core electrodynamics

Core Electrodynamics

Sandra C. Chapman

June 29, 2010

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Preface

This monograph is based on a final year undergraduate course in Electro-dynamics that I introduced at Warwick as part of the new four year physicsdegree. When I gave this course my intention was to engage the students inthe elegance of electrodynamics and special relativity, whilst giving themthe tools to begin graduate study. Here, from the basis of experimentwe first derive the Maxwell equations, and special relativity. Introducingthe mathematical framework of generalized tensors, the laws of mechanics,Lorentz force and the Maxwell equations are then cast in manifestly covari-ant form. This provides the basis for graduate study in field theory, highenergy astrophysics, general relativity and quantum electrodynamics.

As the title suggests, this book is ”electrodynamics lite”. The journeythrough electrodynamics is kept as brief as possible, with minimal diversioninto details, so that the elegance of the theory can be appreciated in aholistic way. It is written in an informal style and has few prerequisites;the derivation of the Maxwell equations and their consequences is dealtwith in the first chapter. Chapter 2 is devoted to conservation equations intensor formulation, here Cartesian tensors are introduced. Special relativityand its consequences for electrodynamics is introduced in Chapter 3 andcast in four vector form, here we introduce generalized tensors. Finally inChapter 4 Lorentz frame invariant electrodynamics is developed.

Supplementary material and examples are provided by the two sets ofproblems. The first is revision of undergraduate electromagnetism, to ex-pand on the material in the first chapter. The second is more advancedcorresponding to the remaining chapters, and its purpose is twofold: toexpand on points that are important, but not essential, to derivation ofmanifestly covariant electrodynamics, and to provide examples of manip-ulation of cartesian and generalized tensors. As these problems introducematerial not covered in the text they are accompanied by full worked solu-tions. The philosophy here is to facilitate learning by problem solving, aswell as by studying the text.

Extensive appendices for vector relations, unit conversion and so forth

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are given with graduate study in mind. As SI units are used throughouthere, conversions for units and equations are also given.

There have been many who have contributed to the existence of thisbook. Thanks in particular go to Nick Watkins for valuable discussionsand to George Rowlands for his insightful reading of the final draft. DavidBetts, the editor of this series, has also shown remarkable patience andtenacity as the many deadlines have come and gone. The completion of thebook was also much assisted by a PPARC personal fellowship.

Finally, my thanks go to the students themselves; their lively receptionof the original course and their insightful questions were the inspirationfor this book. If the reader finds that this book provides a shortcut toexperience the beauty of electrodynamics, without sacrificing the rigourneeded for further study, then I have suceeded.

Sandra Chapman, June, 1999

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Contents

3.5 Four Vectors and Four Vector Calculus. . . . . . . . . . . . 603.5.1 Some mechanics, Newton’s laws. . . . . . . . . . . . 603.5.2 Some four vector calculus. . . . . . . . . . . . . . . . 63

3.6 A Frame Invariant Electromagnetism . . . . . . . . . . . . . 643.6.1 Charge conservation. . . . . . . . . . . . . . . . . . . 643.6.2 A manifestly covariant electromagnetism . . . . . . . 65

4 The Field Tensors 674.1 Invariant Form for E and B: The EM Field Tensor. . . . . 674.2 Maxwell’s Equations in Invariant Form. . . . . . . . . . . . 704.3 Conservation of Energy-Momentum. . . . . . . . . . . . . . 734.4 Lorentz Force. . . . . . . . . . . . . . . . . . . . . . . . . . . 74

4.4.1 Manifestly covariant electrodynamics. . . . . . . . . 754.5 Transformation of the Fields . . . . . . . . . . . . . . . . . . 764.6 Field from a Moving Point Charge. . . . . . . . . . . . . . . 774.7 Retarded Potential. . . . . . . . . . . . . . . . . . . . . . . . 813.5 Four Vectors and Four Vector Calculus. . . . . . . . . . . . 60

3.5.1 Some mechanics, Newton’s laws. . . . . . . . . . . . 603.5.2 Some four vector calculus. . . . . . . . . . . . . . . . 63



4.4.1 Manifestly covariant electrodynamics. . . . . . . . . 754.5 Transformation of the Fields . . . . . . . . . . . . . . . . . . 764.6 Field from a Moving Point Charge. . . . . . . . . . . . . . . 77

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4.7 Retarded Potential. . . . . . . . . . . . . . . . . . . . . . . . 81

1 A Brief Tour of Electromagnetism. 91.1 The Building Blocks. . . . . . . . . . . . . . . . . . . . . . . 91.2 Maxwell I and II . . . . . . . . . . . . . . . . . . . . . . . . 11

1.2.1 Flux of a vector field . . . . . . . . . . . . . . . . . . 131.2.2 Flux of E . . . . . . . . . . . . . . . . . . . . . . . . 151.2.3 Flux of B . . . . . . . . . . . . . . . . . . . . . . . . 181.2.4 Conservative and nonconservative fields. . . . . . . . 19

1.3 Maxwell III and IV . . . . . . . . . . . . . . . . . . . . . . . 211.3.1 Faraday’s law and Galilean invariance. . . . . . . . . 211.3.2 Ampere’s law and conservation of charge. . . . . . . 25

1.4 Electromagnetic Waves. . . . . . . . . . . . . . . . . . . . . 261.5 Conservation Equations. . . . . . . . . . . . . . . . . . . . . 28

2 Field Energy and Momentum. 292.1 Tensors and Conservation Equations. . . . . . . . . . . . . . 30

2.1.1 Momentum flux density tensor. . . . . . . . . . . . . 302.1.2 Momentum flux, gas pressure and fluid equations. . 332.1.3 Cartesian tensors, some definitions. . . . . . . . . . . 34

2.2 Field Momentum and Maxwell Stress . . . . . . . . . . . . . 372.2.1 Energy conservation: Poynting’s theorem . . . . . . 372.2.2 Momentum conservation: Maxwell stress . . . . . . . 39

2.3 Radiation Pressure. . . . . . . . . . . . . . . . . . . . . . . . 41

3 A Frame Invariant Electromagnetism 453.1 The Lorentz Transformation . . . . . . . . . . . . . . . . . . 463.2 The Moving Charge and Wire Experiment. . . . . . . . . . 503.3 Maxwell in Terms of Potentials. . . . . . . . . . . . . . . . . 553.4 Generalized Coordinates. . . . . . . . . . . . . . . . . . . . 563.5 Four Vectors and Four Vector Calculus. . . . . . . . . . . . 60

3.5.1 Some mechanics, Newton’s laws. . . . . . . . . . . . 603.5.2 Some four vector calculus. . . . . . . . . . . . . . . . 63



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4.4.1 Manifestly covariant electrodynamics. . . . . . . . . 754.5 Transformation of the Fields . . . . . . . . . . . . . . . . . . 764.6 Field from a Moving Point Charge. . . . . . . . . . . . . . . 774.7 Retarded Potential. . . . . . . . . . . . . . . . . . . . . . . . 81

A Revision Problems 91A.1 Static Magnetic Fields. . . . . . . . . . . . . . . . . . . . . . 91A.2 Static Electric Fields. . . . . . . . . . . . . . . . . . . . . . 93A.3 Conservation and Poynting’s Theorem. . . . . . . . . . . . . 93A.4 The Wave Equation: Linearity and Dispersion. . . . . . . . 94A.5 Free Space EM Waves I. . . . . . . . . . . . . . . . . . . . . 95A.6 Free Space EM Waves II. . . . . . . . . . . . . . . . . . . . 95A.7 EM Waves in a Dielectric. . . . . . . . . . . . . . . . . . . . 95A.8 Dielectrics and Polarization. . . . . . . . . . . . . . . . . . . 96A.9 EM Waves in a Conductor: Skin Depth. . . . . . . . . . . . 96A.10 Cavity Resonator. . . . . . . . . . . . . . . . . . . . . . . . 97

B Solutions to Revision Problems 99B.1 Static Magnetic Fields. . . . . . . . . . . . . . . . . . . . . . 99B.2 Static Electric Fields. . . . . . . . . . . . . . . . . . . . . . 101B.3 Conservation and Poynting’s Theorem. . . . . . . . . . . . . 102B.4 The Wave Equation: Linearity and Dispersion. . . . . . . . 103B.5 Free Space EM waves I. . . . . . . . . . . . . . . . . . . . . 104B.6 Free Space EM Waves II. . . . . . . . . . . . . . . . . . . . 105B.7 EM Waves in a Dielectric. . . . . . . . . . . . . . . . . . . . 105B.8 Dielectrics and Polarization. . . . . . . . . . . . . . . . . . . 106B.9 EM Waves in a Conductor: Skin Depth. . . . . . . . . . . . 107B.10 Cavity Resonator. . . . . . . . . . . . . . . . . . . . . . . . 109

C Some Advanced Problems 111C.1 Maxwell Stress Tensor. . . . . . . . . . . . . . . . . . . . . . 111C.2 Liouville and Vlasov Theorems: a Conservation Equation for

Phase Space. . . . . . . . . . . . . . . . . . . . . . . . . . . 112C.3 Newton’s Laws and the Wave Equation under Galilean Trans-

formation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112C.4 Transformation of the Fields. . . . . . . . . . . . . . . . . . 112C.5 Metric for Flat Spacetime.1 . . . . . . . . . . . . . . . . . . 113C.6 Length of the EM field Tensor in Spacetime. . . . . . . . . . 114C.7 Alternative Form for the Maxwell Homogenous Equations. . 114C.8 Lorentz Transformation of the EM Field Tensor. . . . . . . 114

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D Solution to Advanced Problems 117D.1 Maxwell Stress Tensor. . . . . . . . . . . . . . . . . . . . . . 117D.2 Liouville and Vlasov Theorems: a Conservation Equation in

Phase Space. . . . . . . . . . . . . . . . . . . . . . . . . . . 118D.3 Newton’s Laws and the Wave Equation under Galilean Trans-

formation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119D.4 Transformation of the Fields. . . . . . . . . . . . . . . . . . 121D.5 Metric for Flat Spacetime. . . . . . . . . . . . . . . . . . . . 123D.6 Length of the EM Field Tensor in Spacetime. . . . . . . . . 126D.7 Alternative Form for the Maxwell Homogenous Equations. . 127D.8 Lorentz Transformation of the EM Field Tensor. . . . . . . 127

E Vector identities 129E.1 Di!erential Relations . . . . . . . . . . . . . . . . . . . . . . 129E.2 Integral Relations . . . . . . . . . . . . . . . . . . . . . . . . 130

F Tensors 133F.1 Cartesian Tensors . . . . . . . . . . . . . . . . . . . . . . . 133F.2 Special Tensors . . . . . . . . . . . . . . . . . . . . . . . . . 135F.3 Generalized Tensors . . . . . . . . . . . . . . . . . . . . . . 135

F.3.1 General properties of spacetime . . . . . . . . . . . . 136F.3.2 Flat spacetime . . . . . . . . . . . . . . . . . . . . . 137

G Units and Dimensions 139G.1 SI Nomenclature . . . . . . . . . . . . . . . . . . . . . . . . 140G.2 Metric Prefixes . . . . . . . . . . . . . . . . . . . . . . . . . 141

H Dimensions and Units 143H.1 Physical Quantities . . . . . . . . . . . . . . . . . . . . . . . 143H.2 Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146

I Physical Constants (SI) 147

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List of Figures

4.1 Charge q is at rest at the origin of the S! frame and is movingwith velocity vx1 w.r.t. frame S. The origins of the S andS! frames coincide at t = 0. . . . . . . . . . . . . . . . . . . 78

4.2 Sketch of E1 and E2 at point P versus x!1/! = x1 ! vt. . . . 79

4.3 Sketch of the E! field around a charge at rest, and the E fieldaround a charge moving at relative velocity v, in the x1, x2

plane. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 804.4 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 Sframe. In the S! frame the volume element contains charge"!dx!

1dx!2dx!

3. The origins of the two frames are coincidentat t = t! = 0. . . . . . . . . . . . . . . . . . . . . . . . . . . 81






1dx!2dx!


1.1 Surface element dS on surface S spanning curve C with lineelement dl. . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.2 Surface element dS on volume V . . . . . . . . . . . . . . . 101.3 The right hand rule . . . . . . . . . . . . . . . . . . . . . . 11

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8 LIST OF FIGURES

1.4 Location of charges q1 and q2 . . . . . . . . . . . . . . . . . 121.5 The electric field due to charge qi = "(r!)dV ! . . . . . . . . 131.6 Flux across dS is maximal when v is parallel to dS, and zero

when v is perpendicular. . . . . . . . . . . . . . . . . . . . . 141.7 Volume vdt containing particles that cross dS in time dt . . 141.8 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. . . . . . . 15

1.9 Side view of cone enclosing single charge q. The surfaceelement vector dS and field E lie in the plane of the paper. 16

1.10 Cone enclosing single charge q in cartesian and cylindricalpolar coordinates. . . . . . . . . . . . . . . . . . . . . . . . . 17

1.11 The closed surface S has zero nett flux through it. . . . . . 191.12 E · dl is integrated around the closed curve C in the vicin-

ity of a point charge. The radial sections of the path givecontributions of equal size and opposite sign and the sec-tions along contours of the potential (dotted lines) give zerocontribution. The integral around the closed curve is zero. . 21

1.13 The observer is at rest in frame 1 (top) and moves with thewire in frame 2 (bottom). . . . . . . . . . . . . . . . . . . . 22

1.14 The observer is at rest and the wire moves on conducting rails. 231.15 Wire element dl sweeps out surface element dS in time dt. . 241.16 Charge flows out of volume V . . . . . . . . . . . . . . . . . 261.17 An ideal discharging capacitor. . . . . . . . . . . . . . . . . 27

2.1 Forces acting on fluid element dV form the components of apressure tensor . . . . . . . . . . . . . . . . . . . . . . . . . 30

2.2 Vector r in the x, y, z and x!, y!, z! coordinate systems. . . . 35

3.1 The light clock is oriented perpendicular to its direction ofmotion. Top: observer rest frame ie the clock moves past us;bottom: clock rest frame ie we move with the clock. . . . . 47

3.2 The light clock is oriented parallel to its direction of motion.Top: observer rest frame ie the clock moves past us; bottom:clock rest frame ie we move with the clock. . . . . . . . . . 48

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 iscarried by the protons, the test charge q is at rest. . . . . . 51

3.4 The (dashed) loop over which we integrate B · dl is a circleof radius r centred on the wire, with r transverse to thedirection of motion u. Also shown is the direction of u " B. 53

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LIST OF FIGURES 9

3.5 The (dashed) surface over which we integrate E · dS is acylinder of radius r centred on the wire, with r transverse tothe direction of motion u. . . . . . . . . . . . . . . . . . . . 53






1dx!2dx!


4.5 Volume element dV ! and point P located w.r.t. a single fixedorigin O. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

A.1 Problem 1a: A straight wire carrying a steady current I. . . 92A.2 Problem 1b: A straight wire carrying a steady current I. . . 92

C.1 A pair of wires . . . . . . . . . . . . . . . . . . . . . . . . . 113

D.1 Surface S enclosing element of a wire of length dl. . . . . . 122D.2 Curve C enclosing the wire with element along the curve dl. 122

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Chapter 1

A Brief Tour ofElectromagnetism.

So far you will have encountered various expressions from electrostaticsand magnetostatics which we will show can then be synthesised into fourequations, the Maxwell equations which, with the Lorentz force law area complete description of the behaviour of charged particles and electro-magnetic fields. Electromagnetism is usually presented in this way for tworeasons , first this is how electromagnetism was first discovered experimen-tally and second, these expressions (Coulomb’s law, Lenz’ law, Faraday’slaw, Biot Savart law and so on) are useful in particular circumstances.

Here we will first take a look at how the Maxwell equations are con-structed, from the experimentally determined expressions and by using themathematics of vector fields. The Maxwell equations are a unification ofelectric and magnetic fields that are inferred experimentally.

The unified equations yield an important prediction: the existence ofelectromagnetic waves, which then compels us to develop a formalism thatis consistent with special relativity. This leads to a form for the Maxwellequations, the Lorentz force law and the laws of mechanics that are frameinvariant and thus consistent with the requirement that physical laws arethe same in all frames of reference, anywhere in the universe.

1.1 The Building Blocks.

We will use the experimentally determined:

• Coulomb’s Law

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C

dl

dS

Figure 1.1: Surface element dS on surface S spanning curve C with lineelement dl.

dS

V

Figure 1.2: Surface element dS on volume V

• Faraday’s Law

• Ampere’s Law

and some mathematics:

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

!

CA · dl =

"

S# " A · dS (1.1)

where surface S spans curve C as shown in figure 1.1

• Gauss’ or Divergence Theorem:

!

SA · dS =

"

V# · AdV (1.2)

where surface S encloses volume V as shown in figure 1.1.

• and the right hand rule

i " j = k

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1.2. MAXWELL I AND II 13

v

F

B

k

ji

Figure 1.3: The right hand rule

which is built in to the way the cross product is defined, so that in theLorentz force law the magnetic force

FM = qv " B

is positive for protons (figure 1.1).If we had used a left handed cross product so that i " j = !k then

Lorentz force law would be written FM = !qv "B for protons. We havealso defined the electric (vector) field to point away from positive charge sothat FE = qE for protons.

1.2 Maxwell I and II

Coulomb experimentally determined the force between two static chargesto be:

F21 =1

4#$0q1q2

| r1 ! r2 |3 (r1 ! r2) (1.3)

where the force is in Newtons (N), charge q is in Coulombs (C), r in metersand $0 = 8.85.10"12C2N"1m"2 is the permittivity of free space (see figure1.2).

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/r2

In addition we can show that the force obeys the principle of superpo-sition so that for a collection of n charges; the force on the jth charge qj

is

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12r r

r-rq

0

q211 2

Figure 1.4: Location of charges q1 and q2

Fj =1

4#$0

n#

i#=j

qiqj

| ri ! rj |3 (ri ! rj) (1.4)

We then define the electric field at the position of the jth charge as

E =Fj

qj(1.5)

that is, the Lorentz force law for electric field only, so that E is in NC"1

(we will see that it is also in V m"1 from energy considerations).The vector field E(rj) is then defined at the position r = rj of the test

charge qj :

E(rj) =1

4#$0

#

i#=j

qi

| ri ! rj |3 (ri ! rj) (1.6)

This can be expressed in terms of a scalar field, the charge density "(r)provided that the collection of point charges can be treated as a smoothlyvarying function, that is

qi = "(r!)dV ! (1.7)

This description will therefore be valid on length and timescales overwhich 1.4 to 1.7 hold. When we consider small length and timescales thereare two distinct considerations. First, we require that the charge den-sity 1.7, and all other quantities that will be described by scalar and vectorfields here, such as electromagnetic fields, energy and momentum densitiesand so forth, are still describable by continuous functions. Second, we needto use the correct mechanics in the equations of motion for the charges andhence the Lorentz force. Here we will develop electrodynamics in terms

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dV’

0

’

’

E(r)

r r

r-r

Figure 1.5: The electric field due to charge qi = "(r!)dV !

of mechanics that is consistent with special (and general) relativity, butis classical. At some length and timescale quantum mechanics is neededto replace classical mechanics. The field equations may also need to bequantized (to give Quantum Electrodynamics)1. In this sense, the Maxwellequations, the Lorentz force and the mechanics of special relativity that wewill discuss are classical .

