic function. Comments: • The time-independent Hamilton–Jacobi equation is a nonlinear differential equation of first order in f variables Q; the solution S(q, P ) contains the integration constants E and P . • The strategy for the solution is the following: The canonical momenta P are constant already. The coordinates one determines from ∂H ∂E Q˙ l = = = Vl = const, ∂Pl ∂P l from which one obtains Ql = Vl t + αl . • The transformation to the coordinates is performed as follows: From Ql = ∂S/∂Pl = Ql (q, P ) one obtains ql = ql (Q, P ) by inversion. This one uses in pl = ∂S(q, P )/∂ql = pl (q, P ) = pl (q(Q, P ), P ). 9.6.3 Examples Example: (A) Free Particle The Hamiltonian function is H=

p2 , 2m

such that (9.70) takes the form (∇S)2 =E 2m or (∇S)2 = 2mE = p2 or ∇S = p

300

9 Hamiltonian Dynamics



and thus S(r, p) = p ·

dr = p · r + α.

The subsequent procedure is given in Sect. 9.6.2 and shall not be followed any further here. Comment: The constant E or the modulus of p is the first integration constant. The three components of p are two additional integration constants. The last integration constant is the irrelevant constant α. Example: (B) Particle in a Central Potential V (r) The Hamiltonian function is H=

p2ϕ p2r + V (r), + 2m 2mr2

such that (9.70) takes the form (∂S/∂r)2 (∂S/∂ϕ)2 + + V (r) = E. 2m 2mr2 This equation can be separated: After multiplication with 2mr2 one obtains # $2 # $2 ∂S ∂S 2 2 r − 2mr [V (r) − E] = − . ∂r ∂ϕ The right side is a function only of ϕ and the left side only of r, and since this has to hold for all ϕ and/or r, each side must be equal to a constant (−l2 ). l is the second integration constant. With ∂S/∂ϕ = l



S = lϕ + g(r)

one obtains for the left side # $2   ∂S − 2mr2 V (r) + l2 /(2mr2 ) − E = 0 r2 ∂r or

∂S 1/2 = ± {2m [E − Veff (r)]} ∂r and with this (except for an irrelevant constant)  1/2 S(r, ϕ) = ± dr {2m [E − Veff (r)]} + lϕ with the integration constants l and E. The subsequent procedure is given in Sect. 9.6.2 and shall not be followed any further here.

9.6 * Hamilton–Jacobi Equation

301

Example: (C) Harmonic Oscillator The Hamiltonian function is H(q, p) =

1 2 mω 2 2 p + q . 2m 2

The time-independent Hamilton–Jacobi equation (9.70) takes the form28 # $ ∂S 2 mω 2 2 1 q = E = ωP. + 2m ∂q 2 With this one obtains  dS = ± 2mωP − m2 ω 2 q 2 dq 

and S=±

dq

 2mωP − m2 ω 2 q 2 = S(q, P ).

(9.71)

The equations analogous to (9.52) and (9.53) with the function S = F2 are from (9.57)  ∂S p= = ± 2mωP − m2 ω 2 q 2 = p(q, P ) ∂q and

 ∂S ∂  Q= = ± dq 2mωP − m2 ω 2 q 2 ∂P ∂P  mω = ± dq  2mωP − m2 ω 2 q 2 mωq = Q(q, P ). = ± arcsin √ 2mωP

This is the laborious point of this example. The rest is rewriting: The reverse of Q(q, P ) is  2P q(Q, P ) = ± sin Q (9.72) mω and with this one obtains for p(q, P ) = p(q(Q, P ), P )   √ √ p = ± 2mωP − m2 ω 2 q 2 = ± 2mωP 1 − sin2 Q = ± 2mωP cos Q. (9.73) 28

The definition of this momentum P is chosen with regard to the results and for the comparison with the results of example (B) of Sect. 9.5.3. See also the generating function (9.56) of this transformation.

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9 Hamiltonian Dynamics

With this the transformation equations are found. One verifies in addition (9.35)

H = ωP. The solution of the equations of motion is P = const. and

∂H Q˙ = = ω → Q(t) = ωt + ϕ. ∂P With this one obtains  2P q(t) = ± sin(ωt + ϕ) mω √ p(t) = ± 2mωP cos(ωt + ϕ) as in Sect. 9.5.6. The sign and the phase ϕ are fixed with the use of the initial conditions. From the function S(q, P ) = F2 one can thus determine the function F1 (q, Q) with (9.66): For (9.66) one needs p(q, Q, t). This one obtains from the dividing the two equations (9.72) and (9.73), p = mωq cot Q, and with this one obtains



F1 (q, Q, t) =

p(q, Q, t) dq + g(Q, t) 

= mω cot Q

q dq + g(Q, t)

= 12 mωq 2 cot Q + g(Q, t).

(9.74)

With P =

∂F1 ∂Q

d cot Q (9.74) 1 = 2 mωq 2 dQ

+

on the one hand and P

(9.72)

=

dg(Q, t) mω 2 q 2 dg(Q, t) = + 2 dQ dQ sin Q

mω 2 q 2 sin2 Q

on the other hand one finds by comparison that g(t) depends only upon the time and is thus an irrelevant function. Also one has F1 (q, Q) = 12 mωq 2 cot Q in accordance with the given ansatz of the example in Sect. 9.5.6.

Summary: Hamiltonian Dynamics

303

Comments: • For the examples treated here the solution with the help of the Hamilton– Jacobi equations appears to be more complicated than the direct solution of the canonical equation. The advantage of the Hamilton–Jacobi method becomes apparent for more difficult problems, as is insured in the text books. • In the examples (A) and (B) the function S(q, P ) has been determined. In fact, one does not need S itself, as (9.71) in example (C) shows; there namely, the integration has not been performed explicitly.

Summary: Hamiltonian Dynamics Hamiltonian function (9.1)

H =



pl q˙l − L

l

Gauge transformation ∂ % (9.8) H = H − F (q, t) ∂t Poisson brackets (9.10)

{f, g}p,q =

$  # ∂f ∂g ∂g ∂f − . ∂pl ∂ql ∂pl ∂ql l

Equation of motion of Observable G dG dt

(9.11)

=

∂G + {H, G} ∂t

in particular ∂H ∂pl ∂H (9.11) (9.3) p˙ l = {H, pl } = − . ∂ql (9.11)

(9.3)

q˙l = {H, ql } =

Particle (mass m, charge q) in an electromagnetic field (9.7)

H =

1 2 (p − qA) + qφ 2m

with

A=

[c] A. c

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9 Hamiltonian Dynamics

Problems 9.1. Poisson Brackets I. Let f and g be functions of the canonical coordinates {ql } and {pl }. Prove the validity of the following relations, ∂g ∂pl ∂g {pl , g} = + ∂ql {ql , g} = −

d {f, g} = {f˙, g} + {f, g}. ˙ dt 9.2. Poisson Brackets II. Determine the Poisson brackets which are formed by the Cartesian components of the momentum p and of the angular momentum L = r × p of a point mass. Also determine the Poisson brackets of the components of L. 9.3. Canonical Equations. Given be a mechanical system with the Hamiltonian function H(q1 , q2 , p1 , p2 ) = q1 p1 − q2 p2 − a q12 + b q22 . Solving the canonical equations show that F1 =

p2 − b q2 q1

and

F2 = ln q2 + t

are constants of the motion. 9.4. Canonical Transformation. Let q and p be canonically conjugate variables. New coordinates q¯, p¯ are defined by the transformation √ q¯ = ln(1 + q cos p), √ √ p¯ = 2(1 + q cos p) q sin p. Prove that the transformation is canonical. 9.5. Point Transformation. A point transformation q = q(Q, t) is given in configuration space. Prove ∂qj ∂ q˙j = , ˙ ∂Q ∂ Qi i  ∂qj ∂L Pi := = . ∂Qi ∂ Q˙ i j 9.6. Canonical Transformation and Poisson Brackets. Prove that the fundamental Poisson brackets for P and Q follow from the fundamental Poisson brackets for p and q using the transformation (9.34).

Problems

305

9.7. Canonical Transformation and Generating Function. Prove that the transformation (9.33) is generated from the generating function F1 (q, Q, t) = qQ. 9.8. Phase Portrait of the Plane Pendulum. Introduce the variables q = ϕ and p = ϕ/ω ˙ with ω 2 = g/l and show that the equations of motion are q˙ = p p˙ = − sin q. Prove that the trajectories are given by 1 2 2p

+ (1 − cos q) = ε

with the reduced energy ε = E/(mgl). Plot the potential energy and the phase-space trajectories for different values of the total energy ε = E/mgl from ε = 0.5 to ε = 3.0 in steps of ε = 0.5. Discuss the energies E  2mgl, E < 2mgl, E > 2mgl, E  2mgl, and E = 2mgl. In the last case the trajectory is called separatrix, why? 9.9. Bloch oscillations. The phenomenon of the Bloch oscillations can be observed, if an electron moves in a periodic potential (period length d) superimposed by a constant electric field. In this case the energy–momentum relation (kinetic energy) is given by # $ Δ pd T (p) = 1 − cos . 2  (a) Write down the Hamiltonian function, and formulate the canonical equations. (b) Solve the canonical equations for the initial conditions p(t) = 0 and x(t) = 0 for t = 0. Hint: First solve the equation for p(t), and then use this solution to obtain x(t). What kind of motion does the electron execute in the x-direction? (c) With the help of the canonical equation for x give an expression for the acceleration x ¨ and the rate of change T˙ (p) of the kinetic energy T (p), and eliminate the dependence upon p and p˙ using T and the canonical equation for p. ˙ The expression for x ¨ can be brought into the form of Newton’s equation of motion if one defines a mass m(T ) which depends upon T . Write down m(T ).

306

9 Hamiltonian Dynamics

(d) Introduce friction by sensibly amending the equations of motion from (c) for x and T . Hint: Here, x˙ and T must tend exponentially to zero without the electrical field. (e) Search for a stationary solution (i.e., independent of t) for v(E) and T (E) which develops under counter-acting accelerations due to the electrical field and the friction. Write down v(E) as a function of Ω/α, where Ω = eEd/ is the frequency of the Bloch oscillations and α the damping constant introduced in (d). Sketch v(E) and v(Ω), and discuss the result. 9.10. Population Dynamics: The Prey–Predator Problem. In the population dynamics the prey–predator problem is a basic and mathematically interesting question. The Volterra system of differential equations describes as simple model of the given situation for the populations of the prey and of the predator, x˙ = +kb x − kzb xy y˙ = −kr y + kzr xy

with

kb , kr , kzb , kzr > 0.

Without predators (y) the prey animals can multiply without restrain (exponential growth x˙ = kb x). Without prey the predators must starve (y˙ = −kr y). The probability of prey and predators meeting is proportional to the respective populations x and y and leads to a reduction of the one (−kzb xy) and to an increase of the other (kzr xy) species. Not only in biology but also in other fields like, e. g., in chemistry (chemical reactions) and in physics (semiconductor physics) equivalent rate equations are found. (a) Determine the singular points of the system, and characterize these according to their type. Sketch the trajectories in phase space. (b) Describe the development for the initial conditions x = 0 and y = 0, respectively. (c) Determine the orbits y(x) by integration. Hint: First, determine the tangent dy/dx to the orbit. 9.11. Vibrational Period of the Mathematical Pendulum. The time period T (b) for the amplitude 0 < φ0 < π of a pendulum with b = sin( φ20 ) is given by integral  π2 dθ 4  . T (b) = ω 0 1 − b2 sin2 θ (a) Prove that T (b) is a monotonically increasing function of the amplitude φ0 . (b) Determine and sketch the solution φ(t) √ on the separatrix. ˙ Hint: Employ the equation φ(φ) = ±ω  + 2 cos φ, the identity cos φ = 2 cos2 φ2 − 1, and

Problems



1 dφ = ln tan cos(a φ) a

#

aφ π + 2 4

307

$ +C .

Comment: Solutions of this form are typical for nonlinear differential equations and occur in many fields of physics. They are denoted as solitons or kinks. (c) Employ the result of (b) to show that the pendulum cannot reach the singular point (φ = π) in finite time.

10 * Introduction to the Mechanics of Continua1

So far we have treated only point-like particles; an exception was the treatment of the rigid body in Chap. 8. In this chapter we will investigate deformable bodies, for a twofold reason: (1) If a body is not rigid but deformable, it has its own internal dynamics, which is independent of its state of motion (disregarding inertial forces), as was investigated, e.g., in Chap. 8. (2) The continuum mechanics is a first example of a field theory. A more detailed treatment of a field theory is the subject of the Maxwell theory of the electromagnetic fields.2 Comment: Here, we have elastic (and not plastic) deformations in mind. The former apply to those systems which return to their original state after the external forces are released, while the latter remain in a deformed state.

10.1 Fields If one wants to treat for example the mechanics of a vibrating violin string, then one is not interested in the atomistic character of the string, but rather in its mass density, which, in combination with the stress, determines the pitch, i.e., the frequency. Differently from Chap. 8 one deals here not with a rigid, but with a deformable body. In the mechanics of point masses (index i) one investigates the time dependence of the coordinates ri (t); there is but one variable: the time t. The dynamics of a mass element dm at the position r of the continuum is described by the time dependence of each of the three components of the displacement field u(r, t); here the variables are the time t and the position r. One must thus 1 2

Goldstein [1], Chap. 12 (2nd ed.); in most of the other text books on Classical Mechanics the mechanics of the continua is omitted. See the Course on Electromagnetism.

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10 * Introduction to the Mechanics of Continua

distinguish: In point mechanics r is a coordinate, and in continuum mechanics r is a variable. The velocity is r˙ in point mechanics, and the velocity field is ˙ u(r, t), etc. The generalized coordinates are q(t) and q(r, t), respectively. Densities Essential properties of the continua are (position and possibly time dependent) densities, like the mass density, momentum density, energy density, etc. The dynamics of fields can be derived alternatively from the Newtonian equations or from a variational principle. In the latter case the Lagrangian and Hamiltonian densities are the starting point of a (classical and possibly also nonclassical) field theory. The theory leads to equations of motion for the fields (thus to the displacement field in the example of the string). The transition from (atomistic) point masses to the continuous mass density is made as in Sect. 8.3. Agreement • For reasons of simplification the four variables t = x0 and xi (i = 1, 2, 3) shall be combined into a vector-like (contravariant) quantity xμ with an upper Greek index μ (similar to what is done in Special Relativity3 ), (t, r) = xμ

μ = 0, 1, 2, 3.

