Classical electrodynamics [2 ed.] 9783031050985, 9783031050992

Table of contents :
Cover
Title
Preface to the second edition
Preface to the first edition
Contents
1 Classical electrodynamics: a short review
2 Multipole expansion of the electrostatic potential
3 The method of image charges
4 Image charges in dielectrics
5 Series of image charges
6 Functions of complex variables and electrostatics
7 Conformal mapping in electrostatics
8 Separation of variables in Laplace equation
9 Relativistic transformation of E and B fields
10 Relativistic covariance of electrodynamics
11 Energy and momentum of the electromagnetic field
12 The Feynman paradox
13 The resonant cavity
14 Fields and radiation
15 Test of the Coulomb's law and limits on the mass of the photon
16 Magnetic monopoles
A Orthogonal curvilinear coordinates
Index

Citation preview

Undergraduate Lecture Notes in Physics Series Editors Neil Ashby, University of Colorado, Boulder, CO, USA William Brantley, Department of Physics, Furman University, Greenville, SC, USA Matthew Deady, Physics Program, Bard College, Annandale-on-Hudson, NY, USA Michael Fowler, Department of Physics, University of Virginia, Charlottesville, VA, USA Morten Hjorth-Jensen, Department of Physics, University of Oslo, Oslo, Norway Michael Inglis, Department of Physical Sciences, SUNY Suffolk County Community College, Selden, NY, USA Barry Luokkala , Department of Physics, Carnegie Mellon University, Pittsburgh, PA, USA

Undergraduate Lecture Notes in Physics (ULNP) publishes authoritative texts covering topics throughout pure and applied physics. Each title in the series is suitable as a basis for undergraduate instruction, typically containing practice problems, worked examples, chapter summaries, and suggestions for further reading. ULNP titles must provide at least one of the following: • An exceptionally clear and concise treatment of a standard undergraduate subject. • A solid undergraduate-level introduction to a graduate, advanced, or non-standard subject. • A novel perspective or an unusual approach to teaching a subject. ULNP especially encourages new, original, and idiosyncratic approaches to physics teaching at the undergraduate level. The purpose of ULNP is to provide intriguing, absorbing books that will continue to be the reader’s preferred reference throughout their academic career.

Francesco Lacava

Classical Electrodynamics From Image Charges to the Photon Mass and Magnetic Monopoles Second Edition

Francesco Lacava “Sapienza” Università di Roma Roma, Italy

ISSN 2192-4791 ISSN 2192-4805 (electronic) Undergraduate Lecture Notes in Physics ISBN 978-3-031-05098-5 ISBN 978-3-031-05099-2 (eBook) https://doi.org/10.1007/978-3-031-05099-2 1st edition: © Springer International Publishing Switzerland 2016 2nd edition: © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, 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. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

To the memory of my parents Biagio and Sara and my brother Walter

All royalties and fees received by the author for this book will be donated to cancer and genetic disease research institutes.

Preface to the Second Edition

The second edition continues to pursue the aim of the original book by examining some topics of special interest for those who want to deepen the study of Electrodynamics. It features four new subjects: series of image charges, conformal mapping, method of separation of variables, fields and radiation for moving charges. As in the first edition, the topics of the problems often complete the subjects examined in the text. New problems have also been added in those chapters already present in the first edition. Some problems are more advanced than those usually featured in introductory textbooks to Electrodynamics and need more elaborate calculations. The preface to the first edition in the following pages has been supplemented to take account of the new arguments. I am very grateful to Prof. S. Petrarca for suggestions on topics of mathematical methods, and to Dr. L. Patrizii for pointing at the most recent results on searches for magnetic monopoles. Special thanks go to Dr. F. Safai Tehrani for his useful advice and for the careful reading of the entire book. Rome, Italy March 2022

Francesco Lacava

vii

Preface to the First Edition

In the undergraduate program of Electricity and Magnetism emphasis is given to the introduction of fundamental laws and to their applications. Many interesting and intriguing subjects can be presented only shortly or are postponed to graduate courses on Electrodynamics. In the last years I examined some of these topics as supplementary material for the course on Electromagnetism for the M.Sc. students in Physics at the University Sapienza in Roma and for a series of special lectures. This small book collects the notes from these lectures. The aim is to offer to the readers some interesting study cases useful for a deeper understanding of the Electrodynamics and also to present some classical methods to solve difficult problems. Furthermore, two chapters are devoted to the Electrodynamics in relativistic form needed to understand the link between the electric and magnetic fields. In the two final chapters two relevant experimental issues are examined. This introduces the readers to the experimental work to confirm a law or a theory. References of classical books on Electricity and Magnetism are provided so that the students get familiar with books that they will meet in further studies. In some chapters the worked out problems extend the text material. Chapter 1 is a fast survey of the topics usually taught in the course of Electromagnetism. It can be useful as a reference while reading this book and it also gives the opportunity to focus on some concepts as the electromagnetic potential and the gauge transformations. The expansion in terms of multipoles for the potential of a system of charges is examined in Chap. 2. Problems with solutions are proposed. Chapter 3 introduces the elegant method of image charges in vacuum. In Chap. 4 the method is extended to problems with dielectrics. This last argument is rarely presented in textbooks. In Chap. 5 the method of images is extended to configurations which need an infinite series of image charges. This yields, for instance, the field for two conductive spheres. These three chapters feature several examples and problems with solutions. Analytic complex functions can be used to find the solutions for the electric field in two-dimensional problems. After a general introduction of the method, Chap. 6 discusses some examples. Chapter 7 introduces the application of the ix

x

Preface to the First Edition

conformal mapping and presents transformations to get the field in some interesting configurations. Problems are proposed for these subjects. Chapter 8 presents the method of separation of variables and applies it using Cartesian, spherical and cylindrical coordinates. This allows to solve the Laplace equation in many electrostatic configurations, and also to reproduce some results previously found using the method of image charges or the complex variables. Problems at the end of the chapter provide further examples. Chapter 9 aims at introducing the relativistic transformations of the electric and magnetic fields by analyzing the force on a point charge moving parallel to an infinite wire carrying a current. The equations of motion are formally the same in the laboratory and in the rest frame of the charge but the forces acting on the charge are seen as different in the two frames. This example introduces the transformations of the fields in special relativity. In Chap. 10 a short historical introduction mentions the difficulties of the classical physics at the end of the 19th century in explaining some phenomena observed in Electrodynamics. The problem of invariance in the Minkowsky spacetime is examined. The formulas of Electrodynamics are written in covariant form. The electromagnetic tensor is introduced and the Maxwell equations in covariant form are given. The energy and momentum conservation in the presence of an electromagnetic field are considered in Chap. 11. The Poynting’s vector is introduced and some simple applications to the resistor, to the capacitor and to the solenoid are presented. The transfer of energy in an electric circuit in terms of the flux of the Poynting’s vector is also examined. Then the Maxwell stress tensor is introduced. Some problems with solutions complete the chapter. The Feynman paradox or paradox of the angular momentum is very intriguing. It is very useful to understand the dynamics of the electromagnetic field. Chapter 12 presents the paradox with comments. An original example of a rotating charged system in a damped magnetic field is discussed. Chapter 13 presents a lecture by Feynman on the capacitor at high frequency. The effects produced by iterative corrections due to the induction law and to the displacement current are considered. For very high frequency of the applied voltage, the capacitor becomes a resonant cavity. This is a very interesting example for the students. Chapter 14 studies the fields for moving particles. First, we compute the fields and potentials for a uniformly moving point charge using Lorentz transformations. Then, we derive the retarded potentials and find the fields for a point particle in arbitrary motion. The radiation emission is considered for the most interesting types of acceleration. The need to test the dependence on the inverse square of the distance for the Coulomb’s law was evident when the law was stated. The story of these tests is presented in Chap. 15. The most sensitive method, based on the Faraday’s cage, was introduced by Cavendish and was used until the half of last century. After that time the test was interpreted in terms of a test on a non-null mass of the photon. The theory is shortly presented and experiments and limits are reported.

Preface to the First Edition

xi

Chapter 16 introduces the problem of the magnetic monopoles. In a paper Dirac showed that the electric charge is quantized if a magnetic monopole exists in the Universe. The Dirac relation is derived. The properties of a magnetic monopole crossing the matter are presented. Experiments to search the magnetic monopoles and their results are mentioned. In the Appendix the general formulas of the differential operators used in Electrodynamics are derived for orthogonal systems of coordinates and the expressions for spherical and cylindrical coordinates are given. I wish to thank Professors L. Angelani, M. Calvetti, A. Ghigo, S. Petrarca, and F. Piacentini for useful suggestions. A special thank is due to Professor M. Testa for helpful discussions and encouragement. I am grateful to Dr. E. De Lucia for reviewing the English version of this book and to Dr. L. Lamagna for reading and commenting this work. Rome, Italy May 2016

Francesco Lacava

Contents

1

2

Classical Electrodynamics: A Short Review . . . . . . . . . . . . . . . . . . . . . . 1.1 Coulomb’s Law and the First Maxwell Equation . . . . . . . . . . . . . 1.2 Charge Conservation and Continuity Equation . . . . . . . . . . . . . . . 1.3 Absence of Magnetic Charges in Nature and the Second Maxwell Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Laplace’s Laws, Biot and Savart Law and the Steady Fourth Maxwell Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Faraday’s Law and the Third Maxwell Equation . . . . . . . . . . . . . 1.6 Displacement Current and the Fourth Maxwell Equation . . . . . . 1.7 Maxwell Equations in Vacuum . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.8 Maxwell Equations in Matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.9 Electrodynamic Potentials and Gauge Transformations . . . . . . . 1.10 Electromagnetic Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Multipole Expansion of the Electrostatic Potential . . . . . . . . . . . . . . . . 2.1 The Potential of the Electric Dipole . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Interaction of the Dipole with an Electric Field . . . . . . . . . . . . . . 2.3 Multipole Expansion for the Potential of a Distribution of Point Charges . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Properties of the Electric Dipole Moment . . . . . . . . . . . . . . . . . . . 2.5 The Quadrupole Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 A Bidimensional Quadrupole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Appendix A: Higher Order Terms in the Multipole Expansion of the Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Appendix B: Expansion in Terms of Spherical Harmonics . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 4 5 5 6 7 8 8 9 14 16 17 17 18 19 22 23 24 25 26 28 29 33

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3

4

5

6

Contents

The Method of Image Charges . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 The Method of Image Charges . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Point Charge and Conductive Plane . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Point Charge and Conducting Sphere . . . . . . . . . . . . . . . . . . . . . . . 3.3.1 Force on a Point Charge Near a Charged Conducting Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Conducting Sphere in a Uniform Electric Field . . . . . . . . . . . . . . 3.5 A Charged Wire Near a Cylindrical Conductor . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

35 35 36 38

Image Charges in Dielectrics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Electrostatics in Dielectric Media . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Point Charge Near the Plane Separating Two Dielectric Media . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Dielectric Sphere in an External Uniform Electric Field . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

63 63

Series of Image Charges . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Point Charge Between Two Grounded Plates . . . . . . . . . . . . . . . . 5.2 Two Separated Charged Conductive Spheres . . . . . . . . . . . . . . . . 5.3 Force Between Two Charged Spherical Conductors . . . . . . . . . . 5.4 A Charged Conductive Sphere and a Conductive Plate at Ground . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 Dielectric Sphere and Point Charge . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

83 83 84 89

Functions of Complex Variables and Electrostatics . . . . . . . . . . . . . . . 6.1 Analytic Functions of Complex Variable . . . . . . . . . . . . . . . . . . . . 6.2 Electrostatics and Analytic Functions . . . . . . . . . . . . . . . . . . . . . . 6.3 The Function f (z) = z µ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.1 The Quadrupole: f (z) = z 2 . . . . . . . . . . . . . . . . . . . . . . . 6.3.2 The Conductive Wedge at Fixed Potential . . . . . . . . . . . 6.3.3 Edge of a Thin Plate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 The Charged Wire: f (z) = log z . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

101 101 104 104 104 105 108 110 110 112 115

41 42 45 47 49 61

64 68 71 72 81

90 92 93 93 99

Contents

7

8

9

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Conformal Mapping in Electrostatics . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Conformal Mapping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Conformal Mapping and Harmonic Functions . . . . . . . . . . . . . . . 7.3 Conformal Mapping and Bilinear Electrostatic Problems . . . . . . 7.3.1 Charged Wire Between Two Grounded Plates . . . . . . . . 7.4 The Linear-Fractional Function . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4.1 The Split Cylinder: Two Opposite Half Cylinders at Different Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5 The Schwarz-Christoffel Transformation . . . . . . . . . . . . . . . . . . . 7.5.1 Conformal Transformation for a Single Corner . . . . . . . 7.5.2 The Field Near the Edge of a Parallel-Plate Capacitor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

117 117 118 119 120 122

Separation of Variables in Laplace Equation . . . . . . . . . . . . . . . . . . . . . 8.1 The Method of Separation of Variables . . . . . . . . . . . . . . . . . . . . . 8.2 Orthogonal and Complete Sets of Functions . . . . . . . . . . . . . . . . . 8.3 Separation of Variables in the Laplace Equation . . . . . . . . . . . . . 8.4 Separation of Variables in Cartesian Coordinates . . . . . . . . . . . . . 8.4.1 Box with a Side at Given Potential and Five Conductive Grounded Sides . . . . . . . . . . . . . . . . . . . . . . . 8.5 Separation of Variables in Spherical Coordinates with Azimuthal Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5.1 A Dielectric Sphere in a Uniform Electric Field . . . . . . 8.5.2 Potential for a Charged Ring . . . . . . . . . . . . . . . . . . . . . . 8.5.3 Conducting/Dielectric Sphere and Point Charge . . . . . . 8.6 Separation of Variables in Spherical Coordinates . . . . . . . . . . . . . 8.7 Separation of Variables in Polar (Cylindrical) Coordinates . . . . 8.7.1 Wire and Cylindrical Capacitor . . . . . . . . . . . . . . . . . . . . 8.7.2 Conducting Wedge (Corner) . . . . . . . . . . . . . . . . . . . . . . . 8.7.3 Split Cylinder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.7.4 Cylinder in a Uniform Electric Field . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

137 137 139 139 140

Relativistic Transformation of E and B Fields . . . . . . . . . . . . . . . . . . . . 9.1 From Charge Invariance to the 4-Current Density . . . . . . . . . . . . 9.2 Electric Current in a Wire and a Charged Particle in Motion . . . 9.3 Transformation of the E and B Fields . . . . . . . . . . . . . . . . . . . . . . 9.4 The Total Charge in Different Frames . . . . . . . . . . . . . . . . . . . . . . 9.5 Force Between Wires Carrying Currents . . . . . . . . . . . . . . . . . . . . 9.6 Electromagnetic Induction and Relative Motion of Circuits . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

179 179 182 184 185 187 190 193

123 125 128 129 132 132 136

140 145 146 151 153 156 158 159 159 160 163 166 167 177

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Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198 10 Relativistic Covariance of Electrodynamics . . . . . . . . . . . . . . . . . . . . . . 10.1 Electrodynamics and Special Theory of Relativity . . . . . . . . . . . 10.2 4-Vectors, Covariant and Contravariant Components . . . . . . . . . 10.3 Relativistic Covariance of the Electrodynamics . . . . . . . . . . . . . . 10.4 4-Vector Potential and the Equations of Electrodynamics . . . . . . 10.5 The Continuity Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.6 The Electromagnetic Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.7 Lorentz Transformation for Electric and Magnetic Fields . . . . . . 10.8 Maxwell Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.8.1 Inhomogeneous Equations . . . . . . . . . . . . . . . . . . . . . . . . 10.8.2 Homogeneous Equations . . . . . . . . . . . . . . . . . . . . . . . . . 10.9 Potential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.10 Gauge Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.11 Phase of the Wave . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.12 The Equations of Motion for a Charged Particle in the Electromagnetic Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

199 199 201 205 206 206 207 208 209 209 210 210 211 211

11 Energy and Momentum of the Electromagnetic Field . . . . . . . . . . . . . 11.1 Poynting’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2.1 Resistor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2.2 Solenoid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2.3 Capacitor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3 Energy Transfer in Electrical Circuits . . . . . . . . . . . . . . . . . . . . . . 11.4 Momentum Conservation in a System of Charges and Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.5 The Maxwell Stress Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.6 Radiation Pressure on a Surface . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.7 Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.8 The Covariant Maxwell Stress Tensor . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

215 215 218 218 219 220 221

12 The Feynman Paradox . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.1 The Paradox . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.2 A Charge and a Small Magnet . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.3 Analysis of the Angular Momentum Present in the System . . . . 12.4 Two Cylindrical Shells with Opposite Charge in a Vanishing Magnetic Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Appendix: Vector Potential for a Spinning Charged Sphere . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

247 247 248 250

212 213

223 224 229 231 231 232 234 245

251 260 263

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xvii

Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275 13 The Resonant Cavity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.1 The Capacitor at High Frequency . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2 The Resonant Cavity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

277 277 282 284 285 287

14 Fields and Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.1 Fields of a Point Charge in Uniform Motion . . . . . . . . . . . . . . . . . 14.1.1 Fields from the Lorentz Transformed Potentials for a Point Charge in Uniform Motion . . . . . . . . . . . . . . 14.2 Potentials and Fields for a Point Charge in Arbitrary Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.2.1 The Retarded Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . 14.2.2 Liénard-Wiechert Potentials . . . . . . . . . . . . . . . . . . . . . . . 14.2.3 Comment to the Liénard-Wiechert Potentials . . . . . . . . 14.2.4 Covariant Form of the Liénard-Wiechert Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.2.5 Electric and Magnetic Fields . . . . . . . . . . . . . . . . . . . . . . 14.2.6 Calculation of the Electric and Magnetic Fields . . . . . . 14.3 Radiation by an Accelerated Charge . . . . . . . . . . . . . . . . . . . . . . . 14.3.1 Radiation by a Charged Particle with Velocity v  c .......................................... 14.3.2 Radiation by a Charged Particle with the Acceleration Parallel to the Velocity . . . . . . . . 14.3.3 Power Radiated by a Charged Particle in Arbitrary Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.3.4 Radiation by a Charged Particle with the Acceleration Normal to the Velocity . . . . . . . . 14.4 Radiation Reaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.5 Electric Dipole Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

289 289

15 Test of the Coulomb’s Law and Limits on the Mass of the Photon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.1 Gauss’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.2 First Tests of the Coulomb’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . 15.3 Proca Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.4 The Williams, Faller and Hill Experiment . . . . . . . . . . . . . . . . . . . 15.5 Limits from Measurements of the Magnetic Field of the Earth and of Jupiter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

292 294 294 296 298 299 300 301 306 307 307 310 311 315 317 323 324 328 329 330 330 332 334 336

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Contents

15.6 The Lakes Experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.7 Other Measurements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.8 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Appendix: Proca Equations from the Euler-Lagrange Equations . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

337 338 339 340 341

16 Magnetic Monopoles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.1 Generalized Maxwell Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.2 Generalized Duality Transformation . . . . . . . . . . . . . . . . . . . . . . . 16.3 Symmetry Properties for Electromagnetic Quantities . . . . . . . . . 16.4 The Dirac Monopole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.5 Magnetic Field and Potential of a Monopole . . . . . . . . . . . . . . . . 16.6 Quantization Relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.7 Quantization from Electric Charge-Magnetic Pole Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.8 Properties of the Magnetic Monopoles . . . . . . . . . . . . . . . . . . . . . 16.8.1 Magnetic Charge and Coupling Constant . . . . . . . . . . . . 16.8.2 Energy Losses for Monopoles in Matter . . . . . . . . . . . . . 16.8.3 Magnetic Monopoles in Magnetic Field . . . . . . . . . . . . . 16.9 Searches for Magnetic Monopoles . . . . . . . . . . . . . . . . . . . . . . . . . 16.9.1 Detection of Magnetic Monopoles . . . . . . . . . . . . . . . . . 16.9.2 Searches for Dirac Monopoles at Accelerators . . . . . . . 16.9.3 Search for Cosmic Monopoles . . . . . . . . . . . . . . . . . . . . . 16.9.4 Bounds on the Flux of Cosmic Magnetic Monopoles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.9.5 Direct Searches for Cosmic Massive Monopoles . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

343 343 344 346 347 348 349 351 352 352 352 354 354 355 355 357 358 358 361

Appendix A: Orthogonal Curvilinear Coordinates . . . . . . . . . . . . . . . . . . . . 363 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371

Chapter 1

Classical Electrodynamics: A Short Review

The aim of this chapter is to review shortly the main steps of Classical Electrodynamics and to serve as a fast reference while reading the other chapters. This book collects selected lectures on Electrodynamics for readers who are studying or have studied Electrodynamics at the level of elementary courses for the degrees in Physics, Mathematics or Engineering. Many text books are available for an introduction1 to Classical Electrodynamics or for more detailed studies.2

1.1 Coulomb’s Law and the First Maxwell Equation Only two kinds of charges exist in Nature: positive and negative. Any charge3 is a negative or positive integer multiple of the elementary charge e = 1.602 × 10−19 Coulomb, that is equal to the absolute value of the charge of the electron. The law of the force between two point charges was stated4 by Coulomb in 1785. In vacuum the force F21 on the point charge q2 , located at r2 , due to the point charge q1 , located at r1 , is:

1

For instance: Griffiths, Introduction to Electrodynamics [1]; Feynman et al. The Feynman Lectures on Physics [2]; Bettini, Electromagnetism [3]; Purcell and Morin, Electricity and Magnetism [4]; Purcell, Electricity and Magnetism, Berkeley Physics Course [5]. 2 For instance: Jackson, Classical Electrodynamics [6]; Zangwill, Modern Electrodynamics [7]; Landau and Lifshitz, The Classical Theory of Fields [8] ; and the other books cited in the chapters of this book. 3 Quarks have charge 1 e or 2 e but are confined in the hadrons. The charge quantization is examined 3 3 in Chap. 12. 4 A short historical note on the discovery of the Coulomb’s law is given in Sect. 15.2.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 F. Lacava, Classical Electrodynamics, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-031-05099-2_1

1

2

1 Classical Electrodynamics: A Short Review

F21 =

r2 − r1 1 q1 q2 4π 0 |r2 − r1 |3

(1.1)

with the permittivity constant 0 = 8.564 × 10−12 F/m (Farad/meter). The force on the charge q1 is F12 = −F21 as required by Newton’s third law. It is experimentally proved that in a system of many charges the force exerted on any charge is equal to the vector sum of the forces acted by all the other charges (superposition principle). If F is the net electrostatic force on the point charge q located at r, the electric field E0 in vacuum in that position is defined5 by the relation: E0 = lim

q→0

F . q

The electric field E0 (r ), due to a point charge Q located in the frame origin, is: E0 (r ) =

1 Q rˆ 4π 0 r 2

(1.2)

and from the superposition principle the electric field at a point r due to a system of point charges qi , located at ri , is equal to the vector sum of the fields produced at the position r, by all the point charges: i=N 1  (r − ri ) E0 (r) = qi . 4π 0 i=1 |r − ri |3

For a continuous distribution with charge density ρ(r ) = d Q/dτ over a volume τ , the electric field is: E0 (r) =

1 4π 0

 τ

(r − r ) ρ(r ) dτ  . |r − r |3

The electric field (1.2) of a charge Q is radial and then it is conservative, so the field can be written as the gradient6 of a scalar electric potential V0 that depends on the position: E0 = −grad V0 = −∇V0

5

(1.3)

The limit is needed to avoid the point charge q modifies the field due to charges induced in conductors or from the polarization of the media. 6 In the book the nabla operator will be used for the differential operators: grad f = ∇ f , div v = ∇ · v and curl v = ∇ × v.

1.1 Coulomb’s Law and the First Maxwell Equation

3

For a point charge at the origin the potential is: 1 Q +C 4π 0 r

V0 (r ) =

with an arbitrary additive constant C that becomes null if V0 (∞) = 0 is assumed. From the superposition principle for the electric field, the electric field of a system of charges is also conservative, and the potential at a given point, is equal to the sum of the potentials at that point from all the charges in the system. Thus it follows: i=N 1  qi + C V0 (r) = 4π 0 i=1 |r − ri |

1 V0 (r) = 4π 0

 τ

ρ(r ) dτ  + C |r − r | (1.4)

where C is an arbitrary constant. For a confined distribution of charges the potential can be fixed null at infinity and thus C = 0. The closed curve line integral of E0 is null and, for the Stokes’ theorem, the field E0 is irrotational: ∇ × E0 = 0 that is the local form to state the electric field is conservative. The Gauss’s law7 is very relevant in electrostatics. It states that the flux  S of E0 through a closed surface S is equal to the total charge inside the surface, divided by 0 : 

E0 · nˆ d S  =

 S (E) = S

Q 0

(1.5)

with nˆ the outward-pointing unit normal at each point of the surface, and the charges outside the surface do not contribute to the flux  S . From this law the Coulomb’s theorem: near the surface of a conductor with surface charge density σ (x, y, z), the electric field is equal to: E0 =

σ nˆ 0

with nˆ the versor with direction outside of the conductor at that point. If ρ(r) is the charge density in the volume τ enclosed by S, the total charge is Q:  Q=

7

See also Sect. 15.1 of this book.

τ

ρ(r ) dτ 

(1.6)

4

1 Classical Electrodynamics: A Short Review

and substituting this expression in the relation (1.5) and using the divergence theorem, we find the first Maxwell equation in vacuum: ∇ · E0 =

ρ 0

(1.7)

that is the differential (or local) expression of the Gauss’s law. After the substitution of the relation (1.3) in the last equation, the Poisson equation for the potential is found: ∇ 2 V0 = −

ρ . 0

For a given distribution of charges and for fixed boundary conditions, due to the unicity of the solution, the solution of Poisson equation is given by (1.4).

1.2 Charge Conservation and Continuity Equation Charge conservation in isolated systems is experimentally proved. The charge in a volume τ , enclosed by the surface S, changes only if an electric current I, positive if outgoing, flows through S. So: −

dQ =I dt

(1.8)

and I is the flux of the electric current density J = ρv, with v the velocity of the charge, through the surface S: 

J · nˆ d S 

I =

(1.9)

S

with nˆ the outward-pointing unit normal to the element of surface d S  of the closed surface. By substituting the relations (1.6) and (1.9) in Eq. (1.8) and applying the divergence theorem, the continuity equation is found: ∂ρ +∇·J=0 ∂t that is the local expression of the charge conservation.

(1.10)

1.4 Laplace’s Laws, Biot and Savart Law and the Steady Fourth Maxwell Equation

5

1.3 Absence of Magnetic Charges in Nature and the Second Maxwell Equation The field lines of the magnetic induction B are always closed. This is easily seen by tracking with a small magnetic needle the field lines of B around a circuit carrying a current. Indeed in Nature no source (magnetic monopole) of magnetic field has ever been observed.8 Thus the flux  S (B) through a closed surface S is always null: 

B · nˆ d S  = 0 .

 S (B) =

(1.11)

S

By applying the divergence theorem to this relation the second Maxwell equation can be found: ∇·B=0 where the null second member corresponds to the absence of magnetic sources.

1.4 Laplace’s Laws, Biot and Savart Law and the Steady Fourth Maxwell Equation H.C. Oersted in 1820 discovered that an electric current produces a magnetic field. This observation led physicists to write the laws of magnetism in a short time. The force acting on the element dl of a circuit carrying a current I, in the direction of dl, located in a magnetic field B, is: dF = I dl × B

(Second Laplace formula)

from which the force on a point charge q moving with velocity v in a magnetic field B: F = qv × B

(Lorentz force).

The contribution to the field B0 (r) in vacuum at a point P(r), given by dl , an element of a circuit, with a current I flowing in the direction of dl , located at r , is: dB0 (r) =

μ0 I dl × (r − r ) 4π |r − r |3

(First Laplace formula)

(1.12)

called Biot and Savart law, with μ0 the permeability constant of free space (μ0 = 4π · 10−7 H/m (Henry/meter)). 8

Chap. 12 is devoted to the theory and the search of magnetic monopoles.

6

1 Classical Electrodynamics: A Short Review

The field B0 (r) from a circuit is: μ0 B0 (r) = 4π



I dl × (r − r ) . |r − r |3

By integrating along a closed line the last formula the Ampère’s law can be found:  B0 · dl = μ0

i=N 

Ii

i=1

with the algebraic sum of all the currents Ii enclosed by the loop. Then applying the Stokes’ theorem and the relation (1.9), the differential law is derived: ∇ × B0 = μ0 J

(1.13)

that is the steady-state fourth Maxwell equation in vacuum.

1.5 Faraday’s Law and the Third Maxwell Equation Faraday’s law of induction says that the induced electromotive force f in a circuit is equal to the negative of the rate at which the flux of magnetic induction (B) through the circuit is changing: f =−

d dt

(Faraday-Neumann-Lenz law)

(1.14)

with f the closed line integral of an induced non conservative electric field Ei :  f =

Ei · dl .

(1.15)

The induction is observed in different situations. If the shape of the circuit is changed or some of its parts move in a steady magnetic field, the electromotive force and the induced electric field Ei can be related to the Lorentz’s force acting on the free charges in the conductor. This is the case of a flux of magnetic field lines cut by parts of the circuit. Differently if the sources of the magnetic field (circuits carrying currents or permanent magnets) move while the circuit is at rest, the induced electric field is determined by the (relativistic) transformations of the fields E and B between different reference frames as discussed in Chaps. 6 and 7. But when the circuit and the sources of the field are at rest and the magnetic field changes (for instance due to the change of the current in one of the circuits used as sources) the induction effect implies a new physical phenomenon. By applying to the (1.15) the Stokes’ theorem and substituting that in the first member of Eq. (1.14) while at the second member is

1.6 Displacement Current and the Fourth Maxwell Equation

7

written the flux of B, as given in (1.11), the third Maxwell equations is found: ∇ × Ei = −

∂B . ∂t

Thus in general the electric field is the superposition of the irrotational field Ee from the electric charges and a non irrotational electric field Ei induced by the rate of change of the magnetic field: E = Ee + Ei .

1.6 Displacement Current and the Fourth Maxwell Equation Taking the divergence of the Eq. (1.13), while the first member is always null because ∇ · ∇ × v = 0 is null for any vector v, the second member ∇ · J is null only in steadystate situations. To solve this fault Maxwell suggested to replace the charge density ρ in the continuity equation (1.10) with the expression for ρ from the first Maxwell equation (1.7). The result is the sum of two terms with always a null divergence:   ∂E +J = 0. ∇ · 0 ∂t Replacing J in the (1.13) with the sum of the two terms, the Eq. (1.13) becomes:   ∂E ∇ × B = μ0 J + 0 ∂t that is the correct fourth Maxwell equation with both members having a null divergence also for time varying fields and currents. The term added to the current density J at the second member, is the displacement current density J S : J S = 0

∂E ∂t

associated to the rate of change of the electric field E.

8

1 Classical Electrodynamics: A Short Review

1.7 Maxwell Equations in Vacuum The four Maxwell equation in vacuum are: ∇ · E0 =

∇ × E0 = −

ρ 0

∂B0 ∂t

(I )

∇ · B0 = 0

(I I I )

(I I )

∇ × B0 = μ0 J + μ0 0

∂E0 ∂t

(I V ).

1.8 Maxwell Equations in Matter In the presence of an external electric field the atomic and molecular dipoles in the media are polarized. The electric polarization P is the average dipole moment per unit volume. Charges due to the polarization are present on the surface of the ˆ with nˆ the outward-pointing unit normal dielectrics with charge density σ P = P · n, vector to the surface, and in the volume with density ρ P = −∇ · P. So in the first Maxwell equation (1.7) the polarization charges have also to be taken into account: ∇·E=

ρ + ρP . 0

By introducing the displacement vector D = 0 E + P this equation becomes: ∇·D=ρ where only the free charges are present. The magnetization of the media can be described by the vector magnetization M that is the average magnetic moment per unit volume. The microscopic currents, associated to the magnetization, flow on the surface with current density Jms = ˆ with nˆ the outward-pointing unit normal vector to the surface, and in the M × n, volume with current density Jmv = ∇ × M. The current density Jmv has to be added to the free current density J in the fourth Maxwell equation (1.13): ∇ × B = μ0 (J + Jmv ) and defining the magnetic field H: H=

B − μ0 M μ0

1.9 Electrodynamic Potentials and Gauge Transformations

9

the steady fourth equation becomes: ∇×H=J with only the free current at the second member. In matter the four Maxwell equations are: ∇·D=ρ

∇×E=−

∇·B=0

∂B ∂t

∇×H=J+

∂D . ∂t

Of course to find the fields the constitutive relations D = D(E) and H = H(B) have to be known. In homogeneous and isotropic media these relations are: D = E

P = 0 (κ − 1)E

 = 0 κ

with  the permittivity and κ the dielectric constant of the medium, and: B = μH

M = (κm − 1)H

μ = μ0 κm

with μ the permeability and κm the relative permeability of the medium.

1.9 Electrodynamic Potentials and Gauge Transformations The Maxwell equations are four first-order equations that, with assigned boundaries conditions, can be solved in simple situations. It is often convenient to introduce potentials, that while are defined to satisfy directly the two homogenous equations, are determined by only two second-order equations. In an isotropic and homogeneous medium with permittivity  and permeability μ, the four Maxwell equations are: ∇·E=

ρ 

∇·B=0

∇×E=−

∂B ∂t

(I )

(1.16)

(I I )

(1.17)

(I I I )

(1.18)

10

1 Classical Electrodynamics: A Short Review

∇ × B = μJ + μ

∂E ∂t

(I V ) .

(1.19)

Electrodynamic Potentials The divergence of a curl is always null (∇ · ∇ × v = 0), so the second equation is satisfied if B is the curl of a vector potential A: B = ∇ × A.

(1.20)

With this definition the third equation becomes:   ∂A ∇× E+ =0 ∂t and since the sum of the two terms has a null curl, it can be the gradient of a scalar potential V with a change of sign as in electrostatics: E+

∂A = −∇V ∂t

and so, in terms of the potentials, the electric field is: E = −∇V −

∂A . ∂t

(1.21)

Thus homogeneous equations are used to introduce a vector potential A and a scalar potential V that have to be determined. Gauge Transformations The vector potential A is determined up to the gradient of a scalar function ϕ. Indeed under the transformation: A → A = A + ∇ϕ

(1.22)

the field B, given by the relation (1.20), since ∇ × ∇ϕ is always null, is left unchanged: B = ∇ × A = ∇ × A + ∇ × ∇ϕ = ∇ × A = B . It is easy to see that in order for the field E (1.21) to be also unchanged, the scalar potential has to transform as: V → V = V − Indeed we find:

∂ϕ . ∂t

(1.23)

1.9 Electrodynamic Potentials and Gauge Transformations

E = −∇V  −

11

∂A ∂ϕ ∂A ∂∇ϕ ∂A = −∇V + ∇ − − = −∇V − = E. ∂t ∂t ∂t ∂t ∂t

The relations (1.22) and (1.23) are the gauge transformations of the electrodynamic potentials. Equations of the Electrodynamic Potentials To determine the potentials A and V we have to consider the two inhomogeneous Maxwell equations. Substituting the relations (1.20) and (1.21) in these equations we get9 the two coupled equations: ∇ 2 A − μ

∂ 2A ∂V − ∇(∇ · A + μ ) = −μJ 2 ∂t ∂t

∇2V +

ρ ∂ ∇·A=− . ∂t 

(1.25)

(1.26)

Lorenz Gauge The freedom in the definition of the potentials from the gauge transformations (1.22) and (1.23) gives the possibility to choose A and V in order they satisfy the Lorenz condition: ∇ · A + μ

∂V =0 ∂t

(1.27)

useful to decouple the two Eqs. (1.25) and (1.26). If not satisfied by A and V, this relation can be satisfied by two new potentials A and V  that are gauge transformed of A and V by the (1.22) and (1.23). The (1.27) for the new potentials gives an equation for the scalar function ϕ used in the transformation:   ∂ 2ϕ ∂V 2 (1.28) ∇ ϕ − μ 2 = − ∇ · A + μ ∂t ∂t that, with assigned boundaries conditions, at least in principle, can be solved. Since this gauge transformation is always possible, we can assume that the potentials satisfy the (1.27). We say we choose to work in the Lorenz gauge. Gauge transformations of potentials which satisfy the Lorenz condition, give new potentials which observe the Lorenz condition if the function ϕ satisfies the equation:

9

Remember the relation: ∇ × ∇ × ∇v = ∇(∇ · v) − ∇ 2 v

See also note 2 at pag. 19 in this book.

(1.24)

12

1 Classical Electrodynamics: A Short Review

∇ 2 ϕ − μ

∂ 2ϕ = 0. ∂t 2

Uncoupled Equations and Retarded Potentials With the Lorenz conditions the two Eqs. (1.25) and (1.26) are decoupled and become: ∇ 2 A − μ

∂ 2A = −μJ ∂t 2

(1.29)

∇ 2 V − μ

∂2V ρ =− ∂t 2 

(1.30)

or with the d’Alambertian operator:  = ∇ 2 − μ

∂2 ∂t 2

in a more compact form: A = −μJ

V = −

ρ . 

The equations for A and V are four second-order scalar equations. Their particular solutions are the retarded potentials: μ A(r, t) = 4π

1 V (r, t) = 4π 







J(r , t − |r −v r | )  dτ + C |r − r |

(1.31)



ρ(r , t − |r −v r | )  dτ + C  |r − r |

(1.32)

with the constants C and C  null if the potentials are zero at infinity. The potentials at the point r at time t depend on the values of the sources at r at time t  = t − (|r − r |)/v, where Δt = (|r − r |)/v is the time interval for the electromagnetic signal propagates from the source at r to the point of observation at r with velocity v = √1μ . To get the solutions of the Eqs. (1.29) and (1.30), homogeneous solutions have to be added to the particular solutions (1.31) and (1.32). It is evident that the homogeneous solutions are waves that propagate. Coulomb Gauge Another possible choice is the Coulomb gauge with the condition:

1.9 Electrodynamic Potentials and Gauge Transformations

13

∇·A=0 thus the Eq. (1.26) becomes the Poisson equation: ∇2V = −

ρ 

with the instantaneous Coulomb potential as solution: 1 V (r, t) = 4π 

 τ

ρ(r, t) dτ  |r − r |

(1.33)

while the Eq. (1.25) becomes: ∇ 2 A − μ

  ∂V ∂ 2A = −μJ + μ∇ . ∂t 2 ∂t

(1.34)

The instantaneous potential (1.33) does not take into account the propagation time of the electromagnetic signal and seems in contrast with the time interval required to propagate the information from the source to the position where the signal is observed. Actually the observable quantities are the fields E and B which depend also on the non instantaneous potential A given by the Eq. (1.34) thus their changes are also delayed with respect to the changes of the sources. This gauge is also called transverse gauge. Indeed the density current J can be written as the sum10 of a longitudinal or irrotational component Jl , (such that ∇ × Jl = 0), and a transverse or solenoidal component Jt , (∇ · Jt = 0). From the continuity equation for the current we have that the term:  ∇

∂V ∂t



in Eq. (1.34) is equal to the longitudinal component of the density current. Thus with only the transverse component Jt at the second member, the equation is: ∇ 2 A − μ

∂ 2A = −μJt ∂t 2

and the vector potential A depends only on the transverse component of the current density and is parallel to that. The gauge ∇ · A = 0 is useful in the absence of sources (ρ = 0 and J = 0). In this gauge, called radiation gauge, V = 0 or constant and the Eq. (1.25) becomes the wave equation:

10

For this subject see Jackson [6], Sect. 6.5.

14

1 Classical Electrodynamics: A Short Review

∇ 2 A − μ

∂ 2A =0 ∂t 2

while the fields are given by the relations: E=−

∂A ∂t

B = ∇ × A.

1.10 Electromagnetic Waves In the absence of electric charges in a non conducting homogeneous and isotropic infinite medium, of permittivity  and permeability μ, the Maxwell equations are: ∇·E=0

∇×E=−

(I )

∂B ∂t

(I I I )

∇·B=0

∇ × B = μ

(I I ) ∂E ∂t

(I V )

and non null solutions are possible for time-varying fields E and B. Taking the curl of the third equation, and substituting the fourth in the right hand side, by the first equation and the relation (1.24) the equation for the field E is: ∇ 2 E − μ

∂ 2E = 0. ∂t 2

From the curl of the fourth equation after the third is substituted at the right hand side, a similar equation is found for the field B: ∇ 2 B − μ

∂ 2B = 0. ∂t 2

In the radiation gauge the same equation is found for the vector potential A. These are the equations for electromagnetic waves propagating in the medium with speed v: 1 1 c c 1 =√ √ = v=√ √ μ 0 μ0 κκm n κ

(κm  1)

where n is the refractive index of the crossed medium while: 1 c= √ 0 μ0 is the speed of light in vacuum.

1.10 Electromagnetic Waves

15

Plane Electromagnetic Waves The plane electromagnetic waves have the fields E and B with same components over a plane surface. If this surface is parallel to the place yz the components depend only on the x component and on the time and the wave equations become: ∂ 2E ∂ 2E − μ =0 ∂x2 ∂t 2

∂ 2B ∂ 2B − μ = 0. ∂x2 ∂t 2

The most general solution is a function of the type: f = f 1 (x − vt) + f 2 (x + vt)

with

1 v=√ μ

that corresponds to the superposition of a wave f 1 (x − vt) travelling in the positive direction of the x axis and of a wave f 2 (x + vt) travelling in the opposite direction with phase velocity ±v. For these two waves the fields E and B have the same components on the planes given by the relation: x = ±vt + const and the fields are: E = E0 f (x ∓ vt)

B = B0 f (x ∓ vt).

Substituting these relations in the Maxwell equations it is easy to find: E=B×v

(1.35)

so in a plane wave the fields E and B are orthogonal each other and also to the direction of propagation vˆ . These waves are transverse waves. For a monochromatic wave of frequency ω propagating in the kˆ direction, the most general solution for the fields is of the type: f (r, t) = Aei(k·r−ωt) + Bei(k·r+ωt) where the wave vector k is: k= and λ is the wavelength.

ω 2π vˆ = vˆ v λ

16

1 Classical Electrodynamics: A Short Review

References 1. D.J. Griffiths, Introduction to Electrodynamics, 4th edn. (Cambridge University Press, Cambridge, 2017) 2. R.P. Feynman, R.B. Leighton, M. Sands, The Feynman Lectures on Physics, Vol. II 3. A. Bettini, A Course in Classical Physics 3—Electromagnetism (Springer, 2016) 4. E.M. Purcell and D.J. Morin, Electricity and Magnetism, 3rd edn. (2013) 5. E.M. Purcell, Electricity and Magnetism, Berkeley Physics Course, Vol. II. (McGraw-Hill) 6. D.J. Jackson, Classical Electrodynamics, 3rd edn. (John Wiley & Sons, NewYork, 1999) 7. A. Zangwill, Modern Electrodynamics (Cambridge University Press, Cambridge, 2013) 8. L.D. Landau and E.L. Lifshitz, The Classical Theory of Fields, Course of Theoretical Physics, Vol. 2. (Elsevier)

Chapter 2

Multipole Expansion of the Electrostatic Potential

The potential of a localized charge distribution at large distance can be expanded as a series of multipole terms.1 The terms of the series depend on the charge spatial distribution in the system and have different dependence from the distance. In this chapter we will first examine the electric dipole, the simplest system after the point charge. We will write the dipole potential and obtain the expressions of the electrostatic energy, the force and the torque acting on the dipole in an external field. Then we will derive the first terms of the multipole expansion for the potential from a charge distribution. Finally we will write the general expression for the multipole expansion together the formula for the expansion in terms of spherical harmonics.

2.1 The Potential of the Electric Dipole The electric dipole is a rigid system of two point charges of opposite sign +q and −q separated by the distance δ. It is characterised by the dipole moment p = qδ with δ oriented from the negative to the positive charge. The potential generated by the electric dipole at point P at position r from its center, is the sum of the potentials of the two point charges: V (P) =

1 4π0



q q − r+ r−

 =

q 4π0



r− − r+ r+ r−



where r+ and r− are the distances of P from +q and −q respectively (see Fig. 2.1).

1

For this subject see for instance: Griffiths [1], Sect. 3.4; Panofsky and Phillips [2], Sects. 1.7–1.8.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 F. Lacava, Classical Electrodynamics, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-031-05099-2_2

17

18

2 Multipole Expansion of the Electrostatic Potential

Fig. 2.1 The potential of the electric dipole at a point P at distance r, is the sum of the potentials of the two opposite point charges. The distances of the point P from the two charges in the approximation r  δ are r+ = r − 2δ cos θ and r− = r + 2δ cos θ

When r  δ first order approximation can be used: r−  r +

δ cos θ 2

r+  r −

δ cos θ 2

r− − r+  δ cos θ

with θ the angle between r and δ, and: r+ r− = r 2 −

δ2 cos2 θ  r 2 4

and the potential at a large distance from the dipole becomes: V (P) =

1 qδ cos θ 1 p·r = . 4π0 r 2 4π0 r 3

(2.1)

For increasing r this potential decreases as 1/r 2 function, faster than the 1/r dependence of the point charge potential.

2.2 Interaction of the Dipole with an Electric Field We can consider the interaction of an electric dipole p with an external electric field E that in general can be non-uniform. For the force F, the Fx component of the total force on the dipole is the sum of the x-component of the forces on the two point charges: Fx = −q E x (x, y, z) + q E x (x + δx , y + δ y , z + δz )

2.3 Multipole Expansion for the Potential of a Distribution of Point Charges

19

with E(x, y, z) the electric field at the position of the charge −q and E(x + δx , y + δ y , z + δz ) that at the position of +q respectively. Writing the second term as: E x (x + δx , y + δ y , z + δz ) = E x (x, y, z) + ∇ E x · δ Fx becomes: Fx = q∇ E x · δ = (p · ∇)E x . Similar expressions can be derived for Fy and Fz and therefore we have: F = (p · ∇)E = ∇(p · E)

(2.2)

where we have used relation2 (p · ∇)E = ∇(p · E) − p × (∇ × E) = ∇(p · E) with ∇ × E = 0 for the electrostatic field. The potential energy of the dipole is equal to the sum of the potential energies of the two point charges: U = −q V (x, y, z) + q V (x + δx , y + δ y , z + δz ) = qd V = q∇V · δ = −p · E. The work δW = −dU done by the electric field when the dipole is displaced by ds by a force F and is rotated by δθ around an axis θˆ by a torque M, is: δW = F · ds + M · dθ = −dU = −∇U · ds −

∂U δθ ∂θ

and from the potential energy expression: F = ∇(p · E)

M=p×E

(2.3)

where the expression for F is that already found in (2.2). Supplemental problems are available at the end of this chapter as additional material on the interaction between two dipoles.

2.3 Multipole Expansion for the Potential of a Distribution of Point Charges The potential V0 (P) at the point P(x, y, z) due to a distribution of N point charges qi (see Fig. 2.2), is equal to the sum of the potentials at P from each charge of the It is easy to prove this relation directly. Note that, since ∇ is a vector operator, we can get the relation substituting ∇ to B in the vector relation A × (B × C) = B(A · C) − (A · B)C.

2

20

2 Multipole Expansion of the Electrostatic Potential

Fig. 2.2 The distribution of point charges. The position of the point P relative to the point charge qi is ri = r − di

system (principle of superposition of the electric potentials): V0 (P) =

i=N 

V0i (P) =

i=1

i=N 1  qi 4π0 i=1 ri

(2.4)

where ri is the distance between P and the ith charge. If the distance between P and the system of the charges is much larger than the dimensions of the system, it is useful to approximate the potential at P as done for the electric dipole. In the reference frame with origin in close proximity to the point charges system, we can write: r = ri + di

and

ri = r − di

where di and r are the vector positions of the charge qi and point P, respectively, and ri the vector from the ith charge to P. Then: 1 1 1 1 1 1 = = = = · . 1 1 2 − 2r·d ) 1 (d ri |r − di | r 2 [(r − di ) · (r − di )] 2 [r 2 − 2r · di + di ] 2 [1 + i 2 i ] 2 r d 2 − 2r·d

If we define α = i r 2 i , since r  di , α is very small and the fraction in the last equation can be expanded in a power series: 3 1 15 = 1 − α + α2 − α3 + · · · 2 8 48 [1 + α] 1

1 2

and keeping terms only to second order in α:

2.3 Multipole Expansion for the Potential of a Distribution of Point Charges

21

  1 1 1 (di2 − 2r · di ) 3 (di2 − 2r · di )2 = · 1− + ri r 2 r2 8 r4 and neglecting terms with higher power than di /r squared we get: 1 1 (di · rˆ ) 1 = + + 3 2 ri r r r



 3 1 (di · rˆ )2 − di2 . 2 2

(2.5)

Using this relation in Eq. 2.4, the potential in the point P becomes:   i=N i=N i=N 1  qi di · rˆ 1  qi 1  qi 3 1 2 2 + (di · rˆ ) − di V0 (P) = + 4π0 i=1 r 4π0 i=1 r 2 4π0 i=1 r 3 2 2 that can be written as:   i=N i=N i=N 3 1 1  1 1  1 1  1 2 2 (di · rˆ ) − di . V0 (P) = qi + qi di · rˆ + qi 4π0 r 4π0 r 2 4π0 r 3 2 2 i=1

i=1

i=1

If we define the total charge of the system: QT OT =

i=N 

(2.6)

qi

i=1

the electric dipole moment: P=

i=N 

(2.7)

qi di

i=1

and the electric quadrupole moment relative to the direction rˆ : Q quadr =

i=N  i=1

 qi

3 1 (di · rˆ )2 − di2 2 2

 (2.8)

the potential of the system of point charges at the point P can be written in the form:

V0 (P) =

1 P · rˆ 1 QT OT 1 Q quadr + + . 4π0 r 4π0 r 2 4π0 r 3

This relation represents the multipole expansion for the potential of the system of point charges, truncated to the second order. The first term depends on the total charge Q T O T of the system and behaves as the 1/r potential of a point charge; the second term is related to the dipole moment P of the system and behaves as the 1/r 2

22

2 Multipole Expansion of the Electrostatic Potential

potential of an electric dipole; the third term depends on the quadrupole moment and decreases as 1/r 3 . The contribution of these terms to the total potential decreases at higher terms. If the total charge Q T O T is equal to zero, the most relevant term is the dipole term, and if also this term is null, the potential is determined by the quadrupole moment. If also this term is zero the multipole expansion should be extended to include terms of higher order. For a continuous distribution of charge limited to a volume τ , described by the density ρ(r ), the summation in (2.6), (2.7) and (2.8) has to be replaced by an integral:  QT OT =  P=  Q quadr =

τ

ρ(r )

τ



τ

ρ(r ) dτ 

r ρ(r ) dτ 

3  1 (r · rˆ )2 − (r  )2 2 2



dτ  .

The measurement of the terms in the potential expansion of a charged structure gives information on the distribution of the charge. Molecules, in which the centers of the positive and negative charge do not coincide, have a permanent electric dipole moment. The dipole moment for molecules of H2 O in its vapour state is 6.1 × 10−30 C m, for HCl is 3.5 × 10−30 C m and for CO is 0.4 × 10−30 C m. In atoms and molecules, also with a null dipole moment, the action of an external field may separate the centers of positive and negative charges and produce an induced electric dipole. Uncharged atoms and molecules can therefore have dipole-dipole or dipole-induced dipole interactions.

2.4 Properties of the Electric Dipole Moment When the total charge of a system is null, the dipole moment is independent from the point (or the reference frame) chosen for the calculation of the moment, and the dipole moment becomes an intrinsic feature of the system.3 Indeed if the vector −−→ a = O  O determines the position of the origin O of the first reference frame in a new frame with origin O  (see Fig. 2.3) we can write: di = a + di , and the dipole moment P is: 3

This is a general property of the first non null term in the multipole expansion.

