Core Electrodynamics 9783030668167, 9783030668181

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Undergraduate Lecture Notes in Physics

Sandra Chapman

Core Electrodynamics

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

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.

More information about this series at http://www.springer.com/series/8917

Sandra Chapman

Core Electrodynamics

123

Sandra Chapman Department of Physics University of Warwick Centre for Fusion, Space, Astrophysics Coventry, UK

ISSN 2192-4791 ISSN 2192-4805 (electronic) Undergraduate Lecture Notes in Physics ISBN 978-3-030-66816-7 ISBN 978-3-030-66818-1 (eBook) https://doi.org/10.1007/978-3-030-66818-1 © Springer-Verlag GmbH Germany, part of Springer Nature 2021 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, 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

Preface to the Springer Edition

Electrodynamics is often the only example of a field unification that physics undergraduates have the opportunity to work through fully and in detail. It provides a striking example of the character of physical law, both in its elegance and all-encompassing nature. Its building blocks, electromagnetism and special relativity, are typically first encountered in distinct courses. Its applications are far-reaching and extend across multiple courses of study, both at undergraduate and graduate level, which are well served by several classic texts in the literature. Here, the development of electrodynamics is stripped down to its essentials and is presented in an informal manner whilst retaining sufficient rigour to support further study. This journey begins with the experimentally determined properties of electricity and magnetism, through special relativity and the fundamental constraint that the speed of light is constant, to the unification of the electric and magnetic fields to give frame-invariant electromagnetism. Since the publication of the first edition, the pedagogical trend has been to package knowledge in progressively smaller, ‘bite sized’ pieces that are studied and examined in a piece-wise manner. As this edition is being released, we are in the midst of a pandemic which is driving a step-change to online learning which can only accelerate this trend. This points to a need for works that offer synthesis whilst retaining brevity, as here. This monograph, whilst being brief, develops electrodynamics, from the founding observations to a unified field theory, in a manner that allows it to be appreciated in its entirety. Kenilworth, UK

Sandra Chapman

v

Preface

This monograph is based on a final year undergraduate course in Electrodynamics that I introduced at Warwick as part of the new four-year physics degree. When I gave this course, my intention was to engage the students in the elegance of electrodynamics and special relativity whilst giving them the tools to begin graduate study. Here, from the basis of experiment, we first derive the Maxwell equations and special relativity. Introducing the mathematical framework of generalized tensors, the laws of mechanics, the Lorentz force and the Maxwell equations are then cast in a manifestly covariant form. This provides the basis for graduate study in field theory, high energy astrophysics, general relativity and quantum electrodynamics. As the title suggests, this book is “electrodynamics lite”. The journey through electrodynamics is kept as brief as possible, with minimal diversion into details, so that the elegance of the theory can be appreciated in a holistic way. It is written in an informal style and has few prerequisites; the derivation of the Maxwell equations and their consequences is dealt with in the first chapter. Chapter 2 is devoted to conservation equations in tensor formulation; here Cartesian tensors are introduced. Special relativity and its consequences for electrodynamics are introduced in Chap. 3 and cast in four vector form; here we introduce generalized tensors. Finally in Chap. 4, the Lorentz frame-invariant electrodynamics is developed. Supplementary material and examples are provided by the two sets of problems. The first is revision of undergraduate electromagnetism, to expand on the material in the first chapter. The second is more advanced, corresponding to the remaining chapters, and its purpose is twofold: to expand on points that are important, but not essential, for the derivation of manifestly covariant electrodynamics and to provide examples of manipulation of cartesian and generalized tensors. As these problems introduce material not covered in the text, they are accompanied by full worked solutions. The philosophy here is to facilitate learning by problem-solving, as well as by studying the text. Extensive appendices for vector relations, unit conversion and so forth are given with graduate study in mind. As SI units are used throughout here, conversions for units and equations are also given. vii

viii

Preface

There have been many who have contributed to the existence of this book. Thanks, in particular, go to Nick Watkins for valuable discussions and to George Rowlands for his insightful reading of the final draft. David Betts, the editor of this series, has also shown remarkable patience and tenacity as the many deadlines have come and gone. The completion of the book was also much assisted by a PPARC personal fellowship. Finally, my thanks go to the students themselves; their lively reception of the original course and their insightful questions were the inspiration for this book. If the reader finds that this book provides a shortcut to experience the beauty of electrodynamics, without sacrificing the rigour needed for further study, then I have succeeded. Kenilworth, UK

Sandra Chapman

Contents

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1 1 3 5 6 10 11 12 13 16 18 19

2 Field Energy and Momentum . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Tensors and Conservation Equations . . . . . . . . . . . . . . . . . 2.1.1 Momentum Flux Density Tensor . . . . . . . . . . . . . . 2.1.2 Momentum Flux, Gas Pressure and Fluid Equations 2.1.3 Cartesian Tensors, Some Definitions . . . . . . . . . . . . 2.2 Field Momentum and Maxwell Stress . . . . . . . . . . . . . . . . 2.2.1 Energy Conservation: Poynting’s Theorem . . . . . . . 2.2.2 Momentum Conservation: Maxwell Stress . . . . . . . . 2.3 Radiation Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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21 22 22 24 26 28 29 30 33

3 A Frame Invariant Electromagnetism . . . . . . . 3.1 The Lorentz Transformation . . . . . . . . . . . 3.2 The Moving Charge and Wire Experiment . 3.3 Maxwell in Terms of Potentials . . . . . . . . . 3.4 Generalized Coordinates . . . . . . . . . . . . . .

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37 38 42 46 47

1 A Brief Tour of Electromagnetism . . . . . . . . . . . . . . . 1.1 The Building Blocks . . . . . . . . . . . . . . . . . . . . . . . 1.2 Maxwell I and II . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.1 Flux of a Vector Field . . . . . . . . . . . . . . . . 1.2.2 Flux of E . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.3 Flux of B . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.4 Conservative and Nonconservative Fields . . 1.3 Maxwell III and IV . . . . . . . . . . . . . . . . . . . . . . . 1.3.1 Faraday’s Law and Galilean Invariance . . . . 1.3.2 Ampère’s Law and Conservation of Charge 1.4 Electromagnetic Waves . . . . . . . . . . . . . . . . . . . . . 1.5 Conservation Equations . . . . . . . . . . . . . . . . . . . . .

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ix

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Contents

3.5 Four Vectors and Four Vector Calculus . . . . . . . . 3.5.1 Some Mechanics, Newton’s Laws . . . . . . . 3.5.2 Some Four Vector Calculus . . . . . . . . . . . 3.6 A Frame Invariant Electromagnetism . . . . . . . . . . 3.6.1 Charge Conservation . . . . . . . . . . . . . . . . 3.6.2 A Manifestly Covariant Electromagnetism .

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51 51 54 55 55 56

Field Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . Invariant Form for E and B: The EM Field Tensor Maxwell’s Equations in Invariant Form . . . . . . . . . Conservation of Energy-Momentum . . . . . . . . . . . Lorentz Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.1 Manifestly Covariant Electrodynamics . . . . 4.5 Transformation of the Fields . . . . . . . . . . . . . . . . . 4.6 Field from a Moving Point Charge . . . . . . . . . . . . 4.7 Retarded Potential . . . . . . . . . . . . . . . . . . . . . . . .

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59 59 62 65 66 67 68 69 72

Suggested Texts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

77

Appendix A: Revision Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

79

Appendix B: Solutions to Revision Problems . . . . . . . . . . . . . . . . . . . . . .

85

Appendix C: Some Advanced Problems . . . . . . . . . . . . . . . . . . . . . . . . . .

97

4 The 4.1 4.2 4.3 4.4

Appendix D: Solution to Advanced Problems . . . . . . . . . . . . . . . . . . . . . . 101 Appendix E: Vector Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 Appendix F: Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117 Appendix G: Units and Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 Appendix H: Dimensions and Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 Appendix I: Physical Constants (SI) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

List of Figures

Fig. 1.1 Fig. Fig. Fig. Fig. Fig.

1.2 1.3 1.4 1.5 1.6

Fig. 1.7 Fig. 1.8

Fig. 1.9 Fig. 1.10 Fig. 1.11 Fig. 1.12

Fig. 1.13 Fig. 1.14 Fig. Fig. Fig. Fig.

1.15 1.16 1.17 2.1

Surface element dS on surface S spanning curve C with line element dl . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Surface element dS on volume V . . . . . . . . . . . . . . . . . . . . . . The right hand rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Location of charges q1 and q2 . . . . . . . . . . . . . . . . . . . . . . . . The electric field due to charge qi ¼ qðr0 ÞdV 0 . . . . . . . . . . . . Flux across dS is maximal when v is parallel to dS, and zero when v is perpendicular . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Volume containing particles that cross dS in time dt . . . . . . . Surfaces enclosing single charge q. The surface element dS lies on the arbitrary surface S and forms the end of the cone. There is no flux of E across the sides of the cone . . . . . . . . . Side view of cone enclosing single charge q. The surface element vector dS and field E lie in the plane of the paper . . Cone enclosing single charge q in cartesian and cylindrical polar coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The closed surface S has zero nett flux through it . . . . . . . . . E  dl is integrated around the closed curve C in the vicinity of a point charge. The radial sections of the path give contributions of equal size and opposite sign and the sections along contours of the potential (dotted lines) give zero contribution. The integral around the closed curve is zero . . . The observer is at rest in frame 1 (top) and moves with the wire in frame 2 (bottom). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The observer is at rest and the wire moves on conducting rails . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Wire element dl sweeps out surface element dS in time dt . . Charge flows out of volume V . . . . . . . . . . . . . . . . . . . . . . . . An ideal discharging capacitor . . . . . . . . . . . . . . . . . . . . . . . . Some of the forces acting on fluid element dV that form the components of a pressure tensor . . . . . . . . . . . . . . . .

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2 2 3 3 4

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List of Figures

Fig. 2.2 Fig. 3.1

Fig. 3.2

Fig. 3.3

Fig. 3.4

Fig. 3.5

Fig. 4.1

Fig. 4.2 Fig. 4.3

Fig. 4.4

Fig. 4.5 Fig. Fig. Fig. Fig. Fig.

A.1 A.2 C.1 D.1 D.2

Vector r in the x; y; z and x0 ; y0 ; z0 coordinate systems . . . . . . . The light clock is oriented perpendicular to its direction of motion. Top: observer rest frame i.e., the clock moves past us; bottom: clock rest frame i.e., we move with the clock . . . . . . The light clock is oriented parallel to its direction of motion. Top: observer rest frame i.e., the clock moves past us; bottom: clock rest frame i.e., we move with the clock . . . . . . . . . . . . . In frame S1 the current in the wire is carried by the electrons, the test charge q moves at u. In frame S2 the current is carried by the protons, the test charge q is at rest. . . . . . . . . . . . . . . . The (dashed) loop over which we integrate B  dl is a circle of radius r centred on the wire, with ^r transverse to the direction ^. Also shown is the direction of u ^ B . . . . . . . . . of motion u The (dashed) surface over which we integrate E  dS is a cylinder of radius r centred on the wire, with ^r transverse to the ^ ............................... direction of motion u Charge q is at rest at the origin of the S0 frame and is moving with velocity v^x1 w.r.t. frame S. The origins of the S and S0 frames coincide at t ¼ 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Sketch of E1 and E2 at point P versus x01 =c ¼ x1  vt . . . . . . Sketch of the E0 field around a charge at rest, and the E field around a charge moving at relative velocity v, in the x1 ; x2 plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A volume element is located at rest at the origin of the S0 frame which is moving at þ v in the x1 direction w.r.t. the S frame. In the S0 frame the volume element contains charge q0 dx01 dx02 dx03 . The origins of the two frames are coincident at t ¼ t0 ¼ 0 . . . Volume element dV 0 and point P located w.r.t. a single fixed origin O . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problem 1a: A straight wire carrying a steady current I . . . . . Problem 1b: A straight wire carrying a steady current I . . . . . A pair of wires . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Surface S enclosing element of a wire of length dl . . . . . . . . . Curve C enclosing the wire with element along the curve dl . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Chapter 1

A Brief Tour of Electromagnetism

So far you will have encountered various expressions from electrostatics and magnetostatics which we will show can then be synthesised into four equations, the Maxwell equations which, with the Lorentz force law, are a complete description of the behaviour of charged particles and electromagnetic fields. Electromagnetism is usually presented in this way for two reasons, first this is how electromagnetism was first discovered experimentally and second, these expressions (Coulomb’s law, Lenz’ law, Faraday’s law, Biot Savart law and so on) are useful in particular circumstances. Here we will first take a look at how the Maxwell equations are constructed, from the experimentally determined expressions and by using the mathematics of vector fields. The Maxwell equations are a unification of electric and magnetic fields that are inferred experimentally. The unified equations yield an important prediction: the existence of electromagnetic waves, which then compels us to develop a formalism that is consistent with special relativity. This leads to a form for the Maxwell equations, the Lorentz force law and the laws of mechanics that are frame invariant and thus consistent with the requirement that physical laws are the same in all frames of reference, anywhere in the universe.

1.1 The Building Blocks We will use the experimentally determined: • Coulomb’s Law • Faraday’s Law • Ampère’s Law

© Springer-Verlag GmbH Germany, part of Springer Nature 2021 S. Chapman, Core Electrodynamics, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-030-66818-1_1

1

2

1 A Brief Tour of Electromagnetism

Fig. 1.1 Surface element dS on surface S spanning curve C with line element dl

dS

C dl Fig. 1.2 Surface element dS on volume V

dS

V

and some mathematics: • Stokes’ Theorem (for any vector field A): 

 A · dl =

C

∇ ∧ A · dS

(1.1)

S

where surface S spans curve C as shown in Fig. 1.1 • Gauss’ or Divergence Theorem: 

 A · dS = S

∇ · Ad V

(1.2)

V

where surface S encloses volume V as shown in Fig. 1.2. • and the right hand rule: ˆi ∧ ˆj = kˆ which is built in to the way the cross product is defined, so that in the Lorentz force law the magnetic force F M = qv ∧ B is positive for protons (Fig. 1.3). If we had used a left handed cross product so that ˆi ∧ ˆj = −kˆ then Lorentz force law would be written F M = −qv ∧ B for protons. We have also defined the electric (vector) field to point away from positive charge so that FE = qE for protons.

1.2 Maxwell I and II

3

Fig. 1.3 The right hand rule

F k i

j

v

B

1.2 Maxwell I and II Coulomb experimentally determined the force between two static charges to be F21 =

q1 q2 1 (r1 − r2 ) 4π0 | r1 − r2 |3

(1.3)

where the force is in Newtons (N ), charge q is in Coulombs (C), r in meter s and 0 = 8.85.10−12 C 2 N −1 m −2 (or in terms of the Farad, Fm −1 ) is the permittivity of free space (see Fig. 1.4). Coulomb’s law embodies three empirically known properties of F: 1. the force is proportional to the charges F ∼ q1 , q2 2. the force points radially away from the positive charges 3. the force is inverse square F ∼ 1/r 2 In addition we can show that the force obeys the principle of superposition so that for a collection of n charges; the force on the j th charge q j is Fj =

n qi q j 1  (ri − r j ) 4π0 i= j | ri − r j |3

(1.4)

We then define the electric field at the position of the j th charge as E=

Fig. 1.4 Location of charges q1 and q2

Fj qj

(1.5)

q

r-r 1 2

1

r2

q

r1

0

2

4

1 A Brief Tour of Electromagnetism

Fig. 1.5 The electric field due to charge qi = ρ(r )d V 

dV’

r-r’ E(r)

r’

r

0 that is, the Lorentz force law for electric field only, so that E is in N C −1 (we will see that it is also in V m −1 from energy considerations). The vector field E(r j ) is then defined at the position r = r j of the test charge q j E(r j ) =

qi 1  (ri − r j ) 4π0 i= j | ri − r j |3

(1.6)

This can be expressed in terms of a scalar field, the charge density ρ(r) provided that the collection of point charges can be treated as a smoothly varying function, that is (Fig. 1.5) qi = ρ(r )d V  (1.7)

This description will therefore be valid on length and timescales over which (1.4)– (1.7) hold. When we consider small length and timescales there are two distinct considerations. First, we require that the charge density (1.7), and all other quantities that will be described by scalar and vector fields here, such as electromagnetic fields, energy and momentum densities and so forth, are still describable by continuous functions. Second, we need to use the correct mechanics in the equations of motion for the charges and hence the Lorentz force. Here we will develop electrodynamics in terms of mechanics that is consistent with special (and general) relativity, but is classical. At some length and timescale, quantum mechanics is needed to replace classical mechanics. The field equations may also need to be quantized (to give Quantum Electrodynamics).1 In this sense, the Maxwell equations, the Lorentz force and the mechanics of special relativity that we will discuss are classical. 1 The

need for a field theory that permits both a continuous, classical limit, and discrete quanta (photons in the case of electromagnetic fields) was highlighted by the discovery of the photoelectric effect.

1.2 Maxwell I and II

5

With these caveats, we can write the electric field in terms of a volume integral over the charge density 1 E(r) = 4π0

 V

ρ(r )(r − r )  dV | r − r  |3

(1.8)

hence the electric field retains the experimentally determined properties: it points radially away from an element of positive charge, its magnitude is inverse square with distance, it is proportional to the charge and it obeys the principle of superposition. We will now take these properties and phrase them in a more profound form, in terms of the flux of E.

1.2.1 Flux of a Vector Field Flux is mathematically and conceptually the same for any vector field. We can explore the concept with a simple example; a cold gas2 comprised of particles with number density n(r) per unit volume all of which have the same velocity v(r) at any position r. We then define Flux through surface element dS = number of particles crossing dS per second. Notice that the flux depends upon the angle between dS and v(r) as in Fig. 1.6. To find the flux across dS we just need to identify the volume containing all particles that will cross the surface element per second. This volume is sketched in Fig. 1.7, where we have rotated our point of view such that both v and dS are in the plane of the paper (we can always do this, since the two vectors will define a plane). The projection of dS in this plane is l and perpendicular to the plane is a so that | dS |= la. The angle between v and dS is θ. If the surface element dS is sufficiently small that n and v are constant across it, then in time dt all particles in the cold gas in volume la cos θvdt =| dS | cos θvdt = v · dSdt cross the surface element dS. The number of particles crossing dS in time dt is then nv · dSdt, so that a flux of nv · dS particles crosses dS per second.

2 We

will make use of the “cold gas” model several times in this book, in all cases it is defined as here: a population of identical particles which all have the same velocity v(r, t) at any given position and time.

6

1 A Brief Tour of Electromagnetism

v(r)

v(r) dS

dS

Fig. 1.6 Flux across dS is maximal when v is parallel to dS, and zero when v is perpendicular Fig. 1.7 Volume containing particles that cross dS in time dt

v(r)

v(r)

θ l

dS

vdt l cos θ

Over an arbitrarily large surface S, where across the surface n(r) and v(r) now vary with r, the total flux of particles is obtained by the surface integral:  nv · d S

Flux =

(1.9)

S

The flux of the vector field nv across the surface S is given by Eq. 1.9.

1.2.2 Flux of E Now we can write down the flux of any vector field. For the electric field we will find the flux due to a single positive charge q which is located somewhere inside the closed surface S  (1.10) Flux of E = E · dS S

We will utilize the fact that we can choose any convenient S as long as q is located inside, and will “build in” the three experimentally determined properties of E = F/q from Coulomb’s law.

1.2 Maxwell I and II

7

Fig. 1.8 Surfaces enclosing single charge q. The surface element dS lies on the arbitrary surface S and forms the end of the cone. There is no flux of E across the sides of the cone

dS r

q

S

We start by taking a surface element dS on an arbitrary surface enclosing q. We can then consider the cone shaped surface with sides formed by radius vectors from q and an end formed by dS as shown in Fig. 1.8. From Coulomb’s law: 1. E points radially out from the charge: that is, E points in the direction of r. There is then no flux of E out of the sides of the cone. The flux of E must emerge from the end of the cone dS. We can again sketch the surface element dS choosing the plane of the paper to be the plane defined by the vectors dS and E and the cone is shown cut by this plane in the Fig. 1.9. The flux of E out of the cone is then E · dS =| E | d Sn

(1.11)

where d Sn is the projection of dS parallel to E (i.e. E is normal to the plane in which surface d Sn lies) as shown in Fig. 1.9. The element d Sn has a very useful property: it lies on the surface of a sphere centred on the charge q. The element d Sn is sketched in cylindrical polar coordinates with q at the origin in Fig. 1.10, where the element on the surface of the sphere at radius r is d Sn = r dθ × r sin θdφ We can now exploit 2. E is inverse square and 3. E is proportional to q so that: | E |=

1 q 4π0 r 2

(1.12)

8

1 A Brief Tour of Electromagnetism

Fig. 1.9 Side view of cone enclosing single charge q. The surface element vector dS and field E lie in the plane of the paper

dS α

E(r)

r q dS n Fig. 1.10 Cone enclosing single charge q in cartesian and cylindrical polar coordinates

z rdθ θ

dθ

φ dφ

x

y

r sin θ d φ then the flux of E through the cone E · dS = E × d Sn =

1 q × r dθ × r sin θdφ 4π0 r 2 q = dθ sin θdφ 4π0

which is independent of r (E is inverse square and the surface area of the sphere is proportional to r 2 ). Now the evaluation of the flux of E over the entire closed surface is simply the integral over the solid angle 4π 

q E · dS = 4π 0 S



π 0

 sin θdθ 0

2π

dφ =

q 0

1.2 Maxwell I and II

9

For a collection of charges we just add the electric field from each one, by principle of superposition. So the flux through S from n charges will be 

 E · dS = S

(E1 + E2 + E3 + · · · ) · dS 

S

=

 E2 · dS +

S

= =



E1 · dS + S

E3 · dS + · · · S

q1 q2 q3 + + + ··· 0 0 0 n 

qi =

i=1

Q 0

Thus to obtain the total flux of E (unlike E itself from Coulomb’s law), we don’t need the locations of the charges, just the total charge enclosed. If we now write the total charge in terms of the charge density ρ integrated over the volume V enclosed by S  ρd V

Q=

(1.13)

V

then we have Gauss’ law in integral form  E · dS = S

1 0

 ρd V

(1.14)

V

The differential form follows immediately from the Divergence theorem (1.2)  V

1 ∇ · Ed V = 0

 ρd V

(1.15)

V

which we can write as MAXWELL I :

∇ ·E=

ρ 0

(1.16)

In going from (1.14) to (1.16) we are implicitly treating the fields as classical, and our field theory will not hold on the quantum scale. It is in this sense that the fields E(r), ∇ · E(r) are defined.

10

1 A Brief Tour of Electromagnetism

1.2.3 Flux of B Similarly we can write a Maxwell equation for magnetic flux. We define the magnetic flux through any surface S as  B · dS = Φ B

(1.17)

S

Again, experimentally, it can be shown that the magnetic flux through any closed surface is zero so that  B · dS = 0 (1.18) S

which immediately from Divergence theorem (1.2) gives MAXWELL II ∇ · B = 0

(1.19)

Field lines are drawn passing through a closed surface S in Fig. 1.11. The electric field has nonzero divergence so that closed surfaces can be found that yield zero or nonzero nett flux from (1.14) depending upon whether charge is enclosed in the surface. The magnetic field has zero divergence so that (1.18) always yields zero nett flux as far as we are able to determine experimentally. Lines of E end on electric charge whereas lines of B are continuous. No particle has yet been identified that is a source of magnetic field, but as we shall see in Chap. 4, Sect. 4.2, the Maxwell equations themselves and special relativity do not preclude the existence of such particles, that is, of magnetic monopoles.