Similarly the space derivatives ∂ ≡ ∂i ∂xi

i = 1, 2, 3

and the time derivative

∂ ≡ ∂0 ∂t shall be combined in the (covariant) quantity ∂μ with lower index, $ # ∂ ∂μ ≡ ,∇ . ∂t • The derivative with respect to xμ is also occasionally denoted by an index with a preceding comma, ∂f ≡ ∂μ f ≡ f,μ ∂xμ ∂μ ql ≡ ql,μ , etc. 3

In Relativity Theory one defines x0 = ct.

10.2 Lagrangian Density

311

• The Greek indices (μ = 0, 1, 2, 3) shall always refer to all four variables, the Latin indices (i = 1, 2, 3) only to the three space variables. • Einstein summation convention: Without explicit notation of the summation symbol, summation is understood over those indices which occur twice, one index being a lower one and the other an upper one. Comment: The notation ∂μ is meant to be a total, i.e., explicit and implicit derivative with respect to xμ ;4 the implicit dependence comes from the dependence of the displacement fields or of the generalized coordinates upon the variables. The derivative of a function f with respect to a variable xμ shall always be understood with the other variables xν (ν = μ) kept fixed. The prescription for an exemplary system with just the two variables x and t is thus # $ ∂f (u(x, t), x, t) ∂f (u(x, t), x, t) ≡ ∂t ∂t # $ # $ x # $ ∂f (u(x, t), x, t) ∂f (u(x, t), x, t) ∂u(x, t) = + . ∂u ∂t ∂t x,t x u,x

10.2 Lagrangian Density 10.2.1 Kinetic Energy Density The kinetic energy of a point mass mi at the position r i is T = 12 mi r˙ 2i and that of a mass element dm at the position r is 2

˙ dT = 12 dm [u(r)] . Here it is assumed that all parts of a mass element dm (consisting, e.g., of atomistic masses) in a small volume dV around the position r have the same ˙ displacements u(r) and thus the same velocity u(r). The total kinetic energy of a mass distribution (mass density) ρ(r) =

dm dV



is then T =

 2 1 ˙ 2 dm [u(r)]

Here

=

1 2

2 ˙ . d3r ρ(r) [u(r)]

2 1 ˙ 2 ρ(r) [u(r)]

is the kinetic energy density. 4

The mere explicit derivative will be marked accordingly.

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10 * Introduction to the Mechanics of Continua

Example: Vibrating String One considers the string as a one-dimensional system, which is displaced by u(x) at the position x. With this one obtains the kinetic energy  1 T = 2 dx ρu˙ 2 (x).

10.2.2 Potential Energy Density For point masses (index i) one can write the potential energy in the form V = V (r i (t), r˙ i (t), t). Similar to the kinetic energy also the potential energy can be written in many cases as an integral over a potential-energy density V. Here the potentialenergy density is a function of the position r and of the time t, of the (position and time dependent) displacements u(r, t) as well as of the time and space derivatives of the displacements, ∂u/∂t (velocity) and ∂u/∂r ≡ ∇⊗u (strain), respectively, # $   ∂u ∂u i 3 , x , t = d3r V (u, ∂ν u(xμ ), xμ ) . V = d r V u, j , ∂x ∂t

Example: (A) Mass Density in an External Force Field In the atomistic picture one has dV = −



F (r i ) · dr i .

i

If one introduces a force density f=

dF , dV

one obtains with the replacement dr i = du(ri ) → du(r)  dV = − d3r f (r) · du(r). In this case the (potential) energy density V with dV(r, t) = −f (r, t) · du(r, t) is a functional of the displacement as a function of the position and of the time (not, however, a function of the derivatives of the displacement).

10.2 Lagrangian Density

313

Example: (B) Vibrating String Let the stress be S. Under the condition of constant stress (at small displacements) the potential energy is # $2  ∂u 1 V = S dx . 2 ∂x In this example the (potential) energy density depends upon the strain ∂u/∂x.

ln un xn–1

Δxn

xn

un+1 xn+1

Fig. 10.1. Left: Schematic displacement pattern of a vibrating string. Right: Microscopic picture of the encircled area on the left

Proof: Imagine the string at rest to be composed of single (small) pieces of length Δxn . In the displaced state these pieces are extended to ln , see Fig. 10.1. For constant stress the potential energy is proportional to the stress and to the change in length (Hooke’s law),  (ln − Δxn )S V = n

(the change of the stress with the displacements is an effect of higher order). For the change in length one obtains for small displacements un ≡ u(xn )  ln − Δxn = Δx2n + (un+1 − un )2 − Δxn ( ' $2 # 1 un+1 − un = Δxn 1 + + . . . − Δxn 2 Δxn # $2 un+1 − un 1 ≈ Δxn . 2 Δxn Thus one obtains for the potential energy # $2 un+1 − un 1  V = S Δxn 2 n Δxn # $2 u(xn + Δxn ) − u(xn ) 1  = S Δxn . 2 n Δxn In the limit Δxn → 0 one obtains the result given above.



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10 * Introduction to the Mechanics of Continua

10.2.3 Lagrangian Density Since one can write the kinetic and potential part in the Lagrangian function L=T −V as an integral over energy densities, one can thus write the Lagrangian function as an integral over a Lagrangian density L (ql = ui in the preceding section), $ #  ∂ql (r, t) ∂ql (r, t) 3 , r, t L = d r L ql (r, t), , ∂xj ∂t   ≡ d3r L (ql (xμ ), ∂ν ql (xμ ), xμ ) = d3r L (ql , ql,ν , xμ ) . Example: Vibrating String In the example of the vibrating string (q ≡ u) one has ρ L= 2

#

∂u ∂t

$2

S − 2

#

∂u ∂x

$2 ≡

ρ 2 S 2 u˙ − u 2 2

with u ≡ ∂u/∂x.

10.3 Hamilton’s Principle and Lagrangian Equations As in Sect. 3.10.1 the action is  t2   t2 dt L = dt d3r L (ql (xμ ), ql,ν (xμ ), xμ ) . S= t1

t1

(10.1)

V

From the Hamilton principle δS = 0 with the boundary conditions δql (t1 ) = 0,

δql (t2 ) = 0

one obtains the equations of motion ∂L ∂L − ∂μ = 0. ∂ql ∂ql,μ

(10.2)

Comment: In comparison to the Lagrangian equations of the point masses here there are space derivatives (strains) in addition to the time derivatives (velocities).

10.3 Hamilton’s Principle and Lagrangian Equations

315

Proof: The variation of the action one obtains from the variation of the (generalized) coordinates ql (of the “trajectory”), 



t2

δS =

d3r [L(ql + δql , (ql,ν + δql,ν ), xμ ) − L(ql , ∂ν ql , xμ )]

dt t1



V





t2

d3r

dt

= t1



V

=





t2

3

dt t1

dr V

∂L ∂L δql + δql,ν ∂ql ∂ql,ν



 ∂L ∂L ∂L δql + δ q˙l + δql,j . ∂ql ∂ q˙l ∂ql,j

The last two terms can be integrated in parts with respect to time and with respect to position, respectively. For the term in the middle one obtains with the use of the boundary conditions 



t2

d3r

dt t1

V

&t # $  t2  ∂ ∂L ∂L && 2 δql & − dt d3r δql ∂ q˙l ∂t ∂ q˙l V t1 V t1 # $  t2  ∂L 3 = 0− dt d r ∂0 δql , ∂ q˙l t1 V

∂L δ q˙l = ∂ q˙l



d3r

and for the last term one obtains with the Gauß theorem  t2  ∂L ∂ dt d3r δql ∂ql,j ∂xj t1 V # $  t2  $ #  t2  ∂L ∂ ∂ ∂L δql = dt d3r j δql − dt d3r ∂x ∂ql,j ∂xj ∂ql,j t1 V t1 V # # $  t2  $  t2  ∂L ∂L = dt dfj δql − dt d3r ∂j δql ∂ql,j ∂ql,j t1 ∂V t1 V # $  t2  ∂L 3 =0− dt d r ∂j δql . ∂ql,j t1 V The closed surface integral vanishes for a sufficiently large system, since the Lagrangian density of a finite physical system vanishes outside of a volumes V if chosen sufficiently large (or for an infinitely extended system decreases faster than the surface increases). Then remains 

3

dt

δS = t1





t2

dr V

 ∂L ∂L ∂L − ∂0 − ∂j δql . ∂ql ∂ q˙l ∂ql,j

Since the displacements δql are linearly independent and arbitrary, the contents of the square brackets must vanish. If one also combines the time and space derivatives, one obtains the equation of motion. 

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10 * Introduction to the Mechanics of Continua

Comments: • In the case of vanishing space derivatives the form of the Euler–Lagrangian equation of the point mechanics is recovered (trivial in that contex). • Like the Hamiltonian function also the Hamiltonian density is not unambiguous.5 Example: Vibrating String The Lagrangian density is L = 12 ρu˙ 2 − 12 Su2 with u ≡ ∂u/∂x. The Lagrangian equation is ∂L ∂ ∂L ∂ ∂L − − ∂u ∂t ∂ u˙ ∂x ∂u ∂ ∂ Su = 0 − ρu˙ − ∂t ∂x = −ρ¨ u + Su .

0=

(10.3)

This is a (one-dimensional) wave equation. Expecting vibrations (waves), one makes the ansatz of a plane wave for the solution of this linear differential equation, u(x, t) ∝ ei(kx−ωt) , and obtains (ρω 2 − Sk 2 )u = 0. For nontrivial solutions u one obtains thus a relation between frequency ω and wave vector k S ω=± k = ±vk ρ (i.e., right and left running waves, respectively) with the (transverse) sound velocity S v= ρ for transverse waves (with displacements perpendicular to the propagation direction). This one should compare with the continuum approximation (ka  1) for the transverse vibrations of a linear chain (mass m, spring constant f , lattice constant a, stress S),  f fa sin ka ≈ k, ω= m m/a plotted in Fig. 10.2. 5

Compare Sect. 9.1.3.

317

ω

10.4 The Energy–Momentum Tensor

k

π/a

Fig. 10.2. Dispersion relation of a linear chain (full line) and continuum approximation for long waves (broken straight line)

The general solution is the superposition of the right and left running wave, u(x, t) = a ei(kx−ωt) + b ei(kx+ωt) .

10.4 The Energy–Momentum Tensor 10.4.1 The Conservation Theorem Conservation laws as for example for the (total) energy E have the (global) form E = const. With the Hamiltonian (energy) density H ≡ j 0 and the energy current density j = {j i } the energy conservation can be written in differential (local) form as a continuity equation H˙ + ∇ · j = 0, (10.4) what one can also write in the simple form ∂μ j μ = 0. In its integral form the relation reads   d3r H˙ + V

df · j = 0,

∂V

the flux j of energy through the areal element ∂V of a volumes V is equal to the time rate of change of the energy H in the volume V . From the Lagrangian density one obtains an analogous continuity equation & ∂μ T μ ν = −∂ν L&expl

ν = 0, 1, 2, 3

(10.5)

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10 * Introduction to the Mechanics of Continua

with the energy–momentum tensor T μν =

∂L ql,ν − δνμ L ∂ql,μ

(10.6)

1 0

(10.7)

with the Kronecker symbol δνμ =

+

for μ = ν otherwise.

In particular, for each of the (cyclic) coordinates xν ,& which does not appear in the Lagrangian density L of the system, i.e., ∂ν L&expl = 0, one obtains a conservation theorem in the form of a continuity equation ∂μ T μ ν = 0

ν = 0, 1, 2, 3

(∂ν L = 0).

(10.8)

Proof: Consider the (four-dimensional) gradient ∂ν L ≡

∂L ∂xν

& ∂L ∂L ∂L & ∂ν ql + ∂ν ql,μ + ν && ∂ql ∂ql,μ ∂x expl & # $ ∂L ∂L && ∂L (10.2) = ∂μ ∂ν ql,μ + ν & ql,ν + ∂ql,μ ∂ql,μ ∂x expl & # $ ∂L ∂L & = ∂μ ql,ν + ν && . ∂ql,μ ∂x expl =

If one combines the derivatives, one obtains & # $ ∂L ∂L & ∂μ ql,ν − Lδνμ = − ν && . ∂ql,μ ∂x expl  10.4.2 Energy Density For the time component (ν = 0) of the continuity equation (10.5) one obtains & $ # ∂ ∂L ∂L ∂ ∂L && q˙l = − q˙l − L + i . (10.9) ∂t ∂ q˙l ∂x ∂(∂ql /∂xi ) ∂t &expl The term in brackets is the component T 0 0 , and comparison with the Hamiltonian function in Sect. 9.1.1 shows that it is the energy density (Hamiltonian density), ∂L (9.1) T 00 = q˙l − L = H. (10.10) ∂ q˙l

10.4 The Energy–Momentum Tensor

319

Example: Vibrating String From the Lagrangian density L = 12 ρu˙ 2 − 12 Su2 one obtains

& ∂L && =0 ∂t &expl ∂L = ρu˙ ∂ u˙ ∂L = −Su ∂u ∂L u˙ − L = 12 ρu˙ 2 + 12 Su2 = H. ∂ u˙

10.4.3 Energy–Current Density If, in the point-mechanics case, the Lagrangian function has no explicit time & dependence, (∂L/∂t)&expl = 0, the Hamiltonian function (the energy in the case of velocity-independent potentials) is a conserved quantity.6 Analogously, for a system with an explicitly time-independent Lagrangian density one has the continuity equation (10.4) for the Hamiltonian density (or for the energy density). Thus, the three components of the second term in (10.9) ∂L q˙l = T i 0 = j i ∂(∂ql /∂xi )

(10.11)

must be the components of the energy–current density. Example: Vibrating String The current density can be made plausible with the help of the example of the vibrating string. In the atomistic picture the change of the potential energy at a displacement at the position xn is dVn = −Su dun



dV = −Su du,

where Su = S

un+1 − un un+1 − un ≈S ln Δxn ∂u = Su ⇒ Su = S ∂x

is the component of the force (tension) in the direction of u. For the notation see Fig. 10.3. 6

See Sect. 3.11.3.