2.5 The Quadrupole Tensor

23

Fig. 2.3 Position of the point charge relative to two different frames

Fig. 2.4 Example of a charge distribution symmetric relative to a point



P =

i=N  i=1

qi di

=

i=N 

i=N i=N   qi (a + di ) = qi a + qi di = Q T O T a + P = P

i=1

i=1

i=1

because Q T O T = 0. If a system of charges has a symmetry center, then the dipole moment is null. For instance for the system of three charges in Fig. 2.4: P=

i=3 

qi di = qr + q(−r) + (−2q)0 = 0 .

i=1

2.5 The Quadrupole Tensor A rank two tensor can be associated to the quadrupole moment. The Eq. (2.8):

24

2 Multipole Expansion of the Electrostatic Potential

Q quadr =

  i=N 3 1  1 2 2 2 q · r) − r (d d i i r 2 i=1 2 2 i

can be written as: Q quadr

⎡ ⎛ ⎞  ⎤ μ=3 ν=3 μ=3 i=N ν=3    1  ⎣ 3 ⎝ 1 = 2 qi xμ diμ ⎠ · xν diν − di2 xμ xν δμν ⎦ r i=1 2 μ=1 2 ν=1 μ=1 ν=1

where δμν is the Kronecker delta (δμν = 1 if μ = ν, and δμν = 0 when μ = ν). Assuming the sum over any index that appears twice in: Q quadr

  i=N 3 1  1 2 diμ xμ diν xν − di δμν xμ xν = 2 qi r i=1 2 2   i=N 1  3 1 2 = 2 diμ diν − di δμν xμ xν qi r i=1 2 2

we can finally write: Q quadr =

1 Q μν xμ xν r2

where we have introduced a symmetric rank two tensor, the quadrupole tensor Q μν given by: Q μν =

i=N  i=1

 qi

 3 1 diμ diν − di2 δμν . 2 2

The quadrupole term in the multipole expansion can be then expressed in the form: 1 1 Q μν xμ xν . 4π0 r 5

(2.9)

2.6 A Bidimensional Quadrupole As an example we can calculate the potential at large distance from the quadrupole4 formed by four point charges as shown in Fig. 2.5.

4

Note that the words dipole, quadrupole, etc. are used in two ways: to describe the charge distribution and secondly to designate the moment of an arbitrary charge distribution.

Appendix A: Higher Order Terms in the Multipole Expansion of the Potential

25

Fig. 2.5 The bidimensional electric quadrupole

The total charge is zero and the dipole moment is null because the point charges are placed symmetrically with respect to the origin. The first non null term in the multipole expansion is the quadrupole term. For the tensor Q μν it is easy to find Q x x = Q yy = Q zz = Q x z = Q zx = Q yz = Q zy = 0 and the only non null components are Q x y = Q yx = 23 qd 2 . Then from (2.9) the potential at a point P(x, y, z) is: V0 (x, y, z) =

xy 3qd 2 4π0 (x 2 + y 2 + z 2 ) 25

that is zero in any point on the z axis.

Appendix A: Higher Order Terms in the Multipole Expansion of the Potential We have already seen the multipole expansion of the potential limited to the second order. To get the general expression5 with all the terms of the expansion, we write the potential at a point P(r) = P(x, y, z) from a continuous charge distribution, limited in space, described by the density ρ(r ) = ρ(x  , y  , z  ). The distance of the point P from the elementary volume dτ  in the point r is:  |r − r | = Δr = (x − x  )2 + (y − y  )2 + (z − z  )2 . 

We set the origin of the frame inside the volume of the charge distribution or nearby. For a distance r large compared with the dimensions of the volume, we can expand 5

For this expansion see for instance: Panofsky and Phillips [2], Sect. 1.7.

26

2 Multipole Expansion of the Electrostatic Potential

the distance |r − r | as a Taylor series:       ∂2 ∂ 1 1 1 1 1    = + xα + x x |r − r | r ∂ xα Δr xα =0 2! α β ∂xα ∂xβ Δr    ∂n 1 1    · · · + xα xβ xγ . . . n! ∂xα ∂xβ ∂xγ . . . Δr

+ ··· xα =xβ =0

xα =xβ =xγ ...=0

where we assume the sum over α, β, γ, . . . = 1, 2, 3, with x1 = x, x2 = y, x3 = z. The potential due to the charge distribution is: V0 (r) =

1 4π0

 τ

ρ(r ) dτ  |r − r |

(2.10)

and by substituting the expression for 1/|r − r | given before, we get: V0 (r) =

1 1 4π0 r

1 1 + 4π0 2!

1 1 + 4π0 n!







τ

ρ(r  )dτ  +

∂2 ∂xα ∂xβ

∂n ∂xα ∂xβ ∂xγ . . .



1 Δr



1 4π0





1 Δr



∂ ∂ xα 

xα =xβ =0

τ

1 Δr



 xα =0

τ

xα ρ(r  )dτ 

xα xβ ρ(r  )dτ  + · · ·



 xα =xβ =xγ ...=0

τ

xα xβ xγ . . . ρ(r  )dτ  .

In this expression we can recognise the terms corresponding to the total charge, to the dipole and to the quadrupole moments that we have already seen and then the general form of the 2n -pole.

Appendix B: Expansion in Terms of Spherical Harmonics The multipole expansion of the potential from a charge distribution limited in space, can be also expressed in series of spherical harmonics.6 If r gives the position of a point inside a sphere of radius R, and r that of a point outside, we can write for 1/|r − r | the expansion in terms of the spherical harmonics 6

For an exhaustive presentation see Jackson [3], Chaps. 3 and 4.

Appendix B: Expansion in Terms of Spherical Harmonics

27

Ylm (θ, ϕ): m=l ∞   1 1 (r  )l ∗     Y θ , ϕ Ylm (θ, ϕ) . = 4π |r − r | 2l + 1 r l+1 lm l=0 m=−l

If the charge distribution ρ(r  ) is confined inside the sphere of radius R we can substitute this expansion in (2.10) and we get: V0 (r)=

⎧ ⎨

1 4π 4π0 ⎩

∞   m=l l=0



1

1

m=−l 2l + 1 r l+1

⎫  ⎬   ∗ θ  , ϕ (r  )l ρ(r ) dτ  Y (θ, ϕ) Ylm lm ⎭

and introducing the multipole moments:  qlm =

∗ θ  , ϕ  (r  )l ρ(r ) dτ  Ylm

we get the expansion:

V0 (r) =

⎧ ⎨

1 4π 4π0 ⎩

⎫ ⎬ 1 1 qlm Ylm (θ, ϕ) . l+1 ⎭ 2l + 1 r

∞ m=l   l=0 m=−l

Exercise With the formulas for the first spherical harmonics reported below, write the first three terms of the multipole expansion in spherical coordinates and compare with those expressed in cartesian coordinates. 1 Y00 = √ 4π

l=0

 l=1

Y11 = −

 Y10 =

l=2

Y22 =

 Y21 = −

3 sin θ eiϕ 8π

3 cos θ 4π

1 4



15 sin2 θ e2iϕ 2π

15 sin θ cos θ eiϕ 8π

28

2 Multipole Expansion of the Electrostatic Potential

 Y20 =

5 4π



3 1 cos2 θ − 2 2



with the relation: ∗ (θ, ϕ) . Yl−m (θ, ϕ) = (−1)m Ylm

Problems 2.1 Find the dipole moment of the system of four point charges q at (a, 0, 0), q at (0, a, 0), −q at (−a, 0, 0) and −q at (0, −a, 0). 2.2 Write the potential for the system of three point charges: two charges +q in the points (0, 0, a) and (0, 0, −a), and a charge −2q in the origin of the frame. Find the approximate form of this potential at distance much larger than a. Compare the result with the potential from the main term in the multipole expansion. 2.3 Two segments cross each other at the origin of the frame and their ends are at the points (±a, 0, 0) and (0, ±a, 0). They have a uniform linear charge distribution of opposite sign. Write the quadrupole term for the potential at a distance r  a. 2.4 Write the potential from a ring of radius R and charge Q = 2π R λ in the approximation up to the quadrupole term. Find the same result by the expansion in spherical harmonics. 2.5 Calculate the quadrupole term of the expansion for the potential from two concentric coplanar rings charged with q and −q and with radii a and b. 2.6 Write the interaction energy of two electric dipoles p1 and p2 with their centers at distance r. 2.7 Using the result of the previous problem write the force between the electric dipoles p1 and p2 at distance r. Then consider the force when the dipoles are coplanar oriented normal to their distance and they are parallel or antiparallel. Determine also the force when the dipoles are on the same line and oriented in the same or in the opposite direction. 2.8 Two coplanar electric dipoles have their centers a fixed distance r apart. Say θ and θ the angles the dipoles make with the line joining their centers and show that if θ is fixed, they are at equilibrium when 1 tan θ = − tan θ . 2

Solutions

29

Solutions 2.1 The dipole moment of the system has components: px =

4 

qi xi = 2qa

py =

1

4 

qi yi = 2qa

pz =

4 

1

qi z i = 0 .

1

√ The dipole moment is p = (2qa, 2qa, 0) with module p = 2q 2a. This is the moment of an elementary dipole with opposite√charges 2q located in the centers of the positive and the negative charges which are distant 2a. 2.2 In spherical coordinates the potential depends only on the distance r from the origin and on the angle θ. Adding the potentials from the three charges we have:   q 1 q 2q + V (r, θ) = + − 1 1 4π0 r [r 2 − 2ra cos θ + a 2 ] 2 [r 2 + 2ra cos θ + a 2 ] 2 and expanding in power series as in (2.5) we find: V (r, θ) =

 1 qa 2  2θ−1 . 3 cos 4π0 r 3

(2.11)

In the multipole expansion for the system of the three charges the first non null term is the quadrupole moment. It is easy to find the components: Q x z = Q yz = Q x y = 0, Q x x = Q yy = −qa 2 and Q zz = 2qa 2 so that the potential is: V (x, y, z) =

=

 1 1  Q x x x 2 + Q yy y 2 + Q zz z 2 5 4π0 r  1 qa 2  2 2 − y2 2z − x 4π0 r 5

that is the formula (2.11) written in cartesian coordinates. 2.3 For the given charge distribution it is easy to see that Q x z = Q yz = Q x y = 0 and by simple calculations we find: Qxx =

1 2 qa 3

1 Q yy = − qa 2 3

Q zz = 0

so that the quadrupole potential is: V (x, y, z) =

(x 2 − y 2 ) 1 qa 2 . 4π0 3 (x 2 + y 2 + z 2 ) 25

30

2 Multipole Expansion of the Electrostatic Potential

2.4 We place the ring in the x − y plane with its center in the origin. Of corse in the expansion we have the term associated to the total charge Q. The charge distribution is symmetric with respect to the origin so the dipole moment is null. We have to calculate the components of the quadrupole tensor: Q μν =

  2π  3 1 Rμ Rν − R 2 δμν λR dϕ 2 2 0

where Rμ is the component along the μ axis of the vector R from the origin to the element R dϕ of the ring.   2π  3 1 1 2 2 (R cos ϕ) − R λR dϕ = Q R 2 Qxx = 2 2 4 0 Q yy =

  2π  3 1 1 (R sin ϕ)2 − R 2 λR dϕ = Q R 2 2 2 4 0

Q zz =

  2π  1 1 − R 2 λR dϕ = − Q R 2 2 2 0

and it is easy to verify that the non diagonal elements of the tensor are null: Q x y = Q x z = Q yz = 0. The expansion is: V (x, y, z) =

1 Q R2 1 2 1 Q + [x + y 2 − 2z 2 ] 4π0 r 4π0 4 r 5

V (x, y, z) =

V (r, θ) =

1 Q R2 1 2 1 Q + [r − 3z 2 ] 4π0 r 4π0 4 r 5

1 Q 1 Q R2 1 [1 − 3 cos2 θ] + 4π0 r 4π0 4 r 3

For the expansion in spherical harmonics we have to calculate the moments:  qlm =

∗ (θ, ϕ)r l ρ(r) dτ Ylm

with the charge density ρ(r):  π ρ(r) = ρ0 δ(r − R) δ θ − 2 Q=

 ∞  π  2π 0

0

0

 π 2 r dr sin θ dθ dϕ ρ0 δ(r − R) δ θ − 2

(2.12)

Solutions

31

Q = 2π ρ0 R 2

ρ0 =

Q 2π R 2

so for instance:   ∞  π  2π  ρ0 15 2 π 2 r sin2 θ e−2iϕ δ(r − R) δ θ − r dr sin θdθ dϕ q22 = 4 2π 2 0 0 0 ρ0 q22 = 4



15 4 i −2iϕ 2π R e =0 2π 2 0

and also the moments q21 , q2 −1 , q2 −2 , q11 , q10 , q1 −1 are null, but q00 = 

 ∞  π  2π 0

∞ π

q20 = 0

0

0



2π

0

 ρ0 r

2

0

q20 =

 2π 0

 ρ0 Q π 2 r dr sin θdθ dϕ = √ √ δ(r − R) δ θ − 2 4π 4π 5 4π



 ρ0

3 1 cos2 θ − 2 2

5 4π



 π 2 δ(r − R) δ θ − r dr sin θdθ dϕ 2

   1 5 1 − R 4 dϕ = − Q R2 2 2 4π

The expansion in spherical harmonics up to the quadrupole term is: V (r, θ) =

!  " 1 1 1 1 q00 Y00 + 4π q Y 20 20 4π0 r 5 r3

1 V (r, θ) = 4π0

#

Q 1 Q R2 + (1 − 3 cos2 θ) r 4 r3

$

that is the potential (2.12). 2.5 We consider the two rings on the plane z = 0 with their centers in the origin. The total charge of the system is zero and, for the symmetry of the charge distribution with respect to the origin, also the dipole moment is null. The quadrupole term is the first non null term. It is evident that Q x z = Q yz = 0 and by simple integrals we can get: Q x x = Q yy =

q 2 (a − b2 ) 4

Q zz =

q 2 (b − a 2 ) 2

Qxy = 0

so that the first term of the potential expansion is: V (x, y, z) =

1 q(a 2 − b2 ) (x 2 + y 2 − 2z 2 ) . 5 4π0 4 (x 2 + y 2 + z 2 ) 2

32

2 Multipole Expansion of the Electrostatic Potential

2.6 The interaction energy of the two dipoles is equal to the potential energy of p2 in the field of p1 . Saying r the vector from the center of p1 to that of p2 , we can write:   1 p1 · r U21 = −p2 · E1 (r) = p2 · ∇V1 = p2 · ∇ 4π0 r 3   ∇(p1 · r) 1 1 p2 · + (p · r)∇ = 1 4π0 r3 r3 and since: 1 r ∇ 3 = −3 5 r r

∇(p1 · r) = p1

1 r (note that ∇ n = −n n+2 ) r r

we get: U21 = U12 =

1 4π0



p 1 · p2 (p1 · r)(p2 · r) −3 r3 r5



(2.13)

that is symmetric in p1 and p2 . 2.7 From the solution of the previous problem and from the formula (2.3) for the force on a dipole in an electric field we get: F2 = −∇U21 = ∇(p2 · E1 (r))   1 1 1 ∇[(p1 · r)](p2 · r)] (p1 · p2 )∇ 3 − 3(p1 · r)(p2 · r)∇ 5 − 3 =− 4π0 r r r5 and then: F2 =

  (p1 · p2 )r + p1 (p2 · r) + (p1 · r)p2 1 (p1 · r)(p2 · r)r 3 . − 15 4π0 r7 r5

This formula is symmetric in p1 and p2 but r is directed from p1 to p2 so we get F1 changing r to −r and we have F1 = −F2 as expected. For two coplanar parallel dipoles normal to the line joining their centers: with same direction(↑ ↑) : F2 =

with opposite direction(↑ ↓) :

3 p1 p2 rˆ 4π0 r 4

F2 = −

3 p1 p2 rˆ 4π0 r 4

a repulsive force

an attractive force

for the dipoles on the same line with same direction(→ →) : F2 = −

6 p1 p2 rˆ 4π0 r 4

an attractive force

Solutions

33

Fig. 2.6 Coplanar dipoles p1 and p2 with their centers at fixed distance r and orientations θ and θ with respect to r

with opposite direction(→ ←) : F2 =

6 p1 p2 rˆ 4π0 r 4

a repulsive force.

2.8 For the two coplanar dipoles shown in Fig. 2.6 the interaction energy (2.13) becomes: Uint =

& 1 p1 p2 %  − θ) − 3 cos θ cos θ  . cos(θ 4π0 r 3

At fixed θ we find the minimum of this energy solving the equation: & ∂U 1 p1 p2 % − sin(θ − θ) + 3 cos θ sin θ = 0 . =  3 ∂θ 4π0 r The solution is: 1 tan θ = − tan θ 2

with the condition

∂2U ∂θ 2

=

& 1 p1 p2 % 2 cos θ cos θ − sin θ sin θ > 0 . 3 4π0 r

For θ = π/2 the minimum is at θ = −π/2, for θ = 0 at θ = 0 and for θ = π at θ = −π.

References 1. D.J. Griffiths, Introduction to Electrodynamics, 4th edn. (Cambridge University Press, Cambridge, 2017) 2. W.K.H. Panofsky and M. Phillips, Classical Electricity and Magnetism (Addison-Wesley Pub., 1953) 3. D.J. Jackson, Classical Electrodynamics, 3rd edn. (John Wiley & Sons, NewYork, 1999)

Chapter 3

The Method of Image Charges

The potentials and the electric field in the presence of point charges and conductors are determined solving the Poisson equation and sometimes this may be very difficult. However in some particular configurations the equipotential surfaces of the conductors can be reproduced adding to the real point charges some image charges inside the conductors.1 Using this technique the solution can be easily written as the sum of well known potentials of point charges. After the introduction of the image charges method, some relevant examples will be presented: (i) the point charge in front of a conductive plane, (ii) the point charge near a conductive sphere, (iii) a conductive sphere immersed in a uniform external electric field, (iv) a charged wire parallel to a cylindrical conductor. Additional examples will be available in the problems.

3.1 The Method of Image Charges In a system of point charges and conductors at fixed potentials, Poisson equation has to be solved using as boundary conditions the potentials on the conductor surfaces. Sometimes by adding to the system of real charges some point charges, called image charges, placed in suitable positions inside the conductors, the sum of the potentials due to both real and image charges can generate equipotential surfaces with the same shapes and potentials of the conductors. Outside the space of the conductors the Poisson equation for the system of real and image charges depends on the real charges and has the same potentials of the 1 For the image charges method see also: Jackson [1], Sects. 2.1–2.2; Feynman et al. [2], Vol. II, Sects. 6.7–6.9; Landau and Lifshitz [3], Chap. 1, Sect. 1.3 with problems; Griffiths [4], Sect. 3.2. The method is examined also in Maxwell [5], Vol. I, Chap. 11.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 F. Lacava, Classical Electrodynamics, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-031-05099-2_3

35

36

3 The Method of Image Charges

real problem boundary conditions at the geometrical surfaces of the conductors. In the two cases therefore we have the same equation with the same conditions on the surfaces of the conductors, thus, given the uniqueness of the solution, the solution for the point charges (real and image) is also the solution of the problem with the real charges and the conductors. Outside the conductors the potential is then given by the sum of the potentials due to real and image charges. Inside the conductors voltages are constant and equal to the values on the surfaces. It is worth noting that, outside the conductors, the sum of the potentials of the real charges represents the particular solution of the Poisson equation while the sum of the potentials from image charges is the solution of the associated Laplace equation.

3.2 Point Charge and Conductive Plane A point charge q at a distance d from an infinite grounded conductive plane, shown in Fig. 3.1, represents the simplest example of the image charges method. The z axis can be defined as the line passing on the point charge and perpendicular to the plane located at z = 0. The Poisson equation for the point charge configuration has to be solved with boundary condition the null potential on the plane and as a consequence the electric field perpendicular to the plane. Consider instead the electric field given by a charge q located at (0, 0, d) and by the image charge −q at (0, 0, −d), mirror image of the charge q with respect to the xy plane. The potential by the two charges at the point P(x, y, z) is: q V0 (x, y, z) = 4π 0



1

1

 − r 2 + (z − d)2 r 2 + (z + d)2

 (3.1)

Fig. 3.1 Point charge near an infinite conductive plane connected to ground. Vector sum of the fields from the two point charges

3.2 Point Charge and Conductive Plane

37

where r 2 = x 2 + y 2 and the components of the electric fields are: q E 0x (x, y, z) = −∇x V = 4π 0

q 4π 0

E 0y (x, y, z) = −∇ y V =

q E 0z (x, y, z) = −∇ y V = 4π 0



x 3

−

3

−

(r 2 + (z − d)2 ) 2 

y (r 2 + (z − d)2 ) 2



z−d 3

(r 2 + (z − d)2 ) 2

−



x 2

(r 2 + (z + d)2 ) 3 

y 2

(r 2 + (z + d)2 ) 3 

z+d 2

(r 2 + (z + d)2 ) 3

.

The potential and the field components on the plane z = 0 are: V0 (x, y, z = 0) = 0

E 0x (x, y, z = 0) = 0

E 0z (x, y, z = 0) = −

E 0y (x, y, z = 0) = 0

2d q . 4π 0 [r 2 + d 2 ] 23

As expected by symmetry the potential is null and the field is normal to the plane. The z-component of the field can be written in the form: E 0z (x, y, z = 0) = − 2

d 1 q = − 2 E ∗ cos θ 4π 0 [r 2 + d 2 ] [r 2 + d 2 ] 21

where the two charge contributions are evident (see Fig. 3.1) given the strength of the fields of the two charges in a point on the plane: E∗ =

1 q . 2 4π 0 [r + d 2 ]

The potential of the two charges for z > 0 is a solution of the Poisson equation and on the plane we have the same configuration required for the point charge and the conductive plane: null potential and normal field. Then the potential V0 (x, y, z), given in (3.1), for z > 0 is also the solution for the problem of the point charge near the conductive plane. From the Coulomb’s theorem the charge density induced on the conductive plane is: σ (r ) = 0 E 0z (r, z = 0) = − and its integral:

d q 2π [r 2 + d 2 ] 23

(3.2)

38

3 The Method of Image Charges

∞ σ (r ) 2πr dr = −q d 0

1 (d 2 + r 2 )

∞   = −q 1  2 0

is equal to −q, the image charge located at z = −d, as expected from the Gauss theorem: the total flux out of the point charge has to enter the plane. In the position of the charge q, the electrostatic field from the charge induced on the plane is equal to the field from the image charge, consequently the charge is attracted by the plane with a force: F=

q2 1 . 4π 0 (2d)2

(3.3)

The electrostatic energy for the system of the point charge and the conductive plane is limited only at z > 0 and therefore is one half of the interaction energy of the two opposite charged points (see Problem 3.1). The reason is because moving the charges from infinite distance to their position on the grounded plate they are always at the same potential so the work is null.

3.3 Point Charge and Conducting Sphere An interesting case is that of a point charge q at a distance d from the center of a conductive sphere with radius R < d. First we consider a grounded sphere as in Fig. 3.2. A charge is induced on the spherical surface and the field outside the sphere is determined by both the point charge and the induced charge distribution. To reproduce this field, we can introduce

Fig. 3.2 Point charge near a grounded conductive sphere

3.3 Point Charge and Conducting Sphere

39

an image charge q  inside the sphere located, by symmetry, on the line from the point charge to the center of the sphere and at a distance x from the center of the sphere.2 The potential from the charges q and q  has to be zero on the spherical surface: 1 V0 (R) = 4π 0



q q + R1 R2

 = 0.

From R12 = d 2 + R 2 − 2Rd cos θ and R22 = R 2 + x 2 − 2Rx cos θ : 1 V0 (R) = 4π 0



q 1

[d 2 + R 2 − 2Rd cos θ ] 2

+



q

=0

1

[R 2 + x 2 − 2Rx cos θ ] 2

and replacing q  = −yq we have: q V0 (R) = 4π 0



1 1

[d 2 + R 2 − 2Rd cos θ ] 2

−



y 1

[R 2 + x 2 − 2Rx cos θ ] 2

=0

which gives: (y 2 d 2 + y 2 R 2 − R 2 − d 2 ) = 2R(y 2 d − x) cos θ . This relation has to be verified for any cos θ value, then the two expressions in parentheses have to be zero and therefore y 2 = dx and for x the two solutions: x1 = d 2

An elegant solution is given in The Feynman Lectures on Physics [2], Sect. 6.9. We have to remember that the spherical surface is the locus of all points for which the distances from two given points are in a constant ratio. Defining R1 and R2 the distances of the points on the sphere from the point charges q and q  , the relation for a null potential on the spherical surface is:

q 1 q V0 (r ) = + =0 4π 0 R1 R2 from which the condition: q R2 = − = const R1 q equal, as pointed out, to that for a spherical surface. Using the ratios of the two distances from the points A and B in Fig. 3.2 and given the radius R of the sphere, we can find the distance x of the charge q  from the center of the sphere: R2 (A) R2 (B) = R1 (A) R1 (B)

R+x R−x = d+R d−R

and the value of q  : q  = −q

R . d

R2 R = R1 d

x=

R2 d

40

3 The Method of Image Charges

and x2 = Rd . The solution x1 = d adds a charge q  = −q to the charge q with a null potential everywhere, which represents a trivial solution. For the second solution an 2 image charge q  = −q Rd is placed at a distance x = Rd from the center of the sphere. Outside the sphere (r > R) the potential will be the sum of the potentials from both the charge q and the image charge q  , with q  representing the total charge induced on the sphere: 2

q V0 (r, θ ) = 4π 0



1

1 R 1 − 2 2 2 2 d [r + x − 2r x cos θ ] 21 [d + r − 2r d cos θ ] 2



From this potential it is possible to write the components of the electric field. In the limit r → R E 0θ = 0 for any θ . while from the Coulomb’s theorem we find the density charge σ (θ ) induced on the conductive sphere: σ (θ ) = −

d 2 − R2 q 4π R (d 2 + R 2 − 2d R cos θ ) 23

which at θ = 0, π/2 and = π takes the values: σ (0) = −

d+R q 4π R (d − R)2

σ (π ) = −

σ

π 2

=−

q d 2 − R2 4π R (d 2 + R 2 ) 23

d−R q 4π R (d + R)2

and its integral on the spherical surface is equal to the image charge q  = −q Rd that is the total charge induced on the sphere. Inside the conductor the potential will be zero as on the surface. If the sphere is insulated and initially uncharged, after the point charge q is brought near by, the total charge on the sphere has still to be zero. With the charge q  , a charge q  = −q  has also to be present on the sphere. To have an equipotential surface this charge has to be uniformly distributed on the surface and the potential is that from the same charge located in the center of the sphere: V0 =

1 q 1 q  = 4π 0 R 4π 0 d

that is precisely the potential produced by the charge q in the point occupied by the center of the sphere. If on the isolated sphere is already present a charge Q and a charge q  is induced, a charge q  = Q − q  has to be uniformly distributed on the surface and determines the potential as in the previous configuration.

3.3 Point Charge and Conducting Sphere

41

For a conductive sphere at the potential V0 , fixed by a generator, a second charge q  = 4π 0 R V0 , located in the center, has to be added to the image charge q  . In all these configurations the force between the point charge and the conductive sphere is equal to the sum of the forces between the charge q and the charges q  and q  .

3.3.1 Force on a Point Charge Near a Charged Conducting Sphere For a point charge q and an insulated conducting sphere with charge Q of same sign the force is: ⎡ ⎤ R R ⎥ Q+q q q ⎢ ⎢ ⎥ d − d (3.4) F= ⎢

2 ⎥ 4π 0 ⎣ d2 R2 ⎦ d− d This force is usually repulsive and at large distance d >> R becomes the Coulomb’s force between two point charges, but if the point charge is close enough to the spherical surface the force can be attractive. At such distances the attractive force from the charge induced on the conductor, described by the image charge, becomes stronger than the repulsion between the same sign charges3 . Thus if a small charge leaves a charged conductor it is attracted back in the conductor and can be moved away only if the external force is strong enough. The work to extract an electron from a charged conductor (work function of a metal) is in large part done against the attractive force from the image charge. For a point charge q near a conductive sphere of radius R with charge Q the force is attractive at a distance: 

1 q d  R 1+ 2 Q from the center of the sphere. If the sphere is at voltage V0 , the charge Q at the center of the sphere has to be Q = 4π 0 R V0 and the distance becomes: 

q 1 d  R 1+ 2 4π 0 R V0 If the sphere is connected to ground the force is always attractive.

3

See Jackson [1], Sects. 2.3–2.4.

42

3 The Method of Image Charges

Fig. 3.3 Point charges to simulate a sphere in a uniform electric field

3.4 Conducting Sphere in a Uniform Electric Field Two point charges of opposite sign produce a uniform electric field close to their midpoint. To solve the case of a sphere of radius R in a uniform electric field4 E 0 , the conductive sphere is placed between two opposite point charges at large distance with respect to the radius of the sphere and then the point charges are moved to an infinite distance. Consider the center of the sphere in the origin of the axis, a charge q on the z axis at the distance −d and symmetrically a charge −q at distance d as in Fig. 3.3. The two charges produce an induced charge on the spherical surface. From the result for a sphere in presence of a point charge, we know that the effect of the induced charge can be taken into account adding two image charges q1 and q2 in the points z 1 and z 2 inside the sphere, with: q1 = −q

R d

z1 = −

R2 d

q2 = q

R d

z2 =

R2 . d

In the point P outside the sphere, at distance r from the origin and at an angle θ with the z axis (see the Fig. 3.3), the potential V is the sum of the potential V  from the charges at large distances and of the potential V  from the two image charges. In the limit d → ∞ the two charges have to produce a uniform field E 0 so that their potential has to be: V  = −E 0 z = −E 0 r cos θ .

4

In Sect. 8.5.1 the potential is found by solving the Laplace equation.

3.4 Conducting Sphere in a Uniform Electric Field

43

and the two image charges q1 and q2 become an electric dipole of infinitesimal dimension with dipole moment:





2

R2 R R R3 R − + q = 2q 2 D = −q d d d d d where the charge q has to be determined. The potential V  from this dipole is: V  =

1 D·r 1 R 3 cos θ = 2 q 4π 0 r 3 4π 0 d2 r 2

so that the total potential is: V (r, θ ) = −E 0 r cos θ +

1 R 3 cos θ 2q 2 . 4π 0 d r2

Requiring a null potential on the surface of the sphere (at r = R), we get for q and D: D = 4π 0 R 3 E 0

q = 2π 0 d 2 E 0 and the total potential5 becomes:

5

The potential can also be calculated directly as the sum of the four potentials of the point charges. The potential V  from the two charges at large distances is:   1 q 1  V0 (r, θ) = − 1 4π 0 [d 2 + r 2 + 2r d cos θ] 21 [d 2 + r 2 − 2r d cos θ] 2

V0 (r, θ)

⎧ ⎨ q = 4π 0 d ⎩ 

1 + ( dr )2 +

⎫ ⎬

1

1 2r d

cos θ

 21 −  1 + ( dr )2 −

2r d

cos θ

 21 ⎭

Moving the two charges to an infinite distance, in the limit d → ∞, ( dr ) is infinitesimal and the potential can be expanded in series:      q 1  r 2 2r 1  r 2 2r V0 = + − 1− cos θ − 1 + cos θ 4π 0 d 2 d d 2 d d and becomes: V0 = − where z = r cos θ.

 z q r q cos θ = − 2 2π 0 d 2π 0 d 2

44

3 The Method of Image Charges



R3 cos θ V (r, θ ) = −E 0 r − 2 r that is null on the surface of the sphere (r = R). In spherical coordinates the components of the electric field are: E 0r = −

∂V = E0 ∂r



R3 1 + 2 3 cos θ r

From the electric field E 0 : E0 = −

1 ∂V  q = ∂z 2π 0 d 2

we get the value of the charge q to get a field E 0 : q = 2π 0 d 2 E 0 and the potential is V0 = −E 0 z = −E 0 r cos θ. We consider now the potential V  from the two image charges (see Fig. 3.3). The distances d1 and d2 of the point P from the two charges, are:  d1 =

 21 1  2 2 4 2 r d + R + 2r R d cos θ d2

 d2 =

 21 1  2 2 4 2 r d + R − 2r R d cos θ . d2

So that the potential is: V0 (r, θ) = −

qR 4π 0

V0 (r, θ) = −



1 1

[r 2 d 2 + R 4 + 2r R 2 d cos θ] 2 ⎧ ⎪ ⎪ ⎨

qR ! 4π 0 dr ⎪ ⎪ ⎩ 1+

1 R4 r 2d2

+

2R 2 rd

−



1 1

[r 2 d 2 + R 4 − 2r R 2 d cos θ] 2

"1 − ! 2 cos θ 1+

⎫ ⎪ ⎪ ⎬

1 R4 r 2d2

−

2R 2 rd

cos θ

"1 ⎪. 2 ⎪ ⎭

In the limit d → ∞, ( Rd ) becomes infinitesimal and the potential expanded in power series is: V0 (r, θ) = −

qR 4π 0 dr

 1−

    1 R4 2R 2 2R 2 1 R4 + − cos θ − 1 + cos θ 2 r 2d2 rd 2 r 2d2 rd

that can be simplified in the form: V0 (r, θ) =

q 2π 0



 R3 R3 cos θ = E 0 2 cos θ 2 2 d r r

where we have used the q found before. The total potential at the point P is: V (r, θ) = V  + V  = −E 0



R3 r− 2 cos θ . r

3.5 A Charged Wire Near a Cylindrical Conductor

E 0θ

1 ∂V =− = −E 0 r ∂θ

E 0ϕ = −

45



R3 1 − 3 sin θ r

1 ∂V = 0. r sin θ ∂ϕ

For r → R E 0θ = 0 and only the radial component E 0 r = 3E 0 cos θ is non null, as expected for an electric field normal to the conductive spherical surface. From this component and the Coulomb’s theorem, we get the density of the induced charge on the spherical surface: σ (θ ) = 3 0 E 0 cos θ .

3.5 A Charged Wire Near a Cylindrical Conductor An infinite wire, with linear charge density λ is placed, at a distance d, parallel to the axis of an infinite charged cylindrical conductor of radius R < d with a linear density −λ. We can imagine to write the potential outside the cylinder as the superposition of the potentials from the real wire and from an image wire, with charge −λ, located at a distance x < R from the axis of the cylinder, as in Fig. 3.4. The potentials V (r1 ) and V (r2 ) due to the two wires with respect to the potential, that we can assume zero, on a line6 at the same distance r0 from both wires, are: V (r1 ) = −

λ r1 log 2π 0 r0

V (r2 ) =

λ r2 log 2π 0 r0

where r12 = r 2 + d 2 − 2r d cos θ and r22 = r 2 + x 2 − 2r x cos θ . At the point r(r, θ ) the potential is: V (r, θ ) = V (r1 ) + V (r2 ) = −

λ r1 λ r2 log = − log 12 . 2π 0 r2 4π 0 r2

(3.5)

The lateral surface of the cylinder has to be equipotential and the electric field has to be normal to this surface, that is with a radial direction with respect to the axis of the cylinder. This means that the component E θ (R, θ ) has to be zero for any value of θ . 6

This line can be any line, parallel to the two wires, that lies on the plane with respect to which the two wires are symmetrical.

46

3 The Method of Image Charges

Fig. 3.4 Charged wire near a cylindrical conductor

The component E θ (r, θ ) is: E θ (r, θ ) = −

  d sin θ λ x sin θ 1 ∂V = − r ∂θ 2π 0 r 2 + d 2 − 2r d cos θ r 2 + x 2 − 2r x cos θ

and in the limit r → R, from E θ (R, θ ) = 0 we get the equation: [R 2 + x 2 ]d − [R 2 + d 2 ]x = 0 with the solutions: x = d (a solution of no interest) and x =

R2 . d

By substituting in (3.5) this last value for x, we get the potential of the cylinder: V (R) = −

λ r2 λ d log 12 = − log . 4π 0 2π 0 R r2

The negative potential depends on the choice of a null voltage on a plane at the same distance from the two wires. Of course the two wires define two equal cylindrical equipotential surfaces which are symmetrical with respect to the plane but with opposite potentials. This is clear from (3.5) where the constant ratios r1 /r2 and r2 /r1 define two equal circles on a plane perpendicular to the wires as seen in the Feynman’s solution to the case of the point charge and the grounded sphere. The radial component of the electric field is:

Problems

47

  r − d cos θ λ r d 2 − R 2 d cos θ ∂V = − 2 2 Er (r, θ ) = − ∂r 2π 0 r 2 + d 2 − 2r d cos θ r d + R 4 − 2R 2 r d cos θ which for r = R becomes: Er (R, θ ) = −

d 2 − R2 λ 1   2π 0 R R 2 + d 2 − 2Rd cos θ

and so the the surface density is: σ (θ ) = −

d 2 − R2 λ 1   2π R R 2 + d 2 − 2Rd cos θ

with the values: σ (θ = 0) = 0 E 0 (R, 0) = −

λ d+R a. Calculate the potential and the induced charge density on the inner surface of the conductor. 3.8 A charged wire, with linear density λ, is parallel to a grounded conductive plane. Consider how to write the potential in the half space with the wire. 3.9 An infinite wire with linear charge density λ, is parallel to the axis of a conductive cylindrical surface of infinite length, with linear charge −λ. The radius of the conductor is R and the distance of the wire from its axis is a < R. Say how to find the electric field inside and outside of the cylindrical surface. 3.10 An infinite uncharged conductive cylinder is located in a uniform electric field E 0 perpendicular to its axis. Write the electric field and the charge density induced on the cylinder. 3.11 Find the capacitance per unit length of a capacitor formed by two infinite cylindrical conductors of radius R with parallel axes at distance c (c > 2R) as in Fig. 3.5c. The configuration studied here is a particular case of the next problem.

Solutions

49

Result: C=

log

where:

ξ=

π 0 √ c+ c2 −4R 2 2R

=

c2 − 1 . 2R 2

2π 0 2π 0 =  acosh ξ log ξ + ξ 2 − 1 

Note that:

  acosh ξ = ln ξ + ξ 2 − 1 .

3.12 Determine the capacitance of two conductive cylinders of radii a and b with their parallel axes at a distance c > a + b (see Fig. 3.5d). Result: C=

2π 0 acosh ξ

where

ξ=

c2 − a 2 − b2 . 2ab

3.13 Calculate the capacitance of a capacitor with the two cylindrical electrodes of radii a and b with parallel axes at a distance c < |a − b| (see Fig. 3.5e). Note that c = 0 for the usual coaxial cylindrical capacitor. Result: C=

2π 0 acosh ξ

where

ξ=

a 2 + b2 − c2 . 2ab

Solutions 3.1 (a) We have to consider only the component of the force along the direction normal to the plane. An elementary ring of radius between r and r + dr attracts the charge with a force: dF =

d 1 qσ (r )2πr dr . 4π 0 (d 2 + r 2 ) (d 2 + r 2 ) 21

By using the expression (3.2) for σ (r ) and integrating on the whole plane we find: F =−

q2 d2 4π 0 2

 0

∞

(r 2

dr 2 q2 1 =− 2 3 +d ) 4π 0 (2d)2

that is the force (3.3) between the charge q and its image charge. (b) to move the charge to an infinite distance we have to balance with an external force Fext the force F applied on the charge by the conductive plane. The work is:

50

3 The Method of Image Charges

Fig. 3.6 Point charge near a conductive corner

 W = d

∞

Fext dr =

q2 4π 0

 d

∞

1 1 1 q2 . dx = 2 (2x) 2 4π 0 (2d)

This energy is one half the interaction energy between the charge and its image charge. As pointed out in the text the electrostatic energy for the charge near the conductive plane is one half of that of two opposite charge at the same distance. 3.2 Remembering the case of a point charge near a conductive plane, we can realize that by setting three image charges as in Fig. 3.6, the orthogonal planes are at zero potential. So the conductive planes which form the corner are at zero potential. The potential outside of the conductive corner is the sum of the potentials of the four point charges. The field is the vector sum of the fields of the four charges and the force on the charge q is the vector sum of the forces from the three images charges. 3.3 For a point charge at distance d from the center of a conductive sphere, adding a image charge q  = −q(R/d), at distance R 2 /d from the center, between the charge q and the center of the spherical surface, the potential on the sphere is null. If we add this charge the hemispherical bump that lies on the sphere is at zero potential. Adding two more image charges q  = −q and q  = −q  symmetric to q and q  relative to the plane (see Fig. 3.7), both the bump and the plane are at a null potential as demanded in the problem. 3.4 We have to introduce an image dipole symmetric to the dipole with respect to the plane as in Fig. 3.8. The two dipoles are coplanar and their interaction energy was found in Prob. 2.7. Now θ  = −θ , p = pimag and r = 2d so the interaction energy is: U=

  1 p2  p2  1 2 cos 2θ − 3 cos 1 + cos2 θ . θ = − 3 3 4π 0 (2d) 4π 0 (2d)

Solutions

51

Fig. 3.7 Hemispherical bump on a conductive plane

Fig. 3.8 Electric dipole in front of a grounded conductive plane

The force on the dipole is: F =−

3 p2 ∂U =− ∂(2d) 4π 0



1 + cos2 θ (2d)4



and the torque: M =−

1 p2 ∂U =− sin 2θ ∂θ 4π 0 (2d)4

M < 0 for 0 < θ < π/2 and M > 0 for π/2 < θ < π. For the dipole oriented parallel and normal to the plane the attractive forces are: F parallel = −

p2 3 nˆ 4π 0 (2d)4

Fnor mal = −

p2 6 nˆ 4π 0 (2d)4

where nˆ is the normal versor out of the plane. The force between two dipoles was found also in Prob. 2.7. Here the image dipole pimag replaces p1 , the real dipole p is p2 and r, going from pimag to p, has length 2d.

52

3 The Method of Image Charges

So the force on the dipole is:   (pimag · p)r + pimag (p · r) + (pimag · r)p (pimag · r)(p · r)r 1 3 . − 15 F= 4π 0 r5 r7 From θ  = −θ , p = pimag and noting that: pimag p r cos θ + pimag p r cos θ  = 2 pimag p r cos2 θ = 2 p 2 r cos2 θ the force acting on the dipole becomes: p2 F= 4π 0



   3 cos 2θ − 9 cos2 θ 3 p 2 1 + cos2 θ nˆ = − nˆ . (2d)4 4π 0 (2d)4

3.5 We start from the solution for a point charge q at distance d from the center of a grounded conducting sphere of radius R. The image charge q  and its distance x from the center of the sphere are: R q = − q d

x=

R2 d

If we say a the distance of the charge q from the sphere, we have d = a + R. The distance b of the image charge from the border of the sphere is: b= R−x = R−

R2 aR R2 = R− = d a+R a+R

In the limit R → ∞ the part of the sphere between the charges becomes an infinite conducting plane, and the image charge and its distance from the plane are: q  = lim − R→∞

R q = −q a+R

and

b = lim

R→∞

aR =a a+R

as for the point charge and the infinite plane connected to ground. 3.6 We can replace the dipole with two point charges q and −q at a distance d and d + a as in Fig. 3.9 with a such that lim a→0 qa = p. From the result for a point charge and a conducting sphere, to have equipotential the sphere we need to introduce two image charges: R d

at x1 =

R2 d

R d +a

at x2 =

R2 d +a

q1 = −q q2 = q

Solutions

53

Fig. 3.9 The electric dipole in front of the sphere decomposed in two opposite charges a distance a apart

The sum of these two charges is: Q = q1 + q2 = −qa

R R = −p d(d + a) d(d + a)

The distance between these charges is: x2 − x1 =

R2 R2 R2 − = −a d +a d d(d + a)

so the their dipole momentum with respect to the point x1 is: p  = (x2 − x1 ) q2 = −qa

R3 d (d + a)2

In the limit a → 0 x2 = x1 =

R2 d

Q = −p

R d2

p = − p

R3 d3

The potential in the space external to the sphere is the sum of the potentials of the charge Q and of the two electric dipoles p and p  . Using spherical coordinates it is easy to write the potential in a point P at a distance r from the center of the sphere and at an angle θ with the line from the center to the electric dipole. For the vector R1 from the dipole p to the point P we have: R12 = d 2 + r 2 − 2r d cos θ while for R2 from the dipole p  (and the charge Q) to the point P:

54

3 The Method of Image Charges

R22 =

R4 R2 2 cos θ + r − 2r d2 d

we can write: V (r, θ ) =

1 Q 1 p · R1 1 p · R2 + + 4π 0 R2 4π 0 R13 4π 0 R23

where p = − p zˆ and p = p  zˆ = − p

R3 zˆ . d3

Now it is easy to see:

p · R1 = p R1 cos α = p (d − r cos θ ) R3 R3 p · R2 = p R2 cos(π − β) = − p 3 R2 cos β = − p 3 d d 





R2 r cos θ − d

so the potential is:

R2 r cos θ − 1 pR 1 p (d − r cos θ ) 1 R3 d V (r, θ ) = − + − p 3 3 2 3 4π 0 R2 d 4π 0 4π 0 d R1 R2 If the sphere is not grounded and its charge is null, we have to add in its center a charge Q  = −Q and its potential is: V =

1 pR 1 1 p 1 Q = = 2 4π 0 R 4π 0 d R 4π 0 d 2

that is the potential in the field of the dipole in a point at the same position of the center of the sphere. 3.7 This problem is the inverse of that with a point charge q at distance d from the center of a conductive sphere connected to ground. There, to get a null potential on the spherical surface, we have added an image charge q  = −q R/d inside the sphere at distance a = R 2 /d from the center. It is evident that if a charge q is inside the sphere, at distance a from the center, we can get the spherical surface at null potential adding outside of the sphere an image charge q ∗ = −qd/R at distance d = R 2 /a from the center of the sphere. The inner charge q induces a charge −q on the inner surface of the conductor. In the case of a sphere isolated and uncharged, a charge q would be uniformly distributed on the outer surface of the conductor and the potential of the conductor would be: V =

1 q . 4π 0 R

and outside of the spherical surface, at a distance r > R from the center, the potential generated by the charge q uniformly distributed on the outer surface would be:

Solutions

55

V =

1 q . 4π 0 r

For the Coulomb’s theorem the charge density induced on the inner surface, in a point with the radius at an angle θ with respect to the line from the center of the conductor to the charge q, is: σ (θ ) = −0 Er = −

R2 − a2 q 4π R # R 2 + a 2 − 2Ra cos θ $ 23

(3.6)

where Er is the radial component of the sum of the fields from the charges q and q ∗ on the inner surface. The charge densities at θ = 0 and θ = π are: σ (θ = 0) = −

R+a q 4π R (R − a)2

σ (θ = π ) = −

R−a q . 4π R (R + a)2

It is easy to verify that the integral of the charge density (3.6) is equal to total induced charge −q on the inner surface. 3.8 The problem is a simple extension of the point charge and the conductive plane. An image wire with charge density λ = −λ can be set symmetric to the real wire with respect to the plane. The potential and the field in the half space with the wire are given by the superposition of the potentials and the fields of the two wires. 3.9 For a charged wire with linear density λ at distance d > R from the axis of a conductive cylinder, the surface of the cylinder is equipotential if we place an image wire, with opposite linear charge, inside the cylinder at distance a = R 2 /d from the axis. From this result we can infer that for a charged wire with density λ at distance a < R from the axis of the cylindrical surface, we can write the field inside the cylindrical surface by setting an image wire with charge density −λ at distance d = R 2 /a > R from the axis. The charge per unit length on the inner surface of the cylindrical surface is −λ. Its distribution can be found from the component of the electric field normal to the surface. On the outer surface of the cylindrical conductor the charge is null, thus also the electric field is null outside the conductor. 3.10 The problem is similar to that of conductive sphere in a uniforme electric field. If we dispose the axis of the cylinder on the z axis, the field E0 = E 0 xˆ can be produced by two wires on the xz plane, parallel to the axis of the cylinder: one with density λ at x = −d and the other with density −λ at x = d. It easy to show that in the limit d → ∞ the two wires produce nearby the z axis a uniform electric field: E0 =

λ xˆ . π 0 d

The potential and the cylindrical components of this field are:

(3.7)

56

3 The Method of Image Charges

V = −E 0 x = −E 0 r cos θ

E 0r = E 0 cos θ

E 0θ = −E 0 sin θ

Ez = 0

where θ is the angle used in Fig. 3.4. To have an equipotential cylindrical surface, from the result for a charged wire near a conductive cylinder, we have to add two image wires: one with charge density −λ at x = −R 2 /d and the other with charge density λ at x = R 2 /d. In the limit d → ∞ the distance between the image wires become infinitesimal and for r > R they produce the potential:

λ x λ R2 −λ 2x cos θ  cos θ = cos θ log 1 − V (r, θ ) = 2π 0 r π 0 r π 0 d r 

(3.8)

to which is associated the field E with components: Er =

λ R2 cos θ π 0 d r 2

E θ =

λ R2 sin θ. π 0 d r 2

By substituting in these two expressions the density λ found from the relation (3.7), for r > R the total field Etot = E0 + E has components:

R2 E totr = E 0 1 + 2 cos θ r



R2 E totθ = −E 0 1 − 2 sin θ . r

For r = R the component E totθ is zero, as expected for the cylindrical equipotential surface, and from the value of E totr the Coulomb’s theorem gives the charge density induced on the conductive cylinder: σ = 20 E 0 cos θ . 3.11 We apply the solution for a charged wire parallel to the axis of a conductive cylinder. Considering the Fig. 3.10, if the wire f 1 , with linear charge density λ, is at distance d from the axis of the cylinder C2 , to get the cylindrical surface of C2 equipotential we have to introduce an image wire f 2 with charge −λ, at distance R 2 /d from the axis of C2 . The potential of C2 is: V2 = −

λ d log . 2π 0 R

Symmetrically to have the equipotential surface of C1 in presence of the wire f 2 with charge density −λ at distance d from its axis, we have to add just the image wire f 1 with charge density λ at distance R 2 /d from the axis of C1 . The potential of C1 is V1 = −V2 and the potential difference between the two cylindrical surfaces is: ΔV = V1 − V2 =

λ d log . π 0 R

Solutions

57

Fig. 3.10 Conductive cylinders of same radius with parallel axis

If the distance of the axis of the cylinders is c, for the distance d we can write the relation: √ R2 c + c2 − 4R 2 c=d+ and we get: d = d 2 thus we get: c + λ log ΔV = π 0

√

c2 − 4R 2 2R

and the capacitance: C=

λ = ΔV

π 0 √ c + c2 − 4R 2 log 2R

We can write ΔV also in the form: ΔV = V1 − V2 =

  λ log ξ + ξ 2 − 1 2π 0

where:

ξ=

and the capacitance per unit length: C=

2π 0 λ 2π 0  = = .  ΔV acosh ξ log ξ + ξ 2 − 1

c2 − 1. 2R 2

58

3 The Method of Image Charges

Fig. 3.11 Conductive cylinders of different radii with parallel axis at distance c > a + b

3.12 We have to find the relative position of two parallel wires f a , with charge density −λ, and f b , with charge density λ, which produce two cylindrical equipotential surfaces Ca and Cb with radii a and b and with their axes at distance c. Looking at Fig. 3.11: if the wire f b is at distance da from the axis of Ca , the wire f a has to be at distance a 2 /da from the axis of Ca . Symmetrically if the wire f a is at distance db from the axis of Cb , the wire f b has to be at distance b2 /db from the axis of Cb . The potentials of the two cylindrical surfaces are: Va = −

λ da log 2π 0 a

Vb =

λ db log . 2π 0 b

We have to determine da and db . If d is the distance of the two wire, we can write the system: ⎧ a2 b2 ⎪ ⎪ ⎪ d+ + =c ⎪ ⎪ da db ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ a2 d = da − ⎪ da ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ b2 ⎪ ⎪ ⎩ d = db − db and solving we find: db =

c2 + b2 − a 2 +



(c2 − a 2 − b2 )2 − 4a 2 b2 2c

Solutions

59

da =

c2 + a 2 − b2 +



(c2 − a 2 − b2 )2 − 4a 2 b2 . 2c

The potential difference becomes: Vb − Va =



λ λ db da c2 − a 2 − b2 + (c2 − a 2 − b2 )2 − 4a 2 b2 · = log log 2π 0 b a 2π 0 2ab

that can be rewritten as: Vb − Va =



% λ λ log ξ + ξ 2 − 1 = acosh ξ 2π 0 2π 0

where ξ =

c2 − a 2 − b2 . 2ab

Finally the capacitance per unit length is: C=

2π 0 2π 0 = .  acosh ξ log ξ + ξ 2 − 1 

If the two radii are equal (a = b) this result becomes that of the previous problem. 3.13 We have to find two parallel opposite charged wires that produce the two equipotential cylindrical surfaces Ca and Cb , inside each other, with radii a and b (a > b) and their axes at distance c < a − b. A charged wire f 1 has to be located inside the inner cylinder and a wire f 2 outside the outer one as in Fig. 3.12. To get Ca equipotential f 1 and f 2 have to be respectively at distance a 2 /da and da from the axis

Fig. 3.12 Cylindrical conductors of radii a and b, inside each other, with distance of the axis c < |a − b|

60

3 The Method of Image Charges

of Ca . At the same time to have equipotential Cb the two wires have to be distant b2 /db and db from the axis of Cb . To find da and db we can write the system: ⎧ a2 b2 ⎪ ⎪ ⎪ c= − = da − db ⎪ ⎨ da db ⎪ ⎪ a2 b2 ⎪ ⎪ ⎩ da − = db − da db with solutions: a 2 − b2 + c2 +

da =

db =

a 2 − b2 − c2 +

 (a 2 + b2 − c2 )2 − 4a 2 b2 2c

 (a 2 + b2 − c2 )2 − 4a 2 b2 . 2c

The potentials of Ca and Cb are: Va = −

λ da log 2π 0 a

Vb = −

λ db log 2π 0 b

and the difference is: Va − Vb =

λ λ a db a 2 + b2 − c2 + = log log 2π 0 da b 2π 0



(a 2 + b2 − c2 )2 − 4a 2 b2 2ab

that can be written in the form: Va − Vb =

  λ log ξ + ξ 2 − 1 2π 0

where:

ξ=

a 2 + b2 − c2 . 2ab

The capacitance per unit length of the two cylindrical conductors is: C=

λ 2π 0 2π 0  = = .  ΔV acosh ξ log ξ + ξ 2 − 1

(3.9)

For two cylindrical conductors of different radius (a = b) which are coaxial (c = 0) the capacitance (3.9) becomes that of a simple cylindrical capacitor C = 2π 0 / log(a/b).