With these caveats, we can write the electric field in terms of a volumeintegral over the charge density:

E(r) =1

4#$0

"

V

"(r!)(r ! r!)| r! r! |3 dV ! (1.8)

hence the electric field retains the experimentally determined properties:it points radially away from an element of positive charge, its magnitude isinverse square with distance, it is proportional to the charge and it obeysthe principle of superposition. We will now take these properties and phrasethem 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. Wecan explore the concept with a simple example; a cold gas2 comprised of

1The need for a field theory that permits both a continuous, classical limit, anddiscrete quanta (photons in the case of electromagnetic fields) was highlighted by thediscovery of the photoelectric e!ect

2We will make use of the ”cold gas” model several times in this book, in all cases itis defined as here: a population of identical particles which all have the same velocityv(r, t) at any given position and time.

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dS

dS

v(r)v(r)

Figure 1.6: Flux across dS is maximal when v is parallel to dS, and zerowhen v is perpendicular.

v(r)

dS

v(r)

vdtl cos

!

l

!

Figure 1.7: Volume vdt containing particles that cross dS in time dt

particles with number density n(r) per unit volume all of which have thesame 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 infigure 1.2.1. To find the flux across dS we just need to identify the volumecontaining all particles that will cross the surface element per second. Thisvolume is sketched in figure 1.2.1, where we have rotated our point of viewsuch that both v and dS are in the plane of the paper (we can always dothis, since the two vectors will define a plane). The projection of dS in thisplane is l and perpendicular to the plane is a so that | dS |= la. The anglebetween v and dS is %. If the surface element dS is su"ciently small thatn and v are constant across it, then in time dt all particles in the cold gas

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Sq

r

dS

Figure 1.8: Surfaces enclosing single charge q. The surface element dS lieson the arbitrary surface S and forms the end of the cone. There is no fluxof E across the sides of the cone.

in volumela cos %vdt =| dS | cos %vdt = v · dSdt

cross the surface element dS. The number of particles crossing dS in timedt is then nv ·dSdt, so that a flux of nv ·dS particles crosses dS per second.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:

Flux ="

Snv · dS (1.9)

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

1.2.2 Flux of E

Now we can write down the flux of any vector field. For the electric fieldwe will find the flux due to a single positive charge q which is locatedsomewhere inside the closed surface S

Flux of E =!

SE · dS (1.10)

We will utilize the fact that we can choose any convenient S as long as qis located inside, and will ”build in” the three experimentally determinedproperties of E = F/q from Coulomb’s law. We start by taking a surfaceelement dS on an arbitrary surface enclosing q. We can then consider thecone shaped surface with sides formed by radius vectors from q and an endformed by dS as shown in figure 1.2.2. From Coulomb’s law:

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r

qdS

n

E(r)"

dS

Figure 1.9: Side view of cone enclosing single charge q. The surface elementvector dS and field E lie in the plane of the paper.

1. E points radially out from the charge: that is, E points in the rdirection. There is then no flux of E out of the sides of the cone. Theflux of E must emerge from the end of the cone dS.We can again sketch the surface element dS choosing the plane of thepaper to be the plane defined by the vectors dS and E and the coneis shown cut by this plane in the figure 1. The flux of E out of thecone is then

E · dS =| E | dSn (1.11)

where dSn is the projection of dS parallel to E (i.e. E is normalto the plane in which surface dSn lies) as shown in figure 1. Theelement dSn has a very useful property: it lies on the surface of asphere centred on the charge q.The element dSn is sketched in cylindrical polar coordinates with q atthe origin in figure 1, where the element on the surface of the sphereat radius r is

dSn = rd% % r sin %d& (1.12)

We can now exploit

2. E is inverse square and

3. E is proportional to q so that

| E |= 14#$0

q

r2

then the flux of E through the cone

E · dS = E % dSn =1

4#$0q

r2% rd% % r sin %d&

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!dr sin #

rd!

z

xy

#

!

d#

d!

Figure 1.10: Cone enclosing single charge q in cartesian and cylindricalpolar coordinates.

=q

4#$0d% sin %d&

which is independent of r (E is inverse square and the surface area ofthe sphere is proportional to r2). Now the evaluation of the flux ofE over the entire closed surface is simply the integral over the solidangle 4#:

!

SE · dS =

q

4#$0

" !

0sin %d%

" 2!

0d& =

q

$0

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

!

SE · dS =

!

S(E1 + E2 + E3 + ...) · dS

=!

SE1 · dS +

!

SE2 · dS +

!

SE3 · dS + ...

=q1

$0+

q2

$0+

q3

$0+ ...

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=n#

i=1

qi =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 integratedover the volume V enclosed by S

Q ="

V"dV (1.13)

then we have Gauss’ law in integral form!

SE · dS =

1$0

"

V"dV (1.14)

The di!erential form follows immediately from the Divergence theorem 1.2"

V# · EdV =

1$0

"

V"dV (1.15)

which we can write as:

MAXWELL I : # · E = "#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 sensethat the fields E(r), # · E(r) are defined.

1.2.3 Flux of B

Similarly we can write a Maxwell equation for magnetic flux. If we definethe magnetic flux through any surface S as

"

SB · dS = #B (1.17)

Again, experimentally, it can be shown that the magnetic flux through anyclosed surface is zero so that

!

SB · dS = 0 (1.18)

which immediately from Divergence theorem 1.2 gives

MAXWELL II # · B = 0 (1.19)

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S

EB B

S

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

Field lines are drawn passing through a closed surface S in figure 1.2.3. Theelectric field has nonzero divergence so that closed surfaces can be foundthat yield zero or nonzero nett flux from 1.14 depending upon whethercharge is enclosed in the surface. The magnetic field has zero divergence sothat 1.18 always yields zero nett flux as far as we are able to determineexperimentally. Lines of E end on electric charge whereas lines of B arecontinuous. No particle has yet been identified that is a source of magneticfield, but as we shall see in Chapter 4 section 2, the Maxwell equationsthemselves and special relativity does not preclude the existence of suchparticles, that is, of magnetic monopoles.

1.2.4 Conservative and nonconservative fields.

Maxwell I was obtained from Coulomb’s law by considering the force ona test charge and then using this to define the electric field as E = F/qNC"1. Another definition is in terms of the work done on the charge as itis moved around in the electric field. The energy gained by charge q fromthe field as it is moved along path C is

W ="

CF · dl = q

"

CE · dl (1.20)

from the Lorentz force law (this is minus the work done by the field on thecharge). The work done per unit charge W/q (in units JC"1) is defined asthe 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 electrostaticfield via 1.20. Knowing that the field is radial and is any function of r, for

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the field from a single point charge we have"

CE · dl =

" r2

r1

E(r)dr (1.21)

Now we can choose C(r, %,& ) (in spherical polar coordinates) to have any%,& dependence, 1.21 will yield the same result which will depend only onthe endpoints r1 and r2. It then follows that 1.21 is zero if r1 = r2, thatis, if the path C is closed. The electrostatic field is conservative. This willalso hold for the field from any collection of charges, since by the principleof superposition we can write 1.20 as a sum of integrals due to the electricfield from each point charge.

For the electrostatic field, taking 1.21 around a closed path and usingStokes’ theorem 1.1 then immediately gives

#" E = 0 (1.22)

then since for any scalar field &

#" (#&) = 0

we can write the conservative electrostatic field in terms of a potential:

E = !#& (1.23)

For magnetic fields, Maxwell II implied that magnetic field lines formclosed loops. From figure 1.2.3 one might expect the magnetic field tohave curl whereas the electrostatic field does not; this is what is foundexperimentally and Ampere’s law of magnetostatics is:

!

CB · dl = µ0I (1.24)

where the current I is in Amperes (A & C/s) from which we can nowdefine the units of B as Tesla (T ) and the permeability of free spaceµ0 = 4#.10"7TmsC"1. Again, if the collection of (moving) charges canbe treated as a smoothly varying function, the current that they carry Iflowing across surface S can be expressed in terms of a vector field, thecurrent density J:

I ="

SJ · dS (1.25)

This, along with 1.1 immediately gives, for magnetostatics:

#" B = µ0J (1.26)

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1.3. MAXWELL III AND IV 23

E

dl

C

#

+q

Figure 1.12: E·dl is integrated around the closed curve C in the vicinity of apoint charge. The radial sections of the path give contributions of equal sizeand opposite sign and the sections along contours of the potential (dottedlines) give zero contribution. The integral around the closed curve is zero.

Then B is nonconservative unless there are no currents. The special case ofcurrent free systems (or regions) can be treated by defining a magnetostaticscalar potential in analogy to 1.23. Generally we define a magnetic vectorpotential

B = # " A (1.27)

Since for any vector field # · (# " A) = 0 this satisfies # · B = 0. Thepotentials & and A are an equivalent representation for the electrostatic andmagnetostatic fields. In section 1.3 we will complete this representation byexplicitly 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 consideringsystems that change with time.

1.3.1 Faraday’s law and Galilean invariance.

It is found experimentally that if the magnetic flux passing through a loopof wire changes for any reason, a voltage is induced across the wire. This

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B

B

v

v

v B^

Figure 1.13: The observer is at rest in frame 1 (top) and moves with thewire in frame 2 (bottom).

is Faraday’s law:!

CE · dl = !d#B

dt= ! d

dt

"

SB · dS (1.28)

where the magnetic flux is through surface S spanned by the wire loopforming 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 themagnetic field, so that the magnetic flux changes, and we will look at whathappens to the charges in the wire in two frames 3. The two frames areshown in 1.3.1.

1. We are in a frame where the magnetic field is time independent, andthe wire moves through the field.

2. We transform frames to move with the wire, so that the magneticfield depends on time.

3Note that this is a non relativistic, or Galilean, frame transformation. The relativistictreatment is in Chapter 3

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Bdlc

v

a

b

d

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

In frame (1) the Lorentz force acts on the charges (electrons) in thewire so that Fe = !ev "B (when a steady state is reached the ends ofthe wire become charged, and a back e.m.f is induced, that is, an electricfield that acts opposite to Fe). When we transform to frame (2) we (theobserver) see a stationary wire. The electrons still respond to the Lorentzforce however, so we now conclude that Fe = !eE. So for the Lorentzforce law to work in both frames, v " B in one frame is just equivalent toE in another. E and B are frame dependent and form a single quantity,the electromagnetic field; a form of the Maxwell Equations must thereforealso exist that is frame invariant, and we will derive it later.

This electric field that is implied by the frame transformation modifiesthe conservative, curl free electrostatic field. We will now calculate its curlfor the moving wire.

To stop the charges ”piling up” at the ends of the wire we will com-plete the circuit by running the wire on conducting rails as sketched infigure 1.3.1; the rails and the rest of the circuit are at rest w.r.t. the ob-server.

The wire and rails then form a closed loop shown in figure 1.3.1. Thework done on the electrons around the loop is:

!F · dl =

" b

aF · dl +

" c

bF · dl +

" d

cF · dl +

" a

bF · dl (1.29)

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

!F · dl =

" b

aF · dl = !e

" b

a(v " B) · dl (1.30)

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dl

dS

v dt

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

in the observer’s rest frame. To evaluate 1.30 consider the small element ofwire dl shown in 1.3.1. The area element swept out by wire element dl intime dt is dS = vdt " dl. Rearranging the r.h.s. of 1.30 gives!

F·dl = e

" b

aB·(v " dl) = e

" b

aB· dS

dt= e

d

dt

" b

aB·dS = e

d#B

dt(1.31)

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

!e

!E · dl & e

d

dt

"

SB · dS (1.32)

ie, 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 di!erent frames. To obtain a relationship between E and B we needto work in a single frame: so let us consider a loop of wire a rest w.r.t theobserver so that dS is fixed, and B changes with time. In this frame dS isfixed and v = dr/dt = 0 so from chain rule:

dBdt

='B't

+'B'r

· v (1.33)

then 1.32 becomes !E · dl = !

"

S

'B't

· dS (1.34)

Using Stokes’ theorem 1.1 this is just Maxwell III in di!erential form:

MAXWELL III # " E = !$B$t (1.35)

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Maxwell III ( 1.35) gives the nonconservative part of the electric field thatarises when the electromagnetic fields are time varying. It also expressesthe equivalence of E and B implied by the Galilean frame transformation.Implicit in Maxwell III is the frame transformation:

E2 = E1 + v "B1 (1.36)

where the subscripts refer to frames 1 and 2 and v is the transformationvelocity. Using the principle of superposition we have added the (arbitrary)electric field in frame 1, E1, which to simplify the above discussion we as-sumed to be zero. The nonrelativistic 1.36 was needed to make the Lorentzforce law Galilean frame invariant; in section 4.5 equation 1.36 will begeneralized for Lorentz frame invariance.

Since #"E is no longer zero we cannot describe this field as the gradientof a scalar potential. To retain B = # " A we can use

E = !#&! 'A't

(1.37)

which is consistent with 1.35 and B = # " A.

1.3.2 Ampere’s law and conservation of charge.

Recall from section 1.2.4 that when the fields and currents are steady wehave Ampere’s law 1.26,

#" B = µ0J (1.38)

which will now be amended to apply to time dependent situations.To work out what happens to the fields when charge and current den-

sities change with time we will make one important assumption, that nettcharge is conserved that is, single charges are neither created or destroyed.Experimentally this has been verified to high precision, but we will assumethat 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 arbitraryfinite volume containing some total charge that is varying with time Q(t),as charges flow out of the volume as in figure 1.3.2.

There is a total current flowing out of volume V :

I = !'Q't

= !"

V

'"

't(1.39)

and as I flows across the surface S which encloses V we can also writefrom 1.25:

I =!

SJ · dS = !

"

V

'"

't(1.40)

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q

q

q Q(t)

S

V

dSq

q

q

Figure 1.16: Charge flows out of volume V .

This can be written in di!erential form using 1.2:

# · J = !'"'t

(1.41)

Equation 1.41 immediately shows the problem with 1.26 in time dependentsituations; the divergence of 1.26 gives # · J = 0, that is, currents mustclose in steady state. When we obtained 1.41 we allowed nett current toflow out of a closed surface, so that figure 1.3.2 shows a divergence of J.The correction needed for 1.26 can be found using 1.16 to rewrite 1.41 as

# · J +'"

't= # ·

$J + $0

'E't

%= 0

equating this with # · (# " B) = 0 then gives:

MAXWELL IV#" B = µ0J + µ0$0'E't

(1.42)

which is Ampere corrected for time dependent fields. We have added adisplacement current to the r.h.s. of 1.26.

How does this work in practice? A simple example of the fields aroundan ideal capacitor that is discharging are sketched in figure 1.3.2, where# " B is given by the conduction current flowing in the circuit outsideof the capacitor plates, and by the displacement current due to the timedependent electric field between the plates.

1.4 Electromagnetic Waves.

It is now straightforward to show that Maxwell’s equations support freespace waves. We use the vector relation:

# " (# "A) = #(# ·A) !#2A (1.43)

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1.4. ELECTROMAGNETIC WAVES. 29

++

+

+

+B

(t)B B(t)

I(t) I(t)

E(t)

++

+

----

----

-Q(t) +Q(t)

(t)

Figure 1.17: An ideal discharging capacitor.

and take the curl of Maxwell IV 1.42 to give

#(# · B) !#2B = µ0#" J + µ0$0'

't#" E (1.44)

which, with Maxwell II 1.19 and III 1.35 and in free space where there areno currents J = 0 gives

#2B =1c2

'2B't2

(1.45)

where µ0$0 = 1c2 . If instead we take the curl of Maxwell III 1.35 a similar

procedure gives (with no charges in free space " = 0)

#2E =1c2

'2E't2

(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 ( and wavenumber k of the form:

B,E $ f((t ! k · r) (1.47)

is a solution to 1.45 and 1.46. These waves will propagate at phasespeed (/k = c. Crucially, we identify these with light waves in freespace. Since 1.45 and 1.46 are linear, we can superpose any solutions of theform 1.47 into a wave group or packet, and this will be nondispersive. Thewavegroup will have speed c and will carry energy and momentum throughfree space4.

4The properties of these waves are explored in the revision problems

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In the next chapter we will discuss momentum flux in the cold gas(defined in section 1.2.1), in terms of distributed bulk properties (such asthe momentum flux density tensor) rather than the motions of individualparticles. Electromagnetic fields can be quantized, that is, treated as acollection of photons which carry energy and momentum. We would thenexpect the free space electromagnetic fields to have equivalent energy andmomentum flux (the latter given by the Maxwell stress tensor).

1.5 Conservation Equations.

Equation 1.41 is a conservation equation and perhaps not surprisingly, allconservation equations are of this form; they embody the premise thatthe particles and the quantity that they carry (in this case, charge) isneither created nor destroyed. If we recall the cold gas from section 1.2.1of number density n(r, t) with each particle moving with the same velocityv(r, t) carrying charge q, then " = 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 di!erent populations of particles asnecessary, each with a di!erent q and/or v(r), to represent a gas with finitetemperature composed of several particle species of di!erent charge. Thenett result will still be an equation of the form of 1.48. If all the equationsdescribing our system (ie the Maxwell equations) are also linear then anyquantity, such as energy, mass, momentum, that can be envisaged as beingcarried by particles will have a conservation equation of the form 1.48. Atthis point it could be argued that the electromagnetic fields are known tobe particulate, ie composed of photons, then conservation equations canbe found to include field energy and momentum as well as that of thecharges. However, to obtain 1.48 from its integral form we represented theensemble of particles in the gas at position r and time t with the numberof particles in elemental volume n(r, t)dV and the flux of particles acrossan elemental surface n(r, t)v(r, t).dS. The assumption of smoothness 1.7has been made, that is, we are on spatiotemporal scales over which fieldsand charge densities behave as smoothly varying functions. So in our fieldtheory here, it is the charges that have been ”smoothed out”, rather thanthe fields treated as photons.