320

10 * Introduction to the Mechanics of Continua

S

un

Δx

un+1

Fig. 10.3. Notation for the stress in the string

The (potential) energy density is H = ∂V /∂x. The energy–current density (as a product of energy density and velocity) has only a space component in this example, jx = j = Hx˙ =

∂V ∂u ∂x ∂V ∂u ∂V ∂x = = = −Su u˙ = −Su u. ˙ ∂x ∂t ∂u ∂x ∂t ∂u ∂t

This result one obtains thus from the energy–momentum tensor, T x0 =

∂L u˙ = −Su u. ˙ ∂u

10.4.4 Energy Conservation In the case of vanishing explicit time dependence the (local) conservation theorem of the energy is obtained from the time component (ν = 0) of (10.8), ∂μ T μ 0 = 0. Example: Vibrating String In the example of the vibrating string one has T 0 0 = H = 12 ρu˙ 2 + 12 Su2 T x 0 = jx = −Su u˙ ∂ ∂ ∂μ T μ 0 = ( 12 ρu˙ 2 + 12 Su2 ) + (−Su u) ˙ ∂t ∂x      = (ρu¨ ˙ u + Su u˙ ) − S(u u˙ + u u˙ ) (10.3)

= u(ρ¨ ˙ u − Su ) = 0. The energy is thus a conserved quantity. 10.4.5 The Strain–Momentum Density Strain is the reaction of a system to stress. Connected with a strain is a displacement, and while the time component (ν = 0) of the continuity equation describes the flux of energy, the space components (ν = 1, 2, 3) describe the flux of the (canonical) momentum connected with the strain, and in particular T j i is the j-component of the current density for the i-component of the momentum.

10.4 The Energy–Momentum Tensor

321

Example: Vibrating String In the example of the vibrating string one has T 0x =

∂L ∂u = ρuu ˙ . ∂ u˙ ∂x

The factor ρu˙ is the momentum density; the second factor u is the strain. All together one has thus the momentum density T 0 x connected with the strain, while the momentum density connected with the motion (velocity q˙l = u) ˙ is proportional to T x 0 . If, in the general case, one identifies πl =

∂L ∂ q˙l

(10.12)

with the canonical momentum density π l analogous to that in Sect. 3.6.1 and ∂i ql with a strain, then one has T 0 i = π l ∂i ql .

(10.13)

One can interpret the individual components as follows: For the notation see Fig. 10.4. With the motion of a piece of the body a volume element dV = dx dy dz at the point r is displaced by ql (r, t). If the body were rigid, the shape of the volume element would remain unchanged; only the position and orientation would change; the mass density would remain unchanged, since the volume elements do not change. In the deformable medium, the shape changes (and with it the value), since ql (r, t) in general is different from ql (r + dr, t).

Fig. 10.4. Notations for the strain

One considers two surface elements dA = dy dz, see Fig. 10.4, with the normal oriented in the x-direction at the position r and at r + dr in the

322

10 * Introduction to the Mechanics of Continua

interior of the deformed medium oriented oppositely. The change of the volume elements by the length dx connected with the strain is [ql (x + dx) − ql (x)]dy dz =

∂ql dx dy dz; ∂x

the relative volume change is thus ∂ql . ∂x The relative density change is then −

∂ql = −ql,x . ∂x

With the canonical momentum density πl from above thus the change of the (canonical) momentum density in the x-direction is T 0x = −

∂L ql,x ∂ q˙l

and analogously for the other directions. 10.4.6 The Stress Tensor According to the Newtonian equation of motion the change of the momentum density π connected with the strain is given by the force density f , π˙ = f ,

f=

dF . dV

The surface element da and the force F acting on the surface are connected to each other via the stress tensor T, dF = T · da dK i = T i j dAj . The elements of the stress tensor are depicted in Fig. 10.5. One considers a volume element dV = dx dy dz; the resulting force is the difference of the forces on opposite surfaces (properly speaking the sum, where, however, the opposite surface normals are oriented oppositely). For the ith component of the force and the surface elements dA = ±dydz in the x-direction one has [T i x (x + dx, y, z) − T i x (x, y, z)]dy dz =

∂T i x dx dy dz ∂x

10.4 The Energy–Momentum Tensor

323

z

Tzx (x+dx)

Tzx(x) Tyx(x)

Tyx (x+dx)

Txx(x)

dz

Txx (x+dx) dy

dx

x

Fig. 10.5. The elements of the stress tensor

and analogously for the other surfaces, the force is thus (no summation convention)  dT i j dK i = dV dxj j and thus (with summation convention) π˙ i = k i =

∂T i j ∂xj

or

∂T i j ∂π i + = 0. ∂t ∂xj One can thus identify the matrix elements −

T i j = T ij .

(10.14)

Example: Vibrating String For the example of the vibrating string one has for the only component (u = u,x ) ∂L  u −L ∂u   = −Su2 − 12 ρu˙ 2 − 12 Su2

T xx =

= − 21 ρu˙ 2 − 12 Su2 .

10.4.7 Conservation of the Strain Momentum The space components of (10.8) describe the conservation of the strain momentum.

324

10 * Introduction to the Mechanics of Continua

Example: Vibrating String In particular for the example of the vibrating string one obtains ∂μ T μ x = ∂0 T 0 x + ∂x T x x ∂ ∂  1 2 1 2  (ρu u) = ˙ − ρu˙ + 2 Su ∂t ∂x 2 ¨) − (ρu˙ u˙  + Su u ) = ρ (u˙  u˙ + u u (10.3)

= u (ρ¨ u − Su ) = 0. The strain–momentum is thus a conserved quantity.

Summary: Mechanics of Continua Action (10.1)





t2

S =

t1



t2

dt L =

d3r L (ql (xμ ), ql,ν (xμ ), xμ ) .

dt t1

V

Equations of motion ∂L ∂L − ∂μ ∂ql ∂ql,μ

(10.2)

= 0.

Continuity equation ∂μ T μ ν

& = −∂ν L&expl

(10.5)

ν = 0, 1, 2, 3

πl

(10.12)

∂L ∂ q˙l

T 00

(10.10)

=

H

Energy–current density T i 0

(10.11)

ji

Stress tensor

T ij

(10.14)

T ij

Strain–momentum density

T 0i

(10.13)

π l ∂i ql

Canonical momentum density Hamiltonian density

=

= = =

A Physical Constants1

Constants in Mechanics Gravitational constant

G = 6.672 59(85) × 10−11 m3 kg−1 s−2 (1996) = 6.673(10) × 10−11 m3 kg−1 s−2 (2002)

Gravitational acceleration

g = 9.806 65 m s−2

Solar mass

M = 1.988 92(25) × 1030 kg

Solar equatorial radius

R = 6.96 × 108 m

Earth mass

M⊕ = 5.973 70(76) × 1024 kg

Earth equatorial radius

R⊕ = 6.378 140 × 106 m

Moon mass

M = 7.36 () × 1022 kg

Moon radius

R = 1.738 × 106 m

Distance of Earth from Sun

Rmax = 0.152 1 × 109 m Rmin = 0.147 1 × 109 m Rmean = 0.149 6 × 109 m

Distance of Moon from Earth

Rmean = 0.380 × 108 m

Period of Earth w.r.t. Sun

T⊕ = 365.25 d = 3.16 × 107 s

Period of Moon w.r.t. Earth

T = 27.3 d = 2.36 × 106 s

1

From Particle Data Group, American Institute of Physics 2002 http://pdg.lbl.gov/2002/contents http://physics.nist.gov/constants

326

A Physical Constants

Constants in Electromagnetism def.

c = 299, 792, 458 m s−1

Velocity of light

Vacuum dielectric constant Elementary charge

μ0 = 4π × 10−7 N A−2 = 12.566 370 614 . . . × 10−7 N A−2 1 ε0 = = 8.854 187 817 . . . × 10−12 F m−1 μ0 c2 e = 1.602 177 33 (49) × 10−19 C def.

Vacuum permeability

e2 = 1.439 × 10−9 eV m = 2.305 × 10−28 J m 4πε0

Constants in Thermodynamics kB = 1.380 650 3 (24) × 10−23 J K−1

Boltzmann constant

= 8.617 342 (15) × 105 eV K−1 NA = 6.022 136 7(36) × 1023 mole−1

Avogadro number Derived quantities :

Gas constant R = NL kB Particle number N , mole number n N = NL n

Constants in Quantum Mechanics Planck’s constant

Bohr radius Rydberg energy

h = 6.626 068 76 (52) × 10−34 J s h = 1.054 571 596 (82) × 10−34 J s = 2π = 6.582 118 89 (26) × 10−16 eV s 4πε0 2 = 0.529 177 208 3 (19) × 10−10 m me2 # 2 $2 e m 2 1 e2 = = E0 = 2 2 4πε0  2ma0 2 4πε0 a0

a∞ =

= 12 mc2 α2 = 13.605 691 72 (53) eV

A Physical Constants

Electron rest mass

327

me = 9.109 381 88 (72) × 10−31 kg = 0.510 998 902 (21) MeVc−2

Myon rest mass

mμ = 1.883 566() × 10−28 kg

Proton rest mass

mp = 1.672 621 58 (13) × 10−27 kg = 938.271 998(38) MeV c−2 mn = 1.674 954 3 () × 10−27 kg

Neutron rest mass

Atomic mass unit (amu) mu =

1g NA mole

= 1.660 538 73 (13) × 10−27 kg

= 931.494 013(37) MeV c−2 Bohr magneton

μB =

Fine-structure constant

α=

e[c] 2mc

= 5.788 381 749 (43) × 10−5 eV T−1

e2 4πε0 c

= 1/137.035 999 76 (50)

Conversion Factors 1 eV = 1.602 176 462 (63) × 10−19 J

B Scalars, Vectors, Tensors

In this appendix some relations are collected, which could be useful in nonrelativistic mechanics.

B.1 Definitions and Simple Rules B.1.1 Definitions (Mathematical) definition: (i) Tensors (of a given rank) are elements of a vector space (i.e., they obey certain operation rules.1 (ii) Under transformation, a tensor of nth rank (with n indices) has, depending on the type of the index, the behavior of the position vectors. (Sloppy, physical) definition: A vector is a quantity with modulus and direction. Comments: • A coordinate transformation (from unprimed to primed coordinates) of a vector a with the (cartesian) components ai is given by  U i j aj . (B.1) ai = j

Correspondingly, the behavior of a tensor nthr rank under a transformation is given by  Ti1 ...in = Ui1 j1 · · · Uin jn Tj1 ...jn . (B.2) j1 ...jn

1

See Sect. B.2.1 for vectors.

330

B Scalars, Vectors, Tensors

• Scalars and vectors are tensors of zeroth and first rank, respectively. • A scalar is a number. But not each number is a scalar: For example the x component of a vector is a number, but no scalar, because it changes with a coordinate transformation. • Tensors of second rank can be represented by matrices. But not each matrix is a tensor: For example the transformation matrix U of (B.1) with the elements Ui j is not a tensor, since it does not refer to a given coordinate system, but gives the relation between two different systems. B.1.2 Behavior Under Inversion Under inversion a tensor of even-numbered rank (in particular thus a scalar) transforms into itself; a tensor of odd-numbered rank (thus in particular a vector) transforms into its negative. Pseudo tensors are likewise elements of vector spaces, have but the wrong behavior under transformation: Under inversion a pseudo tensor of evennumbered rank (in particular a pseudo scalar) transforms into its negative; a pseudo tensor of odd-numbered rank (in particular a pseudo vector) transforms into itself. Notation: In order to discriminate, a vector is denoted alternatively as a polar vector and a pseudo vector also as an axial vector. Comments: • The cross product of two polar vectors (e.g., in the case of the angular momentum) is an axial vector (pseudo vector).2 • The tensor product (dyadic product) of two polar vectors is a tensor.3 • The scalar product of a vector with a pseudo vector (i.e. the triple scalar product of three vectors) is a pseudo scalar.4

B.2 Vectors B.2.1 Rules for Vectors One has the following axioms for vectors as elements of a vector space: A sum of vectors a ∈ Rd and b ∈ Rd is defined, which results in a vector c ∈ Rd , a + b = c ⇔ a i + b i = ci . A multiplication is defined of a vector a ∈ Rd with a scalar α ∈ K; the space is linear: d = αa ⇔ di = αai . 2 3 4

See Appendix B.2.3. See Appendix B.2.2. See Appendix B.2.4.

B.2 Vectors

331

There is an inner product (scalar product), of two vectors resulting in a scalar s, a · b = s. B.2.2 Dyadic Product (Tensor Product) of Vectors The dyadic product (tensor product) ab ≡ a ⊗ b of two vectors a, b is a tensor T = a b = a ⊗ b with     c·T = c· a⊗b = c·a b     T·d = a⊗b ·d = a b·d . Comments: • The behavior of the tensor T under transformation is apparent. • In contrast to the scalar product a · b = b · a the tensor product is not commutative, a b = b a. • The symbol ⊗ is rarely used and is used here only for distinction from the dot product (Fig. B.1). • It is thus of importance, for the distinction from the tensor product to notate the product symbol (dot) in the case of the scalar product!! • The multiplication rule “row times column” for matrices holds likewise for the scalar product of vectors. For the scalar product this is a sum of products (of two terms), while for the tensor product each tensor element consists of just one product. )* . 8 .9: ; . . 9 · .. : ;8

)* .. . ⊗ 8: .9: . . ;9 ;8

Fig. B.1. The scalar product (left) and the tensor product (right) of two vectors with the rule “row times column”

B.2.3 Vector Product (Cross Product) of Vectors The vector product (cross product) is restricted to vectors in R3 ; by c=a×b

332

B Scalars, Vectors, Tensors

Fig. B.2. The outer product of two vectors

a third vector c is assigned to two vectors a, b with a · c = 0 =b · c & & & & & & & & &a × b& = &a · a&2 &b · b&2 − &a · b&2 a, b, and c = a × b forming a right-handed system (Fig. B.2). The resulting vector c is perpendicular to the vectors a and b; the modulus is equal to the area spanned by the vectors a and b, |a × b| = ab sin α, where α is the angle (in the mathematically positive sense from a to b) between the vectors a and b. The cross product has the following properties: a × b = −b × a a×a = 0 a×b = 0



a=0

or b = 0 or a = αb.

B.2.4 Triple Scalar Product of Vectors Like the external product thus the triple scalar product is restricted to vectors in R3 ; with     a· b×c = a×b ·c according to sections B.2.1 and B.2.2 a (pseudo) scalar is assigned to three (polar) vectors. The triple product is the volume of the parallelepiped spanned by the three vectors and positive if the three vectors form a right-handed system. B.2.5 Multiple Products of Vectors The following relations for vectors in R3 are of use occasionally:       a× b×c = a·c b− a·b c           a×b · c×d = a·c b·d − a·d b·c . Comment:

    a × b × c = a × b × c.

(B.3) (B.4)

C Rectangular Coordinate Systems

C.1 Definitions Let a point in R3 be represented by the (generally curvilinear) coordinates ξi (i = 1, 2, 3). Definition: The coordinate sheet fi is the sheet fi (ξi ) = 0 (i = 1, 2, 3). Definition: The coordinate line si is the intersecting line of the sheets fj (ξj ) = 0 and fk (ξk ) = 0 (with i, j, k cyclic). Definition: The unit vector ei is the vector tangential to the coordinate line si in the direction of increasing value of ξi with ei · ei = 1.