References

61

References 1. D.J. Jackson, Classical Electrodynamics, 3rd edn. (John Wiley & Sons, NewYork, 1999) 2. R.P. Feynman, R.B. Leighton, M. Sands, The Feynman Lectures on Physics, Vol. II 3. L.D. Landau and E.L. Lifshitz, Electrodynamics of continuous media, Course of Theoretical Physics, Vol. 8. (Elsevier) 4. D.J. Griffiths, Introduction to Electrodynamics, 4th edn. (Cambridge University Press, Cambridge, 2017) 5. J.C. Maxwell, A Treatise on Electricity and Magnetism (1873)

Chapter 4

Image Charges in Dielectrics

The method of image charges is extended to systems with dielectric media. In some problems it is easy to find the solution for the Poisson equation with the boundary conditions on the fields derived from the constitutive relations. First we consider a point charge near a planar surface separating two dielectric media. Then we study the case of a dielectric sphere embedded in a different dielectric where a uniform electric field is present. More examples are examined in the problems at the end of this chapter.

4.1 Electrostatics in Dielectric Media The field in an isotropic and uniform dielectric is provided by the solution of the Poisson equation provided that the permittivity 0 of the empty space is replaced with the permittivity  of the medium: ΔV (x, y, z) = −

ρ(x, y, z) . 

The solutions are those for the empty space but with the potential and the fields reduced by the factor κ = /0 , the dielectric constant of the medium. In the presence of different dielectric media the problem becomes more complicated: the fields E and D are not continuous at the interfaces between the media, so that the fields are not differentiable and the Poisson equation cannot be written across these surfaces. The recipe is to solve the Poisson equation inside each of the dielectrics (i, j, ...) with the boundary conditions on the fields E(i) and D(i) derived by the Maxwell equations and the constitutive relations D(i) = (i) E(i) . These are that the parallel component of the electric field E and the normal component of the © The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 F. Lacava, Classical Electrodynamics, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-031-05099-2_4

63

64

4 Image Charges in Dielectrics

displacement field D are continuous at the (uncharged) interfaces between the media i and j: ( j)

( j)

E (i) = E 

(i) D⊥ = D⊥

(4.1)

In this chapter we apply the method of image charges in presence of two dielectric media1 in order to satisfy the Poisson equation with the boundary conditions.

4.2 Point Charge Near the Plane Separating Two Dielectric Media A point charge q is located at distance d apart from a plane separating two dielectric media. The z axis can be fixed normal to the plane placed at z = 0 and passing on the charge. The dielectric constants are κ1 for z > 0 and κ2 for z < 0. We have to find a solution for the potential in each of the two media requiring that the boundary conditions (4.1) are satisfied on the plane of separation. The solution seen for a point charge in front of a conductive plane, suggests to write the potential in the semispace z > 0 assuming all of the space of constant κ1 and adding2 an image charge q  on the z axis at z = −d (see Fig. 4.1). For the potential in the semispace z < 0 we consider all the space of constant κ2 and we position only an image charge q  at z = d. For z > 0 we have to find the solution of the Poisson equation: while the potential due to q is the particular solution, the potential from the charge q  is a solution of the homogeneous (Laplace) equation. For z < 0 we need the solution of the Laplace equation and this can be given by potential of the charge q  at z > 0. For a point P at z > 0 and at a distance r from the z axis, the potential is: 1 V1 = 4π1



q q + R1 R2



1 1   where R1 = (z − d)2 + r 2 2 and R2 = (z + d)2 + r 2 2 . The cylindrical components z and r of the electric field are: E z(1)

1

1 ∂V1 = =− ∂z 4π1



q(z − d) q  (z + d) + R13 R23



For the image charges method in dielectrics media see also: Jackson [1], Sect. 4.4; Landau and Lifshitz [2], Chap. 2, Sect. 2.7, with problems; Smythe [3], Chap. 5, Sect. 5.05, p. 125. 2 This choice includes the possibility of the same dielectric filling all the space. In fact in this case the charges have to be q  = 0 and q  = q, as we will find in the general solution for q  and q  .

4.2 Point Charge Near the Plane Separating Two Dielectric Media

65

Fig. 4.1 Image charges for a point charge q in front of a plane separating two dielectric media. For the field in the semispace z > 0 an image charge q  is located at z = −d in a space of dielectric 1 ; for the field in z < 0 only an image charge q  at z = d in a space of dielectric 2

Er(1)

1 ∂V1 = =− ∂r 4π1



qr q r + R13 R23

 .

For a point P at z < 0 the potential is: V2 =

1 4π2



q  R3



1  with R3 = (z − d)2 + r 2 2 and the components of the field are:

66

4 Image Charges in Dielectrics

E z(2) = −

∂V2 1 q  (z − d) = ∂z 4π2 R33

Er(2) = −

∂V2 1 q r . = ∂r 4π2 R33

In every point of the plane separating the two dielectric media, the fields have to (1) (2) = D⊥ that in this problem satisfy the boundary relations (4.1) E (1) = E (2) and D⊥ (1) (2) (1) (2) become Er = Er and 1 E z = 2 E z . For z = 0 R1 = R2 = R3 , and from the previous relations we find: q + q q  = 1 2

q − q  = q 

and finally the values for the image charges: q  = −q

2 − 1 2 + 1

q  = q

22 . 2 + 1

The field lines for 2 > 1 and 2 < 1 are sketched in Fig. 4.2. It is easy to verify from the found components of the fields E and D, the known law of refraction for the field lines: tan θ1 1 = tan θ2 2 where θ1 and θ2 are the angles between the fields and the normal to the separating plane. For 2 = 1 the same dielectric medium is present in the two semispaces. The point charge is embedded in a dielectric that fills all the space and from the previous relations we have: q  = 0 and q  = q as expected (see the footnote in a previous page). The polarization surface charge density on the two sides of the plane between the two media, is found by the relation σ P = P · nˆ where P = 0 (κ − 1)E is the induced polarization and nˆ is the versor, out of the dielectric, normal to the plane: (1) σ (1) P = −0 (κ1 − 1)E z =

2 qd 1 − 0 2π R 3 1 2 + 1

(2) σ (2) P = 0 (κ2 − 1)E z = −

where R = R1 = R2 = R3 .

qd 2 − 0 2π R 3 2 + 1

4.2 Point Charge Near the Plane Separating Two Dielectric Media

67

Fig. 4.2 Field lines for a point charge in front of a plane separating two dielectric media: top for 2 > 1 , bottom for 2 < 1

The total surface polarization charge density and the total polarization charge are:

σP =

qd 0 1 − 2 2π R 3 1 1 + 2

Q pol = q

0 1 − 2 1 2 + 1

and are  0 for 1  2 . For 2  1 the dielectric at z < 0 becomes a conductor and from the previous formulas q  = −q, the components of the electric field E (2) go to zero and the field lines are normal to the plane of separation (Er(1) = 0 and θ1 = 0). The force on the charge q due to the polarization charge on the plane at z = 0 can be written as the force between the charge q and the image charge q  in the dielectric with 1 :

68

4 Image Charges in Dielectrics

F =−

q 2 2 − 1 1 . 4π1 (2d)2 2 + 1

This force is attractive if 2 > 1 and repulsive if 2 < 1 .

4.3 Dielectric Sphere in an External Uniform Electric Field A sphere of dielectric with permittivity 2 , of radius R, is embedded in a dielectric medium with permittivity 1 , where it is already present an external uniform electric field3 E = E zˆ . We suppose, as proved correct a posteriori by the solution, that inside the sphere the electric field E  is uniform and with the same direction of the external field E, while outside the field is the superposition of the uniform field E and of a field produced by the polarization of the dielectric media. Furthermore we assume this last field is that produced by an image electric dipole p = pˆz placed at the center of the sphere. We position the center of the sphere in the origin of the frame. The potential associated to the external field E is: Ve = −E z = −E r cos θ and, in spherical coordinates, the net potential outside the sphere is: V (r, θ) = −E r cos θ +

1 p cos θ . 4π1 r 2

(4.2)

with p to be determined. The components of the field are: Er = −

1 p cos θ p sin θ ∂V 1 ∂V = E cos θ + = −E sin θ + Eθ = − . 3 ∂r 2π1 r r ∂θ 4π1r 3 (4.3)

Inside the sphere the potential is: V  = −E  z = −E  r cos θ and the components of the field are: Er = −

∂V  = E  cos θ ∂r

E θ = −

1 ∂V  = −E  sin θ . r ∂θ

From the boundary conditions (4.1) for the fields E and D on the surface of the sphere (at r = R), we get two independent equations: 3

In Sect. 8.5.1 the potential is found by solving the Laplace equation.

4.3 Dielectric Sphere in an External Uniform Electric Field

E  = E 

 D⊥ = D⊥

E θ = E θ

=⇒

=⇒

→

1 Er = 2 Er

→

E−

69

1 p = E 4π1 R 3

1 E +

1 p = 2 E  2π R 3

and solving the system we find: E =

31 E 21 + 2

p=

2 − 1 1 4π R 3 E . 21 + 2

(4.4)

If 2 > 1 (see Fig. 4.3a): p > 0, the dipole has the same orientation of E; inside the sphere the field is E  < E, that is a lower field due to the field E P produced by the polarization charges, opposite to E. Outside the sphere, the field lines tend to converge around the sphere. For 1 > 2 (see Fig. 4.3b): p < 0, the dipole is opposite to E; inside the sphere E  > E, the strength of the field is larger due to the field E P , from the polarization

Fig. 4.3 Charge of polarization, image dipole p, E  internal electric field, E P field from the polarization charges in a dielectric sphere of permittivity 2 embedded in a dielectric of permittivity 1 with an external electric field E: a2 > 1 , b2 < 1

70

4 Image Charges in Dielectrics

charges, that adds to the field E; outside the sphere the field lines tend to diverge from the sphere. If 1 = 2 , the same dielectric fills the space, and the relations (4.4) clearly give p = 0 and E  = E. Replacing the values of p and E  from (4.4) in (4.2) and (4.3) we get the potential and the field components outside the sphere: 

1 − 2 R 3 V (r, θ) = − 1 + · 21 + 2 r 3

 E r cos θ

  2 − 1 R 3 E cos θ Er (r, θ) = 1 + 2 · 21 + 2 r 3 

1 − 2 R 3 E θ (r, θ) = − 1 + · 21 + 2 r 3

 E sin θ.

At the surface of the sphere (r → R) the potential is: V (R, θ) = −

31 E R cos θ 21 + 2

and the components of the electric field:  Er (R, θ) =

32 21 + 2

 E cos θ

E θ (R, θ) = −

31 E sin θ . 21 + 2

The total charge density of polarization on the surface is: σ p = 30

2 − 1 E cos θ = σ cos θ 21 + 2

where

σ = 30

2 − 1 E. 21 + 2

For a dielectric sphere of permittivity  placed in an external field E 0 in vacuum, we substitute 2 = , 1 = 0 and we get: p=

 − 0 0 4π R 3 E 0 > 0 20 + 

σ p = σ cos θ

σ = 30

E =

30 E0 < E0 20 + 

 − 0 E0 > 0 . 20 + 

For a conductive sphere in a uniform field E 0 in the empty space: 2 → ∞ and 1 = 0 :

Problems

71

p = 4π0 R 3 E 0

E = 0

σ p = σ cos θ

σ = 30 E 0

as found in Sect. 3.4. For a spherical cavity in a dielectric with permittivity  in an external field E, 1 = , 2 = 0 and we find: p=

0 −  4π R 3 E < 0 2 + 0

E =

σ p = σ cos θ

0 −  E < 0. 2 + 0

σ = 30

3 E> E 2 + 0

Problems 4.1 Determine the electric field produced by a charged infinite wire parallel to the plane separating two different dielectric media. Write the force acting on the wire. 4.2 A grounded conductor consists of a plane sheet, lying in the yz plane, with an hemispherical bump of radius R centered at the origin in the region x > 0. The space in front of the conductor is filled with a material of permittivity 1 for z > 0 and with a different material of permittivity 2 for z < 0. Locate the image charges in the case where a charge q lies at x0 , y0 , z 0 (z 0 > 0) with x02 + y02 + z 02 = d 2 . See Fig. 4.4. 4.3 Consider the problem of an infinite wire with linear charge density λ in a dielectric of permittivity 1 parallel, at distance d, to the axis of a cylinder of infinite length

Fig. 4.4 The plane conductor with a boss embedded in two dielectrics

72

4 Image Charges in Dielectrics

and radius R < d, composed by a medium of permittivity 2 . Find the force acting on the wire. 4.4 Find the field for an infinite wire with linear charge density λ, placed at distance h < R parallel to the axis of a infinite long cylinder, of radius R and permittivity 2 , embedded in an other medium with permittivity 1 . Give the force acting on the wire. 4.5 Study the case of a cylinder of permittivity 2 , embedded in a medium of permittivity 1 , where in its absence an electric field is present that is uniform and with direction normal to the axis of the cylinder. Assume, as seen for the dielectric sphere, that the field inside the cylinder is uniform and oriented as the external field at large distance.

Solutions 4.1 We say z = 0 the plane separating the dielectric with 1 in the semispace z > 0 from that with 2 at z < 0. Then we position the infinite wire f with linear charge density λ at distance d on the line (z = d, x = 0) as shown in Fig. 4.5. From the solution for a point charge near the plane of separation, we can infer to write the field in the medium with 1 as the superposition of the field from the wire f and of a field due to an image wire f 1 with linear charge density λ1 , located symmetric to f with respect to the plane. Furthermore we write the field in the other dielectric (at z < 0) as produced by an image wire f 2 with density λ2 located at z = d in the same position of f. The potential at a point P in the semispace z > 0 depends on its distances r and r1 from the wires f and f 1 4 and is: V (P) = −

λ r λ1 r1 λ (z − d)2 + x 2 λ1 (z + d)2 + x 2 log − log =− log − log . 2 2π1 r0 2π1 r0 4π1 4π1 r r2 0

0

The components of the electric field at z > 0 are: (1)

Ex

=−

λ x x ∂V λ1 = + ∂x 2π1 (z − d)2 + x 2 2π1 (z + d)2 + x 2

E z(1) = −

4

(1)

Ey

=−

∂V =0 ∂y

λ (z − d) (z + d) ∂V λ1 = + . 2 2 ∂z 2π1 (z − d) + x 2π1 (z + d)2 + x 2

We can assume a reference potential at a point on the plane so that the distance r0 is the same for the two wires.

Solutions

73

Fig. 4.5 Charged wire near the plane, at z = 0, separating two dielectric media. At top the wire f with charge density λ and the image wire f 1 with charge density λ1 for the field in the semispace z > 0; at bottom image wire f 2 with density charge λ2 for the field in the semispace z < 0

For a point P in the semispace z < 0 the potential is: V = −

λ2 r2 λ2 (z − d)2 + x 2 log = − log 2π2 r0 4π2 r02

with r2 the distance of P from the wire f 2 . The components of the field are: E x(2) = −

λ2 x ∂V  = ∂x 2π2 (z − d)2 + x 2

E z(2) = −

E y(2) = −

λ2 (z − d) ∂V  = . ∂z 2π2 (z − d)2 + x 2

∂V  =0 ∂y

74

4 Image Charges in Dielectrics

The boundary conditions for the fields E and D on the plane give the relations between the charge densities: E (1) = E (2)

(1) (2) D⊥ = D⊥

=⇒

=⇒

E x(1) = E x(2)

λ + λ1 λ2 = 1 2

→

1 E z(1) = 2 E z(2)

→

λ − λ1 = λ2

and from these, the values for λ1 and λ2 are: λ1 =

1 − 2 λ 1 + 2

λ2 =

22 λ. 1 + 2

For the same dielectric filling all the space 1 = 2 thus λ2 = λ and λ1 = 0. The force per unit length on the wire due to the polarization is equal to the force from the wire f 1 : F=

λ2 1 1 − 2 2π1 2d 1 + 2

attractive for 1 < 2 . 4.2 The problem is similar to Problem 3.3 but, as seen in Sect. 4.2, the presence of two different dielectric media, divided by the plane z = 0, requires now more image charges to reproduce the grounded conductor and to satisfy the boundary conditions at z = 0. For the field at z > 0, we have to suppose the image charges in the whole space with permittivity 1 , as shown in Fig. 4.6. They are: R q =− q d 

q1 = −q q1

R = q d

 2  2  2 R R R at x0 , y0 , z0 d d d at −x0 , y0 , z 0  2  2  2 R R R at − x0 , y0 , z0 d d d

q2 = −α q

at x0 , y0 , −z 0

R q2 = α q d

at

q3 = α q R q3 = −α q d

 2  2  2 R R R x0 , y0 , − z0 d d d

at −x0 , y0 , −z 0 at −

 2  2  2 R R R x0 , y0 , − z0 d d d

Solutions

75

Fig. 4.6 Image charges for a plane conductor with a boss embedded in two dielectrics projected on the plane y = 0

where: α=

2 − 1 2 + 1

For the field at z < 0, we have to consider the whole space with permittivity 2 and the charges q, q  , q1 and q1 in their positions but multiplied by a factor β : β=

22 . 2 + 1

4.3 We locate the axis of the cylinder on the z axis and the wire f with linear charge density λ at distance d from the axis. The solution for the dielectric sphere in an external uniform field suggests to take into account the effect of the polarization for r > R with two image wires f 1 and f 2 , parallel to the z axis, respectively with charge densities λ1 and λ2 . The first is located between the wire f and the axis at a distance x < R from the axis, the second on the axis of the cylinder. To describe the field inside the cylinder we can place an image wire f 3 with charge density λ3 in the same place of the wire f. See Fig. 4.7. The potential at a distance r > R from the axis is: 1 V =− 4π1



λ log

r 2 + d 2 − 2r d cos θ

and inside the cylinder is:

r02



+ λ1 log

r 2 + x 2 − 2r x cos θ r02



r2

+ λ2 log 2 r0



76

4 Image Charges in Dielectrics

Fig. 4.7 Wire f with linear charge density λ external to a cylinder of dielectric permittivity 2 embedded in a medium of permittivity 1 : top wire f and image wires f 1 and f 2 for the field outside the cylinders; bottom image wire f 3 for the field inside the cylinder

   2 1 r + d 2 − 2r d cos θ λ3 log . V =− 4π2 r02 

The electric field components outside are: Er =

  r − d cos θ r − x cos θ 1 1 λ 2 + λ + λ 1 2 2 2π1 r + d 2 − 2r d cos θ r + x 2 − 2r x cos θ r

Eθ =

  d sin θ x sin θ 1 λ 2 + λ 1 2 2π1 r + d 2 − 2r d cos θ r + x 2 − 2r x cos θ

Solutions

77

and inside: Er =

  r − d cos θ 1 λ3 2 2π2 r + d 2 − 2r d cos θ

E θ =

  d sin θ 1 . λ3 2 2π2 r + d 2 − 2r d cos θ

From the continuity of the tangent component of the electric field: E θ = E θ on the surface of the cylinder for any value of θ, we have: x=

R2 d

λ λ3 λ2 − + = 0. 1 2 1

and

From the continuity of the normal component of the displacement field: Dr = Dr → 1 Er = 2 Er for any value of θ, and from the previous relations, finally we get: λ1 =

1 − 2 λ 2 + 1

λ2 = −λ1

λ3 =

22 λ. 2 + 1

If 1 = 2 : λ3 = λ and λ1 = λ2 = 0. The force per unit length on the wire is the sum of the forces exerted by the wires f 1 and f 2 : F=

λ 2π1



λ1 λ2 + d−x d

 =

R2 λ2 1 − 2 2π1 1 + 2 d(d 2 − R 2 )

repulsive for 1 > 2 . 4.4 We locate the axis of the cylinder on the z axis. The result of the previous problem suggests to write the field inside the cylinder (at r < R) with an image wire f 1 with charge density λ1 at distance d > R from the axis to be determined (see Fig. 4.8). For the field outside we can introduce an image wire f 2 with charge density λ2 in the position of the wire f and a second image wire f 3 with charge density λ3 on the axis of the cylinder. The potential inside the cylinder (at r < R) is: V =−

  2    2 r + h 2 − 2r h cos θ 1 r + d 2 − 2r d cos θ λ log + λ log 1 4π2 r02 r02

and outside (r > R): V = −

   2  1 r + h 2 − 2hd cos θ r2 λ2 log + λ log + . 3 4π1 r02 r02

The components of the electric field inside are:

78

4 Image Charges in Dielectrics

Fig. 4.8 Wire f with linear charge density λ inside a cylinder of permittivity 2 embedded in a medium of permittivity 1 : top wire f and image wire f 1 for the field inside the cylinder; bottom image wires f 2 and f 3 for the field outside

Er =

  r − h cos θ r − d cos θ 1 λ 2 + λ 1 2 2π2 r + h 2 − 2r h cos θ r + d 2 − 2r d cos θ

Eθ =

  h sin θ d sin θ 1 λ 2 + λ 1 2 2π2 r + h 2 − 2r h cos θ r + d 2 − 2r d cos θ

Solutions

79

and outside: Er =

  r − h cos θ 1 1 + λ λ2 2 3 2π1 r r + h 2 − 2r h cos θ

E θ =

  d sin θ 1 . λ2 2 2π1 r + h 2 − 2r h cos θ

From the continuity of the tangent component of the electric field E θ = E θ on the lateral surface for any value of θ we find: d=

R2 h

and

λ λ2 λ1 − + = 0. 2 1 2

From these relations and from the continuity of the normal component of the displacement field for any value of θ: Dr = Dr → 2 Er = 1 Er we find: λ2 =

21 λ 1 + 2

λ 1 = λ3 =

2 − 1 λ. 1 + 2

If the same medium fills the space 1 = 2 : λ2 = λ, λ1 = λ3 = 0. The force per unit length on the wire is that exerted by the wire f 1 : F=

λ2 2 − 1 h λλ1 1 = 2 2π2 d − h 2π2 1 + 2 R − h 2

repulsive for 2 > 1 (wire pushed to the center). 4.5 The problem is similar to the case of the dielectric sphere in an external uniform field seen in Sect. 4.3. The solution is also suggested by the solutions of Problem 3.9 and Problem 4.3. We choose as z axis the axis of the cylinder of radius R. We can consider the uniform field due to two wires with opposite charge densities (of absolute value λ), located symmetric with respect the z axis, at distance d  R: with density λ at x = −d and

Fig. 4.9 Wires with linear charge density λ to produce the external field E and relative images wires with density λ for a dielectric cylinder embedded in another dielectric medium

80

4 Image Charges in Dielectrics

−λ at x = d as shown in Fig. 4.9. To these wires are associated two parallel image wires, in the dielectric 1 , with opposite charge densities: λ at x = −δ and −λ at x = δ. The field outside the cylinder is the superposition of the uniform field E, produced by the wires at large distance in the limit d → ∞, and of the field E im from the image wires. Inside the cylinder, as proved correct by the result, the field E  has the same direction of the uniform external field. The potential of the image wires at distance r from the axis of the cylinder is: Vim

λ =− 4π1

  r 2 + δ 2 − 2r δ cos θ r 2 + δ 2 + 2r δ cos θ log − log r02 r02

with: λ =

1 − 2 λ from the result of Problem 4.3, 2 + 1

and in the limit δ r the potential and the components of the electric field E im become: Vim =

λ δ cos θ π1 r

E imr =

λ δ cos θ π1 r 2

E imθ =

λ δ sin θ . π1 r 2

The components of the field E are: Er = E cos θ

E θ = −E sin θ .

The components for the electric field outside the cylinder are: E est,r = Er + E imr = E cos θ +

λ δ cos θ π1 r 2

E est,θ = E θ + E imθ = −E sin θ +

λ δ sin θ π1 r 2

and inside: Er = E  cos θ

E θ = −E  sin θ .

From the continuity of the tangent component of the electric field and of the normal component of the displacement field on the surface of the cylinder, we have the equations: E = E −

λ δ π1 R 2

2 E  = 1 E +

λ δ π R2

References

81

with the solution for E  and λ δ: E =

21 E 2 + 1

λ δ =

(2 − 1 ) 1 π R 2 E . 2 + 1

The components of the electric field outside the cylinder are:   1 − 2 R 2 E est,r = E 1 − cos θ 2 + 1 r 2 

E est,θ

1 − 2 R 2 = −E 1 + 2 + 1 r 2

 sin θ .

Note that the two image wires form a sort of dipole of image wires of moment5 : p = 2(−λ )δ = 2

(2 − 1 ) 1 π R 2 E . 2 + 1

It is possible to get the same result if we start considering a dipole of wires image of the wires at distance d in the limit d → ∞. To have a field E from the distant wires the density has to be λ = π1 d E and then from the result of Problem 4.2, the moment of the dipole of image wires is: p = (−λ )2δ =

(2 − 1 ) (2 − 1 ) R2 =2 λ2 (π1 E)R 2 2 + 1 d 2 + 1

that is the result already found.

References 1. D.J. Jackson, Classical Electrodynamics, 3rd edn. (John Wiley & Sons, New York, 1999) 2. L.D. Landau and E.L. Lifshitz, Electrodynamics of continuous media, Course of Theoretical Physics, Vol. 8. (Elsevier) 3. W.R. Smythe, Static and dynamic electricity, 3rd edn. (McGraw-Hill)

The negative sign in the formula for p comes from the choice of a positive λ to produce a field E oriented as the z axis.

5

Chapter 5

Series of Image Charges

Sometime the potential for a distribution of charges and conductors cannot be reproduced by a finite number of image charges. Frequently in presence of many conductors it is possible to introduce interior mirror charges which make equipotential only the surface of one of the conductors. To make equipotential also the surfaces of the other conductors, new image charges have to be introduced inside these conductors but the new image charges distort the first equipotential surface. New image charges added inside the first conductor can restore its equipotential surface but distort the other surfaces. So by adding new image charges it is possible to correct the new distortions and so forth. If the new introduced charges and their effects become smaller and smaller or cancel, it is possible to make equipotential the surfaces of all the conductors with an infinite series of image charges. The potentials can be usually approximated by a finite number of terms in the series.

5.1 Point Charge Between Two Grounded Plates The simplest configuration of a series of image charges is that of a point charge q located halfway between two infinite parallel grounded plates L 1 and L 2 at a distance d. We set the x axis normal to the plates with the origin on the charge as in Fig. 5.1. To keep at ground the plate L 1 we introduce an image charge q1 = −q symmetrical with respect to L 1 at x = d and similarly for L 2 at ground an image charge q1 = −q at x = −d. After these two charges in order to keep the two layers at null potential we need to introduce two more image charges q2 = q in x = 2d e q2 = q in x = −2d and to balance their effects two more charges q3 = −q in x = 3d e q3 = −q in x = −3d. Of course we have to continue with an infinite series of infinite image © The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 F. Lacava, Classical Electrodynamics, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-031-05099-2_5

83

84

5 Series of Image Charges

Fig. 5.1 A point charge q halfway between two parallel plates L 1 and L 2 and the first image charges of the series to get the plates at null potential

charges qn = (−1)n q at x = nd for n positive or negative integer. The potential between the plates is the sum of the potentials of the point charge q and of all the image charges: V (x, y, z) =

n=∞ (−1)n q   4π0 n=−∞ (x − nd)2 + y 2 + z 2

5.2 Two Separated Charged Conductive Spheres The charges Q 1 and Q 2 are deposited respectively on two conducting spheres of radii R1 and R2 , with their centers at distance d > R1 + R2 as shown in Fig. 5.2. We want to find the image charges and their positions to have equipotential spherical surfaces. We can start by placing two point charges q10 and q20 in the centers of the two spheres1 . In absence of mutual induction these two charges determine for the two spheres the potentials: V01 =

1

1 q10 4π0 R1

and

V02 =

1 q20 4π0 R2

(5.1)

The problem of the two conductive spheres can be solved by placing a sphere at ground and the other at a unit potential. After the charges are found as the sums of the induced charges, the potential coefficients are known and can be used for any value of the potentials and of the charges on the two spheres as discussed in a following note in this section. See Smythe [1], Chap. 5, Sect. 5.08, p. 128.

5.2 Two Separated Charged Conductive Spheres

85

Fig. 5.2 Spheres of radii R1 and R2 with centers at distance d > R1 + R2 . The positions x1i and x2i of the image charges q1i and q2i are shown

To take into account the induction we have to remember the case of the point charge and the conducting sphere. We can keep equipotential the first sphere introducing a point charge q11 , image of q20 , at the distance x11 from its center with x11 and q11 given by the relations: x11 =

R12 d

q11 = −

R1 x1 q20 = − 1 q20 d R1

Similarly, to keep the second sphere equipotential, a charge q21 must be placed inside at a distance x21 from its center in the direction of the first sphere: x21 =

R22 d

q21 = −

R2 x2 q10 = − 1 q10 d R2

However, to keep the two surfaces still equipotential after the introduction of these two image charges, it is necessary to introduce two more image charges q12 and q22 at positions x12 and x22 given by: x12 =

R12 d − x21

q12 = −

R1 x1 q2 = − 2 q21 d − x21 1 R1

x22 =

R22 d − x11

q22 = −

R2 x2 q11 = − 2 q11 d − x11 R2

and one must continue with an infinite series of image charges q1i and q2i located in positions x1i and x2i , for i → ∞, given by:

86

5 Series of Image Charges

x1i =

R12 d − x2i−1

q1i = −

R1 x1 q2 = − i q2i−1 d − x2i−1 i−1 R1

(5.2)

x2i =

R22 d − x1i−1

q2i = −

R2 x2 q1i−1 = − i q1i−1 d − x1i−1 R2

(5.3)

with x10 = x20 = 0. The image charges become smaller and smaller as i increases and their positions remain always inside the two spheres. The net charge on each sphere is the sum of all the charges introduced inside: Q1 =

∞ 

R1 q2 d 0

(5.4)

R2 q1 + α4 q20 d 0

(5.5)

q1i = α1 q10 + α2

i=0

Q2 =

∞ 

q2i = α3

i=0

with the coefficients αi which depend2 only on the geometry of the system, that is R1 , R2 and d. It is easy to verify that α1 > 0, α4 > 0 and α2 = α3 < 0 as expected (see footnote in this page). The addition of the image charges does not modify the potentials given by the relations (5.1) and after replacing these quantities in the formulas (5.4) and (5.5) we get: Q 1 = 4π0 R1 α1 V01 + 4π0 Q 2 = 4π0

R1 R2 α2 V02 d

R1 R2 α3 V01 + 4π0 R2 α4 V02 d

The coefficients αi can be written in the form:  i   ∞   R1 R2 α1 = 1 + · d − x22k+1 d − x12k

2

i=0

k=0

α2 = − 1 + α3 = − 1 +



∞ i    i=0 ∞  i=0



k=0 i  

k=0

α4 = 1 +

 i  ∞   i=0

k=0

R1 R2 · d − x22k+2 d − x12k+1 R1 R2 · d − x12k+2 d − x22k+1

R2 R1 · d − x12k+1 d − x22k







and by substituting the expressions (5.2) and (5.3) for x1i and x2i it is easy to show that α2 and α3 are equal as expected (see next footnote).

5.2 Two Separated Charged Conductive Spheres

87

which are the relations3 between the charges and the potentials of the spheres. For two conductors these relations are usually written in the form: Q 1 = c11 V01 + c12 V02

Q 2 = c21 V01 + c22 V02

(5.6)

so for two conductive spheres the capacitance coefficients ci j are: c11 = 4π0 R1 α1 c12 = c21 = 4π0

c22 = 4π0 R2 α4

R1 R2 R1 R2 α2 = 4π0 α3 < 0 d d

or in a more useful form4 : c11 = 4π0 R1 R2 sinh ξ

∞ 

[R1 sinh nξ + R2 sinh(n + 1)ξ]−1

n=0

3

For a system of N conductors with charges Q i and potentials Vi in empty space it is possible to write (see Jackson [3], Sect. 1.11 or Zangwill [4], Sect. 5.4.3) the relations: Qi =

N 

ci j V j

j=1

with |ci j | the capacitance matrix. The diagonal elements are named capacitances or capacities while the off-diagonal ones are the coefficients of induction. It is easy to show that these coefficients satisfy the Maxwell inequalities:  ci j = c ji cii > 0 ci j < 0 ck j ≥ 0 j

Since det |ci j | = 0 these relations can be inverted to give the potentials of each conductor in terms of the charges of all the conductors: N  Vi = pi j Q j j=1

with pi j the elements of the potential matrix | pi j | which of course depend only on the geometry of the system. The energy of the system of conductors is: W =

N N N 1 1 Q i Vi = ci j Vi V j 2 2 i=1

i=1 j=1

The electrostatic force applied on the ith conductor, in the direction of a move si can be found by virtue of the principle of the virtual works by the relation: Fis = − 4

See Smythe [1] and Lekner [2]

∂W ∂si

88

5 Series of Image Charges

Fig. 5.3 Field lines for two opposite charged spherical conductors of radius r with their centers a distance d = 14 r apart. The sphere on the left can be the image of a real sphere at right in front of a conductive plate at ground (dashed line)

c22 = 4π0 R1 R2 sinh ξ

∞ 

[R2 sinh nξ + R1 sinh(n + 1)ξ]−1

n=0

c12 = c21 = −4π0

∞  R1 R2 sinh ξ [sinh nξ]−1 d n=1

cosh ξ =

d 2 − R12 − R22 2R1 R2

Of course, if the potentials V01 e V02 are known, the charges Q 1 and Q 2 can be found from the relations (5.6) and vice versa from the potential matrix. By centering the first sphere at the origin of the axes and the second at the coordinate point (d, 0, 0), the potential outside the two spheres is given by the series: ∞ 1  V0 (x, y, z) = 4π0 i=0

 

q1i (x − x1i )2 + y 2 + z 2

q2i



+ (d − x2i − x)2 + y 2 + z 2

It is easy to verify that with a fairly limited number of terms the potential approaches the asymptotic value. The field lines for two spheres with opposite charge are shown in Fig. 5.3. The case of a charged spherical conductor inside a larger spherical conductor is examined in Problem 5.4.

5.3 Force Between Two Charged Spherical Conductors

89

5.3 Force Between Two Charged Spherical Conductors The force between two charged spherical conductors is the sum of all the forces between the two real or mirror charges of all the pairs formed by combining one of the charges inside a sphere with one of the charge inside the other sphere: F=

∞ ∞ 1   q1i q2 j 4π0 i=1 j=1 di2j

where di j is the distance of the two charges of the pairs. In a different approach one can write the electrostatic energy of the system using the capacitance (or potential) coefficients and then derive the force on a conductor in the s direction from the relation: ∂W (5.7) F =− ∂s In the case of a grounded conductive sphere, if a positively charged conductive sphere is approached, the negative charge induced on the grounded sphere causes an attraction between the two spheres. This is evident also from the relations (5.6) and the Maxwell inequalities: for V01 = 0 and Q 20 > 0 we have V20 = Q 20 /c22 > 0 because c22 > 0 and Q 01 = c12 V02 < 0 because c12 < 0. In Sect. 3.3.1 we have seen the attractive force between a charged sphere and a like sign point charge when they are close enough. A more general and rather interesting case worth considering is the force between two conductive spheres with same-sign charges. The problem, already studied by Kelvin, was recently reviewed by Lekner in [2] where previous works are mentioned and examples from real phenomena are also presented. Kelvin already found that, due to electrostatic induction, the repulsive force acting between two charges Q deposited on two conductive spheres of same radius R in contact, is smaller than the repulsive force between two point charges Q at a distance equal to 2R: Q 2 4 log 2 − 1 Q2 1 1 < (5.8) F= 4π0 (2R)2 6(log 2)2 4π0 (2R)2 where the multiplicative factor is equal to  0.6149. For this result see Problem 5.3. Lekner shows that the force between two spherical conductors with same-sign charge, close enough, is always attractive, see Fig. 5.4, but when the charges on the conductors are in the ratio they would have if the two conductors had been in contact and shared their charges. For two positive charged conductors the attractive force is caused by the negative charge induced on the pole of one of the spherical conductor closest to the other conductor. The attractive force between the negative induced charge and the positive charge of the other conductor is stronger than the repulsive force between the positive charges of the two conductors.

90

5 Series of Image Charges

Fig. 5.4 Force between two conductive spheres with same charge and radii r1 = 4 and r2 = 2 versus the distance between their centers divided by the force between two point charges at a distance d = r1 + r2 . The force is attractive when the spheres are very close

5.4 A Charged Conductive Sphere and a Conductive Plate at Ground We want the electric field for a charged conductive sphere with radius R2 and its center a distance c > R2 from an infinite conductive plate at ground as in Fig. 5.5. Similarly to the problem of the point charge and the conductive plane at ground, we may describe the system as two equal conductive spheres with opposite charges, symmetrically located with respect to the conductive plane, with their centers a distance 2c away, thus the field is that in the real half space as in Fig. 5.3. Another approach is to consider two spherical conductors: one (conductor 2) of radius R2 and the other one (conductor 1) at ground with radius R1 , and then taking the limit R1 → ∞ so that the spherical surface of 1 transforms in the conductive plate5 . We follow the second solution and we begin by setting a charge q20 = 0 in the center of conductor 2. The distance between the centers is d = R1 + c and we have to suppose the conductor 1 at ground and with no charge (q10 = 0) except the induced one. To keep the first sphere at null potential, we have to introduce inside this sphere an image charge q11 at a distance x11 from its center that is at distance s11 from its surface on the line passing through the two centers: 5

It is also possible to consider a conductive spherical shell of radius R2 inside a conductive spherical shell of radius R1 > R2 at ground, with their centers a distance c apart. Then given the distance s of the center of the smaller sphere from the conductive plane, so that R1 = c + s, it is possible to find a series of image charges in the limit c → ∞ (R1 → ∞).

5.4 A Charged Conductive Sphere and a Conductive Plate at Ground

91

Fig. 5.5 The charged conductive spherical shell and the conductive plate at ground. The problem is solved by considering two spherical shells of radius R1 and R2 and transforming the pole of the first sphere, towards the other sphere, in a plane in the limit R1 → ∞

x11 =

R12 R12 = d R1 + c

q11 = −

x11 q2 R1 0

s11 = R1 − x11 =

R1 c R1 + c

In the limit R1 → ∞ the charge q11 becomes: q11 = −q20

at distance

s11 = c

In order to have equipotential the conductor 2, this charge requires the introduction of a new charge q22 inside this conductor6 in the position x22 at distance s22 from the plane: x22 =

R22 R 2 (R1 + c) = 2 d − x11 2R1 c + c2 s22 = R2 − x22 =

q22 = −

x22 R2 (R1 + c) q20 q11 = R2 2R1 c

2R1 c2 + c3 − R22 R1 − R22 c 2R1 c + c2

and in the limit R1 → ∞: q22 =

R2 q2 2c 0

s22 = c −

R22 2c

and continuing in the same way it is possible to find the non-zero charges and their positions:

6

Of course the charge q21 is null because it is the image of q10 at the center of the sphere 1 which is null. So the charge q12 image of q21 is also null and thus all the charges q1i with i > 1 even and q2i with i > 1 odd are null.

92

5 Series of Image Charges

q13 =

R2 q2 = −q22 2c 0

q24 =

R22 q2 4c2 − R22 0 q15 = −q24

q26 =

R22 R2 q2 2 2c 4c − 2R22 0 q17 = −q26

s13 = c − s24 = c

R22 = s22 2c

4c2 − 3R22 4c2 − R22

s15 = s24 s26 =

8c4 − 8R 2 c2 + R24 8c3 − 4R22 c

s17 = s26

and so forth. The charges q1i at the left and q2i at the right of the plane are symmetrical with respect to the conductive plane and of course they are the real and image charges of two conductive spherical shells of radius R2 carrying opposite charges symmetrically located with respect to the real conductive plane. So we have found the first approach mentioned at the beginning of this section. The potential in the free space between the charge sphere and the conductive plate at ground is the sum of the potentials of all these charges. The charge carried by the conductor 2 is the sum of all the point charges inside, the induced charge on the plane is opposite to the total charge on the conductive sphere and is equal to the charge on the image sphere.

5.5 Dielectric Sphere and Point Charge For a sphere of dielectric constant r and radius R in front of a point charge Q at distance d > R from its center the potentials inside and outside the sphere are found from the Laplace equation7 as the superposition of the potentials from 2n poles residing at the center of the sphere. The series converges slowly at times and it is not easy to extrapolate to special cases (for instance to the conductive sphere). I.V. Lindell8 in [6] has shown that this solution is equal to the potentials given by two image linear charge distributions and when written in this form it overcomes the mentioned difficulties. In a coordinate system with the origin in the center of the sphere and the point charge in the position z = d, the external potential is generated by an image charged line: λext = − 7 8

r − 1 R r − 1 Q Q δ(z − d K ) + r + 1 d (r + 1)2 R

See Problem 8.6 and Stratton [5], Sect. 22. See also Bussemer [7].



z dK

− ( +1) r r

Solutions

93

between the origin and d K = R 2 /d, the point of the image charge for a conductive sphere. For the internal potential the image line charge density, distributed from z = d to +∞, is: λint =

r (r − 1) Q 2 r Q δ(z − d) + r + 1 (r + 1)2 d

  ( 1+1) d r z

Problems 5.1 Show that if a point charge is located between two conducting half planes at ground which intersect at an angle α = π/n where n is an integer, the field can be described by a finite number of image charges (see Maxwell [8], Chap. 11, 165). Problem 3.2 is for n = 2. 5.2 A wire of radius r0 with charge per unit length λ is located in the middle of two grounded conductive planes at distance d >> r0 apart. Write the potential between the planes as the sum of the potentials of an infinite series of image wires. 5.3 Reproduce the result (5.8) found by Kelvin for the force between two equal spherical conductors with same-charge in contact. 5.4 A conductive spherical shell of radius R2 is located inside another spherical shell of radius R1 > R2 . The distance of the two centers is c < R1 − R2 as Fig. 5.9. A charge Q 2 is present on the inner conductor while the outer one is connected to ground (it is at null potential). Determine the potential in the free space between the spherical shells. Find the elements of the capacitance matrix for the system of two conductors.

Solutions 5.1 In a plane containing the point charge P and perpendicular to the line of intersection of the two planes, we draw a circle with its center on the line and radius equal to the distance of the point charge P from the line ( see Fig. 5.6). The two segments AO and BO are parts of two diameters AC and BD. To keep at ground the extended planes crossing the diameters AC and BD, we have to introduces some image charges of P both with respect to the planes AC and BD. These images charges lie on the circle and are at the ends of chords normal to the two diameters AC and BD. The angular distances between two same-sign charges on the circle are equal and are twice the angle α between the conductive planes. The condition to have a finite number of charges is:

94

5 Series of Image Charges

Fig. 5.6 Point charge between two grounded conductive intersecting planes (AO and BO) with an angle α = π/4. The point charge P and its image charges have to be introduced to keep at ground the extended planes AC and BD

2α n = 2π

→

α=

π n

The number of image charges with sign equal to the charge in P are n − 1 and with opposite sign are n. 5.2 The voltage at distance r from the center of a wire of radius r0 is: V (r ) = −

r λ log + V (r0 ) 2π0 r0

To have a null voltage on the plates at ground we have to introduce an infinite series of image wires, parallel to the wire, with charge densities λn = (−1)n · λ at distances ± d · n where n = 1, 2, ..., ∞ as shown in Fig. 5.7. For a wire along the z axis, the potential in the point with coordinates (x, y) is: 1 1 ∞ λ (x 2 + y 2 ) 2 λ  [x 2 + (y − nd)2 ] 2 n V (x, y) = − log − (−1) log + 2π0 r0 2π0 n=1 r0

−

1 ∞ ∞  λ  [x 2 + (y + nd)2 ] 2 (−1)n log + V (r0 ) + 2 (−1)n V (r0 ) 2π0 n=1 r0 n=1

(5.9)

The value of V (r0 ) is determined by the condition V (x, ±d/2) = 0. An example of a wire between grounded plates is given in Fig. 5.8. 5.3 Consider two equal conductive spheres of radius R with same charges Q in contact. We begin with two equal point charges q11 and q22 located in the centers of the two spheres. We use the formulas 5.2 and 5.3 for the image charges in two conductive spheres and we write their distances s1n = R − x1n = s2m = R − x2m from the point

Solutions

95

Fig. 5.7 Wire between two grounded plates and the series of image wires

Fig. 5.8 Wire of 50 μm radius charged with linear density λ = 10 nC/m between two grounded conductive plates distant 1 cm. The voltage of the wire is 1620 V. The equipotential contours for V = 600, 350, 200, 100, 40, 15, 5, 2 V are shown (40 iterations in (5.9))

of contact: q1n = (−1)n+1

q11 n

s1n =

R n

n = 1, 2, ..., ∞

q2m = (−1)m+1

q21 m

s2m =

R m

m = 1, 2, ..., ∞

The force between the two spheres is: F=

∞ ∞ ∞ ∞ 1  q1n q2m nm 1 q11 q21   = (−1)n+m 4π0 n=1 m=1 (s1n + s2m )2 4π0 R 2 n=1 m=1 (n + m)2

The charge on each conductor is equal to the sum of the inner charges:

96

5 Series of Image Charges

  1 1 1 1 Q = 1 − + − + + ... q11 = q11 log 2 2 3 4 5

q11 = q21 =

Q log 2

We have to evaluate the double summation. By setting k = n + m we get: ∞  ∞ 

(−1)n+m

n=1 m=1

∞ k−1   nm nk − n 2 k = (−1) (n + m)2 k2 n=1 k=2

and from Gradshteyn and Ryzhik [9] 0.121.1 and 0.121.2: k−1 

n=

n=1

k(k − 1) 2

k−1 

and

n2 =

n=1

k(k − 1)(2k − 1) 6

  ∞ k−1  1 nk − n 2 1 k (−1) = (−1) k − k2 6 k=2 k n=1 k=2

∞ 

k

then by using Gradshteyn and Ryzhik [9] 1.122.2: ∞ 

(−1)k k = 1 −

k=2

∞ 

(−1)k−1 k = 1 −

k=1

1 3 = 2 2 4

and Gradshteyn and Ryzhik [9] 1.511: ∞ 

∞

(−1)k

k=2

 1 =1− (−1)k−1 k = 1 − log 2 k k=1

for the double summation we get: ∞  ∞ 

(−1)n+m

n=1 m=1

nm 11 (4 log 2 − 1) = 2 (n + m) 64

and from this result and the relations for the charges Q, for the force we find: F=

Q 2 4 log 2 − 1 1 4π0 (2R)2 6(log 2)2

5.4 Also for this system we can write an infinite series of image charges. Using the solution for the point charge and the conductive sphere, we start by setting a point charge q20 , say positive, in the center of the inner shell, which in absence of any other charge and of the outer conductor, gives to this conductor a potential:

Solutions

97

Fig. 5.9 Two spherical conductors of radius R1 and R2 with centers at distance c < R1 − R2 . In the figure is shown the position of the first image charge q11 at a distance d11 away from the center of the outer sphere

V02 =

1 q20 . 4π0 R2

(5.10)

To keep the shell of radius R1 at null potential, in presence of the charge q20 , we have to locate an external negative image charge q11 at distance d11 from its center as in Fig. 5.9: d1 R2 q11 = − 1 q20 d11 = 1 R1 c After the addition of this charge, to keep equipotential the inner shell a second positive image charge q21 has to be located a distance x21 away from its center and d21 from the center of the larger spherical surface. These are given by the relations: q21 = −

R2 q1 d11 − c 1

x21 =

R22 d11 − c

d21 = x21 + c

And so forth to keep equipotential the two conductive surfaces, an infinite series of image charges must be added. Negative charges q1i (i ≥ 1) outside the larger conductor at distance d1i from its center: q1i = −q2i−1

d1i R1

d1i =

R12 d2i−1

with d20 = c and positive charges q2i (i ≥ 1) inside the smaller shell, at distance x2i from its center and d2i from the center of the larger shell: q2i = −q1i

R2 d1i − c

x2i =

R22 d1i − c

d2i = x2i + c

The potential in a point between the spherical shells is the sum of the potentials produced in that point by all the charges inside the smaller conductor and outside the larger one. The sum of all the positive charges inside the smaller shell has to be

98

5 Series of Image Charges

equal to the charge Q 2 on that conductor. Q2 =

∞ 

q2i = α q20

(5.11)

i=0

with α dependent9 only on the geometry of the system, that is on R1 , R2 and c. Of course the sum of all the image charges outside the larger conductor is the net charge Q 1 induced on the outer shell and so it has to be equal but opposite in sign to Q2: Q1 =

∞ 

q1i = −Q 2

i=1

It is easy now to extract the elements of the capacitance matrix. The image charges we have introduced do not alter the potentials of the two spherical surfaces, so introducing the relation (5.10) in (5.11) we get: Q 2 = 4π0 α R2 V02 = c22 V02

c22 = 4π0 α R2

(5.12)

From Q 1 = −Q 2 we have: Q 1 = −Q 2 = −4π0 α R2 V02 = c12 V02 and so: c12 = c21 = −4π0 α R2 = −c22 If there is a null charge on the inner surface and a charge Q 1 is deposited on the outer one, the potential of the two conductors is:

9

The α coefficient can be written: α=1+

 ∞  n  R2 d1i R1 d1i − c n=1 i=1

which for c → 0 becomes a simple geometrical series: α=1+

 ∞  R1 R2 n = R1 R1 − R2 n=1

and from (5.12) we have the capacitance of a spherical capacitor: C=

R1 R2 Q2 = 4π0 α R2 = 4π0 V02 R1 − R2

5.5 Dielectric Sphere and Point Charge

99

  V01 = V02 =

Thus:

1 Q 1 4π0 R1

   + c12 V02 = (c11 + c12 ) V01 Q 1 = c11 V01

c11 + c12 = 4π0 R1

c11 = −c12 + 4π0 R1 = c22 + 4π0 R1

The element c22 of the capacitance matrix is equal10 to: c22 = 4π0 R1 R2 sinh ξ

∞ 

[R1 sinh nξ − R2 sinh(n − 1)ξ]−1

n=1

cosh ξ =

R12 + R22 − c2 2R1 R2

In general for a system of two conductors if the potentials V01 and V02 are known, from the relations: Q 1 = c11 V01 + c12 V02

Q 2 = c21 V01 + c22 V02

(5.13)

it is possible to get the charges Q 1 and Q 2 while if the charges Q 1 and Q 2 are known, the relations (5.13) can be inverted and the potentials can be found. If a charge Q ∗ is deposited on the free outer sphere and and a charge Q on the inner one, the potential of the outer conductor is: ∗ = V01

1 Q∗ + Q 4π0 R1

When a charge Q is deposited on the inner shell to have a potential V ∗∗ for the floating outer shell, a charge Q ∗∗ − Q has to be added to this conductor with Q ∗∗ given by the relation: 1 Q ∗∗ ∗∗ = . V01 4π0 R1

References 1. W.R. Smythe, Static and Dynamic Electricity, 3rd edn. (McGraw-Hill) 2. J. Lekner, Electrostatics of Two Charged Conducting Spheres, Proceedings of Royal Society A, https://doi.org/10.1098/rspa.2012.0133 3. D.J. Jackson, Classical Electrodynamics, 3rd edn. (John Wiley & Sons, NewYork, 1999) 4. A. Zangwill, Modern Electrodynamics (Cambridge University Press, Cambridge, 2013) 10

See Smythe [1], Chap. 5, Sect. 5.081.