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Chapter 2

Field Energy andMomentum.

So far we have discussed the free space macroscopic electromagnetic fieldsand the field equations: the Maxwell equations that describe how the fieldsevolve in space and time. Charges are included in this description as macro-scopic charge density and current density. However point charges carryenergy and momentum, the Maxwell equations have wave solutions and wemight expect this to imply that the fields carry energy and momentum also.Indeed, the energy and the radiation pressure of waves, and of photons hasbeen measured experimentally.

In this chapter we will formalize the concept of the energy and momen-tum of the electromagnetic fields. This is most easily achieved throughconservation equations, and we will use the cold gas model from Chapter 1in this context. In a gas, individual particles carry momentum which is avector quantity, the gas as a whole, when described by macroscopic or fluidvariables has a corresponding tensor pressure. We will first use the simplercold gas model to introduce Cartesian tensors, and their role in equationsfor conservation of momentum flux. We will then find the Maxwell StressTensor that is, the ”ram pressure tensor” for the electromagnetic field.

The cold gas model will allow us to treat a system with free charges andelectromagnetic fields (ie E and B). This is readily generalized to linearmedia1.

1The Maxwell equations and conservation equations for free space and charges dis-cussed here are linear. Our approach then generalizes to media in which these equationsremain linear. This is the case if the fields induced in the medium are linearly propor-tional to those in the surrounding free space. See the revision problems for examples.

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32 CHAPTER 2. FIELD ENERGY AND MOMENTUM.

PxxPxy

Pxz

Pyx

Pyy

Pyz

z

x

y

Figure 2.1: Forces acting on fluid element dV form the components of apressure tensor

2.1 Tensors and Conservation Equations.

2.1.1 Momentum flux density tensor.

Tensors arise in macroscopic or bulk descriptions such as gases, fluids, andsolids where we have vector fields describing distributed bulk propertiesrather than properties at a vanishingly small point (such as at a particle).A cube shaped elemental volume dV in a gas is sketched in figure 2.1.1 andthe possible forces that can act on the three faces of the cube are labelled.If we just consider the x faces, we can compress the cube by exerting forcesnormal to the surface, in the ±x direction (Pxx), we can also twist thecube by exerting shear forces tangential to the surface, in the ±y direction(Pxy), and in the ±z direction (Pxz). The same is true of the y and z faces,so that in our three dimensional gas we have nine numbers describing theforces on the gas. These nine elements constitute the pressure tensor forthe gas and we can write them as a matrix:

Pij =

&

'Pxx Pxy Pxz

Pyx Pyy Pyz

Pzx Pzy Pzz

(

) (2.1)

If there were no shear forces the tensor has three independent elements

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2.1. TENSORS AND CONSERVATION EQUATIONS. 33

and becomes:

Pij =

&

'Pxx 0 00 Pyy 00 0 Pzz

(

) (2.2)

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

Pij =

&

'P 0 00 P 00 0 P

(

) (2.3)

We will now obtain the momentum flux tensor for the cold gas in which allparticles move with the same velocity v(r) at a given position r. In a givendirection r the momentum of one particle is

pr = mvr = mv · r (2.4)

and so momentum flowing across surface element dS in time dt due to thiscomponent is

"(v · r)(v · dSdt) = "vrvndsdt (2.5)

where the term in the first bracket of the l.h.s. of 2.5 is the momentumdensity due to the r component, and the term in the second bracket isthe volume containing the particles that flow across dS in time dt. Thesubscripts r and n denote components along r, and the normal to dS, nrespectively.

The momentum flow per unit area per unit time from equation 2.5 isthe scalar

Prn = "vrvn (2.6)

We could write the momentum flux flowing in n due to all componentsof v by summing 2.6 over r = 1, 3 (that is, r = 1, 2, 3 2 the x, y, z directions)as the vector

Pn = "(vxx + vyy + vz z)vn = "vvn (2.7)

If we then sum 2.7 over n = 1, 3 (again, the x, y, z directions), we havethe momentum flux flowing in all directions due to all components as thetensor:

P = "(vxx + vyy + vz z)(vxx + vyy + vzz) = "vv (2.8)

Tensors written as vector outer products as in equation 2.8 are known asdyadics.

2Throughout this book we will use the index notation i, j to mean all values betweeni and j inclusive.

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The dyadic vv can be written out in full:

vv =vxvxxx +vxvyxy +vxvzxz

+vyvxyx +vyvyyy +vyvzyz+vzvxzx +vzvyzy +vzvz zz

= (xyz)

&

'vxvx vxvy vxvz

vyvx vyvy vyvz

vzvx vzvy vzvz

(

)

*

++++,

x

y

z

-

..../

(2.9)For any orthogonal coordinate system we can use the following shorthand:

vv =

&

'vxvx vxvy vxvz

vyvx vyvy vyvz

vzvx vzvy vzvz

(

) (2.10)

For most purposes we don’t really need to write out all the elements of thematrix 2.10. Instead, we can simply write vivj where it is assumed in thisthree dimensional, Cartesian world i = 1, 3 and j = 1, 3. The momentumflux density tensor is then written as:

Pij = "vivj (2.11)

The notation in 2.11 is useful, for example we can immediately deducefrom 2.11 that vv and hence P is symmetric as vivj = vjvi (there are only6 independent components for the cold gas) without writing out all nineelements of the matrix 2.10.

We can now calculate the force on the cube due to the momentum fluxof the cold gas. The gas particles can deliver momentum to the cube eitherby flowing out of the cube (the rocket e!ect), or by slowing down in thecube. The force on the cube of arbitrary volume V due to flow out ofparticles is minus the rate at which momentum flows out. The force dueto particles slowing down in the cube is again minus the momentum inflowrate as dS points outwards on the surface.

If we consider one component of v in the r direction, the rate at whichmomentum is delivered through dS is given by

dFr = !"(v · r)(v · ds) (2.12)

where the first bracket is the momentum density of the r component. Sowe have from all three components:

dF = !"vv · ds = !P · ds (2.13)

Thus the force on the whole volume is

Fv ="

dF = !"

SP ·ds = !

"

S"vv ·ds = !

"

V# ·P dv = !

"

V# ·"vvdv

(2.14)

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2.1.2 Momentum flux, gas pressure and fluid equa-tions.

The above then implies that the momentum flux density corresponds to aforce per unit volume:

Fv

V=

1V

"

VdF = !# · P (2.15)

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 ofmomentum in the gas. For this we will make an important restriction: wewill consider the force over a small volume dV in which the total number ofparticles is constant. This means we are considering the rate of change ofmomentum due to a local ensemble of particles slowing down, or speedingup, and we will choose dV accordingly; if this dV contains a constant massM of particles then it experiences a force

FV =dMv

dt= M

dvdt

(2.16)

Then the force per unit volume

FV

dV= "m

dvdt

(2.17)

One possibility is to choose dV to contain the same particles by movingwith the fluid at velocity v, then 2.17 becomes (using chain rule)

"mdvdt

= "m

0'v't

+ (v.#)v1

(2.18)

This is equivalent to Liouville’s theorem3

The momentum flux density tensor that we have derived is not the sameas the pressure tensor Pth. The pressure of a warm gas is the momentumflux due to the thermal or random motions of the gas particles in the restframe 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 velocityv(r, t) with which the gas as a whole moves, so that the average < u >= vand < c >= 0.

3Liouville’s theorem expresses conservation of probability density along a trajectoryin phase space. In a system with no sources or sinks of particles, we can follow the phasespace trajectory r(t), v(t) of any particle and along that trajectory the probability offinding the particle in elemental phase space volume drdv is constant. See advancedproblem 2.

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In our cold gas, the velocity due to random motions is zero so thatfor each particle u = v. From equation 2.8 above, we can find the nettmomentum flux density in the warm gas due to an average over an ensembleof particles in small volume element dV at position r at time t.

P = " < uu >= "vv + " < cc > (2.19)

In the rest frame of the gas v = 0 and the warm gas has thermal pressurePth = " < cc >. Generally then,

P = "vv + Pth (2.20)

The momentum flux density due to gas-fluid motion as a whole, "vv is theram pressure of the gas, and is just the momentum flux density of the coldgas.

If we move with the gas (so that in our frame v = 0), equation 2.18 canbe written as a fluid equation for the gas:

"mdvdt

= "m

0'v't

+ (v ·#)v1

= !# · Pth (2.21)

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

2.1.3 Cartesian tensors, some definitions.

We can write P or Pij for a tensor of the second rank where in the discussionso far i = 1, 3 and j = 1, 3 i.e. there are nine possible indices (or ninecomponents). 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 thusa tensor of rank zero and has one component. We can extend this to asmany indices as we wish, for example Pijkl is a tensor of rank four.

A rank two dyadic would be ab which can also be written aibj . Notethat all the indices are ”down” (written as subscripts) in this Cartesiansystem so that a dyadic is always of the form aibjck..., later, when weintroduce generalized coordinates as opposed to Cartesian coordinates wewill come across ajaj which has a di!erent meaning - we will cover this inChapter 3 section 4.

A key property of the tensor formalism is that it embodies coordinatetransformations. Figure 2.1.3 shows a vector r w.r.t two orthogonal coordi-nate systems: x, y, z and x!, y!, z!. The primed axes are simply the unprimedaxes rotated in space (x, y, z ' x!, y!, z!). A vector in the unprimed frame:

r =

&

'xyz

(

) (2.22)

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z

x’

r

x

z’

y

y’

Figure 2.2: Vector r in the x, y, z and x!, y!, z! coordinate systems.

will have coordinates in the primed frame:

x!

= a11x +a12y +a13zy

!= a21x +a22y +a23z

z!

= a31x +a32y +a33z

(2.23)

The tensor aij contains the direction cosines of the rotation x, y, z 'x!, y!, z!. This can be written as

r! = A · r (2.24)

where

r!=

&

'x

!

y!

z!

(

) and A =

&

'a11 a12 a13

a21 a22 a23

a31 a32 a33

(

) (2.25)

Instead of the column vector notation we can use indices; since x = r1,y = r2 and z = r3, so equation 2.24 can be represented by

r!

j = ajiri (2.26)

which is shorthand (the Einstein summation convention) for

r!

j =#

i=1,3

ajiri (2.27)

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The sum over index i on the r.h.s. of 2.27 contracts the number of indicesby one (the l.h.s. only has single index j). Equations 2.26 and 2.27 arean example of a tensor dot (or inner) product. Again, when we switch toindex notation, we drop all reference to the basis vectors (or axes) x, y, z.Hence equation 2.28 is also a tensor dot product:

a · T = a1T11 + a2T21 + a3T31 + a1T12 + a2T22 + ...= xjaiTij

(2.28)

which is aiTij if we again drop all reference to the basis vectors xj.So we now can write down # · T in Cartesian coordinates which is

xj'

'xiTij = # · T (2.29)

There are other special tensors. One is the trace which is )ij = 1 wheni = j and 0 when i (= j:

)ij =

&

'1 0 00 1 00 0 1

(

) (2.30)

Contracting )ij over i and j with two vectors extracts the dot product:

)ijaibj = a1b1 + a2b2 + a3b3 = a · b = aibi (2.31)

Another special tensor is the alternating tensor: $ijk = 1 when ijk =123, 231, 312 and !1 when ijk = 321, 132, 213 but 0 when any two of theindices are alike. Contracting $ijk over j and k extracts the following vectorfrom a dyadic:

$ijkajbk = $i11a1b1 +$i12b1a2 +$i13b1a3

+$i21a2b1 +$i22b2a2 +$i23b3a2

+$i31a3b1 +$i32b2a3 +$i33b3a3

(2.32)

If we consider the i = 1 terms we find that all but $123a2b3 and $132a3b2 arezero. These last two give a2b3!a3b2 which we recognise as the x componentof the vector cross product. Similarly, i = 2, 3 gives the y, z componentsrespectively, so that contracting with $ijk extracts the cross product.

In this way, vector operations, vector calculus, and coordinate transfor-mations can all be performed using index notation. This Cartesian formal-ism works fine for the three space dimensions, for six phase space dimen-sions, etc. Later, when we consider special relativity we will work in space-time coordinates, and will need to generalize this Cartesian formalism.

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2.2. FIELD MOMENTUM AND MAXWELL STRESS 39

2.2 Field Momentum and Maxwell Stress

We can now derive equations of conservation of electromagnetic energy andmomentum. Again, we will consider a volume element dV that containsa fixed number of charges, and field energy and momentum that can flowin and out of this volume element. In this sense, the charges act as asource or sink of electromagnetic energy and momentum. At a microscopiclevel, photons move in and out of dV and carry energy and momentum.Here in our field theory, we can only discuss the ”smoothed” behaviour inelement dV , and all quantities are obtained from fluid variables so that thenumber of charges is n(r, t)dV , the energy in the electromagnetic fields isU(r, t)dV and in the particles is *(r, t)dV where n(r, t) is number densityand U(r, t) and *(r, t) are the energy densities carried by the fields andparticles respectively. Strictly speaking, we would derive the conservationequations in integral form, by considering integrals over some volume V ,in a similar manner to the derivation of Maxwells’ equations in Chapter1. Here we will take the short cut and work with the di!erential formsdirectly, by assuming that all bulk quantities are well defined per unit areaor per unit volume. For field energy density for example this means thatthe energy density:

1V

"

VU(r, t)dV = U(r, t)

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 par-ticles. From the Lorentz force law, a single positive charge moving in thefields gains energy

v.mdvdt

=d

dt

$12mv2

%= qv ·E (2.33)

then n charges per unit volume gain energy density * at a rate

d*

dt= J · E (2.34)

This energy density gain by the charges has to be balanced by an energydensity decrease in the fields and an energy flux into the (unit) volumecontaining the particles. Using 1.42 to substitute for J:

J · E = E · # "Bµ0

! $0'

't

E2

2(2.35)

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Now from the form of the conservation equation in Chapter 1 (equation1.48) 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) (2.36)

2.35 can, using 1.35 be rearranged as

J ·E = !#·$

E "Bµ0

%! '

't

0$0

E2

2+

1µ0

B2

2

1(2.37)

equation 2.37 is Poynting’s theorem for the case of an ensemble of freecharges in an electromagnetic field, where the field energy density

U = $0E2

2+

1µ0

B2

2(2.38)

in Jm"3 and the Poynting flux

S =E "B

µ0(2.39)

which has units of energy per unit area per unit time (Jm"2s"1). For somearbitrary volume V , enclosed by surface S the divergence theorem 1.2 gives:

"

V# · SdV =

!

SS · dS (2.40)

so that the surface integral of the Poynting flux S gives the energy carriedout of the volume by the electromagnetic fields per unit time (dS alwayspoints outwards); energy gain by the particles is balanced by minus thePoynting flux. Rearranging Poyntings theorem:

!d*

dt='U

't+ # · S = !J ·E (2.41)

Poyntings theorem 2.41 expresses conservation of energy between the en-semble of charges and the electromagnetic fields4.

4In linear media, that is, where the response from the bound charges in the mediumis linearly proportional to the applied field, B = µrµ0H, D = !r!0E and S = E ! Hand U = 1

2 [E ·D +B ·H] then conservation of energy is still given by 2.41. In nonlinear

media 2.41 is no longer valid. See revision problem 8.

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2.2. FIELD MOMENTUM AND MAXWELL STRESS 41

2.2.2 Momentum conservation: Maxwell stress

In order for the fields to do work on the particles a force must be exertedon the particles, again given by the Lorentz force law. Again we assumethat the number of charges in volume element dV is fixed, and that we canalways 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 theparticles from the fields. For one particle with momentum pq

dpq

dt= q(E + v " B) (2.42)

then for a collection of charges with momentum density Pq = npq

dPq

dt= "qE + J " B (2.43)

This expresses the rate of change of momentum density of the charges.From an understanding of electromagnetism based on experiment we mightalso expect a conservation equation to contain a term describing the rateof change of momentum density of the fields; either because waves carrymomentum and Maxwell’s equations support wave solutions, or because ofthe particle- like behaviour of the electromagnetic wavefields. On that basislet’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 * = mc2

and momentum p = mv. Then the energy flux Sp is just

Sp = nv* = nvmc2 (2.44)

(the flux across surface element dS will be Sp · dS see section 1.2.1). Themomentum density is just

P = nmv (2.45)

so thatSp = Pc2 (2.46)

This is true for any particles, including photons. In the case of photons, theenergy flux is just the Poynting flux 2.39 that we obtained from conservationof energy. Our ”guess” for the momentum density of the fields is then:

Pf =Sc2

= $0E "B (2.47)

To obtain the conservation equation we then just rearrange 2.43 to looklike a conservation equation for momentum:

!dPp

dt='Pf

't!# · T (2.48)

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This is in (almost) the same form as 2.41 except that the terms are nowvectors; the rate of change of momentum density to the charges and fieldsare 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 "and J in 2.43 to give

"E + J " B = $0E(# · E) ! B " (#" B)µ0

+ $0B " 'E't

(2.49)

then since by expanding the l.h.s. and using Maxwell III 1.35:

'

't

$Sc2

%= !$0

0B " 'E

't+ E " (#" E)

1(2.50)

equation 2.49 is

!("E+J"B) ='

't

Sc2

! $0E(# ·E)+ $0E" (#"E)+B " (# " B)

µ0(2.51)

Now the remaining terms on the r.h.s. of 2.51 must give the divergence ofa tensor. Looking at the terms in E only, we can write

E(# · E) ! E " (# " E) = E(# · E) + (E ·#)E !#E2

2(2.52)

by using vector identity E.11. In index notation the r.h.s. of 2.52 is

Ej'

'xiEi + Ei

'

'xiEj ! )ij

'

'xi

E2

2=

'

'xiEiEj ! )ij

'

'xi

E2

2(2.53)

which is just the divergence of a tensor:

T Eij = EiEj ! )ij

E2

2(2.54)

we can use the same manipulation for the B terms in 2.51, by simply addingthe (zero) term (from Maxwell II 1.19)

!B(# · B)µ0

(2.55)

The Maxwell stress tensor in the conservation of momentum 2.48 is then

Tij = $0EiEj +1µ0

BiBj ! )ij12($0E2 +

1µ0

B2) (2.56)

Notice that we wrote 2.48 with a term !# · T where Poynting’s theoremhas a term +# · S. This is a matter of convention: a Poynting flux into avolume (# ·S negative) will correspond to a radiation pressure acting uponit (# · T positive).