C.2 Cartesian Coordinates The position vector is r = x ex + y ey + z ez . The line element is (Fig. C.1) ds = dx ex + dy ey + dz ez .

z dz dx

dy

y

x

Fig. C.1. The volume element in cartesian coordinates

334

C Rectangular Coordinate Systems

The surface element is da = dy dz ex + dz dx ey + dx dy ez . The volume element is (Fig. C.1) d3r = dx dy dz.

C.3 Spherical Polar Coordinates (Spherical Coordinates) The transformation from cartesian coordinates to spherical polar coordinates is given by ⎞⎛ ⎞ ⎛ ⎞ ⎛ sin ϑ cos ϕ sin ϑ sin ϕ cos ϑ ex er ⎝ eϑ ⎠ = ⎝ cos ϑ cos ϕ cos ϑ sin ϕ − sin ϑ ⎠ ⎝ ey ⎠ eϕ ez − sin ϕ cos ϕ 0

Fig. C.2. Spherical polar coordinates (“spherical coordinates”)

The position vector is (Fig. C.2) r = r er . The line element is ds = dr er + r dϑ eϑ + r sin ϑ dϕ eϕ . The surface element is da = r2 sin ϑ dϑ dϕ er + r dr sin ϑ dϕ eϑ + r dr dϑ eϕ . The volume element is d3r = r2 dr sin ϑ dϑ dϕ = −r2 dr d(cos ϑ) dϕ.

C.5 Plane Polar Coordinates

335

C.4 Cylindrical Coordinates The coordinate transformation is ⎛ ⎞ ⎛ ⎞⎛ ⎞ e cos ϕ sin ϕ 0 ex ⎝ eϕ ⎠ = ⎝ − sin ϕ cos ϕ 0 ⎠ ⎝ ey ⎠ ez ez 0 0 1

Fig. C.3. Cylindrical coordinates

The position vector is (Fig. C.3) r =  e + z ez . The line element is ds = d e +  dϕeϕ + dz ez . The surface element is da =  dϕ dz e + d dz eϕ +  d dϕ ez . The volume element is d3r =  d dϕ dz.

C.5 Plane Polar Coordinates The plane polar coordinates one obtains from the cylindrical coordinates with z = 0 or from the spherical polar coordinates with ϑ = π/2, # $ # $# $ er cos ϕ sin ϕ ex = . − sin ϕ cos ϕ eϕ ey

336

C Rectangular Coordinate Systems

C.6 Inverse Relations The (rarely needed) inverse of the relations between the unit vectors in shorthand notation is given by e = T · e



e = T T · e

with the transpose TT = T−1 of the matrix T.

Problems C.1. Vectors and Coordinate Systems. (a) Write down examples for physical quantities which can be described as vectors. (b) Write down the decomposition of a vector in Cartesian, plane, and spherical polar coordinates. What is the difference between the Cartesian and the polar coordinates? What is the relation between spherical and plane polar coordinates? C.2. The Line Element in Spherical Coordinates. For a motion in R3 one can employ Cartesian coordinates r(t) = {x(t), y(t), z(t)} or spherical coordinates {r(t), θ(t), ϕ(t)}. Determine the infinitesimal line element (ds)2 = (dx)2 + (dy)2 + (dz)2 in spherical coordinates, and write down the squared modulus of the velocity in these coordinates. C.3. Vector Functions, Trajectory. (a) What is a vector function? Write down examples. (b) The trajectory of a point mass is described by the time-dependent position vector r(t). Write down the decomposition of r(t) in Cartesian and polar coordinates. How are these decompositions different from each other with respect to the time dependence? (c) The set of all points through which r(t) runs for all times t one denotes as space curve or trajectory of the point mass. For the description of the space curve one uses a coordinate system (local trihedron), which is defined by following set of orthogonal unit vectors: tangent unit vector:

dr(s) =: et (s) ds

where s is the arc length (along the space curve), x0 Comment on (H.4): With the condition that g(x) can be inverted (i.e., that g(x) is injective), g(x) ⇒ x(g), one has    1 dx dg = f (x(g)) dg δ(g) dg f (x) δ(g(x)) dx = f (x) δ(g(x)) dg dx and contributions to the integral come from the points with g = 0, with δ(g(x)) = δ(−g(x)) = δ(|g(x)|).

H.2 Representation of the δ-Function by Functional Sequences The Dirac δ function can be represented as the limit of various functional sequences, δ(x) = lim δn (x). n→∞

The following functional sequences are the most common ones, see Fig. H.2,  1 δn (x) = n |x| < 2n box function 0 const

H.5 The δ-Function in R3 2 2 n δn (x) = √ e−n x π

δn (x) =

1 n π 1 + n2 x2

δn (x) =

1 sin nx = πx 2π

363

Gauß function Lorentz function 

n

dk e±ikx

cf. (H.7)

−n

H.3 Integral Representation of the δ-Function

δ(x) =

1 2π





dt e±ixt .

(H.7)

−∞

H.4 Periodic δ-Function 

δ(x − nL) = lim

M→∞

n

1 sin[(2M + 1)πy/L] L sin(πy/L)

(H.8) 

Proof: See (I.30).

H.5 The δ-Function in R3 Coordinate-Free Representation The three-dimensional generalization of (H.3) is   f (r0 ) for r ∈ V d3rf (r) δ(r − r 0 ) = 0 for r ∈ V. V

(H.9)

Integral Representation

δ(r − r0 ) =

1 (2π)3

 d3k eik·(r−r0 ) .

(H.10)

Representation in Cartesian Coordinates δ(r − r0 ) = δ(x − x0 ) δ(y − y0 ) δ(z − z0 ).

(H.11)

364

H Dirac δ-Function and Heaviside Step Function

Representation in Spherical Polar Coordinates δ(r − r0 ) δ(ϑ − ϑ0 ) δ(ϕ − ϕ0 ) r2 sin ϑ δ(r − r0 ) =− δ(cos ϑ − cos ϑ0 ) δ(ϕ − ϕ0 ). r2

δ(r − r0 ) =

(H.12)

Representation in Cylindrical Coordinates

δ(r − r 0 ) =

δ( − 0 ) δ(ϕ − ϕ0 ) δ(z − z0 ). 

(H.13)

H.6 The δ-Function as an Inhomogeneity of the Poisson Equation

Δ

1 = −4πδ(r − r  ). |r − r  |

(H.14)

Proof: Case r = r : If one places the origin of the coordinate system at the point r , one has for r = 0 1 r 1 ∇·∇ r ∇

(D.13)

=

= =

1 r er = − 3 r2 r r (D.25) 1 1 −∇ · 3 = − 3 ∇ · r − r · ∇ 3 r r r 1 −3r − 3 3 − r · 5 = 0. r r −

Case r = r  : Integration over a (small) sphere K with its center at the origin yields with the Gauß theorem    1 (D.35) 1 er d3r ∇ · ∇ = da · ∇ = − da · 2 . r r r K ∂K ∂K With da = r2 er dΩ (in spherical polar coordinates) one obtains thus on the one hand   1 d3r ∇ · ∇ = − dΩ = −4π r K ∂K and on the other hand

 −4π

d3r δ(r) = −4π. K



I Fourier Transformation

I.1 The Transformation: Fourier Integral Let f (x) be a function with the properties 1. f (x) is piecewise smooth; 2. at discontinuities xi one has f (xi ) = 12 [f (xi + 0) + f (xi − 0)]; 3. f (x) is absolute integrable, i.e., the integral  ∞ |f (x)|dx −∞

exists. Then one has 1 f (x) = 2π

∞ dk eikx F (k) −∞

∞ F (k) =

(I.1)

dx e−ikx f (x).

(I.2)

−∞

Notation: F is called Fourier transform of f (and vice versa). Comments: • The choice of the factor 2π is the common choice in (solid-state) physics. In mathematical treatments mostly the symmetrical convention is used,1 1

See also the comment in footnote 2 on p. 375 further below.

366

I Fourier Transformation

 1 dy eixy F (y) f (x) = √ 2π  1 F (y) = √ dx e−ixy f (x). 2π • F (k = 0) is the mean value of the function f (x). • For the generalization to functions of multiple variables (to higher dimensions) see Sect. I.2. I.1.1 Examples and Applications (i) Real Functions f (x) = f ∗ (x)

F (k) = F ∗ (−k).



Proof: From





F (k) =

dx e−ikx f (x)

−∞

one obtains with f (x) = f ∗ (x)  ∗ F (k) =



dx e+ikx f (x) = F (−k).

−∞

 (ii) Real Even and Odd Functions

f (x) = ±f (−x)

⇔ ⇔

Proof:





F (k) =

F (k) = ±F (−k) = ±F ∗ (k)  1  ∞ cos kx F (k) = 2 dxf (x) × −i sin kx 0

dxf (x) e−ikx

∞  ∞

=

dxf (x) e 0





= 

0



−ikx



0

+

dxf (x) e−ikx +

−∞  ∞

dxf (x) e−ikx dxf (−x) eikx

0

  dxf (x) e−ikx ± eikx  1 0 ∞ 2 cos kx dxf (x) × . = −2i sin kx 0

=



I.1 The Transformation: Fourier Integral

367

(iii) Dirac δ-Function f (x) = δ(x − x0 ) Proof:

∞ F (k) =

F (k) = e−ikx0 .



dx e−ikx δ(x − x0 ) = e−ikx0 .

−∞

 In particular one has thus the reverse  ∞ dk ik(x−x )  e δ(x − x ) = . −∞ 2π

(I.3)

(iv) A Constant  ⇔

f (x) = 1



F (k) =

(I.3)

dx e−ikx = 2πδ(k).

(I.4)

−∞

Comment: The (infinitely extended) function f (x) = 1 does not fulfill con∞ dition (3), namely that the integral −∞ |f (x)|dx exists. Thus the result is not a usual function (but a so-called distribution). (v) Lorentz Function F (t) = iθ(t) e−iω0 t e−γt Proof:





f (ω) = i



f (ω) = 

dt θ(t) ei(ω−ω0 +iγ)t = i

−∞

=i



1 ω0 − ω − iγ

(I.5)

dt ei(ω−ω0 +iγ)t

0

1 0−1 = . i(ω − ω0 + iγ) ω0 − ω − iγ

For the reverse of the proof the integral  ∞ dω e−iωt −∞ 2π ω − ω0 + iγ has to be evaluated. This is done most comfortably with the integral theorem of Cauchy. The pole lies at ω = ω0 − iγ in the lower complex ω half-plane. Case t < 0: For t < 0 one closes the integration path as in Fig. I.1 by a semi-circle in the upper half-plane with ω = R eiϕ

368

I Fourier Transformation

=

-

Fig. I.1. The integration path in the complex frequency plane for t < 0

=

-

Fig. I.2. The integration path in the complex frequency plane for t > 0

and lets the radius R of the semi-circle tend towards ∞,  ∞  1 1 e−iωt dω e−iωt dω = 2π −∞ ω − ω0 + iγ 2π ω − ω0 + iγ  π −iReiϕ t iϕ 1 e R e i dϕ − lim iϕ R→∞ 2π 0 R e − ω0 + iγ  π iϕ i = 0 − lim e−iRe t dϕ R→∞ 2π 0  π i = 0 − lim eR(−i cos ϕ+sin ϕ)t dϕ = 0. R→∞ 2π 0 From the Cauchy integral theorem the closed integral vanishes, since no pole is encircled, and the second integral vanishes, since the integrand with eR sin ϕt , sin ϕ > 0 and t < 0 with increasing R vanishes exponentially. Case t > 0: For t > 0 one closes the integration path as in Fig. I.2 by a semi-circle in the lower half-plane,  ∞  1 dω e−iωt 1 e−iωt dω = 2π −∞ ω − ω0 + iγ 2π ω − ω0 + iγ  −π −iR eiϕ t iϕ 1 e R e i dϕ − lim iϕ R→∞ 2π 0 Re − ω0 + iγ = I1 + I2 . For the closed integral (the path taken in negative sense) one obtains  dω e−iωt 1 1 = (−2πi) e−i(ω0 −iγ)t = −i e−iω0 t e−γt . I1 = 2π ω − ω0 + iγ 2π

I.1 The Transformation: Fourier Integral

=

+

369

+

Fig. I.3. Deformation of the integration path in the complex z-plane; the path on the left is the sum of the paths on the right

The integral over the semi-circle vanishes for the reasons analogous to the case t < 0:  −π −iReiϕ t iϕ e R e i dϕ 1 lim −I2 = 2π R→∞ 0 Reiϕ − ω0 + iγ  −π i lim = eR(−i cos ϕ+sin ϕ)t dϕ = 0. 2π R→∞ 0 The integral vanishes, since the integrand with eR sin ϕt , sin ϕ < 0 and t > 0 vanishes exponentially.  (vi) Gauß Function 2 1 2 1 f (x) = √ e− 2 x /a a 2π

1

F (k) = e− 2 k



2 2

a

.

(I.6)

The Fourier transform of a Gauß function is thus again a Gauß function. Proof:  ∞ 2 1 1 2 F (k) = √ dx e−ikx e− 2 x /a a 2π −∞  ∞ 2 1 1 1 2 2 = √ dx e− 2 (x/a−ika) e− 2 k a , a 2π −∞ where in the exponent one has made a quadratic complement. Now one makes a transition to the complex z-plane. The integrand is an analytic function; the integration path can thus be deformed and partitioned as sketched in Fig. I.3. For the first integral one obtains 1

F1 (k) = e− 2 k

2

a2

1 √ a 2π



∞+ika2

1

dz e− 2 (z/a)

2

−∞+ika2



2

∞+ika 2 1 1 √ =e dz e− 2 (z/a−ika) a 2π −∞+ika2  ∞ 2 1 2 2 1 1 2 2 1 = e− 2 k a √ dz e− 2 (z/a) = e− 2 k a . a 2π −∞ − 12 k2 a2

370

I Fourier Transformation

Since the closed integral does not encircle a pole, the second integral vanishes. The two last integrals along the paths parallel to the imaginary axis vanish, if one writes z = z  + iz  in the exponent, 2

1

1

e− 2 (z/a−ika) = e− 2 [z



/a+i(z  /a−ka)]2

,

and takes the limit z  → ∞.