B

B

E

S S

Fig. 1.11 The closed surface S has zero nett flux through it

1.2 Maxwell I and II

11

1.2.4 Conservative and Nonconservative Fields Maxwell I was obtained from Coulomb’s law by considering the force on a test charge and then using this to define the electric field as E = F/q N C −1 . Another definition is in terms of the work done on the charge as it is moved around in the electric field. The energy gained by charge q from the field as it is moved along path C is   F · dl = q E · dl (1.20) W = C

C

from the Lorentz force law (this is minus the work done by the field on the charge). The work done per unit charge W/q (in units J C −1 ) is defined as the potential of the electric field (in V ) giving units of E as V m −1 . Coulomb’s law then reveals an interesting property of the electrostatic field via (1.20). Knowing that the field is radial and is any function of r, for the field from a single point charge we have 

 E · dl = C

r2

E(r )dr

(1.21)

r1

Now we can choose C(r, θ, φ) (in spherical polar coordinates) to have any θ, φ dependence, (1.21) will yield the same result which will depend only on the endpoints r1 and r2 . It then follows that (1.21) is zero if r1 = r2 , that is, if the path C is closed (Fig. 1.12). The electrostatic field is conservative. This will also hold for the field from any collection of charges, since by the principle of superposition we can write (1.20) as a sum of integrals due to the electric field from each point charge. For the electrostatic field, taking (1.21) around a closed path and using Stokes’ theorem (1.1) then immediately gives ∇ ∧E=0 Fig. 1.12 E · dl is integrated around the closed curve C in the vicinity of a point charge. The radial sections of the path give contributions of equal size and opposite sign and the sections along contours of the potential (dotted lines) give zero contribution. The integral around the closed curve is zero

(1.22)

E

φ

dl +q

C

12

1 A Brief Tour of Electromagnetism

then since for any scalar field φ ∇ ∧ (∇φ) = 0 we can write the conservative electrostatic field in terms of a potential: E = −∇φ

(1.23)

For magnetic fields, Maxwell II implied that magnetic field lines form closed loops. From Fig. 1.11 one might expect the magnetic field to have curl whereas the electrostatic field does not; this is what is found experimentally and Ampère’s law of magnetostatics is  B · dl = μ0 I

(1.24)

C

where the current I is in Amperes (A ≡ C/s) from which we can now define the units of B as Tesla (T ) and the permeability of free space μ0 = 4π.10−7 T msC −1 (or in terms of the Henry, H m −1 ). Again, if the collection of (moving) charges can be treated as a smoothly varying function, the current that they carry I flowing across surface S can be expressed in terms of a vector field, the current density J  J · dS

I =

(1.25)

S

This, along with (1.1) immediately gives, for magnetostatics ∇ ∧ B = μ0 J

(1.26)

Then B is nonconservative unless there are no currents. The special case of current free systems (or regions) can be treated by defining a magnetostatic scalar potential in analogy to (1.23). Generally we define a magnetic vector potential A B=∇ ∧A

(1.27)

Since for any vector field ∇ · (∇ ∧ A) = 0 this satisfies ∇ · B = 0. The potentials φ and A are an equivalent representation for the electrostatic and magnetostatic fields. In Sect. 1.3 we will complete this representation by explicitly considering the general case where the fields vary with time.

1.3 Maxwell III and IV We will now complete the set of Maxwell equations by explicitly considering systems that change with time.

1.3 Maxwell III and IV

13

1.3.1 Faraday’s Law and Galilean Invariance It is found experimentally that if the magnetic flux passing through a loop of wire changes for any reason, a voltage is induced across the wire. This is Faraday’s law  E · dl = − C

dΦ B d =− dt dt

 B · dS

(1.28)

S

where the magnetic flux is through surface S spanned by the wire loop forming curve C. For this to be true, it has to hold in all frames of reference. To check Faraday’s law, we will consider a wire moving with respect to the magnetic field, so that the magnetic flux changes, and we will look at what happens to the charges in the wire in two frames.3 The two frames are shown in Fig. 1.13. 1. We are in a frame where the magnetic field is time independent, and the wire moves through the field. 2. We transform frames to move with the wire, so that the magnetic field depends on time. In frame (1) the Lorentz force acts on the charges (electrons) in the wire so that Fe = −ev ∧ B (when a steady state is reached the ends of the wire become charged, and a back e.m.f is induced, that is, an electric field that acts opposite to Fe ). When we transform to frame (2) we (the observer) see a stationary wire. The electrons

Fig. 1.13 The observer is at rest in frame 1 (top) and moves with the wire in frame 2 (bottom)

v B ^

B v

B

v 3 Note

that this is a non relativistic, or Galilean, frame transformation. The relativistic treatment is in Chap. 3.

14

1 A Brief Tour of Electromagnetism

Fig. 1.14 The observer is at rest and the wire moves on conducting rails

c

B

b

dl

v d

a

still respond to the Lorentz force however, so we now conclude that Fe = −eE. So for the Lorentz force law to work in both frames, v ∧ B in one frame is just equivalent to E in another. E and B are frame dependent and form a single quantity, the electromagnetic field; a form of the Maxwell Equations must therefore also exist that is frame invariant, and we will derive it later. This electric field that is implied by the frame transformation modifies the conservative, curl free electrostatic field. We will now calculate its curl for the moving wire. To stop the charges “piling up” at the ends of the wire we will complete the circuit by running the wire on conducting rails as sketched in Fig. 1.14; the rails and the rest of the circuit are at rest w.r.t. the observer. The wire and rails then form a closed loop shown in Fig. 1.14. The work done on the electrons around the loop is 

 F · dl =

b



c

F · dl +

a



d

F · dl +

b



a

F · dl +

c

F · dl

(1.29)

d

Now in this frame only the wire is moving, so that all the terms in the path integral (1.29) are zero except between a and b (note that in the presence of an additional electrostatic field, the contribution to the integral around the closed loop would still be zero). This leaves 

 F · dl =

b

 F · dl = −e

a

b

(v ∧ B) · dl

(1.30)

a

in the observer’s rest frame. To evaluate (1.30) consider the small element of wire dl shown in Fig. 1.15. The area element swept out by wire element dl in time dt is dS = vdt ∧ dl. Rearranging the r.h.s. of (1.30) gives 

 F · dl = e a

b

 B · (v ∧ dl) = e a

b

B·

dS d =e dt dt

 a

b

B · dS = e

dΦ B dt

(1.31)

1.3 Maxwell III and IV

15

Fig. 1.15 Wire element dl sweeps out surface element dS in time dt

dS dl v dt

since in the observer rest frame the magnetic field is time independent, so that work is done by the rate of change of magnetic flux. Now we can make the following assertion: the Lorentz force law yields the same force on the electrons in both frames. Hence the work done on the electrons in the moving wire in frame 1 (Eq. 1.31) must be equivalent to that done by an electric field in frame 2 where the wire is at rest  −e

E · dl ≡ e

d dt

 B · dS

(1.32)

S

I.e., Faraday’s law (1.28). The electric field in the l.h.s. of (1.32) and the magnetic field in the r.h.s. are in different frames. To obtain a relationship between E and B we need to work in a single frame: so let us consider a loop of wire a rest w.r.t the observer so that dS is fixed, and B changes with time. In this frame dS is fixed and v = dr/dt = 0 so from chain rule ∂B ∂B dB = + ·v (1.33) dt ∂t ∂r then (1.32) becomes



 E · dl = − S

∂B · dS ∂t

(1.34)

Using Stokes’ theorem (1.1) this is just Maxwell III in differential form: MAXWELL III ∇ ∧ E = − ∂B ∂t

(1.35)

Maxwell III (1.35) gives the nonconservative part of the electric field that arises when the electromagnetic fields are time varying. It also expresses the equivalence of E and B implied by the Galilean frame transformation. Implicit in Maxwell III is the frame transformation (1.36) E2 = E1 + v ∧ B1 where the subscripts refer to frames 1 and 2 and v is the transformation velocity. Using the principle of superposition we have added the (arbitrary) electric field in frame 1, E1 , which to simplify the above discussion we assumed to be zero. The

16

1 A Brief Tour of Electromagnetism

nonrelativistic (1.36) was needed to make the Lorentz force law Galilean frame invariant; in Sect. 4.5 Eq. 1.36 will be generalized for Lorentz frame invariance. Since ∇ ∧ E is no longer zero we cannot describe this field as the gradient of a scalar potential. To retain B = ∇ ∧ A we can use ∂A ∂t

E = −∇φ −

(1.37)

which is consistent with (1.35) and B = ∇ ∧ A.

1.3.2 Ampère’s Law and Conservation of Charge Recall from Sect. 1.2.4 that when the fields and currents are steady we have Ampère’s law (1.26), (1.38) ∇ ∧ B = μ0 J which will now be amended to apply to time dependent situations. To work out what happens to the fields when charge and current densities change with time we will make one important assumption, that nett charge is conserved that is, single charges are neither created or destroyed. Experimentally this has been verified to high precision, but we will assume that it is exactly true and later, that it is true in all frames of reference. Charge and current densities can then be related if we consider an arbitrary finite volume containing some total charge that is varying with time Q(t), as charges flow out of the volume as in Fig. 1.16. There is a total current flowing out of volume V I =−

∂Q =− ∂t

 V

∂ρ dV ∂t

(1.39)

Fig. 1.16 Charge flows out of volume V

q q

dS V q

q

Q(t) q q

S

1.3 Maxwell III and IV

17

and as I flows across the surface S which encloses V we can also write from (1.25): 



I =

J · dS = − S

V

∂ρ dV ∂t

(1.40)

This can be written in differential form using (1.2) ∇ ·J=−

∂ρ ∂t

(1.41)

Equation 1.41 immediately shows the problem with (1.26) in time dependent situations; the divergence of (1.26) gives ∇ · J = 0, that is, currents must close in steady state. When we obtained (1.41) we allowed nett current to flow out of a closed surface, so that Fig. 1.16 shows a divergence of J. The correction needed for (1.26) can be found using (1.16) to rewrite (1.41) as   ∂ρ ∂E = ∇ · J + 0 =0 ∇ ·J+ ∂t ∂t equating this with ∇ · (∇ ∧ B) = 0 then gives MAXWELL IV : ∇ ∧ B = μ0 J + μ0 0

∂E ∂t

(1.42)

which is Ampère corrected for time dependent fields. We have added a displacement current to the r.h.s. of (1.26). How does this work in practice? A simple example of the fields around an ideal capacitor that is discharging are sketched in Fig. 1.17, where ∇ ∧ B is given by the conduction current flowing in the circuit outside of the capacitor plates, and by the displacement current due to the time dependent electric field between the plates.

Fig. 1.17 An ideal discharging capacitor

I(t)

B(t)

+Q(t)

-Q(t) -

E(t) + + + +

-

I(t)

+ + + + B(t) B(t)

18

1 A Brief Tour of Electromagnetism

1.4 Electromagnetic Waves It is now straightforward to show that Maxwell’s equations support free space waves. We use the vector relation ∇ ∧ (∇ ∧ A) = ∇(∇ · A) − ∇ 2 A

(1.43)

and take the curl of Maxwell IV (1.42) to give ∇(∇ · B) − ∇ 2 B = μ0 ∇ ∧ J + μ0 0

∂ ∇ ∧E ∂t

(1.44)

which, with Maxwell II (1.19) and III (1.35) and in free space where there are no currents J = 0 gives 1 ∂2B ∇2B = 2 2 (1.45) c ∂t where μ0 0 = c12 . If instead we take the curl of Maxwell III (1.35) a similar procedure gives (with no charges in free space ρ = 0) ∇2E =

1 ∂2E c2 ∂t 2

(1.46)

Equations 1.45 and 1.46 are wave equations for E and B, and are linear. This means that any wave with frequency ω and wavenumber k of the form B, E ∼ f (ωt − k · r)

(1.47)

is a solution to (1.45) and (1.46). These waves will propagate at phase speed ω/k = c. Crucially, we identify these with light waves in free space. Since (1.45) and (1.46) are linear, we can superpose any solutions of the form (1.47) into a wave group or packet, and this will be nondispersive. The wavegroup will have speed c and will carry energy and momentum through free space.4 In the next chapter we will discuss momentum flux in the cold gas (defined in Sect. 1.2.1), in terms of distributed bulk properties (such as the momentum flux density tensor) rather than the motions of individual particles. Electromagnetic fields can be quantized, that is, treated as a collection of photons which carry energy and momentum. We would then expect the free space electromagnetic fields to have equivalent energy and momentum flux (the latter given by the Maxwell stress tensor).

4 The

properties of these waves are explored in the revision problems.

1.5 Conservation Equations

19

1.5 Conservation Equations Equation 1.41 is a conservation equation and perhaps not surprisingly, all conservation equations are of this form; they embody the premise that the particles and the quantity that they carry (in this case, charge) is neither created nor destroyed. If we recall the cold gas from Sect. 1.2.1 of number density n(r, t) with each particle moving with the same velocity v(r, t) carrying charge q, then ρ = nq and J = nqv and (1.41) will become (cancelling q from both sides) ∇ · (nv) = −

∂n ∂t

(1.48)

Since (1.48) is linear we can use as many different populations of particles as necessary, each with a different q and/or v(r), to represent a gas with finite temperature composed of several particle species of different charge. The nett result will still be an equation of the form of (1.48). If all the equations describing our system (i.e., the Maxwell equations) are also linear then any quantity, such as energy, mass, momentum, that can be envisaged as being carried by particles will have a conservation equation of the form (1.48). At this point it could be argued that the electromagnetic fields are known to be particulate, i.e., composed of photons, then conservation equations can be found to include field energy and momentum as well as that of the charges. However, to obtain (1.48) from its integral form we represented the ensemble of particles in the gas at position r and time t with the number of particles in elemental volume n(r, t)dV and the flux of particles across an elemental surface n(r, t)v(r, t).dS. The assumption of smoothness (1.7) has been made, that is, we are on spatiotemporal scales over which fields and charge densities behave as smoothly varying functions. So in our field theory here, it is the charges that have been “smoothed out”, rather than the fields treated as photons.

Chapter 2

Field Energy and Momentum

So far we have discussed the free space macroscopic electromagnetic fields and the field equations: the Maxwell equations that describe how the fields evolve in space and time. Charges are included in this description as macroscopic charge density and current density. However point charges carry energy and momentum, the Maxwell equations have wave solutions and we might expect this to imply that the fields carry energy and momentum also. Indeed, the energy and the radiation pressure of waves, and of photons has been measured experimentally. In this chapter we will formalize the concept of the energy and momentum of the electromagnetic fields. This is most easily achieved through conservation equations, and we will use the cold gas model from Chap. 1 in this context. In a gas, individual particles carry momentum which is a vector quantity, the gas as a whole, when described by macroscopic or fluid variables has a corresponding tensor pressure. We will first use the simpler cold gas model to introduce Cartesian tensors, and their role in equations for conservation of momentum flux. We will then find the Maxwell Stress Tensor that is, the “ram pressure tensor” for the electromagnetic field. The cold gas model will allow us to treat a system with free charges and electromagnetic fields (i.e., E and B). This is readily generalized to linear media.1

1 The

Maxwell equations and conservation equations for free space and charges discussed here are linear. Our approach then generalizes to media in which these equations remain linear. This is the case if the fields induced in the medium are linearly proportional to those in the surrounding free space. See the revision problems for examples. © Springer-Verlag GmbH Germany, part of Springer Nature 2021 S. Chapman, Core Electrodynamics, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-030-66818-1_2

21

22

2 Field Energy and Momentum

Fig. 2.1 Some of the forces acting on fluid element d V that form the components of a pressure tensor

z

P yz

P xz

Pyy

P xx x

P xy

P yx

y

2.1 Tensors and Conservation Equations 2.1.1 Momentum Flux Density Tensor Tensors arise in macroscopic or bulk descriptions such as gases, fluids, and solids where we have vector fields describing distributed bulk properties rather than properties at a vanishingly small point (such as at a particle). A cube shaped elemental volume d V in a gas is sketched in Fig. 2.1 and the possible forces that can act on two of the three faces of the cube are labelled. If we just consider the x faces, we can compress the cube by exerting forces normal to the surface, in the ±ˆx direction (Px x ), we can also twist the cube by exerting shear forces tangential to the surface, in the ±ˆy direction (Px y ), and in the ±ˆz direction (Px z ). The same is true of the y and z faces, so that in our three dimensional gas we have nine numbers describing the forces on the gas. These nine elements constitute the pressure tensor for the gas and we can write them as a matrix: ⎡ ⎤ Px x Px y Px z Pi j = ⎣ Pyx Pyy Pyz ⎦ (2.1) Pzx Pzy Pzz If there were no shear forces the tensor has three independent elements and becomes: ⎤ Px x 0 0 Pi j = ⎣ 0 Pyy 0 ⎦ 0 0 Pzz ⎡

(2.2)

and if these normal forces on all sides of the cube are equal, the pressure is isotropic: ⎡

⎤ P0 0 Pi j = ⎣ 0 P 0 ⎦ 0 0 P

(2.3)

2.1 Tensors and Conservation Equations

23

We will now obtain the momentum flux density tensor for the cold gas in which all particles move with the same velocity v(r) at a given position r. In a given direction rˆ the momentum of one particle is pr = mvr = mv · rˆ

(2.4)

and so momentum flowing across surface element dS in time dt due to this component is, for mass density ρ: ρ(v · rˆ )(v · dSdt) = ρvr vn dsdt

(2.5)

where the term in the first bracket of the l.h.s. of (2.5) is the momentum density due to the rˆ component, and the term in the second bracket is the volume containing the particles that flow across dS in time dt. The subscripts r and n denote components along rˆ , and along dS, nˆ respectively. The momentum flow per unit area per unit time from Eq. 2.5 is the scalar Pr n = ρvr vn

(2.6)

We could write the momentum flux flowing in nˆ due to all components of v by summing (2.6) over r = 1, 3 (that is, r = 1, 2, 32 the x, y, z directions) as the vector Pn = ρ(vx xˆ + v y yˆ + vz zˆ )vn = ρvvn

(2.7)

If we then sum (2.7) over n = 1, 3 (again, the x, y, z directions), we have the momentum flux flowing in all directions due to all components as the tensor: P = ρ(vx xˆ + v y yˆ + vz zˆ )(vx xˆ + v y yˆ + vz zˆ ) = ρvv

(2.8)

Tensors written as vector outer products as in Eq. 2.8 are known as dyadics. The dyadic vv can be written out in full ⎤⎛ ⎞ ⎡ xˆ v x v x v x v y v x vz vx vx xˆ xˆ +vx v y xˆ yˆ +vx vz xˆ zˆ vv = +v y vx yˆ xˆ +v y v y yˆ yˆ +v y vz yˆ zˆ = (ˆxyˆ zˆ ) ⎣ v y vx v y v y v y vz ⎦ ⎝ yˆ ⎠ zˆ +vz vx zˆ xˆ +vz v y zˆ yˆ +vz vz zˆ zˆ vz v x vz v y vz vz

(2.9)

For any orthogonal coordinate system we can use the following shorthand ⎡

⎤ v x v x v x v y v x vz vv = ⎣ v y vx v y v y v y vz ⎦ vz v x vz v y vz vz

2 Throughout

inclusive.

(2.10)

this book we will use the index notation i, j to mean all values between i and j

24

2 Field Energy and Momentum

For most purposes we don’t really need to write out all the elements of the matrix (2.10). Instead, we can simply write vi v j where it is assumed in this three dimensional, Cartesian world i = 1, 3 and j = 1, 3. The momentum flux density tensor is then written as (2.11) Pi j = ρvi v j The notation in (2.11) is useful, for example we can immediately deduce from (2.11) that vv and hence P is symmetric as vi v j = v j vi (there are only 6 independent components for the cold gas) without writing out all nine elements of the matrix (2.10). We can now calculate the force on the cube due to the momentum flux of the cold gas. The gas particles can deliver momentum to the cube either by flowing out of the cube (the rocket effect), or by slowing down in the cube. The force on the cube of arbitrary volume V due to flow out of particles is minus the rate at which momentum flows out. The force due to particles slowing down in the cube is again minus the momentum inflow rate as dS points outwards on the surface. If we consider one component of v in the rˆ direction, the rate at which momentum is delivered through d S is given by d Fr = −ρ(v · rˆ )(v · dS)

(2.12)

where the first bracket is the momentum density of the rˆ component. So we have from all three components dF = −ρvv · dS = −P · dS

(2.13)

Thus the force on the whole volume is







FV = dF = − P · dS = − ρvv · dS = − ∇ · Pd V = − ∇ · ρvvd V S

S

V

V

(2.14)

2.1.2 Momentum Flux, Gas Pressure and Fluid Equations The above then implies that the momentum flux density corresponds to a local force per unit volume

FV 1 dF = −∇ · P (2.15) = V V V We can now obtain a conservation equation for momentum in the gas. To do this we need to relate the l.h.s. of (2.15) to the rate of change of momentum in the gas. For this we will make an important restriction: we will consider the force over a small volume d V within which the total number of particles is constant. This means

2.1 Tensors and Conservation Equations

25

that we are considering the rate of change of momentum due to a local ensemble of particles slowing down, or speeding up, and we will choose d V accordingly; if this d V contains a constant mass M of particles then it experiences a force FV =

dv d Mv =M dt dt

(2.16)

Then the force per unit volume FV dv =ρ dV dt

(2.17)

One possibility is to choose d V to contain the same particles by moving with the fluid at velocity v, then (2.17) becomes (using chain rule) ρ

 ∂v dv =ρ + (v.∇)v dt ∂t

(2.18)

This is equivalent to Liouville’s theorem.3 The momentum flux density tensor that we have derived is not the same as the pressure tensor Pth . The pressure of a warm gas is the momentum flux due to the thermal or random motions of the gas particles in the rest frame of the gas. In the warm gas, a particle will have total velocity u = v + c where c is the random velocity relative to the average or bulk velocity v(r, t) with which the gas as a whole moves, so that the average < u >= v and < c >= 0. In our cold gas, the velocity due to random motions is zero so that for each particle u = v. From Eq. 2.8 above, we can find the nett momentum flux density in the warm gas due to an average over an ensemble of particles in small volume element d V at position r at time t: P = ρ < uu >= ρvv + ρ < cc > (2.19) In the rest frame of the gas v = 0 and the warm gas has thermal pressure Pth = ρ < cc >. Generally then, (2.20) P = ρvv + Pth The momentum flux density due to gas-fluid motion as a whole, ρvv is the ram pressure of the gas, and is just the momentum flux density of the cold gas.

3 Liouville’s theorem expresses conservation of probability density along a trajectory in phase space.

In a system with no sources or sinks of particles, we can follow the phase space trajectory r(t), v(t) of any particle and along that trajectory the probability of finding the particle in elemental phase space volume drdv is constant. See advanced problem 2.

26

2 Field Energy and Momentum

Fig. 2.2 Vector r in the x, y, z and x  , y  , z  coordinate systems

z

z’

r y’

y x x’ If we move with the gas (so that in our frame v = 0), Eq. 2.18 can be written as a fluid equation for the gas:  ∂v dv =ρ + (v · ∇)v = −∇ · Pth ρ dt ∂t

(2.21)

and in this frame there is no ram pressure. If we are at rest w.r.t. the gas bulk flow then we experience both ram and thermal pressure.

2.1.3 Cartesian Tensors, Some Definitions We can write P or Pi j for a tensor of the second rank where in the discussion so far i = 1, 3 and j = 1, 3 i.e., there are nine possible indices (or nine components). A tensor of the first rank (vector) would be written a or ai where i = 1, 3, just three possibilities or three components. A scalar is thus a tensor of rank zero and has one component. We can extend this to as many indices as we wish, for example Pi jkl is a tensor of rank four. A rank two dyadic would be ab which can also be written ai b j . Note that all the indices are “down” (written as subscripts) in this Cartesian system so that a dyadic is always of the form ai b j ck ..., later, when we introduce generalized coordinates as opposed to Cartesian coordinates we will come across a j a j which has a different meaning—we will cover this in Chap. 3, Sect. 3.4.

2.1 Tensors and Conservation Equations

27

A key property of the tensor formalism is that it embodies coordinate transformations. Figure 2.2 shows a vector r w.r.t two orthogonal coordinate systems: x, y, z and x  , y  , z  . The primed axes are simply the unprimed axes rotated in space (x, y, z → x  , y  , z  ). A vector in the unprimed frame: ⎡ ⎤ x r = ⎣y⎦ z will have coordinates in the primed frame 

x = a11 x +a12 y +a13 z  y = a21 x +a22 y +a23 z  z = a31 x +a32 y +a33 z The tensor ai j contains the direction cosines of the rotation x, y, z → x  , y  , z  . This can be written as (2.22) r = A · r where

⎡ ⎤  ⎤ x a11 a12 a13   r = ⎣ y ⎦ and A = ⎣ a21 a22 a23 ⎦  a31 a32 a33 z ⎡

Instead of the column vector notation we can use indices; since x = r1 , y = r2 and z = r3 , so Eq. 2.22 can be represented by 

r j = a ji ri

(2.23)

which is shorthand (the Einstein summation convention) for 

rj =

a ji ri

(2.24)

i=1,3

The sum over index i on the r.h.s. of (2.24) contracts the number of indices by one (the l.h.s. only has single index j). Equations 2.23 and 2.24 are an example of a tensor dot (or inner) product. Again, when we switch to index notation, we drop all reference to the basis vectors (or axes) x, y, z. Hence Eq. 2.25 is also a tensor dot product: a · T = a1 T11 + a2 T21 + a3 T31 + a1 T12 + a2 T22 + ... (2.25) = xˆ j ai Ti j which is ai Ti j if we again drop all reference to the basis vectors xˆj .