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2.3. RADIATION PRESSURE. 43

2.3 Radiation Pressure.

To illustrate how Maxwell stress works, lets look at an example: a monochro-matic plane electromagnetic wave propagating in free space. This is a so-lution to the free space wave equations introduced in section 1.4.

We can show that the operators

# & !ik$$t & i(

(2.57)

apply in the case of the plane wave solution

E = E0 ei(%t"k·r)

B = B0 ei(%t"k·r) (2.58)

The operators 2.57 can now be used to save some algebra. The relationshipbetween E,B and k follows from the Maxwell equations directly. MaxwellIII 1.35 for the plane wave becomes:

!ik "E = !i(B (2.59)

which rearranges to:

B =1ck " E (2.60)

Similarly, Maxwell I and II ( 1.16 1.19) give

k ·E = 0 (2.61)k ·B = 0 (2.62)

Maxwells equations then reveal that the free space plane wave E,B,k forman orthogonal set of vectors. Again for simplicity, lets choose the directionof k to be along the z axis. The wave propagates in the z direction i.e.k = kz and

E = (Ex, Ey, 0)B = (Bx, By, 0) (2.63)

We will now write down the Maxwell stress tensor for free space planewaves. Generally

Tij = $0(EiEj ! )ijE2

2) +

1µ0

(BiBj ! )ijB2

2) (2.64)

and for the z propagating plane wave 2.63 any term with a z index includingTxz, Tzx, Tzy is zero. This gives

Tij =

&

2'$0E2

x + B2x

µ0! PEM $0ExEy + BxBy

µ00

$0EyEx + ByBx

µ0$0E2

y + B2y

µ0! PEM 0

0 0 !PEM

(

3) (2.65)

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wherePEM = +$0

E2

2+

B2

2µ0(2.66)

from 2.60 and with c = 1$µ0#0

= %k we have Bx = Ey

)µ0$0 and By =

Ex)

µ0$0 so that ByBx = !µ0$0EyEx, and the terms Txy and Tyz are zero.These relationships also simplify 2.66 to

PEM =B2

µ0(2.67)

As a result the Maxwell stress tensor for the free space plane wavebecomes diagonal with terms

Txx = $0E2x + B2

xµ0

! PEM

Tyy = $0E2y + B2

y

µ0! PEM

Tzz = !PEM

(2.68)

We can now calculate the force due to the radiation pressure, that is, therate of change of momentum delivered by the free space wave per unit area,from the momentum flux conservation equation 2.48 this is just

# · T = xj'

'xiTij (2.69)

For the z propagating wave, the fields are just functions of z and t (E,B(z, t)only) so that the only contributions to 2.69 will be from xi = z and i = j.All other (i.e the o! axis) terms are zero. We are left with

# · T = z'

'zTzz = z

'

'z[!PEM ] = !z

'

'z

$B2

µ0

%(2.70)

for the plane wave solution, using 2.57 the radiation pressure is just:

# · T =1µ0

2ikB2 (2.71)

Lets compare this with the rate of change of momentum flux in the fields:

'Pf

't='

't

E " Bµ0c2

(2.72)

Maxwell IV in free space for the plane wave solution gives, using 2.57

!k " B = µ0$0(E (2.73)

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2.3. RADIATION PRESSURE. 45

hence

Pf =(!k " B) "B

µ0c=

B2

µ0ck (2.74)

where we have used 2.62. We then have

'Pf

't= 2i(

B2

µ0ck = 2ik

B2

µ0(2.75)

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

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Chapter 3

A Frame InvariantElectromagnetism

So far we have the field equations, the Maxwell equations, and using theLorentz force law have constructed equations describing the conservationof energy and momentum between the fields and charges. Our next task isto 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 allinertial frames of reference.

• The speed of light is the same in all inertial frames.

The second of these yields the Lorentz transformation between one framemoving at constant velocity w.r.t. another.

Our new formalism for electromagnetism will tie up a number of looseends that have already appeared so far. In 1.3.1 Galilean invariance wasimposed on Faraday’s Law to give Maxwell III ( 1.35) and this showed thatE and B are equivalent: B in one frame can look like E in another( 1.36).We will in Chapter 4 obtain a single object, the Electromagnetic FieldTensor that describes the fields E and B and which gives a Lorentz trans-formation of the fields valid for relativistic frame transformations, usingthe formalism developed in this Chapter. In 1.4 we found that the Maxwellequations predict light waves in free space. Since the speed of light is con-stant under Lorentz transformation we must be able to find a form for thewave equations, and the Maxwell equations, that is invariant under Lorentztransformation. So far we have used the Lorentz force law to give the forceon individual charges but have not concerned ourselves with the equations

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48 CHAPTER 3. A FRAME INVARIANT ELECTROMAGNETISM

specifying the particle motion under that force; we will now incorporaterelativistic rather than Newtonian mechanics. Finally, we have found con-servation equations for charge, mass, energy and momentum, between col-lections of charges and the electromagnetic fields; a Lorentz invariant formmust also exist for these. Some individual quantities (such as energy) willbe di!erent between one frame and another, but other properties, suchas charge and particle number, will be (inertial) frame invariant and thisleads us to form conservation equations that are invariant under Lorentztransformation.

To construct an electromagnetic field theory that is invariant underthe Lorentz frame transformation will require a generalization of Cartesiantensors that were introduced in Chapter 2. We have seen that coordinatetransformations in x, y, z space can be written as tensor operators (or 3%3rotation matrices). The Lorentz transformation can be written as a tensoroperator in space-time (a 4 % 4 rotation matrix), and the key to unifyingelectromagnetism with special relativity is to construct four vectors andfour tensors in which the three orthogonal space coordinates of the Carte-sian system are replaced with four coordinates (three space and one time).The generalization to space-time led Einstein to the tensor formulationof general relativity where space-time is curved by the gravitational field.Here for special relativity we neglect gravity (and therefore do not treataccelerating frames), so we consider inertial frame transformation only andspace-time is flat.

3.1 The Lorentz Transformation

Special relativity arises if we insist that the speed of light is the same in allinertial frames and that physical laws look the same in all inertial frames.The consequences of this are immediately clear if we consider the followingexperiment: a ”light clock”1 is composed of a light beam bouncing betweentwo lossless mirrors. The clock is moving past us at speed u. Lets first lookat what happens when the clock is oriented perpendicular to the directionof motion as shown in figure 3.1. In our rest frame the time taken for thelight to travel from one mirror to the other is$ t!, the clock moves past andthe light beam travels a distance:

(c$t!)2 = (u$t)2 + $y2 (3.1)

1This experiment was actually performed by Michelson and Morley in 1887 to attemptto measure our relative velocity w.r.t. the aether which was believed to fill the vacuumto allow the propagation of electromagnetic waves.

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3.1. THE LORENTZ TRANSFORMATION 49

c t’

y

u

$

t’$

$

u

$y

u

Figure 3.1: The light clock is oriented perpendicular to its direction ofmotion. Top: observer rest frame ie the clock moves past us; bottom: clockrest frame ie we move with the clock.

In the clock’s rest frame the time taken for the light to travel from onemirror to the other is$ t, and the light path is simply

c$t = $y (3.2)

Now let’s compare the time in the two frames. If the distance perpendicularto the direction of motion$ y is the same in both frames then

$t!2(c2 ! u2) = c2$t2 (3.3)

so that there is time dilation

$t! = !$t (3.4)

where! =

1) 4

1 ! u2

c2

5 (3.5)

in other words ”moving clocks go slow”.Now let’s turn the clock through 90o so that it is oriented along the

direction of motion as in 3.1. In our rest frame the clock moves away fromthe forward going light pulse. When the light moves in the direction of theclock motion it travels a longer distance

c$t!f = $x! + u$t!f (3.6)

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$x

u

x’

x’$

$

u

u$t’

u$t’

b

f

Figure 3.2: The light clock is oriented parallel to its direction of motion.Top: observer rest frame ie the clock moves past us; bottom: clock restframe ie we move with the clock.

The clock moves towards the backwards going light pulse so that when thelight moves in the opposite direction to the clock motion it travels a shorterdistance

c$t!b = $x! ! u$t!b (3.7)

the total round trip time of the light pulse in the moving clock is then

$t! = $t!f + $t!b =2c$x!

c2 ! u2= !$t (3.8)

from 3.4. In the clock’s rest frame the round trip time is simply given by

c$t = 2$x (3.9)

Eliminating$ t from 3.8 and 3.9 gives the Lorentz contraction of the lengthof the clock

$x! =$x

!(3.10)

in the moving frame.In terms of distances x, x! and times t, t! measured from origins in the

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3.1. THE LORENTZ TRANSFORMATION 51

two (observer) frames, the Lorentz transformation is:

t! = !4t ! vx

c2

5

x! = !(x ! vt)y! = yz! = z

(3.11)

where the primed frame moves at velocity +vx 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 oflight is the same (c) in all frames of reference. So if we consider a point lightsource located at the origin of our x, y, z coordinate system, generating aspherical wavefront at t = 0, the wavefront must propagate a distance:

x2 + y2 + z2 = c2t2 (3.12)

at time t. If we use the Lorentz transformation 3.11 on this expression 3.12,we obtain:

x!2 + y!2 + z!2 = c2t!2 (3.13)

Whatever frame we are in, the distance propagated by the wavefront al-ways obeys an expression of the form of 3.12. This looks almost like thelength squared of a vector in x, y, z, ct space. We can formalize this idea bywriting 3.12 as

s2 = c2t2 ! x2 ! y2 ! z2 (3.14)

where for light waves s = 0 and for objects moving slower than light s2 >0 so that s is real.2 The length of s is the same in all frames, Lorentztransforming from one moving frame to another simply corresponds to arotation of the axes x, y, z, ct. We have already developed notation forcoordinate rotations in Cartesian x, y, z space. Lets see what happens if westick to Cartesian rules. If our four vector is sj and

s = (ct,x) =

&

22'

ctxyz

(

33) (3.15)

then the Lorentz transformation will be of the form:

s!j = %jksk (3.16)

2This is known as a spacelike interval, if s is imaginary then it describes a timelikeinterval, that is, an object travelling faster than c.

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where 3.16 is the shorthand for the tensor operation s! = % · s, and thespacetime coordinate rotation matrix is:

%jk =

&

22'

! ! vc! 0 0

! vc! ! 0 00 0 1 00 0 0 1

(

33) (3.17)

then if we work out the coordinate rotation we have:

s! = % · s =

&

22'

! ! vc ! 0 0

! vc! ! 0 00 0 1 00 0 0 1

(

33)

&

22'

ctxyz

(

33) =

&

22'

c!(t ! xvc2 )

!(x ! vt)yz

(

33) =

&

22'

ct!

x!

y!

z!

(

33)

(3.18)so this Cartesian operation 3.18 just gives the Lorentz transformation of(x, y, z, ct) as required. There is a problem however. Lets calculate thelength of our Cartesian four vector:

s · s & sjsj = [ct, x, y, z]

&

22'

ctxyz

(

33) = x2 + y2 + z2 + c2t2 (= s2 (3.19)

so we have a problem! The Cartesian system isn’t consistent when we try toextend our coordinate system from three space dimesions to four spacetimedimensions. The solution is to generalize the coordinate system, and we willdo this in section 3.4. First, we will look at a simple example to illustratejust why we need generalized coordinates and four vectors in a completedescription of electromagnetism.

3.2 The Moving Charge and Wire Experi-ment.

The following ”thought experiment” is a neat way to understand the con-sequences of special relativity in electromagnetism. An infinitely long wireis oriented along the direction of motion of the observer (the x direction);a section of the wire is sketched in figure 3.2. In the S1 frame the observeris at rest and the current in the wire is carried by the electrons moving tothe right with velocity ux. In the S2 frame the observer moves to the rightwith velocity ux, now the electrons in the wire are at rest and the currentis carried by the protons moving to the left at !ux. We can find out what

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3.2. THE MOVING CHARGE AND WIRE EXPERIMENT. 53

-v=0

v=0+

v=u-

q

q

J

u

v=-u+

S

S

1

J1

2

2

u

x

z

y

Figure 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 theprotons, the test charge q is at rest.

happens to the electromagnetic fields in both frames by releasing a testcharge 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 itstarts by moving to the right with velocity ux.

Now here is the dilemma: if in the S1 frame, we have zero charge densityin the wire ("1 = 0) then the only force acting on the test charge is due tothe magnetic field, ie

F1 = qu " B1 (3.20)

and E1 = 0. This will cause the charge to move transverse to the wire in they, z plane, that is, transverse to the x direction of the frame transformationvelocity. In the S2 frame, the test charge is initially at rest, so there can beno v"B force. But any motion of the charge in y or z, that is, transverse tothe direction of the frame transformation velocity will appear to be exactlythe same in both frames, from the Lorentz frame transformation 3.11. Sowhat causes the force in the S2 frame?

To resolve this, we need to calculate the electric and magnetic fieldsin the two frames. We only know the total charge density in S1. Tocalculate it in S2 we will assume that charge is invariant under Lorentztransformation, that is, that charges are not created or destroyed by movingfrom one inertial frame to the other. Then all we need to do to find thecharge density is to consider a bunch of charge Q and find the volume that

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it occupies in the two frames. Lets assume that the charge Q occupies abox with sides of lengths$ x, $y, and$ z in a frame where it is at restw.r.t. the observer. The observer measures a charge density:

" =Q

$x$y$z(3.21)

If the observer now moves in the x direction w.r.t. the charge, distanceswill Lorentz contract in the x direction from 3.10 and the observer willmeasure a charge density

"! =Q

$x!$y$z= !

Q

$x$y$z= !" (3.22)

If charges carry a current, then the corresponding charge density is al-ways larger than if the charges are at rest3 from 3.22. In our moving wireexperiment:

"+2 = !"+1""2 = ""1

&

(3.23)

then the total charge density in the two frames is:

"1 = "+1 + ""1 = 0"2 = "+2 + ""2 = "+1

6! ! 1

&

7

= "+1 !u2

c2 (= 0(3.24)

so the uncharged wire in frame S1 is charged up in frame S2. As a conse-quence there will be a (radial) electric field in S2 and this will provide theforce on the test charge q:

F1 = qu " B1

F2 = qE2(3.25)

Let’s calculate the forces in the two frames by using Maxwell’s equationsto calculate the fields.

In S1 the force is due to the magnetic field which can be obtained fromthe integral form of 1.42; we perform a line integral around the curve Cenclosing the wire as shown in figure 3.2. Then since

!

CB · dl =

"µ0J · dS = µ0I (3.26)

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

cross sectional area carrying 1 A have an average drift speed of " 10!4ms!1 which iswhy household wiring doesn’t charge up.

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A

rdl

J vB B

u Bv

u

C

Figure 3.4: The (dashed) loop over which we integrate B · dl is a circle ofradius r centred on the wire, with r transverse to the direction of motionu. Also shown is the direction of u " B.

r

A

dl

S

Figure 3.5: The (dashed) surface over which we integrate E·dS is a cylinderof radius r centred on the wire, with r transverse to the direction of motionu.

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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 thegeometry B = B(r) and has the same magnitude on all points on C):

!

CB · dl = 2#rB (3.27)

Since the current I is carried by the electrons moving at speed u throughthe wire which has cross section A, I = ""1 Au. The ”u " B” force 3.25 inframe 1 then has magnitude

F1 =q""1 A

2#$0ru2

c2(3.28)

which acts to repel the test charge as shown in figure 3.2.In S2 the force is due to the electric field which we calculate from the in-

tegral form of 1.16; we perform a surface integral over the closed cylindricalsurface shown in figure 3.2:

!

SE · dS =

"

V# · EdV =

"

V

"2$0

dV =dQ

$0(3.29)

From 1.35 E is directed radially out from the wire so there is no contributionto the surface integral from the ends of the cylinder. From the geometryE = E(r) so that !

SE · dS = 2#rEdl (3.30)

and excluding the test charge, the charge enclosed in the cylinder dQ ="2Adl. The force due to the electric field in frame S2 then has magnitude

F2 =q"2A

2#$0r=

q"+1 A

2#$0ru2

c2! (3.31)

where we have used 3.24, this again acts in the +r direction to repel thetest charge. Comparing 3.28 and 3.31 we find that the forces acts in thesame (r) direction in both frames but the magnitude di!ers by a factor of!;

F2 = !F1 (3.32)

Now recall that any motion transverse to the direction of motion u mustbe the same in both frames. What is happening here? Lets look at theequation of motion of the test charge. During the first (infinitesimallysmall) time interval$ t after it is released from rest it will gain transversemomentum

$p = F1$t1 (3.33)

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3.3. MAXWELL IN TERMS OF POTENTIALS. 57

if observed in frame S1. But in S1 the test charge is moving, and so thetime interval will be dilated w.r.t. frame S2

$t1 = !$t2 (3.34)

thenF1$t1 = F1!$t2 = F2$t2 (3.35)

so that the change in transverse momentum$ p = F$t in any frame, andthe transverse motion will be frame independent, as we would expect.

Special relativity implies that length, time, force etc. change as weLorentz transform from one frame to another. This example has shownhow the electromagnetic fields, current and charge densities are also notinvariant under Lorentz transformation. We have anticipated that spaceand time must form a single coordinate system, space-time, and in thiscoordinate system, the (3) space coordinates and and (1) time coordinateof a point combines to form a single four-vector which has length thatis invariant under Lorentz transformation. In the same way, current andcharge density are in themselves not invariant, but can be combined to forma four vector that has invariant length. This leads to a frame invariantelectromagnetism 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 rank1 tensors (four vectors) if we instead work in terms of the scalar and vectorpotentials &,A, we’ll do that first in this section. In Chapter 1 we foundthat the fields could be written in terms of potentials:

B = # " A (3.36)

E = !#&! '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 tothe currents or charges. It is easy to see that the potentials are not unique:since #"#+ is zero for any + we can add any #+ to A where + is a scalarfunction and still satisfy 3.36. This will then give new potentials:

A! = A + #+&! = &+ $'

$t

(3.38)

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which also satisfy 3.37. The operation 3.38 is a gauge transformation.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 interms of A and &. Using 3.37 to substitute for E in Maxwell I gives:

!#2&! '

't# · A =

"

$0(3.39)

and using 3.37 and 3.36 to substitute for E and B in Maxwell IV gives:

!#2A + #(# ·A) +1c2

0'

't#&+

'2A't2

1= µ0J (3.40)

We can now choose a gauge that reveals a nice symmetry in 3.39 and 3.40.This is the Lorentz gauge4:

# ·A = ! 1c2

'&

't(3.41)

and substituting 3.41 into 3.39 and 3.40 gives:

#2&! 1c2

$2($t2 = ! "

#0

#2A! 1c2

$2A$t2 = !µ0J

(3.42)

in free space (" = 0,J = 0) equations 3.42 are simply wave equations forA and & and predict electromagnetic waves with speed c. They containthe same information as the Maxwell equations, all we need is to put 3.42and the Lorentz gauge 3.41 in Lorentz invariant form and we will have adescription of electromagnetism that incorporates special relativity.