(vii) Periodic Functions with the Periodicity L The functions f (x) = f (x + L) likewise do not fulfill condition (3). See Appendix I.3 on Fourier series. The result is F (k) =

∞ 

2πδ(k − kn )Fkn

n=−∞

Fk

=

1 L

L

e−ikx f (x) dx

0

kn = ⇒

(I.24)

f (x) =

2π n L ∞ 

eikn x Fkn .

n=−∞

I.1.2 Convolution Theorem Let f (t), g(t), and h(t) be functions with the properties (1)–(3) with the Fourier transforms F (ω), G(ω), and H(ω), respectively, (and vice versa) with  F (ω) = dt eiωt f (t) (I.7)  G(ω) = dt eiωt g(t) (I.8)  H(ω) = dt eiωt h(t). (I.9) Then one has the convolution theorem F (ω) = G(ω) H(ω)  f (t) = dt g(t − t ) h(t ) = dt g(t ) h(t − t ).

(I.10)

I.1 The Transformation: Fourier Integral

Proof:

371

 dω −iωt dω −iωt e e F (ω) = G(ω)H(ω) 2π 2π      dω −iωt = e dt eiωt g(t ) dt eiωt h(t ) 2π    dω −iω(t−t −t ) e = dt g(t ) dt h(t ) 2π          = dt g(t ) dt h(t ) δ(t − t − t ) = dt g(t ) h(t − t ). 

f (t) =

Here the presumptions have made sure that the functions have a sufficiently nonpathological behavior, such that the sequence of the integrations can be interchanged.  Comments: • The Fourier transform of a product of functions is thus a convolution of the Fourier transforms (and vice versa). • Example: Forced oscillator (response theory). Let the displacement x(t), the Green function g(t), and the driving force k(t) be given with  x(t) = dt g(t − t ) k(t ). Then the corresponding Fourier transforms are given by X(ω), G(ω), K(ω) with X(ω) = G(ω)K(ω). • The theorem can easily be generalized to higher dimensions. I.1.3 Parseval’s Equation Let f (t) and g(t) be functions with the properties (1)–(3) and F (ω) and G(ω) their Fourier transforms as in (I.7) and (I.8). Then one has Parseval’s theorem   dω ∗ F (ω) G(ω). (I.11) dt f ∗ (t) g(t) = 2π Proof:







  dω ∗ dω  F (ω) eiωt G(ω  ) e−iω t 2π 2π     dω ∗ dω  F (ω) G(ω  ) dt ei(ω−ω )t = 2π 2π   dω ∗ dω  = F (ω) G(ω  )2π δ(ω − ω  ) 2π 2π  dω ∗ F (ω) G(ω). = 2π

dt f ∗ (t) g(t) =

dt

372

I Fourier Transformation

With the presumptions one has made sure that the functions have a sufficiently nonpathological behavior, such that the sequence of the integrations can be interchanged.  Comments: • The theorem can be generalized easily to higher dimensions. • The “scalar product” of two functions is thus independent of the representation (here in (I.11) in t-space and ω-space, respectively). • In generalization to three dimensions and in application to the quantummechanical wave functions as a function of three coordinates the Parseval theorem expresses the equality of the scalar product of the wave function in different representations. I.1.4 Uncertainty Relation From the examples (v) of the Lorentz function and (vi) of the Gauß function as well as from the examples (iii) and (iv) one can observe that functions, the main contributions of which are concentrated in a narrow regime, lead to Fourier transforms, which are distributed over a broad regime, and vice versa. The “widths” of the Gauß function and its Fourier transform 2 1 2 1 f (x) = √ e− 2 x /a a 2π

(I.6)



1

F (k) = e− 2 k

2

a2

.

are

√ 1√ bx ≈ a 2 and bk ≈ 2 a with a product of the order of 1, independent of a. bx bk ≈ 2; For more details see the following. – The widths of the imaginary part of the Lorentz function and its Fourier transform F (t) = −iθ(t) e−iω0 t e−γt are bt ≈

1 γ

(I.5)



f (ω) =

1 ω − ω0 + iγ

and bω ≈ γ,

respectively, with a product of the order of 1, independent of γ. bt bω ≈ 1. Definition : The mean square deviation Δx from the mean value x of the one-dimensional function f (x) is

I.1 The Transformation: Fourier Integral

 1/2  1/2 2 (x − x ) = x2 − x 2  1 n dx xn |f (x)|2 x = N  N = dx |f (x)|2 Δx =

with

373

(a) (b)

(I.12)

(c)

Example: Gauß Function For the mean square deviations Δx and Δk of the Gauß function and its Fourier transform, respectively, one finds Δx · Δk = 12 .

(I.13)

Proof: With (I.6)

f (x) =

1 2 2 1 √ e− 2 x /a a 2π

one finds √  ∞ 1 1 1 a π −x2 /a2 (GR 3.321.3) = √ dx e = 2 2 2πa −∞ 2πa 2 4a π  ∞ 2 2 1 Nx x = dx xe−x /a = 0 2πa2 −∞ √  ∞ a 1 1 a3 π 2 −x2 /a2 (GR 3.461.2) = √ Nx x2 = dx x e = 2πa2 −∞ 2πa2 4 8 π √ a ⇒ (Δx)2 = 4a π √ = 12 a2 8 π Nx =

With (I.6)

F (k) = e−k

2 2

a

one obtains by the replacement a → 1/a √  ∞ 2 2 π dk e−k a = Nk = 2a −∞ Nk k = 0 √  ∞ π 2 2 −k2 a2 dk k e = 3 Nk k = 4a −∞ √ π 2a 1 ⇒ (Δk)2 = √ = 2 π 4a3 2a

(I.14)

(I.15)

374

I Fourier Transformation

With this one obtains (Δx)2 (Δk)2 =

1 . 4 

Comments: • While one can determine the wave length and thus the corresponding wave vector for an infinitely extended plane wave precisely, one can determine the wave vector of a Gauß wave packet of the (finite) width Δx only up to an error (called uncertainty in quantum mechanics) determined by Δk = (2Δx)−1 (and vice versa). A similar relation holds for other wave packets and/or other observables (like frequency and time). • The precision of a measurement is subject to an uncertainty relation found here, which relates the minimum precision of two observables interconnected by Fourier transformation. In the present context this turns out to be a classical phenomenon. In quantum mechanics the uncertainty relation is applied to observables, which are conjugate to each other (like position and momentum).

I.2 Fourier Transformation in R4 : Plane Waves I.2.1 The Whole R3 The canonical convention, at least within solid-state physics, for the Fourier transformation in space (R3 ) and time is   d3k ik·r ∞ dω −iωt f (r, t) = e e F (k, ω). (I.16) 3 −∞ 2π R3 (2π) Notice the different sign in the phases. The time dependence is of no further interest here (but may be in another context). The reverse is   ∞ F (k, ω) = d3r e−ik·r dt eiωt f (r, t). (I.17) R3

−∞

Comment: F and f differ by the units of a volume (and time). One has in particular analogously to (I.3)   ∞ dω −iωt e dt eiωt = 2πδ(ω) ⇔ = δ(t) 2π −∞   d3k ik·r 3 −ik·r 3 d re = (2π) δ(k) ⇔ e = δ(r) (2π)3 R3

(I.18)

(I.19)

I.2 Fourier Transformation in R4 : Plane Waves

375

with the one- and three-dimensional Dirac δ-functions δ(t), δ(ω) and δ(r), δ(k), respectively. I.2.2 Normalization Volume V Often, instead of the R3 , a large but finite (possibly periodically continued) volume V is considered. Then one has discrete wave vectors k, and one has to replace the integration over k by a summation over k,  1  d3k F (k) = F (k) (I.20) (2π)3 V k

for a function F (k).2 Analogously to (I.3) one has  d3r e−ik·r = V δk,0

(I.21)

V

1  ik·r e = δ(r). V

(I.22)

k

For more details on Fourier series in R3 see Sect. I.3.6. Example: Coulomb and Yukawa potential For the Yukawa potential one has 1 −αr 1  4π e = eik·r . r V k 2 + α2

(I.23)

k

The Coulomb potential is obtained taking the limit α → 0. Proof: 1  F (k) eik·r V k  1 F (k) = d3r e−ik·r e−αr . r f (r) =

The representation of the integral in spherical polar coordinates (with the direction of k as the specific direction) leads to  ∞  1 F (k) = dr r2 dΩr e−ikr cos ϑ e−αr . r 0 2

The density of states of the wave vectors √ in k-space is V /(2π)3 . It is this point, at which the symmetrical use of the factor 2π as in the mathematical treatments would be unpractical.

376

I Fourier Transformation

The integration over the angles yields   2π  −ikr cos ϑ dΩr e = dϕ

1

dt e−ikrt = 2π

−1

0

e−ikr − eikr −ikr

and thus

 " 2π  −1 −1  2π ∞ ! (ik−α)r − F (k) = dr e − e−(ik+α)r = ik 0 ik ik − α −ik + α 4π 2π 2ik = 2 . = ik k 2 + α2 k + α2 

Comments: • Corresponding to (I.10) and (I.11), the convolution theorem and Parseval’s theorem for functions of one variable are generalized to functions of several variables. • As an example and in generalization to three dimensions one obtains, with the Coulomb potential 1 e2 , V (r) = 4πε0 |r| the relation between charge density ρ(r) and electrostatic potential φ(r),  % φ(r) = d3r V (r − r ) ρ(r  ) ⇔ φ(k) = V% (k) ρ%(k). • In the case of the Coulomb potential the Fourier coefficient F (k) is not defined for k = 0. This is because in R3 d3r/r the potential decreases with r slower than the volume increases. In application to solid-state physics, screening effects remove this divergence.3

I.3 Fourier Series In solid-state physics one must distinguish between quantities, which are defined in the continuous space, for example the electron density, and those, which are defined only on lattice points, for example nuclear positions or atomic displacements. I.3.1 The Series Let f (x) be a periodic function with the properties 1. f (x) is periodic, f (x) = f (x + L); 2. f (x) is piecewise smooth; 3

See the Course on Solid-State Physics.

I.3 Fourier Series

377

3. At discontinuities xi one has f (xi ) =

1 [f (xi + 0) + f (xi − 0)]; 2

Then the function f can be expanded into a Fourier series, ∞ 

f (x) =

cn eikn x

(I.24)

n=−∞ ∞

alternatively

a0  (an cos kn x + bn sin kn x) + 2 n=1

=

2 an = L

with

2 bn = L 1 cn = L



L

dx f (x) cos kn x = cn + c−n

(I.26)

dx f (x) sin kn x = i(cn − c−n )

(I.27)

dx f (x) e−ikn x

(I.28)

0



L

0



L

0

2π n. L

kn =

(I.25)

(I.29)

Proof: One performs a Fourier transformation, cuts the integration regime into the single periodicity intervals, and obtains 



F (k) = = = = (I.30)

=

(H.5)

=

dx e−ikx f (x)

−∞ ∞ 



(m+1)L

dx e−ikx f (x)

m=−∞ mL  L ∞ 



dx e−ikx e−ikmL f (x + mL)

m=−∞ ∞  m=−∞ ∞ 

0

e−ikmL



L

dx e−ikx f (x)

0



L

2πδ(kL − n2π)

n=−∞ ∞ 

dx e−ikx f (x)

0

2π δ(k − n2π/L) L n=−∞

 0

L

dx e−ikx f (x)

(x = x − mL) (x → x)

378

I Fourier Transformation

with y = kL in (I.30) of Sect. I.3.2 further below. Thus one obtains  ∞ 1 dk eikx F (k) f (x) = 2π −∞  ∞  L ∞   2π 1 δ(k − n2π/L) = dk eikx dx e−ikx f (x ) 2π −∞ L 0 n=−∞  L ∞   1 eikn x dx e−ikn x f (x ) (kn = n2π/L) = L 0 n=−∞ =

∞ 

eikn x cn

n=−∞

with cn =

1 L



L

dx e−ikn x f (x).

0



Comments: • The coefficients an and bn are real, if f is real, while the coefficients cn are complex. • The relation to the Fourier integral is given by F (k) =

∞ 

2πδ(k − kn )cn

cn =

n=−∞

1 F (kn ). 2π

• The convergence is uniform for f (x) continuous. The definition of uniform convergence of a sum N  SN (x) = sn (x) n=−N

is: |SN (x) − S(x)| < 

for

N > N0

is independent of x. Physically meaningful functions are continuous. For strongly simplified models (e.g., the quantum-mechanical potential barrier) functions are possibly discontinuous, and the convergence of the Fourier series of discontinuous functions shows, among others, the so-called Gibbs phenomenon (this is an oscillation around the original function, the amplitude of which cannot be reduced by extending the interval.) • For the generalization to higher dimensions see Sect. I.3.6. I.3.2 Examples and Applications (i) Even Periodic Functions f (x) = f (−x)



bn = 0.

I.3 Fourier Series

379

(ii) Odd Periodic Functions f (x) = −f (−x)



an = 0.

(iii) Periodic δ-Function ∞ 

eimy = 2π

m=−∞

∞ 

δ(y − n2π).

(I.30)

n=−∞

Proof: Consider the (finite) sum SM =

M 

eimy = e−iMy

2M 

eimy .

m=0

m=−M

This is a geometrical series, which can be summed, SM = e−iMy

2M 

eimy

m=0 1

1

e−i(M+ 2 )y − ei(M+ 2 )y − ei(2M+1)y =e = 1 1 iy 1−e e−i 2 y − ei 2 y sin((2M + 1) 12 y) = sin( 12 y) ∞  δ2M+1 ( 12 y − nπ), cf. Sect. H.2. =π −iMy 1

n=−∞

Now one takes the limit ∞  m=−∞

eimy = lim SM = π lim M→∞



∞  n=−∞

M→∞

δ( 12 y

∞ 

δ2M+1 ( 12 y − nπ)

n=−∞

− nπ) = 2π

∞ 

δ(y − n2π).

n=−∞

 Comment: The quantity SM is the representation of the periodic δ-function by a functional sequence similar to the one in Appendix H.2, the behavior of which at y = 0 is the same as that of the last example of the functional sequence of the non-periodic δ function of Appendix H.2. The expression is a periodic function of y.

380

I Fourier Transformation

(iv) Orthogonality (m ∈ Z and n ∈ Z) 1 2π





dy e

iy(m−n)

= δm,n

1 L



0



L



dz eiz L (m−n) = δm,n .

(I.31)

0

Proof: In the case m = n one has  2π 1 dy = 1, 2π 0 and in the case m = n one has  2π 1 1 ei2π(m−n) − 1 = 0. dy eiy(m−n) = 2π 0 2π i(m − n)  (v) Completeness For y and y  both from the same periodicity interval (of the length 2π) one has from example (iii) ∞ 



ei(y−y )m =

m=−∞



2πδ(y − y  − 2πn).