28

2 Field Energy and Momentum

So we now can write down ∇ · T in Cartesian coordinates which is xˆ j

∂ Ti j = ∇ · T ∂ xi

(2.26)

There are other special tensors. One is the trace which is δi j = 1 when i = j and 0 when i = j: ⎡ ⎤ 100 δi j = ⎣ 0 1 0 ⎦ 001 Contracting δi j over i and j with two vectors extracts the dot product δi j ai b j = a1 b1 + a2 b2 + a3 b3 = a · b = ai bi Another special tensor is the alternating tensor: i jk = 1 when i jk = 123, 231, 312 and −1 when i jk = 321, 132, 213 but 0 when any two of the indices are alike. Contracting i jk over j and k extracts the following vector from a dyadic: i jk a j bk =

i11 a1 b1 +i12 a1 b2 +i13 a1 b3 +i21 a2 b1 +i22 a2 b2 +i23 a2 b3 +i31 a3 b1 +i32 a3 b2 +i33 a3 b3

If we consider the i = 1 terms we find that all but 123 a2 b3 and 132 a3 b2 are zero. These last two give a2 b3 − a3 b2 which we recognise as the x component of the vector cross product. Similarly, i = 2, 3 gives the y, z components respectively, so that contracting with i jk extracts the cross product. In this way, vector operations, vector calculus, and coordinate transformations can all be performed using index notation. This Cartesian formalism works fine for the three space dimensions, for six phase space dimensions, etc. Later, when we consider special relativity we will work in space-time coordinates, and will need to generalize this Cartesian formalism.

2.2 Field Momentum and Maxwell Stress We can now derive equations of conservation of electromagnetic energy and momentum. Again, we will consider a volume element d V that contains a fixed number of charges, and field energy and momentum that can flow in and out of this volume element. In this sense, the charges act as a source or sink of electromagnetic energy and momentum. At a microscopic level, photons move in and out of d V and carry energy and momentum. Here in our field theory, we can only discuss the “smoothed” behaviour in element d V , and all quantities are obtained from fluid variables so that the number of charges is n(r, t)d V , the energy in the electromagnetic fields

2.2 Field Momentum and Maxwell Stress

29

is U (r, t)d V and in the particles is ε(r, t)d V where n(r, t) is number density and U (r, t) and ε(r, t) are the energy densities carried by the fields and particles respectively. Strictly speaking, we would derive the conservation equations in integral form, by considering integrals over some volume V , in a similar manner to the derivation of Maxwell’s equations in Chap. 1. Here we will take the short cut and work with the differential forms directly, by assuming that all bulk quantities are well defined per unit area or per unit volume. For field energy density for example this means that the energy density 1 V

U (r, t)d V = U (r, t) V

is well defined for arbitrary (small) V .

2.2.1 Energy Conservation: Poynting’s Theorem Energy is lost from the electromagnetic fields if work is done on the particles. From the Lorentz force law, a single positive charge moving in the fields gains energy d dv = v.m dt dt



1 2 mv 2

 = qv · E

(2.27)

then n charges per unit volume gain energy density ε at a rate dε =J·E dt This energy density gain by the charges has to be balanced by an energy density decrease in the fields and an energy flux into the (unit) volume containing the particles. Using (1.42) to substitute for J J·E=E·

∇ ∧B ∂ E2 − 0 μ0 ∂t 2

(2.28)

Now from the form of the conservation equation in Chap. 1 (Eq. 1.48) we expect the r.h.s. to be the divergence of energy flux (a vector) minus rate of change of energy density. Using the vector identity ∇ · (E ∧ B) = B · (∇ ∧ E) − E · (∇ ∧ B) Equation 2.28 can, using (1.35) be rearranged as

30

2 Field Energy and Momentum

 J · E = −∇ ·

E∧B μ0



 ∂ 1 B2 E2 − 0 + ∂t 2 μ0 2

(2.29)

Equation 2.29 is Poynting’s theorem for the case of an ensemble of free charges in an electromagnetic field, where the field energy density U = 0

E2 1 B2 + 2 μ0 2

in J m −3 and the Poynting flux S=

E∧B μ0

(2.30)

(2.31)

which has units of energy per unit area per unit time (J m −2 s −1 ). For some arbitrary volume V , enclosed by surface S the divergence theorem (1.2) gives 

∇ · Sd V = V

S · dS

(2.32)

S

so that the surface integral of the Poynting flux S gives the energy carried out of the volume by the electromagnetic fields per unit time (dS always points outwards); energy gain by the particles is balanced by minus the Poynting flux. Rearranging Poynting’s theorem dε ∂U − = + ∇ · S = −J · E (2.33) dt ∂t Poynting’s theorem (2.33) expresses conservation of energy between the ensemble of charges and the electromagnetic fields.4

2.2.2 Momentum Conservation: Maxwell Stress In order for the fields to do work on the particles a force must be exerted on the particles, again given by the Lorentz force law. Again we assume that the number of charges in volume element d V is fixed, and that we can always integrate over V to find the value of any quantity per unit volume. The Lorentz force law then implies a rate of change of momentum to the particles from the fields. For one particle with momentum p dp = q(E + v ∧ B) dt 4 In

linear media, that is, where the response from the bound charges in the medium is linearly proportional to the applied field, B = μr μ0 H, D = r 0 E and S = E ∧ H and U = 21 [E · D + B · H] then conservation of energy is still given by (2.33). In nonlinear media (2.33) is no longer valid. See revision problem 8.

2.2 Field Momentum and Maxwell Stress

31

then for a collection of charges with momentum density P p = np p dP p = ρE + J ∧ B dt

(2.34)

This expresses the rate of change of momentum density of the charges. From an understanding of electromagnetism based on experiment we might also expect a conservation equation to contain a term describing the rate of change of momentum density of the fields; either because waves carry momentum and Maxwell’s equations support wave solutions, or because of the particle- like behaviour of the electromagnetic wavefields. On that basis let’s make a guess as to what this might be, from the following argument. If we consider the cold gas, all particles will have rest energy ε = mc2 and momentum p = mv. Then the energy flux S p is just S p = nvε = nvmc2

(2.35)

(the flux across surface element dS will be S p · dS see Sect. 1.2.1). The momentum density is just (2.36) P p = nmv so that S p = P p c2

(2.37)

This is true for any particles, including photons. In the case of photons, the energy flux is just the Poynting flux (2.31) that we obtained from conservation of energy. Our “guess” for the momentum density of the fields is then Pf =

S = 0 E ∧ B c2

(2.38)

To obtain the conservation equation we then just rearrange (2.34) to look like a conservation equation for momentum −

dP p ∂P f = −∇ ·T dt ∂t

(2.39)

This is in (almost) the same form as (2.33) except that the terms are now vectors; the rate of change of momentum density to the charges and fields are balanced by the divergence of a momentum flux density tensor T . To find T , Maxwell I and III (1.16), (1.35) are used to substitute for ρ and J in (2.34) to give ρE + J ∧ B = 0 E(∇ · E) −

B ∧ (∇ ∧ B) ∂E + 0 B ∧ μ0 ∂t

(2.40)

32

2 Field Energy and Momentum

then since by expanding the l.h.s. and using Maxwell III (1.35) ∂ ∂t



S c2



 ∂E = −0 B ∧ + E ∧ (∇ ∧ E) ∂t

(2.41)

Equation 2.40 is: − (ρE + J ∧ B) =

∂ S B ∧ (∇ ∧ B) − 0 E(∇ · E) + 0 E ∧ (∇ ∧ E) + (2.42) 2 ∂t c μ0

Now the remaining terms on the r.h.s. of (2.42) must give the divergence of a tensor. Looking at the terms in E only, we can write E(∇ · E) − E ∧ (∇ ∧ E) = E(∇ · E) + (E · ∇)E − ∇

E2 2

(2.43)

by using vector identity (E.11). In index notation the r.h.s. of (2.43) is Ej

∂ ∂ ∂ ∂ E2 ∂ E2 = Ei + Ei E j − δi j E i E j − δi j ∂ xi ∂ xi ∂ xi 2 ∂ xi ∂ xi 2

(2.44)

which is just the divergence of a tensor Ti Ej = E i E j − δi j

E2 2

(2.45)

We can use the same manipulation for the B terms in (2.42), by simply adding the (zero) term (from Maxwell II (1.19)) −

B(∇ · B) μ0

(2.46)

The Maxwell stress tensor in the conservation of momentum (2.39) is then   1 1 1 2 2 Ti j = 0 E i E j + 0 E + Bi B j − δi j B μ0 2 μ0

(2.47)

Notice that we wrote (2.39) with a term −∇ · T where Poynting’s theorem has a term +∇ · S. This is a matter of convention: a Poynting flux into a volume (∇ · S negative) will correspond to a radiation pressure acting upon it (∇ · T positive).

2.3 Radiation Pressure

33

2.3 Radiation Pressure To illustrate how Maxwell stress works, lets look at an example: a monochromatic plane electromagnetic wave propagating in free space. This is a solution to the free space wave equations introduced in Sect. 1.4. We can show that the operators ∇ ≡ −ik ∂ ≡ iω ∂t

(2.48)

apply in the case of the plane wave solution E = E0 ei(ωt−k·r) B = B0 ei(ωt−k·r)

(2.49)

The operators (2.48) can now be used to save some algebra. The relationship between E, B and k follows from the Maxwell equations directly. Maxwell III (1.35) for the plane wave becomes − ik ∧ E = −iωB (2.50) which rearranges to B=

1ˆ k∧E c

(2.51)

Similarly, Maxwell I and II (1.16), (1.19) give k·E=0

(2.52)

k·B=0

(2.53)

Maxwells equations then reveal that the free space plane wave E, B, k form an orthogonal set of vectors. Again for simplicity, lets choose the direction of k to be along the zˆ axis. The wave propagates in the z direction i.e., k = k zˆ and E = (E x , E y , 0) B = (Bx , B y , 0)

(2.54)

We will now write down the Maxwell stress tensor for free space plane waves. Generally     1 E2 B2 Bi B j − δi j + (2.55) Ti j = 0 E i E j − δi j 2 μ0 2 and for the zˆ propagating plane wave (2.54) any term with a z index including Tx z , Tzx , Tzy is zero. This gives

34

2 Field Energy and Momentum

⎡

0 E x2 +

Bx2 μ0

− PE M

⎢ Ti j = ⎣ 0 E y E x + 0

B y Bx μ0

0 E x E y + 0 E y2 +

B y2 μ0

Bx B y μ0

− PE M

0

where PE M = +0

0

⎤

⎥ 0 ⎦ −PE M

E2 B2 + 2 2μ0

(2.56)

(2.57)

√ √ from (2.51) and with c = √μ10 0 = ωk we have Bx = −E y μ0 0 and B y = E x μ0 0 so that B y Bx = −μ0 0 E y E x , and the terms Tx y and Tyz are zero. These relationships also simplify (2.57) to PE M =

B2 μ0

(2.58)

As a result the Maxwell stress tensor for the free space plane wave becomes diagonal with terms B2 Tx x = 0 E x2 + μx0 − PE M B2 (2.59) Tyy = 0 E 2 + y − PE M y

Tzz = −PE M

μ0

We can now calculate the force due to the radiation pressure, that is, the rate of change of momentum delivered by the free space wave per unit area, from the momentum flux conservation Eq. 2.39 this is just ∇ · T = xˆ j

∂ Ti j ∂ xi

(2.60)

For the zˆ propagating wave, the fields are just functions of z and t (E, B(z, t) only) so that the only contributions to (2.60) will be from xi = z and i = j. All other (i.e., the off axis) terms are zero. We are left with ∂ ∂ ∂ ∇ · T = zˆ Tzz = zˆ [−PE M ] = −ˆz ∂z ∂z ∂z



B2 μ0

 (2.61)

For the plane wave solution, using (2.48) the radiation pressure is just ∇ ·T =

1 2ikB 2 μ0

(2.62)

Lets compare this with the rate of change of momentum flux in the fields ∂P f ∂ E∧B = ∂t ∂t μ0 c2

(2.63)

2.3 Radiation Pressure

35

Maxwell IV in free space for the plane wave solution gives, using (2.48) − k ∧ B = μ0 0 ωE hence Pf =

(−kˆ ∧ B) ∧ B B2 ˆ k = μ0 c μ0 c

(2.64)

(2.65)

where we have used (2.53). We then have ∂P f B2 ˆ B2 = 2iω k = 2ik ∂t μ0 c μ0

(2.66)

so that in the absence of free charges, the divergence in radiation pressure balances the rate of increase in momentum density in the fields.

Chapter 3

A Frame Invariant Electromagnetism

So far we have the field equations, the Maxwell equations, and using the Lorentz force law have constructed equations describing the conservation of energy and momentum between the fields and charges. Our next task is to cast these equations in a form that is consistent with special relativity. Special relativity has two fundamental predictions: • Frame Invariance, that is, the laws of physics are the same in all inertial frames of reference. • The speed of light is the same in all inertial frames. The second of these yields the Lorentz transformation between one frame moving at constant velocity w.r.t. another. Our new formalism for electromagnetism will tie up a number of loose ends that have already appeared so far. In Sect. 1.3.1 Galilean invariance was imposed on Faraday’s Law to give Maxwell III (1.35) and this showed that E and B are equivalent: B in one frame can look like E in another (1.36). We will in Chap. 4 obtain a single object, the Electromagnetic Field Tensor that describes the fields E and B and which gives a Lorentz transformation of the fields valid for relativistic frame transformations, using the formalism developed in this chapter. In Sect. 1.4 we found that the Maxwell equations predict light waves in free space. Since the speed of light is constant under Lorentz transformation we must be able to find a form for the wave equations, and the Maxwell equations, that is invariant under Lorentz transformation. So far we have used the Lorentz force law to give the force on individual charges but have not concerned ourselves with the equations specifying the particle motion under that force; we will now incorporate relativistic rather than Newtonian mechanics. Finally, we have found conservation equations for charge, mass, energy and momentum, between collections of charges and the electromagnetic fields; a Lorentz invariant form must also exist for these. Some individual quantities (such as energy) will be different between one frame and another, but other properties, © Springer-Verlag GmbH Germany, part of Springer Nature 2021 S. Chapman, Core Electrodynamics, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-030-66818-1_3

37

38

3 A Frame Invariant Electromagnetism

such as charge and particle number, will be (inertial) frame invariant and this leads us to form conservation equations that are invariant under Lorentz transformation. To construct an electromagnetic field theory that is invariant under the Lorentz frame transformation will require a generalization of Cartesian tensors that were introduced in Chap. 2. We have seen that coordinate transformations in x, y, z space can be written as tensor operators (or 3 × 3 rotation matrices). The Lorentz transformation can be written as a tensor operator in space-time (a 4 × 4 rotation matrix), and the key to unifying electromagnetism with special relativity is to construct four vectors and four tensors in which the three orthogonal space coordinates of the Cartesian system are replaced with four coordinates (three space and one time). The generalization to space-time led Einstein to the tensor formulation of general relativity where space-time is curved by the gravitational field. Here for special relativity we neglect gravity (and therefore do not treat accelerating frames), so we consider inertial frame transformation only and space-time is flat.

3.1 The Lorentz Transformation Special relativity arises if we insist that the speed of light is the same in all inertial frames and that physical laws look the same in all inertial frames. The consequences of this are immediately clear if we consider the following experiment: a “light clock”1 is composed of a light beam bouncing between two lossless mirrors. The clock is moving past us at speed u. Lets first look at what happens when the clock is oriented perpendicular to the direction of motion as shown in Fig. 3.1. In our rest frame the time taken for the light to travel from one mirror to the other is Δt  , the clock moves past and the light beam travels a distance (cΔt  )2 = (uΔt  )2 + Δy 2

(3.1)

In the clock’s rest frame the time taken for the light to travel from one mirror to the other is Δt, and the light path is simply cΔt = Δy

(3.2)

Now let’s compare the time in the two frames. If the distance perpendicular to the direction of motion Δy is the same in both frames then Δt 2 (c2 − u 2 ) = c2 Δt 2

1 This

(3.3)

experiment was actually performed by Michelson and Morley in 1887 to attempt to measure our relative velocity w.r.t. the aether which was believed to fill the vacuum to allow the propagation of electromagnetic waves.

3.1 The Lorentz Transformation Fig. 3.1 The light clock is oriented perpendicular to its direction of motion. Top: observer rest frame i.e., the clock moves past us; bottom: clock rest frame i.e., we move with the clock

39

c Δ t’ Δy u Δ t’ u

Δy u Fig. 3.2 The light clock is oriented parallel to its direction of motion. Top: observer rest frame i.e., the clock moves past us; bottom: clock rest frame i.e., we move with the clock

Δ x’ uΔt’f uΔt’b

Δ x’ u

Δx u

so that there is time dilation

where

Δt  = γΔt 1 γ = √ 1−

u2 c2

(3.4)



(3.5)

in other words “moving clocks go slow”. Now let’s turn the clock through 90◦ so that it is oriented along the direction of motion as in Fig. 3.2. In our rest frame the clock moves away from the forward going

40

3 A Frame Invariant Electromagnetism

light pulse. When the light moves in the direction of the clock motion it travels a longer distance (3.6) cΔt f = Δx  + uΔt f The clock moves towards the backwards going light pulse so that when the light moves in the opposite direction to the clock motion it travels a shorter distance cΔtb = Δx  − uΔtb

(3.7)

the total round trip time of the light pulse in the moving clock is then Δt  = Δt f + Δtb =

2cΔx  = γΔt c2 − u 2

(3.8)

from (3.4). In the clock’s rest frame the round trip time is simply given by cΔt = 2Δx

(3.9)

Eliminating Δt from (3.8) and (3.9) gives the Lorentz contraction of the length of the clock Δx (3.10) Δx  = γ in the moving frame. In terms of distances x, x  and times t, t  measured from origins in the two (observer) frames, the Lorentz transformation is:   t  = γ t − vx c2 x  = γ(x − vt) y = y z = z

(3.11)

where the primed frame moves at velocity +v xˆ w.r.t. the unprimed frame (so that v → −v in (3.11) gives the inverse transform). Now Lorentz contraction and time dilation arose because the speed of light is the same (c) in all frames of reference. So if we consider a point light source located at the origin of our x, y, z coordinate system, generating a spherical wavefront at t = 0, the wavefront must propagate a distance (squared): x 2 + y 2 + z 2 = c2 t 2

(3.12)

at time t. If we use the Lorentz transformation (3.11) on this expression (3.12), we obtain: (3.13) x 2 + y 2 + z 2 = c2 t 2

3.1 The Lorentz Transformation

41

Whatever frame we are in, the distance propagated by the wavefront always obeys an expression of the form of (3.12). This looks almost like the length squared of a vector in x, y, z, ct space. We can formalize this idea by writing (3.12) as s 2 = c2 t 2 − x 2 − y 2 − z 2

(3.14)

where for light waves s = 0 and for objects moving slower than light, s 2 > 0, so that s is real.2 The length of s is the same in all frames, Lorentz transforming from one moving frame to another simply corresponds to a rotation of the axes x, y, z, ct. We have already developed notation for coordinate rotations in Cartesian x, y, z space. Lets see what happens if we stick to Cartesian rules. If our four vector is s j and ⎡

⎤ ct ⎢x⎥ ⎥ s = (ct, x) = ⎢ ⎣y⎦ z

(3.15)

then the Lorentz transformation will be of the form s j = Λ jk sk

(3.16)

where (3.16) is the shorthand for the tensor operation s = Λ · s, and the spacetime coordinate rotation matrix is ⎡ ⎤ γ − vc γ 0 0 ⎢−vγ γ 0 0 ⎥ c ⎥ Λ jk = ⎢ (3.17) ⎣ 0 0 1 0⎦ 0 0 01 then if we work out the coordinate rotation we have ⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎡ ⎤ γ − vc γ 0 0 ) cγ(t − xv ct ct c2 v  ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ γ γ 0 0 − x γ(x − vt) c ⎥⎢ ⎥ = ⎢ ⎥ = ⎢ x ⎥ s = Λ · s = ⎢ ⎣ 0 ⎦ ⎣y ⎦ y 0 1 0⎦⎣ y ⎦ ⎣ z z z 0 0 01

(3.18)

so this Cartesian operation (3.18) just gives the Lorentz transformation of (x, y, z, ct) as required.

2 This is known

as a spacelike interval, if s is imaginary then it describes a timelike interval, that is, an object travelling faster than c.

42

3 A Frame Invariant Electromagnetism

There is a problem however. Lets calculate the length of our Cartesian four vector: ⎡

⎤ ct ⎢x⎥ 2 2 2 2 2 2 ⎥ s · s ≡ s j s j = [ct, x, y, z] ⎢ ⎣ y ⎦ = x + y + z + c t = s z

(3.19)

so we have a problem! The Cartesian system isn’t consistent when we try to extend our coordinate system from three space dimensions to four spacetime dimensions. The solution is to generalize the coordinate system, and we will do this in Sect. 3.4. First, we will look at a simple example to illustrate just why we need generalized coordinates and four vectors in a complete description of electromagnetism.

3.2 The Moving Charge and Wire Experiment The following “thought experiment” is a neat way to understand the consequences of special relativity in electromagnetism. An infinitely long wire is oriented along the direction of motion of the observer (the xˆ direction); a section of the wire is sketched in Fig. 3.3. In the S1 frame the observer is at rest and the current in the wire is carried by the electrons moving to the right with velocity u xˆ . In the S2 frame the observer moves to the right with velocity u xˆ , now the electrons in the wire are at rest and the current is carried by the protons moving to the left at −u xˆ . We can find out what happens to the electromagnetic fields in both frames by releasing a test charge q located outside of the wire (q doesn’t contribute to the fields). The test charge starts at rest in the S2 frame, so that in the S1 frame it starts by moving to the right with velocity u xˆ . Now here is the dilemma: if in the S1 frame, we have zero charge density in the wire (ρ1 = 0) then the only force acting on the test charge is due to the magnetic field, i.e., (3.20) F1 = qu ∧ B1 and E1 = 0. This will cause the charge to move transverse to the wire in the y, z plane, that is, transverse to the xˆ direction of the frame transformation velocity. In the S2 frame, the test charge is initially at rest, so there can be no v ∧ B force. But any motion of the charge in y or z, that is, transverse to the direction of the frame transformation velocity, will appear to be exactly the same in both frames from the Lorentz frame transformation (3.11). So what causes the force in the S2 frame? To resolve this, we need to calculate the electric and magnetic fields in the two frames. We only know the total charge density in S1 . To calculate it in S2 we will assume that charge is invariant under Lorentz transformation, that is, that charges are not created or destroyed by moving from one inertial frame to the other. Then all we need to do to find the charge density is to consider a bunch of charge Q and find the volume that it occupies in the two frames. Lets assume that the charge Q occupies a box with sides of lengths Δx, Δy, and Δz in a frame where it is at rest w.r.t. the observer.

3.2 The Moving Charge and Wire Experiment

43

q

Fig. 3.3 In frame S1 the current in the wire is carried by the electrons, the test charge q moves at u. In frame S2 the current is carried by the protons, the test charge q is at rest

u

S

1

+ v=0

J1

v=u z y x

q S2 J

+ v=-u

2

v=0 u The observer measures a charge density: ρ=

Q ΔxΔyΔz

(3.21)

If the observer now moves in the xˆ direction w.r.t. the charge, distances will Lorentz contract in the x direction from (3.10) and the observer will measure a charge density Q Q =γ = γρ (3.22) ρ = Δx  ΔyΔz ΔxΔyΔz If charges carry a current, then the corresponding charge density is always larger than if the charges are at rest3 from (3.22). In our moving wire experiment + ρ+ 2 = γρ1 ρ− − ρ2 = γ1

(3.23)

then the total charge density in the two frames is − ρ1 = ρ+ 1 + ρ1 = 0

ρ2 =

ρ+ 2

+

= ρ+ 1γ

ρ− 2

=

ρ+ 1



1 γ− γ

(3.24)

2

u = 0 c2

3 In everyday experience this effect is small, the electrons in a copper wire of 1 mm2

cross sectional area carrying 1 A have an average drift speed of ∼10−4 ms−1 which is why household wiring doesn’t charge up.