3.4 Generalized Coordinates.

From equation 3.19 we found that space-time is not Cartesian. We need ageneralized coordinate system that has the following properties:

4In electrostatics this reduces to the Coulomb gauge # · A = 0 which gives Poissonequations for both " and A

$#2" =#

!0and

$#2A = µ0J

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3.4. GENERALIZED COORDINATES. 59

1. The space-time interval

s2 = c2t2 ! x2 ! y2 ! z2 (3.43)

is invariant under Lorentz transformation.

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 fourvector s in space-time.

The system that we devise will work for any four tensor (four vectors beingthe special case of four tensors of rank 1).

We will start by defining two ”versions” of our four vector5 and for thevector s these look like:

Covariant:

x) = (ct,!x) =

&

22'

ct!x!y!z

(

33) (3.47)

and Contravariant:

x) = (ct, +x) =

&

22'

ctxyz

(

33) (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 in-dices down), contravariant (all indices up) or mixed (indices up and down).We are also going to use the summation convention introduced for Cartesiantensors 2.27. Now, if we simply define the scalar (dot, or inner) product of

5In general if we have a well defined transformation that gives x"! = x"!(x0, x1, x2, x3)a covariant vector transforms as:

a"! =

$x0

$x"! a0 +$x1

$x"! a1$x2

$x"! a2$x3

$x"! a3 =$x"

$x"! a" (3.45)

and a contravariant vector transforms as

a"! =$x"!

$x"a" (3.46)

See the appendix on tensors for details.

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two four vectors as the product of a covariant and a contravariant vector,then the length of the spacetime interval s2 is:

s2 = x)x) = x · x = xx = [ct,!x,!y,!z]

&

22'

ctxyz

(

33) = (ct)2 ! x2 ! y2 ! z2

(3.49)as required. Then we just need to know how to turn a covariant tensor intoa contravariant one, and vice versa, and from inspection of 3.47 and 3.48we simply need an operation that changes all the signs on the spacelikecomponents x1,3 = x, y, z, whilst leaving the timelike component x0 =ct unchanged. We can use the same operation as 3.49, that is, a dot orinner product or contraction over a pair of contravariant indices, this timebetween a rank 1 tensor (vector) and a rank 2 tensor:

x) = g)*x* (3.50)

and

x) = g)*x* (3.51)

where

g)* =

&

22'

1 0 0 00 !1 0 00 0 !1 00 0 0 !1

(

33) = g)* (3.52)

g)* is known as the metric of the spacetime. The spacetime interval is thenwritten as

s2 = x)g)*x* = g)*x)x* (3.53)

which in matrix notation is

s2 = (x, gx) = (gx, x) = xgx (3.54)

The spacetime metric 3.52 is defined by the form of the spacetime in-terval s2, via 3.53. This particular metric then is just what is needed toembody the flat spacetime of special relativity, in which the length of aspacetime interval (the length of s) has the same constant value anywherein spacetime, i.e. spacetime itself is uniform. The formalism that we are de-veloping here can just as easily be applied where spacetime is nonuniformor curved by gravity. The metric for curved spacetime embodies generalrelativity.

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3.4. GENERALIZED COORDINATES. 61

The dot product operation that we defined to obtain the desired be-haviour for s2 has to be the same as the dot product between two fourtensors generally. For two four tensors of any rank and index (ie covariant,contravariant, or a mixture of the two), the dot or inner product will be:

a · b = a......)b...)

... (3.55)

that is, a contraction over the index ,. Contraction with g)* or g)* changesa index from contravariant to covariant (up to down) or vice versa (downto up). It works like this:

Ccontravariant to covariant (up to down):

a......) = g)*a

...*

... (3.56)

and covariant to contravariant (down to up):

a...)... = g)*a...

...* (3.57)

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

x!

) = %)*x* (3.58)

on the contravariant form and

x)!= %)*x* (3.59)

on the covariant form. The contractions 3.58 and 3.59 change a contravari-ant four vector into a covariant one and vice versa. The rotation matrix6

is (almost) the same as that discussed in Cartesian formalism and is:

%)* =

&

22'

! ! vc! 0 0

+ vc! !! 0 0

0 0 !1 00 0 0 !1

(

33) (3.60)

where% )* is the transpose of% )* . The rotation of coordinates in spacetimewritten out in full is:

x!

) = %)*x* =

&

22'

ct!

!x!

!y!

!z!

(

33) =

&

22'

! ! vc! 0 0

+ vc ! !! 0 0

0 0 !1 00 0 0 !1

(

33)

&

22'

ctxyz

(

33) =

&

22'

c!(t ! vxc2 )

!!(x ! vt)!y!z

(

33)

(3.61)which just give the Lorentz transformation 3.11 as required (the inversetransformation matrix is then just obtained by putting v ' !v in 3.60).

6This is a member of the Lorentz Group of transformations, see e.g. Classical Elec-trodynamics 2nd Ed., J. D. Jackson, (Wiley, 1975) for details.

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3.5 Four Vectors and Four Vector Calculus.

The rules of flat spacetime were defined from the Lorentz transformationof the four vector x) but will apply to any four vector. If we can writethe laws of mechanics and electromagnetism in four vector form, then wewill have a description of mechanics and electromagnetism that is Lorentzinvariant (i.e. manifestly covariant). The required four vectors will alwayshave the same length in spacetime in all frames, and will Lorentz transformsimply by coordinate rotation in spacetime.

3.5.1 Some mechanics, Newton’s laws.

As we have shown, length and time form the four-vector:

x) =

&

22'

ct!x!y!z

(

33) (3.62)

Another example is the energy- momentum four vector:

p) =

&

22'

+c

!px

!py

!pz

(

33) = (*

c,!p) (3.63)

The length of p) should be constant. In full it is:

p)p) =8*c,!px,!py,!pz

9&

22'

+cpx

py

pz

(

33) =*2

c2! p2 (3.64)

Now recall some expressions from special relativity:

*2 = p2c2 + m20c

4 (3.65)

The energy-momentum four vector then has length

p)p) = m20c

2 (3.66)

This just gives the rest energy of a particle (p)p)/m0); if we can find aframe where the particle is at rest (ie it is an electron, proton but nota photon) then since the length of p) is Lorentz invariant, its length in

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3.5. FOUR VECTORS AND FOUR VECTOR CALCULUS. 63

all frames must be just that found in the rest frame. If the particle is atrest the spacelike components of p) are zero and the timelike component isp0 = +

c = m0c.Now from special relativity the relativistic mass m = !m0 of a moving

particle is larger than the rest mass m0, so we have in the moving frame,the timelike component given by:

p0c = * = mc2 = !m0c2 (3.67)

and the spacelike components p1,3 are given by:

p = mu = !m0u (3.68)

We can write these as a single equation in terms of four vectors:

p) = m0u) (3.69)

where we have introduced another four vector, the four velocity:

u) =p)

m0=

&

22'

c!!ux

!uy

!uz

(

33) (3.70)

The four velocity u) must transform in the same way as the energy- mo-mentum four vector as we have simply divided by m0 which is invariantunder Lorentz transformation. The expression 3.69 is manifestly covariant,that is, under Lorentz transformation it will retain the same form and thecomponent four vectors will Lorentz transform, with invariant lengths.

What about Newton’s laws? In a nonrelativistic world, frame trans-formations are Galilean and correspond to rotations of Cartesian positionvector x in x, y, z plus translations x! = x + ut, where u is the constantvelocity 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 velocityunless subject to some force.

2. Momentum is conserved (”for every action there is an equal and op-posite reaction”) under Galilean transformation.

3. The equations of motion F = dpdt and u = dx

dt are invariant underGalilean transformation.

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From (1) and (2) there will be a preferred rest frame in the sense thatx = (x, y, z) is constant and the total p = (px, py, pz) = 0.

The four vectors we have already written down for x) and p) will allowus to rewrite laws (1) and (2). We just need to look at di!erentiation w.r.t.time in (3) before we can rewrite all of Newton’s laws.

Intuitively, we might expect that since time is dilated in the movingframe 3.4, an invariant interval to replace dt in the derivative would bethe proper time dt/!. We can formalize this by considering the coordinatesof 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 havespacetime coordinates (ct, x, y, z). As the particles move apart at constantvelocity u = (ux, uy, uz), at time t+$t, they will be separated by a distance:

$x = ux$t$y = uy$t$z = uz$t

(3.71)

the spacetime interval between the two particles is then

$s2 = c2$t2 ! $x2 ! $y2 ! $z2

= (c2 ! u2)$t2(3.72)

so that an invariant interval is

$s

c=

$t

!(3.73)

The time derivative in Newton’s laws is then to be replaced by the invariantspacetime derivative

d

ds= !

d

dt(3.74)

The spacetime derivative of (four) position in spacetime is then

dx)

ds= !

d

dt

&

22'

ctxyz

(

33) =

&

22'

c!! dx

dt

! dydt! dz

dt

(

33) = u) (3.75)

i.e. the four- velocity as we would expect.Newton’s laws of motion then become:

1. The spacetime interval x) = (ct, x, y, z) has constant length underLorentz transformation.

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3.5. FOUR VECTORS AND FOUR VECTOR CALCULUS. 65

2. The energy- momentum four- vector p) = ( +c , px, py, pz) has constantlength under Lorentz transformation.

3. The equations of motion f) = dp!

ds and u) = dx!

ds are invariant underLorentz transformation.

Then f), the ”four - force”, must also be a four vector; in Chapter 4 wewill find the four vector force on a charged particle in an electromagneticfield, that is, the covariant version of the Lorentz force law.

3.5.2 Some four vector calculus.

Since we know how to define intervals in space time we can do some calculusin terms of four vectors.

First let’s consider the four gradient of some scalar field &(t, x, y, z) =&(x)) defined in spacetime. The scalar field itself is an invariant, the co-ordinates transform as the components of a four vector should. Since wehave defined & as a function of contravariant position in spacetime x) wecan use chain rule to obtain the change in scalar field d& on some spacetimeinterval ds given the contravariant coordinates of the interval dx):

d& = dt'&

't+ dx

'&

'x+ dy

'&

'y+ dz

'&

'z=

d&

dsds (3.76)

The quantity d& is also invariant, so the r.h.s. of 3.76 must simply be inthe form of the dot product of two four vectors:

d& = ')(&)dx) (3.77)

This means that the covariant four gradient is

') =

&

22'

1c$$t$$x$$y$$z

(

33) = (1c

'

't,#) (3.78)

which is almost what we would expect, except that the spacelike part ispositive. We could have instead written the scalar field as a function ofcovariant position x), ie & = &(ct,!x,!y,!z); the invariant interval wouldthen need to be written as a four vector dot product

d& = ')(&)dx) (3.79)

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so to generate 3.76 from 3.79 we need a contravariant four gradient

') =

&

22'

1c$$t

! $$x

! $$y

! $$z

(

33) = (1c

'

't,!#) (3.80)

which again has spacelike components with opposite signs than we wouldexpect.

As a consistency check, if covariant ') is a four vector then the con-travariant ') should also be defined via the spacetime metric:

') = g)*'* =

&

22'

1c$$t

! $$x

! $$y

! $$z

(

33) = (1c

'

'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 ofa dot product:

')a) = ')a) =1c

'

'ta0 + # · a (3.82)

(where a is the spacelike part of four vector field a)).Finally, the length of the four vector ') is the ”four #2” D’Alembertian:

')') =

1c2

'2

't2!#2 = ! (3.83)

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

3.6 A Frame Invariant Electromagnetism

3.6.1 Charge conservation.

Charge conservation was introduced in Chapter 1 to obtain 1.42 and so is”built in” to Maxwell’s equations. This is expressed by:

# · J +'"

't= 0 (3.84)

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which was derived in 1.3.2 by assuming that charge is conserved. If wenow insist that charge is Lorentz invariant, ie it is the same in all frames,we should have a manifestly covariant form of 3.84. Equation 3.84 has theform of the four divergence of some four vector:

')J) = 0 (3.85)

and by inspection, knowing ') we have the four current:

J) = (cp,J) =

&

22'

c"Jx

Jy

Jz

(

33) (3.86)

The length of J) is invariant under Lorentz transformation so that itscomponents, the charge and current density are not, as we found when weconsidered the moving charge and wire experiment in section 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 electromag-netic fields. Recall from section 3.3 that the Maxwell equations can bewritten in terms of scalar potential & and vector potential A, and take on(what we will see is a very useful) symmetry if we also choose the Lorentzgauge:

# ·A +1c2

'

't& = 0 (3.87)

from Maxwell. This has the same form as the charge conservation equationso again we can write

')A) = 0 (3.88)

where the new four vector is the four vector potential

A) = (&

c,A) =

&

22'

(cAx

Ay

Az

(

33) (3.89)

To obtain a manifestly covariant form of the Maxwell equations, let’s ex-amine the two inhomogenous Maxwell equations written in terms of A and& using the Lorentz gauge 3.42. The equation in terms of & (originallyMaxwell I 1.16) has been divided by c to give:

!#2&

c+

1c2

'2

't2&

c="

$0c= c"µ0 (3.90)

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and the l.h.s. of this looks like the ”four #2” of the timelike part of A).The equation in terms of A (originally Maxwell IV 1.42) is:

!#2A +1c2

'

'tA = µ0J (3.91)

and the l.h.s. of this looks like the ”four #2” of the spacelike part of A).Together they are a single 4 vector equation:

!(&

c,A) = µ0(c",J) (3.92)

or!A) = µ0J

) (3.93)

which is our equation for electromagnetism in manifestly covariant form.Since the invariance of charge implies that J) is a four vector from 3.85,3.93 shows that A) must also be a four vector. So, to Lorentz transformthe electromagnetic fields, all we need to do is write the fields in terms of Aand &, transform (ie rotate the four vectors in spacetime) A) and J), thenwork out the electromagnetic fields again from the new A and & using 3.36and 3.37. In the next Chapter we will find the electromagnetic field tensorsthat lead to a direct transformation of the fields. For the moment, wehave already shown that Maxwell’s equations are consistent with specialrelativity. The Maxwell equations can be written in terms of four vectors.In free space J) = 0 and we are left with the free space wave equations forA and & in manifestly covariant form:

!A) = 0 (3.94)

which predicts light waves moving at speed c, and holds in all frames ofreference.

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Chapter 4

The Field Tensors

In Chapter 3 we developed a formalism in which Maxwell’s equations andthe electromagnetic fields, as embodied in the scalar and vector potentialdescription, could be written in manifestly covariant form, that is, in termsof four vectors which Lorentz transform by a coordinate rotation in space-time. We now wish to write the Maxwell equations, the Lorentz force law,and the equations of conservation of field energy and momentum, explicitlyin terms of the fields E and B. This will require us to develop four tensorsof rank 2 (as opposed to the rank 1 four tensors, or four vectors, of Chapter3). The generalization of 3 coordinate Cartesian space into 4 coordinatespacetime will still apply. The Maxwell equations in the form that we haveseen in Chapter 1 have an assymetry which we will be able to explore morefully in the framework of spacetime; to gain a glimpse of what the universemight look like if monopoles exist. Finally, we can obtain general solutionto the Maxwell equations.

4.1 Invariant Form for E and B: The EMField Tensor.

Let’s begin by trying to write down an invariant form for E and B. InCartesian 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 & hasbeen replaced with the four vector

A) = (&

c, Ax, Ay, Az) = (A0, A1, A2, A3) (4.2)

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70 CHAPTER 4. THE FIELD TENSORS

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

') =$

1c

'

't,! '

'x,! '

'y,! '

'z

%= ('0, '1, '2, '3) (4.3)

to construct a ”four- curl” of A).Writing out the components of # " A gives:

Bx = $Az$y ! $Ay

$z = !('2A3 ! '3A2) = !F 23

By = $Az$x ! $Ax

$z = '1A3 ! '3A1 = +F 13

Bz = $Ay

$x ! $Ax$y = !('1A2 ! '2A1) = !F 12

(4.4)

We can arrange the components of # "A to look like combinations of thecomponents of the four vectors if we use terms such as '2A3. From ournotation these are not a contraction over an index as in the case of ”fourdiv” which had terms such as '2A2, instead, they look like components ofa four tensor of rank 2. All the three components of 4.4 have the samepattern, that is, they look like they are part of a single rank 2 tensor F)* .We might guess that more terms of F)* originate in the electric field, sinceit too is represented by components of A):

E = !#&! 'A't

(4.5)

Writing this in component form gives

Ex = !$($x ! $Ax

$t = !c('0A1 ! '1A0) = !F 01c

Ey = !$($y ! $Ay

$t = !c('0A2 ! '2A0) = !F 02c

Ez = !$($z ! $Az

$t = !c('0A3 ! '3A0) = !F 03c

(4.6)

So, the three components of B in 4.4 and the three components of E in 4.6give six terms of a rank 2 tensor

F)* = ')A* ! '*A) (4.7)

The other terms of the 4 % 4 ”four- tensor F)* can be obtained frominspection of 4.7. The on axis terms are when , = -:

F)) = ')A) ! ')A) = 0 (4.8)

that is, all four are zero (, = 0, 1, 2, 3). The o! axis terms are when , (= -,and are for example when , = 0, - = 1

F 01 = '0A1 ! '1A0 = !('1A0 ! '0A1) = !F 10 (4.9)

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4.1. INVARIANT FORM FOR E AND B: THE EM FIELD TENSOR.71

so that F)* is antisymmeric: F)* = !F *). The 4 % 4 = 16 terms in F)*

is comprised of the four on axis terms which are zero, plus the six di!erentterms from the 3E ( 4.6) and 3B ( 4.4) components, each used twice:

F)* =

&

22'

0 !Exc !Ey

c !Ezc

Exc 0 !Bz By

Ey

c Bz 0 !BxEzc !By Bx 0

(

33) (4.10)

This is the electromagnetic field tensor F)* which has ”packaged” E andB in in terms of four vectors from 4.7. This means that we can use F)*

to describe laws of physics directly in terms of E and B in a form thatis ”manifestly covariant”, that is, invariant in form under Lorentz trans-formation. We will in the next section write down the Maxwell equationsin manifestly covariant form directly, instead of in terms of the scalar andvector potentials (the four vector A)) as in the previous Chapter.