100

5 Series of Image Charges

5. J. A. Stratton, Electromagnetic Theory (IEEE PRESS SERIES), https://onlinelibrary.wiley.com/ doi/book/10.1002/9781119134640 6. I.V. Lindell, Am. J. Phys. 61, 39 (1993). https://doi.org/10.1119/1.17407 7. P. Bussemer, Am. J. Phys. 62, 657 (1994). https://doi.org/10.1119/1.17487 8. J.C. Maxwell, A Treatise on Electricity and Magnetism (1873) 9. I.S. Gradshteyn and I.M. Ryzhik, Tables of Integrals, Series and Products, 7th edn. (Academic Press, Elsevier, London 2007)

Chapter 6

Functions of Complex Variables and Electrostatics

A useful and elegant indirect approach to solve some two-dimensional problems in Electrostatics, comes from the analytic functions of a complex variable. The real and imaginary parts of these functions are solutions of the Laplace equation and if they satisfy the boundary-values of the problems, they offer the solutions to these problems. After a brief introduction to the analytic functions, some interesting examples are discussed: the potentials for a quadrupole, a wedge, the edge of a thin plate and a wire. At the end of the chapter the same potentials are derived from the direct solution of the Laplace equation.

6.1 Analytic Functions of Complex Variable The functions of complex variables are studied in the courses in Mathematics and Mathematical Methods of Physics. Some notions useful for applying these functions to the solution of some electrostatic problems will be briefly discussed1 . A complex number can be represented in the Cartesian plane O(x, y) with a correspondence between the complex number z = x + i y and the position of the point P with coordinates (x, y), see Fig. 6.1. A complex number can also be expressed using polar coordinates (ρ, ϕ): z = ρeiϕ

z = ρ(cos ϕ + i sin ϕ)

so that x = ρ cos ϕ and y = ρ sin ϕ.

1

For the functions of a complex variable see for instance: Sveshnikov and Tikhonov [1]; Morse and Feshbach [2], Chap. 4; Arfken and Weber [3], Chap. 11; Bernardini et al. [4].

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 F. Lacava, Classical Electrodynamics, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-031-05099-2_6

101

102

6 Functions of Complex Variables and Electrostatics

Fig. 6.1 A complex number z = x + i y represented by a point on the complex plane

Fig. 6.2 Functions u(x, y) = x 2 − y 2 = A (solid line) and v(x, y) = 2x y = B (dashed line)

A function f (z) of the complex variable z can always be written in the form f (z) = u + iv with u = u(x, y) and v = v(x, y) two real functions of the real variables x and y. As an example, the function f (z) = z 2 can be written as: f (z) = (x + i y)2 = (x + i y)(x + i y) = x 2 − y 2 + i2x y with u(x, y) = x 2 − y 2 and v(x, y) = 2x y (see Fig. 6.2). From the theory of the functions of complex variable it is known that a function f (z) = u(x, y) + iv(x, y) is differentiable in a simply connected domain if the Chauchy-Riemann or monogenity equations: ∂v ∂u = ∂x ∂y

∂u ∂v =− ∂y ∂x

(6.1)

are satisfied with u(x, y) and v(x, y) two continuous functions with continuous partial derivatives. A function which satisfies these equations is called analytic or holomorphic function and the existence of its first derivative implies the existence of the derivatives at

6.1 Analytic Functions of Complex Variable

103

any order and, as a direct consequence, the function can be expanded in power series. Moreover the integral of the function along a closed path on the complex plane is null and as a consequence its integral along an arbitrary line depends only on the extremes of the paths (γ1 = γ2 ): 



 f (z) dz = 0

γ1

f (z) dz =

γ2

f (z) dz .

Taking the derivative of the first relation in (6.1) with respect to x(y) and of the second relation with respect to y(x) and adding (subtracting) the two relations, we get: ∂2u ∂2u + 2 =0 2 ∂x ∂y

∂2v ∂2v + 2 =0 2 ∂x ∂y

and then the functions u(x, y) and v(x, y) of any analytic function, satisfy the Laplace equations: Δu = 0

Δv = 0 .

If one of the two functions u or v is given, the other is obtained from (6.1) up to an arbitrary constant. Indeed if u is known, we can write: dv =

∂v ∂v ∂u ∂u dx + dy = − d x + dy ∂x ∂y ∂y ∂x

which can be integrated to find v. The vectors:   ∂u ∂u u , ∂x ∂ y

 and

v

∂v ∂v , ∂x ∂ y



respectively normal to u(x, y) = const and v(x, y) = const  , as a consequence of (6.1), are normal: u·v =0 and thus also the curves u(x, y) = const and v(x, y) = const  are locally normal to each other in their intersection point.

104

6 Functions of Complex Variables and Electrostatics

6.2 Electrostatics and Analytic Functions For a system of N conductors of known geometry and with fixed potentials Vi (i = 1, . . . , N ), in order to find the potential V (x, y, z) at each point in space, we have to solve the Laplace equation ΔV = 0 in the region outside the conductors, assuming the potential at the surfaces of the conductors as boundary conditions. This is the Dirichelet’s problem. The functions u(x, y) and v(x, y), associated to an analytic function, are solutions of the Laplace equation and can be the solution for the potential when the charge distributions depend on two coordinates only and are conformal in all the planes normal to the third axis. These are two-dimensional electrostatic problems. Indeed, if for the system of the examined conductors it is possible to find a function u(x, y) (or v(x, y)) with the value Vi on the surface of the ith conductor, then, for the uniqueness of the solution of the Laplace equation, u(x, y) (or v(x, y)) is the function that describes the potential in the space outside the conductors. The equipotential curves u(x, y) = Vi (or v(x, y) = Vi ) describe the equipotential borders of the conductors and the equipotential surfaces outside the conductors are given by the same function by changing the value of the potential. Moreover the field lines are described by the family of curves v(x, y) (or u(x, y)) normal in each point to u(x, y) (or v(x, y)). The subject of this chapter is presented in the Feynman Lectures on Physics [5], Vol. II, Sect. 7.1.2. with an interesting introduction. A very short but effective presentation is given by Pauli in his book Electrodynamics [6], Sect. 11b.

6.3 The Function f (z) = z µ 6.3.1 The Quadrupole: f (z) = z 2 Let us consider the function f (z) = z 2 in Electrostatics. This function describes the electric field between the four polar expansions of a quadrupole as sketched in Fig. 6.3. If the functions u(x, y) = x 2 − y 2 = ±A are the equipotential surfaces of the poles, the equipotential surfaces inside the quadrupole are similarly described by x 2 − y 2 = A with |A | < A and the field lines are represented by the function v(x, y) = 2x y = B with different values for B (positive or negative). This is evident from Figs. 6.2 and 6.3.

6.3 The Function f (z) = z μ

105

Fig. 6.3 Positive and negative poles in an electric quadrupole and electric field lines

In order to get the potential from the function u(x, y), this has to be multiplied by a constant C = V /d 2 with ±V being the potentials of the polar expansions and d their distance from the center.2 The electric field is:      2V V  2 2V 2 E : − 2 x, 2 y x −y E = −∇ d2 d d focusing in the horizontal direction a positive particle crossing the quadrupole while defocusing it in the vertical direction (Fig. 6.3). The field for a quadrupole rotated by 45◦ can be obtained simply by exchanging the functions u and v. It is easy to see that the function v describes the field in the corner formed by two conductive planes crossing each other on the z axis at a 90◦ angle.

6.3.2 The Conductive Wedge at Fixed Potential Consider the more general function: f (z) = z μ = ρμ eiμϕ = ρμ (cos μϕ + i sin μϕ) , its imaginary part: v = ρμ sin μϕ

(6.2)

The function u = x 2 − y 2 is also the potential for the electric field at the center of a quadrupole composed by four charged wires parallel to the z axis: two with linear charge density λ which cross the plane xy in the points (a, 0) and (−a, 0), and two with linear charge density √ −λ located at (0, a) and (0, −a). The field is that between the poles of the quadrupole if a = d (λ/π0 V ).

2

106

6 Functions of Complex Variables and Electrostatics

Fig. 6.4 Charged wedge of opening angle α

gives the solution for the potential3 outside a conductive wedge with an opening angle α (see Fig. 6.4) extended to infinity along the direction orthogonal to the ρ, ϕ plane:

3

The formula (6.2) is only a mathematical one, in order to describe a potential it has to be multiplied by a constant C having dimension [V l −μ ] and then an arbitrary constant can be added: where VC is the voltage of the conductive wedge while the constant C depends on the strength of the field near the wedge. We can fix the potential V1 at distance ρ1 on the field line ϕ = 2π 2− α , μϕ = π2 , the bisector of the angle external to the wedge (note that we cannot choose a null potential at an infinite distance because the charge extends to infinity). We get: −μ

C = (V1 − VC )ρ1 and the formula for the potential becomes:  V = (V1 − VC )

ρ ρ1

μ

sin μϕ + VC .

On the bisector line we can write: V − VC V1 − VC = μ ρμ ρ1 and if we fix also the potential V2 at distance ρ2 , from: V1 − VC V2 − VC = μ μ ρ2 ρ1 we find the potential to be applied to the wedge: μ

VC =

μ

ρ2 V1 − ρ1 V2 μ μ ρ2 − ρ1

and the value of the constant C: C=

V2 − V1 μ μ . ρ2 − ρ1

The voltage of the wedge is therefore determined by the voltage difference V2 − V1 between two points at distances ρ2 and ρ1 from the edge, that is by the strength of the electric field near the conductive wedge (C > 0 for a negative charge and C < 0 for a positive charge). For a conductive charged semispace (see in the following) μ = 1 and if positive/negative C = ∓E 0 .

6.3 The Function f (z) = z μ

107

Fig. 6.5 Special configurations for the function z μ : a corner with α = π/2, b corner with α = 3π/2, c plate (α = π)

V = Cρμ sin μϕ + VC .

(6.4)

For the equipotential conductor it has to be: V (ϕ = 0) = V (ϕ = 2π − α) and from this: μ=

π . 2π − α

(6.5)

Special configurations are: α = π2 μ = 23 corner with inner right angle (Fig. 6.5a), α = 23 π μ = 2 corner with outer right angle (Fig. 6.5b), α = π μ = 1 conductive plate (semispace) (Fig. 6.5c). The cylindrical components of the electric field are: E ρ = −∇ρ V = −

E ϕ = −∇ϕ V = −

∂V = −μCρμ−1 sin μϕ ∂ρ

1 ∂V = −μCρμ−1 cos μϕ ρ ∂ϕ

with E ρ = 0 for ϕ = 0 and ϕ = (2π − α) as required for the two equipotential ˆ the electric field conductive planes. For ϕ = 0 and E ϕ = −μCρμ−1 (opposite to ϕ), enters normal into the plane at ϕ = 0, while for ϕ = (2π − α) and E ϕ = μCρμ−1 (oriented as ϕ), ˆ the field enters normal into the plane at ϕ = 2π − α. The surface charge density can be obtained using the Coulomb theorem: σ = − 0 μCρμ−1 . The charge per unit length at distances 0 gives the solution of the same equation for the configuration of the two plates in the plane (x, y). The solution for a wire parallel to a conductive plane can be found by introducing an image wire of opposite charge in the position D  (0, −i) symmetrical of D with respect to the plane. So the solution is: V (u, v) = −

  λ w−i λ [log(w − i) − log(w + i)] = − log 2π0 2π0 w+i

  πz λ λ ed − i λ =− V (u, v) = − log log(ρeiϕ ) = − (log ρ + iϕ) π z d 2π0 2π0 2π0 e +i

3

Also in Zangwill [5], Example 8.5. This function is derived at the end of this chapter using a Schwarz-Christoffel transformation. 5 Table of correspondance between the points in the (x, y) plane and in the (u, v) plane as shown also in Fig. 7.2. 4

x

A +∞

B +∞

C −∞

C −∞

y

d

0

d

0

u v

−∞ 0

+∞ 0

0 0

0 0

D 0 d 2 0 1

E 0

F 0

d

0

-1 0

1 0

7.3 Conformal Mapping and Bilinear Electrostatic Problems

121

Fig. 7.2 Mapping of the (x, y) plane onto the (u, v) plane: (a) the wire in the position D between the two grounded plates in (x, y) plane, (b) the wire in D with only one grounded plate in the (u, v) plane

Fig. 7.3 Equipotential surfaces and field lines for a charged wire between two grounded conductive plates

 π π e d x cos πd y + i(e d x sin πd y − 1) λ V (u, v) = V (x, y) = − log π π 2π0 e d x cos πd y + i(e d x sin πd y + 1) The real part of this function gives the potential V (x, y): π

[e d x cos2 πd y + (e d x sin πd y − 1)2 ] λ λ log ρ = − log 2π π 2π0 4π0 [e d x cos2 πd y + (e d x sin πd y + 1)2 ] 2π

V (x, y) = −

while the imaginary part with different values of ϕ gives the field lines: π

tan ϕ = −

2 e d x cos πd y 2π

ed x −1

The equipotential surfaces and the field lines are shown in Fig. 7.3. The equipotentials for a real example were shown in Fig. 5.8. The MWPC (Multi Wire Proportional Chamber), a detercor used in Particle Physics, is an array of many parallel wires of radius 20–50 μm, few millimeters apart, operated at positive voltage between two grounded plates (d  1 cm) in a gas mixture. This detector can measure the position of crossing charged particles with a resolution ∼ 100 μm or better. In the past [6] the complex method was used to study the field in this type of detectors.

122

7 Conformal Mapping in Electrostatics

7.4 The Linear-Fractional Function A linear-fractional function (or Möbius transformation) is a function of a complex variable z of the form: a + bz (7.1) w = f (z) = c + dz where a, b, c and d are complex constants which satisfy the condition: a b = c d otherwise it would be a constant. The function can be also written:

where: λ=

b d

w = f (z) = λ

α+z β+z

a b

c d

α=

The inverse function: z=

β=

α = β

λα − βw −λ + w

is a linear-fractional function defined on the w-plane. The function (7.1) is a conformal map of the z-plane onto the w-plane which extends also at points z or w at infinity. It depends on three independent parameters a, b and c or α, β and λ thus there are an infinite number of linear-fractional transformations mapping the z-plane onto the w-plane. There are three relevant theorems for fractional linear functions: (a) A linear-fractional function is uniquely defined by the correspondence between three points of the z-plane and three points of the w-plane. The choice of the three points on the plane defines the three complex constants. (b) A linear-fractional function transforms circles in the z-plane into circles in the w-plane. Including in the circles the straight lines regarded as circles of infinite radius. Three points on the plane define a circle. (c) In a mapping by a linear-fractional function, points symmetrical 6 with respect to any circle are transformed into symmetrical points with respect to the image of the circle. This last property adds three more points to determine the geometry of the transformation.

Two points P and P  are symmetrical with respect to a circle if they lie on a common ray passing through the center O of the circle, and the product of their distances from the center is equal to the square of the radius R of the circle: O P · O P  = R 2

6

7.4 The Linear-Fractional Function

123

Fig. 7.4 Conformal mapping of the inner point in the split cylinder in (x,y) plane and the points at u ≥ 0 in the plane (u, v)

The external points for a circle in the z-plane are mapped onto external points also for the circle on the w-plane.

7.4.1 The Split Cylinder: Two Opposite Half Cylinders at Different Potentials A cylindrical surface of infinite length and unitary radius, is divided in two opposite semicylindrical shells. Its axis lies on the z axis, as shown in Fig. 7.4, and the upper shell (y > 0) is at potential V0 while the lower one (y < 0) is at zero potential7 . We want to find the potential V (x, y) in the free space between the two shells8 . We have to solve the Laplace equation ΔV (x, y) = 0 with the boundary conditions the values of the potential on the semicylindrical surfaces. To find the solution we can look for a linear-fractional function which maps the points of the complex plane z inside a unit circle |z| < 1 onto the half-plane

(w) > 0 of the complex w = u + iv plane as in Fig. 7.4. The correspondence of three boundary points: z D = i → wD = 1 z A = 1 → wA = 0 z E = −i → w E = −1

7

Sveshnikov and Tikhonov [1], Sect. 6.2; Zangwill [5], Sects. 7.10.4–7.10.5, and Bernardini et al. [4], Sect. 1.5. 8 In Sect. 8.7.3 the potential is found by solving the Laplace equation using the separation of variables.

124

7 Conformal Mapping in Electrostatics

is enough, after some algebra, to determine the function: w = f (z) = i

1−z 1+z

This is a conformal mapping of the points inside the cylinder in the (x,y) plane onto the half plane w = u + iv with v ≥ 0 with the exclusion of the origin. The points of the upper half circle at potential V0 are associated to the positive semiaxis u > 0 and those of the lower part at zero potential to the negative semiaxis u < 0. The relation of the relevant points A-E is shown in Fig. 7.4. So if we find a solution V (u, v) of the Laplace equation with the given boundary conditions on the (u, v) plane, the same function after a transformation w = f (z) is also the solution V (x, y) on the (x, y) plane. Imposing the boundary conditions on the real axis V = 0 for u < 0 and V = V0 for u > 0 in the most general solution of the Laplace equation9 , the solution on the (u, v) plane is: V (w) = V (ρ, ϕ) = V0



  1 ϕ = V0 1 − arg w 1− π π

The conformal transformation is: w = u + iv = f (z) =

2y − i(x 2 + y 2 − 1) (x + 1)2 + y 2

and taking the imaginary part of V (x, y) = V (u, v) as the solution of the V(x,y) in the inner points of the cylinder, we have  π 1 − (x 2 + y 2 ) = (V0 − V ) ϕ = arctan 2y V0 

and the equipotential curves ϕ = constant or tan ϕ = c are the circles: 1 − (x 2 + y 2 ) =c 2y

x 2 + (y + c)2 = 1 + c2

having the center in the points (0, −c) and crossing the x axis in the points (−1, 0) e (1, 0). The half cylindrical equipotential surfaces are given for c = 0. For the upper one: 9

As seen in Sect. 8.7 of this book the most general solution in cylindrical coordinates is (8.35): V (ρ, ϕ) = (a0 + b0 log ρ)(A0 + B0 ϕ) + (cρ A + dρ−A )(a cos Aϕ + b sin Aϕ)

The boundary conditions in the (u, v) plane, V = V0 at ϕ = 0 and V = 0 at ϕ = π for any value of ρ, impose null values for b0 , c, and d, while a0 A0 = V0 and a0 B0 = −V0 /π.

7.5 The Schwarz-Christoffel Transformation

125

Fig. 7.5 Equipotential surfaces (solid lines) and field lines (dashed) for the half semicylinders (V0 > 0)

c → 0+

tan ϕ → 0+

ϕ=0

V = V0

tan ϕ → 0−

ϕ=π

V =0

and for the lower one: c → 0−

The analytic function10 having (w) = ϕ is: log w = log ρ + iϕ so the field lines on the (u, v) are given by ρ = const, normal to ϕ = const, which become on the (x, y) plane:  ρ = u 2 + v2 =

2y (x + 1)2 + y 2

2

 +

x 2 + y2 − 1 (x + 1)2 + y 2

2  21 = const

The equipotential and field lines for the split cylinder are shown in Fig. 7.5.

7.5 The Schwarz-Christoffel Transformation A Schwarz-Christoffel tranformation11 maps the upper half y > 0 of the z = x + i y plane onto the interior of a polygon in the w = u + iv plane (although the polygon need not be closed). The transformation is defined by the differential equation:

Given the function v = ϕ, the function u = log ρ is determined from the Cauchy-Riemann conditions (see Sect. 6.1). 11 Sveshnikov and Tikhonov [1], Sect. 6.4; Binns et al. [7], Chap. 6; Panofsky and Phillips [8], Sect. 4.6; Morse and Feshbach [2], Sect. 4.7. 10

126

7 Conformal Mapping in Electrostatics

Fig. 7.6 Schwarz-Christoffel transformation. Mapping of the (z) > 0 points in the z plane onto the interior of a polygon in the w plane n

dw = C1 (z − ai )βi dz i=1

(7.2)

where C1 is a constant and the ai are real numbers arranged in increasing order on the x axis, which are associated to the corners of the polygonal in the points Ai on the w plane as shown in Fig. 7.6. The transformation is analytic everywhere except at the points ai which are real and to be determined. Taking the argument of the relation (7.2) we have:  arg

dw dz

 = arg C1 + β1 arg(z − a1 ) + β2 arg(z − a2 ) + ...

(7.3)

+βn arg(z − an ). Moving along the real x axis for an interval dz = d x from the point ai , we find:  arg

dw dx

 = arg (dw) = tan

−1



dv du

 = θi

(7.4)

where θi is the angle between the direction of the side Ai Ai+1 of the polygonal with the real axis of the w plane. In the relation (7.3) for z on the real axis arg(z − ai ) = 0 if z > ai while arg(z − ai ) = π if z < ai and so from (7.3) and (7.4): θi = arg C1 + π(βi+1 + βi+2 + ... + βn ) θi+1 − θi = −π βi+1 . After replacing i + 1 → i the interior angle in the corner Ai is: αi = π + π βi

7.5 The Schwarz-Christoffel Transformation

127

taken from the direction Ai Ai+1 to the direction Ai−1 Ai as positive if clockwise and negative if counterclockwise. For the interior angles αi of a polygon with n (n ≥ 3) sides there is the condition: i=n 

αi = (n − 2) π.

(7.5)

i=1

For the ith corner we can write: βi =

αi −1 π

and the relation (7.2) becomes: n

αi dw = C1 (z − ai )( π −1) dz i=1

(7.6)

The transformation is found by integrating the (7.6):  z  α1 α2 αn (ζ − a1 )( π −1) (ζ − a2 )( π −1) · ... · (ζ − an )( π −1) dζ + C2 w = f (z) = C1 z0

(7.7) In this relation the modulus and the argument of C1 determine the relative scale and the orientation of the polygonal with respect to the real axis on the w plane while the constant C2 gives a traslation. We have to mention three important properties12 : (1) When mapping the upper half plane (y > 0) onto a given polygon with n corners Ai and angles αi in the w plane, it is possible to specify only three values of the ai , a j , ak on the real x axis, which are associated to three chosen corners Ai , A j , Ak of the polygon, the other constants in (7.7) are defined uniquely. (2) The angles αi are assumed positive. If an angle αk is negative, then for z → ai the integral (7.7) diverges. This means that the corner Ak lies at a point at infinity w = ∞. (3) If a point ai is at infinite, the corresponding term is not present in the product of terms in the integral (7.7). All the angles of the polygon are completely defined by those at the corners corresponding at finite points on the z plane; if a point at z = ±∞ corresponds with a corner, the angle at that corner is fixed by the remaining angles since the sum of the interior angles of a polygon is (n − 2) π where n is the number of corners13 .

12

For more details on these properties see Sveshnikov and Tikhonov [1], Sect. 6.4; Binns et al. [7], Sect. 6.2.1. 13 The number of factors in (7.6) is (n − 1) when the point z = ∞ corresponds with a corner and is n when z = ∞ corresponds to a point on the side of the polygon.

128

7 Conformal Mapping in Electrostatics

Fig. 7.7 Schwarz-Christoffel transformation for a single corner

7.5.1 Conformal Transformation for a Single Corner For a single corner transformation14 as shown in Fig. 7.7 the equation (7.6) becomes: dw α = C1 (z − a)( π −1) dz and after integration:

π α w = C1 (z − a) π + C2 α

The choice a = 0 places the point associated to the corner at the origin in the z plane and taking C2 = 0 the corner is fixed in the origin also in the w plane. If C1 is real the direction of the second side is coincident with the real axis in w plane and introducing the parameter μ: μ= we have the function:

α π

e

C3 = C1

π α

w = C3 z μ .

already analyzed15 in Sect. 6.3.2. For x > 0 we have the points on the side O A while for x < 0 those on the side O B. The polygonal with two corners and their interior angles α = π/2 gives the function w = sin z analyzed in Problems 7.2 and 7.3.

14

This configuration can be seen as a triangle with the corners corresponding to the points on the real axis of the z plane: O at z = a = 0 with interior angle α O = μπ, A at z = +∞ and B at z = −∞. 15 In Sect. 6.3.2 the angle α was the exterior angle equal to (2π − α) where α is the interior angle used here.

7.5 The Schwarz-Christoffel Transformation

129

Fig. 7.8 Conformal mapping of the inner space of a parallel-plate capacitor onto the space at edge of the capacitor

7.5.2 The Field Near the Edge of a Parallel-Plate Capacitor The fringing field at the edge of a parallel-plate capacitor can be described by the conformal transformation16 : z = f (w) =

h (1 + w + ew ) 2π

(7.8)

which associates the edge of a capacitor in the z = x + i y plane, shown in Fig. 7.8 right, to a capacitor with unlimited extended plates in the w = u + iv plane in Fig. 7.8 left. It is easy to verify that the real and the imaginary parts of the function (7.8): h h (1 + u + eu cos v) (v + eu sin v) y= x= 2π 2π map the plates of the capacitor at v = ±π in the (u, v) plane onto the edges of the plates a distance h apart in the (x, y) plane: for u : − ∞ → 0 for u :

0→∞

x : −∞→0 x : 0 → −∞

So the analytic function z = f (w) maps the equipotential surfaces and the field lines, well known for the parallel-plate capacitor, to those at the edge of the capacitor as shown in Fig. 7.9. The function (7.8) can be written as the combination of two Schwarz-Christoffel transformations17 . In the first transformation we consider the space between the plane 16

This function can be found in Maxwell [9], Vol. I, 202, Fig. XII; mentioned also in Zangwill [5], Sect. 7.10.6. 17 See for instance Sveshnikov and Tikhonov [1], Sect. 7.2.

130

7 Conformal Mapping in Electrostatics

Fig. 7.9 Electric field lines and equipotential surfaces near the edge of the parallel-plate capacitor

at the center of the capacitor (y = 0) and the edge of the upper plate18 at a distance h/2. For this we make the Schwarz-Christoffel transformation of the upper half plane of a ζ complex plane onto the z plane as in Fig. 7.10. To the points at ζ1 = 0, ζ2 = +∞, ζ3 = −1 on the ζ real axis, we associate on the plane z the corners A1 , A2 , A3 of a triangle in the points z 1 = −∞, z 2 = +∞, z 3 = i h/2 with the angles α1 = 0 and α3 = 2π while at x = +∞ the angle has to be α2 = −π from the sum of the interior angles in a triangle. The integral (7.7) becomes:  z = f (ζ) = C1

ζ

ζ0

ζ −1 (ζ + 1) dζ + C2

Assuming ζ0 = −1, the condition z = i h/2 for ζ = −1 determines C2 = i h/2. Moving from positive to negative values of the real axis on the plane ζ = ρ eiφ , along a semicircle of radius ρ around the origin, in the limit ρ → 0 the increase of the z variable is Δz = i h/2. From the previous integral, substituting ζ = ρ eiφ , we find: 

π

Δz = iC1

dφ = iC1 π = i

0

so: z=

h 2π



ζ

−1

h 2

ζ −1 (ζ + 1) dζ + i

C1 =

h 2π

h 2

and after integration the transformation is:

The field in the space between the central plane and the lower edge at y = −h/2 is symmetric to the upper one.

18

7.5 The Schwarz-Christoffel Transformation

131

Fig. 7.10 In the limit  → 0 the half plane (ζ) ≥ 0 is mapped onto the half plane y ≥ 0 at the edge of the capacitor

z=

h (1 + ζ + log ζ) 2π

(7.9)

The negative real semi-axis of the plane ζ corresponds to the upper plate of the capacitor at potential V while the positive part of the axis to the equipotential surface at zero potential at the center of the capacitor with the two plates at V and −V potentials. We have now to find a transformation which maps the inner space of an infinite plate capacitor, for which we know the potential surfaces and the field lines, onto the upper part of the ζ plane. The plate at V voltage and the plate at zero potential have to be mapped respectively on the negative and positive parts of the real ζ axis. This transformation is given by the well known function f (ζ) = ew = eu+iv , already used in Sect. 7.3.1, which maps the strip in the (u, v) plane between v = 0 and v = π, shown in Fig. 7.11, onto the upper part of ζ plane with the required conditions on the ζ axis. It is easy to see that this transformation can be found by a Schwarz-Christoffel transformation which, as shown in Fig. 7.11, maps the upper part of the plane ζ onto the space between the plates of an ideal capacitor on the plane w = u + iv with the potentials19 V at v = π and 0 at v = 0. Similarly to what was done in the first transformation, we can choose on the real axis of the plane ζ the points ζ1 = 0, ζ2 = +∞, ζ3 = −1 associated on the plane w to a triangle with the corners A1 , A2 , A3 in the points w1 = −∞, w2 = +∞, w3 = iπ with angles α1 = 0, α2 = 0, α3 = π. So the integral (7.7) becomes:  w = f (ζ) = C1

ζ ζ0

ζ −1 dζ + C2

Taking ζ0 = −1 corresponding to w3 = iπ it follows C2 = iπ and moving from the positive semiaxis to the negative one on a semicircle of radius ρ around the origin, in the limit ρ → 0 from Δw = iπ, we find C1 = 1 and the integral becomes:

Here we write the formulas for 0 ≤ v ≤ π. The potentials are given by v multiplied by a factor V /π.

19

132

7 Conformal Mapping in Electrostatics

Fig. 7.11 Conformal mapping of the half-plane (ζ) ≥ 0 onto the strip 0 ≤ (w) ≤ π on the plane w

 w = f (ζ) = and so:

ζ ζ0

ζ −1 dζ + iπ = log ζ

ζ = ew = eu (cos v + i sin v)

After substitution of this last relation in the (7.9) we finally find the transformation (7.8): h (1 + w + ew ) z = f (w) = 2π

Problems 7.1 Show that the function: f (z) =



z2 − a2

gives the potential and field lines for an infinite conductive strip wide 2a (crossing the (x,y) plane at y = 0 between x = −a and x = a) embedded in a otherwise uniform electric field parallel to the x axis. (It is interestig to find the induced charge density on the strip. See Zangwill [5], Sect. 7.10.3.) 7.2 Find the configurations of conductors mapped by the analytic function w = sin z between the planes w = u + iv and z = x + i y (−π/2 < x < π/2). 7.3 Show that the function w = sin z, analyzed in Problem 7.2, corresponds to a Schwarz-Christoffel transformation with two corners having right internal angles.

Solutions 7.1 It is easy to rewrite the function as f (z) = u(x, y) + iv(x, y).

Solutions

133

f (z) =



z2 − a2 =

ρ eiϕ

ρ eiϕ = (x + i y)2 − a 2 = x 2 − y 2 − a 2 + 2i x y ρ=

(x 2 − y 2 − a 2 )2 + 4x 2 y 2

tan ϕ = 1

ϕ

1

f (z) = u(x, y) + i v(x, y) = ρ 2 ei 2 = ρ 2 (cos

2x y x 2 − y2 − a2

ϕ ϕ + i sin ) 2 2

2 1 1  2 (x − y 2 − a 2 )2 + 4x 2 y 2 2 + (x 2 − y 2 − a 2 ) u(x, y) = √ 2

1

2 1 1  2 (x − y 2 − a 2 )2 + 4x 2 y 2 2 − (x 2 − y 2 − a 2 ) v(x, y) = √ 2

1

The function u(x, y) gives the potential while v(x, y) gives the field lines as shown in Fig. 7.12. For x = 0 the potential V (0, y) = u(0, y) = 0. For y = 0 and |x| < a the potential is zero. At large distance from the strip (|x| >> a or |y| >> a) the field becomes uniform E = −1 xˆ and the field lines are v(x, y) = y parallel to the x axis. 7.2 The function w = sin z can be written in the form: w = u + iv = sin z = sin x cosh y + i cos x sinh y u = sin x cosh y = sin x

e y + e−y 2

v = cos x sinh y = cos x

e y − e−y 2

It is easy to verify the mapping between the points in the two planes as reported in the table and shown in Figs. 7.13 and 7.14.

x y u v

A

B

C

+∞ +∞ 0

0 1 0

−∞ +∞ 0

π 2

π 2

π 2

P 0 +∞ 0 +∞

O 0 0 0 0

Q 0 −∞ 0 −∞

D − π2 +∞ −∞ 0

E − π2 0 -1 0

F − π2 −∞ −∞ 0

If we assume in the (x, y) plane the lines x = const (−π/2 < x < π/2) as equipotential surfaces (V = x), it is easy to recognize, as in Fig. 7.13, two parallel opposite sign conductive plates at x = π2 and x = − π2 with a uniform field E = −1 xˆ and field lines y = const  . In the u, v plane the two parallel plates are transformed in

134

7 Conformal Mapping in Electrostatics

Fig. 7.12 Conductive strip embedded in a otherwise uniform electric field

two coplanar opposite sign conductive half-planes at v = 0: one for u ≥ 1 and the other for u ≤ −1, separated by a distance Δu = 2. In the u, v plane (see Fig. 7.15) the equipotential surfaces are hyperbolas (x = const): v2 u2 =1 − 2 cos2 x sin x while the field lines are ellipses (y = const): u2 v2 + =1 cosh2 y sinh2 y Otherwise if we assume in the (x, y) plane the lines y = const for −π/2 ≤ x ≤ π/2 as equipotential segments, we have in the u, v plane an equipotential segment at v = 0 for −1 ≤ u ≤ 1 which represents an infinite equipotential strip in three dimensions as shown in Fig. 7.14. In this configuration the ellipses (y = const) in Fig. 7.15 are the equipotential surfaces and the hyperbolas (x = const) are the field lines. 7.3 For the Schwarz-Christoffel transformation we can assume a polygonal in the z-plane (x, y) with three corners A1 , A2 e A3 corresponding to the points on the real axis of the ζ plane: ζ1 = +∞ for A1 with angle α1 = 0, ζ2 = −a for A2 with α2 = π/2 and ζ3 = a for A3 with α2 = π/2. The equation (7.6) becomes:

Solutions

135

Fig. 7.13 Conformal mapping for the function w = sin z when the equipotentials are V = x. Two parallel opposite sign conductive plates in the (x, y) plane are transformed in two coplanar opposite sign conductive half-planes in the (u, v) plane

Fig. 7.14 Conformal mapping for the function w = sin z when in the (x, y) plane the equipotentials are V = y. In the (u, v) plane the equipotentials are the surfaces produced by an infinite strip which crosses the (u, v) plane at v = 0 for −1 ≤ u ≤ 1

dz C1 −i C1 1 1 = C1 (ζ + a)− 2 (ζ − a)− 2 = = 2 2 dζ ζ −a a2 − ζ 2 and after integration we find: z = −i C1 arcsin

ζ + C2 a

The choice to associate the point z = 0 to the point ζ = 0 and to limit the domain for z between −π/2 and π/2, gives C2 = 0 and C1 = i. Taking the inverse function: ζ = a sin z and exchanging the names of the variable ζ to w we have: w = a sin z The choice of the constants makes a correspondence between the points at z = ±π/2 and the points w = ±a.

136

7 Conformal Mapping in Electrostatics

Fig. 7.15 Equipotential surfaces and field lines for the conformal mapping of the function w = sin z. The equipotential surfaces V = x between the plates of a capacitor in the (x, y) plane, are transformed in hyperbolic equipotential surfaces for two coplanar opposite sign conductive halfplanes in the (u, v) plane. The equipotential surfaces V = y in the (x, y) plane become in the (u, v) plane the elliptic equipotential surfaces produced by an infinite strip along the z axis

References 1. A.G. Sveshnikov and A.N. Tikhonov, The Theory of Functions of a Complex Variable (MIR Publishers, 1973) 2. P.M. Morse, H. Feshbach, Methods of Theoretical Physics (McGraw-Hill Book C, Part I, 1953) 3. G.B. Arfken and H.J. Weber, Mathematical Methods for Physicists, 6th Edn. (Elsevier Academic Press, 2005) 4. C. Bernardini, O. Ragnisco, P. Santini, Metodi Matematici della Fisica (La Nuova Italia Scientifica, Roma, 1993) 5. A. Zangwill, Modern Electrodynamics (Cambridge University Press, Cambridge, 2013) 6. G.A. Erskine, Electrostatic Problems in Multiwire Proportional Chambers. Nucl. Inst. Methods 105, 565 (1972) 7. K.J. Binns, P.J. Lawrenson and C.W. Trowbridge, The Analytical and Numerical Solutions of Electric and Magnetic Fields (John Wiley & Sons, 1992) 8. W.K.H. Panofsky and M. Phillips, Classical Electricity and Magnetism (Addison-Wesley Pub., 1953) 9. J.C. Maxwell, A Treatise on Electricity and Magnetism (1873)

Chapter 8

Separation of Variables in Laplace Equation

The separation of variables is a powerful technique to solve the Laplace equation. After a brief introduction, the method is applied to the equation for the electrostatic potential by using systems of coordinates which appropriately exploit the geometrical symmetries of the configurations to be studied. We start with a simple problem in Cartesian coordinates and then we use the separation of variables in spherical coordinates with azimuthal symmetry and in cylindrical (polar) coordinates. These subjects offer the possibility to introduce some of the most common functions used in Physics. Sometime we reproduce the solutions already found in previous chapters by the image charge method or by conformal mapping.

8.1 The Method of Separation of Variables The electrostatic potential is given by the solution of the Poisson equation: ΔV = −

ρ 

where ρ is the charge density or, in charge free regions, by the solution of the Laplace equation: ΔV = 0

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 F. Lacava, Classical Electrodynamics, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-031-05099-2_8

137

138

8 Separation of Variables in Laplace Equation

with assigned boundary conditions1 . The boundary conditions fix on the surfaces the values of the potential V (Dirichlet conditions) or the values of the normal derivative ∂ V /∂n (Neumann conditions) or the two mixed (Dirichlet over part of surfaces and Neumann on the remaining part). These conditions determine the solution of the equations when fixed on a closed surface. A surface is closed if it surrounds completely the volume with the field. The surface can extend to infinity2 . The uniqueness of the solutions of the Poisson and Laplace equations can be easily proved. The solution of the Poisson or Laplace equations is easier when using curvilinear orthogonal coordinates3 which take advantage of the symmetries present in the geometry of the problems. In these coordinates the equations are simpler and also the boundary conditions can be more easily applied. In systems of orthogonal coordinates u, v, w, appropriate to the symmetry of the problems, the method of separation of variables easily gives the solution of the Laplace equation as the product of three functions of one variable: V (u, v, w) = Vu (u) Vv (v) Vw (w)

(8.1)

This technique splits the Laplace partial (second order) differential equation in three (second order) ordinary equations, one for each variable. When, as in the Laplace equation, in the equation there are no products of differential operators of different variables, the separation method usually works. In the separation process, three sets of non independent separation constants have to be introduced. For each equation the values of the separation constants are determined by the boundary conditions and are the eigenvalues of the equations. The associated eigenfunctions are the solutions of the equations and are a set of orthogonal and complete functions as mentioned in next section. Of course any linear combination of the solutions is also a solution of the equation. So the solution of the Laplace equation is the sum of the products of the solutions found for each of the three functions Vu (u), Vv (v), Vw (w) in (8.1) and we can write: V (u, v, w) =

 i

j

Ci jk Vu i (u) Vv j (v) Vwk (w)

k

where the coefficients Ci jk are determined by the boundary conditions. The examples presented in the following sections will clarify the process of variable separation. 1 For a discussion on the boundary conditions and on the uniqueness of the solutions see Jackson [1], Sect. 1.8–1.10 and references therein. For a more extensive analysis see: Morse and Feshbach [2], Chaps. 5 and 6. 2 To assign arbitrarily on a closed surface both the values of the potential and the values of the normal derivative (Cauchy conditions) is an overspecification of the problem and the solution generally does not exist. This is not the case for the boundary conditions on an open surface that does not enclose completely the field. 3 For the curvilinear orthogonal coordinates see Appendix A.

8.3 Separation of Variables in the Laplace Equation

139

8.2 Orthogonal and Complete Sets of Functions The set of eigenfunctions ϕk (ξ ), solution of an eigenvalue problem in one dimension with appropriate boundary conditions, is orthogonal and complete. Usually the functions are also normalized and the orthogonality relation is:  a

b

ϕk∗ (ξ ) ϕk (ξ ) dχ = δk  k

If the functions ϕk (ξ ) are defined in the interval (a, b) of ξ , for the completeness of this set of functions, any function F(ξ ), defined in the same interval and there square integrable, can be expanded in a series of the ϕk (ξ ) functions: F(ξ ) =



ak ϕk (ξ )

(8.2)

k

with the coefficients ak determined by the orthogonality of the ϕk (ξ ) functions. Multiplying the relation (8.2) by ϕk∗ (ξ ) and integrating over (a, b) we get: 

b

ak = a

ϕk∗ (ξ ) F(ξ ) dξ

Upon substitution of this expression in (8.2) we can write: F(ξ ) =

 k

a

b

ϕk∗ (ξ  ) ϕk (ξ ) F(ξ  )dξ 

and from this relation we find the completeness or closure relation for the ϕk (ξ ) functions:  ϕk∗ (ξ  ) ϕk (ξ ) = δ(ξ  − ξ ) k

8.3 Separation of Variables in the Laplace Equation In the following sections we introduce the method of separation of variables for the Laplace equation in the most common sets of orthogonal coordinates4 . In some simple configurations it will be possible to show that the solutions are the potentials

4

The method of separation of variables for the Laplace equation with many examples and problems can be found in many books. See for instance: Jackson [1], Chaps. 2 and 3; Griffiths [3], Sect. 3.3; Zangwill [4], Chap. 7; Stratton [5], Chap. 3; Smythe [6], Chap. 4; Panowfsky and Phyllips [7], Chap. 5; Arfken and Weber [8], Chap. 9.

140

8 Separation of Variables in Laplace Equation

already found by intuitive and practical approaches (image charges or conformal mapping).

8.4 Separation of Variables in Cartesian Coordinates The Laplace equation in Cartesian coordinates is: ∂2V ∂2V ∂2V + + =0 ∂x2 ∂ y2 ∂z 2

(8.3)

The method of separation of variables looks for a solution given by the product of three functions of a single variable: V (x, y, z) = X (x) Y (y) Z (z)

(8.4)

thus substituting (8.4) in (8.3) and dividing by (8.4) we get: 1 ∂ 2Y 1 ∂2 Z 1 ∂2 X + + = 0. 2 2 X ∂x Y ∂y Z ∂z 2 In this equation one variable can change while the other two are fixed, so we conclude that the equation can be split in the three ordinary second order equations: 1 d2 X = C1 X dx2

1 d 2Y = C2 Y dy 2

1 d2 Z = C3 Z dz 2

with the three separation constants C1 , C2 and C3 bound by the condition: C1 + C2 + C3 = 0

8.4.1 Box with a Side at Given Potential and Five Conductive Grounded Sides For example we want to find the potential inside the empty box of dimensions a, b and c, shown in Fig. 8.1, with a given potential V (x, y, 0) on the side at z = 0 and the other five conductive sides at zero potential. The boundary conditions are: a null potential on the sides at x = 0, x = a, y = 0, y = b and at z = c while on the side at z = 0 the potential is given by the function V (x, y, 0) = 0. The solution of the Laplace equation can be a function of the type (8.4) and the three ordinary equations are:

8.4 Separation of Variables in Cartesian Coordinates

141

Fig. 8.1 The box with a side at voltage V (x, y, 0) and the other five conductive sides at zero potential

1 d2 X = −k x2 X d x2

1 d2 Y = −k 2y Y d y2

1 d2 Z = k z2 Z d z2

with positive separation variables5 k x2 , k 2y and k z2 which have to satisfy the relation: k x2 + k 2y − k z2 = 0

.

The general solution for Z (z) is: Z (z) = A ekz z + B e−kz z with the coefficients A and B which can depend on x and y only through k z . Defining at z = 0 the coefficients Vkz = Z (0), functions only of k z , we have the two conditions: A + B = Vkz

A ekz c + B e−kz c = 0

and we find: Z (z) = Vkz

ekz z − ekz (2c−z) ekz (c−z) − e−kz (c−z) = V k z 1 − e2kz c ekz c − e−kz c Z (z) = Vkz

sinh[k z (c − z)] sinh[k z c]

(8.5)

The general solutions of the equations for X (x) and for Y (y) are: X (x) = C cos k x x + D sin k x x

k z = 0 has to be excluded because it implies both k x = 0 and k y = 0 and then a constant value for the potential. Negative values for k z2 would give not acceptable oscillating solutions along z.

5

142

8 Separation of Variables in Laplace Equation

Y (y) = E cos k y y + F sin k y y The conditions V = 0 at x = 0 and y = 0 give C = 0 and E = 0 while V = 0 at x = a and at y = b demand: nπ kx = a

mπ ky = b

k z = knm =



k x2

+

k 2y

=



nπ 2  mπ 2 + a b

with n and m integer positive numbers. The solutions of the equations for X (x) and Y (y) can be linear combinations of sinusoidal functions with any value for n and m: X (x) =

∞ 

dn sin

n=1

nπ x a

Y (y) =

∞ 

f m sin

m=1

mπ y b

which are Fourier expansions of the potential respectively along lines at fixed y and z and at fixed x and z. Thus by substitution of X (x), Y (y) and Z (z) in (8.4), the general solution for the Laplace equation is: ∞  ∞ 

V (x, y, z) =

Cnm

n=1 m=1

nπ x mπ y sinh[knm (c − z)] sin sin sinh[knm c] a b

(8.6)

where the constant coefficients Cnm = dn f m Vknm have to be determined. The solution (8.6) is a double Fourier series of the potential in the variables x and y on a plane at fixed z. The coefficients Cnm are fixed by the potential at z = 0: V (x, y, 0) =

∞  ∞  n=1 m=1

Cnm sin

mπ y nπ x sin a b

After multiplication of both members of the equation for: sin

mπ y nπ x sin a b

and integration on the side of the box at z = 0 (0 ≤ x ≤ a , 0 ≤ y ≤ b): 

a



0 ∞  ∞  n=1 m=1

b

V (x, y, 0) sin

0



a



b

Cnm

sin 0

0

nπ x mπ y sin dx dy = a b

nπ x nπ x mπ y mπ y sin sin sin dx dy a a b b

(8.7)

8.4 Separation of Variables in Cartesian Coordinates

143

from the orthogonality relations: 

a 0



nπ x nπ x a sin sin d x = δnn  a a 2

b

sin 0

mπ y mπ y b sin dy = δmm  b b 2

for the second member of (8.7) we have: ∞  ∞ 

Cnm

n=1 m=1

a b ab δnn  δmm  = Cn  m  2 2 4

and so, if n = n  and m = m  , the coefficients for the solution are given by the relation:  a  b mπ y 4 nπ x sin dx dy (8.8) V (x, y, 0) sin Cnm = ab 0 0 a b If V (x, y, 0) = V0 constant, the integrals give: 

a

sin 0



b

sin 0

nπ x dx = a

mπ y dy = b

thus: Cnm =

⎧ 2a ⎪ ⎪ ⎨ nπ ⎪ ⎪ ⎩ 0 ⎧ 2b ⎪ ⎪ ⎨ mπ ⎪ ⎪ ⎩ 0

for n odd for n even

for m odd for m even

4 2a 2b 16 1 V0 = 2 V0 a b nπ mπ π nm

for both n and m odd and: Cnm = 0

for n or m even.

Finally the potential in the box with one side at V0 is: V (x, y, z) =

∞ ∞ nπ x mπ y 16 V0   1 sinh[knm (c − z)] sin sin (8.9) 2 π n=1,3,5 m=1,3,5 nm sinh[knm c] a b

The ratio V /V0 on the plane at x = a/2 inside the box is shown in Fig. 8.2.

144

8 Separation of Variables in Laplace Equation

Fig. 8.2 The voltage ratio V /V0 on the plane at x = a/2 inside the box (a = 6, b = 6, c = 10) with 20 iterations

For a grounded pipe of infinite length, terminated at z = 0 by a plate at voltage V (x, y, 0), we can use the solution (8.6) in the limit c → ∞: lim

c→∞

sinh[knm (c − z)] = e−knm z sinh[knm c]

and the solution for the potential becomes: V (x, y, z) =

∞  ∞ 

Cnm e−knm z sin

n=1 m=1

mπ y nπ x sin a b

with the coefficients Cnm given by the relations (8.8), and if V (x, y, 0) = V0 constant also from (8.9) in the limit c → ∞: V (x, y, z) =

16 V0 π2

∞ 

∞ 

n=1,3,5 m=1,3,5

mπ y 1 −knm z nπ x e sin sin nm a b

(8.10)

The charge densities on the sides of an infinite pipe are the subject of Problem 8.1. If a grounded pipe of length c is terminated at z = 0 and z = c by two conductive plates at voltage V (x, y, 0) = V (x, y, c), we can use the solution (8.6) but in place of the relation (8.5) the new boundary conditions give for the function Z (z) the relation: Z (z) = Vknm

sinh[knm z] + sinh[knm (c − z)] sinh[knm c]

8.5 Separation of Variables in Spherical Coordinates with Azimuthal Symmetry

145

8.5 Separation of Variables in Spherical Coordinates with Azimuthal Symmetry The Laplace equation in spherical coordinates6 is: Δ V (r, θ, ϕ) =

1 ∂ r 2 ∂r





1 ∂ ∂V 1 ∂2V ∂V r2 + 2 sin θ + 2 2 =0 ∂r r sin θ ∂θ ∂θ r sin θ ∂ϕ 2

that in presence of azimuthal symmetry becomes: 1 ∂ Δ V (r, θ ) = 2 r ∂r





1 ∂ ∂V 2 ∂V r + 2 sin θ =0 ∂r r sin θ ∂θ ∂θ

(8.11)

Searching for a solution as a product of two separate functions, one of r and the other one of θ : V (r, θ ) = R(r ) Θ(θ ) we have two second order ordinary equations: d dr



dR r2 = k2 R dr

1 d sin θ dθ

sin θ

dΘ dθ

= −k 2 Θ

The second equation can be rewritten in the form: d d cos θ d dz



dΘ sin2 θ + k2 Θ = 0 d cos θ



dΘ (1 − z 2 ) + k2 Θ = 0 dz (1 − z 2 )

z = cos θ

d2 Θ dΘ + k2 Θ = 0 − 2z 2 dz dz

and with the substitution k 2 = l(l + 1) this equation becomes: (1 − z 2 )

d2 Θ dΘ + l(l + 1) Θ = 0. − 2z dz 2 dz

(8.12)

The solutions7 are the Legendre polynomials Pl (cos θ ) = Pl (z) given by the Rodrigues formula:

6 7

See Appendix A See for instance: Gradshteyn and Ryzhik [9] 8.910.1-2, 8.912, and also 8.820.6, 7.112.1.