44

3 A Frame Invariant Electromagnetism

C A J

B

B

v

u

r dl

u B v

Fig. 3.4 The (dashed) loop over which we integrate B · dl is a circle of radius r centred on the ˆ Also shown is the direction of u ∧ B wire, with rˆ transverse to the direction of motion u.

so the uncharged wire in frame S1 is charged up in frame S2 . As a consequence there will be a (radial) electric field in S2 and this will provide the force on the test charge q F1 = qu ∧ B1 (3.25) F2 = qE2 Let’s calculate the forces in the two frames by using Maxwell’s equations to calculate the fields. In S1 the force is due to the magnetic field which can be obtained from the integral form of (1.42); we perform a line integral around the curve C enclosing the wire as shown in Fig. 3.4. Then since 

 B · dl =

μ0 J · dS = μ0 I

(3.26)

C

and the curve C is a circle of radius r centred on the middle of the wire, then (using (1.19) to determine that B is directed along dl; and from the geometry B = B(r ) and has the same magnitude on all points on C)  B · dl = 2πr B

(3.27)

C

Since the current I is carried by the electrons moving at speed u through the wire which has cross section A, I = ρ− 1 Au. The “u ∧ B” force (3.25) in frame 1 then has magnitude qρ− A u 2 F1 = 1 (3.28) 2π0 r c2 which acts to repel the test charge as shown in Fig. 3.4.

3.2 The Moving Charge and Wire Experiment Fig. 3.5 The (dashed) surface over which we integrate E · dS is a cylinder of radius r centred on the wire, with rˆ transverse to the direction of motion uˆ

45

A

r S dl In S2 the force is due to the electric field which we calculate from the integral form of (1.16); we perform a surface integral over the closed cylindrical surface shown in Fig. 3.5    ρ2 dQ E · dS = ∇ · Ed V = dV = (3.29) 0 S V V 0 From (1.35) E is directed radially out from the wire so there is no contribution to the surface integral from the ends of the cylinder. From the geometry E = E(r ) so that  E · dS = 2πr Edl

(3.30)

S

and excluding the test charge, the charge enclosed in the cylinder d Q = ρ2 Adl. The force due to the electric field in frame S2 then has magnitude F2 =

qρ+ A u 2 qρ2 A = 1 γ 2π0 r 2π0 r c2

(3.31)

where we have used (3.24), this again acts in the +ˆr direction to repel the test charge. Comparing (3.28) and (3.31) we find that the forces acts in the same (ˆr) direction in both frames but the magnitude differs by a factor of γ F2 = γ F1

(3.32)

Now recall that any motion transverse to the direction of motion u must be the same in both frames. What is happening here? Lets look at the equation of motion of the test charge.

46

3 A Frame Invariant Electromagnetism

During the first (infinitesimally small) time interval Δt after it is released from rest it will gain transverse momentum Δp = F1 Δt1

(3.33)

if observed in frame S1 . But in S1 the test charge is moving, and so the time interval will be dilated w.r.t. frame S2 Δt1 = γΔt2 (3.34) then F1 Δt1 = F1 γΔt2 = F2 Δt2

(3.35)

so that the change in transverse momentum Δp = FΔt in any frame, and the transverse motion will be frame independent, as we would expect. Special relativity implies that length, time, force etc. change as we Lorentz transform from one frame to another. This example has shown how the electromagnetic fields, current and charge densities are also not invariant under Lorentz transformation. We have anticipated that space and time must form a single coordinate system, space-time, and in this coordinate system, the (3) space coordinates and (1) time coordinate of a point combine to form a single four-vector which has length that is invariant under Lorentz transformation. In the same way, current and charge density are in themselves not invariant, but can be combined to form a four vector that has invariant length. This leads to a frame invariant electromagnetism that we will introduce next.

3.3 Maxwell in Terms of Potentials Working with E and B directly needs rank 2 tensors to represent the fields. A frame invariant electromagnetism can be written down in terms of rank 1 tensors (four vectors) if we instead work in terms of the scalar and vector potentials φ, A, we’ll do that first in this section. In Chap. 1 we found that the fields could be written in terms of potentials B=∇ ∧A (3.36) E = −∇φ −

∂A ∂t

(3.37)

which we showed always satisfy the homogenous Maxwell equations II (1.19) and III (1.35), that is, the Maxwell equations which do not refer to the currents or charges. It is easy to see that the potentials are not unique: since ∇ ∧ ∇ψ is zero for any ψ we can add any ∇ψ to A where ψ is a scalar function and still satisfy (3.36). This will then give new potentials A = A + ∇ψ (3.38) φ = φ + ∂ψ ∂t which also satisfy (3.37). The operation (3.38) is a gauge transformation.

3.3 Maxwell in Terms of Potentials

47

In principle any arbitrary gauge can be used. Let’s look at what happens to the other two inhomogenous Maxwell equations I (1.16) and IV (1.42) (which contain reference to currents and charges) when we write them in terms of A and φ. Using (3.37) to substitute for E in Maxwell I gives − ∇2φ −

∂ ρ ∇ ·A= ∂t 0

and using (3.37) and (3.36) to substitute for E and B in Maxwell IV gives   ∂2A 1 ∂ 2 ∇φ + 2 = μ0 J − ∇ A + ∇(∇ · A) + 2 c ∂t ∂t

(3.39)

(3.40)

We can now choose a gauge that reveals a nice symmetry in (3.39) and (3.40). This is the Lorentz gauge4 1 ∂φ ∇ ·A=− 2 (3.41) c ∂t and substituting (3.41) into (3.39) and (3.40) gives ∇ 2 φ − c12 ∂∂tφ2 = − ρ0 2 ∇ 2 A − c12 ∂∂tA2 = −μ0 J 2

(3.42)

in free space (ρ = 0, J = 0) Eq. 3.42 are simply wave equations for A and φ and predict electromagnetic waves with speed c. They contain the same information as the Maxwell equations, all we need is to put (3.42) and the Lorentz gauge (3.41) in Lorentz invariant form and we will have a description of electromagnetism that incorporates special relativity.

3.4 Generalized Coordinates From Eq. 3.19 we found that space-time is not Cartesian. We need a generalized coordinate system that has the following properties: 1. The space-time interval s 2 = c2 t 2 − x 2 − y 2 − z 2

(3.43)

is invariant under Lorentz transformation. electrostatics this reduces to the Coulomb gauge ∇ · A = 0 which gives Poisson equations for both φ and A ρ −∇ 2 φ = 0 and −∇ 2 A = μ0 J. 4 In

48

3 A Frame Invariant Electromagnetism

2. The space-time interval is just the length of the four vector: s2 = s · s

(3.44)

3. The Lorentz transformation corresponds to a rotation of the four vector s in spacetime. The system that we devise will work for any four tensor (four vectors being the special case of four tensors of rank 1). We will start by defining two “versions” of our four vector5 and for the vector s these look like: Covariant: ⎡ ⎤ ct ⎢ −x ⎥ ⎥ xα = (ct, −x) = ⎢ (3.47) ⎣ −y ⎦ −z ⎡

⎤ ct ⎢x⎥ ⎥ x α = (ct, +x) = ⎢ ⎣y⎦ z

and Contravariant:

(3.48)

Covariant rank 1 tensors (vectors) such as (3.47) always have a single “down” index (subscript) and contravariant vectors such as (3.48) always have an “up” index (superscript). Rank 2 or more tensors can be covariant (all indices down), contravariant (all indices up) or mixed (indices up and down). We are also going to use the summation convention introduced for Cartesian tensors (2.24).

In general if we have a well defined transformation that gives x α = x α (x 0 , x 1 , x 2 , x 3 ) a covariant vector transforms as:

5

aα =

∂x 2 ∂x 3 ∂x 0 ∂x 1 ∂x β a0 + a1 α a2 α a3 = aβ α α ∂x ∂x ∂x ∂x ∂x α

(3.45)

and a contravariant vector transforms as a α = See the appendix on tensors for details.

∂x α β a ∂x β

(3.46)

3.4 Generalized Coordinates

49

Now, if we simply define the scalar (dot, or inner) product of two four vectors as the product of a covariant and a contravariant vector, then the length of the spacetime interval s 2 is ⎡ ⎤ ct ⎢x⎥ 2 α ⎢ s = xα x = x · x = x˜ x = [ct, −x, −y, −z] ⎣ ⎥ = (ct)2 − x 2 − y 2 − z 2 y⎦ z (3.49) as required. Then we just need to know how to turn a covariant tensor into a contravariant one, and vice versa, and from inspection of (3.47) and (3.48) we simply need an operation that changes all the signs on the spacelike components x 1,3 = x, y, z, whilst leaving the timelike component x 0 = ct unchanged. We can use the same operation as (3.49), that is, a dot or inner product or contraction over a pair of contravariant indices, this time between a rank 1 tensor (vector) and a rank 2 tensor

and

xα = gαβ x β

(3.50)

x α = g αβ xβ

(3.51)

⎡

where gαβ

⎤ 1 0 0 0 ⎢ 0 −1 0 0 ⎥ αβ ⎥ =⎢ ⎣ 0 0 −1 0 ⎦ = g 0 0 0 −1

(3.52)

g αβ is known as the metric of the spacetime. The spacetime interval is then written as (3.53) s 2 = xα g αβ xβ = gαβ x α x β which in matrix notation is ˜ s 2 = (x, gx) = (gx, x) = xgx

(3.54)

The spacetime metric (3.52) is defined by the form of the spacetime interval s 2 , via (3.53). This particular metric then is just what is needed to embody the flat spacetime of special relativity, in which the length of a spacetime interval (the length of s) has the same constant value anywhere in spacetime, i.e., spacetime itself is uniform. The formalism that we are developing here can just as easily be applied where spacetime is nonuniform or curved by gravity. The metric for curved spacetime embodies general relativity. The dot product operation that we defined to obtain the desired behaviour for s 2 has to be the same as the dot product between two four tensors generally. For two

50

3 A Frame Invariant Electromagnetism

four tensors of any rank and index (i.e., covariant, contravariant, or a mixture of the two), the dot or inner product will be ... ...α b... a · b = a...α

(3.55)

that is, a contraction over the index α. Contraction with gαβ or g αβ changes a index from contravariant to covariant (up to down) or vice versa (down to up). It works like this: Contravariant to covariant (up to down): ... ...β = gαβ a... a...α

(3.56)

and covariant to contravariant (down to up): ...α ... = g αβ a...β a...

(3.57)

Now we can write down the Lorentz transformation -rotation matrix as a 4- tensor. A rotation of four vector s will be a operation of the form 

xα = Λαβ x β on the contravariant form and



x α = Λαβ xβ

(3.58) (3.59)

on the covariant form. The contractions (3.58) and (3.59) change a contravariant four vector into a covariant one and vice versa. The rotation matrix6 is (almost) the same as that discussed in Cartesian formalism and is ⎡ ⎤ γ − vc γ 0 0 ⎢ + v γ −γ 0 0 ⎥ c ⎥ Λαβ = ⎢ (3.60) ⎣ 0 0 −1 0 ⎦ 0 0 0 −1 where Λαβ is the transpose of Λαβ . The rotation of coordinates in spacetime written out in full is ⎡  ⎤ ⎡ ⎤⎡ ⎤ ct γ − vc γ 0 0 ct  ⎢ −x ⎥ ⎢ + v γ −γ 0 0 ⎥ ⎢ x ⎥  β ⎥ ⎢ c ⎥⎢ ⎥ xα = Λαβ x = ⎢ ⎣ −y  ⎦ = ⎣ 0 0 −1 0 ⎦ ⎣ y ⎦  z 0 0 0 −1 −z (3.61) ⎤ ⎡ vx cγ(t − c2 ) ⎢ −γ(x − vt) ⎥ ⎥ =⎢ ⎦ ⎣ −y −z which just give the Lorentz transformation (3.11) as required (the inverse transformation matrix is then just obtained by putting v → −v in (3.60)). 6 This is a member of the Lorentz Group of transformations, see e.g., Classical Electrodynamics 2nd

Ed., J. D. Jackson, (Wiley 1975) for details.

3.5 Four Vectors and Four Vector Calculus

51

3.5 Four Vectors and Four Vector Calculus The rules of flat spacetime were defined from the Lorentz transformation of the four vector xα but will apply to any four vector. If we can write the laws of mechanics and electromagnetism in four vector form, then we will have a description of mechanics and electromagnetism that is Lorentz invariant (i.e., manifestly covariant). The required four vectors will always have the same length in spacetime in all frames, and will Lorentz transform simply by coordinate rotation in spacetime.

3.5.1 Some Mechanics, Newton’s Laws As we have shown, length and time form the four-vector ⎡

⎤ ct ⎢ −x ⎥ ⎥ xα = ⎢ ⎣ −y ⎦ −z

(3.62)

Another example is the energy-momentum four vector ⎡

ε c

⎤

 ⎢ − px ⎥  ε ⎥= , −p pα = ⎢ ⎣ − py ⎦ c − pz

(3.63)

The length of pα should be constant. In full it is ⎡ pα p α =

ε c px

⎤

 ⎢ ⎥ ε2 2 ⎥ , − p x , − p y , − pz ⎢ ⎣ p y ⎦ = c2 − p c pz

ε

(3.64)

Now recall some expressions from special relativity ε2 = p 2 c2 + m 20 c4

(3.65)

The energy-momentum four vector then has length pα p α = m 20 c2

(3.66)

This just gives the rest energy of a particle ( pα p α /m 0 ); if we can find a frame where the particle is at rest (i.e., it is an electron, proton but not a photon) then since the

52

3 A Frame Invariant Electromagnetism

length of pα is Lorentz invariant, its length in all frames must be just that found in the rest frame. If the particle is at rest the spacelike components of pα are zero and the timelike component is p0 = εc = m 0 c. Now from special relativity the relativistic mass m = γm 0 of a moving particle is larger than the rest mass m 0 , so we have in the moving frame, the timelike component given by (3.67) p 0 c = ε = mc2 = γm 0 c2 and the spacelike components p 1,3 are given by p = mu = γm 0 u

(3.68)

We can write these as a single equation in terms of four vectors pα = m 0 u α

(3.69)

where we have introduced another four vector, the four velocity ⎤ cγ ⎢ γu x ⎥ pα ⎥ =⎢ uα = ⎣ γu y ⎦ m0 γu z ⎡

(3.70)

The four velocity u α must transform in the same way as the energy-momentum four vector as we have simply divided by m 0 which is invariant under Lorentz transformation. The expression (3.69) is manifestly covariant, that is, under Lorentz transformation it will retain the same form and the component four vectors will Lorentz transform, with invariant lengths. What about Newton’s laws? In a nonrelativistic world, frame transformations are Galilean and correspond to rotations of Cartesian position vector x in x, y, z plus translations x = x + ut, where u is the constant velocity between the unprimed and the primed frames. The nonrelativistic (Newton) laws of motion are: 1. An object is at rest or moves in a straight line at constant velocity unless subject to some force. 2. Momentum is conserved (“for every action there is an equal and opposite reaction”) under Galilean transformation. and u = dx are invariant under Galilean trans3. The equations of motion F = dp dt dt formation. From (1) and (2) there will be a preferred rest frame in the sense that x = (x, y, z) is constant and the total p = ( px , p y , pz ) = 0. The four vectors we have already written down for xα and pα will allow us to rewrite laws (1) and (2). We just need to look at differentiation w.r.t. time in (3) before we can rewrite all of Newton’s laws.

3.5 Four Vectors and Four Vector Calculus

53

Intuitively, we might expect that since time is dilated in the moving frame (3.4), an invariant interval to replace dt in the derivative would be the proper time dt/γ. We can formalize this by considering the coordinates of two particles that are moving apart at constant velocity in spacetime. At time t, the particles are both at the same position (x, y, z) and have spacetime coordinates (ct, x, y, z). As the particles move apart at constant velocity u = (u x , u y , u z ), at time t + Δt, they will be separated by a distance Δx = u x Δt Δy = u y Δt (3.71) Δz = u z Δt the spacetime interval between the two particles is then Δs 2 = c2 Δt 2 − Δx 2 − Δy 2 − Δz 2 = (c2 − u 2 )Δt 2 so that an invariant interval is

Δs Δt = c γ

(3.72)

(3.73)

The time derivative in Newton’s laws is then to be replaced by the invariant spacetime derivative d d =γ (3.74) ds dt The spacetime derivative of (four) position in spacetime is then ⎡

⎤ ⎡ cγ ct ⎥ ⎢ γ dx dxα d ⎢ x dt ⎥ = ⎢ dy =γ ⎢ ds dt ⎣ y ⎦ ⎣ γ dt z γ dz dt

⎤ ⎥ ⎥ = uα ⎦

(3.75)

i.e., the four- velocity as we would expect. Newton’s laws of motion then become: 1. The spacetime interval x α = (ct, x, y, z) has constant length under Lorentz transformation. 2. The energy-momentum four-vector p α = ( εc , px , p y , pz ) has constant length under Lorentz transformation. α α and u α = ddsx are invariant under Lorentz 3. The equations of motion f α = dp ds transformation. Then f α , the “four-force”, must also be a four vector; in Chap. 4 we will find the four vector force on a charged particle in an electromagnetic field, that is, the covariant version of the Lorentz force law.

54

3 A Frame Invariant Electromagnetism

3.5.2 Some Four Vector Calculus Since we know how to define intervals in space-time we can do some calculus in terms of four vectors. First let’s consider the four gradient of some scalar field φ(ct, x, y, z) = φ(x α ) defined in spacetime. The scalar field itself is an invariant, the coordinates transform as the components of a four vector should. Since we have defined φ as a function of contravariant position in spacetime x α we can use chain rule to obtain the change in scalar field dφ on some spacetime interval ds given the contravariant coordinates of the interval d x α dφ = dt

∂φ ∂φ ∂φ dφ ∂φ + dx + dy + dz = ds ∂t ∂x ∂y ∂z ds

(3.76)

The quantity dφ is also invariant, so the r.h.s. of (3.76) must simply be in the form of the dot product of two four vectors dφ = ∂α (φ)d x α

(3.77)

This means that the covariant four gradient is ⎡1 ∂ ⎤ ⎢ ⎢ ∂α = ⎢ ⎢ ⎣

c ∂t ∂ ∂x ∂ ∂y ∂ ∂z

⎥

⎥ ⎥ = 1 ∂ ,∇ ⎥ c ∂t ⎦

(3.78)

which is almost what we would expect, except that the spacelike part is positive. We could have instead written the scalar field as a function of covariant position xα , i.e., φ = φ(ct, −x, −y, −z); the invariant interval would then need to be written as a four vector dot product (3.79) dφ = ∂ α (φ)d xα so to generate (3.76) from (3.79) we need a contravariant four gradient ⎡ α

∂ =

1 ∂ c ∂t ⎢ ∂ ⎢ − ∂x ⎢ ⎢− ∂ ⎣ ∂y ∂ − ∂z

⎤ ⎥

⎥ ⎥ = 1 ∂ , −∇ ⎥ c ∂t ⎦

(3.80)

which again has spacelike components with opposite signs than we would expect.

3.5 Four Vectors and Four Vector Calculus

55

As a consistency check, if covariant ∂α is a four vector then the contravariant ∂ α should also be defined via the spacetime metric ⎡ ∂ α = g αβ ∂β =

1 ∂ c ∂t ⎢ ∂ ⎢ − ∂x ⎢ ⎢− ∂ ⎣ ∂y ∂ − ∂z

⎤ ⎥

⎥ ⎥ = 1 ∂ , −∇ ⎥ c ∂t ⎦

(3.81)

Having found the “four gradient” operator of a scalar field we can define “four divergence” of a four vector field (and more generally, a four tensor). We would expect a “four-divergence” of a four vector to be of the form of a dot product ∂α a α = ∂ α a α =

1 ∂ 0 a +∇ ·a c ∂t

(3.82)

(where a is the spacelike part of four vector field a α ). Finally, the length of the four vector ∂ α is the “four ∇ 2 ” D’Alembertian: ∂α ∂ α =

1 ∂2 − ∇2 =  c2 ∂t 2

(3.83)

We will derive “four curl” later in Chap. 4 when it is needed. First, we can use the four divergence and four ∇ 2 to develop a frame invariant electromagnetism in terms of scalar and vector potentials.

3.6 A Frame Invariant Electromagnetism 3.6.1 Charge Conservation Charge conservation was introduced in Chap. 1 to obtain (1.42) and so is “built in” to Maxwell’s equations. This is expressed by ∇ ·J+

∂ρ =0 ∂t

(3.84)

which was derived in (1.3.2) by assuming that charge is conserved. If we now insist that charge is Lorentz invariant, i.e., it is the same in all frames, we should have a manifestly covariant form of (3.84). Equation (3.84) has the form of the four divergence of some four vector ∂α J α = 0 (3.85)

56

3 A Frame Invariant Electromagnetism

and by inspection, knowing ∂α we have the four current ⎤ cρ ⎢ Jx ⎥ ⎥ J α = (cp, J) = ⎢ ⎣ Jy ⎦ Jz ⎡

(3.86)

The length of J α is invariant under Lorentz transformation so that its components, the charge and current density are not, as we found when we considered the moving charge and wire experiment in Sect. 3.2.

3.6.2 A Manifestly Covariant Electromagnetism We can finally write down the laws of electromagnetism in four vector form, if we do so in terms of the scalar and vector potentials for the electromagnetic fields. Recall from Sect. 3.3 that the Maxwell equations can be written in terms of scalar potential φ and vector potential A, and take on (what we will see is a very useful) symmetry if we also choose the Lorentz gauge ∇ ·A+

1 ∂ φ=0 c2 ∂t

(3.87)

from Maxwell. This has the same form as the charge conservation equation so again we can write ∂α A α = 0 (3.88) where the new four vector is the four vector potential

Aα =



⎡ φ ⎤ c

⎢ Ax ⎥ φ ⎥ ,A = ⎢ ⎣ Ay ⎦ c Az

(3.89)

To obtain a manifestly covariant form of the Maxwell equations, let’s examine the two inhomogenous Maxwell equations written in terms of A and φ using the Lorentz gauge (3.42). The equation in terms of φ (originally Maxwell I (1.16)) has been divided by c to give 1 ∂2 φ ρ φ = cρμ0 (3.90) − ∇2 + 2 2 = c c ∂t c 0 c and the l.h.s. of this looks like the “four ∇ 2 ” of the timelike part of Aα .

3.6 A Frame Invariant Electromagnetism

57

The equation in terms of A (originally Maxwell IV (1.42)) is: − ∇2A +

1 ∂ A = μ0 J c2 ∂t

(3.91)

and the l.h.s. of this looks like the “four ∇ 2 ” of the spacelike part of Aα . Together they are a single 4 vector equation

φ , A = μ0 (cρ, J)  c

(3.92)

Aα = μ0 J α

(3.93)



or

which is our equation for electromagnetism in manifestly covariant form. Since the invariance of charge implies that J α is a four vector from (3.85), (3.93) shows that Aα must also be a four vector. So, to Lorentz transform the electromagnetic fields, all we need to do is write the fields in terms of A and φ, transform (i.e., rotate the four vectors in spacetime) Aα and J α , then work out the electromagnetic fields again from the new A and φ using (3.36) and (3.37). In the next chapter we will find the electromagnetic field tensors that lead to a direct transformation of the fields. For the moment, we have already shown that Maxwell’s equations are consistent with special relativity. The Maxwell equations can be written in terms of four vectors. In free space J α = 0 and we are left with the free space wave equations for A and φ in manifestly covariant form: (3.94) Aα = 0 which predicts light waves moving at speed c, and holds in all frames of reference.

Chapter 4

The Field Tensors

In Chap. 3 we developed a formalism in which Maxwell’s equations and the electromagnetic fields, as embodied in the scalar and vector potential description, could be written in manifestly covariant form, that is, in terms of four vectors which Lorentz transform by a coordinate rotation in spacetime. We now wish to write the Maxwell equations, the Lorentz force law, and the equations of conservation of field energy and momentum, explicitly in terms of the fields E and B. This will require us to develop four tensors of rank 2 (as opposed to the rank 1 four tensors, or four vectors, of Chap. 3). The generalization of 3 coordinate Cartesian space into 4 coordinate spacetime will still apply. The Maxwell equations in the form that we have seen in Chap. 1 have an asymmetry which we will be able to explore more fully in the framework of spacetime; to gain a glimpse of what the universe might look like if monopoles exist. Finally, we can obtain a general solution to the Maxwell equations.