What if we had used the covariant four vectors to form a ”four curl” ofA? The starting point would have been:

A) = (&

c,!Ax,!Ay,!Az) = (A0, A1, A2, A3) (4.11)

and') =

$1c

'

't,'

'x,'

'y,'

'z

%= ('0, '1, '2, '3) (4.12)

then # " A would look like:

Bx ='Az

'y! 'Ay

'z= !'2A3 + '3A2 = !('2A3 ! '3A2) = !F23 (4.13)

so Bx = !F23 = !F 23, and similarly for By = 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 = !'&'x

! 'Ax

't= !c'1A0 + c'0A1 = c('0A1 ! '1A0) = F01c (4.14)

and Ex/c = F01 = !F 01, that is, it has the opposite sign to the contravari-ant case 4.6, and similarly for Ey and Ez .

We can then writeF)* = ')A* ! '*A) (4.15)

with the covariant form of the electromagnetic field tensor:

F)* =

&

22'

0 Exc

Ey

cEzc

!Exc 0 !Bz By

!Ey

c Bz 0 !Bx

!Ezc !By Bx 0

(

33) (4.16)

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To get from the covariant to the contravariant forms of the electromagneticfield tensor we just need to use the spacetime metric g)& to change the signson the timelike part of the tensor, that is, the F 0) and F)0 terms (in thiscase, the ”E” terms)

F)* = g)&F&,g,* (4.17)

which is equivalent to E ' !E, B ' B. Finally, if F)* is manifestlycovariant we would expect it to have constant ”length”, this just turns outto be zero for electromagnetic waves in free space (see advanced problem6).

The usefulness of F)* will now become clear as we use it to write downmanifestly 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, solet’s begin by taking the ”4 div” of F)* .

')F)* =01c

'

't,'

'x,'

'y,'

'z

1&

22'

0 !Exc !Ey

c !Ezc

Exc 0 !Bz By

Ey


(

33) =

&

22'

%·Ec

(# " B)x ! 1c2

$Ex$t

(# " B)y ! 1c2

$Ey

$t(# " B)z ! 1

c2$Ez$t

(

33)

(4.18)i.e a single four vector containing the (slightly rearranged) left hand side ofthe Maxwell inhomogeneous equations with timelike component giving

# · Ec

= µ0c" (4.19)

and with spacelike components giving:

#" B ! 1c2

'E't

= µ0J (4.20)

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

J) = (c",J) (4.21)

so the inhomogeneous Maxwell equations are just the single equation

')F)* = µ0J* (4.22)

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We would like to use the same trick to get the homogenous Maxwell equa-tions in manifestly covariant form. In this case, the ”four div” of sometensor must yield a single four vector, this time with timelike component

# · B = 0 (4.23)

and spacelike components

#" E +'B't

= 0 (4.24)

so we need some transformation that swaps the position of the E and Bterms in F)* so that the timelike components involve B and the spacelikecomponents involve E. This is known as the duality transformation and isE ' Bc2 and B ' !E; performing the transformation on F)* gives theDual field tensor

F)* =

&

22'

0 !Bxc !Byc !BzcBxc 0 Ez !Ey

Byc !Ez 0 Ex

Bzc Ey !Ex 0

(

33) (4.25)

The ”four div” of F)* then yields the required homogenous Maxwell equa-tions:

')F)* =

&

22'

c# ·B!(#" E)x ! $Bx

$t

!(#" E)y ! $By

$t!(#" E)z ! $Bz

$t

(

33) (4.26)

so that the homogenous Maxwell equations are

')F)* = 0 (4.27)

But what is the significance of F)*? First, we need to relate it directly toF)* to demonstrate that just like F)* , F)* is indeed a ”four tensor” andLorentz transforms in the same way. This is done via the ”four alternatingtensor” (the 4 %4% 4 version of the Cartesian 3% 3 alternating tensor thatwe first met in Section 2.1.3):

F)* =c2

2$)*&,F&, (4.28)

Second, the dual tensor has an interesting physical significance. If weapply the duality transformation to Maxwell’s equations we have:

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Maxwell equations Dual equations# · E = "

#0# · B = µ0"

# ·B = 0 # · E = 0# "E = !$B

$t #" B = $E$t

#" B = µ0J + 1c2

$E$t #" E = !µ0J ! $B

$t

(4.29)

The Dual Maxwell equations 4.29 describe a world in which we wouldidentify " as magnetic charge and J as magnetic current. In this dualworld, 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 varywith time.

Two properties of the Maxwell equations implied that they must em-body 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 un-der Lorentz transformation. If we put " = 0 and J = 0 in 4.29 then theMaxwell and Dual equations are identical, ie they will both predict lightwaves moving at invariant speed c in free space. Charge conservation isimplicit in Maxwell IV, we can look at of conservation of magnetic chargeusing the dual equations. The Dual IV equation is

# "E = !µ0J! 'B't

(4.30)

Taking the divergence of 4.30 gives

# · (# "E) = 0 = !µ0# · J ! '

't(# ·B) (4.31)

which from Dual I (# · B = µ0") gives

# · J +'"

't= 0 (4.32)

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

Writing the Maxwell equations in Lorentz invariant form then tells usthat the existence of ”electric charge” or ”magnetic charge” is not excludedby special relativity. The only reason we wrote down the original Maxwellequations with electric charge and zero magnetic charge is that experimen-tally the magnetic charge is known to be zero to very high precision. Theinteresting question is whether all particles have the same ratio of electricto magnetic charge; if so, then we can always perform a duality transforma-tion similar to that discussed here to make the Maxwell equations appearas if the magnetic charge is zero.

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4.3. CONSERVATION OF ENERGY-MOMENTUM. 75

4.3 Conservation of Energy-Momentum.

In section 3.6.1 we found a manifestly covariant form for charge conser-vation. We also found in section 3.5.1 that the energy and momentumof individual particles forms a single four vector, which will also have in-variant length under Lorentz transformation. The energy- momentum ofan ensemble of particles, a mixture of charged particles and photons, say,must also form a four vector. We might then guess that the conservationequations for field energy (Poynting’s theorem 2.37, 2.41) and momentum( 2.48) could be combined to form an equation for the conservation of fieldand charged particle energy-momentum that is manifestly covariant.

To demonstrate this we need to ”package”Energy conservation (Poynting’s Theorem)

1c

'U

't+ # · S

c= !J · E

c(4.33)

andMomentum conservation

1c

'

't

S

c!# · T = !("E + J "B) (4.34)

into one object. The r.h.s. of these equations should then form a fourvector, with timelike component 4.33 and spacelike components 4.34. Thel.h.s. of the desired expression must be the four divergence of a four tensor')T)* ; we can guess that this is the case since it involves ') and must berank 1, that is, a four vector, to match the r.h.s.

Looking at the l.h.s. first then, since

') = (1c

'

't,'

'x,'

'y,'

'z) (4.35)

we need a 4 %4 energy- momentum field tensor:

T)* =

&

22'

U s1c

s2c

s3c

s1c !T11 !T12 !T13s2c !T21 !T22 !T23s3c !T31 !T32 !T33

(

33) (4.36)

where the Tij are the 9 components of the 3 % 3 Cartesian Maxwell stresstensor 2.56. Then ')T)* 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, whichmeans that we need either the electromagnetic field tensor F)* or its dual.We can guess which one by inspection of the r.h.s. terms; the spacelike part

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needs a J"B term, and we formed #"B by taking the ”four div” of F)*

( 4.22). So if we replace the ”four div” in ')F)* with a dot product withthe four current J) we have

J)F)* = (c",!Jx,!Jy,!Jz)

&

22'

0 !Exc !Ey

c !Ezc

Exc 0 !Bz By

Ey


(

33) =

&

22'

! 1cJ ·E

!("E + J " B)x

!("E + J " B)y

!("E + J " B)z

(

33)

(4.37)which gives the r.h.s. as required. Our expression for energy-momentumconservation is then

')T)* = J)F)* (4.38)

The r.h.s of equation 4.38 is constructed from four tensors and ') is afour vector, so the only remaining term T)* must be a four tensor and theexpression is manifestly covariant.

4.4 Lorentz Force.

We have one equation left to express in manifestly covariant form, theLorentz force law. In section 3.5.1 the laws of mechanics were written inLorentz invariant form giving manifestly covariant equations of motion:

f) = dp!

dsu) = dx!

ds

(4.39)

We now just need to find the four- force f) for a charged particle in anelectromagnetic field. Since the Lorentz force contains a +v " B term, wecan again guess that since J)F)* gives !J"B terms we require somethingof the form

F)*u* =

&

22'

0 !Exc !Ey

c !Ezc

Exc 0 !Bz By

Ey


(

33)

&

22'

c!!!vx

!!vy

!!vz

(

33) =

&

22'

1c!v · E

!(E + v " B)x

!(E + v " B)y

!(E + v " B)z

(

33)

(4.40)(we have swapped the order of the u) and the F)* around to get + insteadof ! on the components of the r.h.s. in 4.40). The spacelike parts of 4.40just look like the Cartesian components of the Lorentz force:

F =d(mv)

dt= e(E + v "B) (4.41)

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4.4. LORENTZ FORCE. 77

multiplied by !.The timelike part of 4.40 just looks like the Cartesian particle energy

equation obtained from 4.41

F · v =d(1

2mv2)dt

=d$

dt= ev ·E (4.42)

multiplied by !/c. Taking 4.41 and 4.42 together (and taking care of thee), we then have

eF)*u* =

&

22'

! ddt

4#c

5

! dmvxdt

! dmvy

dt! dmvz

dt

(

33) =

&

22'

dds

4#c

5dmvx

dsdmvy

dsdmvz

ds

(

33) (4.43)

that is,

eF)*u* =dp)

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 equa-tions of motion, all in manifestly covariant form. The rest of this chapterwill 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:

')F)* = µ0J*

• The homogenous Maxwell equations:

')F)* = 0

• Charge -current conservation:

')J) = 0

• Energy-momentum conservation:

')T)* = J)F)*

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• The equations of motion:

f) = dp!

dsu) = dx!

ds

• The Lorentz force law:

eF)*u* =dp)

ds

• and with the Lorentz gauge:

')A) = 0

• we have the Maxwell equations in terms of the four vector potential:

!A) = µ0J)

4.5 Transformation of the Fields

The electromagnetic field tensor F)* allows us to ”package” the compo-nents of E and B in terms of four vectors as in 4.7. The Lorentz trans-formation of a four vector is simply a rotation of its spacetime coordinatesas defined by 3.58 and 3.59. We can then find out how to Lorentz trans-form F)* by applying the coordinate rotation to all of its’ component fourvectors (see avanced problem 8):

F !)* = '!)A!* ! '!*A!)

= (%),',)(%*&A&) ! (%*&'&)(%),A,) =% ),F,&%&*

To obtain a simple formula to transform individual field components wewould then just need to compare terms in F)* and F !)* . This (not di"cult,but rather long) operation gives:

E!& = E&

B!& = B&

E!' = !(E + v " B)'

B!' = !

4B! v(E

c2

5'

(4.45)

where subscripts ” * ” and ” + ” mean components parallel and per-pendicular to the transformation velocity v. The inverse transform is justobtained by putting v ' !v in 4.45.

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4.6. FIELD FROM A MOVING POINT CHARGE. 79

We have seen a simplified version of 4.45 before. For the moving chargeand wire experiment in section 3.2, we considered a special case: Frame 1in which the test charge moved at velocity +v1 = v = +v& and Frame 2in which the charge was at rest (v2 = 0). Life had been made simple bychoosing the charge velocity to be just the transformation velocity +v.

In Frame 2 the Lorentz force law gives:

F&2 = e(E2 + v2 " B2)& = eE&2 (4.46)

and we now know from 4.45 that the Lorentz force should be:

F&2 = eE&2 = eE&1 = 0 (4.47)

as we chose E1 to be zero (by choosing "1 = 0). Also the Lorentz force lawgives:

F'2 = e(E2 + v2 " B2)' = eE'2 (4.48)

which from 4.45 transforms to

F'2 = e!(E1 + v1 " B1)' = e!v1 " B1 (4.49)

Hence F2 = !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)* also contains the electromagneticfields in Lorentz transformable form we could have used it to find the trans-formation of the fields. We can find out what this would yield by applyingthe duality transformation E ' Bc2 and B ' !E to 4.45, and it is easyto verify that we obtain the same transformations for the fields.

4.6 Field from a Moving Point Charge.

In nonrelativistic situations (ie under Galilean frame transformation) weknow that the field from a positive point charge just points radially out-wards, and this was used in Chapter 1 to discuss Gauss’ law. Now that weknow how to transform the fields we can look at what happens to the fieldsfrom charges moving at relativistic speeds.

Lets consider a positive charge moving past us at speed v as shown infigure 4.6. If we are in the rest frame of the charge (the S! frame), there isno magnetic field and the electric field is radial giving (from Chapter 1):

B! = 0E! = q

4!#0r!

r!3(4.50)

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x

q

S’S x’2

v

3 x’3

x1

x’1

r

Px2

a

vt

Figure 4.1: Charge q is at rest at the origin of the S! frame and is movingwith velocity vx1 w.r.t. frame S. The origins of the S and S! framescoincide at t = 0.

We now transform to the S frame by moving in !vx1 so that in the S framethe 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 theS frame:

E& = E!&

B& = 0E' = !E!

'B' = &

c2 (v " E!)' = v(Ec2

(4.51)

We will now obtain the fields at some location P = (ct, x1, a, 0) in theS frame (we can always choose a convenient orientation of x2, x3 suchthat x3 = 0). To do this we need to express E! in 4.51 as a functionof ct, x1, x2, x3 instead of x!

1, x!2, x

!3. In S! the charge is at rest and so the

fields are independent of t!, we just then need to transform the spacelikecoordinates using the Lorentz transformation. At point P this gives:

x!1 = !(x1 ! vt)

x!2 = x2 = a

x!3 = x3 = 0

(4.52)

Substituting for x!1, x

!2, x

!3 in 4.51 then gives the electric field at P from the

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4.6. FIELD FROM A MOVING POINT CHARGE. 81

x-vt x-vt

%=1

>1%

% >1

%=1

E2E1

1 1

Figure 4.2: Sketch of E1 and E2 at point P versus x!1/! = x1 ! vt.

moving point charge:

E1 = E!1 = q

4!#0

x!1

(x!21 +x!2

2 +x!23 )

32

= q4!#0

&(x1"vt)

(&2(x1"vt)2+a2)32

E2 = !E!2 = q

4!#0

&x!1

(x!21 +x!2

2 +x!23 )

32

= q4!#0

&qa

(&2(x1"vt)2+a2)32

E3 = !E!3 = 0

(4.53)

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 4.6 shows the electric field components in the S frameE1 and E2 at point P , plotted versus x!

1/! = x1!vt. As the relative speed ofthe charge becomes relativistic, and ! > 1, the x!

1/! axis Lorentz contracts.The magnitude of E1 doesn’t change (as E1 = E!

1) instead the E1(x!1/!)

profile becomes compressed. The magnitude of E2 = !E!2 increases, and

again, the profile is compressed as ! > 1. The electric field occupies a moredisc like region as ! is increased. This is sketched in figure 4.6 which showsthe E field vectors in the x1, x2 plane.

We expect the moving charge to have an accompanying magnetic fieldwhether or not it is relativistic as it is carrying a current. In the nonrela-tivistic limit ! ' 1, 4.51 becomes:

B =v "E

c2=

µ0dI " r4#r3

(4.54)

where dI = qv is the current carried by the charge in its direction of relativemotion. This it just the Biot Savart law1 in magnetostatics. Generally, thespatial distribution of B field just follows that of E from 4.51, so that when

1Strictly speaking the total magnetic field from a closed current loop C in static

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E

E’

v

Figure 4.3: Sketch of the E! field around a charge at rest, and the E fieldaround a charge moving at relative velocity v, in the x1, x2 plane.

the charge moves at relativistic speeds, B will also be compressed into amore disc like region. As the charge moves closer to the speed of light v ' cand

B ' v " Ec

(4.56)

We can compare this to the free space electromagnetic wave solution toMaxwell’s equations 1.4. In section 2.4 we found that if we take a planewave solution of the form

E,B $ ei(%t"k·r) (4.57)

and substitute into Maxwell III we obtain 2.60:

B =k " 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 aroundthe relativisitic charge, the fields start to look like a pulse of light centeredon the charge.

situations is obtained by adding up the contributions due to all the current elements dIalong the loop:

B =

!

C

µ0dI ! r

4%r3(4.55)

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4.7. RETARDED POTENTIAL. 83

S’

r r’S

3xx’

3

Px’2x2

& ’dx’dx’dx’1 2 3

x’1

x1

v

Figure 4.4: A volume element is located at rest at the origin of the S! framewhich is moving at +v in the x1 direction w.r.t. the S frame. In the S!

frame the volume element contains charge "!dx!1dx!

2dx!3. The origins of the

two frames are coincident at t = t! = 0.

4.7 Retarded Potential.

In the previous section we found that Lorentz contraction and time dilationgave us a good idea of what happens to the fields around a single pointcharge as it moves at relativistic speeds. We will now obtain a generalsolution to Maxwell’s equations by finding the fields from a collection ofcharges described in terms of the (frame dependent) charge and currentdensity. To make life simple we can work in terms of the scalar and vectorpotentials rather than the field transformations directly. Everything neededfor the calculation then forms four vectors; the scalar and vector potentialsA), the charge and current densities J), and spacetime x), these then arestraightforward to transform using the Lorentz transformation.

The problem is posed in figure 4.7. We will first consider the scalarand vector potential due to the charge dQ contained in a small (elemental)volume element; since all our equations are still linear under the Lorentztransformation we can find the scalar and vector potential from chargesdistributed over a larger volume later by doing a volume integral.

To simplify the algebra, the volume element containing dQ is located atthe origin of the S! frame, and in this frame the charges are at rest. The S!

frame then moves at speed +v along the x!1 direction w.r.t. the S frame,

and again for simplicity we will arrange for the origins of both frames tocoincide at t = t! = 0. We now observe the scalar and vector potentialdA) at some point P due to the charges when the frames coincided. Since

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disturbances in the electromagnetic fields travel at speed c in all frames, inthe S frame the ”signal” takes a time

t =rc

(4.59)

to reach P from the S origin. By the time it has done so, the charges havemoved; they are now located at the S! origin which is distance r! from P .

We can find dA) from dA!) in the S! frame where the charges are atrest, this is:

cdA!0 = d&! = dQ4!#0r! = "!dx!