146

8 Separation of Variables in Laplace Equation

1 Pl (cos θ ) = Pl (z) = l 2 l!



d dz

l (z 2 − 1)l

These are normalised so as Pl (1) = 1 and are orthogonal over the interval [−1, 1]: 

1

−1

Pi (z) P j (z) dz =

2 δi j 2i + 1

(8.13)

The first six polynomials are: P0 (z) = 1 P3 (z) =

P1 (z) = z

1 (5 z 3 − 3 z) 2 P5 (z) =

P2 (z) =

1 (3 z 2 − 1) 2

1 (35 z 4 − 30 z 2 + 3) 8

P4 (z) =

1 (63 z 5 − 70 z 3 + 15 z) 8

If the function Θ(θ ) which satisfies the equation (8.12) is a Legendre polynomial Pl (z), the associated equation for R(r ) becomes: d dr



dR r2 = l(l + 1) R dr

It is straightforward to verify that a function R = A r α is a solution of this equation if α = l or α = −(l + 1). So for a given l the solution is a linear combination of two functions: Bl Rl (r ) = Al r l + l+1 r and the most general solution for the Laplace equation (8.11) is: V (r, θ ) =

∞  l=0

Rl (r ) Pl (cos θ ) =

∞   l=0

Al r + l

Bl r l+1

 Pl (cos θ )

(8.14)

8.5.1 A Dielectric Sphere in a Uniform Electric Field We want to study the field for a dielectric sphere of radius Rs and permittivity 2 , surrounded by a dielectric of permittivity 1 , embedded in a uniform electric field E = E e zˆ and potential V = −E e r cos θ . We write the most general solution of the Laplace equation inside the sphere:

8.5 Separation of Variables in Spherical Coordinates with Azimuthal Symmetry

Vin (r, θ ) =

∞  

Al r +

l=0

while outside:

Bl

l

147

 Pl (cos θ )

r l+1

 ∞   Dl l Vout (r, θ ) = Cl r + l+1 Pl (cos θ ) r l=0

where Al , Bl , Cl and Dl have to be determined by the boundary conditions. At large distance from the sphere, the potential has to be Vout (r >> R, θ ) = −E r cos θ thus: and Cl = 0 for l > 1 C1 = −E e while C0 is a constant. The inclusion of the point at r = 0 in Vin implies Bl = 0 for any l. At the spherical border between the two dielectrics we have to apply the continuity relations: Vin (Rs , θ ) = Vout (Rs , θ ) (or E in = E out that is E in θ = E outθ ) and Din⊥ = Dout⊥ that is Dinr = Doutr . So from Vin (Rs , θ ) = Vout (Rs , θ ) we have: ∞  l=0

Al Rsl

 ∞   Dl l Cl Rs + l+1 Pl (cos θ ) Pl (cos θ ) = Rs l=0

(8.15)

and using the orthogonality of the Legendre polynomials (8.13): 

1 −1

Pl (z) Pl  (z) dz =

we get for l = 1: A1 = C 1 + and for l = 1: Al =

2 δll  2l + 1

D1 Rs3

(8.16)

Dl

(8.17)

Rsl+1

From the continuity relation Dr1 = Dr2 we have:  ∞ ∞      Dl l−1 (l + 1) l+2 Pl (cos θ ) l Al R s Pl (cos θ ) = 1 C1 cos θ − 1 2 Rs l=0 l=0 and from the orthogonality of the Legendre polynomials, for l = 1: 2 A1 = 1 C1 − 21

D1 Rs3

(8.18)

148

8 Separation of Variables in Laplace Equation

and for l = 1: 2 l Al Rsl−1 = −1 (l + 1) from (8.17) and (8.19):

Dl Rsl+2

→

Dl = −

2 l Al Rs2l+1 1 l + 1

(8.19)



2 l l R =0 Dl 1 + 1 l + 1 s

and so Dl = 0 and Al = 0. From (8.16) and (8.18): ⎧ D1 ⎪ ⎪ ⎨ A1 = C 1 + 3 Rs D1 ⎪ ⎪ ⎩2 A1 = 1 C1 − 21 3 Rs

A1 = −

31 Ee 2  1 + 2

D1 =

 2 − 1 3 R Ee 2 − 2 1 s

For l = 0 we have A0 = 0 and D0 = 0 and from (8.15): C 0 = A0 −

D0 =0 Rs

So the potentials inside and outside the sphere are: Vin = −

3 1 3 1 E e r cos θ = − Ee z 2  1 + 2 2  1 + 2

Vout = −E e r cos θ +

2 − 1 Rs3 E e cos θ 2 + 2 1 r 2

(8.20)

(8.21)

Note that the field inside the sphere is uniform and oriented in the z direction: Ez = −

3 1 ∂ Vin = Ee ∂z 2  1 + 2

(8.22)

and the potential outside is the superposition of the external field E e and the field from an electric dipole p: Vout = −E e r cos θ +

1 p cos θ 4π 1 r 2

p = 4π 1

2 − 1 3 R Ee 2 + 2 1 s

(8.23)

8.5 Separation of Variables in Spherical Coordinates with Azimuthal Symmetry

149

Fig. 8.3 Field lines for a dielectric sphere of permittivity 2 in a medium of permittivity 1 where a uniform electric field is present for a ratio 2 /1 = 5

Fig. 8.4 Field lines for a dielectric sphere of permittivity 2 in a medium of permittivity 1 where a uniform electric field is present for a ratio 2 /1 = 0.2

The dielectric sphere in a dielectric medium was already discussed in Sect. 4.3. The field lines for two configurations with ratios of permittivity 2 /1 = 5 and 2 /1 = 0.2 are shown in Figs. 8.3 and 8.4. The shape of the field lines depends on the relative polarization of the media as pointed out in the comments to Fig. 4.3 in Sect. 4.3. - Dielectric sphere in a uniform electric field in vacuum We can use formulas (8.20), (8.22) and (8.23) with 1 = 0 and 2 =  and we have: Vin = −

3 0 Ee z 2 0 + 

Ez =

3 0 Ee 2 0 + 

150

8 Separation of Variables in Laplace Equation

Fig. 8.5 Field lines for a conductive sphere in a uniform electric field

Fig. 8.6 Field lines for a spherical cavity in a medium of dielectric constant r = 2 in presence of a uniform electric field

Vout = −E e r cos θ +

1 p cos θ 4π 0 r 2

p = 4π 0

 − 0 R3 Ee  + 2 0 s

- Conductive sphere in a uniform electric field For the conductive sphere in vacuum we put 2 = ∞ and 1 = 0 and relations (8.20) and (8.23) give: Vin = 0

Vout = −E e r cos θ +

1 p cos θ 4π 0 r 2

p = 4π 0 Rs3 E e

that is the solution already found by the image charge method in Sect. 3.4. The field lines for a conductive sphere in a uniform electric field are shown in Fig. 8.5.

8.5 Separation of Variables in Spherical Coordinates with Azimuthal Symmetry

151

- Hollow spherical cavity in a dielectric medium with a uniform electric field For 1 =  and 2 = 0 we have: Vin = −

3 Ee z 2  + 0

Vout = −E e r cos θ +

Ez =

1 p cos θ 4π  r 2

3 Ee 2  + 0

p = 4π 

0 −  R3 Ee 0 + 2  s

The field lines for a spherical cavity in a dielectric medium of dielectric constant r = 2 in presence of a uniform external electric field are shown in Fig. 8.6.

8.5.2 Potential for a Charged Ring The general solution (8.14) can be used to find the potential for a ring of radius R with uniform charge density λ placed in the z = 0 plane and with its center on the z axis as shown in Fig. 8.7b. The solution has to extend over the full range of r . At large r the solution cannot include r l terms while to extend the solution at r = 0 we have to excludes 1/r l+1 terms. We can write two separate solutions for r < R and for r > R and then connect the two solutions at the distance r = R as suggested by the geometry of the problem. It is useful to write (8.14) as a function of the ratio r/R: for r ≤ R: ∞  r l  al Pl (cos θ ) (8.24) V< (r, θ ) = R l=0 and for r ≥ R: V> (r, θ ) =

∞  l=0

l+1 R bl Pl (cos θ ) r

(8.25)

Matching the two solutions at r = R gives the relations: al = bl . The uniqueness of the solution of the Laplace equation suggests sometime to determine the coefficients al and bl from boundary conditions on the potential in limited regions where the potential is known. For the charged ring we can use the potential on its axis as a function of the position r (0, 0, z): V (z) =

1 1 Q Q = √ 2 2 4π 0 z + R 4π 0 |r − r |

Q = 2π R λ.

(8.26)

152

8 Separation of Variables in Laplace Equation

Fig. 8.7 Distance between two points: (a) in general as in relation (8.27) and (b) for a point on the z axis from a point on the ring of radius R as in (8.26)

where r is the position of any point on the ring. The expansion of the inverse of the distance8 between two points located at r and r as Fig. 8.7a, is: ∞  1 r where γ is the angle between the two vectors r and r and r< (r> ) is the smaller (larger) between |r| and |r |. Upon substituting in (8.26) the expansion (8.27) with γ = π/2, the potential becomes: ∞ Q  zl Pl (0) for z < R V< (z) = 4π 0 l=0 R l+1 for z > R

∞ Q  Rl V> (z) = Pl (0) 4π 0 l=0 z l+1

and comparing these two expressions with (8.24) and (8.25) on the z axis (at cos θ = 1), where for the normalisation of the Legendre polynomials Pl (1) = 1, we have: for z < R

V< (z) =

∞ 

al

 z l

l=0

for z > R

V> (z) =

∞  l=0

See Jackson [1], Sect. 3.3.

∞ Q  zl Pl (0) 4π 0 l=0 R l+1

l+1 ∞ R Q  Rl bl = Pl (0) z 4π 0 l=0 z l+1

thus for the coefficients al and bl we find: 8

R

=

8.5 Separation of Variables in Spherical Coordinates with Azimuthal Symmetry

al = bl =

Q Pl (0) 4π 0 R

153

(8.28)

and only coefficients with even l are non null because9 Pl (0) = 0 for odd l. Finally the potential for a ring is: for r < R

∞  rl Q V< (r, θ ) = Pl (0) Pl (cos θ ) 4π 0 l=0,2,4 R l+1

for r > R

∞  Rl Q Pl (0) Pl (cos θ ) V> (r, θ ) = 4π 0 l=0,2,4 r l+1

This potential can also be found from the charge distribution on the ring as in Problem 8.4. The first two terms of the last series give the potential: V (r, θ ) =

Q 1 R2 Q 1 + (1 − 3 cos2 θ ) 4π 0 r 4π 0 4 r 3

which is the sum of the potentials from the total charge and from the quadrupole moment of the ring found by an expansion in terms of multipoles in Problem 2.4.

8.5.3 Conducting/Dielectric Sphere and Point Charge It is interesting to find the potential for a conductive or dielectric sphere and a point charge. In this section we consider a conductive sphere of radius R at potential VR , centered in the origin of the axes, in presence of a point charge Q at distance z > R on the z-axis as shown in Fig. 8.8. The potential for a dielectric sphere and a point charge is a simple extension and is studied in Problem 8.6. The symmetry of the system suggests to use spherical coordinates with azimuthal symmetry. We expect for the potential a superposition of the potential VQ from the point charge and a potential Vsph (r, θ ) that depends on the charge distribution on the sphere: V (r, θ ) = VQ (r, θ ) + Vsph (r, θ ) The potential of the charge Q expanded in spherical harmonics was already given in (8.27) and for r < z it is:

The polynomials Pl (cos θ) for odd l are linear combination of powers of cos θ so for odd l Pl (0) = 0.

9

154

8 Separation of Variables in Laplace Equation

Fig. 8.8 A conductive sphere and a point charge Q at distance z from the center

VQ =

∞ 1 Q Q  rl Pl (cos θ ) = 4π 0 |r − r | 4π 0 l=0 z l+1

(8.29)

where r (0, 0, z) is the position of the charge Q. For the potential Vsph (r, θ ) we can start from the most general solution of the Laplace equation in terms of Legendre polynomials given by the relation (8.14) with Al = 0 for any l in order to have a finite solution at infinity. We can write: Vsph (r, θ ) =

∞  Bl Pl (cos θ ) r l+1 l=0

with the coefficients Bl to be determined. So for the potential V (r, θ ) we can write: V (r, θ ) =

∞ ∞  Q  rl Bl P (cos θ ) + Pl (cos θ ) l l+1 4π 0 l=0 z r l+1 l=0

that on the surface of the sphere (r = R) becomes: VR =

 ∞   Q Rl Bl Pl (cos θ ) + 4π 0 z l+1 R l+1 l=0

(8.30)

Of course the potential VR on the sphere cannot depend on θ so the coefficients of the Pl (cos θ ) have to be null10 for any l = 0 and we get the relations: This follows also from a direct calculation. By multiplying the (8.30) by Pl  (cos θ) and its first member by P0 (cos θ) = 1 we have:

10

8.5 Separation of Variables in Spherical Coordinates with Azimuthal Symmetry

VR =

B0 Q 1 + 4π 0 z R

→

Q Rl Bl + l+1 = 0 4π 0 z l+1 R

B0 = −

→

155

Q R + VR R 4π 0 z

Bl = −

Q R 2l+1 4π 0 z l+1

and the potential for r < z is: V (r, θ ) =

 ∞  R Q  rl R 2l+1 1 Pl (cos θ ) VR + − r 4π 0 l=0 z l+1 z l+1 r l+1

In this last formula the potential of the charge point Q is expanded as in (8.29) at r < z, without expansion the general solution for r ≥ R is: ∞ Q  R 2l+1 1 1 R Q − V (r, θ ) = VR + Pl (cos θ ) r 4π 0 |r − r | 4π 0 l=0 z l+1 r l+1

The surface charge density on the sphere is:  ∞ Q  ∂ V  0 V R R l−1 − = (2l + 1) Pl (cos θ ) σ (θ ) = 0 E R (R, θ ) = −0 ∂r r =R R 4π l=0 z l+1 (8.31) The total charge is:  π  2π σ (θ ) R 2 sin θ dθ dϕ qtot = 0

0

From the integral (8.13) when i = 0 and j = l = 0 (also in last footnote): 

π

Pl (cos θ ) sin θ dθ = 0

0

so only the term l = 0 in (8.31) contributes to the integral of σ (θ ) on the spherical shell and we find: 

π

VR



π

P0 (cos θ)Pl  (cos θ) d cos θ =

0

0



 Q Rl Bl + Pl (cos θ)Pl  (cos θ) d cos θ 4π 0 z l+1 R l+1

from the normalisation of the polynomials: VR and for l  = 0:

  2 2 Rl Q Bl  = + δ δll  0l 2l  + 1 4π 0 z l+1 R l+1 2l + 1 

Q Rl Bl + l+1 4π 0 z l+1 R

 =0

156

8 Separation of Variables in Laplace Equation



π

qtot = 0



2π

σ (θ ) R 2 sin θ dθ dϕ = 4π 0 RVR − Q

0

R z

(8.32)

If the sphere is connected to ground VR = 0 and qtot = −Q R/z = q  is the image of the charge Q inside the sphere as found in Sect. 3.3 and the potential becomes:  ∞  Q  rl R 2l+1 1 V (r, θ ) = − l+1 l+1 Pl (cos θ ) 4π 0 l=0 z l+1 z r where the first term is the potential (8.29) of the charge Q in the position z while the second term written in the form: 1 V (r, θ ) = 4π 0 



R −Q z

 ∞ 2 l R 1 P (cos θ ) l+1 l z r l=0

(8.33)

is the expansion in spherical harmonics (8.27) at r > R 2 /z for the potential of the image charge q  at the distance z = R 2 /z from the center of the conductive sphere as found in Sect. 3.3. For a null net charge qtot = 0, from (8.32) we have the potential of the sphere: VR =

1 Q 4π 0 z

given by the opposite of the image charge located in the center of the sphere as seen in Sect. 3.3. Finally the potential for r ≥ R is: Vout (r, θ) =

Q 1 1 + 4π 1 |r − r | 4π 1

∞ l

1 Pl (cos θ) R  R2 R 1 + −Q Q z z r l+1 4π 1 z r l=0

(8.34) the superposition of the potentials from the charge Q, from the image charge q =   R 2  −Q z in the position R /z, as pointed out in (8.33), and from a charge −q in the center of the uncharged sphere.

8.6 Separation of Variables in Spherical Coordinates The Laplace equation in spherical coordinates is: Δ V (r, θ, ϕ) =

1 ∂ r 2 ∂r





1 ∂ ∂V 1 ∂2V ∂V r2 + 2 sin θ + 2 2 =0 ∂r r sin θ ∂θ ∂θ r sin θ ∂ϕ 2

8.6 Separation of Variables in Spherical Coordinates

157

and we want to find a solution V (r, θ, ϕ) = R(r ) Θ(θ ) Φ(ϕ) that is the product of three functions of one variable. As in the case of azimuthal symmetry, we can separate the radial part of the equation by writing:

d 2 dR r = l(l + 1) R dr dr so the angular part is: 1 ∂ sin θ ∂θ



∂Y 1 ∂ 2Y sin θ + = −l(l + 1) Y ∂θ sin2 θ ∂ϕ 2

where Y (θ, ϕ) = Θ(θ )Φ(ϕ). Then we separate this equation in two ordinary differential equations: d 2Φ = −m 2 Φ dϕ 2 1 d sin θ dθ



  dΘ m2 sin θ + l(l + 1) − Θ=0 dθ sin2 θ

The last equation, substituting z = cos θ , becomes:   d dΘ m2 Θ=0 (1 − z 2 ) + l(l + 1) − dz dz 1 − z2 which is the generalized Legendre equation11 . The solutions of this equation over the interval −1 ≤ z ≤ 1 are the associated Legendre functions12 Plm (z) with l integer and positive and m integer between −l and l. Using the Rodrigues formula, these functions are: m (−1)m m d (1 − z 2 ) 2 m (z 2 − 1)l Plm (z) = l 2 l! dz with the orthogonality relation13 : 

1 −1

Plm (z) Plm (z) dz =

2 (l + m)! δll  2l + 1 (l − m)!

The normalized solutions for Φ(ϕ) are: 1 Φm (ϕ) = √ eimϕ 2π 11

Gradshteyn and Ryzhik [9] 8.700.1, 8.81. Gradshteyn and Ryzhik [9] 8.752-1. 13 Gradshteyn and Ryzhik [9] 7.112.1. 12

158

8 Separation of Variables in Laplace Equation

with m integer to have the same value for ϕ and ϕ + 2π . The orthogonality condition is:  2π  2π 1  Φm∗  Φm dϕ = e−im ϕ e−imϕ dϕ = δmm  2π 0 0 thus we can introduce the spherical harmonics Ylm (θ, ϕ) = Plm (z) Φm (ϕ):  Ylm (θ, ϕ) =

2l + 1 (l − m)! imϕ m e Pl (z) 4π (l + m)!

with the orthogonality relation:  4π

and the relation:

Yl∗ m  (θ, ϕ) Ylm (θ, ϕ) dΩ = δll  δmm 

∗ (θ, ϕ) Yl −m (θ, ϕ) = (−1)m Ylm

We have already introduced the spherical harmonics in the multipole expansion in the Appendix to Chap. 2.

8.7 Separation of Variables in Polar (Cylindrical) Coordinates For two-dimensional configurations infinitely extended in the other dimension, as for a wire, conductive corners or cylindrical shells, the invariance along a direction (usually the z axis) suggests the use of polar (cylindrical) coordinates. The Laplace equation given in Appendix A becomes: 1 ∂ ρ ∂ρ



∂V 1 ∂2V ρ + 2 =0 ∂ρ ρ ∂ϕ 2

and it can be solved using the separation of variables. By the substitution: V (ρ, ϕ) = R(ρ)Φ(ϕ) and upon multiplication for ρ 2 /V we can write the two equations: d ρ dρ



dR ρ dρ

= A2 R

d 2Φ = −A2 Φ dϕ 2

8.7 Separation of Variables in Polar (Cylindrical) Coordinates

159

For A2 < 0 the solutions are exponential in ϕ and not acceptable for our problems, so we can have only A2 ≥ 0. It is easy to show that the solutions are: for A2 = 0

R(ρ) = a0 + b0 log ρ and Φ(ϕ) = A0 + B0 ϕ

for A > 0

R(ρ) = c ρ A + d ρ −A and Φ(ϕ) = a cos Aϕ + b sin Aϕ.

2

Thus the most general solution is:   V = (a0 + b0 log ρ)(A0 + B0 ϕ) + c ρ A + d ρ −A (a cos Aϕ + b sin Aϕ) (8.35)

8.7.1 Wire and Cylindrical Capacitor The electric field around a charged wire or between the conductive shells of a cylindrical capacitor presents a symmetry around an axis. This feature can be exploited using cylindrical coordinates. The exclusion of a ϕ dependence demands A2 = 0 and B0 = 0 in the solution (8.35). Therefore the potential is: V (ρ) = A0 a0 + A0 b0 log ρ with the coefficient A0 b0 = −λ/2π 0 where λ is the linear density charge and A0 a0 an additive constant potential.

8.7.2 Conducting Wedge (Corner) We have already studied the potential for a conductive corner in Sects. 6.3.2 and 7.5.1. Now we want to find the potential from the solution (8.35) of the Laplace equation. For a conductive wedge of angle14 α, as in Fig. 8.9, we have to consider the solutions for A2 ≥ 0. To include in the solution the point at ρ = 0 we have to take b0 = 0 and d = 0. To have the same potential V0 for any value of ρ at ϕ = 0 and at ϕ = α, the other coefficients have to be a = 0, a0 B0 = 0, a0 A0 = V0 and Aϕ = nπ

A=

nπ = nμ α

where μ is the coefficient defined in 6.5. So the solution is reduced to: V = V0 +

∞ 

Vn ρ nμ sin nμϕ.

n=1

14 In Sect. 6.3.2 the angle α was the exterior angle equal to (2π − α) where α is the interior angle used here and in Sect. 7.5.1.

160

8 Separation of Variables in Laplace Equation

Fig. 8.9 Two plates with a corner of interior angle α

It is evident that terms with n > 1 are of no interest for a simple conductive wedge. Only the term with n = 1 is left and so: A=μ=

π α

and finally the solution is: V = V0 + V1 ρ μ sin μϕ that was already found in Sects. 6.3.2 and 7.5.1.

8.7.3 Split Cylinder We want to find the potential inside a split cylinder, that is an infinite cylinder of radius R, around the z axis with one half circular shell (0 < ϕ < π at potential V0 and the other half shell (π < ϕ < 2π ) at −V0 as shown in Fig. 8.10. The two shells are separated by a very thin strip of insulator. We have already considered this potential by means of conformal mapping in Sect. 7.4.1. For the potential outside the split cylinder see Problem 8.7. We start from the general solution (8.35) of the Laplace equation in cylindrical coordinates. To include the points at ρ = 0 we have to exclude the terms with log ρ and with ρ −A (A > 0). The boundary conditions on the potential at ρ = R, V0 for

Fig. 8.10 The split cylinder with the two shells at V0 and at −V0 potentials

8.7 Separation of Variables in Polar (Cylindrical) Coordinates

161

0 < ϕ < π and −V0 for π < ϕ < 2π exclude in the solutions terms proportional to ϕ or with cos Aϕ, thus the solutions have to be of the type: V (ρ, ϕ) = C ρ A sin Aϕ that we can write: V (ρ, ϕ) = C A

 ρ A R

sin Aϕ .

The boundary condition V (A(ϕ + π )) = −V (Aϕ) determines A = n odd integer, thus the solution is the sum: V (ρ, ϕ) =

∞ 

Cn

 ρ n

sin nϕ

R

n=1,3,5

with the coefficients Cn to be determined from the boundary condition on V (ρ, ϕ) at ρ = R. After multiplication by sin mϕ and integration over the interval [0, 2π ]: 

π

 V0 sin mϕ dϕ −

2π

π

0



π

2



∞ 

V0 sin mϕ dϕ =

n=1,3,5 ∞ 

V0 sin mϕ dϕ =

0

2V0 Cn = π

2π

Cn

sin nϕ sin mϕ dϕ 0

Cn π δnm

n=1,3,5



π

sin nϕ dϕ =

0

we find: Cn =

2V0 (− cos nϕ)π0 nπ

4V0 nπ

thus the solution is: 4V0 V (ρ, ϕ) = π

∞ ∞  1  ρ n 1  ρ 2k−1 4V0  sin nϕ = sin(2k − 1)ϕ n R π k=1 2k − 1 R n=1,3,5

and from the sum of the series15 : ρ 2 sin ϕ 2V0 −1 R tan V (ρ, ϕ) = π ρ2 1− 2 R 15

See for instance Gradshteyn and Ryzhik [9] 1.448.3.

162

8 Separation of Variables in Laplace Equation

Fig. 8.11 The equipotentials for the split cylinder. Shell at voltage V0 upper thick, shell at voltage −V0 lower thick; equipotentials inside and outside: dashed for |V | = 45 V0 , dashed-dotted for |V | = 21 V0 , dotted for |V | = 14 V0 and thin continuous for V = 0

An equipotential surface of voltage V is associated to the value of the parameter a:

ρ 2 sin ϕ Vπ = R 2 a = tan α = tan 2V0 ρ 1− 2 R

and in Cartesian coordinates its equation is: x 2 + y2 + 2

R y = R2. a

This last equation describes  circles on the plane z = 0, which cross the x axis at x = ±R, with radii Rc = R 1 + 1/a 2 and centers on the y axis at yc = −R/a, negative if a > 0 and positive if a < 0. This equation is equal to the solution16 found in Sect. 7.4.1 by a conformal mapping through a Möbius transformation where the parameter c is just the inverse of the parameter a used in this section. For V = V0 : α = π/2, a = +∞, Rc = R and yc = 0− , that is the positive half cylinder; for V = −V0 : α = −π/2, a = −∞ Rc = R and yc = 0+ that is the negative half cylinder. √ For V = V0 /2: α = π/4, a =√1, Rc = R 2 and yc = −R, while for V = −V0 /2: α = −π/4, a = −1 Rc = R 2 and yc = R. For V = 0± : a = 0± , Rc = ∞ and yc = ∓∞. Some equipotentials inside the cylinder are shown in Fig. 8.11.

16

The potentials assigned in Sect. 7.4.1 to the two shells of the split cylinder can be reproduced by the present solution by the substitution V0 → V0 /2 and a0 A0 = V0 /2.

8.7 Separation of Variables in Polar (Cylindrical) Coordinates

163

8.7.4 Cylinder in a Uniform Electric Field An infinite conducting cylinder of radius r1 is surrounded by a layer of dielectric of external radius r2 , characterised by a dielectric constant r . We want to find the solution17 for the Laplace equation when the grounded cylinder, with its axis on the z axis, is embedded in an external initially uniform electric field E = E 0 xˆ in vacuum with potential V = −E 0 x = −E 0 ρ cos ϕ. In the general solution of the Laplace equation (8.35) only terms with A2 > 0 have to be considered18 : V (ρ, ϕ) = (c ρ A + d ρ −A )(a cos Aϕ + b sin Aϕ). Terms with sin Aϕ are excluded because of the symmetry with respect to the x axis. A has to be integer because the terms with cos Aϕ have to be periodic. Outside the dielectric, for ρ > r2 , only the term ρ A with A = 1 can be considered because for ρ → ∞ the potential is V = −E 0 x = −E 0 ρ cos ϕ. Thus for the outer potential (ρ > r2 ) we expect: V0 (ρ, ϕ) = −E 0 ρ cos ϕ +

∞ 

An ρ −n cos nϕ

n=1

where An = ad. Inside the dielectric both terms ρ n and ρ −n have to be considered for the potential and we can write: Vi (ρ, ϕ) =

∞ 

(Bn ρ n + Cn ρ −n ) cos nϕ

n=1

To determine the coefficients An , Bn and Cn we use now the boundary conditions on the potential and on the electric fields E in the dielectric and E 0 in the vacuum: E 0ρ = r E ρ and E 0ϕ = E ϕ . At ρ = r2 we have: E 0 cos ϕ +

∞ 

n An r2−(n+1) cos nϕ = r

n=1

− E 0 sin ϕ +

∞ 

n (−Bn r2n−1 + Cn r2−(n+1) ) cos nϕ

n=1 ∞  n=1

n An r2−(n+1) sin nϕ =

∞ 

(8.36) n (Bn r2n−1 + Cn r2−(n+1) ) sin nϕ

n=1

(8.37) 17

See also Smythe [6], Chap. 4.01-3. In the solution we have to exclude a term proportional to ϕ because the field lines have to be symmetrical with respect to the plane x = 0 and a radial term dependent on log ρ in absence of a central source.

18

164

8 Separation of Variables in Laplace Equation

and at ρ = r1 the condition on the potential is: Vi =

∞ 

(Bn r1n + Cn r1−n ) cos nϕ = 0

(8.38)

n=1

If we multiply both sides of the relations (8.36) and (8.38) by cos mϕ and of the relation (8.37) by sin mϕ and integrate from 0 to 2π , using the results: 

2π

 cos nϕ cos mϕ dϕ = π δmn

0

for n = 1 these give:

2π

sin nϕ sin mϕ dϕ = π δmn

0

An r2−n = r (−Bn r2n + Cn r2−n ) An r2−n = Bn r2n + Cn r2−n Bn r1n + Cn r1−n = 0

Subtracting the first relation to the second one: (1 + r ) Bn = (r − 1) Cn r2−2n and by using the third relation, we find: −(1 + r ) Cn r12n = (r − 1) Cn r2−2n which is verified only if Cn = 0 and thus also Bn = 0 and An = 0. For n = 1 the three relations (8.36), (8.37) and (8.38) become: E 0 + A1 r2−2 = r (B1 + C1 r2−2 ) −E 0 + A1 r2−2 = B1 + C1 r2−2 B1 r1 + C1 r −1 = 0 which give: A1 = E 0 r22

r22 (r − 1) + r12 (r + 1) r22 (r + 1) + r12 (r − 1)

8.7 Separation of Variables in Polar (Cylindrical) Coordinates

165

Fig. 8.12 Field lines for a conductive cylinder surrounded by a layer of dielectric with r = 3 in a uniform electric field

B1 = −

C1 =

2 E 0 r22 r22 (r + 1) + r12 (r − 1)

2 E 0 r12 r22 r22 (r + 1) + r12 (r − 1)

and so the potentials are:

A1 cos ϕ V0 (ρ, ϕ) = −E 0 ρ + ρ

Vi (ρ, ϕ) =

B1 ρ +

C1 ρ

cos ϕ

The field lines for a conductive cylinder surrounded by a layer of dielectric with r = 3 in a uniform electric field are shown in Fig. 8.12. - For a conducting cylinder of radius R in vacuum (r = 1) we write in the solution R = r1 = r2 , that is a null thickness of the dielectric, thus A1 = C1 = E 0 R 2 and B1 = −E 0 and so the potential and the field are:

R2 cos ϕ V0 (ρ, ϕ) = −E 0 ρ − ρ

R2 E ρ = E 0 1 + 2 cos ϕ ρ



R2 E ϕ = −E 0 1 − 2 sin ϕ ρ

as found by image charges method in Problem 3.10. The field lines for a conductive cylinder are shown in Fig. 8.13. - For a dielectric cylinder of dielectric constant r and radius R in vacuum we substitute in the general solution r2 = R and r1 = 0 and we find:

166

8 Separation of Variables in Laplace Equation

Fig. 8.13 Field lines for a conductive cylinder in a uniform electric field

A1 = E 0 R 2

r − 1 r + 1

B1 = −

2 E0 r + 1

C1 = 0

the potentials are:

r − 1 R 2 cos ϕ V0 (ρ, ϕ) = −E 0 ρ − r + 1 ρ

Vi (ρ, ϕ) = −

2 E0 2 E0 ρ cos ϕ = − x r + 1 r + 1

and the field inside the dielectric cylinder is uniform and parallel to the x axis: Ex = −

2 E0 ∂ Vi = ∂x r + 1

Ey = −

∂ Vi =0 ∂y

Ez = −

∂ Vi =0 ∂z



while outside it is:

r − 1 R 2 1+ cos ϕ r + 1 ρ 2

r − 1 R 2 sin ϕ = −E 0 1 − r + 1 ρ 2

E est,ρ = E 0 E est,ϕ

The field lines for a dielectric cylinder with dielectric constant r = 5 in vacuum are shown in Fig. 8.14. For a cylinder of permittivity 2 in a medium of permittivity 1 we have to replace r = 2 /1 and we find the result in Problem 4.5.

Problems 8.1 Consider a pipe of infinite length as in Sect. 8.4.1 with the lateral sides grounded and terminated at z = 0 by an insulated conductive plate at voltage V0 . Find the charge densities on the five sides of the box and show that the total charge is null. 8.2 Find the potential in the space limited for z > 0 by two parallel grounded conductive layers at x = 0 and x = a, and at z = 0 by a conductive strip of infinite length, at voltage V0 , parallel to the y axis between x = 0 and x = a.

Solutions

167

Fig. 8.14 Field lines for a dielectric cylinder with r = 5 in a uniform electric field

8.3 Show that the formula (8.41) in the solution of Problem 8.2 is equal to the result (8.40), that is: 4V0 V (x, z) = π



∞  sin πax 2V0 1 − nπ z nπ x −1 a e = tan sin n a π sinh πaz n=1,3,5

8.4 Find the potential for a spherical shell of radius R with charge distribution σ (θ ). Use the result to write the potential for a ring of radius R where the charge Q is uniformly distributed. 8.5 The potential of a spherical shell of radius R is given by the function VS (θ ). Find the potential in the space. Use the result to study the simple configuration with VS (θ ) = V0 for 0 ≤ θ < π/2 and VS (θ ) = −V0 for π/2 < θ ≤ π . 8.6 Find the potential for a dielectric sphere of radius R and permittivity 2 , with its center in the origin of the axes, in a medium of permittivity 1 in the presence of a point charge Q on the z axis at distance z > R from the origin. 8.7 Find the solution of the Laplace equation outside the split cylinder studied in Sect. 8.7.3.

Solutions 8.1 We start from the result (8.10) for the potential inside the pipe: V (x, y, z) =

16 V0 π2

∞ 

∞ 

n=1,3,5 m=1,3,5

where n and m are both odd and

mπ y 1 −knm z nπ x e sin sin nm a b

168

8 Separation of Variables in Laplace Equation

knm =



nπ 2  mπ 2 + a b

The x component of the field is: Ex = −

∞ 

∞ 

16 V0 ∂V =− ∂x π

n=1,3,5 m=1,3,5

mπ y 1 −knm z nπ x e sin cos ma a b

the density charge on the side at x = 0 is: σx=0 = 0 E · xˆ = −0

16 V0 π

∞ 

∞ 

1 −knm z mπ y e sin ma b

n=1,3,5 n=1,3,5

which is equal to that at x = a: σx=a

16 V0 = 0 E · (−x) ˆ = −0 π

∞ 

∞ 

n=1,3,5 m=1,3,5

1 −knm z mπ y e sin ma b

and after integration the induced charges on the two sides are: 

b

Q x=0 = Q x=a = 0



∞

σx=0 dy dz = −

0

∞ ∞ 32 0 V0   b 1 1 π 2 n=1,3,5 m=1,3,5 a m 2 knm

Similar results can be found for the y component of the field: Ey = −

16 V0 ∂V =− ∂y π

∞ 

∞ 

n=1,3,5 m=1,3,5

mπ y 1 −knm z nπ x e cos sin nb a b

and for the charges on the sides at y = 0 and at y = b: σ y=0 = σ y=b = −0

16 V0 π

∞ 

∞ 

n=1,3,5 m=1,3,5

1 −knm z nπ x e sin nb a

After integration, the induced charges are: 

a

Q y=0 = Q y=b = 0



∞ 0

σ y=0 d x dz = −

∞ ∞ 32 0 V0   a 1 1 π 2 n=1,3,5 m=1,3,5 b n 2 knm

and the total charge on the four grounded sides is:

Solutions

169

Q gr ound = −

Q gr ound

∞ ∞ 1 64 0 V0 ab   knm 4 2 π n m2 n=1,3,5 m=1,3,5

∞ ∞ 1 64 0 V0 ab   =− 3 2 π n m2 n=1,3,5 m=1,3,5

(8.39)

  n 2  m 2 + a b

The z component of the field is: Ez = −

16 V0 ∂V = ∂z π2

∞ 

∞ 

n=1,3,5 m=1,3,5

mπ y knm −knm z nπ x e sin sin nm a b

and the density charge on the side at V0 is: σz=0 (x, y) = 0

16 V0 π2

∞ 

∞ 

n=1,3,5 m=1,3,5

nπ x mπ y knm sin sin nm a b

and after integration on the charge on the plate at z = 0 is: Q z=0 =

∞ ∞ 1 64 0 V0 ab   knm 4 2 π n m2 n=1,3,5 m=1,3,5

which is positive and opposite in sign to the total charge induced on the grounded sides given in (8.39). 8.2 The geometry does not depend on y so we expect a potential V (x, z) = X (x) Z (z) with one separation constant k 2 > 0 and the two equations: 1 d2 Z = k2 Z dz 2

1 d2 X = −k 2 X dx2

d2 Z − k2 Z = 0 dz 2

d2 X + k2 X = 0 dx2

with solutions: Z (z) = Aekz + Be−kz

X (x) = C sin kx + D cos kx

We can apply now the boundary conditions:

170

8 Separation of Variables in Laplace Equation

from V = 0 at x = 0 we have D = 0 while from V = 0 at x = a → sin ka = 0 and for k the condition: nπ n = 1, 2, ... k= a From V = 0 for z → ∞ we get A = 0 so that the solution for a given k is: V (x.z) = C e−kz sin kx and the general solution is V (x, z) =

∞ 

cn e−

nπ z a

sin

n=1

nπ x a

For V (x, z) = V0 at z = 0 we have: V0 =

∞ 

cn sin

n=1

nπ x a

Upon multiplication by sin(n  π x/a) and integration in x over the interval [0,a]: 

∞

a

V0

sin 0

 nπ x dx = cn a n=1



a

sin 0

nπ x nπ x sin dx a a

from orthogonality: −V0

cn = 0

 ∞ a n  π x a  a cos = cn δnn  π n a 0 n=1 2

for even n,

and finally: V (x, z) =

4V0 π

which is equal to:

cn =

4V0 nπ

for odd n

∞  1 − nπ z nπ x e a sin n a n=1,3,5

⎞

⎛ V (x, z) =

(8.40)

πx a

2V0 ⎟ ⎜ sin tan−1 ⎝ πz ⎠ π sinh a

(8.41)

Solutions

171

8.3 We consider the relation at the first member: nπ nπ x ∞ ∞   1 − nπ z 1 − nπ z ei a x − e−i a nπ x e a sin = e a n a n 2i n=1,3,5 n=1,3,5

∞  n=1,3,5

∞    1 1  nπ (i x−z) 1 1  n nπ ea ξ1 − ξ2n − e− a (i x+z) = 2i n 2i n n=1,3,5

where we say:

π

ξ1 = e a (i x−z)

π

ξ2 = e− a (i x+z)

(8.42)

After replacing n by 2k + 1: =

∞ ∞   2k+1    1 1 1 (i)2k+1  ξ1 (iξ2 )2k+1 − (iξ1 )2k+1 = − ξ22k+1 = 2i 2k + 1 2i 2k + 1 k=0 k=0

for instance from Gradshteyn and Ryzhik [9] 1.622.3: =

∞  1  −1  1  (−1)k  (iξ2 )2k+1 − (iξ1 )2k+1 = tan (iξ2 ) − tan−1 (iξ1 ) = 2 k=0 2k + 1 2

from Gradshteyn and Ryzhik [9] 1.625.9 and replacing the relations (8.42) we find: ⎛

=

and:

1 tan−1 2

πx ⎞ iξ2 − iξ1 1 ⎜ a ⎟ = tan−1 ⎝ πz ⎠ 1 + ξ1 ξ2 2 sinh a sin

⎛ πx ⎞ ∞ sin  1 1 − nπ z nπ x ⎜ a ⎟ e a sin = tan−1 ⎝ πz ⎠ n a 2 sinh n=1,3,5 a

which finally proves: ⎛ πx ⎞ ∞ sin  2V0 1 − nπ z nπ x 4V0 ⎜ a ⎟ e a sin = tan−1 ⎝ V (x, z) = πz ⎠ π n=1,3,5 n a π sinh a 8.4 We start from the solutions (8.24) and (8.25) with al = bl and we add the condition for components of the fields near the surface: E ⊥out − E ⊥in =

σ 0

(8.43)

172

8 Separation of Variables in Laplace Equation

The components of the fields are: E < (r, θ ) = − and:

∞  ∂ V< r l−1 =− al l l Pl (cos θ ) ∂r R l=0

∞

E > (r, θ ) = −

 ∂ V> R l+1 = al (l + 1) l+2 Pl (cos θ ) ∂r r l=0

Writing z = cos θ and replacing these relations in (8.43) we get: ∞ 

al

l=0

2l + 1 R

Pl (z) =

σ (z) 0

Upon multiplication for Pi (z) and integration over the interval [−1, 1]: ∞   l=0



1

−1

al

2l + 1 R

Pl (z) Pi (z) dz =

1 0



1

−1

σ (z) Pi (z) dz

and using the relation (8.13) for the orthogonality of the Legendre polynomials: R 2 0

al =



1 −1

σ (z) Pl (z) dz.

(8.44)

For a ring with a uniformly distributed charge Q: σ (θ ) =

Q λ δ(cos θ ) = δ(cos θ ) R 2π R 2

and from (8.44) for the coefficients: al =

Q Pl (0) 4π 0 R

already found in (8.28). 8.5 We can use the general solutions (8.24) and (8.25) for the inner and outer spaces of the spherical surface of radius R. The boundary potential VS (θ ) at r = R is: VS (θ ) = V< (R, θ ) = V> (R, θ ) that is: VS (θ ) =

∞  l=0

al Pl (cos θ ) =

∞  l=0

bl Pl (cos θ )

(8.45)

Solutions

173

and so for the coefficients we have al = bl . After multiplication of the (8.45) by Pi (cos θ ) and integration on the interval [-1,1] of cos θ : 

1

−1

VS (θ )Pi (cos θ ) d cos θ =



∞ 

al

1

Pl (cos θ ) Pi (cos θ ) d cos θ

−1

l=0

using the orthogonality condition of the Legendre polynomials: 

1

−1

VS (θ )Pi (cos θ ) d cos θ =

∞ 

al

l=0

2 δil 2l + 1

for the coefficients we find: al = bl =

2l + 1 2



1

−1

VS (θ )Pl (cos θ ) d cos θ.

For the second part of the problem the coefficients are:   0  1 2l + 1 − V0 Pl (cos θ ) d cos θ + V0 Pl (cos θ ) d cos θ 2 −1 0

al = bl =

The first integral can be written:  −

0 −1



−1

Pl (cos θ ) d cos θ =

Pl (cos θ ) d cos θ =

0

and from the relation Pl (− cos θ ) = (−1)l Pl (cos θ ): 

1



1

Pl (− cos θ ) d(− cos θ ) = −(−1)l

0

Pl (cos θ ) d cos θ

0

so for the coefficients to be used in (8.24) and (8.25) we find: al = bl =

 2l + 1  1 + (−1)l+1 V0 2



1

Pl (cos θ ) d cos θ

0

which are null for even l while for odd l they are:  al = bl = (2l + 1) V0 0

1

Pl (cos θ ) d cos θ.

174

8 Separation of Variables in Laplace Equation

The result for this integral19 for l odd is:  1 l−1 Pl (cos θ ) d cos θ = (−1) 2 0

l!! l(l + 1)(l − 1)!!

so: al = bl = (−1)

l−1 2

l!! (2l + 1) V0 l(l + 1)(l − 1)!!

and finally for the potential: VR (r, θ ) =

∞ 

(−1)

l−1 2

l=1,3,5

 r l l!! (2l + 1) V0 Pl (cos θ ) l(l + 1)(l − 1)!! R

l!! (2l + 1) V0 l(l + 1)(l − 1)!!

l+1 R Pl (cos θ ) r

For r >> R the first non null term is for l = 1 and V>R (r, θ ) becomes: 2 R 3 V>R (r, θ ) V0 cos θ 2 r which is the potential for a dipole of moment p = 6π 0 V0 R 2 as can be found by a direct calculation. 8.6 For the potential of the charge Q for r < z we can use the relation (8.29): ∞ Q  rl 1 Q = Pl (cos θ ) VQ = 4π 1 |r − r | 4π 1 l=0 z l+1 where r (0, 0, z) is the position of the charge Q. The most general solution of the Laplace equation in spherical coordinates with azimuthal symmetry is: V (r, θ ) =

∞  

Al r + l

l=0

Bl r l+1

 Pl (cos θ )

Inside the sphere to have a continuous potential in the center we have to exclude solution with r −(l+1) , so for the inner solution we expect: Vin (r, θ ) =

∞ 

Al r l Pl (cos θ )

l=0

19

Formula 49 in https://mathworld.wolfram.com/LegendrePolynomial.html Here we use the notation: l!! = l (l − 2)!! [0!! = 1, 1!! = 1].

Solutions

175

while for the outer solution we exclude terms of the type r l not acceptable at infinity and adding this potential to the potential of the point charge for r < z, we expect: Vout (r, θ ) =

∞ ∞  Bl Q  rl P (cos θ ) + Pl (cos θ ) l l+1 4π 1 l=0 z r l+1 l=0

The condition Vin (R, θ ) = Vout (R, θ ), using the orthogonality of the Legendre polynomials, gives for every l the relation: Al R l =

Q Rl Bl + l+1 l+1 4π 1 z R

while the boundary condition Drin = Drout gives the relation:  2 A l l R

l−1

= 1

Q l R l−1 Bl − (l + 1) l+2 l+1 4π 1 z R



so from these relations for the coefficients Al and Bl we find: Al = Bl =

2l + 1 1 Q l+1 4π 1 (l + 1) + 2 l z

R 2l+1 (1 − 2 ) l Q 4π 1 1 (l + 1) + 2 l z l+1

Finally the potentials outside and inside the dielectric sphere are: Vout (r, θ ) =

∞ Q  (1 − 2 ) l R 2l+1 Pl (cos θ ) 1 Q +  4π 1 |r − r | 4π 1 l=0 1 (l + 1) + 2 l z l+1 r l+1

Vin (r, θ ) =

∞ rl 2l + 1 Q  Pl (cos θ ) 4π l=0 1 (l + 1) + 2 l z l+1

In the limit 2 → ∞ the dielectric sphere is a conductive one and the outer potential becomes: Vout (r, θ) =

Q 1 1 + 4π 1 |r − r | 4π 1

∞ l

R  R2 R 1 1 Pl (cos θ) −Q Q + z z r l+1 4π 1 z r l=0

that is the potential (8.34) for a conductive sphere with null charge found in Sect. 8.5.3.

176

8 Separation of Variables in Laplace Equation

8.7 For the general solution of the Laplace equation (8.35) outside the cylinder the boundary conditions are the same given in Sect. 8.7.3 for the inner potential but we have to exclude terms with ρ A (A > 0) and include terms with ρ −A (A > 0) so we expect solutions of the type: V (ρ, ϕ) = D ρ −A sin Aϕ and we can repeat the steps done in Sect. 8.7.3: V (ρ, ϕ) = D A ∞ 

V (ρ, ϕ) =

n=1,3,5

Dn =

A R sin Aϕ ρ n R Dn sin nϕ ρ 4V0 nπ

thus the solution is: V (ρ, ϕ) =

4V0 π

∞  n=1,3,5

1 n



2k−1 ∞ R n R 1 4V0  sin nϕ = sin(2k − 1)ϕ ρ π 2k − 1 ρ k=1

and from the sum of the series already seen:

V (ρ, ϕ) =

2V0 tan−1 π

R sin ϕ ρ R2 1− 2 ρ

2

The equipotential surfaces of voltage V are determined by the parameter a: Vπ a = tan α = tan = 2V0

R sin ϕ ρ R2 1− 2 ρ

2

with the condition −V0 ≤ V ≤ V0 . In Cartesian coordinates these surfaces have equations: R x 2 + y2 − 2 y = R2 a

References

177

 and intersect the plane z = 0 in circles with radii Rc = R 1 + 1/a 2 and centers on the y axis at yc = R/a, negative if a < 0 (V < 0) and positive if a > 0 (V > 0) and crossing the x axis at x = ±R. For V = V0 : α = π/2, a = +∞, Rc = R and yc = 0+ , that is the positive half cylinder; for V = −V0 : α = −π/2, a = −∞, Rc = R and yc = 0− that is the negative half cylinder. √ For V = V0 /2: α = π/4, a = √ 1, Rc = R 2 and yc = R, while for V = −V0 /2: α = −π/4, a = −1 Rc = R 2 and yc = −R. For V = 0± : a = 0± yc = ±∞ and Rc = ∞. Some equipotentials outside the cylinder are shown in Fig. 8.11. At large distance ρ >> R the potential, given as a function of the angle θ = π/2 − ϕ with respect the axis of symmetry y, is: V (r, θ ) =

4V0 R cos θ π ρ

that is the potential (6.7) of two parallel wires with opposite charge density λ = 80 RV0 /d at a distance d R2 . Initially the charge is Q 0 . Then the capacitor is connected to t a resistor and the charge decreases as Q(t) = Q 0 e− τ . (a) Write the magnetic field and the Poynting vector between the two shells. (b) Show that the initial electrostatic energy in the capacitor flows through the two edges.

Problems

233

Fig. 11.5 The coaxial cable connecting the battery to the resistor

Fig. 11.6 The capacitor of Problem 11.3 with a wire connected to a plate and passing through an in the other plate

11.3 A capacitor has circular parallel plates of radius a at distance h. A wire passing through a small hole in the lower plate connects the upper plate to a resistor R and a battery of voltage V0 as shown in Fig. 11.6. Find the displacement current density and the Poynting vector between the plates while the capacitor is charging. Write the total flux of the Poynting vector at a distance r from the wire and show that the electrostatic energy in the charged capacitor is the integral in time of the flux from the piece of the wire inside the plates. 11.4 Sommerfeld gave the solution for a wire of radius a and conductivity σ , surrounded by a cylindrical hollow conductor with inner radius b and an infinite outer radius. An electrical current, with uniform density J, flows in the central conductor from z = −∞ to z = +∞ and returns in the outer conductor. The wire lies on the z axis. Find the solution for the field and consider the Poynting vector or proceed through the following steps: (a) define the boundaries conditions for the electric field in the space between the two conductors; (b) verify that the solution of the Laplace equation in the space between the conductors is:

V (r, z) = −

J log br z σ log ab

(c) find the charge densities over the surfaces of the conductors; (d) write the Poynting vector between the two conductors; (e) consider the Poynting vector inside the central conductor. 11.5 In the solar system a space vehicle might be propelled by the pressure of the solar radiation on a large fully reflective sail. Determine the minimal area of the sail

234

11 Energy and Momentum of the Electromagnetic Field

to push a satellite of mass m v = 300 kg when it is distant from the Sun as the Earth where the solar radiation power is 1.35 kW/m2 . (M Sun = 1.99 × 1030 kg, distance Sun-Earth d S E = 1.5 × 1011 m, GGravit. = 6.67 × 10−11 Nm2 /kg2 ) 11.6 Find the force on the plates of a charged parallel plate capacitor by using the stress tensor. 11.7 Calculate the pressure on the turns of a long solenoid with a constant current. 11.8 A uniformly charged sphere has charge Q and radius R. Find the total force on a hemisphere when applying the Maxwell stress tensor and by a direct calculation.