4.1 Invariant Form for E and B: The EM Field Tensor Let’s begin by trying to write down an invariant form for E and B. In Cartesian space we have already defined B in terms of a vector field B=∇ ∧A

(4.1)

Now in spacetime, the vector potential A and the scalar potential φ has been replaced with the four vector   φ α , A x , A y , A z = (A0 , A1 , A2 , A3 ) (4.2) A = c

© Springer-Verlag GmbH Germany, part of Springer Nature 2021 S. Chapman, Core Electrodynamics, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-030-66818-1_4

59

60

4 The Field Tensors

so we should be able to use the four grad (here in contravariant form): α



∂ =

∂ ∂ ∂ 1 ∂ ,− ,− ,− c ∂t ∂x ∂y ∂z

 = (∂ 0 , ∂ 1 , ∂ 2 , ∂ 3 )

(4.3)

to construct a “four-curl” of Aα . Writing out the components of ∇ ∧ A gives −

∂ Ay ∂z

= −(∂ 2 A3 − ∂ 3 A2 ) = −F 23

B y = − ∂∂xAz +

∂ Ax ∂z

=

∂ Ax ∂y

= −(∂ 1 A2 − ∂ 2 A1 ) = −F 12

Bx

Bz

=

=

∂ Az ∂y

∂ Ay ∂x

−

∂ 1 A3 − ∂ 3 A1 = +F 13

(4.4)

We can arrange the components of ∇ ∧ A to look like combinations of the components of the four vectors if we use terms such as ∂ 2 A3 . From our notation these are not a contraction over an index as in the case of “four div” which had terms such as ∂2 A2 , instead, they look like components of a four tensor of rank 2. All the three components of (4.4) have the same pattern, that is, they look like they are part of a single rank 2 tensor F αβ . We might guess that more terms of F αβ originate in the electric field, since it too is represented by components of Aα E = −∇φ −

∂A ∂t

(4.5)

Writing this in component form gives E x = − ∂φ − ∂x

∂ Ax ∂t

= −c(∂ 0 A1 − ∂ 1 A0 ) = −F 01 c

E y = − ∂φ − ∂y

∂ Ay ∂t

= −c(∂ 0 A2 − ∂ 2 A0 ) = −F 02 c

∂ Az ∂t

= −c(∂ A − ∂ A ) = −F c

Ez =

− ∂φ ∂z

−

0

3

3

0

(4.6)

03

So, the three components of B in (4.4) and the three components of E in (4.6) give six terms of a rank 2 tensor F αβ = ∂ α Aβ − ∂ β Aα

(4.7)

The other terms of the 4 × 4 “four-tensor F αβ can be obtained from inspection of (4.7). The on axis terms are when α = β F αα = ∂ α Aα − ∂ α Aα = 0 that is, all four are zero (α = 0, 1, 2, 3).

(4.8)

4.1 Invariant Form for E and B: The EM Field Tensor

61

The off axis terms are when α = β, and are for example when α = 0, β = 1 F 01 = ∂ 0 A1 − ∂ 1 A0 = −(∂ 1 A0 − ∂ 0 A1 ) = −F 10

(4.9)

so that F αβ is antisymmetric: F αβ = −F βα . The 4 × 4 = 16 terms in F αβ is comprised of the four on axis terms which are zero, plus the six different terms from the 3E (4.6) and 3B (4.4) components, each used twice ⎡ F

αβ

⎢ ⎢ ⎢ =⎢ ⎢ ⎣

0 − Ecx − Ex c Ey c Ez c

0 Bz

Ey c

− Ecz

⎤

⎥ −Bz B y ⎥ ⎥ ⎥ 0 −Bx ⎥ ⎦

−B y Bx

(4.10)

0

This is the electromagnetic field tensor F αβ which has “packaged” E and B in terms of four vectors from (4.7). This means that we can use F αβ to describe laws of physics directly in terms of E and B in a form that is “manifestly covariant”, that is, invariant in form under Lorentz transformation. We will in the next section write down the Maxwell equations in manifestly covariant form directly, instead of in terms of the scalar and vector potentials (the four vector Aα ) as in the previous chapter. What if we had used the covariant four vectors to form a “four curl” of A? The starting point would have been  Aα =

φ , −A x , −A y , −A z c



and ∂α =

1 ∂ ∂ ∂ ∂ , , , c ∂t ∂x ∂ y ∂z

 = (A0 , A1 , A2 , A3 )

(4.11)

= (∂0 , ∂1 , ∂2 , ∂3 )

(4.12)



then ∇ ∧ A would look like: Bx =

∂ Ay ∂ Az − = −∂2 A3 + ∂3 A2 = −(∂2 A3 − ∂3 A2 ) = −F23 ∂y ∂z

(4.13)

so Bx = −F23 = −F 23 , and similarly for B y = F13 = F 13 , Bz = −F12 = −F 12 , so the form is not unlike the contravariant case (4.4). For the E field, we now have Ex = −

∂φ ∂ A x − = −c∂1 A0 + c∂0 A1 = c(∂0 A1 − ∂1 A0 ) = F01 c ∂x ∂t

(4.14)

and E x /c = F01 = −F 01 , that is, it has the opposite sign to the contravariant case (4.6), and similarly for E y and E z .

62

4 The Field Tensors

We can then write Fαβ = ∂α Aβ − ∂β Aα

(4.15)

with the covariant form of the electromagnetic field tensor: ⎡

Ey c

Ex c

0

⎤

Ez c

⎢ Ex ⎥ ⎢ − c 0 −Bz B y ⎥ ⎥ Fαβ = ⎢ ⎢ Ey ⎥ ⎣ − c Bz 0 −Bx ⎦ − Ecz −B y Bx 0

(4.16)

To get from the covariant to the contravariant forms of the electromagnetic field tensor we just need to use the spacetime metric gαγ to change the signs on the timelike part of the tensor, that is, the F 0α and F α0 terms (in this case, the “E” terms) Fαβ = gαγ F γδ gδβ

(4.17)

which is equivalent to E → −E, B → B. Finally, if F αβ is manifestly covariant we would expect it to have constant “length”, this just turns out to be zero for electromagnetic waves in free space (see advanced problem 6). The usefulness of F αβ will now become clear as we use it to write down manifestly covariant forms of the Maxwell equations.

4.2 Maxwell’s Equations in Invariant Form We know that Maxwell’s equations involve the ∇· and ∇∧ of E and B, so let’s begin by taking the “4 div” of F αβ . ⎡ ∂α F αβ =



1 ∂ ∂ ∂ ∂ , , , c ∂t ∂x ∂ y ∂z

⎡

∇·E c

⎢ (∇ ∧ B)x − ⎢ =⎢ ⎣ (∇ ∧ B) y − (∇ ∧ B)z −

⎢ ⎢ ⎢ ⎢ ⎣ ⎤

1 c2 1 c2 1 c2

∂ Ex ∂t ∂ Ey ∂t ∂ Ez ∂t

0 Ex c Ey c Ez c

Ex c

Ey c

Ez c

⎤

⎥ 0 −Bz B y ⎥ ⎥ ⎥ Bz 0 −Bx ⎦ −B y Bx 0

(4.18)

⎥ ⎥ ⎥ ⎦

i.e a single four vector containing the (slightly rearranged) left hand side of the Maxwell inhomogeneous equations with timelike component giving ∇ ·E = μ0 cρ c

(4.19)

4.2 Maxwell’s Equations in Invariant Form

63

and with spacelike components giving ∇ ∧B−

1 ∂E = μ0 J c2 ∂t

(4.20)

The r.h.s. of (4.19) and (4.20) taken together look like (μ0 times) the four-current J α = (cρ, J)

(4.21)

so the inhomogeneous Maxwell equations are just the single equation ∂α F αβ = μ0 J β

(4.22)

We would like to use the same trick to get the homogenous Maxwell equations in manifestly covariant form. In this case, the “four div” of some tensor must yield a single four vector, this time with timelike component ∇ ·B=0 and spacelike components ∇ ∧E+

(4.23)

∂B =0 ∂t

(4.24)

so we need some transformation that swaps the position of the E and B terms in F αβ so that the timelike components involve B and the spacelike components involve E. This is known as the duality transformation and is E → Bc2 and B → −E; performing the transformation on F αβ gives the Dual field tensor ⎤ 0 −Bx c −B y c −Bz c ⎢ Bx c 0 E z −E y ⎥ ⎥ =⎢ ⎣ B y c −E z 0 Ex ⎦ Bz c E y −E x 0 ⎡

F˜ αβ

(4.25)

The “four div” of F˜ αβ then yields the required homogenous Maxwell equations: ⎡

∂α F˜ αβ

c∇ · B ⎢ −(∇ ∧ E)x − ⎢ =⎢ ⎣ −(∇ ∧ E) y − −(∇ ∧ E)z −

⎤ ∂ Bx ∂t ∂ By ∂t ∂ Bz ∂t

⎥ ⎥ ⎥ ⎦

(4.26)

so that the homogenous Maxwell equations are ∂α F˜ αβ = 0

(4.27)

64

4 The Field Tensors

But what is the significance of F˜ αβ ? First, we need to relate it directly to F αβ to demonstrate that just like F αβ , F˜ αβ is indeed a “four tensor” and Lorentz transforms in the same way. This is done via the “four alternating tensor” (the 4 ×4 × 4 version of the Cartesian 3 × 3 alternating tensor that we first met in Sect. 2.1.3): c2 F˜ αβ = αβγδ Fγδ 2

(4.28)

Second, the dual tensor has an interesting physical significance. If we apply the duality transformation to Maxwell’s equations we have: Maxwell equations ∇ ·E=

ρ 0

Dual equations ∇ · B = μ0 ρ

∇ ·B=0

∇ ·E=0

∇ ∧ E = − ∂B ∂t ∇ ∧ B = μ0 J +

1 ∂E c2 ∂t

c2 ∇ ∧ B =

(4.29)

∂E ∂t

∇ ∧ E = −μ0 J −

∂B ∂t

The Dual Maxwell Eq. (4.29) describe a world in which we would identify ρ as magnetic charge and J as magnetic current. In this dual world, E is nonconservative, that is, since ∇ · E = 0 lines of E must close. In static situations B will be conservative as ∇ ∧ B = 0 if E does not vary with time. Two properties of the Maxwell equations implied that they must embody Lorentz invariance if only we could write them in an invariant form. Both charge and the speed of light waves in free space c are invariant under Lorentz transformation. If we put ρ = 0 and J = 0 in (4.29) then the Maxwell and Dual equations are identical, i.e., they will both predict light waves moving at invariant speed c in free space. Charge conservation is implicit in Maxwell IV, we can look at of conservation of magnetic charge using the dual equations. The Dual IV equation is ∇ ∧ E = −μ0 J −

∂B ∂t

(4.30)

Taking the divergence of (4.30) gives ∇ · (∇ ∧ E) = 0 = −μ0 ∇ · J −

∂ (∇ · B) ∂t

(4.31)

which from Dual I (∇ · B = μ0 ρ) gives ∇ ·J+

∂ρ =0 ∂t

(4.32)

so (in this case magnetic) charge is again conserved, and we would from the Dual equations be able to write charge conservation in invariant form in the same way as in Sect. 3.6.1.

4.2 Maxwell’s Equations in Invariant Form

65

Writing the Maxwell equations in Lorentz invariant form then tells us that the existence of “electric charge” or “magnetic charge” is not excluded by special relativity. The only reason we wrote down the original Maxwell equations with electric charge and zero magnetic charge is that experimentally the magnetic charge is known to be zero to very high precision. The interesting question is whether all particles have the same ratio of electric to magnetic charge; if so, then we can always perform a duality transformation similar to that discussed here to make the Maxwell equations appear as if the magnetic charge is zero.

4.3 Conservation of Energy-Momentum In Sect. 3.6.1 we found a manifestly covariant form for charge conservation. We also found in Sect. 3.5.1 that the energy and momentum of individual particles forms a single four vector, which will also have invariant length under Lorentz transformation. The energy-momentum of an ensemble of particles, a mixture of charged particles and photons, say, must also form a four vector. We might then guess that the conservation equations for field energy, (Poynting’s theorem (2.33), and momentum (2.39) could be combined to form an equation for the conservation of field and charged particle energy-momentum that is manifestly covariant. To demonstrate this we need to “package”: Energy conservation (Poynting’s Theorem) S J·E 1 ∂U +∇ · =− c ∂t c c

(4.33)

1 ∂ S − ∇ · T = −(ρE + J ∧ B) c ∂t c

(4.34)

and Momentum conservation

into one object. The r.h.s. of these equations should then form a four vector, with timelike component (4.33) and spacelike components (4.34). The l.h.s. of the desired expression must be the four divergence of a four tensor ∂α T αβ ; we can guess that this is the case since it involves ∂α and must be rank 1, that is, a four vector, to match the r.h.s. Looking at the l.h.s. first then, since  ∂α =

1 ∂ ∂ ∂ ∂ , , , c ∂t ∂x ∂ y ∂z

 (4.35)

66

4 The Field Tensors

we need a 4 ×4 energy-momentum field tensor: ⎡

U

⎢ s1 c T αβ = ⎢ ⎣ s2 c s3 c

s1 c

s2 c

−T11 −T21 −T31

−T12 −T22 −T32

⎤

s3 c

−T13 ⎥ ⎥ −T23 ⎦ −T33

(4.36)

where the Ti j are the 9 components of the 3 × 3 Cartesian Maxwell stress tensor (2.47). Then ∂α T αβ will give the l.h.s. of (4.33) and (4.34). We have to deal with the r.h.s. in terms of E and B directly, which means that we need either the electromagnetic field tensor F αβ or its dual. We can guess which one by inspection of the r.h.s. terms; the spacelike part needs a J ∧ B term, and we formed ∇ ∧ B by taking the “four div” of F αβ (4.22). So if we replace the “four div” in ∂α F αβ with a dot product with the four current Jα we have ⎡ ⎢ Jα F αβ = (cρ, −Jx , −Jy , −Jz ) ⎢ ⎣ ⎤ − 1c J · E ⎢ −(ρE + J ∧ B)x ⎥ ⎥ =⎢ ⎣ −(ρE + J ∧ B) y ⎦ −(ρE + J ∧ B)z ⎡

0 Ex c Ey c Ez c

− Ecx 0 Bz −B y

E

− cy −Bz 0 Bx

⎤ − Ecz By ⎥ ⎥ −Bx ⎦ 0

(4.37)

which gives the r.h.s. as required. Our expression for energy-momentum conservation is then ∂α T αβ = Jα F αβ (4.38) The r.h.s of Eq. (4.38) is constructed from four tensors and ∂α is a four vector, so the only remaining term T αβ must be a four tensor and the expression is manifestly covariant.

4.4 Lorentz Force We have one equation left to express in manifestly covariant form, the Lorentz force law. In Sect. 3.5.1 the laws of mechanics were written in Lorentz invariant form giving manifestly covariant equations of motion: fα = uα =

dpα ds dxα ds

(4.39)

We now just need to find the four-force f α for a charged particle in an electromagnetic field.

4.4 Lorentz Force

67

Since the Lorentz force contains a +v ∧ B term, we can again guess that since Jα F αβ gives −J ∧ B terms we require something of the form ⎡ ⎢ F αβ u β = ⎢ ⎣

0 Ex c Ey c Ez c

− Ecx 0 Bz −B y

E

− cy −Bz 0 Bx

⎤ ⎡ ⎤ ⎤⎡ 1 − Ecz cγ γv · E c ⎥ ⎢ ⎥ ⎢ By ⎥ ⎥ ⎢ −γvx ⎥ = ⎢ γ(E + v ∧ B)x ⎥ (4.40) ⎦ ⎣ ⎦ ⎣ −γv y γ(E + v ∧ B) y ⎦ −Bx −γvz γ(E + v ∧ B)z 0

(we have swapped the order of the u α and the F αβ around to get + instead of − on the components of the r.h.s. in (4.40)). The spacelike parts of (4.40) just look like the Cartesian components of the Lorentz force F=

d(mv) = e(E + v ∧ B) dt

(4.41)

multiplied by γ. The timelike part of (4.40) just looks like the Cartesian particle energy equation obtained from (4.41) d( 1 mv 2 ) d F·v = 2 = = ev · E (4.42) dt dt multiplied by γ/c. Taking (4.41) and (4.42) together (and taking care of the e), we then have ⎡ d ⎤ ⎡ d ⎤ γ dt c ds c ⎢ dmvx ⎥ ⎢ dmvx ⎥ ⎥ ⎢ ⎥ ⎢ γ dt ⎥ ⎢ ds ⎥ ⎢ (4.43) eF αβ u β = ⎢ dmv ⎥ = ⎢ dmv ⎥ y ⎥ ⎢ γ y ⎥ ⎢ ⎦ ⎣ ds ⎦ ⎣ dt z γ dmv dt

that is, eF αβ u β =

dmvz ds

dp α ds

(4.44)

the Lorentz force law in manifestly covariant form. We now have a complete system, the Maxwell equations, equations for conservation of charge and energy momentum, and a force law and equations of motion, all in manifestly covariant form. The rest of this chapter will be devoted to looking at the implications of what we have found so far.

4.4.1 Manifestly Covariant Electrodynamics To summarize, we have: • The inhomogenous Maxwell equations: ∂α F αβ = μ0 J β

68

4 The Field Tensors

• The homogenous Maxwell equations: ∂α F˜ αβ = 0 • Charge -current conservation: ∂α J α = 0 • Energy-momentum conservation: ∂α T αβ = Jα F αβ • The equations of motion: fα = uα =

dpα ds dxα ds

• The Lorentz force law: eF αβ u β =

dp α ds

• and with the Lorentz gauge: ∂α A α = 0 • we have the Maxwell equations in terms of the four vector potential: Aα = μ0 J α

4.5 Transformation of the Fields The electromagnetic field tensor F αβ allows us to “package” the components of E and B in terms of four vectors as in (4.7). The Lorentz transformation of a four vector is simply a rotation of its spacetime coordinates as defined by (3.58) and (3.59). We can then find out how to Lorentz transform F αβ by applying the coordinate rotation to all of its’ component four vectors (see advanced problem 8) F αβ = ∂ α Aβ − ∂ β Aα = (αδ ∂δ )(βγ Aγ ) − (βγ ∂γ )(αδ Aδ ) = αδ Fδγ γβ

4.5 Transformation of the Fields

69

To obtain a simple formula to transform individual field components we would then just need to compare terms in F αβ and F αβ . This (not difficult, but rather long) operation gives E  = E  B = B (4.45) E ⊥ = γ(E ∧ B)

⊥ + vv∧E  B⊥ = γ B − c2 ⊥ where subscripts “” and “⊥” mean components parallel and perpendicular to the transformation velocity v. The inverse transform is just obtained by putting v → −v in (4.45). We have seen a simplified version of (4.45) before. For the moving charge and wire experiment in Sect. 3.2, we considered a special case: Frame 1 in which the test charge moved at velocity +v1 = v = +v and Frame 2 in which the charge was at rest (v2 = 0). Life had been made simple by choosing the charge velocity to be just the transformation velocity +v. In Frame 2 the Lorentz force law gives F2 = e(E2 + v2 ∧ B2 ) = eE 2

(4.46)

and we now know from (4.45) that the Lorentz force should be F2 = eE 2 = eE 1 = 0

(4.47)

as we chose E1 to be zero (by choosing ρ1 = 0). Also the Lorentz force law gives F⊥2 = e(E2 + v2 ∧ B2 )⊥ = eE ⊥2

(4.48)

which from (4.45) transforms to F⊥2 = eγ(E1 + v1 ∧ B1 )⊥ = eγv1 ∧ B1

(4.49)

Hence F2 = γ F1 which is just what we found in (3.32). Next, we will look at an application of (4.45). Before we do, there is one interesting point. Since the dual of F αβ also contains the electromagnetic fields in Lorentz transformable form we could have used it to find the transformation of the fields. We can find out what this would yield by applying the duality transformation E → Bc2 and B → −E to (4.45), and it is easy to verify that we obtain the same transformations for the fields.

4.6 Field from a Moving Point Charge In nonrelativistic situations (i.e., under Galilean frame transformation) we know that the field from a positive point charge just points radially outwards, and this was used

70 Fig. 4.1 Charge q is at rest at the origin of the S  frame and is moving with velocity v xˆ 1 w.r.t. frame S. The origins of the S and S  frames coincide at t = 0

4 The Field Tensors

x2

x’2

S P r

v

a vt

q

x3

S’

x1

x’1

x’

3

in Chap. 1 to discuss Gauss’ law. Now that we know how to transform the fields we can look at what happens to the fields from charges moving at relativistic speeds. Lets consider a positive charge moving past us at speed v as shown in Fig. 4.1. If we are in the rest frame of the charge (the S  frame), there is no magnetic field and the electric field is radial giving (from Chap. 1): B = 0 q r E = 4π 3 0 r

(4.50)

We now transform to the S frame by moving in −v xˆ 1 so that in the S frame the charge appears to be moving past us in the x1 direction with speed +v. The inverse transform gives us (with B = 0 and v = v ) the fields in the S frame E  = E  B = 0 E ⊥ = γ E ⊥ B⊥ = cγ2 (v ∧ E )⊥ =

(4.51) v∧E c2

We will now obtain the fields at some location P = (ct, x1 , a, 0) in the S frame (we can always choose a convenient orientation of x2 , x3 such that x3 = 0). To do this we need to express E in (4.51) as a function of ct, x1 , x2 , x3 instead of x1 , x2 , x3 . In S  the charge is at rest and so the fields are independent of t  , we just then need to transform the spacelike coordinates using the Lorentz transformation. At point P this gives x1 = γ(x1 − vt) x2 = x2 = a (4.52) x3 = x3 = 0

4.6 Field from a Moving Point Charge

71

E2

E1

γ >1 γ =1

x-vt

γ >1

x-vt

1

1

γ =1 Fig. 4.2 Sketch of E 1 and E 2 at point P versus x1 /γ = x1 − vt Fig. 4.3 Sketch of the E field around a charge at rest, and the E field around a charge moving at relative velocity v, in the x1 , x2 plane

E E’

v

Substituting for x1 , x2 , x3 in (4.51) then gives the electric field at P from the moving point charge: E1 =

E 1 =

E2 =

γ E 2 =

E3 =

γ E 3 =

x1 q q γ(x1 −vt) = 4π 3 4π 0 (x 2 +x 2 +x 2 ) 23 0 (γ 2 (x −vt)2 +a 2 ) 2 1 1 2 3 γx1 q q γa = 4π 0 2 3 4π 0 (x 2 +x 2 +x 2 ) 23 (γ (x1 −vt)2 +a 2 ) 2 1 2 3

(4.53)

0

We can always obtain the magnetic field components at P from E via (4.51). To understand what is happening at relativistic speeds, lets sketch the electric field. Figure 4.2 shows the electric field components in the S frame E 1 and E 2 at point P, plotted versus x1 /γ = x1 − vt. As the relative speed of the charge becomes relativistic, and γ > 1, the x1 /γ axis Lorentz contracts. The magnitude of E 1 doesn’t change (as E 1 = E 1 ) instead the E 1 (x1 /γ) profile becomes compressed. The magnitude of E 2 = γ E 2 increases, and again, the profile is compressed as γ > 1. The electric field occupies a more disc like region as γ is increased. This is sketched in Fig. 4.3 which shows the E field vectors in the x1 , x2 plane. We expect the moving charge to have an accompanying magnetic field whether or not it is relativistic as it is carrying a current. In the nonrelativistic limit γ → 1, (4.51)

72

4 The Field Tensors

becomes: B=

v∧E μ0 dI ∧ r = 2 c 4πr 3

(4.54)

where dI = qv is the current carried by the charge in its direction of relative motion. This it just the Biot Savart law1 in magnetostatics. Generally, the spatial distribution of the B field just follows that of E from (4.51), so that when the charge moves at relativistic speeds, B will also be compressed into a more disc like region. As the charge moves closer to the speed of light v → c and B→

vˆ ∧ E c

(4.56)

We can compare this to the free space electromagnetic wave solution to Maxwell’s equations (1.4). In Sect. 2.3 we found that if we take a plane wave solution of the form E, B ∼ ei(ωt−k·r) (4.57) and substitute into Maxwell III we obtain (2.51): B=

kˆ ∧ E c

(4.58)

which is just (4.56), the magnetic field from the (strongly) relativistic charge. So as v → c, and both E and B become “compressed” into a disc around the relativistic charge, the fields start to look like a pulse of light centered on the charge.