(I.32)

n

The completeness relation expresses the fact that each well-behaved function with the properties (1)–(3) (periodic in an interval) can be expanded into a Fourier series,  f (y) = dy  δ(y − y  )f (y  )    1 ei(y−y )m f (y  ) = dy  2π m    1 iym = dy  e−iy m f (y  ) e 2π m  eiym cn . = m

I.3.3 Convolution Theorem For functions f , g, and h with the properties (1)–(3), with the same period L and with the series ∞  Fn eikn x f (x) = n=−∞

I.3 Fourier Series

381

and analogously for g and h one has h(x) = f (x) g(x) ∞ ∞   ⇔ Hn = Fm Gn−m = Fn−m Gm m=−∞

(I.33)

m=−∞

Hn = Fn Gn   1 L 1 L ⇔ h(x) = dy f (x − y) g(y) = dy f (y) g(x − y). L 0 L 0

(I.34)

Proof: (i) By inserting one obtains   1 1 Hn = dx e−ikn x h(x) = dx e−ikn x f (x) g(x) L L  ∞ ∞   1 dx e−ikn x Fm eikm x Gl eikl x = L m=−∞ ∞ 

=

∞ 

Fm Gl

m=−∞ l=−∞ ∞ 

(I.31)

=

∞ 

l=−∞



1 L

dx e−i(kn −km −kl )x

Fm Gl δn,m+l

m=−∞ l=−∞ ∞ 

=

Fm Gn−m =

m=−∞

∞ 

Fn−l Gl .

l=−∞

(ii) By inserting one obtains h(x) =

∞  n=−∞

=

∞ 

e

n=−∞

= (I.32)

=

=

1 L 1 L 1 L



L

0



−ikn x

1 L

L

0

1 L

e−ikn x Fn Gn

n=−∞



1 dx f (y) L dy f (y)



∞ 

e−ikn x Hn = L

dy e

−ikn y

0



L

dy g(z) 0



∞ 



L

dz e−ikn z g(z)

0

e−ikn (x−y−z)

n=−∞

L

dz g(z)Lδ(x − y − z) 0

L

dy f (y) g(x − y) = 0

1 f (y) L

1 L



L

dz f (x − z) g(z). 0



382

I Fourier Transformation

I.3.4 Parseval’s Equation For two functions f and g with the properties (1)–(3) with equal periods and with the series ∞ 

f (y) = g(y) =

eikn y Fn

n=−∞ ∞ 

eikn y Gn

n=−∞

one has 1 L



L

dy f ∗ (y) g(y) =

0

∞ 

Fn∗ Gn .

(I.35)

n=−∞

Proof: By inserting one obtains 1 L



L



dy f (y) g(y) = 0

=

1 L



dy 0

=

m

=

n

 



∗ e−ikm y Fm

m

 m

(I.31)

L

1 ∗ Fm Gn L





eikn y Gn

n L

dy e−i(km −kn )y

0

∗ Fm Gn δm,n

n ∗ Fm Gm .

m

 I.3.5 Fourier Series in R3 : Lattices Functions, which are defined on a periodic lattice (with N lattice points h in the periodicity volume V ), can be described by wave vectors q defined from within a so-called Brillouin zone (BZ), f (h) =

1  iq·h e F (q) V

(I.36)

V  −iq·h e f (h) N

(I.37)

q∈BZ

F (q) =

h

with the replacement  d3r f (r) = V

V  f (h). N h

I.3 Fourier Series

383

Comments: • Occasionally one diverges from this replacement and defines  f (h) = eiq·h F˜ (q) q∈BZ

1  −iq·h F˜ (q) = e f (h). N h

Then f and F˜ have the same units. • In particular one has 1  iq·h e = δh,0 N

(I.38)

1  −iq·h  e = δq,G =: Δ(q) N

(I.39)

q∈BZ

G

h

with reciprocal lattice vectors G.4 I.3.6 Functions with Lattice Periodicity Functions with lattice periodicity f (r) = f (r + h) with lattice vectors h can be expanded analogously to (I.24)pp. into a Fourier series,  f (r) = F (G) eiG·r (I.40) G

F (G) =

1 Va



d3r f (r) e−iG·r

(I.41)

Va

with reciprocal lattice vectors G and the volume Va = V /N of the unit cell. One has in particular 1 Va

 Va

eiG·h = 1 d3r e−iG·r = δG,0  G

4

(I.42)

eiG·r = N



(I.43) δ(r − h).

(I.44)

h

This is of key importance for the structure determination in crystallography, see the Course on Solid-State Physics.

384

I Fourier Transformation

Summary: Fourier Integral and Fourier Series Notation: r position vector h lattice vector G reciprocal lattice vector k vector from the reciprocal space q vector from the (first) Brillouin zone The reciprocal space is discrete for V finite, continuous for V = R3 Fourier integrals: (I.16)



f (r) =

(I.17)



F (k) = 

d3r e−ik·r

d3k ik·r (I.20) 1  ik·r e F (k) = e F (k) 3 (2π) V k

d3r e−ik·r f (r)

(I.19)

= (2π)3 δ(k) = V δk,0  d3k ik·r 1  ik·r (I.19) δ(r) = e = e . 3 (2π) V k

Fourier series for lattice-periodic functions f (r) = f (r + h): (I.40)

f (r) =

 G

(I.41)

F (G) =

1 Va

 Va

1 Va

eiG·r F (G) 

d3r e−iG·r f (r) Va

(I.42)

1 = eiG·h d3r e−iG·r 

eiG·r

G

(I.43)

= δG,0

(I.44)

=



δ(r − h)

h

also 1  −ik·h (I.39)  e = δk,G =: Δ(k) N h G 1  iq·h (I.38) e = δh,0 . N q∈BZ

Problems

385

Problems I.1. Fourier Transformation of a Rectangular Function. Given be a periodic function f (t) with the period length T ,  h − t21 < t < t21 , 0 < t1 < T f (t) = 0 − T2 < t < − t21 , t21 < t < T2 . (a) Sketch this function. (b) Determine the Fourier coefficients Fn of f (t), +∞ 

f (t) =

Fn eiωn t .

n=−∞

(c) Write down the real Fourier series for t1 = T2 . (d) How does the frequency spectrum change for t1  T ? I.2. Fourier Series: Properties. For the functions f, g, h with the same period L and with the series +∞ 

f (x) =

Fn eikn x

n=−∞

and analogously for g and h one has: (a) Parseval’s equation 1 T



L

f ∗ (x)g(x)dx =

0

+∞ 

Fn∗ Gn

n=−∞

(b) Convolution theorem (i) h(x) = f (x)g(x) ↔ Hn =

+∞ 

+∞ 

Fm Gn−m =

m=−∞

Fn−m Gm

m=−∞

(ii) Hn = Fn Gn ↔ h(x) = Prove these claims.

1 L



L

f (x − y) g(y) dy = 0

1 L



L

f (y) g(x − y) dy 0

J Change of Variables: Legendre Transformation

Here the transition from the Lagrangian function to the Hamiltonian function will be performed, which results in a change of the variables, namely from the generalized velocities q˙ to the canonical momenta p. Let f (x, y, . . .) be a function of multiple variables with # $ # $ # $ ∂f ∂f ∂f dx + dy + dz + . . . df = ∂x y,z,... ∂y x,z,... ∂z x,y,... = X dx + Y dy + Z dz + . . . #

with X=

∂f ∂x

$ ⇒

= X(x, y, z, . . .)

x = x(X, y, z, . . .)

y,z,...

and analogously for y, z, etc. A Legendre transformation is a transformation of variables; for example the Legendre transformation with respect to the variable x has the form g(X, y, . . .) = f (x, y, . . .) − x

∂f = f − xX ∂x

with x = x(X, y, . . .). Namely, then one has dg = df − d(xX) = (X dx + Y dy + Z dz + . . .) − (X dx − x dX) = −x dX + Y dy + Z dz + . . . # # $ # $ $ ∂g ∂g ∂g = dX + dy + dz + . . . ∂X y,z,... ∂y X,z,... ∂z X,y,... #

with

#

and Y =

∂g ∂X

∂g ∂y

$ = −x y,z,...

#

$ = X,z,...

∂f ∂y

$ x,z,...

388

J Change of Variables: Legendre Transformation

etc. While x is one of the variables (in thermodynamics so-called natural variables) of the function f , X is one of the natural variables of the function g. Comments: • Analogous results are obtained for the transformation from the coordinate pair y and Y etc. instead of the coordinate pair x and X. • The two consecutive Legendre transformation (of the same pair of coordinates) leads back to the original function, h=g−X

∂g = g + Xx = f. ∂X

• H = L − pq, ˙

p=

dL . dq˙

• Legendre transformations play a vital role in thermodynamics for the transition of one thermodynamic potential to another. As an example of thermodynamical potentials, the internal energy U and the free energy F are related to each other via the Legendre transformation F = U − T S, and for example compressible systems one has dU (S, V ) = T dS − p dV dF (T, V ) = −S dT − p dV. The canonical pairs of variables here are (T, S) and (p, V ), while in the Lagrangian theory the pair is (p, q). ˙

References

A. Textbooks on Classical Mechanics There is a vast literature on classical mechanics. In the following a subjective selection of text books is given. 1. H. Goldstein, C. Poole, J. Safko, Classical Mechanics, 3rd edn. (Addison-Wesley, San Francisco, 2002) 638 pp. The first and even more so the second edition was the prime mechanics textbook with a vast range of contents. 2. W. Greiner, in Theoretical Physics, vols. 1 and 2. Mechanics, 6th edn. A text book with numerous worked-out examples. 3. L. D. Landau, I. M. Lifshitz, in Theoretical Physics vol. 1. Mechanics, The textbooks by Landau and Lifshitz always present unconventional and exciting views. The first volume starts with the Lagrangian formulation. 4. G. Ludwig, in Einf¨ uhrung in the Grundlagen der Theoretischen Physik vol. 1. Raum, Zeit, Mechanik, 2. Aufl. (Vieweg, Darmstadt, 1978) 449 pp., out of print (in German). First of four volumes with the mathematical and philosophical treatment of theoretical physics. 5. F. Scheck, Mechanics (Springer, Berlin, 1999). A very modern book with treatments extending to deterministic chaos. 6. A. Sommerfeld, in Lectures on Theoretical Physics vol. I. Mechanics reprint (Harri Deutsch, Frankfurt a. M.) B. Texbooks on Special Relativity Concerning the literature on Special Relativity: A series of textbooks on mechanics contains a chapter on Special Relativity theory (SR), in particular the books by Goldstein et al. [1], Greiner [2], and Scheck [5], given above. In addition there is a separate Volume 3A on SR by Greiner and Rafelski [7]. A subjective selection follows, see also the listing with Comments at the end of Chap. 7 (p. 332ff) in Goldstein [1].

390

References

7. W. Greiner, J. Rafelski, Theoretische Physik Band 3A Special Relativity theory This is one of the volumes of the text book series, which I personally do not like particularly; there is material overlapping with that of other volumes (Vol. 1 Mechanics I, Vol. 3 Electrodynamics), such that this volume is thus an (occasionally interesting) collection; with problems and solutions. 8. W. Rindler, Essential Relativity, 2nd edn. (Springer, Berlin, 1979) DM 72, 284 pp. This is a very beautifully and clearly written book, in which Special (104 pp.) and General Relativity theory (180 pp.) is treated in a compact overview. 9. W. Rindler, Introduction to Special Relativity, 2nd edn. (Clarendon, Oxford, 1992) pp. 59–185. A somewhat more elementary version of the first part of the preceding book, it contains in addition continuum mechanics. C. Special References 10. M. Abramowitz, I.A. Stegun, Handbook of Mathematical Functions (Dover, New York, 1965). 11. I.S. Gradshteyn, I.M. Ryzhik (abbreviated by GR), Tables of Integrals, Series, and Products, 2nd edn. (Academic, New York, 1980). 12. G. Herzberg, Molecular Spectra and Molecular Structure II: Infrared and Raman Spectra of Polyatomic Molecules, (van Nostrand, New York, 1945). 13. E.L. Hill, Rev. Mod. Phys. 23, 253 (1951). 14. K. Jung, Figur der Erde, Handbook of Physics, vol. 17 (Springer, Heidelberg, 1956) p. 606. 15. E. Noether, Nachr. Akad. Wiss. G¨ ottingen II, Math. P. Kl. 235 (1918). 16. H. W¨ anke The earth in the planetary system, Handbook of Physics (New Series) vol. V/2a (1984), Fig. 5 (p. 29).

Index

absolute space, 5 absolute time, 5 as principle, 219 absorption, oscillator, 136 acceleration definition, 5 plane polar coordinates, 44 action continuum formulation, 314 definition, 92 function, 296 adjoint matrix, definition, 356 advanced Green function, 134 agreement conservative forces, 11 constant mass, 7 del operator, 11 reference system, 7 standard configuration, 219 time derivatives, 5 velocity-independent forces, 13 amu, numerical value, 327 angular acceleration, 44 angular momentum conservation, 21, 157 central force, 18 central potential, 28 closed system, 17, 50 collision, 185 Kepler problem, 164 scattering, 185, 189, 198 definition, 14 equation of motion, 15

orbital, 240 plane polar coordinates, 45 quantum mechanics, 171 rigid body, 230, 239 rough collision, 190 spin, 240 Steiner’s theorem, 239 aphelion, 166 arc length, 336 areal velocity conservation, 158, 164, 171 atomic mass unit (amu) numerical value, 327 Atwood machine, 86, 104 force of constraint, 87 modified, 105 Avogadro number, numerical value, 326 axial vector, 330 definition, 15 axiom, Newton’s, 207 axioms, Newton’s, 4, 56 ballistic trajectory, 181 basic experience charges in e.m. fields, 12 interacting masses, 10 basis fixed and moving, 41 bead on rotating rod, 107 bead on rotating rod, 59, 67, 99, 106 conserved Hamiltonian function, 99 constraint, 70