Solutions 11.1 The inner conductor is at voltage V while the outer is connected to ground. Say λ the linear charge density on the inner conductor, and −λ that on the outer one. Consider a cylindrical surface of length l and of radius r (a < r < b) coaxial with the cable. The axis of the cylinders is the z axis. From the Gauss theorem we get the electric field inside the cable: λl 0

E · 2πrl =

→

E=

λ 1 rˆ . 2π 0 r

From the voltage between the two conductors we can get λ:  V = a

b

E dr =

λ 2π 0

 a

b

λ 1 b dr = log r 2π 0 a

λ=

2π 0 V log ab

and the electric field becomes: E=

V 1 rˆ . log ab r

A current I = V /R is flowing in the conductor from the battery to the resistor. The field has circular field lines around the axis. From the Ampère law we get the field B at a radius r: B · 2πr = μ0 I B =

μ0 V 1 ϕˆ . 2π R r

The field E and B are perpendicular in any point between the coaxial surfaces and the Poynting vector is:

Solutions

235

S=

1 1 V2 1 |E × B| zˆ = zˆ μ0 2π log ab R r 2

oriented in the direction of the axis from the battery to the resistor. The total flux of this vector through a cross section normal to the axis, is: 

b

(S) =

 2πr Sdr =

a

2πr a

=

b

1 1 V2 1 1 V2 dr = 2π log ab R r 2 log ab R

 a

b

1 dr r

2 b 1 V2  = V . log r a R log ab R

The flux of energy between the cylindrical surfaces is just the power dissipated in the resistor. 11.2 We use cylindrical coordinates with the axis of the capacitor on the z-axis. In the space between the shells the electric field is radial. For symmetry the magnetic field cannot have components in the direction of the axis (Bz = 0) and for the third Maxwell equation has to be perpendicular to the electric field. So B can have only a ϕ component. We consider a cylindrical shell of length l and radius r with R1 < r < R2 at the center of the capacitor. We split this shell in two opposite semicylindrical shells (shell1 and shell2 ) as shown in Fig. 11.7. We calculate the line integral of B along the path ABCDA of shell1 and we use the Stokes’s theorem and the fourth Maxwell equation: 





B · ds= shell1

∇ × B · rˆ d S=μ0 0 shell1

shell1

∂E d · rˆ d S=μ0 0 ∂t dt

 E·ˆr d S shell1

We repeat a similar calculation for shell2 : 

d B · ds = μ0 0 dt shell2

 E · rˆ d S shell2

and we add the two relations:      d B · ds + B · ds = μ0 0 E · rˆ d S + E · rˆ d S dt shell1 shell2 shell1 shell2 At the first member the net result is the integral of B along two circles of radius r in opposite directions because the sum of the line integrals on the four edges parallel to the z-axis is null. At the second member we have the derivative of the flux of the electric field through a cylindrical surface of radius r :

236

11 Energy and Momentum of the Electromagnetic Field

2 (2πr ) Bϕ = μ0 0



d dt

shell1 +shell2

E · rˆ d S

Bϕ is positive when in the ϕ-direction ˆ on the circle at left if the z-axis is oriented at right. From the Gauss theorem: 2 (2πr ) Bϕ = μ0

l d Q(t) μ0 l t =− Q 0 e− τ L dt τ L

Bϕ (r, t) = −

μ0 1 l Q 0 − t e τ 4π r L τ

The electric field between the two conductive shells is: E(r, t) =

1 1 t Q 0 e− τ rˆ 2π 0 L r

The Poynting vector at right is: E×B l Q 20 e− τ 1 zˆ zˆ = 2 μ0 8π 0 L 2 r 2 τ 2t

S(r, t) = The flux of S on each side is:   S (t) =

R2 R1

−τ R2 l 1 2 e log S · 2πr dr = Q 0 2 4π 0 L τ R1 2t

1 l 2 Q 20 e− τ  S (t) = 2 L2 C τ

2t

C=

2π 0 l log

R2 R1

where C is the capacitance of a cylindrical capacitor of lenght l. The total energy flowed through the two sides is: 

∞

Ue = 2 0

11 (S) dt = 2C



l Q0 L

2

that is the electrostatic energy initially stored in part of lenght l of the capacitor. When l → L the total energy flowed from the sides of the capacitor is the initial energy in the capacitor of lenght L: Ue,L =

1 Q 20 2 C

Solutions

237

Fig. 11.7 The cylindrical shell of radius r splitted in two opposite shells (shell1 and shell2 ) and the integration path ABCDA for shell1

11.3 The charge on the (upper) plate at time t is: Q(t) = Q 0 (1 − e− τ ) t

τ = RC

C=

The current in the wire is: i(t) =

π a 2 0 h

Q 0 = C V0

Q0 − t dQ = e τ dt τ

The electric field between the plates is: E 0 (t) = −

Q(t Q0 t = − 2 (1 − e− τ ) 2 πa 0 πa 0

and the displacement current density: Js (t) = 0

1 Q0 − t ∂ E 0 (t) =− e τ ∂t τ π a2

From Ampère’s law taking a circular loop of radius r we find the B(r, t): 2π r B0 (r, t) = μ0 [i(t) + πr 2 Js (t)] B0 (r, t) =

μ0 Q 0 2π τ



1 r − 2 r a



e− τ t

The Poynting vector is: S(r, t) =

E0 × B0 E 0 B0 =− rˆ μo μ0

1 Q 20 1 S(r, t) = 2π πa 2 0 τ



1 r − 2 r a

  t 2t e− τ − e− τ rˆ

238

11 Energy and Momentum of the Electromagnetic Field

The flux of S across a cylindrical surface of radius r around the piece of the wire inside the plates is: (r, t) = 2πr h S(r, t) and the energy that has passed through this surface at time t is:  U (r, t) =

t

0

Q 20 h (r, t) dt = πa 2 0 τ

U (r, t) =

 t    r2 t 2t 1− 2 e− τ − e− τ dt a 0

   Q 20 1 r2 1 − 2t − τt τ − e e h 1 − + πa 2 0 a2 2 2

At t = ∞:



∞

U (r, ∞) = 0

Q 20 h (r, t) dt = πa 2 0 2

  r2 1− 2 a

and the total energy flowed from a cylinder of radius r → 0 around the wire is: U (r = 0, ∞) =

h Q 20 1 1 Q 20 = C V02 = UC = 2 2 πa 0 2 C 2

The electrostatic energy of the charged capacitor flows from the wire between the two plates and fills the space between the plates. 11.4 (a) Nearby the surface of the wire, the outside z component of the electric field has to be equal to the inside z component that is E z = J/σ . Near the surface of the external conductor, at r = b, E z = 0 because inside this conductor the field is null. The radial component of the electric field can be non null on the surface of the wire and on the inner surface of the external conductor. (b) It is easy to verify that the function V (r, z) is a solution of the Laplace equation V = 0. The component E z is: E z (r, z) = −

J log br ∂V = . ∂z σ log ab

Thus for r = a E z = J/σ while for r = b E z = 0 as requested by the boundary conditions. The radial component of the electric field is: Er (r, z) = −

J z 1 ∂V = ∂r σ log ab r

that for r = a gives: Er (a, z) =

J z 1 σ log ab a

Solutions

239

Fig. 11.8 The resistive wire with a coaxial hollow conductor. The dotted lines are electric field lines; the full lines represent the equipotential surfaces (only outside the wire) and also the Poynting vector field lines

and for r = b: Er (b, z) =

J z 1 . σ log ab b

(c) From the Coulomb’s theorem the surface charge densities on the conductors are: at r = a σ =  Er (a, z) = 

J z 1 σ log ab a

at r = b σ = − Er (b, z) = −

J z 1 . σ log ab b

The electric field lines and the equipotential surfaces inside the conductors are shown in the Fig. 11.8. (d) The magnetic field due to the current in the wire is: H=

r J ϕˆ 2

r R. On the plane at z = 0, at the distance r > R from the center, the field is: E(r ) =

1 Q rˆ 4π 0 r 2

Ex =

1 Qx 4π 0 r 2 r

Ey =

1 Qy 4π 0 r 2 r

the stress tensor components are: Tzz = −

0 2



Q 4π 0

2

1 r4

Tzx = 0

Tzy = 0

and the force for r > R is 0 F = 2 

The total force is:



Q 4π 0

2 

∞ R

1 1 1 Q2 2π r dr = r4 8 4π 0 R 2

Ez = 0

11.8 The Covariant Maxwell Stress Tensor

Fz = Fz + F  =

245

3 1 Q2 16 4π 0 R 2

References 1. A. Sommerfeld, Electrodynamics (Academic Press, New York, 1952) 2. I. Galili, E. Goihbarg, Energy transfer in electrical circuits: a qualitative account. Am. J. Phys. 73, 141–144 (2005) 3. F.S. Crawford Jr, Waves, Berkeley Physics Course, vol. 3. (McGraw-Hill, 1965) 4. D.J. Jackson, Classical Electrodynamics, 3rd edn. (John Wiley & Sons, NewYork, 1999) 5. L.D. Landau and E.L. Lifshitz, The Classical Theory of Fields, Course of Theoretical Physics, vol. 2. (Elsevier)

Chapter 12

The Feynman Paradox

The Feynman paradox also known as the paradox of the angular momentum is very intriguing. An insulating disk, with charged spheres at its edges, is at rest in the steady magnetic field produced by a solenoid at its center. The system is isolated, but if the magnetic field vanishes, the disk begins to rotate in apparent contrast with angular momentum conservation. This chapter presents the analysis of the process. The conservation of the angular momentum is verified by taking into account the angular momentum of the electromagnetic field. Then, an original example of two cylindrical shells with opposite charges immersed in a damped magnetic field is examined.

12.1 The Paradox In his lectures1 on the induction laws, Feynman presents the following paradox. Consider a thin plastic disk (see Fig. 12.1) supported by a concentric shaft with excellent bearings so that it is quite free to rotate in the horizontal plane. Some metal spheres are uniformly distributed near the edge of the disk. Each sphere has a charge q and is insulated by the plastic disk. Fixed and coaxial with the disk there is a solenoid that carries a steady current provided by a small battery. The system is isolated and at rest. If the current is interrupted with no intervention from the outside2 , the magnetic field inside the solenoid vanishes and for the induction law an electric field E is generated. This field has a field line tangent to the circle centered on the axis of the system and passing on the small spheres. Thus a force tangent to the circle and going 1

Feynman et al. The Feynman Lectures on Physics [1], Vol. II, Sect. 17.4. We suggest reading Chaps. 16 and 17 on the electromagnetic induction. 2 This can be determined by a suitable device mounted on the disk or the solenoid could be done with a superconducting wire initially at very low temperature but later, the exchange of heat with the ambient, increases the temperature over the transition temperature, so that the wire becomes resistive and the current is brought to zero. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 F. Lacava, Classical Electrodynamics, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-031-05099-2_12

247

248

12 The Feynman Paradox

Fig. 12.1 The disk with the charged spheres and the solenoid at the center

in the same direction is applied on each sphere resulting in a net torque applied to the disk that begins to rotate. If R Sol is the radius of the solenoid, R the distance of the spheres from the center, and B the initial magnetic field inside the solenoid, the field E is given3 by the Faraday-Neumann law:  E · dl = −

dΦ dt

2 2π R E = −π R Sol

dB dt

E =−

2 R Sol dB 2R dt

and the net torque relative to the axis of the system is M = R × N qE. From the equation of the rotation the final angular momentum L is: L = Nq

2 R Sol Bin 2

(12.1)

where Bin is the initial magnetic field inside the solenoid. But we could also say that the system is isolated and for the conservation of the angular momentum that initially is null, the disk should not rotate. Feynman asks which of the two arguments is correct: the disk begins to rotate or not? He also warns that: the solution is not easy, nor is it a trick, and adds: when you figure it out, you will have discovered an important principle of electromagnetism.

12.2 A Charge and a Small Magnet In the lecture4 on the Poynting vector, Feynman proposes the case of a point charge near the center of a small bar magnet both at rest. The relative electric E and magnetic B fields are static and clearly the electric and magnetic energy densities do not change 3

The return flux of B in the region with a distance from the axis between R Sol and R can be neglected in the approximation of a solenoid of length l  R Sol . 4 The Feynman Lectures on Physics [1], Vol. II, Sect. 27.5.

12.2 A Charge and a Small Magnet

249

Fig. 12.2 Point charge Q at the center of a small bar magnet of moment m. Electric and magnetic fields with the associated Poynting vector in a point and their field lines

with time. But the Poynting vector E × B/μ0 is not null and has circular field lines centered on the line of the bar magnet as shown in Fig. 12.2. This can appear strange because we have to conclude that while the electromagnetic energy is conserved, there is a steady circular flow of this energy around the line of the magnet. Moreover, as seen in the previous chapter in relation (11.18), a flow of momentum is associated to the flow of the Poynting vector, then there is also an angular momentum of the electromagnetic field relative to the axis. This is not strange at all because the Ampère currents in the magnet (the spinning electrons inside) or the current in the solenoid seen at the beginning of this chapter, circulate in one of the two possible directions and the charges, carried in the currents, drag with them their electric fields. In the Feynman paradox the initial presence of the magnetic field of the solenoid and of the electric field of the charged spheres, determines a flow of the Poynting vector and an associated angular momentum of the electromagnetic field. When the magnetic field vanishes, the initial angular momentum of the electromagnetic field is transferred (in part) to the disk that begins to rotate and the energy of the magnetic field is (in part) transformed into kinetic energy.5

5

The balance of energy and angular momentum between the field and the rotating system will be clear in the example examined in next section.

250

12 The Feynman Paradox

12.3 Analysis of the Angular Momentum Present in the System An interesting analysis of the Feynman paradox is given by Bettini6 in the approximation of the solenoid with a vanishing small magnetic dipole at the center of the disk. The total flux of the field B at distances from the axis smaller than R, the distance of the spheres, can be easily calculated by observing that it is equal in modulus but opposite to the flux at larger distances when this is evaluated over the plane normal to the dipole and passing through its center. The field B on this plane at a distance r from the axis is: B=−

μ0 m 4π r 3

and the flux for r > R is:  Φ(r >R) (B) =

∞

B(r )2π r dr = −

R

μ0 m. 2R

Then the flux for r < R is: Φ(r R) (B) =

μ0 m 2R

and from this relation, the electromotive field E e associated to the change of dipole moment m, from the Faraday-Neumann law, is: 2π R E e = −

μ0 dm dΦ(r R1 , and length l  R2 , as shown in Fig. 12.3. The electric charges are deposited on the shells, with surface density σ1 = λ/2π R1 and σ2 = −λ/2π R2 where λ > 0 is the charge per unit length. The two shells are inside a solenoid8 of 7

The contribution of the electrostatic force to the torque dMext is null. The solenoid and the two shells can be considered isolated. An internal device can decrease to zero the current in the solenoid.

8

252

12 The Feynman Paradox

Fig. 12.3 The two insulating cylindrical shells with opposite sign surface charge densities in the magnetic field B(t)

radius R  R2 , coaxial with the shells and of length l   R. Initially the current flowing in the solenoid produces a uniform magnetic field Bin parallel to the axis. Then the current decreases slowly and vanishes. As a consequence also the magnetic field B inside the solenoid decreases and, for the Faraday-Neumann law, at distance r from the axis there is an induced electric field E in , tangent to the circle of radius r and centered on the axis, given by: 2πr E in = −πr 2

dB dt

E in = −

r dB 2 dt

so that the fields on the shells at distances R1 and R2 are: E1 = −

R1 d B 2 dt

E2 = −

R2 d B . 2 dt

Thus on each element d of the cylindrical shells is applied a force d F = σ E in d tangent to the surface and there is a torque dM = R × dF relative to the axis. For a unit length of the shells the net torques are: M1 = R1 λE 1

M2 = −R2 λE 2

and after substitution with the fields expressions: M2 + M1 =

 dB λ 2 R2 − R12 . 2 dt

When the system begins to rotate with angular velocity ω (negative for our system) the charges deposited on the two surfaces are equivalent to two surface currents ˆ directed as ϕ:

12.4 Two Cylindrical Shells with Opposite Charge in a Vanishing Magnetic Field

λl ω 2π

i1 =

i2 = −

253

λl ω 2π

and a magnetic field9 appears between the two cylindrical charged surfaces: Ba = −μ0

λω . 2π

(12.3)

While ω changes a self-induced electric field E a is present on the outer shell: E a 2π R2 = −

  d   2 π R2 − R12 Ba dt

  E a = R22 − R12

λ dω 1 μ0 2R2 2π dt

and a force is applied on the element d2 of the outer shell: d F2 = σ2 d2 E a and the net axial torque per unit length of the outer surface is: M2 = −

 λ dω λ 2 R2 − R12 μ0 . 2 2π dt

Thus the equation of the angular momentum L = I ω per unit length of the rigid system is: dL = M2 + M1 + M2 dt

9

As shown at the end of the chapter, the field Ba is a consequence of the motion of the electrostatic field between the two rotating shells when seen in the laboratory frame. From the fourth Maxwell equation, the equation for Ba is: ∇ × Ba = μ0 J + μ0 0

∂ E in r ∂2 B ϕˆ ∇ × Ba = μ0 J − μ0 0 ϕˆ ∂t 2 ∂t 2

that after integration gives: 

∂2 B  λω μ0 0 2 Ba (r ) = −μ0 r − R12 zˆ . + 2π 4 ∂t 2 The last term is null if d B/dt = const but can be neglected if, for usual charges and dimensions, the frequency of rotation is  1011 /B (B in Tesla).

254

12 The Feynman Paradox

I

 dB  λ 2 λ  2 λ dω dω = R2 − R12 − R2 − R12 μ0 dt 2 dt 2 2π dt

that can be written as: I

  dB λ dω dω λ  2 λ 2 R2 − R12 μ0 R2 − R12 + = dt 2 2π dt 2 dt

and integrating over the time that the external magnetic field Bin takes to become null, we get the relation: I ω f in +

  λ ω f in λ λ  2 R2 − R12 μ0 = − R22 − R12 Bin 2 2π 2

(12.4)

  λ  2 λ R2 − R12 Baf in = − R22 − R12 Bin 2 2

(12.5)

I ω f in − f in

where Ba is the final self-induced magnetic field from (12.3). The relation (12.5) is the expression of the conservation of the angular momentum when the initial and final angular momenta of the electromagnetic field are taken into account. From the relation (12.4) the final angular velocity ω f in is:

ω f in

 λ 2 R2 − R12 Bin 2 =−  λ2  2 R2 − R12 I + μ0 4π

and, as expected, it is negative and depends on the moment of inertia of the rigid system. To understand the meaning of Eq. (12.5) we have to consider the electric and magnetic fields present while the system begins to rotate. Inside the outer cylindrical shell (at r < R2 ) there are: • the magnetic field of the external solenoid B = B zˆ , • the electric field induced10 by the change of B: 10

This field can be found from the third Maxwell equation. In cylindrical coordinates the components are:   ∂ Eϕ ∂ Bz 1 ∂ Ez ∂ Er ∂ Er ∂ Ez 1 ∂ =− − =0 − =0 (r E ϕ ) − . r ∂ϕ ∂z ∂z ∂r r ∂r ∂ϕ ∂t Since Er , E ϕ e E z cannot depend on z and on ϕ, from the first two equations we get E z = const = 0, and from the third equation we find: 1 ∂ ∂ Bz (r E ϕ ) = − r ∂r ∂t

that integrated gives:

Eϕ = −

r d Bz C + 2 dt r

12.4 Two Cylindrical Shells with Opposite Charge in a Vanishing Magnetic Field

Ein = −

255

r dB ϕˆ . 2 dt

Between the two cylindrical shells (R1 < r < R2 ) there are: • the electrostatic field from the two surface charges: Es =

λ 1 rˆ , 2π 0 r

• the self induced magnetic field: Ba = Ba zˆ = −μ0

λω zˆ , 2π

• the self induced electric field11 :

R12 λ dω 1 r− μ0 ϕˆ . Ea = 2 r 2π dt Thus the Poynting vector is: S=E×

(B + Ba ) B = (Es + Ein + Ea ) × μ0 μ0

with six components: S = Es ×

B Ba B Ba Ba B + Es × + Ein × + Ea × + Ein × + Ea × μ0 μ0 μ0 μ0 μ0 μ0 S = Sϕ1 ϕˆ + Sϕ2 ϕˆ + Sr 1 rˆ + Sr 2 rˆ + Sr 3 rˆ + Sr 4 rˆ

the first two with direction ϕˆ and the other four with radial direction rˆ . We have to examine now the variation of the energy of the electromagnetic field of the system by analyzing the energy flux through the cylindrical shells at R1 and R2 . where the constant C has to be null since E ϕ (r = 0) = 0, then: Eϕ = − 11

r d Bz . 2 dt

To find this self induced electric field E a we have to solve the same equation seen in the previous note replacing B with the field Ba and since this is limited to the space between the two shells we have to determine the constant C imposing E a is null on the shell at r = R1 . So it follows:   R12 1 λ dω Ea = r− μ0 . 2 r 2π dt

256

12 The Feynman Paradox

The first radial component S1r rˆ : S1r rˆ = Ein ×

B r dB B rˆ =− μ0 2 dt μ0

corresponds to the energy flux, with radial direction, that escapes from the shell of radius R2 while B vanishes. The outgoing energy per unit time over a length l of a cylindrical surface of radius r is: −

1 r d B2 r dB B dU1 2πrl = − = S1r rˆ · rˆ 2πrl = − 2πrl dt 2 dt μ0 μ0 4 dt

and the total energy out from the surface of radius R2 when B is null, is: −U1 =

2 2 1 Bin 1 Bin 2π R22 l = (π R22 l) 4 μ0 2 μ0

(12.6)

that is just the magnetic energy initially inside the cylinder of length l and radius R2 . The second radial component S2r rˆ :

λω λ dω Ba 1 R12 rˆ S2r rˆ = Ea × =− r− μ0 μ0 2 r 2π 2π dt corresponds to a flow of energy with radial direction entering the solenoid. The energy entering per unit time over a length l of the cylindrical shell of radius R2 is: dU2 1 = S2r rˆ · rˆ 2π R2 l = − − dt 2



R22 − R12 R2

μ0

λω λ dω 2π R2 l 2π 2π dt

 1 λ2 dω2 1 2 dU2 = R2 − R12 μ0 l dt 2 2 2π dt and the total energy flowed through the surface of radius R2 is:  1  U2 = π R22 − R12 l μ0 2



λω f in 2π

2

 1 Baf in  = π R22 − R12 l 2 μ0

2

(12.7)

that is the magnetic energy located at the end between the cylindrical shells for a length l. This energy is due to the magnetic field produced by the currents associated to the rotation of the two charged shells.

12.4 Two Cylindrical Shells with Opposite Charge in a Vanishing Magnetic Field

257

Then there is the flow of the component Sr 3 rˆ : r 1 dB Sr 3 rˆ = 2 μ0 dt



λω 2π

rˆ

that has to be integrated over the cylindrical surfaces of radii R1 and R2 : −

2πl dU3 = dt μ0



R22 − R12 2

dB dt



λω 2π

and the flow of the component Sr 4 rˆ :

R12 d λω 1 r− B rˆ Sr 4 rˆ = 2 r dt 2π to be integrated only over the surface of radius R2 (Sr 4 is null at r = R1 ): −

dU4 = 2πl dt



R22 − R12 2

d dt



λω 2π

B . μ0

The sum of these two terms is:



1d  1 dU3 dU4 2πl R22 − R12 d λω + =− B = πl R22 − R12 (2Ba B) dt dt μ0 2 dt 2π 2 dt μ0 and is given by the superposition of the fields B and Ba in the magnetic energy density while the fields are changing. This term integrated over the time from the initial state, when ω = 0 and Ba = 0, to the final state, when B = 0, gives a net null contribution to the magnetic energy inside the two cylindrical shells. Finally we have to consider the components Sϕ1 ϕˆ and Sϕ2 ϕˆ of the Poynting vector: Sϕ1 ϕˆ = Es ×

B λ 1 1 B ϕˆ =− μ0 2π 0 μ0 r

Sϕ2 ϕˆ = Es ×

Ba λ 1 λω ϕˆ = μ0 2π 0 r 2π

both with circular field lines around the axis of the system. From (11.19) a flow of momentum: p=

Sϕ ϕˆ c

is associated to each of these components, with the same field lines of Sϕ ϕˆ and momentum density:

258

12 The Feynman Paradox

g=

Sϕ ϕˆ . c2

The momentum of the electromagnetic field in a volume dτ = r dr dϕ dz is: dp =

Sϕ ϕˆ r dr dϕ dz . c2

and the associated angular momentum is: dL = r × g dτ . In our case, due to the symmetry, there is a non null axial angular momentum that can be found by integration over the volume of height l, from R1 to R2 and for 0 < ϕ < 2π . For the component Sϕ1 ϕˆ we get an angular momentum per unit length of the cylindrical system: 1 L1 = l

R2 2π  l R1 0

Sϕ1 1 r dr dϕdz = 2 c l

0

R2 2π  l R1 0

0

1 c2



λ Bin 1 − r dr dϕdz 2π 0 μ0 r

that, remembering that c2 = 1/0 μ0 , results exactly: L1 = −

 λ 2 R − R12 Bi . 2 2

(12.8)

That is the initial angular momentum of the electromagnetic field between the cylindrical shells per unit length, and is the second member of the Eq. (12.4). Similarly for the component Sϕ2 ϕˆ we get: L2 =

 λ ω f in  λ λ  2 R2 − R12 μ0 = − R22 − R12 Baf in 2 2π 2

(12.9)

that is the final angular momentum of the electromagnetic field between the cylindrical shells per unit length, in the first member of the Eq. (12.5). Comments • From the equation for the motion of the rotating system and the FaradayNeumann law we have found the relation (12.5). This is the expression of the conservation of the total angular momentum of the system: when the external magnetic field Bin decreases and becomes null, the initial angular momentum (12.8) of the electromagnetic field (magnetic field Bin and electrostatic field E s ), is transferred to the rigid system of moment of inertia I and to the final f in angular momentum (12.9) of the electromagnetic field (magnetic field Ba and electrostatic field E s ) that is present between the two shells as a consequence

12.4 Two Cylindrical Shells with Opposite Charge in a Vanishing Magnetic Field

259

of the rotation. At the same time the initial magnetic energy that was inside the cylindrical shell of radius R2 escapes from the outer shell as in (12.6) and the f in magnetic energy (12.7), associated to the self induced magnetic field Ba , that appears between the two shells, enters in the space between the two charged shells. • The final angular momentum (12.9) of the electromagnetic field is due to the rotation of the electrostatic field between the shells. The initial angular momentum (12.8) cannot be associated to the motion of the fields because the shells are at rest. As for the angular momentum in the system with a magnetic dipole and a point charge, the initial angular momentum of the present system is given to the electromagnetic field when assembling the system. • Every point between the two shells can be considered as the origin of a frame S at rest in the laboratory, with the z axis parallel to the axis of the system, the y axis in the radial direction and the x axis to form with the other two a right handed frame. Then we consider the frame S  that at time t = t  = 0 has the origin and the axis coincident with those of S, but moves with the rotating system. At time t = t  = 0 the axis x  is in motion relative to x axis with velocity vx = −ωr . In the system S  the only non null component of the electric field is E y : E y =

λ 1 . 2π 0 r

At the initial time the system S is in motion relative to S  with velocity vx = ωr and transforming the field E y from S  to S, from the relations (10.11) we find: E x = 0 E y = γ E y E z = 0 B x = 0 B y = 0 Bz = γ

−vx  Ey c2

with 1 γ = 1 − β2

β=

ωr . c

Substituting the expressions of vx and of E y in Bz we find:  μ  0 Bz = γ − ωλ . 2π In the limit ωr  c this field becomes Ba and so it is evident that this field originates from the motion of rotation of the electrostatic field seen in the laboratory frame.

260

12 The Feynman Paradox

The Poynting vector: Sϕ2 ϕˆ =

Es × Ba μ0

is also a consequence of the rotation of the electrostatic field seen in the laboratory. ˆ We can associate to this vector a momentum density in the direction ϕ: E y E y S 1 E y Bz 1 β g= 2 = 2 = − γ 2 3 β = −0 γ 2 c c μ0 μ0 c c 2

2

and an energy density12 : 1 1 1 2 1 1 1 2 E y 2 1 2 2 0 E y2 + Bz = 0 E y γ 2 + γ 2 β = 0 E y γ 2 (1 + β 2 ) . 2 2 μ0 2 2 μ0 c 2 2

u=

Appendix: Vector Potential for a Spinning Charged Sphere In the exercises proposed at the end of this chapter one needs to know the vector potential inside and outside a sphere, of radius R with surface charge density σ , which rotates about a diameter with constant angular velocity ω. This problem is proposed in many textbook13 . Here we follow for the benefit of the reader, an elegant and quick solution given in the Griffiths’s book14 but we suggest you solve this problem before reading the solution. Solution In magnetostatics the vector potential A(r) in a point r induced by a density surface current J S (R ) is given by the formula: 12

In the frame S  the electrostatic energy density is: u =

1 2 0 E y 2

and from the previous relations it is easy to verify the relation: u  = u 2 − g 2 c2 2

similarly to the case of a particle of mass m at rest in S  : 2 E  = (mc2 )2 = E 2 − p 2 c2 .

13 14

See for instance Jackson [5], Problem 5.7. Griffiths [3], Example 5.11, p. 245.

Appendix: Vector Potential for a Spinning Charged Sphere

261

Fig. 12.4 Sphere with density charge σ on its surface spinning with angular velocity ω

μ0 A(r) = 4π

 Sph

J S (R ) dS |r − R |

(12.10)

where R is the position of the density current. The clever choice is now to position the point r on the z axis and the angular velocity ω on the x z plane at angle θ from the z axis as in Fig. 12.4. For R we can use spherical coordinates (R, θ  , ϕ  ) and the cartesian coordinates are (R sin θ  cos ϕ  , R sin θ  sin ϕ  , R cos θ  ). For ω the components are respectively (ω, θ, 0) and (ω sin θ, 0, ω cos θ ) while d S = R 2 sin θ  dθ  dϕ  . The density current is J S = σ v and the velocity of a point on the rigid sphere from classical mechanics is v = ω × R so we have for J S : J S = ω R σ [(− cos θ sin θ  sin ϕ  ) iˆ + (cos θ sin θ  cos ϕ  − sin θ cos θ  ) jˆ + (sin θ sin θ  sin ϕ  ) kˆ ]

(12.11)

When the expression (12.11) is introduced in (12.10), the integrals of the x and z components of the current and the first term of the y component give null results because:  2π  2π sin ϕ  dϕ  = 0 cos ϕ  dϕ  = 0 0

0

and the only non null component for A(r) is: A y (r) = −

μ0 4π

 Sph

ω R σ sin θ cos θ  2 R sin θ  dθ  dϕ  |r − R|

μ0 ω R 3 σ sin θ Ay = − 2



1

−1

cos θ  d cos θ  √ r 2 + R 2 − 2 r R cos θ 

262

12 The Feynman Paradox

We introduce ξ = cos θ  and after we integrate by parts: 

1

−1

 1   1  ξ dξ 2 + R 2 − 2r Rξ  = r 2 + R 2 + r Rξ r = 3r 2 R 2 r 2 + R2 − 2 r R ξ −1     1  2 r + R 2 + r R |r − R| − r 2 + R 2 − r R (r + R) 3r 2 R 2

the result is: 2 r 3 R2

for r < R

and

2 R 3 r2

for r > R

thus for the potential we find: for r ≤ R for r ≥ R

A(r) = − A(r) = −

μ0 ω R σ μ0 R σ r sin θ yˆ = (ω × r) 3 3

μ0 R 4 σ 1 μ0 ω R 4 σ sin θ y ˆ = (ω × r) 3 r2 3 r3

where we use the relation ω × r = −ω r sin θ yˆ and we get for A two vectorial relations. If we rotate the system around the y axis so that ω is on the z axis while r has spherical components r, θ, ϕ, from the relation ω × r = ω r sin θ ϕˆ we have the potentials: μ0 ω R σ r sin θ ϕ. ˆ (12.12) for r ≤ R A(r) = 3 for r ≥ R

A(r) =

μ0 ω R 4 σ sin θ ϕˆ 3 r2

(12.13)

The field B = ∇ × A inside the sphere (r < R) is: B=

2 μ0 σ ω R zˆ 3

(12.14)

which is uniform and in the zˆ direction while outside (r > R) the components are: Br =

μ0 σ R 4 ω 2 cos θ 3 r3

Bθ =

μ0 σ R 4 ω sin θ 3 r3

Bϕ = 0

(12.15)

Problems

263

Fig. 12.5 The charged wire with the opposite charged coaxial cylindrical shell in the vanishing magnetic field

Problems 12.1 An insulating cylindrical shell, of radius rc and length l, has its axis on the z axis where also lies an infinite long wire with linear charge density −λ (see Fig. 12.5). The cylindrical shell has moment of inertia I and a uniform surface charge density σ = λ/2πrc . The system, initially al rest, is embedded in an external magnetic field Bext zˆ that at time t = 0 begins to decrease slowly to zero. Determine the final angular velocity of the cylindrical shell: (a) from the equation for the motion of the shell, (b) from the conservation law of the angular momentum of the system. This problem has been extensively considered by many authors. See the article by Belcher and MacDonald: Feynman Cylinder Paradox [4], where relativistic corrections are considered. Many references on the Feynman paradox can be found in this paper. 12.2 Show that the angular momentum Lem of the electromagnetic field for a point charge Q in the origin in presence of a magnetic field B = B(x, y)ˆz is: Lem = −

Q Φ(B) zˆ 2π

where Φ(B) is the flux of B(x, y) through the plane z = 0. 12.3 An insulating spherical shell of radius R rotates with angular velocity ω = ωˆz around its axis. A uniform charge density σ is distributed on the surface. Find the angular momentum L E M of the electromagnetic field.

264

12 The Feynman Paradox

12.4 An insulating spherical shell of radius R and with a charge Q uniformly distributed on its surface is initially at rest in a magnetic field Bext = Bext zˆ . Then the magnetic field is damped. Find the equation of motion and the final angular velocity of the spherical shell. The solution from the angular momentum conservation presents a divergence because the initial magnetic field extends to infinity. Suggest a shape for the Bext field to avoid this difficulty. 12.5 An iron sphere of radius R, moment of inertia I , and uniform magnetization M M = M M zˆ is initially at rest. A charge Q is uniformly distributed on a thin insulating layer surrounding its surface. (a) Compute the angular momentum of the electromagnetic field. If the sphere is demagnetized, for instance by heating it over the Curie point, the sphere starts to rotate. Find the final angular velocity: (b) from the conservation of angular momentum, (c) from the Faraday-Neumann law. 12.6 Two concentric spherical shells fixed together, with radii R A an R B (R B > R A ) and with opposite charges Q and −Q uniformly distributed on their surfaces, are initially in a uniform magnetic field B = B zˆ . The moment of inertia of the two connected spheres is I . Find the angular velocity of the two shells after the magnetic field has been switched off: (a) from angular momentum conservation; (b) from the Faraday-Neumann law.

Solutions 12.1(a) From the equation of motion and the Faraday’s law. While the magnetic field decreases, there is an induced electric field Ei tangent to a circle of radius rc concentric with the shell, that can be found by the Faraday’s law: f =−

dΦ(B) dt

2πrc E i = −πrc2

dB dt

Ei = −

rc d B ϕˆ . 2 dt

The force applied to an element d of the lateral surface of the cylinder is: dF = σ Ei d and its torque relative to the z axis is: dM = rc × dF = rc × σ Ei d = −rc × σ The total torque is:

rc d B ϕˆ d . 2 dt

Solutions

265

M=−

rc2 d B (2πrc lσ ) zˆ 2 dt

and the equation of motion is: I

r2 d B dω r2 d B =− c (2πrc lσ ) = − c (λl) dt 2 dt 2 dt

that integrated gives:  λ  I  ω = − rc2 B f in − Bext 2

B f in = μ0

λω 2π

I =

I l

where B f in is the magnetic field generated by the rotation of the charged shell (see point b). The final angular velocity is: λrc2 Bext 2

ω=

1  . μ0 λ2 rc2  I 1+ 4π I 

(b) From angular momentum conservation. With the system at rest, the electric field and the Poynting vector in the space at r < rc are: E=−

λ 1 rˆ 2π 0 r

Sin =

λ 1 Bext ϕˆ . 2π 0 r μ0

The momentum density associated to Sin from (11.18) is: gin =

Sin λ 1 Bext ϕˆ = 2 c 2π r

and the initial angular momentum of the electromagnetic field per unit length is: Lin = l



rc

r × gin 2πr dr =

0

λBext 2 r zˆ . 2 c

At the end the shell is rotating with an angular velocity ω and the surface charge becomes a surface current: i=

λl λl = ω T 2π

and for r < rc there is a magnetic field:

266

12 The Feynman Paradox

B f in = μ0

λω zˆ . 2π

The final momentum density of the electromagnetic field is: S f in λ 1 = B f in ϕˆ 2 c 2π r

g f in =

and the angular momentum per unit length is: L f in = l



rc

r × g f in 2πr dr = μ0

0

λ2 rc2 ω zˆ . 4π

The angular momentum conservation gives: I  ω + μ0

λBext 2 λ2 rc2 ω= r 4π 2 c

and thus the final angular velocity is: ω=

λrc2 Bext 2

1

. μ0 λ2 rc2  I 1+ 4π I 

12.2 The momentum density of the electromagnetic field is: S 1 E×B = 2 c2 c μ0

g=

and the angular momentum for the field in a small volume dτ is: dLem = r × g dτ =

1 1 r × (E × B) dτ = [E (B · r) − B (E · r)] dτ μ0 c2 μ0 c2

In cylindrical coordinates R, ϕ, z: r ( R, 0, z)

E

R z Q Q 3 , 0, 4π 0 [R 2 + z 2 ] 2 4π 0 [R 2 + z 2 ] 23

B ( 0, 0, B)

and dτ = R d R dϕ dz. The only non null component is along the z axis: L em,z = −

Q 4π



∞ 0

 0

2π



∞

B R2

−∞

[R 2 + z 2 ] 2

3

R d R dϕ dz

Solutions

267



∞ −∞

L em,z = −



Q 2π

1 [R 2 + z 2 ] ∞

0



2π

3 2

2 R2

dz =

B R d R dϕ = −

0

Q Φ(B) 2π

12.3 The angular momentum is present only outside of the shell where the electric field is non null. In spherical coordinates the surface current density associated to the rotating charge density σ is: ˆ Js = σ ω R sin θ ϕ. The magnetic field produced at a point r from the center of the spinning shell by this current is given in the Appendix to this chapter. For r ≥ R the components: Br =

μ0 σ R 4 ω 2 cos θ 3 r3

Bθ =

μ0 σ R 4 ω sin θ 3 r3

Bϕ = 0

(12.15)

For r ≥ R the electric field is: 1 Q rˆ 4π 0 r 2

E=

Q = 4π R 2 σ

thus the momentum density is: g=

S μ0 Q 2 R 2 ω sin θ = ϕˆ c2 (4π )2 3 r5

The angular momentum L E M in the z direction is:  LEM =

μ0 Q 2 R 2 ω g r sin θ dτ = (4π )2 3 LEM =





2π

π

dϕ 0

 sin θ dθ 3

0

41 μ0 Q 2 R ω μ0 Q 2 R 2 ω 2π = (4π )2 3 3R 2π 9

∞ R

1 dr r2 (12.16)

12.4 From equation of motion and Faraday-Neuman law. At the interior of a charged spherical shell spinning with angular velocity ω = ω zˆ around the z axis the magnetic field is uniform and is given by 12.14: Bsph =

2 μ0 σ R ω 3

σ =

Q . 4π R 2

We put the center of the sphere in then origin of the axes. To find the electric field on the charged surface we use the Faraday-Neumann law with a circular loop which

268

12 The Feynman Paradox

is the intersection of the plane z = R cos θ with the spherical shell: 2π R sin θ E i = −

d Φ(B) = − (π R 2 sin2 θ Btot ) dt dt

where Btot = Bext + Bsph . This yields: d Btot 1 E i = − R sin θ 2 dt The torque applied to an element d S = R 2 sin θ dθ dϕ of the surface is: 1 d Btot d Mz = (R sin θ )(σ d S)E i = − π (R sin θ )2 σ R 2 sin θ dθ dϕ 2 dt and after integration: 1 Mz = − R 4 σ 2





2π

π

sin3 θ

dϕ 0

0

4 1 d Btot d Btot d Btot = − π σ R4 = − Q R2 dt 3 dt 3 dt

The equation of the motion for the shell is: I

1 dω d Btot = − Q R2 dt 3 dt

dω d Bext dω 1 1 = − Q R2 − μ0 Q 2 R dt 3 dt 18 π dt

1 1 d Bext dω μ0 Q 2 R = − Q R2 I+ 18 π dt 3 dt

I

and after integration: I+

1 μ0 Q 2 R 18 π

ω=

1 Q R 2 Bext 3

(12.17)

so the final angular velocity is ω=

1 3

Q R 2 Bext 1 μ0 Q 2 R I+ 18 π

In the equation (12.17) the second term at left is the angular momentum of the spinning shell (12.16) already found in Problem 12.3. From angular momentum conservation.

Solutions

269

The final angular momentum is the sum of the momenta of the electromagnetic field of the spinning shell plus the angular momentum of the shell: L f in = I ω +

1 μ0 Q 2 R ω 18 π

We have to find the initial angular momentum of the electromagnetic field. The electric field inside the spherical shell is zero. Outside for r > R is: E=

1 Q rˆ 4π 0 r 2

In spherical coordinates the magnetic field Bext = Bext zˆ has components (Bext cos θ, −Bext sin θ, 0). The Poynting vector is: S=

E × Bext Bext sin θ Q =− ϕˆ μ0 4π 0 μ0 r 2

and the momentum density: g=

S Bext sin θ Q =− ϕˆ . 2 c 4π r2

After integration inside a sphere of radius R∞ , the angular momentum with respect to the z axis is:  L in (R < r < R∞ ) =

R q if the charge moves towards point P while q  < q if the charge moves away. Finally the scalar potential V for a particle in motion can be derived by replacing q with q  in the potential of a charge at rest: 4

See Feynman Lectures on Physics [3], Vol. II, Sect. 21.5; A. Zangwill [4], Sect. 23.2.2.

14.2 Potentials and Fields for a Point Charge in Arbitrary Motion

299

Fig. 14.4 The charge in ABCD moves to A’B’C’D’ and contributes to the potential in P at time t for the retarded time interval dt  = d R  /c

V =

1 q 4π0 R

→

V =

1 q 1 q = 4π0 R 4π0 R (1 − nˆ · β)

that is the potential (14.10).

14.2.4 Covariant Form of the Liénard-Wiechert Potentials An elegant way to derive the Liénard-Wiechert potentials is based on arguments of relativistic covariance. In Minkowski space the scalar and the vector potentials are the components of a 4-potential, so for a charged particle in motion we have to expect a 4-potential Aμ that in the rest frame of the charge has components: V (r, t) =

1 q 1 q = 4 π 0 R 4 π 0 c(t − t  )

A(r, t) = 0

The only available quantities to write this potential are the 4-vector R μ (c(t − t  ), R), that is the distance of the point of observation from the particle, with Rμ R μ = c2 (t − t  )2 − R 2 = 0 and the 4-velocity u μ (cγ, vγ) which accounts for the motion of the particle. The scalar term c(t − t  ) = R in V (r, t) suggests to write in the lower part of V (r, t) the scalar quantity Rν u ν = c2 (t − t  )γ − R · vγ = c Rγ − R · vγ = c R (1 − nˆ · β)γ while in the upper part we need a 4-vector and this can be u μ that in the rest frame has null the spatial components and constant the time component. Thus for Aμ (A0 = V /c, A) we expect: Aμ =

q 1 uμ μ0 q c u μ = 4 π 0 c Rν u ν 4π Rν u ν

Replacing in this formula Rν u ν = c R (1 − nˆ · β)γ, and u μ (cγ, vγ) we have:

300

14 Fields and Radiation

V = c A0 =

q 1 4π0 R (1 − nˆ · β)

Ai =

q vi μ0 4π R (1 − nˆ · β)

which are the formulas (14.10) and (14.11).

14.2.5 Electric and Magnetic Fields The electric and magnetic fields can be calculated directly from the potentials using the relations: ∂A B=∇×A (14.12) E = −∇V − ∂t Here we give only the results of these derivatives, the reader can find the very long calculations in next section. The fields from the relations (14.12) are: q E(r, t) = 4π0 μ0 q B(r, t) = 4π





˙ 1 nˆ × [(nˆ − β) × β] +  3  3 c 1 − nˆ · β R γ 2 1 − nˆ · β R 2 nˆ − β



˙ c β × nˆ nˆ × {nˆ × [(nˆ − β) × β]} +  3  3 2 γ 1 − nˆ · β R 2 1 − nˆ · β R 1 B = nˆ × E c

= E v + Ea r et



(14.13) = Bv + Ba

r et

(14.14) (14.15)

where the values in the curly brackets have to be taken at the retarded time t  . Both fields can be decomposed in two terms. The first ones Ev and Bv are the velocity terms which depend on the velocity of the charge and fall off as R −2 so that in (14.13) these fields are mainly the Coulomb field dragged by the charge in its motion.5 The second ones are the acceleration terms Ea and Ba , which are non null if the charge is accelerated and decrease as R −1 . For the electric and magnetic fields and separately for the velocity and acceleration terms, the following relations hold: E·B=0

|B| ≤

1 |E| c

B · nˆ = 0

At large distance only the acceleration terms are present and Ea , Ba and nˆ form a triad of normal vectors:

5

These fields for a point charge in uniform motion become the fields found by a Lorentz transformation in Sect. 14.1. See Problem 14.2.

14.2 Potentials and Fields for a Point Charge in Arbitrary Motion

1 Ba = nˆ × Ea c

Ea · Ba = 0

Ea · nˆ = 0

301

Ba · nˆ = 0

(14.16)

and as we will see in next section these fields are radiated by the accelerated charge. Their properties are just those expected for electromagnetic waves given by the relation (1.35) and its comments in Sect. 1.10.

14.2.6 Calculation of the Electric and Magnetic Fields The derivatives in (14.12) need some attention because the potentials V and A are function of r and of t  and are implicit functions of r and t through the retarded time: t = t −

|r − r0 (t  )| c

so it is useful to write first the derivatives6 of t  with respect to t and to the position r (that is the gradient of t  ). Starting from: t = t −

R c

and

we find:

R = [(r − r0 (t  )) · (r − r0 (t  ))] 2

1

R·v ∂R =−  ∂t R dt  = dt +

1 ∂R  R·v  dt dt = dt + c ∂t  cR

  R·v = dt dt 1 − cR 

(14.17)

∂t  = ∂t

R R·v R− c

(14.18)

1R·v 1 1 Rx gradx t  = − gradx R = − + gradx t  c c R c R Rx  gradx t  = −  R·v c R− c and from similar relations for the y and z components: In this section grad is the gradient associated to a change in R that depends both on r and t  while ∇ is the gradient which operates only on r.