4.7 Retarded Potential In the previous section we found that Lorentz contraction and time dilation gave us a good idea of what happens to the fields around a single point charge as it moves at relativistic speeds. We will now obtain a general solution to Maxwell’s equations by finding the fields from a collection of charges described in terms of the (frame dependent) charge and current density. To make life simple we can work in terms of the scalar and vector potentials rather than the field transformations directly. Everything needed for the calculation then forms four vectors; the scalar and vector potentials Aα , the charge and current densities J α , and spacetime x α , these then are straightforward to transform using the Lorentz transformation. The problem is posed in Fig. 4.4. We will first consider the scalar and vector potential due to the charge d Q contained in a small (elemental) volume element; 1 Strictly speaking the total magnetic field from a closed current loop C in static situations is obtained by adding up the contributions due to all the current elements dI along the loop:  μ0 dI ∧ r (4.55) B= 4πr 3 C

4.7 Retarded Potential

73

Fig. 4.4 A volume element is located at rest at the origin of the S  frame which is moving at +v in the x1 direction w.r.t. the S frame. In the S  frame the volume element contains charge ρ d x1 d x2 d x3 . The origins of the two frames are coincident at t = t  = 0

x2

S r

x3

P

x’2

x’3

S’ r’

v x1

ρ ’dx’dx’dx’ 1 2 3

x’1

since all our equations are still linear under the Lorentz transformation we can find the scalar and vector potential from charges distributed over a larger volume later by doing a volume integral. To simplify the algebra, the volume element containing d Q is located at the origin of the S  frame, and in this frame the charges are at rest. The S  frame then moves at speed +v along the x1 direction w.r.t. the S frame, and again for simplicity we will arrange for the origins of both frames to coincide at t = t  = 0. We now observe the scalar and vector potential d Aα at some point P due to the charges when the frames coincided. Since disturbances in the electromagnetic fields travel at speed c in all frames, in the S frame the “signal” takes a time t=

r c

(4.59)

to reach P from the S origin. By the time it has done so, the charges have moved; they are now located at the S  origin which is distance r from P. We can find d Aα from d Aα in the S  frame where the charges are at rest, this is cd A0 = dφ = d A1,2,3 = 0

dQ 4π 0 r 

=

ρ d x1 d x2 d x3 4π 0 r 

(4.60)

αβ

The inverse Lorentz transform (i) (since S is moving in −v xˆ 1 w.r.t. S  ) of (4.60) (see Sect. 3.4) is ⎡

γ − vc γ ⎢ + vc γ −γ αβ Aα = (i) Aβ = ⎢ ⎣ 0 0 0 0 so that

0 0 −1 0

⎤ ⎡ dφ 0 c ⎢ 0 ⎥ ⎥⎢ 0 0 ⎦⎣ 0 −1 0

⎡ dφ ⎤  ⎤ γ dφc c  ⎥ ⎢ γ vdφ ⎥ ⎢ d A1 ⎥ ⎥ ⎥ = ⎢ c2 ⎥ = ⎢ ⎦ ⎣ 0 ⎦ ⎣ d A2 ⎦ (4.61) d A3 0 ⎤

dφ = γdφ dA = xˆ 1 γ cv2 dφ

⎡

(4.62)

74

4 The Field Tensors

The four-current J α will inverse transform in just the same way, from charges at rest in the S  frame to the charges moving in the S frame2 ρ = γρ J = xˆ γvρ = xˆ vρ

(4.63)

Next, we can write dφ (d x1 , d x2 , d x3 , r  ) as a function of coordinates in the S frame by using the Lorentz transformation of spacetime coordinates i.e., (4.52) to get expressions for the volume element d x1 , d x2 , d x3 and for r  . If we substitute in the time delay for t (4.59) then  vr  x1 = γ x1 − c

(4.64)

r  = (x12 + x22 + x32 ) 2  21 

2 = γ 2 x1 − vrc + x22 + x32

(4.65)

This gives an expression for r  1

since r 2 = x12 + x22 + x32 this can be written as  vx1  r = γ r − c

(4.66)

The volume element as seen in S is obtained by implicit differentiation of (4.52)  d x1 = γd x1 1 − d x2 = d x2 d x3 = d x3

v dr c d x1



= γd x1 1 −

v x1 c r

(4.67)

where we used the change in r across the volume element, dr/d x1 , obtained by differentiating (4.68) r 2 = x12 + x22 + x32 w.r.t. x1 to give

dr x1 = d x1 r

(4.69)

Putting all this together, in the S frame we then have   dφ = γ

2 This



ρ

d x1 d x2 d x3 4π 0 r 

=γ

ρ γ

d x1 γ 1 − vc xr1 d x2 d x3

4π 0 γ r − vxc 1

just yields what we found for the moving charge and wire experiment in Sect. 3.2.

(4.70)

4.7 Retarded Potential

75

Fig. 4.5 Volume element d V  and point P located w.r.t. a single fixed origin O

dV’

r-r’ P

r’ r O which is just dφ =

ρd x1 d x2 d x3 4π 0 r

(4.71)

This is the potential at P that we would expect from electrostatics if charge d Q were located at distance r instead of r  ; that is, as if d Q were still at the origin of the S frame. So to get a solution of Maxwell’s equations, we have two approaches. One is to “do the relativistic treatment” and Lorentz transform four vectors as above. The other is to insist that charge d Q is frame invariant and always has a potential of the form (4.60) and (4.71), and also that information about d Q travels at frame invariant speed c. Information about the location of the charges takes a time t = r/c to reach point P, so we see a retarded potential consistent with the electrostatics of the charges at their previous location a time t = r/c earlier. The vector potential in the S frame is then (from 4.62 and 4.63) dA = xˆ 1

v μ0 Jd x1 d x2 d x3 dφ = 2 c 4πr

(4.72)

which, with B = ∇ ∧ A will again give the Biot Savart law, consistent with the magnetostatics of the current due to the moving charges at their previous location a time t = r/c earlier. The algebra was simplified by choosing the charges to be at the origin of the S  frame and having the two frames coincident at t = 0. A more general arrangement is shown in Fig. 4.5 where all positions are known relative to a fixed origin. A volume element d V  is at position r , point P is then a distance | r − r | from the charges in d V  . It takes light a time tl =| r − r | /c to propagate from the charges to point P, so the potentials seen at P at time t are given by the charge density in d V  at time t − tl . Integrating over all space, the potentials at P in terms of the charge and current densities in the S frame are:    ρ r , t − |r−r | c 1 dV  (4.73) φ(r, t) = 4π 0 V | r − r | and

76

4 The Field Tensors

μ0 A(r, t) = 4π

  J r , t − V

|r−r | c

| r − r |

 dV 

(4.74)

Why does this work? The whole edifice of the Lorentz transformation and four vectors that we have developed amounts to three things: the speed of light c, the laws of physics, and certain quantities, like charge, are the same in all inertial frames. Hence we can use Gauss’ law of electrostatics, the invariance of charge, and the invariance of the speed of light, to obtain the (retarded) potential in any given frame. Provided that the resulting integrals for the retarded potentials are tractable, that’s all there is to it.

Suggested Texts

Suggested Texts This is not an exhaustive list, they are the books that author has found the most useful in the context of electrodynamics, so if you enjoyed this book, read on. • The Feynman Lectures on Physics, Vol II, R. P. Feynman, R. B. Leighton, M. Sands, Addison-Wesley, 1975: A lively and insightful journey through electrodynamics, best read in one go. Training in the formalism should be sought elsewhere. • Classical Electrodynamics, Second Edition, J. D. Jackson, Wiley, 1975: A must for every working physicists bookshelves. Thorough and rigorous treatment of electromagnetism from first principles to electrodynamics. Includes all the material that I left out. • The six core theories of modern physics, C. F. Stevens, MIT press, 1995: Forgotten your entire physics degree? This is a compact summary. • Mathematical Methods in the Physical Sciences Third Edition, M. L. Boas, Wiley, 1983: A nice introduction to tensors, and to vector calculus. • The Classical Theory of Fields, Fourth Edition, Course of Theoretical Physics Vol. 2, L. D. Landau and E. M. Lifshitz, Pergamon, 1983: Extends the electrodynamics of flat spacetime treated here to curved spacetime, i.e., general relativity. • Classical Field Theory, D. E. Sopher Dover, 2008: Compact yet comprehensive treatment of classical field theory of continuous media, use it to set electrodynamics in the context of the field theory of fluids and solids.

© Springer-Verlag GmbH Germany, part of Springer Nature 2021 S. Chapman, Core Electrodynamics, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-030-66818-1

77

Appendix A

Revision Problems

A.1 Static Magnetic Fields A straight wire carrying a steady current I has radius a as shown in Figs. A.1 and A.2. (a) Inside the wire, i.e., for r ≤ a calculate the current enclosed inside a circle of radius r as shown in Figs. A.1. Use the integral form of ∇ · B = 0 to show that B is always perpendicular to r. From Ampère’s law (the integralform of ∇ ∧ B = μ0 J) calculate B(r ) and hence using Stokes theorem (i.e., from B · dl) calculate ∇ ∧ B. Calculate ∇ ∧ B directly. In a cylindrical coordinate system you can use the identity ∇ ∧ B = (∇ ∧ B)z zˆ =

1 ∂ (r Bθ )ˆz r ∂r (A.1)

ˆ when B = Bθ θ. (b) Outside the wire i.e., for r > a use Ampère’s law to calculate B(r ) using curve C1 shown in Fig. A.2 (i.e., a circle of radius r centred on the wire). ∇ ∧B Use this expression for B to calculate B · dl around curve C2 . Calculate  directly at a point within C2 and compare it with your value for B · dl. Sketch the variation of the magnitude of the current density J(r ) as a function of r and use Maxwell’s IVth equation for steady fields ∇ ∧ B = μ0 J to sketch the variation of ∇ ∧ B as a function of r .

© Springer-Verlag GmbH Germany, part of Springer Nature 2021 S. Chapman, Core Electrodynamics, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-030-66818-1

79

80

Appendix A: Revision Problems

Fig. A.1 Problem 1a: A straight wire carrying a steady current I

I

r a Fig. A.2 Problem 1b: A straight wire carrying a steady current I

I

C 1

C1 C 2

A.2 Static Electric Fields A long bar of radius a carries a staticcharge density ρl per unit length which is spread uniformly through the bar. What is E · dl around any closed path? What does this tell you about the direction of the electric field? Use Gauss’ theorem in integral form on cylindrical surfaces centred on the bar to calculate E(r) inside the bar (r < a) and outside the bar (r > a). Calculate ∇ · E directly for both cases. In a cylindrical coordinate system you can use the identity ∇ ·E= when E = Er rˆ .

1 ∂ (r Er ) r ∂r

(A.2)

Appendix A: Revision Problems

81

A.3 Conservation and Poynting’s Theorem (a) Consider a gas composed of identical particles of number density n(r) and velocity v(r) at position r. Particles are neither created nor destroyed. (i) What is the flux of particles across surface S (give integral form)? (ii) If S encloses a volume V show that the gas obeys a conservation equation ∇ · (nv) = − ∂n ∂t hence obtain an expression for conservation of charge in terms of charge density ρ and current density J. (b) (i) Ampère’s law for steady currents is  B · dl = μ0 I c

Write this in differential form. (ii) Using conservation of charge obtain Maxwell IV for time dependent fields: ∇ ∧ B = μ0 J + μ0 0

∂E ∂t

(c) In general (ρ f = 0, Jf = 0) Poynting’s theorem in media is ∂ ∇ · (E ∧ H) = − ∂t



 1 [E · D + B · H] − Jf · E 2

(A.3)

Given that the energy density U= starting with

∂U ∂t

1 [E · D + B · H] 2

(A.4)

use Maxwell’s equations in media and the identity ∇ · (E ∧ H) = H · (∇ ∧ E) − E · (∇ ∧ H)

(A.5)

to derive Poynting’s theorem.

A.4 The Wave Equation: Linearity and Dispersion A general form for the wave equation is: ∂2ψ = v2 ∇ 2 ψ ∂t 2

(A.6)

82

Appendix A: Revision Problems

Show that ψ1 = ψ01 ei(ω1 t−k1 x)

(A.7)

is a solution of this equation. What is the phase speed and direction of propagation of this wave? If the wave (A.8) ψ2 = ψ02 ei(ω2 t−k2 x) is also a solution of the wave equation, show that the wave resulting from the superposition of these waves (A.9) ψ3 = ψ1 + ψ2 is also a solution. In a dispersive medium the wave equation may look like ∂2ψ = v 2 (ω)∇ 2 ψ ∂t 2

(A.10)

Show that ψ3 is a solution. Give an example of a PDE for which ψ1 and ψ2 are solutions, but ψ3 is not.

A.5 Free Space EM Waves I The E and B fields of a plane EM wave have the form: E = E0 ei(ωt−k.r) B = B0 ei(ωt−k.r) (a) Show that Maxwell’s equations in free space can be rewritten as: k · E0 = 0 k ∧ E0 = ωB0

k · B0 = 0 k ∧ B0 = −μ0 0 ωE0

(b) Show that E, B and k (in that order) form a right-handed orthogonal system. (c) If, with an electric antenna, it is established that E only has a y component, what are the possible directions of propagation of the wave, and the directions of the associated B components. (d) Describe the polarization of this wave. Write down E and B for a wave which is (i) linearly polarized and (ii) circularly polarized.

Appendix A: Revision Problems

83

A.6 Free Space EM Waves II The magnetic field of a uniform plane wave in free space is given by: B(r, t) = 10−6 [1, 2, Bz ]ei(ωt+3x−y−z) Determine: (a) (b) (c) (d)

ˆ wavelength λ, angular frequency ω. The direction of propagation k, The z component of B. The electric field E associated with B. The Poynting vector S.

A.7 EM Waves in a Dielectric In a ‘perfect’ dielectric there are no free currents or charges (ρ f = 0, Jf = 0). Use Maxwell’s equations in media and B = μμ0 H, D = 0 E to derive wave equations for E and B. (Use the identity ∇ ∧ (∇ ∧ A) = ∇(∇ · A) − ∇ 2 A). What is the phase speed v = ωk of the waves in SI units? Show that 1 (A.11) B = kˆ ∧ E v

A.8 Dielectrics and Polarization A parallel plate capacitor with plates of area A, separated by a distance d has a voltage V applied across the plates. Initially there is a vacuum between the plates. A charge of magnitude Q = C V (where C is the capacitance) corresponding to a surface charge density σ f has built up on each plate (so that there is charge +Q on one plate and −Q on the other). (a) Use Gauss’ law to calculate the electric field between the plates. What is the capacitance? (b) The region between the plates is now filled with a dielectric, and the capacitor is again charged up by placing charge of magnitude Q on each plate. A surface charge density σ p is induced on the dielectric. What is the polarization (dipole moment/unit volume) in terms of σ p ? NB: we define polarization as P = np where p is the atomic dipole moment. (c) If the material is linear, P = ( − 1)0 E. What is the electric field inside the dielectric, and the capacitance?

84

Appendix A: Revision Problems

A.9 EM Waves in a Conductor: Skin Depth In an ideal conductor at low frequencies we can assume that there are no free charges (ρ f = 0) and that the free current density obeys a simple Ohm’s law Jf = σE. Use Maxwell’s equations and the identity ∇∧(∇ ∧ A) = ∇(∇.A) − ∇ 2 A to show that the “wave equation” in the conductor is ∇ 2 E = μμ0 σ

1 ∂2E ∂E + 2 2 ∂t v ∂t

(A.12)

Derive the dispersion relation (that is, k as a function of ω) in the low frequency limit (terms O(ω 2 )  terms O(ω)) by examining the properties of plane wave solutions. Hence show that the skin depth (distance waves can penetrate into the conductor before they are appreciably attenuated) is  δ=

2 μμ0 σω

 21 (A.13)

What is the significance of this result for (i) radio propagation (ii) microwave ovens (iii) power transmission?

A.10 Cavity Resonator Find the Poynting flux and the radiation pressure for a standing plane electromagnetic wave (that is, two travelling waves of equal amplitude propagating in opposite directions. Use this to calculate the cycle average Poynting flux and radiation pressure.

Appendix B

Solutions to Revision Problems

B.1 Static Magnetic Fields (a) The current density | J |=

I πa 2

(B.1)

is uniform within the wire. Current enclosed within circle, radius r :  Ir =

J · dS =

I Ir2 2 × πr = πa 2 a2

(B.2)



Then since

B · dS = 0

(B.3)

over a surface defined by a cylinder of radius r , B only has components parallel to the surface i.e., perpendicular to r. From Ampère  B · dl = μ0 I

(B.4)

and using cylindrical symmetry to infer that | B | is independent of r :  B · dl = B × 2πr = μ0 that is, B=

Ir2 a2

μ0 I r 2πa 2

© Springer-Verlag GmbH Germany, part of Springer Nature 2021 S. Chapman, Core Electrodynamics, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-030-66818-1

(B.5)

(B.6)

85

86

Appendix B: Solutions to Revision Problems

The direction of B is obtained from the right hand rule, or ∇ ∧ B = μ0 J. Using Stokes:   B · dl = is μ0

∇ ∧ B · dS

(B.7)

Ir2 =| ∇ ∧ B | πr 2 a2

giving | ∇ ∧ B |= directed in +J. Directly: ∇ ∧ B = zˆ

(B.8)

μ0 I πa 2

(B.9)

1 ∂ μ0 I μ0 I r = zˆ 2 r r ∂r 2πa 2 πa

(B.10)

(b) Outside the total current enclosed by C1 is I then  B · dl = μ0 I

(B.11)

C1

B2πr = μ0 I

(B.12)

i.e., B = μ0 I /2πr . Around C2  B · dl = 0

(B.13)

C2

as no current is enclosed. Evaluating this integral the contributions from the straight line segments of C2 along rˆ are zero and perpendicular to rˆ along the curved segments (at constant r1 and r2 ) give  B · dl = C1

μ0 I × r1 2πr1

 dθ − 1

μ0 I × r2 2πr2

 dθ = 0

(B.14)

2

where the integral over θ just gives the angle subtended by curve C2 at r = 0. Directly

1 ∂ r μ0 I =0 (B.15) ∇ ∧ B = zˆ r ∂r 2πr

Appendix B: Solutions to Revision Problems

87

B.2 Static Electric Fields 

Static hence

E · dl = 0

(B.16)

Choosing a circular path, circle centre at the centre of the bar, cylindrical coordinates z, r, θ so that r · dl = 0 we have either (i) E asymmetric (i.e., with a θ dependence) and E · dl at some θ1 cancels that at some θ2 , or (ii) E has cylindrical symmetry and hence must be perpendicular to dl everywhere. Since E due to a point charge is symmetric and the principle of superposition holds we have case (ii). Thus this also admits a constant component along zˆ . To use Gauss in integral form over a cylinder of radius r < a, length L we have charge enclosed Q = ρV where ρ=

ρl ρl L = 2 πa L πa 2

(B.17)

Q=

ρl × πr 2 L πa 2

(B.18)

i.e.,

The only contribution to E is on the curved surface of the cylinder on which E is constant.  1 (B.19) E · dS = E2πr L = ρl r 2 La 2 0 hence E= Then

1 ∂ ∇ ·E= r ∂r



ρl r rˆ 20 a 2 π

(B.20)

ρl r 2 ρl = 20 a 2 π 0 a 2 π

(B.21)

For cylinder radius r > a the enclosed Q = ρl L so that  E · dS = E2πr L = then E= and

ρl L 0

ρl rˆ 2πr 0

1 ∂ ∇ ·E= r ∂r



ρl r =0 2πr 0

(B.22)

(B.23)

(B.24)

88

Appendix B: Solutions to Revision Problems

B.3 Conservation and Poynting’s Theorem (a) (i) Number through dS in time dt is nv · dSdt since volume containing particles that cross surface in dt is v · dSdt. Then number crossing S in dt is  nv · dSdt

(B.25)

S

1 S

Flux = No/sec/unit area = (ii) Number enclosed in S is

 nv · dS

(B.26)

S

 N=

nd V

(B.27)

V

so number crossing S in unit time is ∂N ∂ = ∂t ∂t



 nd V = V

V

∂n dV ∂t

(B.28)

Now using our expression for the flux across S, the number flowing out of V in time dt is   ∂n d V dt (B.29) nv · dSdt = − S V ∂t From Divergence theorem then −

∂n = ∇ · (nv) ∂t

(B.30)

For charges we have ρ = nq, J = nqv giving −

∂ρ =∇ ·J ∂t

(B.31)

(b) Given in the text. (c) Poynting’s theorem in media: from the energy density U= we have

1 [E.D + B.H] 2



1 ∂D ∂E ∂H ∂B ∂U = E· +D· +B· +H· ∂t 2 ∂t ∂t ∂t ∂t

(B.32)

(B.33)

Appendix B: Solutions to Revision Problems

89

then since D = 0 E, B = μμ0 H we have ∂U ∂D ∂B =E· +H· ∂t ∂t ∂t

(B.34)

which is ∂U = E · (∇ ∧ H − J f ) + H · (−∇ ∧ E) = −J f · E − H · ∇ ∧ E + E · ∇ ∧ H ∂t

(B.35)

Using the identity

we obtain

∇.(E ∧ H) = H.(∇ ∧ E) − E.(∇ ∧ H)

(B.36)

∂U = −J f · E − ∇.(E ∧ H) ∂t

(B.37)

B.4 The Wave Equation: Linearity and Dispersion Substitute ψ1 into the wave equation: (iω1 )2 ψ1 = v 2 (−ik1 )2 ψ1

(B.38)

I.e., if ω1 = vk1 then ψ1 is a solution. v is the phase speed, that is, the speed of a point of constant phase. Consider some point x at phase φ = ω1 t − k1 x at time t. At time t + dt this point has moved to x + d x where d x = vdt. So ω1 (t + dt) − k1 (x + d x) = φ

(B.39)

Hence ωdt = kd x and the point of constant phase moves at v=

ω1 dx = =v dt k1

(B.40)

Now ψ2 is a solution. Substitute ψ3 = ψ1 + ψ2 into the wave equation: ∂2 ∂2 ∂2 ψ3 = 2 ψ1 + 2 ψ2 = v 2 ∇ 2 ψ3 = v 2 [∇ 2 ψ1 + ∇ 2 ψ2 ] 2 ∂t ∂t ∂t and since

∂2 ψ1,2 = v 2 ∇ 2 ψ1,2 ∂t 2

(B.41)

(B.42)

ψ3 is also a solution. This will hold for any linear PDE, hence ψ3 can be shown to be a solution of the dispersive wave equation by the same method.

90

Appendix B: Solutions to Revision Problems

Require nonlinear PDE (with terms in ψ n , (∇ψ)n etc.) to break the principle of superposition.

B.5 Free Space EM Waves I (a) For plane waves it can be shown by calculating the derivatives in any orthogonal coordinate system (e.g., cartesian) that ∇· ≡ −ik · ∇∧ ≡ −ik ∧ ∂ ≡ iω ∂t Then substituting the wave solutions into the free space Maxwell equations: ∇ ·E ⇒ ∇ ·B ⇒ ∂ ∇ ∧E=− B ⇒ ∂t ∂ ∇ ∧ B = μ0 0 E ⇒ ∂t

k · E0 = 0 k · B0 = 0 k ∧ E0 = ωB0 k ∧ B0 = −ωμ0 0 B0

(b) Use result (a): k · E0 = 0 k · B0 = 0 ⇒ E, B ⊥ k 1 B = kˆ ∧ E then ⇒ handedness c (c) k · E0 = 0 so k = xˆ k x + zˆ k z , then since B = 1c kˆ ∧ E, B = xˆ Bx + zˆ Bz . (d) Linearly polarized. (i) Linear polarization: E = E0 ei(ωt−k.r) B = B0 ei(ωt−k.r) plane of polarization is E, k (ii) Circular polarization: π

E = E 0 xˆ ei(ωt−kz) + E 0 yˆ ei(ωt−kz± 2 ) with B = 1c kˆ ∧ E.

(B.43)

Appendix B: Solutions to Revision Problems

91

B.6 Free Space EM Waves II √ (a) k = −3ˆx + yˆ + zˆ , | k |= 11 m −1 so λ = 2π/k = 1.89m. ω = ck = 9.93 × 108 s −1 . (b) Use k · B = 0, k from (a) giving Bz = 10−6 T . (c) Use k ∧ B0 = −ωμ0 0 E0 with above k, B to give 300 E0 = √ (ˆx − 4ˆy + 7ˆz)V m −1 11 E = E0 ei(ωt+3x−y−z) (d) S=

1 ˆ m −2 E ∧ B = 432 cos2 (ωt − k · r)kW μ0

(B.44)

B.7 EM Waves in a Dielectric Using ∇ ∧H=

∂D ∂t

(B.45)

we have using ∇ · H = 0: ∇ ∧ (∇ ∧ H) = −∇ 2 H = ∇ ∧ Then since Maxwell III ∇ ∧E=− can be written as

∂B ∂t

  ∂H ∇ ∧ D = 0 μμ0 − ∂t

we have ∇ 2 H = 0 μμ0

∂D ∂t

(B.46)

(B.47)

(B.48)

∂2H ∂t 2

(B.49)

∂2D ∂t 2

(B.50)

similarly taking ∇ ∧ (∇ ∧ D) yields ∇ 2 D = 0 μμ0

92

Appendix B: Solutions to Revision Problems

Using B = μμ0 H and D = 0 E gives wave equations in terms of E and B: ∇ 2 E = 0 μμ0

∂2E ∂t 2

(B.51)

∇ 2 B = 0 μμ0

∂2B ∂t 2

(B.52)

which by inspection have wave solutions with phase speed ω 1 = v=√ k (0 μμ0 )

(B.53)

As in free space plane waves, Maxwell III yields: B=

1 kˆ k ∧ E = kˆ ∧ E ω v

(B.54)

B.8 Dielectrics and Polarization Far from the edges of the plates, assume that they are approximately infinite in extent. Then E is directed normal to the plates. (a) Gauss in integral form for the surface becomes: 

1 E · dS = 0

Then E=

 σf dS

(B.55)

σf Q = 0 0 A

(B.56)

Since Q = C V and V = Ed (from ∇ ∧ E = 0 and E directed normal to the plates) we have Q Q 0 A C= = = (B.57) V Ed d (b) From slab geometry E is uniform in the capacitor, hence no polarization charge is induced within the volume of dielectric. From the definition of polarization: 



 P · dS = −

σpd S =

Pd S

(B.58)

as P is then also normal to the (approximately infinite) plates. This gives P = −σ p in units of Coulomb m −2 .