1dx!2dx!

34!#0r!

dA!1,2,3 = 0(4.60)

The inverse Lorentz transform% )*(i) (since S is moving in !vx1 w.r.t. S!)

of 4.60 (see section 3.4) is

A) = %)*(i) A

!* =

&

22'

! ! vc ! 0 0

+ vc! !! 0 00 0 !1 00 0 0 !1

(

33)

&

22'

d(!

c000

(

33) =

&

22'

! d(!

c

! vd(!

c2

00

(

33) =

&

22'

d(c

dA1

dA2

dA3

(

33)

(4.61)so that

d& = !d&!dA = x1!

vc2 d&!

(4.62)

The four-current J) will inverse transform in just the same way, fromcharges at rest in the S! frame to the charges moving in the S frame:2

" = !"!J = !v"! = v"

(4.63)

Next, we can write d&!(dx!1, dx!

2, dx!3, r

!) as a function of coordinates inthe S frame by using the Lorentz transformation of spacetime coordinatesie 4.52 to get expressions for the volume element dx!

1, dx!2, dx!

3 and for r!.If we substitute in the time delay for t ( 4.59) then

x!1 = !

6x1 !

vr

c

7(4.64)

This gives an expression for r!:

r! = (x!21 + x!2

2 + x!23 ) 1

2

=8!24x1 ! vr

c

52 + x22 + x2

3

9 12 (4.65)

2This just yields what we found for the moving charge and wire experiment in sec-tion 3.2.

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4.7. RETARDED POTENTIAL. 85

since r2 = x21 + x2

2 + x23 this can be written as:

r! = !8r ! vx1

c

9(4.66)

The volume element as seen in S is obtained by implicit di!erentiationof 4.52:

dx!1 = !dx1

61 ! v

cdrdx1

7= !dx1

41 ! v

cx1r

5

dx!2 = dx2

dx!3 = dx3

(4.67)

where we used the change in r across the volume element, dr/dx1, obtainedby di!erentiating

r2 = x21 + x2

2 + x23 (4.68)

w.r.t. x1 to givedr

dx1=

x1

r(4.69)

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

d& = !"!dx!

1dx!2dx!

3

4#$0r!= !

6"&

7dx1!

41 ! v

cxr

5dx2dx3

4#$0!4r ! vx1

c

5 (4.70)

which is just

d& ="dx1dx2dx3

4#$0r(4.71)

This is the potential at P that we would expect from electrostatics if chargedQ were located at distance r instead of r!; that is, as if dQ were still atthe origin of the S frame. So to get a solution of Maxwell’s equations, wehave two approaches. One is to ”do the relativistic treatment” and Lorentztransform four vectors as above. The other is to insist that charge dQis frame invariant and always has a potential of the form 4.60 and 4.71,and also that information about dQ travels at frame invariant speed c.Information about the location of the charges takes a time t = r/c to reachpoint P , so we see a retarded potential consistent with the electrostatics ofthe charges at their previous location a time t = r/c earlier.

The vector potential in the S frame is then

dA = x1v

c2d& =

µ0Jdx1dx2dx3

4#r(4.72)

which, with B = # " A will again give the Biot Savart law, consistentwith the magnetostatics of the current due to the moving charges at theirprevious location a time t = r/c earlier.

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O

r

r’

dV’ r-r’P

Figure 4.5: Volume element dV ! and point P located w.r.t. a single fixedorigin O.

The algebra was simplified by choosing the charges to be at the originof the S! frame and having the two frames coincident at t = 0. A moregeneral arrangement is shown in figure 4.7 where all positions are knownrelative to a fixed origin. A volume element dV ! is at position r!, point Pis then a distance | r ! r! | from the charges in dV !. It takes light a timetl =| r! r! | /c to propagate from the charges to point P , so the potentialsseen at P at time t are given by the charge density in dV ! at time t ! tl.Integrating over all space, the potentials at P in terms of the charge andcurrent densities in the S frame are:

&(r, t) =1

4#$0

"

V

"6r!, t ! |r"r!|

c

7

| r ! r! | dV ! (4.73)

and

A(r, t) =µ0

4#

"

V

J6r!, t ! |r"r!|

c

7

| r ! r! | dV ! (4.74)

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

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Suggested Texts.

This is not an exhaustive list, they are the books that author has found themost 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 insightfuljourney through electrodynamics, best read in one go. Training inthe formalism should be sought elsewhere.

• Classical Electrodynamics, Second Edition, J. D. Jackson, Wiley,1975: A must for every working physicists bookshelves. Thoroughand rigorous treatment of electromagnetism from first principles toelectrodynamics. 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 sum-mary.

• Quantum Field Theory, C Itzykson and J B Zuber, McGraw Hill,1980. A standard treatment, the first chapter covers classical electro-dynamics.

• Mathematical Methods in the Physical Sciences Second Edition, M.L. Boas, Wiley, 1983: A nice introduction to tensors, and to vectorcalculus.

• The Classical Theory of Fields, Fourth Edition, Course of TheoreticalPhysics Vol. 2, L. D. Landau and E. M. Lifshitz, Pergamon, 1983:Extends the electrodynamics of flat spacetime treated here to curvedspacetime, i.e. general relativity.

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Index

alternating tensor, 36, 71Ampere’s law, 18, 23

di!erential form, 19

basis vectors, 36Biot Savart law, 79

Cartesianinconsistency with spacetime, 50

charge conservation, 23, 75equation of, 24

charge density, 10charge invariance, 51classical limit

fields, 11, 16fluids, 37particles, 26

cold gas, 12conservation equation, 26definition, 11energy flux, 39momentum flux density, 30

conservationenergy-momentum, 74, 75of current, 75of energy density, 38of field-particle momentum, 39of four-current, 65

contractionin Cartesian tensors, 36

contravariant, 57coordinate rotation

in spacetime, 49

coordinate transformationCartesian, 34Lorentz, 49, 59

Coulombunit, 10

Coulomb gauge, 56Coulomb’s law

experiment, 9covariant, 57cross product, 36current density, 18

D’Alembertian, 64direction cosines, 35displacement current, 24

in capacitors, 24Divergence theorem

definition of, 8dual tensor, 71duality transformation, 71, 77dyadic, 29, 34

Einstein summation convention, 35electric field, 10

definition, 17definition of, 10flux, 13in terms of charge density, 11in terms of potentials, 23, 55lines, continuous, 17units, 10work done by, 17

electric potential

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INDEX 89

definition, 18electromagnetic field tensor, 69, 70

length, 70electromagnetic waves, 41

free space, 25from Maxwell equations, 25general solution, 25group velocity, 25phase speed in vacuum, 25relativistic charge, 80

electrostatic fieldconservative, 18

electrostatic potential, 18energy- momentum field tensor, 73Energy-momentum, 75equivalence of electric and magnetic

fields, 21, 23, 51

Faraday’s law, 20, 22from Lorentz force, 21

fieldenergy density, 37, 38from moving point charge, 79momentum density, 39momentum flux density, 40

fieldsclassical limit, 16transformation of, 23

flat spacetime, 58fluid bulk variables, 37fluid equation, 32flux, 13

definition, 12of electric field, 13of energy, 39

forcefrom momentum flux tensor, 30,

31four vector

four current, 75four-curl, 68four-vector, 55

contravariant, 57covariant, 57energy-momentum, 60force, 63, 75, 76four vector potential, 65four-current, 65, 82four-divergence, 64four-gradient, 63four-velocity, 61, 76inner product, 57length-time, 60potential, 68

frame transformationGalilean, 23nonrelativistic, 23

Galilean frame transformation, 23Galilean transformation, 61gas

finite temperature, 26warm, 31

gauge transformation, 56Gauss’ law, 16

integral form, 16general relativity, 58generalized coordinates

inner product, 59generalized coordinates, 56

inner product, 36, 59

kinetic energy, 37

light clock, 46linear media, 38Liouville’s theorem, 31Lorentz contraction, 48Lorentz force, 39Lorentz force law

invariant form, 75Lorentz gauge, 56, 65, 76

in four-vector form, 65, 76

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90 INDEX

Lorentz transformation, 49, 59inverse, 59of charge density, 52retarded potential, 82

Lorentz transformation of fields, 76

magnetic fieldlines, continuous, 17units, 18

magnetic flux, 16magnetostatic scalar potential, 19Maxwell equations

dual, 72homogenous, 55, 71, 75inhomogenous, 56, 66, 71, 75manifestly covariant, 66, 76

Maxwell homogenous equationsinvariant form, 71

Maxwell I, 16integral form, 16

Maxwell II, 17integral form, 16

Maxwell III, 23integral form, 22

Maxwell inhomogenous equationsinvariant form, 71

Maxwell IV, 24Maxwell stress

for free space EM waves, 41tensor, 40

mechanics, invariant form, 74momentum flux density, 30momentum flux tensor, 29monopoles, 72moving charge and wire, 50, 77, 82

Newton’s laws, 61and special relativity, 62

phase space, 31photons, 26

and relativistic charge, 80

Poynting flux, 38, 39units, 38

Poynting’s theorem, 38pressure tensor, 28, 31

isotropic, 29shear free, 29

principle of superposition, 9proper time, 62

radiation pressure, 42rest energy, 60retarded potential, 83right hand rule

in cross product, 8in Lorentz force law, 9

rocket e!ect, 30rotation matrix, 35rotation of four vector, 57

spacelike interval, 49spacetime

deridative, 62interval, 49, 56, 62metric, 58rotation in, 59

special relativity, 45speed of light, 66, 72

from Maxwell’s equations, 25from potentials, 56

speed of light in vacuum, 25Stokes theorem

definition of, 8superposition

in Galilean frame transformation,23

of electromagnetic waves, 25of electrostatic fields, 18

tensoralternating, 36, 71Cartesian, 30coordinate transformation, 35

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INDEX 91

divergence of, 36dot product, 36dual, 71electromagnetic field, 69, 70energy- momentum field, 73ensor

Cartesian, 34field momentum flux density, 40force from, 31inner product, 36Lorentz transformation, 59Maxwell stress, 40, 73momentum flux density, 29pressure, 28, 31rank, 34trace, 36warm gas pressure, 31

Tesla, 18time dilation, 47timelike interval, 49trace, 36transformation

as rotation, 35, 49, 57Cartesian, 34charge invariance, 51duality, 71, 77Galilean, 61gauge, 56Lorentz, 49, 59of charge density, 52of EM field tensor, 76retarded potential, 82rotation in spacetime, 59

vector potential, 55from magnetostatics, 19

warm gas pressure tensor, 32wave equation

for electrmagnetic fields, 25in terms of potentials, 56manifestly covariant, 66

workdone by Lorentz force, 37

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92 INDEX

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Appendix A

Revision Problems

A.1 Static Magnetic Fields.

A straight wire carrying a steady current I has radius a as shown in A.1.

a Inside the wire, ie for r , a calculate the current enclosed inside acircle of radius r as shown. Use the integral form of #.B = 0 to showthat B is always perpendicular to r. From Ampere’s law (the integralform of #" B = µ0J) calculate B(r) and hence using Stokes theorem(ie from

:B · dl) calculate # " B.

Calculate # " B directly. In a cylindrical coordinate system you canuse the identity

#" B = (#" B)z z =1r

'

'r(rB-)z

(A.1)

when B = B-%.

b Outside the wire ie for r > a use Ampere’s law to calculate B(r) usingcurve C1 shown below (ie a circle of radius r centred on the wire).

Use this expression for B to calculate:

B · dl around curve C2. Cal-culate # " B directly at a point within C2 and compare it with yourvalue for

:B · dl. Sketch the variation of the magnitude of the cur-

rent density J(r) as a function of r and use Maxwell’s IVth equationfor steady fields # " B = µ0J to sketch the variation of # "B as afunction of r.

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94 APPENDIX A. REVISION PROBLEMS

I

r

a

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

I

C1

C1

C2

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

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A.2. STATIC ELECTRIC FIELDS. 95

A.2 Static Electric Fields.

A long bar of radius a carries a static charge density "l per unit lengthwhich is spread uniformly through the bar. What is

:E · dl around any

closed path? What does this tell you about the direction of the electricfield? Use Gauss’ theorem in integral form on cylindrical surfaces centredon 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 coordinatesystem you can use the identity

#.E =1r

'

'r(rEr) (A.2)

when E = Er r.

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 nordestroyed.

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 conservationequation

# · (nv) = ! $n$t

hence obtain an expression for conservation of charge in terms ofcharge density " and current density J.

b i) Ampere’s law for steady currents is

!

cB · dl = µ0I

Write this in di!erential form.

ii) Using conservation of charge obtain Maxwell IV for time dependentfields:

#" B = µ0J + µ0$0$E$t

c In general ("f (= 0, Jf (= 0) Poynting’s theorem in media is

# · (E " H) = ! ''t

(12[E · D + B ·H]) ! Jf · E (A.3)

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Given that the energy density

U =12[E ·D + B · H] (A.4)

starting with $U$t 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 Dis-persion.

A general form for the wave equation is:

'2+

't2= v2#2+ (A.6)

Show that+1 = +01e

i(%1t"k1x) (A.7)

is a solution of this equation. What is the phase speed and direction ofpropagation of this wave?

If the wave+2 = +02e

i(%2t"k2x) (A.8)

is also a solution of the wave equation, show that the wave resulting fromthe superposition of these waves

+3 = +1 + +2 (A.9)

is also a solution.In a dispersive medium the wave equation may look like

'2+

't2= v2(()#2+ + b+ (A.10)

where b is a constant. Show that +3 is a solution. Give an example of aPDE for which +1 and +2 are solutions, but +3 is not.

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A.5. FREE SPACE EM WAVES I. 97

A.5 Free Space EM Waves I.

The E and B fields of a plane EM wave have the form:

E = E0ei(%t"k.r)

B = B0ei(%t"k.r)

a Show that Maxwell’s equations in free space can be rewritten as:

k · E0 = 0 k ·B0 = 0k " E0 = (B0 k " B0 = !µ0$0(E0

b Show that E, B and k (in that order) form a right-handed orthogonalsystem.

c If, with an electric antenna, it is established that E only has a ycomponent, what are the possible directions of propagation of thewave, and the directions of the associated B components.

d Describe the polarization of this wave. Write down E and B for awave which is (i) linearly polarized and (ii) circularly polarized.

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(%t+3x"y"z)

Determine:

a The direction of propagation k, wavelength ., angular frequency (.

b The z component of B.

c The electric field E associated with B.

d The Poynting vector S.

A.7 EM Waves in a Dielectric.

In a ’perfect’ dielectric there are no free currents or charges ("f = 0,Jf = 0).Use Maxwell’s equations in media and B = µµ0H, D = $$0E to derive waveequations for E and B. (Use the identity #" (# " A) = #(# ·A)!#2A).

What is the phase speed v = %k of the waves in SI units?

Show thatB =

1vk " E (A.11)

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A.8 Dielectrics and Polarization.

A parallel plate capacitor with plates of area A, separated by a distanced has a voltage V applied across the plates. Initially there is a vacuumbetween the plates. A charge of magnitude Q = CV (where C is thecapacitance) corresponding to a surface charge density /f has built up oneach 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. Whatis the capacitance?

b The region between the plates is now filled with a dielectric, and thecapacitor is again charged up by placing charge of magnitude Q oneach plate. A surface charge density /p is induced on the dielectric.What is the polarization (dipole moment/ unit volume) in terms of/p?

NB: we define polarization as P = np where p is the atomic dipolemoment.

c If the material is linear, P = ($ ! 1)$0E. What is the electric fieldinside the dielectric, and the capacitance?

A.9 EM Waves in a Conductor: Skin Depth.

In an ideal conductor at low frequencies we can assume that there are nofree charges ("f = 0) and that the free current density obeys a simpleOhm’s law Jf = /E.

Use Maxwell’s equations and the identity #"(#" A) = #(#.A)!#2Ato show that the ”wave equation” in the conductor is

#2E = µµ0/'E't

+1v2

'2E't2

(A.12)

Derive the dispersion relation (that is, k as a function of () in the lowfrequency limit (terms O((2) - terms O(()) by examining the propertiesof plane wave solutions. Hence show that the skin depth (distance wavescan penetrate into the conductor before they are appreciably attenuated)is

) = { 2µµ0/(

} 12 (A.13)

What is the significance of this result for i) radio propagation ii) mi-crowave ovens iii) power transmission?

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A.10. CAVITY RESONATOR. 99

A.10 Cavity Resonator.

Find the Poynting flux and the radiation pressure for a standing plane elec-tromagnetic wave (that is, two travelling waves of equal amplitude propa-gating in opposite directions. Use this to caclulate the cycle average Poynt-ing flux and radiation pressure.

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Appendix B

Solutions to RevisionProblems

B.1 Static Magnetic Fields.

a The current density

| J |= I

#a2(B.1)

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

Ir ="

J · dS =I

#a2% #r2 =

Ir2

a2(B.2)

Then since !B · dS = 0 (B.3)

over a surface defined by a cylinder of radius r, B only has componentsparralel to the surface ie perpendicular to r.From Ampere !

B · dl = µ0I (B.4)

and using cylindrical symmetry to infer that | B | is independent ofr: !

B · dl = B % 2#r = µ0Ir2

a2(B.5)

that is,

B =µ0Ir

2#a2(B.6)

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102 APPENDIX B. SOLUTIONS TO REVISION PROBLEMS

The direction of B is obtained from the right hand rule, or # " B =µ0J.

Using Stokes: !B · dl =

"# "B · dS (B.7)

is

µ0Ir2

a2=| # " B | #r2 (B.8)

giving

| #" B |= µ0I

#a2(B.9)

directed in +J.

Directly:

#" B = z1r

'

'r

0rµ0Ir

2#a2

1= z

mu0I

#a2(B.10)

b Outside the total current enclosed by C1 is I then!

C1

B · dl = µ0I (B.11)

B2#r = µ0I (B.12)

ie B = µ0I/2#r. Around C2

!

C2

B · dl = 0 (B.13)

as no current is enclosed. Evaluating this integral the contributionsfrom the straight line segments of C2 along r are zero and perpendic-ular to r along the curved segments (at constant r1 and r2) give

!

C1

B · dl =µ0I

2#r1% r1

"

1d% ! µ0I

2#r2% r2

"

2d% = 0 (B.14)

where the integral over % just gives the angle subtended by curve C2

at r = 0.

Directly

#" B = z1r

'

'r

0rµ0I

2#r

1= 0 (B.15)

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B.2. STATIC ELECTRIC FIELDS. 103

B.2 Static Electric Fields.