392

Index

Lagrange, 73 virtual displacement, 65 behavior under transformation pseudo scalar, 330 pseudo tensor, 330 tensor, 329 vector, 329 binormal, 336 Bloch ansatz, 154 Bloch oscillation, 305 body deformable, 309 rigid, 309 spherical, 243 symmetrical, 243 body-fixed coordinates, 253 Bohr magneton, numerical value, 327 Bohr radius, numerical value, 327 Boltzmann constant, numerical value, 326 Born approximation, 197 bound motion, 33, 157 bound state, 170 brachistochrone, 102, 103 c.c., 119 calculus of residues, 131 canonical equations and conserved quantity, 304 Bloch oscillation, 305 canonical equations of motion, 271, 276 harmonic oscillator, 277 modified Hamilton principle, 282 pendulum, 272 canonical momentum, 78, 81, 269 charged particle, 81, 273 conservation, 96 gauge transformation, 111, 274 canonical momentum density, 321 canonical system, 273 canonical transformation, 282, 304 and time development, 287 definition, 284 magnetic field, 286 oscillator, 285 phase-space volume, 294 cartesian coordinates, 333 Laplace operator, 339 nabla operator, 339

catenary curve, 103 Cauchy integral theorem, 132, 368 causality, 93, 133, 134 center of mass, 14, 185, 228 hemisphere, 233 center-of-mass coordinate, 49 center-of-mass motion, 50, 77, 184 center-of-mass system, 185 central collision, 189, 190 central force, 53, 157 and central potential, 28 and trajectory, 160 angular-momentum conservation, 18 field, definition, 9 motion in a plane, 41 scattering, 189 central potential, 28, 157, 187 and trajectory, 160 angular-momentum conservation, 28 characteristic function, 300 definition, 12 energy conservation, 28 generalized quantities, 79 scattering, 189, 198 centrifugal force, 45, 79, 214 centrifugal potential, 159 centripetal force, 45 cgs system, 10 CH4 molecule, 151 chain, linear, 316 Chandler period, 260 characteristic function, 299 central potential, 300 free particle, 299 harmonic oscillator, 301 charge density, electrical, 232 charged particle canonical momentum, 81, 273 Hamiltonian function, 273 Lagrangian function, 81, 273 closed system, 50 angular-momentum conservation, 17 conservation of momentum, 17 conserved quantities, 50 definition, 2 CO2 -molecule, 138 collinear collision, 190

Index collision, 33, 183, 184 central, 189, 190 collinear, 190 eccentric, 190 elastic, 190 definition, 187 hard spheres, 184, 190 inelastic definition, 187 laboratory and center-of-mass system, 204 oblique, 189, 190 rough, 190 smooth, 190 Special Relativity theory, 189 collision normal, 190 commutator brackets, 278 completeness eigen vectors, 146, 358 Fourier series, 380 configuration space, 8, 61, 278 point transformation, 282 conics, 165 conjugate momentum, 78 conservation angular momentum, 21, 96, 100, 157 central force, 18 central potential, 28 closed system, 17, 50 areal velocity, 158, 164, 171 canonical momentum, 96 energy, 18, 21, 96–98 central potential, 28 closed system, 50 Hamiltonian function, 98 bead, 99 momentum, 17, 21, 96, 100 closed system, 17, 50 collision, 185 scattering, 185, 186 space homogeneity, 96 strain–momentum, 323 string, 324 conservation laws and symmetry, 23 conservation of areal velocity, 158 conservation of parity space inversion, 96 conservation theorems, 28, 317

393

conservative force, 18, 27, 68, 71, 72, 80 definition, 11 one-dimensional, 32 conserved quantity, 17, 276 in static electro-magnetic field, 28 constant of the motion, 17 constraint, 34 and generalized coordinates, 57 Atwood machine, 87 dumbbell, 58 holonomic, 58, 71, 84, 86 motion with, 55 nonholonomic, 58 pendulum, 61 skate, 60 pendulum, 34, 57, 59 reduction of the degrees of freedom, 55, 61 rheonomous, 58, 67, 271 rigid body, 227 scleronomous, 58, 67 sliding bead, 70 spherical pendulum, 62 wheel, 90 continuity equation energy current density, 317 energy–momentum, 317 continuum limit, 232 rigid body, 232 continuum mechanics as field theory, 309 continuum theory, 309 convergence uniform, 378 convolution theorem Fourier integrals, 370 Fourier series, 380 oscillator, 131 coordinate line, 333 coordinate sheet, 333 coordinate systems, 333 coordinate transformation infinitesimal rotation, 210 Legendre, 98, 269, 387 moving, 207 rotation, 209 translation, 208

394

Index

coordinates as observables, 5 body system, 253 cartesian, 333 Laplace operator, 339 nabla operator, 339 cyclic, 96, 272 cylindrical, 335 Laplace operator, 340 nabla operator, 340 generalized, 57, 62, 278 continuum, 310 moving, 83, 207, 283 rotation, 207 spherical, 334 Laplace operator, 340 nabla operator, 340 translation, 207 Coriolis force, 6, 45, 79, 214 meteorology, 224 on earth, 214, 224 corpuscular nature of matter, 5, 232 Coulomb force, 5 Coulomb friction, 13, 53 Coulomb interaction, 6 Coulomb potential, 184 Fourier transformation, 375 coupled pendulums, 150 coupled vibrations, 138, 145, 151 (in)stability, 141 coupling, minimal, 274 crank, 109 cross product, 331 cross-section, 185, 195, 196 differential, 197, 198 Rutherford, 200 sphere, 201 total, 200 Rutherford, 201 crystal, as oscillator, 115 crystallography, 383 cube inertia tensor, 244 with obstacle, 267 cuboid inertia tensor, 236, 237, 241 Steiner’s theorem, 241 moment of inertia, 264

cyclic coordinate, 96, 272, 296 and Hamiltonian function, 272 cycloid pendulum, 152 cyclotron frequency, 287 cylinder in hollow cylinder, 265 on inclined plane, 265 cylindrical coordinates, 335 Laplace operator, 340 nabla operator, 340 d’Alembert’s principle, 67–70, 92 decay, nuclear, 187 deflection towards east, 223 deformation, elastic, 309 deformation, plastic, 309 degree of freedom, 61 delta function, Dirac, 361 in R3 , 363 integral representation, 363 periodic, 363 determinism, 279 quantum mechanics, 279 differential cross-section, 197, 198 Rutherford, 200 differential equation, linear, 113, 351 eigen solution, 352 Dirac δ function, 361 displacement field, 310 displacement, virtual, 64 dissipation function free case with friction, 111 distribution, 361 double pendulum, 62 Lagrange, 74 double star, 179 dumbbell constraint, 58 dyadic product, 331 dynamical momentum, 81 dynamics, 5, 25 E¨ otv¨ os, 9 earth as top, 260 Coriolis force, 214, 224 rotating, 223 earth mass numerical value, 325

Index earth radius numerical value, 325 eccentric collision, 190 eccentricity, conic, 165 effective potential, 159 eigen angular momentum, 240 eigen solution CO2 -molecule, 140 coupled vibrations, 146 inertia tensor, 243 linear differential equation, 352 matrix, 355 oscillator, 118, 121 eigen vector orthogonality, 143 properties, 358 Einstein equivalence principle, 8 summation convention, 288, 311 Einstein mechanics unified, 222 elastic collision, 190 elastic deformation, 309 electro-magnetic field, 9, 109 particle in –, 80, 273 static, conserved quantity, 28 electron mass, numerical value, 327 elementary charge, numerical value, 326 energy equation of motion, 16 inner, 14 kinetic, 14 mass distribution, 311 rigid body, 230 potential, 11, 18 energy conservation, 18, 21, 98, 320 central potential, 28 closed system, 50 collision, 185 in magnetic field, 28 scattering, 185 string, 320 time homogeneity, 96 energy current density continuity equation, 317 energy–momentum tensor, 320 string, 319

395

energy density, 310, 311, 318 potential, 312 string, 312, 313 energy dissipation, oscillator, 134 energy–current density, 319 energy–momentum tensor, 317 continuity equation, 317 energy current density, 320 equation of motion canonical, 271, 276 pendulum, 272 Euler, 257 Hamiltonian, 269, 271 Lagrangian, first kind, 84 Lagrangian, second kind, 70 Newtonian, 4, 56, 71 observable, 276 pendulum, 36 mathematical, 116 Poisson bracket, 276 total angular momentum, 15 total energy, 16 total momentum, 15 equivalence principle, 8, 9 Euklidian space, 5 Euler angles, 254, 255 Euler equation, 92 variational method, 348 Euler equations of motion, 257 Euler period, 260 Euler–Lagrangian equations, 71 external force definition, 2 extremum principle, 55 field, 309 definition, 9 displacement –, 310 electro-magnetic, 9 fine-structure const., num. value, 327 force centrifugal, 45, 214 centripetal, 45 conservative, 11, 18, 27, 68, 71, 72, 80 Coriolis, 6, 45, 79, 214 on earth, 214, 224 Coulomb, 5 definition, 8

396

Index

external definition, 2 frictional, 20, 64, 72 Newtonian, 13 Stokes, 13 generalized (type I), 68 generalized (type II), 78 gravitational, 5, 6 impulsive, 184 inertial, 5, 45, 208, 209, 214 inner definition, 2 Lorentz-, 12 Lagrange, 80 of constraint, 34, 56, 63, 66 Atwood machine, 87 Lagrangian equations, first kind, 84 pendulum, 34 rigid body, 227 virtual work, 66 wheel, 91 rotation-free, 18 velocity dependent, 13, 80 velocity-dependent, 31 Lorentz, 80 force density, 312, 322 force field, 312 form (in)variance Newton equations, 56, 83 form invariance Galilean, 7 Lagrangian equations, 56, 83 Lagrangian function, 81 Foucault pendulum, 214 four-gradient, 318 Fourier coefficients periodic structures, 22 Fourier integral, 384 Fourier series, 376, 384 completeness, 380 convolution theorem, 385 orthogonality, 380 Parseval’s equation, 382, 385 Fourier transformation, 113, 365 Coulomb potential, 375 forced oscillator, 126 functions with lattice periodicity, 383 lattice, 382 phase-space volume, 295

rectangular function, 385 Yukawa potential, 375 free fall from large height, 52 into center, 180 with friction, 111 free motion, 33, 52 free particle characteristic function, 299 friction, 53, 305 Coulomb, 13, 53 driven oscillator, 155 Newton, 53 sliding, 13 static, 13 Stokes, 53, 120 friction coefficient, 13 frictional force, 20, 64 Lagrange, 72 Newtonian, 13 Stokes, 13 frictional heat, 135 oscillator, 134 functional, 347 action as –, 93 definition, 92 FWHM, 137 Galilean invariance, 220 Galilean transformation, 84, 217–219 acceleration, 213 and Lorentz transformation, 217 as limit, 222 center of mass conservation, 96 form invariance, 7 general, 210 special, 210, 219 velocity, 212 gas constant, numerical value, 326 gauge invariance Lagrangian function, 81 gauge transformation, 81 canonical momentum, 111, 274 Hamiltonian function, 274 Lagrangian function, 82, 274 Gauß function, 363 Fourier transformation, 369 uncertainty relation, 373 Gauß, integral theorem, 342

Index general Galilean transformation, 210 General relativity theory, 9, 175, 218 general transformation Galilean and Lorentzian, 218 generalized coordinate, 57, 62, 278 continuum, 310 generalized force (type I), 68 generalized force (type II), 78 generalized momentum, 78 generalized potential, 80 generalized susceptibility, 136 generating function, 290, 297 harmonic oscillator, 291 geosynchronous satellite, 181 Gibbs’ phenomenon, 378 gradient four-, 318 gravitational acceleration numerical value, 325 gravitational constant, 10, 165 numerical value, 325 gravitational force, 5, 6 gravitational mass, 8 gravitational potential, 12, 165, 184 perturbation of the, 175 scattering, 195 Green function, 113, 352 advanced, 134 Laplace operator, 341 oscillator, 127, 129, 131 poles, 130 retarded, 133 Green, integral theorem, 343 half width absorption, oscillator, 137 Hamilton principle, 92, 93, 298, 314 Lagrangian density, 314 modified, 281 Hamilton–Jacobi equation, 296, 297 time independent, 299 Hamiltonian density, 310 continuity equation, 317 string, 319 Hamiltonian equations of motion, 271 Hamiltonian function, 97, 269 as Legendre transformation, 269, 387

397

charged particle, 112, 273 conservation, 98 bead, 99 cyclic variable, 272 gauge transformation, 274 pendulum, 270 separation, 273 Hamiltonian mechanics, 78, 269 hard-core potential, 184 harmonic oscillator, see oscillator, 113 heat, frictional, 135 Heaviside step function, 131, 361 Heisenberg uncertainty relation, 232 hemisphere, center of mass, 233 hermitian matrix, definition, 356 holonomic constraint, 58, 71, 84, 86 homogeneity in time, 21 energy conservation, 96, 97 space, 21 conservation of momentum, 96, 100 homogeneous function, 109 Hooke’s law, 113, 114 Huygens, 152 impact parameter, 189 inclined plane, 53, 251, 263 moving, 88, 108 force of constraint, 89 with cylinder, 265 with mass, 108 inertia moment, 248 inertia tensor, 229, 231 cube, 244 cuboid, 236, 237, 241 Steiner’s theorem, 241 pyramid, 263 sphere, 234, 242, 264 Steiner’s theorem, 242 Steiner’s theorem, 238 inertial force, 5, 45, 69, 79, 207–209, 214 inertial mass, 8 inertial system, 5, 83, 84, 207, 208, 210, 218 definition, 4 examples, 6 inertial systems transformation, 222

398

Index

inner product definition, 331 instability coupled oscillators, 141 definition, 353 symmetrical top, 260 integral of the motion, 17, 23 integral theorem Cauchy, 368 Gauß, 342 Green, 343 Stokes, 343 internal energy, 14 internal force definition, 2 invariance Galilean, 7, 220 lattice translation, 22 space rotations, 21, 96 space translation, 21, 96 time translation, 21, 96 invariance and conservation quantity, 28 isotropy, space angular-momentum conservation, 21, 96 Kepler laws, 163, 178, 179 and force, 178 double star, 179 first, 163 second, 164 third law, 164, 167 as approximation, 172 Kepler problem, 157 angular-momentum conservation, 164 trajectory, 281 Kepler trajectories polar representation, 165 kinematics, 5, 25 kinetic energy definition, 14 mass distribution, 311 relativistic, 69 rigid body, 230 kinetic momentum, 81 charged particle, 273

laboratory system, 185 Lagrange multiplier, 64, 84 Lagrangian density, 96, 310, 311, 314 Hamilton principle, 314 string, 314, 316 Lagrangian equations, first kind, 84, 85 Lagrangian equations, second kind, 67, 70, 71, 94 form invariance, 83 Lagrangian function, 71, 269 charged particle, 81, 273 CO2 molecule, 138 form invariance, 81 gauge invariance, 81 gauge transformation, 82, 274 separability, 76 top, 268 Lagrangian Mechanics, 55 Landau levels, 287 Laplace operator cartesian coordinates, 339 cylindrical coordinates, 340 polar coordinates, 340 spherical coordinates, 340 Laplace–Runge–Lenz vector, 173 lattice translation invariance, 22 lattice vibrations absorption, 136 Laue, 22 law of inertia, 4 law of motion, 4 Legendre transformation, 98, 269, 387 thermodynamics, 388 length contraction, 5, 222 Lenz vector, 173, 182 line element cartesian coord., 333 cylindrical coordinates, 335 spherical coordinates, 334, 336 line integral, 18, 345 closed, 18 linear chain, 316 diatomic, 153 monatomic, 154 Liouville, theorem of –, 295 local trihedron spiral line, 336