Appendix B: Solutions to Revision Problems

93

(c) In the presence of the dielectric  E · dS =

1 0

 (σ f + σ p )d S

(B.59)

which in this slab geometry is E=

σ f + σp σf P = − 0 0 0

(B.60)

then since for linear media P = ( − 1)0 E we have E=

σf 0

(B.61)

This again yields the capacitance from Q = C V and V = Ed: C=

Q 0 A Q = = V Ed d

(B.62)

B.9 EM Waves in a Conductor: Skin Depth In media Maxwell IV is ∇ ∧ H = Jf + which can be written as

∂D ∂t

1 ∂E ∇ ∧ B = Jf + 0 μμ0 ∂t

(B.63)

(B.64)

using Jf = σE this becomes: ∇ ∧ B = μμ0 σE +

1 ∂E v 2 ∂t

(B.65)

We take ∂/∂t of this equation, and use ∇ ∧ (∇ ∧ E) = ∇(∇ · E) − ∇ 2 E

(B.66)

with ∇ · E = 0 since ∇ · D = ρ f = 0 to obtain a modified wave equation: ∇ 2 E = μμ0 σ

1 ∂2E ∂E + 2 2 ∂t v ∂t

(B.67)

94

Appendix B: Solutions to Revision Problems

Assume plane wave solutions of the form E = E0 ei(ωt−k·r)

(B.68)

then substitute in to (B.67) to obtain the dispersion relation: k2 =

w2 − μμ0 σiω v2

(B.69)

The two terms on the r.h.s. are due to the displacement and conduction currents respectively. Taking the low frequency limit of this expression is equivalent to neglecting the displacement currents in comparison to the conduction currents in the conductor and yields: (B.70) k 2 = −μμ0 σiω Clearly k has real and imaginary parts. Writing k = β ± iα gives plane wave solutions of the form: ˆ ˆ (B.71) E = E0 ei(ωt−β k·r) e±αk·r which corresponds to waves that grow (+α) or decay (−α) as they propagate in +r. For a semi- infinite slab of conductor, the waves must have finite amplitude as r → ∞ which excludes growing solutions. The waves then decay and a measure of the distance that they can penetrate into the conductor before they are appreciably attenuated (the skin depth δ) is that required for the amplitude to fall by a factor 1/e: δ=

1 α

(B.72)

α is then obtained by equating coefficients, with k 2 = β 2 − α2 − 2iαβ

(B.73)

giving α = β, 2αβ = μμ0 σω and  δ=

2 μμ0 σω

(B.74)

Implication: lower frequency waves propagate further into conductors. Some examples are: • The earth’s ionosphere (a conductor) will not propagate radio waves at all frequencies, hence longwave can be bounced off the ionosphere to be received at points on the earth’s surface below the horizon. • Re-entering astronauts lose radio contact with the ground as their spacecraft is engulfed in plasma.

Appendix B: Solutions to Revision Problems

95

• A few mm of conductor are sufficient to attenuate microwaves so that it is safe to use microwave ovens. • household power (50 Hz) can be carried over short distances by copper with limited power loss, whereas computer network signals require co axial cable or twisted pair ethernet (computer clock speeds are GHz). • your TV uses copper wire to carry the power, and coaxial cable to carry the TV signal. • over long distances copper wire would dissipate too much power to be efficient. Instead, high voltage, low current transmission lines are used.

B.10 Cavity Resonator The Poynting flux and radiation pressure for a single wave are given in (2.3). The standing wave is just obtained by summing for the two waves of equal amplitude propagating in opposite directions. For this standing wave the cycle average is then zero.

Appendix C

Some Advanced Problems

C.1 Maxwell Stress Tensor Momentum flux conservation in an ideal Magnetohydrodynamic (MHD) plasma gives the following: Du = −∇ · P + J ∧ B (C.1) Dt For equilibrium the l.h.s. of this equation must be zero. 1. Show that at low frequencies J ∧ B can be written as the divergence of a stress tensor TM . 2. For a gyrotropic plasma, the pressure tensor P can be written as: Pi j = P⊥ δi j + (P − P⊥ )

Bi B j B2

(C.2)

where the local magnetic field direction is zˆ . Use the equilibrium condition ∇ · (TM − P) = 0 to obtain the following stability criteria: B2 =K 2μ0 B2 + K P − P⊥ = μ0

P⊥ +

(C.3) (C.4)

where K , K  are constants.

© Springer-Verlag GmbH Germany, part of Springer Nature 2021 S. Chapman, Core Electrodynamics, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-030-66818-1

97

98

Appendix C: Some Advanced Problems

C.2 Liouville and Vlasov Theorems: A Conservation Equation for Phase Space The Vlasov equation describes the time evolution of the phase space probability density f (x, v, t) in 6 dimensional phase space x, v, and is: Df ∂f = + v · ∇ f + a · ∇v f = 0 Dt ∂t

(C.5)

(a) Show (by using Taylor expansion) that this corresponds to the change in f along a particle orbit (Liouville’s theorem). (b) Show that for a charged particle moving under Lorentz force law that the Liouville Theorem is equivalent to a conservation equation in 6 dimensional phase space (hint, treat x and v as independent phase space coordinates).

C.3 Newton’s Laws and the Wave Equation Under Galilean Transformation Determine whether or not the form of the following equations is invariant under the Galilean transformation x = x − vt, t  = t: (a) Newton’s equations of motion for the ith particle in an ensemble where the force between the ith and jth particle is F(xi − x j ). mi

dvi = F(xi − x j ) dt j

(C.6)

(b) The wave equation for some field Ψ (x, t) 

1 ∂ ∇ − 2 2 c ∂t 2

 Ψ =0

(C.7)

For (b) write down a solution for Ψ (x, t) and Ψ (x , t  ) and show that it is consistent with Doppler shift.

C.4 Transformation of the Fields Two charged straight wires carrying a charge density ρ per unit volume are a distance l apart and are at rest. The charges are not free to move within the wires. Calculate the electric field at each wire due to the other from Gauss’ law and the force acting

Appendix C: Some Advanced Problems Fig. C.1 A pair of wires

99

+

+ +

+ +

+

+ +

+ +

+

+

+

+

+

+ +

+

+ +

+

+ +

u

+

on each charge from the Lorentz force law. Give an expression for the total electric field. You now move past the wires at a constant velocity v as shown in Fig. C.1. In this moving frame the wires carry a current. Calculate this current and the magnetic field due to each wire from Ampere’s law. Calculate the force on each wire due to this magnetic field from the Lorentz force law. The total Lorentz force must be the same in both frames of reference. What has happened to the electric field?

C.5 Metric for Flat Spacetime 1 (a) If gαβ is the metric for flat spacetime show that gαγ g γβ = δαβ where δαβ is a 4 × 4 rank 2 tensor with zero off axis terms and trace 1. (b) The Lorentz transformation of covariant four vector xα and contravariant four vector x α can be written as xα = Λαβ x β x α = Λαβ xβ show that the metric for flat spacetime gαβ transforms a contravariant vector into a covariant vector in all frames under Lorentz transformation, i.e., show gαβ x β = xα (c) Show that Fαβ = gαγ F γδ gδβ .

(C.8)

100

Appendix C: Some Advanced Problems

C.6 Length of the EM Field Tensor in Spacetime The tensor scalar product is

T : T = Tαβ T βα

(C.9)

where we sum over both β and α. Show that the tensor scalar product of the electromagnetic field tensor is Fαβ F

βα

 =2

E2 − B2 c2

 (C.10)

And that this is zero for free space electromagnetic waves (no charges present). Hence show that F˜ αβ F˜ βα = −c2 Fαβ F βα (C.11) where F˜ αβ is the dual of Fαβ

C.7 Alternative Form for the Maxwell Homogenous Equations Show that Maxwell’s homogenous equations can be written as ∂ α F βγ + ∂ β F γα + ∂ γ F αβ = 0

(C.12)

C.8 Lorentz Transformation of the EM Field Tensor (a) Show that the Lorentz transformation of the field tensor is

(b) The transpose rule is:

F αβ = ∂ α Aβ − ∂ β Aα

(C.13)

a αβ.. x γδ.. = x ...δγ a ..βα

(C.14)

for both contravariant and covariant tensors. Use the transpose rule and the Lorentz transformation of four vectors to show that the Lorentz transformation of the electromagnetic field tensor is:  = Λαδ Fδγ Λγβ Fαβ

(C.15)

Appendix D

Solution to Advanced Problems

D.1 Maxwell Stress Tensor 1. The magnetic part of the Maxwell stress tensor is: Ti j =

1 2 1 Bi B j − δi j B μ0 2μ0

(D.1)

as shown in (2.2). 2. We can combine this with the pressure tensor as given: Pi j = P⊥ δi j + (P − P⊥ )

Bi B j B2

(D.2)

to obtain  (TM − P)i j =

1 (P − P⊥ ) − μ0 B2



  1 2 Bi B j + − B − P⊥ δi j 2μ0

(D.3)

This is divergence free if

and

B2 − (P − P⊥ ) = K  μ0

(D.4)

1 2 B + P⊥ = K 2μ0

(D.5)

where K , K  are both constants.

© Springer-Verlag GmbH Germany, part of Springer Nature 2021 S. Chapman, Core Electrodynamics, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-030-66818-1

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102

Appendix D: Solution to Advanced Problems

D.2 Liouville and Vlasov Theorems: A Conservation Equation in Phase Space (a) Expanding f in 6 dimensional phase space and time gives ∂f + δx · ∇ f + δv · ∇v f + · · · ∂t (D.6) so that the total change in f as we move from x → x + δx, v → v + δv, and t → t + δt is f (x + δx, v + δv, t + δt) = f (x, v, t) + δt

Df f (x + δx, v + δv, t + δt) − f (x, v, t) = δt Dt

(D.7)

and along a particle orbit such that: δr δt δv a= δt v=

this becomes:

Df ∂f = + v · ∇ f + a · ∇v f Dt ∂t

(D.8) (D.9)

(D.10)

(b) If x and v are independent then (v · ∇ f ) j = v j

∂v j f ∂f = ∂x j ∂x j

(D.11)

that is, v · ∇ f = ∇ · (v f )

(D.12)

Also, from the Lorentz force law a=

q [E(x, t) + v ∧ B(x, t)] m

(D.13)

then again as x and v are independent E·

∂f ∂ = · (E f ) ∂v ∂v

(D.14)

and using an identity we can write ∂ ∂f ∂ · (v ∧ B f ) = (v ∧ B) · + f · (v ∧ B) ∂v ∂v ∂v

(D.15)

Appendix D: Solution to Advanced Problems

103

by expanding into components the second term on the r.h.s. can be shown to be zero. Then ∂ ∂f = (v ∧ B f ) (D.16) (v ∧ B) · ∂v ∂v so that we can finally rewrite the Vlasov equation as ∂f + ∇ · (v f ) + ∇v · (a f ) = 0 ∂t

(D.17)

i.e., in the form of a conservation equation for f in 6 dimensional phase space.

D.3 Newton’s Laws and the Wave Equation Under Galilean Transformation Consider Galilean frame transformation x = x − vt, t  = t, with v constant. (a) For the ith particle the velocity in the transformed frame is vi = vi − v so that the acceleration is dvi dv dvi = = i (D.18) dt dt dt then the force between the ith and jth particle mi

dvi = F(xi − x j ) dt j

(D.19)

transforms as mi

dvi = F(xi + vt − xj − vt) = F(xi − xj )  dt j j

(D.20)

Hence Newton’s equations are Galilean invariant in form. (b) To transform the wave equation 

1 ∂ ∇ − 2 2 c ∂t 2

we need to transform ∇ and the unprimed frame

∂ . ∂t

 Ψ =0

(D.21)

These are defined via the chain rule so that in

∂ Ψ (x, t) dΨ (x, t) = dx · ∇ + dt ∂t

(D.22)

104

Appendix D: Solution to Advanced Problems

so that

dΨ (x, t) |t const dx

(D.23)

∂Ψ (x, t) dΨ (x, t) = |x const ∂t dt

(D.24)

∇Ψ (x, t) = and

we can compare this with the chain rule applied in the primed frame

∂ dΨ (x , t  ) = dx · ∇  + dt   Ψ (x , t  ) ∂t then

dΨ (x , t  ) |t  const dx

(D.26)

∂Ψ (x , t  ) dΨ (x , t  ) = |x const ∂t  dt 

(D.27)

∇  Ψ (x , t  ) = and

(D.25)

Now t = t  so that t  constant is equivalent to t constant which implies that ∇  ≡ ∇. However, x = x, instead we have x constant equivalent to x − vt constant when obtaining the partial derivatives from chain rule. We can write dΨ (x , t  ) in terms of elements in the unprimed frame using dx = dx − vdt, dt  = dt



∂ ∂ dΨ (x , t  ) = dx · ∇  + dt   Ψ = (dx − vdt) · ∇  + dt  Ψ (D.28) ∂t ∂t Then, given that we can write Ψ as a function of variables in either frame Ψ (x, t) ≡ Ψ (x , t  )

(D.29)

this yields

∂Ψ (x , t  ) ∂ dΨ (x , t  )  |x const = = −v · ∇ +  Ψ (x , t  ) dt ∂t ∂t

(D.30)

Then the wave equation in the unprimed frame is also  ∇2 − which transforms to

1 ∂ c2 ∂t 2



Ψ (x , t  ) = 0

(D.31)

Appendix D: Solution to Advanced Problems

105



 1 ∂ 2  ∇ − 2 −v · ∇ +  Ψ (x , t  ) = 0 c ∂t 2

(D.32)

which expands to

1 ∂ 1 2 ∂ ∇ 2 − 2 2 − 2 (v · ∇  )2 + 2 (v · ∇  )  Ψ = 0 c ∂t c c ∂t

(D.33)

Hence the wave equation is not invariant in form. Solutions to the wave equation in the unprimed frame will be of the form Ψ (x, t) ∼ ei(ωt−k·x)

(D.34)

which transformed to the primed frame becomes: 



Ψ (x , t  ) ∼ ei([ω−k·v]t −k·x )

(D.35)

which is consistent with the doppler shift of frequency: ω = ω − k · v

(D.36)

We can verify that this is a solution of (D.33) by direct substitution.

D.4 Transformation of the Fields This question is an extension of the moving charge and wire experiment discussed in (3.2). To in the rest frame calculate the electric field from one of the wires we enclose an elemental length dl of wire in surface S as shown in Fig. D.1. S is a cylinder of radius l. Gauss’ law in integral form then gives 





E · dS = S

E · dS = E2πldl = Sc

V

ρ ρ d V = Adl 0 0

(D.37)

where A is the c.s.a. of the wire. Then E=

ρA 2πl

(D.38)

from each wire and is directed perpendicular to the wire from geometry. The total electric field is just the sum of the contribution from each wire and the force in this frame in which the charges are at rest is just d F = ρAE(l)dl per length element dl of wire.

106

Appendix D: Solution to Advanced Problems

A

Fig. D.1 Surface S enclosing element of a wire of length dl

+ +

S

l

dl

+ + +

A

Fig. D.2 Curve C enclosing the wire with element along the curve dl

+ + + +

C

+

l dl

In the moving frame we can calculate the magnetic field by enclosing the wire with a curve as shown in Fig. D.2 and by using Ampere: 





∇ ∧ B · dS = μ0 S

ρ v · dS

J · dS = μ0 S

S

(D.39)

Appendix D: Solution to Advanced Problems

107

but taking into account the Lorentz contraction along the direction of the wire in order to calculate ρ in the moving frame as in (3.2). The electric field can also be calculated in the moving frame using charge density ρ . Alternatively, the transformation of the fields can be used directly given the electric field in the unprimed (rest) frame where the magnetic field is zero: B = −γ and

v∧E c2

E = γE

(D.40)

(D.41)

where we have exploited the geometry, that is, in the unprimed frame the electric field is perpendicular to the wires. In each frame the Lorentz force is then just given directly from the Lorentz force law.

D.5 Metric for Flat Spacetime (a) To show

gαγ g γβ = Dαβ = δαβ

(D.42)

where we will use the notation Dαβ for the l.h.s. and ⎡

gαβ

⎤ 1 0 0 0 ⎢ 0 −1 0 0 ⎥ αβ ⎥ =⎢ ⎣ 0 0 −1 0 ⎦ = g 0 0 0 −1

(D.43)

Now all off axis α = β terms of gαβ and g αβ are zero, hence all of the off axis terms of the inner product Dαβ will be zero. For example D12 = g1γ g γ2 = 0

(D.44)

for any γ. The remaining terms are: α = β = 0 giving D00 = g0γ g γ0 = g00 g 00 + g01 g 10 + g02 g 20 + g03 g 30 = 1 + 0 + 0 + 0 (D.45) α = β = 1: D11 = g1γ g γ1 = g10 g 01 + g11 g 11 + g12 g 21 + g13 g 31 = 0 + 1 + 0 + 0 (D.46)

108

Appendix D: Solution to Advanced Problems

similarly, D22 = 1 and D33 = 1 Then

⎡

1 ⎢0 β ⎢ Dα = ⎣ 0 0 as required. (b) To show

⎤ 000 1 0 0⎥ ⎥ = δβ α 0 1 0⎦ 001

(D.47)

gαβ x β = xα

(D.48)

gαγ Λγβ xβ

(D.49)

we write the l.h.s. of this as

this is covariant by inspection (the only remaining index after contraction will be “down”). Explicitly this can be calculated as follows: First we can calculate gαγ Λγβ = Λβα

(D.50)

by exploiting the fact that α = β terms will be zero (from the definition of gαβ ). The remaining nonzero terms are β

Λ0 β Λ1 β Λ2 β Λ3 then

= g00 Λ0β = g11 Λ1β = g22 Λ2β = g33 Λ3β

= = = =

Λ0β −Λ1β −Λ2β −Λ3β

gαγ Λγβ xβ = Λβα xβ = xα

(D.51)

(D.52)

is a four vector with components α = 0, 3: β

Λ0 xβ β Λ1 xβ β Λ2 xβ β Λ3 xβ

= Λ0β xβ = = −Λ1β xβ = = −Λ2β xβ = = −Λ3β xβ =

x0 x1 x2 x3

(D.53)

so that if covariant four vector ⎡

⎤ ct ⎢ −x ⎥ ⎥ xβ = (ct, −x) = ⎢ ⎣ −y ⎦ −z

(D.54)

Appendix D: Solution to Advanced Problems

and

then we have

109

⎤ ct  ⎢ x ⎥ ⎥ Λαβ xβ = ⎢ ⎣ y ⎦ z ⎡

⎤ ct  ⎢ −x  ⎥  ⎥ Λβα xβ = ⎢ ⎣ −y  ⎦ = xα −z 

(D.55)

⎡

which is covariant in the primed frame. (c) To show Fαβ = gαγ F γδ gδβ

(D.56)

(D.57) γ

perform the contraction in two steps. First, the nonzero terms of F γδ gδβ = Fβ are γ δ = 0, β = 0 F0 = +F γ0 γ δ = 1, β = 1 F1 = −F γ1 (D.58) γ δ = 2, β = 2 F2 = −F γ2 γ δ = 3, β = 3 F3 = −F γ3 γ

for γ = 0, 3. Then the nonzero terms of gαγ Fβ are α=γ α=γ α=γ α=γ

=0 =1 =2 =3

F0β F1β F2β F3β

= +Fβ0 = −Fβ1 = −Fβ2 = −Fβ3

(D.59)

for β = 0, 3. Here we have exploited the fact that all β = γ terms of gβγ are zero. We now just consider the signs of the terms of Fαβ for all α, β. row 0: α = 0 row 1: α = 1 row 2: α = 2 row 3: α = 3

all change sign except F00 = 0 F10 changes sign only F20 changes sign only F30 changes sign only

110

Appendix D: Solution to Advanced Problems

D.6 Length of the EM Field Tensor in Spacetime We have the electromagnetic field tensor ⎤ E 0 − Ecx − cy − Ecz ⎢ 0 −Bz B y ⎥ ⎥ =⎢ ⎣ Bz 0 −Bx ⎦ −B y Bx 0 ⎡

F αβ

Ex c Ey c Ez c

(D.60)

from Chap. 4, and also Fαβ is obtained via the transformation E i → −E i and Bi → Bi in F αβ . The symmetries of this tensor are then by inspection

and the trace is zero

F βα = −F αβ Fβα = −Fαβ

(D.61)

F αα = Fαα = 0

(D.62)

Fαβ F βα then has the following properties for γ = 0, 3: (i) The on axis terms are zero

F γγ Fγγ = 0

(D.63)

(ii) The timelike terms, that is, those involving E are F0γ F γ0 = −F0γ Fγ0 = −(−F0γ F0γ ) Fγ0 F 0γ = −Fγ0 F0γ = −(−Fγ0 Fγ0 )

(D.64)

contracting these terms over γ gives E2 [Fγ0 ]2 = 2 = [F0γ ]2 c γ γ

(D.65)

so that the total contribution of the timelike terms is 2E 2 /c2 . (iii) The spacelike terms, that is, terms involving B then have the following properties (D.66) Fαγ F γα = −Fαγ F αγ = −Fαγ Fαγ for α = 0, γ = 0. Contracting these terms gives α=1,3 γ=1,3

−(Fαγ )2 = −2B 2

(D.67)

Appendix D: Solution to Advanced Problems

111

The result of the contraction is then Fαβ F βα = 2



E2 − B2 c2

 (D.68)

In free space (photons only) we have E = Bc so that Fαβ F βα vanishes. For the dual tensor F˜ αβ we use the fact that it is obtained from Fαβ via the duality transformation E i → Bi c2 , Bi → −E i (see 4.2). Transforming the result gives F˜ αβ F˜ βα = 2



B 2 c4 − E2 c2



2 =− 2 c



E2 − B2 c2



= −c2 Fαβ F βα

(D.69)

as required.

D.7 Alternative Form for the Maxwell Homogenous Equations This left as an exercise for the student.

D.8 Lorentz Transformation of the EM Field Tensor (a) In (4.1) we obtain

F αβ = ∂ α Aβ − ∂ β Aα

(D.70)

then the Lorentz transformation of this is just the transformation of each of the component four vectors. (b) To evaluate this i.e., (D.71) F αβ = ∂ α Aβ − ∂ β Aα we apply the Lorentz transformation and the transpose rule: F αβ = (Λαδ ∂δ )(Λβγ Aγ ) − (Λβγ ∂γ )(Λαδ Aδ ) = Λαδ ∂δ Aγ Λγβ − Λβγ ∂γ Aδ Λδα = Λαδ ∂δ Aγ Λγβ − Λαδ ∂γ Aδ Λγβ = Λαδ (∂δ Aγ − ∂γ Aδ )Λγβ = Λαδ Fδγ Λγβ as required.