6

302

14 Fields and Radiation

R  grad t  = −  R·v c R− c

(14.19)

With these formulas for the derivatives of t  and R it is possible to find the fields from the relations (14.12). Electric field For the first term of the electric field in (14.12): grad V =

q grad  4π0

  1 1 R·v q  =− grad R −   R·v 4π0 c R·v 2 R− R − c c

(14.20) - for grad R from (14.17) and (14.19) we have: grad R = ∇ R +

∂R R R·v grad t  = +  ∂t R R

R R  = −  R·v R·v R− c R− c c 

- and for grad (R · v/c) using (14.19) : grad

R·v 1 ∂ R·v v 1 =∇ + (R · v) grad t  = − (v 2 − R · v˙ ) grad t   c c c ∂t c c grad

v 1 R·v R = = + (v 2 − R · v˙ )  R·v c c c c R− c

 

 R·v v2 R · v˙ v R− + 2R− 2 R  c c c c

1  R·v R− c

and substituting these results in (14.20): q grad V = −  4π0

     v2 v R·v R · v˙ 1− 2 R− R− + 2 R  c c c c R·v 3

1 R−

c

Consider now the second term of the electric field in (14.12) and use relations (14.17) and (14.18):

14.2 Potentials and Fields for a Point Charge in Arbitrary Motion

μ0 q ∂ ∂A  = ∂t 4π ∂t

=

v v 1 q ∂  = 2 = R·v R·v c 4π0 ∂t R− R− c c

⎧ ⎪ ⎪ ⎨

v˙ 1 q v  − 2 2 ⎪ R · v c 4π0 ⎪ R − R·v ⎩ R− c c

q = 4π0



v˙ c2



303

R·v R− c



v − 2 c

⎫ ⎪ ⎪ ⎬

∂t  ∂R ∂ R·v = − ∂t  ∂t  c ⎪ ∂t ⎪ ⎭

 R · v v2 R · v˙ − + −  R c c

R R−

R·v c

3

Substituting the two relations for grad V and ∂A/∂t in (14.12) for the electric field we have: E=

q  4π0

1 R·v R− c

3 ·



    vR R · v˙ v2 v˙ R·v R · v˙ 2 R− + 2 R− 2 R − 1− 2 R −vR 3 c c c c c c Now using the relation a × (b × c) = b (a · c) − c (a · b) we can write:  1  1 R · v˙ R − v˙ R 2 = 2 R × (R × v˙ ) 2 c c and:

v˙ R · v R · v˙ R R R − v R 3 = 3 R × (˙v × v) = − 3 R × (v × v˙ ) c2 c c c c

and adding the last two relations we have: 1 R× c2

 

 R R − v × v˙ c

so finally for the electric field we find: q E=  4π0

1 R−



 R·v 3 c

or in a more compact form:

v2 1− 2 c



vR R− c



1 + 2 R× c



  R R − v × v˙ c

r et

304

14 Fields and Radiation



q E= 4π0

˙ nˆ − β 1 nˆ × [(nˆ − β) × β] + 2 3 2 3 γ (1 − nˆ · β) R c (1 − nˆ · β) R

r et

Magnetic field For the magnetic field we proceed by considering its components. We begin from Bx : ∂ Ay ∂ Az Bx = − (14.21) ∂y ∂z The y and z components of the magnetic potential are: Ay =

μ0 q  4π

vy  R·v R− c

Az =

μ0 q  4π

vz  R·v R− c

For the first term in (14.21) using (14.19): ∂  ∂y

∂vz vz 1 ∂ 1  =  = + vz R·v R · v ∂y R·v ∂y R− R− R− c c c =

∂vz vz 1  grad y t  −   · R · v ∂t  R·v 2 R− R− c c

 ∂R ∂R ∂ R·v ∂ R·v   +  grad y t − −  grad y t = ∂y ∂t ∂y c ∂t c v˙ z R y vz =−  2 −   · R·v 2 R·v R− c R− c c

⎡ ⎢ Ry R·v ·⎢ ⎣ R + R

Ry vy − − R·v c c R− c 



v2 R · v˙ − c c



⎤ ⎥ Ry ⎥ = R·v ⎦ c R− c 

v˙ z R y vz =−  2 −   · R·v 3 R·v R− c R− c c

14.2 Potentials and Fields for a Point Charge in Arbitrary Motion



Ry R

R·v R− c



vy R · v Ry + − R c c



R·v R− c



305

 v2 R · v˙ − 2 Ry + 2 Ry = c c

     v˙ z R y vy vz v2 R·v R · v˙ =−  − · R R 1 − − R − + y y    c2 c c c2 R·v 2 R·v 3 c R− R− c c

v˙ z R y 1 =−  2 −   · R·v 3 R·v R− c R− c c     

vz v y v2 R·v R · v˙ R− + 2 vz R y vz R y 1 − 2 − c c c c We find ∂ A y /∂z by swapping y and z in the last relation and so for Bx we have: Bx =

∂ Ay ∂ Az − = ∂y ∂z

⎧ ⎪ ⎪ ⎪ ⎨

⎫ ⎪ ⎪ ⎪ 

  ⎬ v˙ z R y − v˙ y Rz R · v˙ μ0 q 1 v2 −  + = − · (v R − v R ) 1 − (v R − v R ) z y y z z y y z    2 3 2 2 ⎪ 4π ⎪ c c R · v R · v ⎪ ⎪ ⎪ ⎪ ⎩ c R− ⎭ R− c c

from this result it is easy to figure out the expressions for B y and Bz and we can write: ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 

  ⎬ R × v˙ R · v˙ μ0 q ⎨ 1 v2 −  = + B= − · (R × v) 1 − (R × v)    2 3 2 2 ⎪ 4π ⎪ c c R · v R · v ⎪ ⎪ ⎪ ⎪ ⎩ c R− ⎭ R− c c

B=

B=

μ0 q  4π

μ0 q  4π

B=

1 R−

R·v c



1 R−

μ0 q  4π

R·v c

3

1 R−

    R·v 1 R · v˙ v˙ −R × R− − (R × v) 2 − 2 (R × v) = c c γ c

3

R·v c

−R ×

 3

 R · v˙ v˙ 1 v˙ R · v R+R× − (R × v) 2 − 2 (R × v) = c c c γ c

1 1 v˙ R × [˙v (R · v) − v (R · v˙ )] − R × R − (R × v) 2 c2 c γ

Now using again the relation a × (b × c) = b (a · c) − c (a · b):

 =

306

14 Fields and Radiation

B=

μ0 q  4π

  1 1 v˙ ˙ − = R − (R × v) R × × (v × v − R × [R ]  c2 c γ2 R·v 3

1 R−

μ0 q B= 4π



c ! ˙ − nˆ × β˙ −nˆ × nˆ × (β × β) (β × n) ˆ c   +  3 γ 2 R 2 1 − nˆ · β 3 R 1 − nˆ · β r et

and with the same vector relation it is easy to verify that: ˙ = −nˆ × β˙ nˆ × [nˆ × (nˆ × β)] so the previous expression for B finally becomes: μ0 q B= 4π



nˆ × nˆ × (nˆ − β) × β˙ c (β × n) ˆ +    3 3 γ 2 R 2 1 − nˆ · β R 1 − nˆ · β

! r et

14.3 Radiation by an Accelerated Charge Electric charges, when accelerated, radiate electromagnetic energy which propagates at large distances. The radiation is determined by the acceleration terms in the fields of the charges given in (14.13) and (14.14) which fall off as R −1 at large distance while the velocity term depend on R −2 . So the relevant part of the electric field is: q 1 E(r, t) = 4π0 c



˙ nˆ × [(nˆ − β) × β]  3 1 − nˆ · β R

(14.22) r et

Replacing in the Poynting vector the relation (14.15) for the magnetic field we have: S=

E×B E × (nˆ × E) 1 1 = ˆ = = [E 2 nˆ − E(E · n)] E 2 nˆ μ0 μ0 c μ0 c μ0 c

(14.23)

where E · nˆ is null because E is normal to nˆ in (14.22) and in (14.16). The flux of energy dΦ radiated in a solid angle dΩ in the direction nˆ is dΦ = S · nˆ R 2 dΩ and we have: dΦ 1 = S · nˆ R 2 = E 2 R2 dΩ μ0 c

(14.24)

14.3 Radiation by an Accelerated Charge

307

14.3.1 Radiation by a Charged Particle with Velocity v  c For a charged particle at a velocity v  c the field in (14.22) can be approximated by:  ˙ q 1 nˆ × [nˆ × β] E(r, t) = 4π0 c R r et

and we have:  E = 2

q 1 4π0 c

2

  q 1 2 1 ˙2 2 1 " ˙ 2  ˙ 2 # β − nˆ · β = β sin θ R2 4π0 c R2

˙ The Poynting with θ the angle between the direction of nˆ and the acceleration βc. vector is: μ0 c q 2 ˙ 2 2 β sin θ nˆ S= 16π 2 R 2 and from relation (14.24) for the flux of energy dΦ in dΩ we can write: dΦ μ0 c q 2 ˙ 2 2 β sin θ = dΩ 16π 2

(14.25)

The flux is symmetric with respect to the normal to the acceleration where the energy radiated has a maximum. After integration on the solid angle we have the Larmor formula for the power radiated by a non relativistic charge subject to an acceleration a = v: ˙ P=

μ0 2 2 μ0 2 2 q v˙ = q a 6πc 6πc

(14.26)

14.3.2 Radiation by a Charged Particle with the Acceleration Parallel to the Velocity A second easy configuration is that of an acceleration in the same direction of the velocity. The field (14.22) becomes: q E(r, t) = 4π0



˙ 1 nˆ × [nˆ × β]  3 c 1 − nˆ · β R

r et

The flux of energy per unit solid angle in the direction nˆ is now:

308

14 Fields and Radiation

Fig. 14.5 Angular distributions of the power radiated by a point charge with acceleration parallel to the velocity for: β  1 (β = 0), β = 0.3 and β = 0.6. The radiated power at θ = 90◦ is the same for all the velocities. The arrows show the directions of maximum radiated power for the three values of β at θ = 90◦ , 54◦ and 32◦

μ0 c q 2 dΦ β˙ 2 sin2 θ = dΩ 16π 2 (1 − β cos θ)6 This flux is the energy dE in dΩ seen by the observer in a time interval dt but if we want the energy radiated in dΩ by the source while it moves we have to consider a time interval dt  so we have: 1 dE 1 dE dt dΦ dt dΦ dΦ  = (1 − nˆ · β) = = =    dΩ dΩ dt dΩ dt dt dΩ dt dΩ where we have used the relation (14.18) so we have: dΦ  μ0 c q 2 β˙ 2 sin2 θ = 2 dΩ 16π (1 − β cos θ)5

(14.27)

This distribution at the numerator depends on the acceleration as in the Larmor formula while at the denominator is connected to the Doppler effect due to the Lorentz transformation7 from the particle rest frame to the laboratory and for relativistic particles the power is emitted near by the direction of the speed. 7

This can be understood simply. As seen in Sect. 10.11 the wave vector is the spatial component of  are the parallel and transverse the 4-vector k (ω/c, k). If in the frame of the charge at rest k and k⊥ components of the wave vector with respect to the direction of the motion, in the frame where the charge has a velocity βc these are:   ω  k = γ k + β with ω  = k  c . k⊥ = k⊥ c If θ and θ are the angles of the wave vector with respect to the speed in the two frames:

14.3 Radiation by an Accelerated Charge

309

The distribution dΦ  /dΩ at an angle θ with respect to the direction of the velocity is shown in Fig. 14.5 for different values of β. The distribution depends on β˙ 2 thus is the same for acceleration or deceleration and it is symmetric around the direction of the motion. At a given acceleration it depends only on the velocity and it is the same at θ = 90◦ for all the velocities. There is a maximum at the angle θmax given by the relation: θmax = cos

−1

$  15β 2 + 1 − 1 3β

lim θmax =

β→1

1 2γ

with the last limit typical of radiation emission at high energy. High-energy charged particles crossing matter when decelerated in Coulomb field of the nuclei (or of the atomic electrons) radiate photons also with a large fraction of their energy. This process, called bremsstrahlung (braking radiation) is very relevant for the energy loss in matter by light particles as electrons and positrons. Its description demands a calculation performed in Quantum Electrodynamics. The angle of emission of the photon by an electron is ∼ m e c2 /Ee = 1/γ with m e the mass and E the energy of the electron. After integration of the flux (14.27) on the solid angle, the power radiated by the charge is:   μ0 2 2 6 μ0 q 2 dp 2 (14.28) P = q v˙ γ = 6πc 6πc m 2 dt where m and p are the mass and the momentum of the particle. This relation can be used to find the power radiated by a charged particle in a linear accelerator. From the relativistic expression for the energy E of the particle we have: (mc2 )2 dE c β˙ = E3 dx where dE/d x is the increase of energy per unit path during acceleration. We can write: μ0 q 2 c4 1 dE dE P = 6πc (mc2 )2 v dt d x and in the limit β → 1 the ratio between the power radiated and the increase of the particle energy per unit of time is: cot θ =

  γ(k + k  β) k β = = γ cot θ +   k⊥ k⊥ sin θ

and so also a wave, emitted at θ = π/2 in the rest frame of the charge, is observed in the laboratory at an angle with the speed: 1 θ= γβ and in the limit β → 1 θ → 0.

310

14 Fields and Radiation

μ0 q 2 c4 1 dE P = dE/dt 6πc (mc2 )2 v d x For a relativistic electron substituting its classical radius (in S.I. units): re = we have:

1 e2 4π0 mc2

P 2 re dE dE = = 3.67 × 10−12 2 dE/dt 3 mec d x dx

where E is in GeV and x in meters and we find that the radiation loss is very negligible during the linear acceleration. In fact in this relation dE/d x is compared with m e c2 /re = 1.8 × 1011 GeV/m, an increase in energy equal to 0.511 MeV in a length equal to the classical radius of the electron re = 2.82 × 10−15 m, that is a value far beyond any possible reach in accelerators. At the LHC at CERN the eight superconducting RF cavities deliver an accelerating field of 5 MV/m, in the future linear colliders (ILC, CLIC) gradients in the range 30–100 MV/m are foreseen. In plasma accelerators driven by laser or electron beam, gradients in the range 10–100 GV/m could be achieved in the future.

14.3.3 Power Radiated by a Charged Particle in Arbitrary Motion To find the power radiated by a particle in arbitrary motion we have to replace the field (14.22) in (14.24): dΦ  μ0 c q 2 1 = · dΩ 16π 2 (1 − nˆ · β)5 %

(14.29) &

˙ 2 (1 − β 2 ) + 2 (β˙ · β) (nˆ · β) (1 − nˆ · β) β˙ 2 (1 − nˆ · β)2 − (nˆ · β)

The power radiated by the charge is the integral of this flux. The result was first given by Liénard. The integration needs some work but the result can be derived more quickly from the Larmor formula (14.26) exploiting the invariance of the power under Lorentz transformations8 In the charge rest frame if the energy dE  is radiated in a time interval dτ , the power is: dE  P = dτ 8

See for instance A. Zangwill [4], Sect. 23.3.4 and D.J. Jackson [2], Sect. 14.2.

14.3 Radiation by an Accelerated Charge

311

and for the symmetry of the flux in (14.25), the momentum of the radiation emitted in dτ is dp = 0. In a frame where the charge moves with velocity v, after the Lorentz transformation we have dE = dE  γ and dt = γ dτ thus: P=

dE  dE = = P dt dτ

So to write the general formula for the power radiated we have to find a scalar expression which in the instantaneous rest frame of the charge is equal to the Larmor result (14.26). The only vectors we have to describe the phenomenon are the velocity v and the acceleration a of the charge. It seems reasonable to replace in (14.26) the acceleration with the 4-acceleration a μ = du μ /dτ where u μ (cγ, vγ)) is the 4velocity (10.13) and then write the expression in terms of the 4-momentum (10.14): μ0 2 q P=− 6πc



du μ du μ dτ dτ



μ0 q 2 =− 6πc m 2



dpμ dp μ dτ dτ



Finally, replacing the components p μ (mcγ, mvγ) of the 4-momentum, after some algebra we get the Liénard formula for the power radiated by a charged particle in motion: μ0 c 2 6 ˙ 2 P= q γ [β − (β˙ × β )2 ] 6π which becomes the Larmor formula (14.26) in the limit v  c.

14.3.4 Radiation by a Charged Particle with the Acceleration Normal to the Velocity If the acceleration is normal to the velocity the (14.29) becomes: % & μ0 c q 2 1 d ˙ 2 (1 − β 2 ) = · β˙ 2 (1 − nˆ · β)2 − (nˆ · β) d 16π 2 (1 − nˆ · β)5 In the limit of small velocity β  1 we have: μ0 c q 2 ˙ 2 dΦ  ˙ 2] = [β − (nˆ · β) dΩ 16π 2 and if we take the z axis in the direction of the acceleration we find: dΦ  μ0 c q 2 ˙ 2 = β sin2 θ dΩ 16π 2

(14.30)

312

14 Fields and Radiation

Fig. 14.6 Frame x yz centered on the charge q with the velocity v on the z axis and the acceleration, normal to v, on the x axis

that is the same angular distribution around the direction of the acceleration found in (14.25). The flux of radiation can be studied in an instantaneous frame centered on the particle with the velocity on the z axis and the acceleration on the x axis (see Fig. 14.6). In this frame the direction of emission is nˆ (sin θ cos ϕ, sin θ sin ϕ, cos θ) and so the distribution is:   sin2 θ cos2 ϕ μ0 c q 2 ˙ 2 1 dΦ  1 − (14.31) = β dΩ 16π 2 (1 − β cos θ)3 γ 2 (1 − β cos θ)2 The distribution of the flow of radiation for β = 0, 0.3, 0.6 and 0.9 is shown in Fig. 14.7. For β → 1 the Doppler effect seen in Sect. 14.3.2 gives a peak at θ = 0, that is in the direction of the velocity, and the distribution can be approximated by the relation:   4γ 2 θ2 cos2 ϕ μ0 c q 2 ˙ 2 8γ 6 dΦ  1−  β dΩ 16π 2 (1 + γ 2 θ2 )3 (1 + θ2 γ 2 )2 In circular particle accelerators (cyclotrons, betatrons, synchrotrons, colliders) the acceleration is normal to the velocity. For β  1 the radiation is distributed as in (14.30) (cyclotron radiation) while for β → 1 the strong emission in the forward direction is the synchrotron radiation. The synchrotron radiation presents interesting properties9 : the possibility to select on a broad spectrum from microwaves to hard X-rays, to deliver high fluxes collimated in space, pulsed in very short times and also polarized. All features that can be exploited in many research domains (solid state physics, chemistry, biology, geology and cultural heritage). Initially the synchrotron radiation was used parasitically in the high energy storage rings but later dedicated synchrotron radiation facilities10 have been built and special devices as wigglers and undulators have been inserted in the rings. 9

A general introduction to this subject can be found for instance in D.J. Jackson [2], Sects. 14.6 and 14.7. 10 See the list at https://en.wikipedia.org/wiki/List_of_synchrotron_radiation_facilities.

14.3 Radiation by an Accelerated Charge

313

Fig. 14.7 Angular distributions of the power radiated in the plane of the velocity and the acceleration (ϕ = 0 in (14.31), full lines) and in the plane normal to the acceleration (ϕ = 90◦ in (14.31), dashed lines) by a charge with the same acceleration normal to the velocity for: β  1 (β = 0), β = 0.3, β = 0.6 and β = 0.9. The arrows show the directions of the velocity and of the acceleration. In the figure the flux for β = 0.6 is scaled by a factor 2 and for β = 0.9 by a factor 100 with respect to the fluxes for β = 0 and β = 0.3

Synchrotron radiation is also a powerful tool in astrophysics wherever charged relativistic particles travel through magnetic fields. Pulsed emission of gamma-rays up to ≥25 GeV has recently been observed from the Crab nebula, probably due to synchrotron emission by electrons trapped in the strong magnetic field around its pulsar. Also the typical polarization of the synchrotron radiation is present in the 0.1–1 MeV photons from this nebula. The observation of synchrotron radiation emitted by relativistic electrons moving through the interstellar or the intergalactic medium allows astronomers to make inferences about the strength and orientation of the magnetic fields in these regions. The total power radiated by an accelerated charge can be found by integration of ˙ = β β: ˙ (14.31) or from the Liénard formula with |β × β| P=

μ0 c 2 4 ˙ 2 q γ β 6π

and substituting: mγ

dv dp = dt dt

314

14 Fields and Radiation

μ0 c q 2 2 P= γ 6πc m 2



dp dt

2

so for a given force equal to (dp/dt), when the acceleration is normal to the velocity the power radiated is a factor γ 2 the power (14.28) radiated when acceleration and velocity are collinear. Replacing for a circular motion:



ωv 1 dv 1 β˙ =



= |ω × v| = c dt c c we have: P=

μ0 2 4 2 2 q γ ω v . 6πc

In circular accelerators a relevant parameter is the energy radiated in an orbit: δE R = P · T =

μ0 2 4 1 2 4 β3 q γ ω v2 = q γ 3c 30 ρ

δE R =

1 2 β3 E 4 q 30 ρ (mc2 )4

(14.32)

where ρ = v/ω is the radius of the orbit. For electrons (positrons) the energy loss δE R in synchrotron radiation per turn is: [δE R (MeV)]  8.83 × 10−2

[E(GeV)]4 ρ(m)

.

In the past e+ e− collider LEP, with a bending radius of 3.5 km, the 100 GeV particles radiated per turn about 2.3 GeV. The dependence on the inverse of the fourth power of the mass in (14.32) reduces drastically the synchrotron radiation loss in proton accelerators. At the LHC collider, in the same LEP tunnel, with 2.8 km bending radius, the 7 TeV protons lose only 6.7 keV per turn. √ The large radiation loss limits the energy of the future e+ e− circular colliders at s ≤ 500 GeV while larger energy are foreseen for e+ e− linear colliders (up to 3 TeV). At the same time while studies for future proton-proton colliders up to 100 TeV are in progress, there is a large interest for the construction of a muon collider. In this machine the cross section for Higgs boson production is about 4 × 104 larger with respect to e+ e− colliders of the same energy but with strongly reduced radiation loss because the mass of the muon is 207 times larger than the electron mass.

14.4 Radiation Reaction

315

14.4 Radiation Reaction When accelerated a charged particle emits radiation which carries off energy, momentum and angular momentum. Clearly this emission generates a recoil effect on the particle, called radiation reaction, that can be described as a force exerted by the radiation on the particle. Writing the equations of motion for a charged particle only the forces acting on the charge in the external fields are considered and the emission of radiation is neglected. This description is usually adequate and agrees very well with experiments. The emission of radiation is considered separately for the accelerated charges as done in this chapter. A complete analysis of the motion of a charge should consider both effects and is connected to the nature and the structure of the elementary particle. This subject is complex and beyond the scope of this book.11 We give here a simple short introduction to the subject. A naive calculation shows when the radiation emission can be neglected. Take a particle with mass m and charge q initially at rest. If it is accelerated with an acceleration a, after a time T its kinetic energy is: Ekin  m(aT )2 and using the non relativistic Larmor formula (14.26), in the same time interval T , it radiates an energy: μ0 2 2 q a T. Erad  PL T = 6π c The effects of the radiation emission are negligible if Erad  Ekin and that happens if the time interval T is much larger than a characteristic time τ : T 

1 μ0 q 2 =τ 6 πc m

or for an electron, introducing its classical radius re = e2 /4π0 m e c2 , if: T 

2 re = τ = 6.26 × 10−24 s 3 c

that is when the time interval of the acceleration is much longer than the time it takes for the light to travel a path comparable with the dimension of the electron. For times shorter than τ the acceleration is applied on dimensions smaller than the structure of the particle and so the particle has to be considered as an extended body. The parts of the particle can be also acted upon by different forces, and the internal electromagnetic forces between the parts of the particle play a role. 11

For this subject see D.J. Jackson [2], Chap. 16; A. Zangwill [4], Sect. 23.6; D.J. Griffiths [5], Sect. 11.2.2-3. An interesting approach from the expansion of the potential is given in L.D. Landau and E.L. Lifshitz [1], Sect. 65.

316

14 Fields and Radiation

The reaction on the particle produced by the radiation can be summarized in a radiation reaction force Frad and the equation of motion for a charged particle in presence of an external force Fext becomes: ma = Fext + Frad

(14.33)

The work of this force in a time interval t1 < t < t2 has to be equal to the negative of the energy radiated in the same interval. In a non relativistic approach, using the Larmor formula (14.26), we can write: 

t2



t2

Frad · v dt = −

t1

t1

μ0 2 2 μ0 2 q a dt = − q 6π c 6π c



t2 t1



dv dv · dt dt

 dt

which integrated by parts becomes: 

t2

Frad t1

 μ0 2 dv

t2 μ0 2 t2 d 2v q · v + q · v dt = − v · 2 dt . 6π c dt 6π c dt t1 t1

If v˙ · v = 0 or it is the same at t1 and t2 , for instance in a periodic motion, we have: 

t2



Frad −

t1

μ0 2  q a˙ · v dt = 0 6π c

and for the radiation reaction force we find: Frad =

μ0 2 q a˙ . 6π c

(14.34)

This relation gives the component of the force along the v direction and it is correct only if the conditions on v˙ · v are satisfied. It can be considered a time-averaged approximation for the radiation reaction on the charged particle. Using the (14.34) for the radiation reaction the equation of motion (14.33) becomes: m v˙ −

μ0 2 q v¨ = Fext 6π c

m(˙v − τ v¨ ) = Fext

(14.35)

and is called Abraham-Lorentz equation of motion. This equation is a second order equation in time and presents runaway solutions. If Fext = 0 the solutions are: ⎧ ⎨a = 0 ⎩

t

a = a0 e τ

14.5 Electric Dipole Radiation

317

The first solution is of no interest, in the second a0 is the initial acceleration, and if a0 = 0 the solution gives an exponential increase of the acceleration. If an external force is applied to the charged particle its acceleration increases due to the radiation emission. The force (14.34) is found from the Larmor relation which integrates the energy escaping from the particle at large distance where only the 1/r terms of the fields, which depend on the acceleration, are present. During the acceleration a charge modifies also the neighboring 1/r 2 fields, which depend on the velocity, so we should expect an exchange of energy and momentum between these fields and the particle. As suggested by the naive example at the beginning of this section, the radiation reaction is relevant when the dimensions of the charge distribution cannot be neglected. We have to consider an extended model of the particle and the fields between the different parts of the particle have to be added to the external fields. The electromagnetic forces between two parts of the particle are not equal and opposite12 and when all these forces are added the result is a net force of the charge on itself. This self-force accounts for the radiation reaction. This mechanism is presented by D.J. Griffiths13 by a simple but effective “dumbell” model that reproduces the radiation reaction force (14.34). Abraham and Lorentz made14 a purely electromagnetic model of a charged particle with spherical symmetry. In the limit of point charged particle they found the Abraham-Lorentz equation (14.35). An electromagnetic contribution to the mass of the particle appears in these extended models. This term of mass can be associated to the electrostatic energy of the particle at rest. It becomes infinite in the limit of zero dimensions of the particle so a mass renormalization should be introduced as in Quantum Electrodynamics.

14.5 Electric Dipole Radiation An electric dipole with time dependent moment p(t) radiates.15 The simplest electric dipole is composed by two small metal spheres connected by a wire of length d as in Fig. 14.8, where an electric current I (t) flows up and down forced by an oscillator. In the following we will find the retarded potentials for this dipole at large distance and from the associated electric and magnetic fields we will write the Poynting vector and the flux of radiation. The opposite charges on the spheres are: q+ (t) = q(t) = q0 sin(ωt) 12

and

q− (t) = −q(t)

See the forces between two charges in motion in Sect. 11.4. See D.J. Griffiths [5], Sect. 11.2.3. 14 See D.J. Jackson [2], Sect. 16.3. 15 For a general case see A. Zangwill [4], Sect. 20.4. 13

318

14 Fields and Radiation

Fig. 14.8 The oscillating dipole

while the current is: I (t) =

dq = q0 ω cos(ωt) = I0 cos(ωt) dt

I 0 = q0 ω

If the dipole is in the origin and aligned with the z axis, the moment is: p = q d = q0 d sin(ωt) zˆ = I0 sin(ωt) zˆ The current can be considered the same along the wire if its period T is much longer with respect to the time taken by the electric signal to travel the length d: T =

d 2π  ω c

and so the wavelength λ of the radiated electromagnetic field will be much longer than the dipole dimension: λ = cT  d If the cross-section of the wire is Σ, the density current J in the wire is: J(r, t) =

I0 I (t) zˆ = cos(ωt) zˆ Σ Σ

and the retarded vector potential:   Δr    J r ,t = t − μ0 c dτ  A(r, t) = 4π |r − r | where Δr = |r − r |, becomes:

A(r, t) =

μ0 4π



d 2

− d2

  Δr  zˆ I t =t− c dz  |r − r |

dτ  = Σ dz 

14.5 Electric Dipole Radiation

319

Fig. 14.9 The frame with the oscillating dipole in the origin and a point P at distance r

In the approximation of large distance from the dipole (r  d), we can replace |r − r | by r and we have: μ0 4π

A(r, t) =

A(r, t) =



d 2

− d2

'  r ( zˆ I0 cos ω t − c dz  r

'  r ( μ0 I0 d cos ω t − zˆ 4π r c

or in a more compact form16 :  r ˙ p t − μ0 c A(r, t) = 4π r

(14.36)

The reference frame with the dipole in the origin and the angles used in this section is shown in Fig. 14.9. The scalar potential can be written using the approximation already seen for the steady electric dipole (see Problem 14.2) or in a shorter form from the relation of the Lorenz gauge: ∂V =0 ∇ · A + μ0 0 ∂t We note that if ξ = ω(t − r/c): ∂ ∂ξ ∂ ∂ = =ω ∂t ∂ξ ∂t ∂ξ

∂ ∂ ∂ξ ω∂ 1∂ = =− =− ∂r ∂ξ ∂r c ∂ξ c ∂t

and so from (14.36) we have: 16

From:

→

'  r ( p = q0 d sin ω t − c '  '  ∂p r ( r ( = I0 d cos ω t − . p˙ = = q0 d ω cos ω t − ∂t c c

(14.37)

320

14 Fields and Radiation

Fig. 14.10 Spherical components Ar and Aθ of the vector potential A

1 ∂V 1 =− ∇·A= ∂t μ0 0 4π0 and after integration: 1 V (r, t) = 4π0





p˙ p¨ + r2 rc

p p˙ + 2 r rc





z r

z r

In the following it is more convenient to work with spherical coordinates. The components of the vector potential (14.36) as shown in Fig. 14.10 are: Ar (r, θ, t) = while:

μ0 p˙ cos θ 4π r

Aθ (r, θ, t) = −

1 V (r, θ, t) = 4π0



μ0 p˙ sin θ 4π r

p p˙ + 2 r rc

Aϕ = 0

(14.38)

 cos θ

(14.39)

with no dependence on ϕ because of the symmetry around the z axis. In the scalar potential V (r, t) the first term is the usual potential of the electric dipole with 1/r 2 dependence while the second term, associated to the change in time of the dipole moment, depends on 1/r and contributes to the radiation of the electromagnetic fields. From the expression of the curl in spherical coordinates (see Appendix A) for B = ∇ × A we find:   p˙ μ0 p¨ + 2 Bθ = 0 Bϕ = Br = 0 4π cr r and for the electric field from: E=−

∂A − ∇V ∂t

14.5 Electric Dipole Radiation

321

using (14.37), (14.38) and (14.39): 2 cos θ Er = 4π0



p˙ p + cr 2 r 3



sin θ Eθ = 4π0



p¨ p˙ p + 2+ 3 c2 r cr r

 Eϕ = 0

The B field has only the ϕ component so its field lines are circles around the z axis, while the field E has null ϕ component. The fields E and B are orthogonal: E · B = 0. The Poynting vector is:

E×B 1 sin2 θ S= = μ0 16π 2 0 r 2

1 2 sin θ cos θ − 16π 2 0 r3

p¨ 2 ( p˙ 2 + p¨ p) 2 p¨ p˙ p˙ p + + + 3 c3 c2 r cr 2 r

 rˆ +

 p¨ p˙ p˙ p ( p˙ 2 + p¨ p) ˆ + 2 + θ c2 r cr

Taking the average over a period: p¨ p˙ = 0

p˙ 2 + p¨ p = 0

p˙ p = 0

and only the term with p¨ 2 in the radial direction is non null. The average Poynting vector is: μ0 p¨ 2 S= sin2 θ rˆ 16 π 2 c r 2 and depends on the inverse of r 2 as expected for an electromagnetic wave propagating toward infinity. The p¨ 2 term in the Poynting vector is the product of the p¨ term in the θ component of the E field and of a similar term in the ϕ component of the B field so the average Poynting vector is radial as shown in Fig. 14.11. The other terms in the vector S are associated to fluxes of energy oscillating back and forth in the radial direction and in the θˆ directions but with null average values in time. The average power radiated per unit solid angle, in the direction dΩ: d P¯ = S · rˆ r 2 dΩ for the oscillating electric dipole is: d P¯ μ0 μ0 = p 2 ω 4 sin2 θ p¨ 2 sin2 θ = 2 dΩ 16 π c 32 π 2 c 0

(14.40)

This distribution is symmetric around the z axis, depends on the fourth power of the frequency f = ω/2π and has a maximum in the plane through the center of the dipole at θ = 90◦ with the direction of the dipole as shown in Fig. 14.12. The total power radiated is:

322

14 Fields and Radiation

Fig. 14.11 The Poynting vector S as a vector product of the electric field E and of the magnetic field B radiated by the oscillating dipole p

P¯ =

μ0 p2 ω4 . 12 π c 0

The power radiated by an oscillating dipole can be related to the Larmor formula (14.26) for the radiation by an accelerated point charge with velocity v  c that is equivalent to the condition d/T  c for the motion of the charge in the dipole. If we write p = qz we have p¨ = qa, which introduced in (14.40) reproduces (14.25) and after integration on the solid angle, gives the total radiated power: P=

μ0 2 2 q a 6πc

that is the power (14.26) for an accelerated particle. The dependence on the frequency of the radiated power in (14.40) is the reason for the blue sky in the morning and the red color at sunset. In electromagnetic waves the electric field is transverse to the direction of propagation.17 The rays from the sun stimulate the oscillations of the dipoles associated to the molecules in the atmosphere. The light from the sun has a broad spectrum in the visible range (white light) but its components are reradiated with a power proportional to the fourth power of the frequency. So the presence of higher frequencies (blue light) is enhanced in the visible light seen by an observer on the earth. At the same time the angular dependence in (14.40) selects the light radiated by the dipoles normal to the line of sight and this determines a polarization of the light reaching the observer (see Fig. 14.13). At sunset the longer path of sunlight in the atmosphere and the stronger radiation of higher frequencies in directions normal to the rays strongly decreases the presence of the blue light and leaves the smaller frequencies (red light) for the observer looking directly at the sun.

17

See (1.35) and Sect. 14.16 with the comments.

Problems

323

¯ Fig. 14.12 d P/dθ for the oscillating dipole versus the angle θ with respect to the direction of the dipole. The length of the arrow in the figure is proportional to the radiated power

Fig. 14.13 The dipoles of the molecules in the atmosphere reradiate the light from the sun. The dipoles a and b are both normal to the sun ray but because of the dependence on the angle of the radiated light only light from the dipole a reaches the observer. This determines a non-zero average polarization of the light

Problems 14.1 Liénard-Wiechert potentials for a charge in uniform motion. For a charge q with speed v the retarded potentials at point P are the Liénard-Wiechert potentials (14.10) and (14.11):

V (r, t) =

1 q qv 1 1   A(r, t) =     4π0 R  − Rc·v t  =t− R 4π0 c2 R  − Rc·v t  =t− R c

c

where R in this problem is the vector from the charge position to the point P, and  R and the speed v in the bracket are taken at retarded time t  = t − Rc . Show that these potentials for a charge with constant speed v are equal to the potentials (14.3) and (14.4) found by Lorentz transformations. (Also in Feynman Lectures on Physics [3], Vol. II, 21.6.) 14.2 Show that the general relations (14.13) and (14.14) for the electric and magnetic fields in Sect. 14.2.5 for a particle in uniform motion (β = const) become the fields found by a Lorentz transformations in Sect. 14.1. Use the relations found in Problem 14.1.

324

14 Fields and Radiation

14.3 Write the scalar potential at large distance for an oscillating electric dipole with opposite charges q(t) = q0 cos ωt a distance d apart in the approximation T = 2π/ω  d/c. This potential is the scalar potential (14.39) found from the vector potential using the Lorenz gauge relation (see text).

Solutions 

14.1 We say R the vector from the charge to the point P at retarded time t  = t − Rc , as shown in Fig. 14.14. In the time interval Δt = t − t  the electromagnetic signal travels for a distance R  = c Δt while the charge moves for a distance v Δt. To determine Δt we say R the vector from the charge to the point P at time t, we can write: R = R + v Δt and squaring:

R  = (c Δt)2 = R 2 + v 2 (Δt)2 + 2R · v Δt 2

we have the equation: (c2 − v 2 ) (Δt)2 − 2R v Δt cos θ − R 2 = 0. whose only acceptable solution is: 1

Δt =

R v cos θ + R [v 2 cos2 θ + c2 − v 2 ] 2 c2 − v 2

From Fig. 14.14 we see R  cos θ = vΔt + R cos θ so for the term in brackets in the Liénard-Wiechert potentials we can write: R −

=

v  v R cos θ = c Δ t − (v Δt + R cos θ) = c c   v v2 = c 1 − 2 Δt − R cos θ = c c

1 v 1 [R v cos θ + R [v 2 cos2 θ + c2 − v 2 ] 2 ] − R cos θ = c c 1

= R [1 − β 2 sin2 θ] 2 =

1 2 1 [γ (x − vt)2 + y 2 + z 2 ] 2 γ

where we have used the relation (14.1). By last relation in the Liénard-Wiechert potentials for a particle with velocity v = v xˆ we get:

Solutions

325

Fig. 14.14 The potential in point P at time t is the potential in P from the charge q in its position at the retarded time t  = t − R  /c. R is the vector from the charge at time t to the point P

V =γ A=γ

1 q 4π0 [γ 2 (x − vt)2 + y 2 + z 2 ] 21

1 1 qv c2 4π0 [γ 2 (x − vt)2 + y 2 + z 2 ] 21

which are the formulas (14.3) and (14.4) found from the Lorentz transformations of the 4-potential. 14.2 We call R  the distance from the position of the particle at the retarded time to the point of observation. From the solution of Problem 14.1: R −

v  1 1 R cos θ = R  (1 − nˆ · β) = [γ 2 (x − vt)2 + y 2 + z 2 ] 2 c γ (1 − nˆ · β) =

1 1 2 1 [γ (x − vt)2 + y 2 + z 2 ] 2  R γ

(14.41)

For β = const the electric field from (14.13) is: q E(r, t) = 4π0



nˆ − β

 3 γ 2 1 − nˆ · β R  2

r et

and substituting the relation (14.41): E(r, t) =

E x = E · xˆ =

=

γ R  (nˆ − β) q 4π0 [γ 2 (x − vt)2 + y 2 + z 2 ] 23 γ R  (nˆ − β) q · xˆ 4π0 [γ 2 (x − vt)2 + y 2 + z 2 ] 23

q γ (R  nˆ · xˆ − R  β · x) ˆ γ (x − vt) q 3 = 2 2 2 2 2 4π0 [γ (x − vt) + y + z ] 2 4π0 [γ (x − vt)2 + y 2 + z 2 ] 23

326

14 Fields and Radiation

Fig. 14.15 Distance r+ and r− of the point P from the two point charges of the dipole

where we have replaced t = R  /c because t  = 0 in Sect. 14.1. For E y and E z : E y = E · yˆ =

γy q 2 4π0 [γ (x − vt)2 + y 2 + z 2 ] 23

E z = E · zˆ =

γz q 4π0 [γ 2 (x − vt)2 + y 2 + z 2 ] 23

Similarly for the B field. 14.3 We want the potential in a point P at position r from the center of the dipole set in the origin and with z direction as in Fig. 14.15. potential is the    The scalar  sum of the retarded potentials of the charges q+ at t+ = t − rc+ and of q− at t− = t − rc− :   r+  r−  t − t − q q + − 1 c + 1 c V (r, t) = (14.42) 4 π 0 r+ 4 π 0 r− In the approximation r  d for the distance from the charges we can write: d cos θ 2

1  r+

1 1  d r cos θ r 1− 2r

d r−  r + cos θ 2

1  r−

1 1   d r cos θ r 1+ 2r

r+  r −



  d 1+ cos θ 2r   d 1− cos θ 2r

and for the charges at the retarded times: q+ =

q0 cos ωt+

  r+  r  ωd  q0 cos ω t − + = q0 cos ω t − cos θ = c c 2c 

14.5 Electric Dipole Radiation

327

       r ωd r ωd cos cos θ − sin ω t − sin cos θ q0 cos ω t − c 2c c 2c

   r−  r  ωd  −q0 cos ω t − − q− = −q0 cos ωt− = −q0 cos ω t − cos θ = c c 2c        r ωd r ωd cos cos θ + sin ω t − sin cos θ −q0 cos ω t − c 2c c 2c thus after substitution of these approximation in (14.42): V (r, t) =

q0 1 4π0 r

q0 1 4π0 r



    r ωd −2 sin ω t − sin cos θ + c 2c

   d cos θ r ωd cos ω t − cos cos θ r c 2c

We apply now the condition T = 2π/ω  d/c so that ωd/2c is very small and we can write:     ωd r  d cosθ r q0 1 − cos θ sin ω t − + cos ω t − V (r, t) = 4π0 r c c r c (14.43) In the limit T → ∞ that is ω → 0 we have a steady electric dipole p = q0 d and the potential becomes the dipole potential (2.1) seen in Sect. 2.1: V (r ) =

1 p·r 4π0 r 3

For the oscillating dipole:  r p(t  ) = q0 d cos ω t − c

 r p(t ˙  ) = −q0 d ω sin ω t − c

and we can write the (14.43) in the form: V (r, t) = that is the scalar potential (14.39).

1 4π0



p p˙ + r2 rc

 cosθ

328

14 Fields and Radiation

References 1. L.D. Landau, E.L. Lifshitz, The Classical Theory of Fields, Course of Theoretical Physics, vol. 2. Elsevier 2. D.J. Jackson, Classical Electrodynamics, 3rd edn. (Wiley, NewYork, 1999) 3. R.P. Feynman, R.B. Leighton, M. Sands, The Feynman Lectures on Physics, vol. II 4. A. Zangwill, Modern Electrodynamics (Cambridge University Press, Cambridge, 2013) 5. D.J. Griffiths, Introduction to Electrodynamics, 4th edn. (Cambridge University Press, Cambridge, 2017)

Chapter 15

Test of the Coulomb’s Law and Limits on the Mass of the Photon

In 1785 Charles Augustin de Coulomb measured the forces between electric charges and deduced1 the law that governs them. For two point charges the force is proportional to the product of the charges and inversely proportional to the square of their distance. The dependence on the inverse square of the distance seems almost obvious to the readers of an introductory textbook on Electricity because usually they have already encountered this dependence in the gravitational force. This was not the case for the physicists who contributed to the formulation of the Electrostatics. Indeed they performed many tests on its validity and gave limits on a possible deviation  from the second power of the distance. Similar tests were performed until the half of ’900. More recently physicists started to interpret the tests of the Coulomb’s law in terms of limits on the mass of the photon. A short review of the subjects of this chapter can be found in D.J. Jackson’s book [1].2 For some topics we refer to the original papers. In 2010 A.S. Goldhaber and M.M. Nieto wrote a paper [2] on the limits for the masses of the photon and the graviton. This is an update of a previous paper [3] of these authors that is useful to consider. All experimental results published prior to 2005 have been summarized in detail by L.C. Tu et al. [4]. For the historical part on the birth of Electrostatics we suggest E. Segrè’s book [5].

1

The dependence on the inverse square of the distance was already found by John Robinson in 1769. In 1767 Joseph Priestley wrote in a book that he had repeated an experiment, already done by Benjamin Franklin, showing that there is no charge inside a charged closed metallic box. From this result he deduced that the repulsion force between same sign charges has to be proportional to the inverse square of the distance. The results by Robinson and Priestley remained unknown to the scientific community. See note at page 283 of the paper by A.S. Goldhaber and M.M. Nieto (1971) [3] and E. Segrè’s book on the history of Physics [5]. See also D. Halliday and R. Resnick textbook [6]. 2 J.D. Jackson [1], Sects. 1.2 and 12.8. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 F. Lacava, Classical Electrodynamics, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-031-05099-2_15

329

330

15 Test of the Coulomb’s Law and Limits on the Mass of the Photon

15.1 Gauss’s Law To introduce the first tests of the Coulomb’s law we recall and comment the Gauss’s law. In order to demonstrate the Gauss’s law one has to start considering a point charge inside a closed surface S. The flux of the field E = E rˆ from the charge through an element dS of the surface is dΦ = E · nˆ d S with nˆ the external versor normal to dS. Since d Sn = rˆ · nˆ d S = r 2 dΩ, we can write: dΦ =

1 Q 2 1 Q d Sn = r dΩ . 2 4π 0 r 4π 0 r 2

(15.1)

After the term r 2 at the numerator cancels out with the same term at denominator and integrating over the solid angle, the total flux through the whole closed surface, is: Φ(E) =

Q . 0

But the simplification done in (15.1) is not trivial at all because while the term at the numerator comes from a correct mathematical relation, the term at the denominator is a consequence of the Coulomb’s law. If this did not depend not on r −2 but on r −(2+) in the result for Φ we should find a dependence on r and the Gauss’s theorem would be wrong. Also the geometrical principle of conservation of the number of field lines coming out from a charge would not be valid. Moreover the first Maxwell equation would lose its validity and at the same time all the consequences of the Gauss’s law would be incorrect like the well known property that in a conductor the charges are distributed on the external surface.

15.2 First Tests of the Coulomb’s Law The law for the force between two charges was found by Coulomb using a torsion balance. However, in his treatise Maxwell3 observes that the measurement errors with this apparatus might be not negligible. A more accurate test of the Coulomb’s law can be performed with the following experiment used by Cavendish to prove the law of inverse square of the distance. A metallic globe was surrounded by two hollow metallic hemispheres that, when closed, formed a spherical surface concentric with the globe but isolated from that (see Fig. 15.1). Initially the globe was electrically connected to the hemispheres by a conductive wire. After having deposited an electric charge on the hemispheres, the wire was removed and the hemispheres were opened and taken away. At that point the residual charge on the globe was measured with an electrometer but was found smaller than the minimal charge detectable by that 3

J.C. Maxwell [7], Cap. II, Sect. 74a,b.

15.2 First Tests of the Coulomb’s Law

331

Fig. 15.1 Cavendish’s experiment to test the inverse square of the distance dependence for the force between point charges. Top the drawing done by Cavendish, bottom the drawing redone by Maxwell (Figures reproduced from H. Cavendish, edited by J.C. Maxwell, [8], by permission of Cambridge University Press.)

instrument. The minimal measurable charge was then measured by depositing on the globe known fractions of electrical charge. From the results of this experiment, Cavendish was able to prove that the dependence of the electrostatic force on the inverse square of the distance was different from a power 2 by a quantity || ≤ 0.02. This result of course is also expected from the Gauss’s law, not stated at that time, for the positioning of an excess charge placed on an isolated conductor. The experiment was performed by Cavendish in 1773, that is well before Coulomb presented his law, however Cavendish was4 “an extraordinary experimenter and a major physicist but his eccentricities were equally extraordinary, ... His relations with other scientists were reduced to a minimum, ...”, so he did not publish his result that became known only about a hundred years later when Maxwell, at that time Cavendish Professor at Cambridge, undertook the publication [8] of the Cavendish’s electrical work. Maxwell repeated the Cavendish’s experiment at the Cavendish Lab4

See E. Segrè [5], p. 115.

332

15 Test of the Coulomb’s Law and Limits on the Mass of the Photon

oratory in Cambridge with some differences and more accurate instruments and found for the deviation from the square power || ≤ 5 × 10−5 .

15.3 Proca Equations In 1936 Alexandre Proca, under the influence of de Broglie, added a mass term to the Lagrangian density5 of the electromagnetic field: 1 1 L = − Fμν F μν − μ0 Jμ Aμ + μ2γ Aμ Aμ 4 2

(15.2)

where μγ =

mγ c 

is the inverse of the Compton length of the photon with a non null mass m γ , while F μν = ∂ μ Aν − ∂ ν Aμ

(15.3)

are the fields E and B in terms of the potentials: E=−

∂A − ∇V ∂t

B=∇×A

(15.4)

and are equivalent to the second and the third Maxwell equations. The first term in the Lagrangian density (15.2) is the energy density of the electromagnetic field, the second is the term of interaction of the field with the charges and the currents while the last is the mass term of the photon. From the Euler-Lagrange equations for the fields with the density given in (15.2) the first and the fourth Proca equations can be derived as shown in the Appendix to this chapter. Thus the Maxwell equation modified by Proca after the introduction of the mass term for the photon are ∇·E=

ρ − μ2γ V 0

∇·B=0

∇×E=− 5

The Lagrangian density is written in SI units.

∂B ∂t

(15.5)

(15.6)

(15.7)

15.3 Proca Equations

333

∇ × B = μ0 (J + 0

∂E ) − μ2γ A . ∂t

(15.8)

These equations are different from the Maxwell equations because of the presence of the pseudo-charge density −μ2γ V 0 in the first equation and the pseudo-current density −μ2γ A/μ0 in the fourth one. These terms vanish if μγ = 0, a null mass for the photon, and the Proca equations reduce to Maxwell equations. The fields given by (15.3) and (15.4) are invariant under a gauge transformation: Aμ → Aμ = Aμ + ∂ μ Λ

A → A + ∇Λ

V → V = V −

∂Λ ∂t

that leaves unchanged the Maxwell equations but not the Proca equations (15.5) and (15.8). The charge conservation is expressed in local form by the continuity equation: ∇·J+

∂ρ =0 ∂t

that is included in the non homogeneous Maxwell equations. For a massive photon from (15.5) and (15.8), we get the relation: μ2γ ∂ρ ∇·J+ = ∂t μ0

  1 ∂V ∇·A+ 2 c ∂t

(15.9)

that is the conservation equation for the sum of the charge and the pseudo-charge:    ∂  A + ρ − μ2γ V 0 = 0 . ∇ · J − μ2γ μ0 ∂t In order for the charge conservation to be separately valid, from (15.9) it follows that also the pseudo-charge has to be conserved and for the potentials the following relation has to be valid: ∇·A+

1 ∂V = 0. c2 ∂t

This relation is just the condition for the Lorenz gauge and it implies that in the gauge transformations the scalar function Λ must satisfy the equation: ∇2Λ −

1 ∂ 2Λ = 0. c2 ∂t 2

Finally, for a non null mass of the photon, the equations for the potentials are: Aμ − μ2γ Aμ = −J μ

334

15 Test of the Coulomb’s Law and Limits on the Mass of the Photon

→ V − μ2γ V = −

ρ 0

A − μ2γ A = −μ0 J .

For a point charge q, in the stationary case, from the last equation for the potential V or from the first Proca equation (15.5) we get the solution: V (r ) =

q e−μγ r . 4π 0 r

This ‘Yukawa’ potential represents a new possible deviation for the electrostatic force from the inverse square of the distance. This point was clear only later, when Yukawa wrote the equation for a scalar particle, the meson π discovered afterwards, that justifies the short range of the strong nuclear interaction with its mass.

15.4 The Williams, Faller and Hill Experiment In 1936, in a test similar to the Cavendish’s experiment, S.J. Plinton and W.E. Lawton6 found a limit || ≤ 1 × 10−9 for the deviation from the inverse square of the distance. Afterwards the last experiment of this type was performed by E.R. Williams, J.E. Faller and H.A. Hill and was published in 1971 [9]. Their experimental apparatus, sketched in Fig. 15.2, consisted of five concentric icosahedrons with an external diameter of about 1.5 m. A 4 MHz radio frequency, 10 kV peak-to-peak voltage was applied to the resonant circuit formed by the outer shells (4 and 5 in figure) and a high-Q water cooled coil. After the intermediate shell 3, the voltage difference across the inductor between the inner shells 1 and 2, was measured by a battery-powered lock-in amplifier located inside the innermost shell and then isolated from the outer of the apparatus. Considering a spherical surface of radius r between conductors 1 and 2 and an approximated7 potential V0 eiωt in the volume inside shell 4, from Eq. (15.5) and Gauss’s law, the electric field is: E(r ) =

q 1 1 ( 2 − μ2γ V0 eiωt r )ˆr 4π 0 r 3

where q is the total charge on the inner shell 1. Then if C is the capacitance between conductors 1 and 2, the voltage across the inductor L connecting these two shells is:

6

S.J. Plinton and W.E. Lawton used two conductive concentric hollow spheres of radii 2.5 and 2.0 ft, both initially at ground. After the outer sphere was connected to a 3000 V generator with a 130 cycles/min frequency, the difference of voltage between the two spheres was measured by detecting the current between them with a galvanometer. 7 A detailed solution is given in the paper.

15.4 The Williams, Faller and Hill Experiment

335

Fig. 15.2 The Williams, Faller and Hill experiment setup (Reprinted figure with permission from E.R. Williams, J.E. Faller and H.A. Hill, [9]. Copyright 1971 by American Physical Society. http:// dx.doi.org/10.1103/PhysRevLett.26.721)

336

15 Test of the Coulomb’s Law and Limits on the Mass of the Photon

ΔV =

1 q − μ2γ V0 eiωt (R22 − R12 ) C 6

that is applied to the LC circuit formed by the two shells and the inductor. The proportionality to V0 for the term with the photon mass suggests to apply an high V0 voltage. The measurements put a limit || ≤ (2.7 ± 3.1) × 10−16 for the deviation from the inverse square of the distance, while if interpreted in terms of a non null photon mass the limit was m γ ≤ 1.6 × 10−50 kg.

15.5 Limits from Measurements of the Magnetic Field of the Earth and of Jupiter E. Schrödinger proposed a method to get a limit on the photon mass from measurements of the Earth’s static magnetic field. The magnetic field B(r) generated by a magnetic dipole of moment D = D zˆ , at the position r, from the fourth Proca equation (15.8) is: μ0 De−μγ r B(r) = 4π r 3



   2 2 2 1 2 2  1 + μγ r + μγ r 3ˆz · rˆ rˆ − zˆ − μγ r zˆ 3 3

(15.10)

while the field due to the same magnetic dipole from the fourth Maxwell equation (μγ = 0 in (15.8)) is: B(r) =

 μ0 D  3ˆz · rˆ rˆ − zˆ . 3 4π r

In the magnetic field in (15.10) there is a component proportional to − 23 μ2γ r 2 zˆ antiparallel to the dipole D. As evident from the ratio of this term and the coefficient of the main dipole component: 2 2 2 μ r 3 γ   1 2 2 1 + μγ r + μγ r 3 its relative contribution to the field increases with the radius r. The measurements performed on the terrestrial surface (at r = RT ) allowed for a long time to set the best limit on the photon mass. This limit was then improved in 1968 by A.S. Goldhaber and M.M. Nieto (1971) [3] who considered the contributions from different origins. Later the same measurement at a larger radius was done using the Pioneer10 probe in the Jupiter magnetic field by L. Davis Jr. et al. in 1975 [10] and the new

15.6 The Lakes Experiment

337

limit was m γ ≤ 8 × 10−52 kg. A more sensitive measure could be performed in the solar magnetic field with r the distance from the Sun.

15.6 The Lakes Experiment In 1998 R. Lakes [11] proposed a very creative method to measure in a laboratory experiment the mass of the photon in the cosmic magnetic potential.8 Before introducing this method, remember that, for the equation ∇ × B = μ0 J, a coil carrying a current I is the source of a field B, and has a dipole moment m d = πr 2 I with I = J Σ, where Σ is the cross section of the wire and r is the radius of the coil. If the coil is embedded in an external magnetic field Bext , a torque τ d = md × Bext acts on the coil as a consequence of the energy density of the magnetic field that is proportional to B 2 . Then in analogy with the coil, since the equations are formally the same, if we consider an iron toroid, wound with turns carrying a current, since ∇ × A = B, the flux of B inside the toroid, Φ = BΣ  (Σ  the cross section of the toroid), becomes the dipolar source of a field A. This source has a dipole moment m d A = πr 2 Φ parallel to the axis of the toroid, and if the toroid is embedded in the cosmic potential Aamb , there is a torque acting on the toroid of momentum τ d A = md A × μ2γ Aamb . The factor μ2γ in the torque is a consequence of the fact that the contribution of the potential A to the energy density is μ2γ A2 /2 while that of the field B is B 2 /2, as evident from (15.2). In the experiment (see Fig. 15.3) a toroid with turns carrying a current, was suspended by the wire of a Cavendish balance to measure the torque generated by the cosmic magnetic potential Aamb . To a first approximation |Aamb | ≈ |B|R where R is the radius of a cylinder with axis in the direction of B, thus even though the cosmic magnetic field is very small, this potential might be very large since R is of the order of galactic or extragalatic dimensions. From the galactic magnetic field (≈1 µG) and its reversal position toward the center of the Milky Way (≈600 pc, 1 pc = 3.08 × 1016 m) the potential would be Aamb ≈ 2 × 109 Tm. This has to be compared with the potential ≈200 Tm from the terrestrial magnetic field and ≈10 Tm from the Sun. If the ambient cosmic vector potential were 1012 Tm, corresponding to the Coma cluster of galaxies (0.2 µG over 1300 kpc diameter), the limit9 would be m γ ≤ 2 × 10−55 kg.

8

A more precise experiment of this type was later performed by J. Luo et al. in 2003 [12] and L.C. Tu et al. in 2006 [13]. 9 L.C. Tu et al. (2006) [13] and A.S. Goldhaber and M.M. Nieto (2010) [2].