Index Lorentz force, 12 in mechanics and electrodynamics, 222 Lagrange, 80 Lorentz function, 136, 363 Fourier transformation, 367 Lorentz transformation, 96, 217, 218, 221 and Galilean transformation, 217 linearity, 221 special, 221 magnetic field canonical transformation, 286 particle in –, 80, 273 magneton, Bohr numerical value, 327 many-body system, 27 mass electron numerical value, 327 gravitational, 8 in the electromagnetic field, 109 inertial, 8 on a cone, 109 on a sphere, 110 on inclined plane, 108 on logarithmic spiral, 106 on parabola, 106 on rotation paraboloid, 111 on spiral, 106 reduced, 49, 157 mass density, 5, 232, 310 mathematical pendulum, 116 period, 306 matrix adjoint, definition, 356 definition, 356 eigen solutions, 113 eigen value problem, 355 hermitian, definition, 356 orthogonal, definition, 356 real, definition, 356 self-adjoint, definition, 356 skew-symmetric, definition, 356 symmetric, definition, 356 transposed, definition, 356 unitary, definition, 356

Maxwell theory, 218 unified, 222 mean value, in time definition, 23 virial theorem, 23 mechanical similarity, 172 meteorology and Coriolis force, 224 minimal coupling, 81, 274 model, oscillator as –, 114 molecular crystal, 117 molecule CH4 , 151 diatomic, 150 molecule, as oscillator, 115 moment of inertia cuboid, 264 momentum canonical, 78, 81, 269 gauge transformation, 111 conjugate, 78 definition, 14 dynamical, 81 equation of motion, 15 generalized, 78 in electromagnetic field, 81 kinetic, 81 quasi-, 22 strain–, 320 uncertainty relation, 374 momentum density, 310, 322 canonical, 321 string, 321 moon mass numerical value, 325 moon radius numerical value, 325 motion bound, 33 free, 33, 183 in a plane, 41, 157 unbound, 33, 183 nabla operator cartesian coordinates, 339 cylindrical coordinates, 340 grad, div, rot, 344 polar coordinates, 340 spherical coordinates, 340

399

400

Index

Newton axiom I, 207 axiom II, 56 axioms, 4, 92 equation of motion, 71 first law, 55 Newtonian equations form (in)variance, 83 Newtonian friction, 13, 53 Newtonian mechanics, 222 Noether’s theorem, 20, 95 invariances, 95 non-holonomic constraint skate, 60 nonholonomic constraint, 58 pendulum, 61 normal coordinate, 147, 149 normal vibration, 147 definition, 143 nuclear excitation, 203 oblique collision, 189, 190 observable definition, 276 equation of motion, 276 open system, 51 definition, 2 operator identity, 213 optics phase-space volume, 295 orbit, 280 orbital angular momentum, 240 orthogonal matrix, definition, 356 orthogonality eigen vectors, 143, 146, 358 Fourier series, 380 plane waves, 380 oscillation period of the general, 34 transient, 135 oscillator n-dimensional, 149 absorption, 136 half width, 137 as model, 114 canonical equations, 277 canonical transformation, 285 characteristic function, 301

coupled, 138, 151 (in)stability, 141 crystal, 115 damped, 120 driven with friction, 155 eigen solutions, 118, 121 energy dissipation, 134 forced, 125 generating function, 291 Green function, 127, 129, 131 half width, 137 harmonic, 113, 149 molecule, 115 overdamped case, 124 phase portrait, 280 power, 135 RLC circuit, 116 simple, 117 three-dimensional, 115 trajectory, 280 with friction, 305 oscillator, harmonic trajectory, 280 overdamped case (damped oscillator), 124 parity conservation of, 96 normal vibration, 142, 144 Parseval’s equation, 135 Fourier integrals, 371 Fourier series, 382 Parseval’s theorem quantum mechanics, 372 particle decay, 203 pendulum, 34 canonical equations of motion, 272 constraint, 34, 57, 59 force of constraint, 46 forces, 34 Foucault, 214 Hamiltonian function, 270 inertial force, 46 large amplitude, 38 mathematical, 116 period, 306 physical, 249 plane, 53

Index spherical constraint, 62 trajectory, 281, 305 with mobile suspension, 152 pendulums coupled, 150 perihelion, 166, 194 conserved quantity, 173 rotation, 163, 171, 175, 181 period Chandler, 260 Euler, 260 general oscillation, 34 periodic delta function, 363 phase portrait oscillator, 280 phase space, 278–280 point transformation, 283 phase-space volume canonical invariance, 294 optics, 295 quantum mechanics, 295 time invariance, 295 physical pendulum, 249 Planck constant, numerical value, 327 plane motion Lagrange, 72 plane pendulum, 53 plane polar coordinates, 335 acceleration, 44 velocity, 44 plane waves orthogonality, 380 planetary motion, 157 plastic deformation, 309 point mass, 5, 17 as approximation, 1 point masses and rigid body, 227 point transformation configuration space, 83, 282 phase space, 283 Poisson brackets, 304 and quantum mechanics, 278 canonical invariance, 288 definition, 275 equation of motion, 276 fundamental, 277 problem, 304

properties, 277 quantum mechanics commutator, 277 Poisson equation, 364 polar coordinates Laplace operator, 340 moving basis, 41 nabla operator, 340 plane, 335 acceleration, 44 velocity, 44 spherical, 334 polar representation Kepler trajectories, 165 polar vector, 330 definition, 15 pole motion of the earth, 260 position uncertainty relation, 374 potential, 12 and trajectory, 178, 180 central, 12 Coulomb, 184 effective, 159 from trajectory, 178, 179 generalized, 80 gravitational, 165, 184 hard-core, 184 scalar, 80 vector, 80 potential energy, 11, 18 density, 312 potential step scattering at a –, 205 power, 16 damped oscillator, 135 definition, 14, 19 principal axes, 231, 243 principal moments of inertia, 243 and (un)stable rotation, 266 circular cone, 264 cube, 264 cylinder, 264 sphere, 264 tetrahedral molecule, 264 principal normal, 336 principle absolute time, 219 d’Alembert, 67–70, 92

401

402

Index

Hamilton, 92, 93 Lagrangian density, 314 relativity, 218, 221 virtual work, 66 process, 279 product cross, 331 dyadic, 331 tensor, 331 triple scalar, 332 vector, 331 pseudo scalar behavior under transformation, 330 pseudo tensor behavior under transformation, 330 pseudo vector, 330 definition, 15 pyramid inertia tensor, 263 quantization, 278 quantization of vibrations, 148 quantum mechanics, 78, 114, 117, 120, 232, 274, 278 and Poisson brackets, 278 angular momentum, 171 commutator, 277 determinism, 279 magnetic field, 287 Parseval’s theorem, 372 phase-space volume, 295 spin, 240 tunneling effect, 34 quarks, 5 quasi-momentum, 22 radius, Bohr numerical value, 327 reaction law, 4 real matrix, definition, 356 reduced mass, 49, 157 reference frame, moving, 207 relative coordinate, 49 relative motion, 50, 77, 184 relativity principle, 218, 221 Relativity theory General, 9, 175, 218 Special, 5, 6, 96, 218, 221 capture reaction, 187

collision, 189 kinetic energy, 69 representation different coordinate systems, 211 acceleration, 213 position, 212 velocity, 212 response function oscillator, 127, 129, 131 rest mass, 6 retarded Green function, 133 rheonomous constraint, 58, 67 rigid body, 227, 309 angular momentum, 230, 239 constraints, 227 continuum, 232 force of constraint, 227 kinetic energy, 230 torque, 239 RLC circuit as oscillator, 116 rocket, 54 rolling pendulum, 266 rotating pendulum, 107 stability, 108 rotation, 210 (un)stable, 266 coordinates, 207 fixed axis, 247 fixed point, 253 space, 20 with friction, 266 rotation vector, 21, 247 rotational motion, 28 stability of the –, 260 rough collision, 190 Runge–Lenz vector, 173 Rutherford scattering, 184, 205 cross-section, 200 scattering angle, 195 Rydberg energy, numerical value, 327 satellite, geosynchronous, 181 scalar, 329 scalar field definition, 9 scalar potential, 80 scalar product, 356 scattering, 33, 183, 184 at a potential step., 205

Index central potential, 198 elastic, 188 definition, 187 from reflecting sphere, 204 gravitational potential, 195 inelastic, 203 definition, 187 Rutherford, 184, 205 with nuclear excitation, 203 scattering angle definition, 194 Rutherford scattering, 195 scattering state, 170 scleronomous constraint, 58, 67 secular determinant, 140 secular equation, 118, 121 self-adjoint matrix, definition, 356 separation Hamiltonian function, 273 Lagrangian function, 76 separation of variables, 31 separatrix, 305 SI system, 10 similarity, mechanical, 172 skate, 60, 90 constraint, 60 skew-symmetric matrix, definition, 356 sliding friction, 13 smooth collision, 190 soccer ball, 266 solar equatorial radius numerical value, 325 solar mass numerical value, 325 solid-state physics, 22, 365, 374, 376 sound velocity, string, 316 space absolute, 5 configuration, 8 space homogeneity conservation of momentum, 96, 100 space inversion angular-momentum conservation, 100 conservation of parity, 96 space isotropy angular-momentum conservation, 96, 100 space rotation, 20 space translation, 20

403

space, Euklidian, 5 special Galilean transformation, 210 special Lorentz transformation, 221 Special Relativity theory, 5, 6, 96, 218, 221 capture reaction, 187 collision, 189 kinetic energy, 69 special transformation Galilean and Lorentzian, 218 sphere inertia tensor, 234, 242, 264 Steiner’s theorem, 242 rolling and sliding, 266 scattering from –, 204 spherical body, 243 spherical coordinates, 334 Laplace operator, 340 line element, 336 nabla operator, 340 spherical top, 244 spin angular momentum, 240, 256 spiral line local trihedron, 336 spool with thread, 264 spring constant, 114 stability coupled oscillators, 141 definition, 353 rotating pendulum, 108 rotation, 260 top, 268 standard configuration, 219 state bound, 170 definition, 278 scattering, 170 static friction, 13 statistical mechanics, 113 Steiner’s theorem, 231, 238, 249 angular momentum, 239 cuboid, 241 sphere, 242 torque, 239 step function, Heaviside, 131, 361 Stokes friction, 53, 120

404

Index

frictional force, 13 integral theorem, 343 strain momentum conservation, 323 string, 324 strain–momentum density, 320 stress, 320 stress tensor, 322 string, 323 string, 309 energy conservation, 320 energy current density, 319 energy density, 312, 313 Hamiltonian density, 319 Lagrangian density, 314, 316 momentum density, 321 sound velocity, 316 strain–momentum conservation, 324 stress tensor, 323 summation convention, Einstein, 288, 311 sun radius numerical value, 325 superposition principle, 4 surface element cartesian coord., 334 cylindrical coordinates, 335 spherical coordinates, 334 susceptibility, generalized, 136 symmetrical body, 243 symmetrical matrix, definition, 356 symmetrical top, 258 (in)stability, 260 symmetry and conservation laws, 23 system canonical, 273 closed, 50 angular-momentum conservation, 17 conservation of momentum, 17 conservation quantities, 50 definition, 2 open, 51 definition, 2 tangent unit vector, 336 tangential space, 280 tangential vector, 333

target, 185, 197 tensor, 329 behavior under transformation, 329 definition, 329 energy-momentum, 317 inertia, 229 stress, 322 tensor product, 331 theorem Noether, 20 Parseval Fourier integrals, 371 Fourier series, 382 thermodynamics, 20, 279, 388 Legendre transformation, 388 throw inclined, 53 vertical, 52 time absolute, 5 as principle, 219 development as canonical transformation, 287 dilation, 5, 222 homogeneity energy conservation, 96, 97 translation, 20 top, 243, 253 earth as, 260 Lagrangian function, 268 spherical, 244 stability, 268 symmetrical, 258 (in)stability, 260 torque, 15, 18, 27 rigid body, 239 Steiner’s theorem, 239 total angular momentum definition, 14 total cross-section, 200 Rutherford, 201 total energy definition, 14 total momentum definition, 14 trajectory, 280, 336 and potential, 178, 179 ballistic, 181 from potential, 178, 180

Index harmonic oscillator, 280 Kepler problem, 281 pendulum, 281, 305 trajectory and central force, 160 trajectory and central potential, 160 trajectory of a particle, 5 transformation canonical, 282 Galilean, 84, 217–219 general Galilean and Lorentzian, 218 Legendre, 98, 269, 387 Lorentz, 96, 217, 218, 221 special Galilean and Lorentzian, 218 transient oscillation, 135 translation coordinates, 207 in time, 20 space, 20 transposed matrix, definition, 356 triple scalar product, 332 tunneling effect, 34, 114 turning points, 33 two-particle interaction, 6 two-particle system, 48 uncertainty relation, 372 Gauß function, 373 Heisenberg, 232 momentum, 374 position, 374 uniform convergence, 378 unit system, electromagnetism, 10 unitary matrix, definition, 356 universe, 54 vacuum diel. constant, num. value, 326 vacuum permeability, numerical value, 326 variational method, 71, 347 brachistochrone, 102, 103 catenary curve, 103 Dido’s problem, 350 Euler equation, 348 variational principle, 310 vector, 329

405

axial, 330 definition, 15 behavior under transformation, 329 decomposition in orthog. coord., 336 polar, 330 definition, 15 pseudo, 330 definition, 15 Runge–Lenz, 173 vector field definition, 9 div, rot, 344 vector potential, 80 vector product, 331 velocity definition, 5 plane polar coordinates, 44 velocity dependent force, 80 velocity of light, 10, 218 as SR principle, 221 numerical value, 326 velocity-dependent force, 31 Lorentz, 80 vibrations, coupled, 145 virial theorem, 23, 28 virtual displacement, 64 definition, 64 virtual work, 99 of the force of constraint, 66 principle of –, 66 Volterra differential equation, 306 volume element cartesian coord., 334 cylindrical coordinates, 335 spherical coordinates, 334 wheel, 90 constraint, 90 force of constraint, 91 wheel and axis, 105 work definition, 14, 19 virtual, 66, 99 world line, 8 Yukawa potential Fourier transformation, 375