(D.72)

Appendix E

Vector Identities

ˆ Assuming the right hand rule relates the orthogonal unit vectors ˆi, ˆj, and k: ˆi ∧ ˆj = kˆ

E.1 Differential Relations a · b = b · a = a1 b1 + a2 b2 + a3 b3 iˆ a ∧ b = −b ∧ a = a1 b1

jˆ a2 b2

kˆ a3 b3

(E.1)

(E.2)

a · (b ∧ c) = (a ∧ b) · c = (c ∧ a) · b

(E.3)

a ∧ (b ∧ c) = (a · c)b − (a · b)c

(E.4)

(a ∧ b) · (c ∧ d) = (a · c)(b · d) − (a · d)(b · c)

(E.5)

(a ∧ b) ∧ (c ∧ d) = [a · (b ∧ d)] c − [a · (b ∧ c)] d

(E.6)

∇(ψφ) = φ∇ψ + ψ∇φ

(E.7)

∇ · (φa) = φ∇ · a + a · ∇φ

(E.8)

∇ ∧ (φa) = φ∇ ∧ a + (∇φ) ∧ a

(E.9)

∇ · (a ∧ b) = b · (∇ ∧ a) − a · (∇ ∧ b)

(E.10)

© Springer-Verlag GmbH Germany, part of Springer Nature 2021 S. Chapman, Core Electrodynamics, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-030-66818-1

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Appendix E: Vector Identities

∇(a · b) = (a · ∇)b + (b · ∇)a + a ∧ (∇ ∧ b) + b ∧ (∇ ∧ a)

(E.11)

∇ ∧ (a ∧ b) = a(∇ · b) + (b · ∇)a − b(∇ · a) − (a · ∇)b

(E.12)

a ∧ (∇ ∧ b) = (∇B) · a − (a · ∇)b

(E.13)

∇ ∧ (∇ ∧ a) = ∇(∇ · a) − ∇ 2 a

(E.14)

∇ · (∇ ∧ a) = 0

(E.15)

∇ ∧ (∇φ) = 0

(E.16)

E.2 Integral Relations For the following integral relations vector surface element d S is directed along the normal nˆ to the surface S, line element dl is directed along curve C. Flux of a vector field a:  Flux =

a · dS

(E.17)

dS ∧ (∇φ)

(E.18)

∇ ∧ a · dS

(E.19)

(dS ∧ ∇) ∧ a

(E.20)

S

If surface S spans curve C: •



 φdl =

C

• Stokes Theorem:

S



 a · dl = C

•

S



 dl ∧ a = C

•

s







dS · (∇φ ∧ ∇ψ) =

φdψ = −

S

C

ψdφ

(E.21)

C

If surface S encloses volume V , dS points outwards: •



 φdS = S

v

(∇φ)d V

(E.22)

Appendix E: Vector Identities

115

• Gauss or Divergence Theorem: 

 a · dS = S

•

(dS ∧ a) =

(∇ ∧ a)d V

(E.24)

V





 2  φ∇ ψ + (∇φ) · (∇ψ) d V

φ(∇ψ) · dS = S

(E.23)



 S

• Green I:

∇ · ad V V

(E.25)

V

• Green II or Green’s Theorem in a plane: 

 (φ∇ψ − ψ∇φ) · dS = S

(φ∇ 2 ψ − ψ∇ 2 φ)d V

(E.26)

V

• Green II in vector form:  [b ∧ (∇ ∧ a) − a ∧ (∇ ∧ b)] · dS = 

S

(a · [∇ ∧ (∇ ∧ b)] − b · [∇ ∧ (∇ ∧ a)]) d V V

(E.27)

Appendix F

Tensors

F.1 Cartesian Tensors A tensor of the first rank (vector) would be written a or ai where i = 1, 3. A scalar is thus a tensor of rank zero (one number). We can generalize to as many indices as we wish, for example Pi jkl is a tensor of rank four. • A Dyadic is formed of two vectors a, b for any orthogonal coordinate system: ⎡

⎤ a x b x a x b y a x bz ab = ⎣ a y bx a y b y a y bz ⎦ az b x az b y az bz

(F.1)

which can, where i = 1, 3 and j = 1, 3 be written as Ai j = ai b j

(F.2)

• Coordinate rotation A vector in the unprimed frame: ⎡ ⎤ x r = ⎣y⎦ z will have coordinates in the primed frame: 

x = a11 x +a12 y +a13 z  y = a21 x +a22 y +a23 z  z = a31 x +a32 y +a33 z

© Springer-Verlag GmbH Germany, part of Springer Nature 2021 S. Chapman, Core Electrodynamics, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-030-66818-1

117

118

Appendix F: Tensors

The tensor ai j contains the direction cosines of the rotation x, y, z → x  , y  , z  . This can be written as (F.3) r = A · r ⎡

⎡ ⎤  ⎤ x a11 a12 a13  r = ⎣ y ⎦ and A = ⎣ a21 a22 a23 ⎦  a31 a32 a33 z

where



• Einstein convention Since x = r1 , y = r2 and z = r3 , Eq. (2.22) is 

r j = a ji ri

(F.4)

which is shorthand (the Einstein summation convention) for 

rj =



a ji ri

(F.5)

i=1,3

The sum over index i on the r.h.s. of (2.24) contracts the number of indices by one (the l.h.s. has single index j). • Inner product. Equations (2.23) and (2.24) are an example of a tensor dot (or inner) product. The switch to index notation drops all reference to the basis vectors (or axes) x, y, z. Hence Eq. (2.25) is tensor dot product: a · T = a1 T11 + a2 T21 + a3 T31 + a1 T12 + a2 T22 + · · · ≡ ai Ti j = xˆ j ai Ti j

(F.6)

dropping all reference to the basis vectors xˆ j . • Then ∇ · T in Cartesian coordinates is xˆ j

∂ ∂ Ti j = ∇ · T ≡ Ti j ∂xi ∂xi

(F.7)

It follows that ∇ · (φT ) = φ(∇ · T ) + (∇φ) · T

(F.8)

∇ · (ab) = (∇ · a)b + (a · ∇)b

(F.9)

• Divergence theorem for tensors of arbitrary rank: 

 T · dS = S

(∇ · T ) d V V

(F.10)

Appendix F: Tensors

119

F.2 Special Tensors • The trace is δi j = 1 when i = j and 0 when i = j ⎡

⎤ 100 δi j = ⎣ 0 1 0 ⎦ 001 Contracting δi j over i and j with two vectors extracts the dot product δi j ai b j = a1 b1 + a2 b2 + a3 b3 = a · b = ai bi This generalizes to tensors of arbitrary rank δi j ai Ti jkl.. = a · T

(F.11)

1 if i jk = 123, 231, 312 i jk = −1 if i jk = 321, 132, 213 0 any two indices alike

(F.12)

• The alternating tensor:

Contracting i jk over j and k extracts the following vector from a dyadic i jk a j bk =

i11 a1 b1 +i12 b1 a2 a1 b2 +i13 a1 b3 +i21 a2 b1 +i22 a2 b2 +i23 a2 b3 +i31 a3 b1 +i32 a3 b2 +i33 b3 a3

If we consider the i = 1 terms, all except 123 a2 b3 and 132 a3 b2 are zero. These last two give a2 b3 − a3 b2 which are the x component of the vector cross product. Similarly, i = 2, 3 gives the y, z components respectively, so that contracting with i jk extracts the cross product. This again holds for tensors of arbitrary rank.

F.3 Generalized Tensors The generalized coordinate system is defined in terms of the space-time interval s 2 = c2 t 2 − x 2 − y 2 − z 2

(F.13)

The space-time interval is the length of the four vector: s 2 = s · s = xα x α

(F.14)

120

Appendix F: Tensors

F.3F.1 General Properties of Spacetime • s has 2 forms: Covariant:

⎡

⎤ ct ⎢ −x ⎥ ⎥ xα = (ct, −x) = ⎢ ⎣ −y ⎦ −z

and Contravariant:

(F.15)

⎡

⎤ ct ⎢x ⎥ ⎥ x α = (ct, +x) = ⎢ ⎣y ⎦ z

(F.16)

• In general (for any spacetime geometry) if we have a well defined transformation x α = x α (x 0 , x 1 , x 2 , x 3 ) a covariant vector transforms as: aα =

∂x 0 ∂x 1 ∂x 2 ∂x 3 ∂x β a + a + a + a = aβ 0 1 2 3 ∂x α ∂x α ∂x α ∂x α ∂x α

(F.17)

and a contravariant vector transforms as a α =

∂x α 0 ∂x α 1 ∂x α 2 ∂x α 3 ∂x α β a + a + a + a = a 0 1 2 3 ∂x ∂x ∂x ∂x ∂x β

(F.18)

• For tensors of rank 2, a covariant tensor G αβ transforms as G αβ =

∂x γ ∂x δ G γδ ∂x α ∂x β

(F.19)

∂x α ∂x β γδ G ∂x γ ∂x δ

(F.20)

a contravariant rank 2 tensor G αβ as: G αβ =

and a mixed tensor of rank 2 G αβ transforms as: G α β = • The dot or inner product is:

∂x α ∂x δ γ G ∂x γ ∂x β δ

... ...α a · b = a...α b...

that is, a contraction over the index α.

(F.21)

(F.22)

Appendix F: Tensors

121

For general spacetime geometry the inner product is invariant under transformation since: a · b =

∂x β ∂x α ... ...γ ∂x β ... ...γ ... ...γ a...β b... = a b = δγβ a...β b... = a · b α γ ∂x ∂x ∂x γ ...β ...

(F.23)

F.3F.2 Flat Spacetime The Lorentz transformation corresponds to a rotation of the four vector s in flat space-time, under which s 2 is constant. • The norm or metric for flat spacetime is then the invariant differential length element: (ds)2 = (d x 0 )2 − (d x 1 )2 − (d x 2 )2 − (d x 3 )2 (F.24) which is a special case of the general differential length element (ds)2 = gαβ d x α d x β

(F.25)

• The spacetime metric for flat space is given by: ⎡

gαβ

⎤ 1 0 0 0 ⎢ 0 −1 0 0 ⎥ αβ ⎥ =⎢ ⎣ 0 0 −1 0 ⎦ = g 0 0 0 −1

(F.26)

• Covariant and contravariant vectors are related by xα = gαβ x β

(F.27)

x α = g αβ xβ

(F.28)

and

• Lorentz transformation -rotation matrix A rotation of four vector s will be a operation of the form 

xα = Λαβ x β on the contravariant form and



x α = Λαβ xβ

on the covariant form where the Lorentz transformation is:

(F.29)

(F.30)

122

Appendix F: Tensors

⎡

− vc γ 0

γ

0

⎤

⎢ + v γ −γ 0 0 ⎥ ⎢ ⎥ Λαβ = ⎢ c ⎥ ⎣ 0 0 −1 0 ⎦ 0

0

(F.31)

0 −1

the inverse transformation matrix is then just obtained by v → −v in (3.60). • The covariant four gradient is ⎡1 ⎢ ⎢ ∂α = ⎢ ⎢ ⎣

∂ c ∂t ∂ ∂x ∂ ∂y ∂ ∂z

⎤ ⎥   ⎥ ⎥ = 1 ∂ ,∇ ⎥ c ∂t ⎦

(F.32)

⎥   ⎥ ⎥ = 1 ∂ , −∇ = g αβ ∂β ⎥ c ∂t ⎦

(F.33)

• The contravariant four gradient is ⎡1 α

∂ =

∂ c ∂t ⎢ ∂ ⎢ − ∂x ⎢ ⎢− ∂ ⎣ ∂y ∂ − ∂z

⎤

• Hence the “four-divergence” of a four vector ∂α a α = ∂ α a α =

1 ∂ 0 a +∇ ·a c ∂t

(F.34)

(where a is the spacelike part of four vector field a α ). • The “4 curl” of four vector a is a rank 2 four tensor F αβ = ∂ α a β − ∂ β a α

(F.35)

• The D’Alembertian invariant operator is then ∂α ∂ α =

1 ∂2 − ∇2 =  c2 ∂t 2

which is just the wave equation operator in vacuum.

(F.36)

Appendix G

Units and Dimensions

G.1 SI Nomenclature Physical Quantity length mass time current temperature amount of substance luminous intensity plane angle solid angle frequency energy force pressure power electric charge electric potential electric resistance electric conductance electric capacitance magnetic flux magnetic inductance magnetic intensity luminous flux illuminance activity (of radioactive source) absorbed dose (of ionising radiation)

Name metre kilogram second Ampere Kelvin mole candela radian steradian Hertz Joule Newton Pascal Watt Coulomb Volt Ohm Siemens Farad Weber Henry Tesla lumen lux Becquerel Gray

Symbol m kg s A K mol cd rad sr Hz J N Pa W C V Ω S F Wb H T lm lx Bq Gy

© Springer-Verlag GmbH Germany, part of Springer Nature 2021 S. Chapman, Core Electrodynamics, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-030-66818-1

123

124

Appendix G: Units and Dimensions

G.2 Metric Prefixes Multiple 10−1 10−2 10−3 10−6 10−9 10−12 10−15 10−18

Prefix deci centi milli micro nano pico femto atto

Symbol d c m μ n p f a

Multiple 10 102 103 106 109 1012 1015 1018

Prefix deca hecto kilo mega giga tera peta exa

Symbol da h k M G T P E

Appendix H

Dimensions and Units

H.1 Physical Quantities SI has been used throughout this book. Gaussian units are still often used however and conversion factors between SI and Gaussian units are given here. To obtain the value of a quantity in Gaussian Units, multiply the value expressed in SI units by the conversion factor. Multiples of 3 in the conversion factor result from approximating the speed of light c = 2.99979 × 108 ms−1  3 × 108 ms−1 . Dimensions SI Gaussian

Physical Quantity mass

Symbol m

m

m

length

l

l

l

time

t

t

t

SI Unit kg

Conversion Factor 103

Unit gram (g)

m

102

centimetre (cm)

s

1

second (s)

© Springer-Verlag GmbH Germany, part of Springer Nature 2021 S. Chapman, Core Electrodynamics, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-030-66818-1

125

126

Appendix H: Dimensions and Units

Physical Quantity

Symbol

Capacitance

C

Charge Charge Density

q ρ

Conductance Conductivity

σ

Current

I, i

Current density

J

Density

ρ

Displacement Electric field Electromotance Energy

D E , emf W

Energy density

U

Force

F

Frequency

f, ν

Dimensions SI Gaussian t2q2 ml 2

q q l3 tq 2 ml 2 tq 2 ml 3 q t q l2t m l3

l m

1 3 2l2

t 1 2

SI Unit

Conversion Factor

Unit

F

9 × 1011

cm

C

3 × 109

stat- coulomb

Cm −3

3 × 103

stat- coulomb cm −3

l2t 1 t

S

9 × 1011

cms −1

1 t

S/m

9 × 109

s −1

A

3 × 109

statampere

Am −2

3 × 105

statampere cm −2

kgm −3

10−3

gcm −3

Cm −2

12π × 105

stat- coulomb cm −2

m 3

1 3

m2l2 t2 m

1 2

1

l 2 t2 m l3 1 2

q l2

m

ml t2q

m2

ml 2 t2q

m2l2 t

ml 2 t2 m lt 2 ml t2 1 t

ml 2 t2 m lt 2 ml t2 1 t

1

l2t 1

1 2

l t

V /m

1 3

× 10−4

statvoltcm −1

V

1 3

× 10−2

statvolt

1 1

J

107

erg

J/m 3

10

ergcm −3

N

105

dyne

Hz

1

Hz

Appendix H: Dimensions and Units

127

Dimensions SI Gaussian

Physical Quantity

Symbol

Impedance

Z

ml 2 tq 2

t l

L

ml 2 q2

t2

Inductance Magnetic intensity Magnetic flux Magnetic induction Magnetic moment Magnetisation Magneto- motance Momentum

H Φ B m, μ M

M p

Momentum density Permeability Permittivity Polarization Potential Power

μ  P

q lt ml 2 tq

1 2

m l t m

l2q

1 2

q lt q t ml t m 2 l t ml q2 t2q2 ml 3 q l2

1 9

× 10−11

scm −1

H

1 9

× 10−11

s 2 cm −1

A/m

4π × 10−3

Oersted

Wb

108

Maxwell

T

104

Gauss

Am 2

103

Oersted cm 3

Am −1

10−3

Oersted

A kgms −1

4π 10 105

gcms −1

kgm −2

10−1

gcm −2 s −1

l2t

m tq

t

Ω

3 2

1 2

1

l2t m l t m

5 2

1 2

1

l2t 1 2

m l t2 ml t m 2 l t

1 2

1

H m −1

1 m

1 2

1

× 107

-

Fm −1

36π × 109

-

Cm −2

3 × 105

stat- coulomb cm −2

1 4π

l2t 1

Gilbert

1

m2l2 t

V

P

ml 2 t3

ml 2 t3

W

107

erg s −1

m lt 3 m lt 2 q2 ml 2 ml 2 tq 2

m lt 3 m lt 2 1 l

W m −3

10

erg cm −3 s −1

Pa

10

dyne cm −2

AW b−1

4π × 10−9

cm −1 scm −1

P

Reluctance

R

Resistance

R

Resistivity

η, ρ

ml 3 tq 2

Thermal conductivity

κ

ml t3

A v

Viscosity

η, μ

Vorticity

ζ

Work

1

Unit

ml 2 t2q

Pressure

Velocity

m

1 2

Conversion Factor

V, φ

Power density

Vector potential

l

SI Unit

W

ml tq l t m lt 1 t ml 2 t2

1 3

× 10−2

statvolt

t l

Ω

1 9

× 10−11

t

Ωm

1 9

× 10−9

ml t3

W m −1 K −1

105

W bm −1

106

Gauss cm

ms −1

102

cms −1

kgm −1 s −1

10

Poise

s −1

1

s −1

J

107

erg

1 2

m l t l t m lt 1 t ml 2 t2

1 2

s erg

cm −1 s −1 K −1

128

Appendix H: Dimensions and Units

H.2 Equations In SI, in media the permittivity  = 0 r and the permeability μ = μ0 μr where 0 and μ0 are the permittivity and permeability in free space, and the relative permittivity and permeability, r and μr are dimensionless. μ0

SI 4π × 10−7 H m −2

Gaussian 1

0

1 c2 μ0

1

D=

0 E + P

E + 4πP

H=

B μ0

−M

B − 4πM

Maxwell I Maxwell II Maxwell III Maxwell IV

∇ ·E=

ρ 0

∇ ·D=ρ

∇ · D = 4πρ

∇ ·B=0

∇ ·B=0

∇ ∧E=

− ∂B ∂t

∇ ∧ B = μ0 J + μ0 0 ∂E ∂t ∇ ∧H =J+

Lorentz force per unit charge

∇ ∧ E = − 1c ∂B ∂t

E+v∧B

∂D ∂t

4π 1 ∂D c J + c ∂t v E+ c ∧B

∇ ∧H=

Appendix I

Physical Constants (SI)

Physical Quantity Free space permittivity Free space permeability Free space speed of light Elementary charge Bohr magneton Electron mass Proton mass Atomic mass unit Planck constant Boltzmann constant Avogadro number

Symbol 0 μ0 c = √μ10 0 e μB me mp mu h k N0

Value 8.854 × 10−12 4π × 10−7 2.998 × 108 1.602 × 10−19 9.274 × 10−24 9.109 × 10−31 1.673 × 10−27 1.661 × 10−27 6.626 × 10−34 1.381 × 10−23 6.022 × 1023

Units Fm −1 H m −1 ms −1 C Am −2 kg kg – Js J K −1 mol −1

© Springer-Verlag GmbH Germany, part of Springer Nature 2021 S. Chapman, Core Electrodynamics, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-030-66818-1

129

Index

A Alternating tensor, 28, 64 Ampère’s law, 12, 16 differential form, 12

B Basis vectors, 27 Biot Savart law, 72

C Cartesian inconsistency with spacetime, 42 Charge conservation, 16, 68 equation of, 17 Charge density, 4 Charge invariance, 42 Classical limit fields, 4, 9 fluids, 28 particles, 19 Cold gas, 5 conservation equation, 19 definition, 5 energy flux, 31 momentum flux density, 24 Conservation energy-momentum, 66, 68 of current, 68 of energy density, 30 of field-particle momentum, 31 of four-current, 56 Contraction in Cartesian tensors, 27 Contravariant, 48 Coordinate rotation

in spacetime, 41 Coordinate transformation Cartesian, 27 Lorentz, 40, 50 Coulomb unit, 4 Coulomb gauge, 47 Coulomb’s law experiment, 3 Covariant, 48 Cross product, 28 Current density, 12

D D’Alembertian, 55 Direction cosines, 27 Displacement current, 17 in capacitors, 17 Divergence theorem definition of, 2 Duality transformation, 63, 64, 69 Dual tensor, 63 Dyadic, 23, 26

E Einstein summation convention, 27 Electric field, 3 definition, 11 definition of, 4 flux, 6 in terms of charge density, 5 in terms of potentials, 16, 46 lines, continuous, 10 units, 4 work done by, 11

© Springer-Verlag GmbH Germany, part of Springer Nature 2021 S. Chapman, Core Electrodynamics, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-030-66818-1

131

132 Electric potential definition, 11 Electromagnetic field tensor, 61, 62 length, 62 Electromagnetic waves, 33 free space, 18 from Maxwell equations, 18 general solution, 18 group velocity, 18 phase speed in vacuum, 18 relativistic charge, 72 Electrostatic field conservative, 11 Electrostatic potential, 12 Energy-momentum, 68 Energy-momentum field tensor, 66 Equivalence of electric and magnetic fields, 14, 15, 42

F Faraday’s law, 13, 15 from Lorentz force, 14 Field classical limit, 9 energy density, 29, 30 from moving point charge, 71 momentum density, 31 momentum flux density, 31 transformation of, 15 Flat spacetime, 49 Fluid bulk variables, 29 Fluid equation, 26 Flux, 6 definition, 5 of electric field, 6 of energy, 31 Force from momentum flux tensor, 24 Four-curl, 60 Four-vector, 46 contravariant, 48 covariant, 48 energy-momentum, 51 force, 53, 67, 68 four-current, 56, 68, 74 four-divergence, 55 four-gradient, 54 four vector potential, 56 four-velocity, 52, 68 inner product, 49 length-time, 51 potential, 59

Index Frame transformation Galilean, 15 nonrelativistic, 16

G Galilean frame transformation, 15 Galilean transformation, 52 Gas finite temperature, 19 warm, 25 Gauge transformation, 46 Gauss’ law, 9 integral form, 9 Generalized coordinates, 47 inner product, 50 General relativity, 49

I Inner product, 27, 50

K Kinetic energy, 29

L Light clock, 38 Linear media, 30 Liouville’s theorem, 25 Lorentz contraction, 40 Lorentz force, 31 Lorentz force law invariant form, 67 Lorentz gauge, 47, 56, 68 in four-vector form, 56, 68 Lorentz transformation, 40, 50 inverse, 50 of charge density, 43 retarded potential, 73 Lorentz transformation of fields, 69

M Magnetic field lines, continuous, 10 units, 12 Magnetic flux, 10 Magnetostatic scalar potential, 12 Maxwell equations dual, 64 homogenous, 46, 63, 68 inhomogenous, 47, 56, 63, 67

Index manifestly covariant, 57, 68 Maxwell homogenous equations invariant form, 63 Maxwell inhomogenous equations invariant form, 63 Maxwell stress for free space EM waves, 33 tensor, 32 Maxwell I, 9 integral form, 9 Maxwell II, 10 integral form, 10 Maxwell III, 15 integral form, 15 Maxwell IV, 17 Mechanics, invariant form, 66 Momentum flux density, 24 Momentum flux tensor, 23 Monopoles, 65 Moving charge and wire, 42, 69, 74

N Newton’s laws, 52 and special relativity, 53

P Phase space, 25 Photons, 18, 19 and relativistic charge, 72 Poynting flux, 30, 31 units, 30 Poynting’s theorem, 30 Pressure tensor, 22, 25 isotropic, 22 shear free, 22 Principle of superposition, 3 Proper time, 53

R Radiation pressure, 34 Rest energy, 51 Retarded potential, 75 Right hand rule in cross product, 2 in Lorentz force law, 2 Rocket effect, 24 Rotation matrix, 27 Rotation of four vector, 48

133 S Spacelike interval, 41 Spacetime deridative, 53 interval, 41, 47, 53 metric, 49 rotation in, 50 Special relativity, 37 Speed of light, 57, 64 from Maxwell’s equations, 18 from potentials, 47 Speed of light in vacuum, 18 Stokes theorem definition of, 2 Superposition in Galilean frame transformation, 15 of electromagnetic waves, 18 of electrostatic fields, 11

T Tensor alternating, 28, 64 Cartesian, 24, 26 coordinate transformation, 27 divergence of, 28 dot product, 27 dual, 63 electromagnetic field, 61, 62 energy- momentum field, 66 field momentum flux density, 31 force from, 24 inner product, 28 Lorentz transformation, 50 Maxwell stress, 32, 66 momentum flux density, 23 pressure, 22, 25 rank, 26 trace, 28 warm gas pressure, 25 Tesla, 12 Time dilation, 39 Timelike interval, 41 Trace, 28 Transformation as rotation, 27, 41, 48 Cartesian, 27 charge invariance, 42 duality, 63, 64, 69 Galilean, 52 gauge, 46 Lorentz, 40, 50 of charge density, 43

134 of EM field tensor, 69 retarded potential, 73 rotation in spacetime, 50

V Vector potential, 46 from magnetostatics, 12

Index W Warm gas pressure tensor, 25 Wave equation for electromagnetic fields, 18 in terms of potentials, 47 manifestly covariant, 57 Work done by Lorentz force, 29