338

15 Test of the Coulomb’s Law and Limits on the Mass of the Photon

Fig. 15.3 The Lakes’s experiment setup (Reprinted figure with permission from R. Lakes, [11]. Copyright 1998 by American Physical Society. http://dx.doi.org/10.1103/PhysRevLett.80.1826)

15.7 Other Measurements For a non null mass of the photon, the electromagnetic waves should have a frequency dependent speed and limits on the photon mass could be found by measurements of the dispersion of the electromagnetic waves. The best limit from this method was deduced by N.M. Kroll [14] with measurements on the Schumann resonances.10 These are low-frequency standing electromagnetic waves traveling through the atmosphere between the Earth’s surface and the ionosphere, which are two conductive layers of a waveguide. If these surfaces were planar and parallel, for a special mode the speed could be independent from the frequency and equal to c also for a non null mass of the photon. But these two surfaces are two concentric spherical surfaces and there is a dispersion relation of the special mode which depends on an effective mass m 2γ e f f = g m 2γ where the dilution factor g is equal to the ratio (1%) of the distance Earth surface—ionosphere (about 60 km) to the sum of the radii of the Earth (6400 km) and of the ionosphere. Considering the low frequencies of the Schumann resonances (the lowest is equal to 8 Hz), Kroll deduced11 the limit m γ  4 × 10−49 kg. At present the strongest and most controlled limit on the photon mass has been found by D.D. Ryutov in 2007 [15] in measurements on the solar wind. In the model 10

See J.D. Jackson [1], Sects. 8.9 and 12.9. Since the mass energy has to be smaller than the total energy of the wave, from the m γ c2 < hν, for the lowest Schumann frequency equal to 8 Hz, considering the dilution factor, the mass limit is m γ < 4 × 10−49 kg.

11

15.8 Comments

339

Table 15.1 Experimental limits for the photon mass m γ and its Compton wavelength ňC = /m γ c [ňC (m) = 3.52 × 10−43 /m(kg)] Experimental method

m γ (kg) 

ňC (m) 

Shumann resonances (Kroll) Williams, Faller and Hill experiment Jupiter magnetic field Lakes’s type experiments Ryutov, plasma of the solar wind Principle of indetermination

4 × 10−49

8.8 × 105 2.2 × 106 3.3 × 108 1.6 × 1012 2 × 1011 1.3 × 1028

1.6 × 10−49 1.1 × 10−51 2.1 × 10−55 1.5 × 10−54 2.7 × 10−69

of solar wind supported by experiments, the plasma moving radially from the Sun, carries with itself the field lines of the magnetic field and, due to the Sun rotation, these lines wind up like Archimedes spirals and at large distance have almost an azimuthal direction. If the photon has a non null mass, to maintain this configuration, a real current of plasma J should balance the pseudo-current −μ2γ A/μ0 in the Proca equation (15.8). As a consequence in the local magnetic field an acceleration of the plasma J × B should be observed. The measurements of density, speed and pressure of the plasma beyond the Pluto orbit allow the exclusion of the existence of this current and provide an upper limit m γ  1.5 × 10−54 kg that is the best accepted limit on the photon mass.

15.8 Comments The daily test of the classical Electromagnetism in many applications and the achievements of the Quantum Electrodynamics in the prediction of phenomena with high accuracy (more than six orders of magnitude) support a null mass of the photon as usually assumed. This tacit assumption implies the hypothesis that an effective mass may be determined only by the uncertainty principle: mγ ≈

 ≈ 2.5 × 10−69 kg ΔtU niv · c2

where ΔtU niv is the age of the Universe.12 The fully experimental measurements mentioned in this chapter are the most sensitive to photon mass and are reported in Table 15.1. Lower limits can be derived from the measurements with theoretical hypotheses. A.S. Goldhaber and M.M. Nieto (2010) [2] provided an exhaustive review on the theory and the experiments. Experiments before 2005 have been reviewed by L.C. Tu et al. [4]. A complete and annually updated report on the mass 12

This limit means a Compton wave length comparable with the dimensions of the Universe.

340

15 Test of the Coulomb’s Law and Limits on the Mass of the Photon

limits of photons can be found13 in the in The Review of Particle Physics by the Particle Data Group [16].

Appendix: Proca Equations from the Euler-Lagrange Equations The Proca equations can be derived from the Euler-Lagrange equations for the fields: ∂

μ



∂L ∂ (∂ μ Aν )

 −

∂L =0 ∂ Aν

using the Lagrangian density of the electromagnetic field with a term to account for a non null mass of the photon: 1 1 L = − Fμν F μν − μ0 Jμ Aμ + μγ 2 Aμ Aμ . 4 2 For the first term of the equation we find:   1 ∂ ∂ ∂L 1 =− Fαβ F αβ = − gαρ gβσ F ρσ F αβ μ ν μ ν μ ν ∂ (∂ A ) 4 ∂ (∂ A ) 4 ∂ (∂ A )  ∂ 1 gαρ gβσ (∂ ρ Aσ − ∂ σ Aρ )(∂ α Aβ − ∂ β Aα ) =− μ ν 4 ∂ (∂ A )  ρ σ 1 = − gαρ gβσ (δμ δν − δμσ δνρ )(∂ α Aβ − ∂ β Aα ) + (δμα δνβ − δμβ δνα )(∂ ρ Aσ − ∂ σ Aρ ) 4 1 = − gαμ gβν F αβ − gαν gβμ F αβ + gμρ gνσ F ρσ − gνρ gμσ F ρσ 4 1 = − Fμν − Fνμ + Fμν − Fνμ = −Fμν 4 and for the second: ∂L ∂ 1 1 = −μ0 Jν + μγ 2 ν [gμα Aα Aμ ] = −μ0 Jν + μγ 2 [gμν Aμ + gνα Aα ] = ν ∂A 2 ∂A 2 −μ0 Jν + μγ 2 Aν thus the Lagrange equation becomes: ∂ μ Fμν − μ0 Jν + μγ 2 Aν = 0 or in the usual form: 13

See γ (photon) in Particle Listing.

References

341

∂μ F μν = μ0 J ν − μγ 2 Aν . This equation corresponds to the two non homogeneous Proca equations (15.5) and (15.8) and for μγ = 0 gives the covariant expression (10.12) of the non homogeneous Maxwell equations.

References 1. D.J. Jackson, Classical Electrodynamics, 3rd edn. (Wiley, NewYork, 1999) 2. A.S. Goldhaber, M.M. Nieto, Photon and graviton mass limits. Rev. Modern Phys. 82, 939–979 (2010) 3. A.S. Goldhaber, M.M. Nieto, Terrestrial and extraterrestrial limits on the photon mass. Rev. Modern Phys. 43, 277–296 (1971) 4. L.C. Tu, J. Luo, G.T. Gilles, The mass of the photon. Rep. Prog. Phys. 68, 77–130 (2005) 5. E. Segrè, From Falling Bodies to Radio Waves. Classical Physicists and Their Discoveries (W.H. Freeman & Company, New York, 1984) 6. D. Halliday, R. Resnick, Physics for Students of Science and Engineering, Part II, 28.4-5, 2nd edn. (Wiley, 1960) 7. J.C. Maxwell, A Treatise on Electricity and Magnetism (1873) 8. H. Cavendish, The Electrical Researches of the Honourable Henry Cavendish, 1771–1781, ed. by J.C. Maxwell (Cambridge University Press, 1879) 9. E.R. Williams, J.E. Faller, H.A. Hill, Phys. Rev. Lett. 26, 721 (1971) 10. L. Davis Jr., A.S. Goldhaber, M.M. Nieto, Phys. Rev. Lett. 35, 1402 (1975) 11. R. Lakes, Phys. Rev. Lett. 80, 1826 (1998) 12. J. Luo et al., Phys. Rev. Lett. 90, 081801–1 (2003) 13. L.C. Tu et al., Phys. Lett. A 352, 267 (2006) 14. N.M. Kroll, Phys. Rev. Lett. 27, 340 (1971) 15. D.D. Ryutov, Plasma Phys. Control. Fusion 49, B429–B438 (2007) 16. R.L. Workman et al. (Particle Data Group), Prog. Theor. Exp. Phys. 2022, 083C01 (2022) (http://pdg.lbl.gov/index.html)

Chapter 16

Magnetic Monopoles

In the absence of electric charges and currents the Maxwell equations are clearly symmetric in the electric and magnetic fields. This symmetry would be conserved in the presence of field sources if, in addition to the electric charges and currents, magnetic charges (monopoles) and currents would exist. In 1931 Dirac showed that the existence in nature of one magnetic monopole could account for the electric charge quantization. Probably the high mass of the monopoles did not allow their production and observation in experiments performed so far at particle accelerators. Modern Grand Unification Theories (GUT) of the fundamental interactions predict the existence of massive magnetic monopoles. These magnetic monopoles are too massive to be produced at the present or future accelerators, but if produced in the early Universe, should be observed as relics because the magnetic charge conservation should prevent their decay in other particles.

16.1 Generalized Maxwell Equations Under the hypothesis of the existence of magnetic charges and currents with densities ρm and jm , similar to the electric charges and currents with densities ρe e je , Maxwell equations1 would be:

1

∇ · D = 4πρe

(16.1)

∇ · B = 4πρm

(16.2)

Gaussian units are used in this chapter.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 F. Lacava, Classical Electrodynamics, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-031-05099-2_16

343

344

16 Magnetic Monopoles

−∇×E=

4π 1 ∂B jm + c c ∂t

∇×H =

4π 1 ∂D je + . c c ∂t

(16.3) (16.4)

In addition to these equations, a continuity equation2 must be formulated to include the conservation of the magnetic charge: ∇ · jm +

∂ρm =0 ∂t

similar to the continuity equation for the electric charge: ∇ · je +

∂ρe =0 ∂t

and a law of interaction of the magnetic charge g with the fields3 : F = g [B −

v × E] c

(16.5)

similar to the Lorentz force: F = q [E +

v × B]. c

(16.6)

16.2 Generalized Duality Transformation The generalized duality transformation: E = E cosϕ + H sinϕ H = −E sinϕ + H cosϕ

D = D cosϕ + B sinϕ B = −D sinϕ + B cosϕ

leaves invariant the quadratic forms of the fields: the Poynting’s vector, the energy density, the Maxwell stress tensor. If a similar transformation is applied to the sources:

2

ρe = ρe cosϕ + ρm sinϕ

je = je cosϕ + jm sinϕ

ρm = −ρe sinϕ + ρm  cosϕ

jm = −je sinϕ + jm cosϕ

This equation determines the sign of jm in Eq. (16.3) This relation, as for the Lorentz force, can be derived directly from the relativistic transformation of the electric and magnetic fields from a frame where the magnetic charge is at rest to a frame where it moves with velocity v.

3

16.2 Generalized Duality Transformation

345

also the Maxwell equations (16.1–16.4) and the Lorentz forces (16.5) and (16.6) are invariant: the fields E  , D  , H  , B  satisfy the same equations of the fields E, D, H, B. This transformation represents a rotation of an angle ϕ in a two dimensional frame with the electric charge and the magnetic charge as axes. The electric and magnetic charges of a particle are determined by the value of the angle ϕ. If the ratio of the magnetic charge to the electric charge is the same for all particles, it is possible to fix the angle ϕ to have always ρm = 0 and jm = 0. With this choice the equations (16.1–16.4) take the usual form of the Maxwell equations in the absence of magnetic charges. Under the assumption ϕe = 0 for the electron (that is qme = 0), any other particle could have a different angle ϕ p given by: p

tan ϕ p = −

qm p qe

.

A body with N p protons and Nn neutrons would have a magnetic charge: Q m = N p qmp + Nn qmn and considering the negligible contribution ( 0

B yf = 0 Bzf = 4π g δ(x)δ(y)

for z < 0.

Thus the magnetic field B is given by the relation: B = ∇ × A − Bf

(16.13)

where the field −B f cancels the singularity on the Dirac string.

16.6 Quantization Relation We have seen that the magnetic monopole located at a point P can be considered as the end point of a Dirac string L (see Fig. 16.1). In the Coulomb gauge its vector potential A is defined by the equations: ∇×A=B

∇·A=0

(16.14)

where B is the radial field given by the relation (16.9) and A(P, L) can be calculated from (16.11). Formally the Eqs. (16.14) are identical to the usual equations for a magnetostatic field b: 4π j ∇·b=0 ∇×b= c and thus the integral (16.11) has the same functional form of the Biot and Savart law: b=

9

1 c



I dl × (r − r ) |r − r |3

(16.15)

Here the sign is positive because associated to the direction of the circular path of the integral that, at θ = π , is opposite to the direction of the outgoing flux.

350

16 Magnetic Monopoles

Fig. 16.1 Two Dirac strings L and L  from infinity to the monopole to calculate the vector potential

The analogy between the two sets of equations can be used10 to derive the Dirac relation (16.7). By substituting 4π j/c with the radial field B given by (16.9), we have a current flux 4π jr 2 = gc = Φ( j) = I and the Dirac string is equivalent to a wire carrying a current I equal to gc that from infinity reaches the monopole. A vector potential A(P, L) is associated to this Dirac string. But it is possible to assume a second Dirac string L  , as in Fig. 16.1, which from infinity goes to the point P and to this line is associated a vector potential A(P, L  ). If we consider the circuit composed by the line L  and the line −L, we have a loop with a current I = gc and, from the similarity between the relations (16.15) and (16.11), the field b associated to this current is: b = A(P, L  ) − A(P, L). From magnetostatics we know that b can be the gradient of a scalar potential U : b = −∇U where U = U0 + 4π m I /c = U0 + 4π mg is a multiple-valued function with m = 0, ±1, ±2, . . . the number of times the path, chosen for the calculation of U from infinity to the point where we want b, circles around the loop composed by the two Dirac strings. Thus from the previous relations the result: A(P, L  ) = A(P, L) − ∇U

10

The quantization proof reported here was given by E. Fermi [9] and is reported also in the paper by E. Amaldi [7], p. 46.

16.7 Quantization from Electric Charge-Magnetic Pole Scattering

351

that is a gauge transformation corresponding to the arbitrary choice of the Dirac string and to its non observability. In a semiclassical approximation the quantum wave function of a charged particle in a potential A is:  ie Ψ = Ψ0 e c A·ds where Ψ0 is the the wave function of the free particle. After the gauge transformation the new wave function is: Ψ → Ψ  = Ψ e− c ie



∇U ·ds

= Ψ e− c U = Ψ e− c (U0 +4πgm) ie

ie

and this function has the same value only if: e · 4π gm = 2π n c

with n = 0, ±1, ±2, . . .

that is the Dirac relation (16.7).

16.7 Quantization from Electric Charge-Magnetic Pole Scattering Consider the scattering of a particle,11 with charge e and velocity v, in the radial field B (16.9) of a magnetic monopole. The Lorentz force (16.6), normal to v and B, acts on the particle and for a large impact parameter b, the deflection of the charge is negligible, and the momentum transferred to the particle is: Δp⊥ =

2eg . cb

The change of its angular momentum is: ΔL e =

2eg c

in the direction of the velocity and is independent of the impact parameter and of the velocity. Since the angular momentum is quantized, the change has to be equal to an integer multiple of  and this gives the Dirac condition (16.7). From the conservation of the angular momentum in the isolated system, the change of the angular momentum L em of the electromagnetic field has to be opposite to that of L e .

This argument was proposed by A.S. Goldhaber [10]. For the calculation of ΔL e see also J.D. Jackson [8] Sect. 6.13.

11

352

16 Magnetic Monopoles

The angular momentum L em of the field E of the electric charge and the field H of the magnetic monopole, can be found by the integral of the momentum r × g where g is the electromagnetic momentum density (11.18). L em is independent of the origin because the total momentum of the fields in the isolated system is null. The result12 is: eg nˆ Lem = c where nˆ is a versor with direction from the charge to the monopole. Of course also this angular momentum has to be quantized but in order to get the Dirac condition it has to be equal to an half-integer multiple of  in some disagreement with the unitary spin of the photon.

16.8 Properties of the Magnetic Monopoles 16.8.1 Magnetic Charge and Coupling Constant The condition (16.7) with n = 1 gives for the elementary magnetic charge g D (Dirac monopole) (16.8): 1 e c 137 = e = 68.5 e = gD = 2e 2 αe 2 where αe is the fine structure constant.13 From this the adimensional magnetic coupling constant to the electromagnetic field 1 g2 αg = D = c 4



c e

2

αe 1 = = 34.25 c 4

that is 1 and so a perturbative approximation cannot be used in computations of processes with magnetic monopoles.

16.8.2 Energy Losses for Monopoles in Matter For the Lorentz transformation of its magnetic field at rest, a moving monopole produces an electric field whose lines of force lie in planes perpendicular to the

12

This result was first given by J.J. Thompson [11] and can be found in E. Amaldi [7], p. 16, and in J.D. Jackson [8], Sect. 6.13. 13 The electric charge of the d quark is e/3 (2e/3 in u quark) thus elementary magnetic charge should be g = 3g D

16.8 Properties of the Magnetic Monopoles

353

monopole trajectory. In matter this field may ionize or excite nearby atoms and molecules so the monopoles lose energy14 For velocity β > 0.05 the energy loss is well understood and can be derived from that of a charged particle. For a particle of charge ze the energy loss for ionization −d E/d x is given by the Bethe and Bloch formula15 Z 1 dE = k z2 − dx A β2



   1 2 m e c2 β 2 γ 2 2 −β ln 2 I2

(16.16)

where k = 4π N A re2 m e c2 = 0.307 MeV cm2 /g and I  Z · 10 eV is the average excitation energy of the atoms in the crossed medium. This expression is the weighted sum of all the energies transferred, per unit path, by the electric field E of the charged particle to the electrons of the medium. For a moving monopole with magnetic charge ng D the electric field is that of a moving charged particle with the charge ze replaced by ng D β. By this substitution in (16.16), the energy loss due to ionization for a monopole is:  ng 2 Z  1  2 m c2 β 2 γ 2  dE D e 2 −β . =k ln − dx e A 2 I2

(16.17)

The differences between this last relation and the (16.16) are evident. For βγ  1 a monopole ionizes heavily as a nucleus of charge Z e = ng D = n 68.5 e, that is if n = 1 about 4900 times a minimum ionizing particle. The range is much shorter than that for a same momentum charged particle. The factor 1/β 2 , relevant at small β for the charged particles, is missing. Of course these features in ionization can be exploited in detectors for the direct search for magnetic monopoles. In Fig. 16.2 the ionization energy losses in air for a proton and a Dirac monopole (n=1) are compared. For the velocity range 10−3 < β < 10−2 in the calculation of the energy loss the material is approximated by a free gas of electrons and the −d E/d x is small and proportional to β. For β < 10−3 monopoles cannot excite atoms and molecules, but can lose energy in elastic collisions with atoms and nuclei through interactions with the atomic or nuclear magnetic moments. At very high energy βγ > 104 , as for charged particle, the energy loss is dominated by single processes of bremsstrahlung, pair production and nuclear interactions with large energy transfer.

14

For more details see the paper by L. Patrizii and M. Spurio (2015) [12]. For this subject and its application to the experiments see C. Bauer et al. [13]. 15 Terms for the density effect and the shell correction have to be added to this simple formula. See Sect. Passage of particle through matter in The Review of Particle Physics by the Particle Data Group [14].

354

16 Magnetic Monopoles

Fig. 16.2 Ionization energy losses in air for a proton and a Dirac monopole, normalized to the value of a minimum ionizing charged particle. (Reprinted figure from C. Bauer et al., [13]. Copyright 2005, with permission from Elsevier.)

16.8.3 Magnetic Monopoles in Magnetic Field A magnetic monopole with charge g D in a magnetic field B is accelerated by the Lorentz force (16.5) and after a path of length L its energy increase is Wm = g D B L or in numbers Wm = 20.50 B L keV G−1 cm−1 . For a cosmic magnetic field B  3 μG and a path of cosmic length L  300 pc the energy is Wm  1011 GeV and the β of the velocity can be approximated by:

β

⎧ ⎨1 ⎩ 10−3



1017 Gev M

1 2

for M  1011 GeV for M  1011 GeV

(16.18)

16.9 Searches for Magnetic Monopoles The searches for magnetic monopoles focus on three types of monopoles, introduced in next sections, which are characterized by a large difference in mass. Classical or Dirac monopoles with mass smaller or in the TeV range which might be produced at the accelerators and are relativistic (β  1). Super Massive GUT Monopoles (MM) with huge mass (1016 ÷ 1017 GeV), which are expected to have been produced in the Early Universe at the breaking of the Grand Unification group about 10−35 s after the Big Bang. These monopoles would be

16.9 Searches for Magnetic Monopoles

355

gravitationally bounded to the galaxy with velocity distribution peaked at β  10−3 or trapped around the Earth or the Sun with β  10−5 and β  10−4 respectively. Intermediate Mass Monopoles (IMM) with masses 105 ÷ 1013 GeV which could be produced in later phase transitions of the Early Universe and after acceleration in cosmic magnetic fields could reach velocities with β  0.1. Extensive reviews of the properties, the detectors and the results of the searches for magnetic monopoles are given in the papers by L. Patrizii et al. (2019) [15], L. Patrizii and M. Spurio (2017) [16], L. Patrizii and M. Spurio (2015) [12] and also G. Giacomelli and L. Patrizii (2005) [17]. A last review in 2020 has been done by N.E. Mavromatos and V.A. Mitsou [18]. A periodically updated review of the searches can be found16 in The Review of Particle Physics by the Particle Data Group [14]. For the first searches see E. Amaldi [7], updated by E. Amaldi and N. Cabibbo [5].

16.9.1 Detection of Magnetic Monopoles In the magnetic field of the experiments a monopole is accelerated and while its energy increases, its trajectory is deflected but it is not helicoidal as for charged particles. A clear signature to identify a magnetic monopole in experiments is its heavy ionization that can be observed in scintillation detectors, in gaseous detectors or in nuclear track detectors (NTD). Moreover a monopole crossing a superconducting coil induces an electromotive force that produces an increase in current that can be detected by a SQUID. So monopoles trapped in rocks or in materials exposed to beams or near the interaction points at colliders, can be detected passing samples of the material through a superconducting coil. In spite of its unique properties very different from those of the usual particles, no signal of magnetic monopole has been observed so far. In the following we shortly report the main results of monopole searches.

16.9.2 Searches for Dirac Monopoles at Accelerators The mass of the monopole is not predicted by the Dirac theory. The relation (16.8) and the naive assumption that the classical radius of the monopole rg = g 2D /m g c2 is equal to the classical radius of the electron re = e2 /m e c2 , give: mg =

g 2 D

e

 me =

1 c 2 e2

2 m e  4700 m e = 2.4 GeV/c2

a large mass but much smaller than the present experimental limits.

16

See Sects. Magnetic Monopoles and Magnetic Monopoles Searches.

356

16 Magnetic Monopoles

Direct searches for monopoles performed in experiments at the accelerators (e+ e− , p p, ¯ pp and ep colliders) have set upper limits on the production cross section17 and lower limits on the mass. To extract these limits assumptions have to be made to model the kinematics of the monopole-antimonopole pair production since perturbative field theory cannot be used to calculate the rate and kinematic properties of the events. The most recent searches were done at the pp collider LHC by the ATLAS and the MoEDAL experiments. The last result published by ATLAS is a search [19] for magnetic monopoles in pp collisions at 13 TeV center of mass energy.18 The signature was based upon high ionization in the transition radiation tracker and large pencil shape energy deposit in the electromagnetic calorimeter. No monopole has been observed and limits have been extracted. The analysis considered a model of Drell-Yan production for spin 0 and 1/2 monopoles and magnetic charges 1g D and 2g D . Upper limits on the cross sections in the mass range from 200 GeV to 4 TeV are: σ ∼ 0.3 fb for spin 1/2 and g = 1g D (σ ∼ 4 fb for g = 2g D ), σ ∼ 0.15 fb for spin 0 and g = 1g D (σ ∼ 10 ÷ 4 fb for g = 2g D ). In the hypothesis of Drell-Yan production cross section the mass limits (95% CL) are: for spin 1/2 and g = 1g D m g > 2370 GeV (m g > 2125 GeV for g = 2g D ) and for spin 0 and g = 1g D m g > 1850 GeV (m g > 1725 GeV for g = 2g D ). The MoEDAL experiment uses two main passive systems deployed around an interaction point at LHC. These are a nuclear track detector (NTD) composed of layers sensitive only to highly ionizing particles and about 800 kg of aluminium for monopole trapping. At the end of each run the NTD and the aluminium are removed for analysis and replaced with new ones. After a chemical etching of the NTD foils, a scanning searches for etch-pits produced in points damaged by the passage of heavily ionizing particles. For the trapping search samples of the aluminium are passed through a superconducting coil of a SQUID magnetometer. No signal of monopole has been observed so far. MoEDAL reported results for searches in pp collisions at 7, 8 and 13 TeV center of mass energy. In the analysis of the trapping material exposed to collisions at 13 TeV [20], the non observation of magnetic monopoles led to upper limits ( 1016 ÷ 1017 GeV) in the phase transition corresponding to the spontaneous symmetry breaking that originated the known interactions at a temperature of the order of 1015 GeV and at an age of the Universe about 10−35 s. The present-day abundance of these monopoles would exceed by many orders of magnitude the critical energy density of the Universe. The subsequent cosmological inflation would have reduced their abundance to values that would make the detection difficult. Magnetic monopoles of intermediate mass (IMM) (M ∼ 1010 GeV) might

358

16 Magnetic Monopoles

have been produced in a later phase transition at a temperature of the order of 109 GeV at a time ∼10−23 s. The structures of these massive monopoles with the surrounding cloud of virtual particles can be found in L. Patrizii et al. (2019) [15]. In super massive monopoles a condensate of fermion-antifermion pairs may contain terms violating the baryon number conservation so the interaction of these monopoles with a nucleon can lead to reactions in which the nucleon decays (catalysis of nucleon decay, e.g. M + p → M + e+ + π 0 ). The very massive GUT monopoles are beyond the possibility of production at existing or foreseen accelerators, but, due to the magnetic charge conservation, the lightest magnetic monopoles, expected to be stable if produced in the early Universe, should be present as cosmic relics in the present Universe. Their kinetic energy would be affected by the Universe expansion and by the interaction with galactic and extragalactic magnetic fields.

16.9.4 Bounds on the Flux of Cosmic Magnetic Monopoles Considering that the acceleration of monopoles in the galactic magnetic field would subtract energy to this magnetic field, an upper bound (Parker bound) on the flux Φ of cosmic monopoles can be derived from the condition that this magnetic energy has to be smaller than that generated by the dynamo effect connected to the rotation of the Galaxy: Φ  10−15 cm−2 s−1 sr−1 for M  1017 GeV and Φ  10−15 × (1017 GeV/M) cm−2 s−1 sr−1 for M  1017 GeV. Similar arguments when applied to the survival of an early seed that should be the progenitor of the galactic magnetic field, gives a tighter bound (Extended Parker bound): Φ  10−16 × (1017 GeV/M) cm−2 s−1 sr−1 . An upper limit on the monopole flux can also be found from the constraint that the contribution Ωmon of the magnetic monopoles to the mass of the Universe has to be smaller than the critical density. For masses ∼1017 GeV it follows Φ < 1.3 × 10−13 Ωmon β cm−2 s−1 sr−1 where Ωmon can be assumed 4 × 10−5 . The upper limits (90% C.L.) for the flux of cosmic MM monopoles from many underground experiments at different depths are shown in Fig. 16.4 versus β and in Fig. 16.5 versus the mass (from L. Patrizi et al. [15]). The ongoing NoVA experiment with an active surface of 4290 m2 has the potential to perform a search beyond the MACRO limits.

16.9.5.2

IMM Monopoles

The SLIM experiment at the Chalcataya laboratory at 5230 m altitude, in Bolivia, with a 4.2-year exposure of a 427 m2 NTD detector, searched for a flux of monopoles with magnetic charge g D and mass as low as 100 GeV. With no candidate monopole observed set an upper limit on the flux ∼1.3 × 10−15 cm−2 s−1 sr−1 for β > 4 × 10−3 . Upper limits (90% C.L.) on the mass are shown in Fig. 16.5.

360

16 Magnetic Monopoles

Fig. 16.5 Upper limits (90% C.L.) on flux of GUT monopoles with magnetic charge g D versus the mass. (Figure reprinted with permission from L. Patrizii et al. (2019) [15].)

Fig. 16.6 Upper limits (90% C.L.) on flux of magnetic monopoles with magnetic charge g D versus the velocity for β > 0.5. (Figure reprinted with permission from L. Patrizii et al. (2019) [15].)

16.9.5.3

Relativistic Monopoles

Monopoles with velocity exceeding the Cherenkov threshold can emit radiation a factor ∼5000 larger than a muon’s yield. The threshold is 0.76 c in water and ice but below the light can be produced by δ−electrons with β > 0.6. Neutrino tele-

References

361

scopes ANTARES, Baikal and IceCube exploited this approach. Limits from these experiments on the monopole flux are given in Fig. 16.6. The best upper limit by ANTARES is 2 × 10−18 cm−2 s−1 sr−1 for β  0.8.

References 1. MoEDAL Experiment, B. Acharya et al., Phys. Rev. Lett. 126, 071801 (2021) 2. P.A.M. Dirac, Quantised singularities in the electromagnetic field. Proc. Roy. Soc. A 133, 60 (1931) 3. P.A.M. Dirac, Electrons and protons. Proc. Roy. Soc. A 126, 360 (1930) 4. C.D. Anderson, The apparent existence of easily deflectable positives. Science 76, 238 (1932) 5. E. Amaldi, N. Cabibbo, On the Dirac Magnetic Poles, in Aspects of Quantum Theory, a volume in honour of P.A.M. Dirac, ed. by Abdus Salam, E.P. Wigner (Cambridge, University Press, 1972) 6. P.A.M. Dirac, The theory of magnetic poles. Phys. Rev. 74, 817 (1948) 7. E. Amaldi, On the Dirac Magnetic Poles, published in the volume Old and new problems in Elementary Particles, in honour of G. Bernardini, ed. by G. Puppi (Academic Press, New York, 1968) 8. D.J. Jackson, Classical Electrodynamics, 3rd edn. (Wiley, NewYork, 1999) 9. E. Fermi, Il monopolo di Dirac, Conferenze di Fisica Atomica, Fondazione Donegani, Accademia Nazionale Lincei, 1950, p. 117 (see E. Fermi, Opera Omnia, Vol. II, p. 780, http:// operedigitali.lincei.it/Fermi/Fermi-omnia2.pdf) 10. A.S. Goldhaber, Phys. Rev. 140 B, 1407 (1965) 11. J.J. Thompson, Elements of the Mathematical Theory of Electricity and Magnetism (Cambridge, University Press, 1900–1904) 12. L. Patrizii, M. Spurio, Status of searches for magnetic monopoles. Ann. Rev. Nuclear Phys. 65, 279–302 (2015). https://doi.org/10.1146/annurev-nucl-102014-022137 13. C. Bauer et al., Nuclear Instr. Methods Phys. Res. A 545, 503–515 (2005) 14. R.L. Workman et al. (Particle Data Group), Prog. Theor. Exp. Phys. 2022, 083C01 (2022) (http://pdg.lbl.gov/index.html) 15. L. Patrizii, Z. Sahnoun, V. Togo, Searches for cosmic magnetic monopoles: past, present and future. Phil. Trans. R. Soc. A 377, 20180328 (2019). https://doi.org/10.1098/rsta.2018.0328 16. L. Patrizii, M. Spurio, The MoEDAL experiment at the LHC: in search of magnetic monopoles. Il Nuovo Saggiatore 33, 5 (2017) 17. G. Giacomelli, L. Patrizii, Magnetic Monopoles Searches (2005), https://arxiv.org/abs/hep-ex/ 0302011 18. N.E. Mavromatos, V.A. Mitsou, Magnetic monopoles revisited: models and searches at colliders and in the cosmos (2020), https://arxiv.org/abs/2005.05100 19. ATLAS Experiment, Aad et al., Phys. Rev. Lett. 124, 031802 (2020) 20. MoEDAL Experiment, B. Acharya et al., Phys. Rev. Lett. 123, 021802 (2019) 21. MoEDAL Experiment, B. Acharya et al., https://arxiv.org/abs/2112.05806 (2021) 22. MoEDAL Experiment, B. Acharya et al., Nature 602, 63–67 (2022)

Appendix A

Orthogonal Curvilinear Coordinates

Given a symmetry for a system under study, the calculations can be simplified by choosing, instead of a Cartesian coordinate system, another set of coordinates which takes advantage of that symmetry. For example calculations in spherical coordinates result easier for systems with spherical symmetry. The use of suitable coordinates gives the possibility to reduce the number of the variables needed in a problem and in the solution of the Laplace equation allows the separation of variables which reduces partial second order equations in ordinary second order equations of easier solution as presented in Chap. 8. In this chapter we write the general form of the differential operators used in Electrodynamics and then we give their expressions in spherical and cylindrical coordinates.

A.1

Orthogonal curvilinear coordinates

A system of coordinates u 1 , u 2 , u 3 , can be defined so that the Cartesian coordinates x, y and z are known functions of the new coordinates: x = x(u 1 , u 2 , u 3 ) y = y(u 1 , u 2 , u 3 ) z = z(u 1 , u 2 , u 3 )

(A.1)

Systems of orthogonal curvilinear coordinates are defined as systems for which locally, nearby each point P(u 1 , u 2 , u 3 ), the surfaces u 1 = const, u 2 = const, u 3 = const are mutually orthogonal. An elementary cube, bounded by the surfaces u 1 = const, u 2 = const, u 3 = const, as shown in Fig. A.1, will have its edges with lengths h 1 du 1 , h 2 du 2 , h 3 du 3 where h 1 , h 2 , h 3 are in general functions of u 1 , u 2 , u 3 . The length ds of the lineelement OG, one diagonal of the cube, in Cartesian coordinates is: ds =

 (d x)2 + (dy)2 + (dz)2

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 F. Lacava, Classical Electrodynamics, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-031-05099-2

363

364

Appedix A: Orthogonal Curvilinear Coordinates

Fig. A.1 The elementary cube in coordinates u1, u2, u3

and in the new coordinates becomes:  ds = (h 1 du 1 )2 + (h 2 du 2 )2 + (h 3 du 3 )2 . The coefficient h 1 , h 2 , h 3 can be determined from these two expressions and the relations (A.1). The volume of the cube is dτ = h 1 h 2 h 3 du 1 du 2 du 3 . We consider a scalar function f = f (u 1 , u 2 , u 3 ) and a vector A with the components A1 , A2 , A3 , relative to the directions uˆ 1 , uˆ 2 and uˆ 3 , which are functions of u 1 , u2, u3.

A.2

Gradient

The differential operator gradient of a scalar function f is the vector grad f = ∇ f defined by the relation: d f = grad f · dl = ∇ f · dl where df is the differential of f along an elementary displacement dl having components (h 1 du 1 , h 2 du 2 , h 3 du 3 ) and has is maximum value when dl is in the same direction of grad f. Moreover we can write: df =

∂f ∂f ∂f du 1 + du 2 + du 3 ∂u 1 ∂u 2 ∂u 3

and comparing the two formulas:

Appedix A: Orthogonal Curvilinear Coordinates

365

(grad f )1 h 1 du 1 + (grad f )2 h 2 du 2 + (grad f )3 h 3 du 3 =

∂f ∂f ∂f du 1 + du 2 + du 3 ∂u 1 ∂u 2 ∂u 3

the gradient components in the new coordinates result: (grad f )i =

A.3

1 ∂f . h i ∂u i

Divergence

To find the general formula for the operator divergence of a vector (div A = ∇ · A), we apply the Gauss’ theorem to the elementary cube in Fig. A.1. The flux dφ of the vector A out of the cube surface is equal to the divergence of the vector multiplied by the volume of the cube: dφ(A) = div A dτ .

(A.2)

The flux out of the cube face OBHC on the surface u 1 = const is1 : −A1 (u 1 ) h 2 (u 1 )du 2 h 3 (u 1 )du 3 = −A1 h 2 h 3 du 2 du 3 and the flux out of the face AFGJ on the surface u 1 + du 1 = const  is: A1 (u 1 + du 1 ) h 2 (u 1 + du 1 )du 2 h 3 (u 1 + du 1 )du 3 . Series expansion at the first order of the three factors of the last expression gives:  A1 +

∂ A1 du 1 ∂u 1

 h2 +

∂h 2 du 1 ∂u 1

 h3 +

∂h 3 du 1 ∂u 1



and neglecting second order terms this flux becomes: A1 h 2 h 3 du 2 du 3 +

∂ (A1 h 2 h 3 ) du 1 du 2 du 3 . ∂u 1

The total flux on the two opposite faces is: ∂ (A1 h 2 h 3 ) du 1 du 2 du 3 ∂u 1

1

The flux can be approximated by the value of the component A1 at the face center multiplied by the face area.

366

Appedix A: Orthogonal Curvilinear Coordinates

and adding the similar fluxes out of the other faces, for the flux (A.2) out of the cube we find:   ∂ ∂ ∂ (A1 h 2 h 3 ) + (A2 h 1 h 3 ) + (A3 h 1 h 2 ) du 1 du 2 du 3 dφ(A) = ∂u 1 ∂u 2 ∂u 3 = div A h 1 h 2 h 3 du 1 du 2 du 3 from which we obtain the formula for the divergence: 1 div A = h1h2h3

A.4



∂ ∂ ∂ (A1 h 2 h 3 ) + (A2 h 1 h 3 ) + (A3 h 1 h 2 ) ∂u 1 ∂u 2 ∂u 3

 .

Curl

To write the general formula for the operator curl (curl A = ∇ × A), we use the Stokes’ theorem:   A · dl = curl A · nˆ d S . We consider the circulation of the vector A over the curve defined by the loop OBHCO of the elementary cube. The contribution from paths OB and HC, neglecting second order terms, is: A2 (u 3 ) h 2 (u 3 )du 2 − A2 (u 3 + du 3 ) h 2 (u 3 + du 3 )du 2 = −

∂ (A2 h 2 )du 2 du 3 . ∂u 3

Similarly from paths BH and CO we have: A3 (u 2 + du 2 ) h 3 (u 2 + du 2 )du 3 − A3 (u 2 ) h 3 (u 2 )du 3 =

∂ (A3 h 3 )du 2 du 3 ∂u 2

and adding the two contributions, taking into account the Stokes’ theorem, we have: 

 ∂ ∂ (A3 h 3 ) − (A2 h 2 ) du 2 du 3 ∂u 2 ∂u 3



 ∂ ∂ (A3 h 3 ) − (A2 h 2 ) . ∂u 2 ∂u 3

(curl A)1 h 2 h 3 du 2 du 3 = from which: (curl A)1 =

1 h2h3

Similar expressions can be found for the other two components of the curl, therefore we can write:

Appedix A: Orthogonal Curvilinear Coordinates

(curl A)i =

A.5

1 h j hk



367

 ∂ ∂ (Ak h k ) − (A j h j ) . ∂u j ∂u k

Laplacian

The Laplacian operator can be written as Δ = div grad = ∇ · ∇ = ∇ 2 and, using the formulas found for the divergence and the gradient, we get: 1 Δf = h1h2h3

A.6



∂ ∂u 1



h2h3 ∂ f h 1 ∂u 1



∂ + ∂u 2



h3h1 ∂ f h 2 ∂u 2



∂ + ∂u 3



h1h2 ∂ f h 3 ∂u 3

 .

Systems of Orthogonal Curvilinear Coordinates

There are thirteen orthogonal coordinate systems listed in books of methods of theoretical physics2 or in tables3 . These systems are: Cartesian rectangular coordinates, circular cylinder coordinates (cylindrical polar), elliptic cylinder coordinates, parabolic coordinates, spherical coordinates (rotational) (spherical polar coordinates), conical coordinates, (rotational) parabolic coordinates, prolate spheroidal coordinates (rotational), oblate spheroidal coordinates (rotational), ellipsoidal coordinates, paraboloidal coordinates, bispherical coordinates and toroidal coordinates. The last two systems can be used under specific conditions4 .

A.7

Spherical Coordinates ⎧ ⎨ x = r sin θ cos ϕ y = r sin θ sin ϕ ⎩ z = r cos θ ds 2 = dr 2 + r 2 dθ 2 + r 2 sin2 θ dϕ 2

2

Morse and Feshbach [1], Part I, Chap. 5, Table of Separable Coordinates in Three Dimensions, pp. 655–666 3 Gradshteyn and Ryzhik [1], Orthogonal curvilinear coordinate, Sects. 10.51–10.61 4 See Morse and Feshbach [2], p. 665.

368

Appedix A: Orthogonal Curvilinear Coordinates

⎧ ⎨ u1 = r u2 = θ ⎩ u3 = ϕ

⎧ ⎨ h1 = 1 h2 = r ⎩ h 3 = r sin θ

grad f :

(grad f )r =

∂f ∂r

(grad f )θ =

1 ∂f r ∂θ

(grad f )ϕ =

∂f 1 r sin θ ∂ϕ

div A: div A =

∂ 1 ∂ 2 1 1 ∂ Aϕ (r Ar ) + (sin θ Aθ ) + 2 r ∂r r sin θ ∂θ r sin θ ∂ϕ

curl A: (curl A)r =

1 r sin θ

(curl A)θ =



∂ ∂ Aθ (sin θ Aϕ ) − ∂θ ∂ϕ



1 ∂ 1 ∂ Ar − (r Aϕ ) r sin θ ∂ϕ r ∂r

(curl A)ϕ =

  1 ∂ ∂ Ar (r Aθ ) − r ∂r ∂θ

Δf : Δf =

1 ∂ r 2 ∂r

    1 ∂ ∂f 1 ∂2 f ∂f r2 + 2 sin θ + 2 2 ∂r r sin θ ∂θ ∂θ r sin θ ∂ϕ 2

Sometime it is useful to replace the first term in the Laplacian with the relation5 :

5

This relation can be easily derived:      2  1 ∂ ∂f 1 ∂f 1 ∂f ∂2 f 2∂f 2∂ f 2r = r = + r + + r r 2 ∂r ∂r r2 ∂r ∂r 2 r ∂r ∂r ∂r 2

=

1 ∂ r ∂r

 f +r

∂f ∂r

 =

1 ∂ ∂ 1 ∂2 (r f ) = (r f ) r ∂r ∂r r ∂r 2

Appedix A: Orthogonal Curvilinear Coordinates

1 ∂ r 2 ∂r

A.8

369

  1 ∂2 2∂f r = (r f ) ∂r r ∂r 2

Cylindrical Coordinates ⎧ ⎨ x = ρ cos ϕ y = ρ sin ϕ ⎩ z=z ds 2 = dρ 2 + ρ 2 dϕ 2 + dz 2 ⎧ ⎨ u1 = ρ u2 = ϕ ⎩ u3 = z

⎧ ⎨ h1 = 1 h2 = ρ ⎩ h3 = 1

grad f : (grad f )ρ =

∂f 1 ∂f ∂f (grad f )ϕ = (grad f )z = ∂ρ ρ ∂ϕ ∂z

div A:

div A =

∂ Az 1 ∂ Aϕ 1 ∂ (ρ Aρ ) + + ρ ∂ρ ρ ∂ϕ ∂z

curl A: (curl A)ρ =

∂ Aϕ 1 ∂ Az − ρ ∂ϕ ∂z

(curl A)ϕ =

(curl A)z =

1 ρ



∂ Az ∂ Aρ − ∂z ∂ρ

∂ ∂ Aρ (ρ Aϕ ) − ∂ρ ∂ϕ



370

Appedix A: Orthogonal Curvilinear Coordinates

Δf : 1 ∂ Δf = ρ ∂ρ

  ∂f 1 ∂2 f ∂2 f ρ + 2 2 + 2 ∂ρ ρ ∂ϕ ∂z

References 1. P.M. Morse, H. Feshbach, Methods of Theoretical Physics (Part I, McGraw-Hill Book C, 1953) 2. I.S. Gradshteyn, I.M. Ryzhik, Tables of Integrals, Series and Products, 7th edn. (Academic Press, Elsevier, London, 2007)

Index

A Abraham-Lorentz equation, 316 Ampère’s law, 6 Angular momentum of the electromagnetic field, 231 for a charged spherical shell, 263 for disk with charges and solenoid, 248, 250 for point charge and magnetic dipole, 248 for two cylindrical shells in damped B field, 251 paradox, 247 ATLAS experiment, 356–357 B Bessel functions of first kind, 280, 281 Biot and Savart law, 5 Boundary conditions for electrostatic problems, 138 C Capacitance matrix, 87 Capacitor at high frequency, 277 Feynman approach, 277 from Laplace equation, 284 Poynting vector, 284 Cavendish, 330–332 Charge in frames in relative motion, 185 Charge invariance, 179, 185 Charge quantization relation, 347–352 Chauchy-Riemann relations, 102 Color of the sky, 322 Complex functions analytic functions, 101

bidimensional problems, 104 field at the center of a quadrupole, 111 function f (z) = z μ , 104–105 quadrupole f (z) = z 2 , 104 1

thin plate f (z) = z 2 , 108 two opposite sign wires, 111 two same sign wires, 111 wire f (z) = log z, 109 Conformal mapping, 117–120 function w = sin z, 132 single corner, 127 split cylinder, 123 wire between plates, 120 Constitutive relations, 63 Continuity equation, 4 in covariant form, 206 Continuity relations between two dielectrics, 63 Coulomb gauge, 12 Coulomb’s law, 1, 329, 330 Coulomb’s theorem, 3 Covariant and contravariant components, 201 for operators, 204 for tensors, 204 for vectors, 202, 204 Current density 4-vector, 181 Cyclotron radiation, 312 Cylinder and wire. see Method of image charges in dielectrics Cylinder in uniform field from charge images, 48 separation of variables, 163

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 F. Lacava, Classical Electrodynamics, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-031-05099-2

371

372 D Dipole, electric, 17–19, 22–23 Dipole pair force between dipoles, 28 interaction energy, 28 Dirac monopole magnetic charge g D , 348 theory, 347 Dirac string, 347 Displacement current, 7 Duality transformation for Maxwell equations, 344

E Einstein theory of relativity, 200 Electric field of a charge in motion. see Fields of a charge Electrodynamic potentials, 10 equations in covariant form, 206, 210 in covariant form, 206 Electromagnetic induction and relative motion of circuits, 190 Electromagnetic tensor, 207 Lorentz transformation of, 208 Electromagnetic waves, 14 Energy transfer in circuits, 221 Equation of motion for a charged particle in electromagnetic field, 212 Ether, 200 Euler-Lagrange equations, 340

F Faraday-Neumann-Lenz law, 6 and relative motion, 190 Fermi, quantization relation, 350 Feynman paradox, 247 Field near the edge of a capacitor, 128 Fields of a charge in arbitrary motion, 294, 300–306 in uniform motion, 289, 292 Force between wires carrying currents, 187 Force on a charge moving near a wire carrying current, 182

G Galileian transformation, 199–200 Gauge transformations, 10 in covariant form, 211 Gauss’s law, 3, 330

Index I Image charges. see Method of image charges Image charges in dielectrics. see Method of image charges in dielectrics

J Jupiter magnetic field limit on the mass of the photon, 336

K Kelvin, force between charged conductive spheres, 89, 93

L Laplace equation, 36, 64, 92, 101–104, 117– 124, 137, 138, 233 in Cartesian coordinates, 140 in spherical azimuthal coordinates, 145 in spherical coordinates, 156 in cylindrical coordinates, 158 in resonant cavity, 284 Laplace’s laws, 5 Larmor formula, 307 Legendre polynomials, 145 associated, 157 Liénard formula, 311 Liénard-Wiechert potentials, 296–300 covariant form of, 299 Linear-fractional function, 121 Lorentz transformation for 4-vector, 203 for E and B fields, 184, 208 Lorenz gauge, 11 in Proca equations, 333 in covariant form, 210

M Magnetic field of a charge in motion. see Fields of a charge Magnetic monopole, 347 detection, 355 energy loss in matter, 352 field and potential, 348 Magnetic monopole search at accelerators, 355 cosmic massive monopole, 358 GUT MM monopole, 354, 359 IMM intermediate mass, 355, 359 Maxwell equations, 3–7 in matter, 8

Index in vacuum, 8 in covariant form, 209–210 Maxwell equations, generalized, 343 Maxwell inequalities, 87, 89 Maxwell stress tensor, 224, 228 covariant form, 231 Method of image charges, 35, 83. see Series of image charges capacitance of two conductive cylinders, 48, 49 charged wire and conductive plane, 48 conductive spherical shell and inner point charge, 48 cylindrical conductor and inner wire, 48 cylindrical conductor and outer wire, 45 cylindrical condutor in uniform field, 48 dipole and conducting sphere, 48 dipole and conductive plane, 47 force on a point charge near a sphere, 41 plane and point charge, 47, 48 point charge and conductive corner, 47, 93 sphere and point charge, 38 sphere in uniform field, 42 point charge and conductive plane, 36 Method of image charges in dielectrics, 63 charged wire at separation of two dielectrics, 71 dielectric cylinder and inner charged wire, 72 dielectric cylinder and outer charged wire, 72 dielectric cylinder in a uniform electric field, 72 dielectric sphere in uniform electric field, 68 point charge at separation of two dielectrics, 64 Möbius transformation. see Linearfractional function MoEDAL experiment, 346, 356–357 Momentum conservation in Electrodynamics, 223, 228 Momentum of the electromagnetic field, 228 density, 229 Multipole expansion for an electric system, 19–22 for a ring, 28 high order terms, 25 in spherical harmonics, 26 N Nodal line. see Dirac string

373 P Parity in electromagnetism, 346 Parker bound, 358 Phase of the wave in covariant form, 211 Photon mass limit from Jupiter magnetic field, 336 Lakes experiment, 337 Schumann resonances, 338 solar wind, 338 Williams, Faller and Hill experiment, 334 Plinton and Lawton test, 334 Poisson equation, 4, 13, 35–37, 63–64, 137, 138, 295 Potentials for a charge in arbitrary motion, 294 in uniform motion, 292 Power radiated in circular accelerator, 312–314 in linear accelerator, 309–310 Poynting’s theorem, 215 Poynting vector, 217 in a battery, 218 in a capacitor, 220 in a resistor, 218 in a solenoid, 219 in electrical circuits, 221 in resonant cavity, 284 Pressure of radiation, 229 Principle of indetermination and mass of the photon, 339 Principle of virtual work, 242 Proca equations, 332 in covariant form, 341 Proca Lagrangian, 332 Pseudo-charge density, 333 Pseudo-current density, 333 Q Quadrupole tensor, electric, 23 R Radiation by an accelerated charge, 306 for a ∞ v, 307 for a ⊥ v, 311 for v ≥ c, 307 in arbitrary motion, 310 Radiation by an electric dipole, 317 Radiation gauge, 13 Radiation reaction, 315 Relativistic covariance of Electrodynamics, 205

374 Resonant cavity, 282. see Capacitor at high frequency Retarded potentials, 12, 294 equations of, 12 Ring, potential for a charged ring, 151, 167

S Schrödinger E., method for photon mass limit, 336 Schumann resonances, 338 Schwarz-Christoffel transformation, 125– 127 Separation of variables, 137 in Cartesian coordinates, 140 in cylindrical coordinates, 158 in spherical coordinates, 156 in spherical coordinates with azimuthal symmetry, 145 in resonant cavity, 284 Series of image charges, 83 conductive sphere and plate, 89, 90 dielectric sphere and point charge, 92 force between two charged conductive spheres, 88 point charge between plates, 83 two separated conductive spheres, 84 wire between plates, 93 conductive sphere inside another one, 93 Sommerfeld, field in resistive coaxial cable, 221, 233 Sphere and point charge from image method, 38, 92 from Laplace equation, 153, 167 Spherical configurations in uniform electric field, 42, 68, 146

Index Spherical shell, potential for σ (θ) distribution, 167 Split cylinder potential from conformal mapping, 123 from Laplace equation, 160, 167 Stokes’ theorem, 3 Synchrotron radiation, 312

T Test of Coulomb’s law. see photon mass limit Theory of relativity and Electrodynamics, 199 Time reversal in electromagnetism, 346 Transformation of E and B fields in relativity, 184 from the electromagnetic tensor transformation, 208 Transverse gauge, 13

V Vector potential of a spinning charge sphere, 260

W Wave 4-vector, 211 Wedge potential from analytic function, 105 conformal mapping, 127 separation of variables, 159 Williams, Faller and Hill experiment, 334

Y Yukawa potential, 334