Static hence !E · dl = 0 (B.16)

Choosing a circular path, circle centre at the centre of the bar, cylintricalcoordinates z, r,% so that r · dl = 0 we have either (i) E assymetric (iewith a % dependence) and E · dl at some %1 cancels that at some %2, or(ii) E has cylindrical symmetry and hence must be perpendicular to dleverywhere. Since E due to a point charge is symmetric and the principleof superposition holds we have case (ii).

Thus this also admits a constant component along z.To use Gauss in integral form over a cylinder of radius r < a, length L

we have charge enclosed Q = "V where

" ="lL

#a2L="l

#a2(B.17)

ieQ =

"l

#a2% #r2L (B.18)

The only contribution to E is on the curved surface of the cylinder on whichE is constant. !

E · dS = E2#rL =1$0"lr

2La2 (B.19)

henceE =

"lr

2$0a2#r (B.20)

Then

# · E =1r

'

'r

0"lr2

2$0a2#

1=

"l

$0a2#=

rho

$0(B.21)

For cylinder radius r > a the enclosed Q = "lL so that!

E · dS = E2#rL ="lL

$0(B.22)

thenE =

"l

2#r$0r (B.23)

and

# · E =1r

'

'r

0"lr

2#r$0

1= 0 (B.24)

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B.3 Conservation and Poynting’s Theorem.

a i) Number through dS in time dt is nv ·dSdt since volume containingparticles that cross surface in dt is v ·dSdt. Then number crossing Sin dt is "

Snv · dSdt (B.25)

Flux = No/sec/unit area =1S

"

Snv · dS (B.26)

ii) Number enclosed in S is

N ="

VndV (B.27)

so number crossing S in unit time is

'N

't='

't

"

VndV =

"

V

'n

'tdV (B.28)

Now using our expression for the flux across S, the number flowingout of V in time dt is

!

Snv · dSdt = !

"

V

'n

'tdV dt (B.29)

From Divergence theorem then

!'n't

= # · (nv) (B.30)

For charges we have " = nq, J = nqv giving

!'"'t

= # · J (B.31)

b Given in the text.

c Poynting theorem in media: from the energy density

U =12[E.D + B.H] (B.32)

we have

'U

't=

12

0E · 'D

't+ D · 'E

't+ B · 'H

't+ H · 'B

't

1(B.33)

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B.4. THE WAVE EQUATION: LINEARITY AND DISPERSION. 105

then since D = $$0E, B = µµ0H we have

'U

't= E · 'D

't+ H · 'B

't(B.34)

which is'U

't= E ·(#"H!Jf )+H ·(!#"E) = !Jf ·E!H ·#"E+E ·#"H

(B.35)Using the identity

#.(E "H) = H.(# " E) ! E.(# " H) (B.36)

we obtain'U

't= !Jf ·E !#.(E " H) (B.37)

B.4 The Wave Equation: Linearity and Dis-persion.

Substitute +1 into the wave equation:

(i(1)2+1 = v2(!ik1)2+1 (B.38)

ie if (1 = vk1 then +1 is a solution. v is the phase speed, that is, the speedof a point of constant phase. Consider some point x at phase & = (1t!k1xat time t. At time t + dt this point has moved to x + dx where dx = vdt.So

(1(t + dt) ! k1(x + dx) = & (B.39)

Hence (dt = kdx and the point of constant phase moves at

v =dx

dt=(1

k1= v (B.40)

Now +2 is a solution. Subsitute +3 = +1 + +2 into the wave equation:

'2

't2+3 =

'2

't2+1 +

'2

't2+2 = v2#2+3 = v2[#2+3 + #2+2] (B.41)

and since'2

't2+1,2 = v2#2+1,2 (B.42)

+3 is also a solution. This will hold for any linear PDE, hence +3 can beshown to be a solution of the dispersive wave equation by the same method.

Require nonlinear PDE (with terms in +n, (#+)n etc) to break theprinciple of superposition.

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B.5 Free Space EM waves I.

a For plane waves it can be shown by calculating the derivatives in anyorthogonal coordinate system (eg cartesian) that

#· & !ik ·#" & !ik "

'

't& i(

Then substituting the wave solutions into the free space Maxwellequations:

# ·E . k ·E0 = 0# ·B . k ·B0 = 0

# "E = ! ''t

B . k " E0 = (B0

# " B = µ0$0'

'tE . k " B0 = !(µ0$0B0

b Use result (a):

k · E0 = 0 k · B0 = 0 . E,B + k

B =1ck " E then . handedness

c k ·E0 = 0 so k = xkx + yky, then since B = 1c k"E, B = xBx + zBz.

d Linearly polarized.

i) Linear polarization:

E = E0ei(%t"k.r)

B = B0ei(%t"k.r)

plane of polarization is E,k

ii) Circular polarization:

E = E0xei(%t"kz) + E0yei(%t"kz± "2 ) (B.43)

with B = 1c k " E.

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B.6 Free Space EM Waves II.

a k = !3x + y + z, | k |=)

11m"1 so . = 2#/k = 1.89m. ( = ck =9.93 % 108s"1.

b Use k ·B = 0, k from (a) giving Bz = 10"6T .

c Use k " B0 = !(µ0$0B0 with above k,B to give

E0 =300)11

(x ! 4y + 7z)V m"1

E = E0ei(%t+3x"y"z)

dS =

1µ0

E " B = 432 cos2((t ! k · r)kWm"2 (B.44)

B.7 EM Waves in a Dielectric.

Using

#" H ='D't

(B.45)

we have using # ·H = 0:

#" (# " H) = !#2H = #" ='D't

(B.46)

Then since Maxwell III# "E = !'B

't(B.47)

can be written as# " D = $$0µµ0

$!'H't

%(B.48)

we have#2H = $$0µµ0

'2H't2

(B.49)

similarly taking #" (# " D) yields

#2D = $$0µµ0'2D't2

(B.50)

Using B = µµ0H and D = $$0E gives wave equations in terms of E and B:

#2E = $$0µµ0'2E't2

(B.51)

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#2B = $$0µµ0'2B't2

(B.52)

which by inspection have wave solutions with phase speed

v =1;

($$0µµ0)=(

k(B.53)

As in free space plane waves, Maxwell III yields:

B =k

(k " E =

1vk " E (B.54)

B.8 Dielectrics and Polarization.

Far from the edges of the plates, assume that they are approximately infi-nate in extent. Then E is directed normal to the plates.

a Gauss in integral form for the surface becomes:!

E · dS =1$0

"/fdS (B.55)

ThenE =

/f

$0=

Q

$0A(B.56)

Since Q = CV and V = Ed (from # " E = 0 and E directed normalto the plates) we have

C =Q

V=

Q

Ed=$0A

d(B.57)

b From slab geometry E is uniform in the capacitor, hence no polar-ization charge is induced within the volume of dielectric. From thedefinition of polarization:

!P · dS =

1$0

"/pdS =

"PdS (B.58)

as P is then also normal to the (approximately infinate) plates. Thisgives P = /p in units of Coulomb m"2.

c In the presence of the dielectric!

E · dS =1$0

"(/f ! /p)dS (B.59)

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B.9. EM WAVES IN A CONDUCTOR: SKIN DEPTH. 109

which in this slab geometry is

E =/f ! /p

$0=/f

$0! P

$0(B.60)

then since for Linear media P = ($! 1)$0E we have

E =/f

$$0(B.61)

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

C =Q

V=

Q

Ed=$0$A

d(B.62)

B.9 EM Waves in a Conductor: Skin Depth.

In media Maxwell IV is

# " H = Jf +'D't

(B.63)

which can be written as

1µµ0

# "B = Jf + $$0'E't

(B.64)

using Jf = /E this becomes:

#" B = µµ0/Jf +1v2

'E't

(B.65)

We take '/'t of this equation, and use

# " (# " E) = #(# · E) !#2E (B.66)

with # ·E = 0 since # ·D = "f = 0 to obtain a modified wave equation:

#2E = µµ0/'E't

1v2

'2E't2

(B.67)

Assume plane wave solutions of the form

E = E0ei(%t"k·r) (B.68)

then substitute in to obtain the dispersion relation:

k2 =w2

v2! µµ0/i( (B.69)

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The two terms on the r.h.s. are due to the displacement and conductioncurrents respectively. Taking the low frequency limit of this expression isequivalent to neglecting the displacement currents in comparison to theconduction currents in the conductor and yields:

k2 = !µµ0/i( (B.70)

Clearly k has real and imaginary parts. Writing k = - ± i, gives planewave solutions of the form:

E = E0ei(%t"*k·r)e±)k·r (B.71)

which corresponds to waves that grow (+,) or decay (!,) as they prop-agate in +r. For a semi- infinate slab of conductor, the waves must havefinite amplitude as r ' / which excludes growing solutions. The wavesthen decay and a measure of the distance that they can penetrate into theconductor before they are appreciably attenuated (the skin depth )) is thatrequired for the amplitude to fall by a factor 1/e:

) =1,

(B.72)

, is then obtained by equating coe"cients, with

k2 = -2 ! ,2 ! 2i,- (B.73)

giving , = -, 2,- = µµ0/( and

) =<

2µµ0/(

(B.74)

Implication: lower frequency waves propagate further into conductors. Someexamples are:

• The earth’s ionosphere (a conductor) will not propagate radio wavesat all frequencies, hence longwave can be bounced o! the ionosphereto be received at points on the earth’s surface below the horizon.

• Re-entering astronauts lose radio contact with the ground as theirspacecraft is engulfed in plasma.

• A few mm of conductor are su"cient to attenuate microwaves so thatit is safe to use microwave ovens.

• household power (50Hz) can be carried over short distances by copperwith limited power loss, whereas computer network signals require coaxial cable or twisted pair ethernet (computer clock speeds are several% $ 100Mhz).

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B.10. CAVITY RESONATOR. 111

• your TV uses copper wire to carry the power, and coaxial cable tocarry the TV signal.

• over long distances copper wire would dissipate too much power tobe e"cient. Instead, high voltage, low current transmission lines areused.

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 equalamplitude propagating in opposite directions. For this standing wave thecycle average is then zero.

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Appendix C

Some Advanced Problems

C.1 Maxwell Stress Tensor.

Momentum flux conservation in an ideal Magnetohydrodynamic (MHD)plasma gives the following:

DuDt

= !#· P + J " B (C.1)

For equilibrium the l.h.s. of this equation must be zero.

1. Show that J " B can be written as the divergence of a stress tensorTM .

2. For a gyrotropic plasma, the pressure tensor P can be written as:

Pij = P')ij + (P& ! P')BiBj

B2(C.2)

where the local magnetic field direction is z.

Use the equilibrium condition #·(TM !P ) = 0 to obtain the followingstability criteria:

P' +B2

2µ0= K (C.3)

P& ! P' =B2

µ0+ K ! (C.4)

where K, K ! are constants.

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114 APPENDIX C. SOME ADVANCED PROBLEMS

C.2 Liouville and Vlasov Theorems: a Con-servation Equation for Phase Space.

The Vlasov equation describes the time evolution of the phase space prob-ability density f(x,v, t) in 6 dimensional phase space x,v, and is:

Df

Dt='f

't+ v ·#f + a ·#vf = 0 (C.5)

a Show (by using Taylor expansion) that this corresponds to the changein f along a particle orbit (Liouville’s theorem).

b Show that for a charged particle moving under Lorentz force law thatthe Liouville Theorem is equivalent to a conservation equation in 6dimensional phase space (hint, treat x and v as independent phasespace coordinates).

C.3 Newton’s Laws and the Wave Equationunder Galilean Transformation.

Determine whether or not the form of the following equations is invariantunder the Galilean transformation x! = x ! vt, t! = t:

a Newton’s equations of motion for the ith particle in an ensemble wherethe force between the ith and jth particle is F(xi ! xj).

midvi

dt=#

j

F(xi ! xj) (C.6)

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

$#2 ! 1

c2

'

't2

%& = 0 (C.7)

For (b) write down a solution for &(x, t) and &(x!, t!) and show that it isconsistent with Doppler shift.

C.4 Transformation of the Fields.

Two charged straight wires carrying a charge density " per unit volume area distance l apart and are at rest. The charges are not free to move within

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+

+ + + +

+++ +

++

+

+

+ + + +

+++ +

++

+

u

Figure C.1: A pair of wires

the wires. Calculate the electric field at each wire due to the other fromGauss’ law and the force acting 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 C.4: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 theforce 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. Whathas happened to the electric field?

C.5 Metric for Flat Spacetime.1

a If g)* is the metric for flat spacetime show that g)&g&* = )*) where)*) is a 4 % 4 rank 2 tensor with zero o! axis terms and trace 1.

b The Lorentz transformation of covariant four vector x) and con-travariant four vector x) can be written as

x!) = %)*x*

x!) = %)*x*

show that the metric for flat spacetime g)* transforms a contravariantvector into a covariant vector in all frames under Lorentz transforma-

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tion, ie showg)*x

!* = x!) (C.8)

c Show that F)* = g)&F &,g,*.

C.6 Length of the EM field Tensor in Space-time.

The tensor scalar product is

T : T = T)*T *) (C.9)

where we sum over both - and ,.Show that the tensor scalar product of the electromagnetic field tensor

is

F)*F *) = 2$

E2

c2! B2

%(C.10)

And that this is zero for free space electromagnetic waves (no chargespresent).

Hence show thatF)*F*) = !c2F)*F *) (C.11)

where F)* is the dual of F)*

C.7 Alternative Form for the Maxwell Ho-mogenous Equations.

Show that Maxwells’ homogenous equations can be written as

')F *& + '*F &) + '&F)* = 0 (C.12)

C.8 Lorentz Transformation of the EM FieldTensor.

a Show that the Lorentz transformation of the field tensor

F !)* = '!)A!* ! '!*A!) (C.13)

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C.8. LORENTZ TRANSFORMATION OF THE EM FIELD TENSOR.117

b The transpose rule is:

a)*..x&,.. = x...,&a..*) (C.14)

for both contravariant and covariant tensors.

Use the transpose rule and the Lorentz transformation of four vectorsto show that the Lorentz transformation of the electromagnetic fieldtensor is:

F !)* = %),F,&%&* (C.15)

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Appendix D

Solution to AdvancedProblems

D.1 Maxwell Stress Tensor.

1. The magnetic part of the Maxwell stress tensor is:

Tij =1µ0

BiBj ! )ij1

2µ0B2 (D.1)

as shown in 2.2.

2. We can combine this with the pressure tensor as given:

Pij = P')ij + (P& ! P')BiBj

B2(D.2)

to obtain

(TM ! P )ij =$

1µ0

!(P& ! P')

B2

%BiBj +

$! 1

2µ0B2 ! P'

%)ij

(D.3)This is divergence free if

B2

µ0! (P& ! P') = K ! (D.4)

and1

2µ0B2 + P' = K (D.5)

where K, K ! are both constants.

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120 APPENDIX D. SOLUTION TO ADVANCED PROBLEMS

D.2 Liouville and Vlasov Theorems: a Con-servation Equation in Phase Space.

a Expanding f in 6 dimensional phase space and time gives

f(x+ )x,v + )v, t + )t) = f(x,v, t) + )t'f

't+ )x ·#f + )v ·#vf + ...

(D.6)so that the total change in f as we move from x ' x+)x, v ' v+)v,and t ' t + )t is

f(x + )x,v + )v, t + )t) ! f(x,v, t))t

=Df

Dt(D.7)

and along a particle orbit such that:

v =)r)t

(D.8)

a =)v)t

(D.9)

this becomes:Df

Dt='f

't+ v ·#f + a ·#vf (D.10)

b If x and v are independent then

(v ·#f)j = vj'f

'xj='vjf

'xj(D.11)

that is,v ·#f = # · (vf) (D.12)

Also, from the Lorentz force law

a =q

m[E(x, t) + v " B(x, t)] (D.13)

then again as x and v are independent

E · 'f'v

='

'v· (Ef) (D.14)

and using an identity we can write

'

'v· (v " Bf) = (v " B) · 'f

'v+ f

'

'v· (v "B) (D.15)

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by expanding into components the second term on the r.h.s. can beshown to be zero. Then

(v " B) · 'f'v

='

'v(v " Bf) (D.16)

so that we can finally rewrite the Vlasov equation as

'f

't+ # · (vf) + #v · (af) = 0 (D.17)

ie in the form of a conservation equation for f in 6 dimensional phasespace.

D.3 Newton’s Laws and the Wave Equationunder 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 v!i = vi!v

so that the acceleration is

dv!i

dt=

dvi

dt=

dv!i

dt!(D.18)

then the force between the ith and jth particle

midvi

dt=#

j

F(xi ! xj) (D.19)

transforms as

midv!

i

dt!=#

j

F(x!i + vt ! x!

j ! vt) =#

j

F(x!i ! x!

j) (D.20)

Hence Newton’s equations are Galilean invariant in form.

b To transform the wave equation$#2 ! 1

c2

'

't2

%& = 0 (D.21)

we need to transform # and $$t . These are defined via the chain rule

so that in the unprimed frame:

d&(x, t) =0dx ·# + dt

'

't

1&(x, t) (D.22)

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so that#&(x, t) =

d&(x, t)dx

|t const (D.23)

and'&(x, t)'t

=d&(x, t)

dt|x const (D.24)

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

d&(x!, t!) =0dx! ·#! + dt!

'

't!

1&(x!, t!) (D.25)

then#!&(x!, t!) =

d&(x!, t!)dx! |t! const (D.26)

and'&(x!, t!)'t!

=d&(x!, t!)

dt!|x! const (D.27)

Now t = t! so that t! constant is equivalent to t constant which impliesthat #! & #. However, x! (= x, instead we have x! constant equivalentto x! vt constant when obtaining the partial deridatives from chainrule.We can write d&(x!, t!) in terms of elements in the unprimed frameusing dx! = dx ! vdt, dt! = dt:

d&(x!, t!) =0dx! ·#! + dt!

'

't!

1& =

0(dx ! vdt) ·#! + dt

'

't!

1&

(D.28)Then, given that we can write & as a function of variables in eitherframe:

&(x, t) & &(x!, t!) (D.29)

this yields:

d&(x!, t!)dt

|x const='&(x!, t!)

't=0!v ·#! +

'

't!

1&(x!, t!) (D.30)

Then the wave equation in the unprimed frame is also$#2 ! 1

c2

'

't2

%&(x!, t!) = 0 (D.31)

which transforms to=#!2 ! 1

c2

0!v ·#! +

'

't!

12>&(x!, t!) = 0 (D.32)

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D.4. TRANSFORMATION OF THE FIELDS. 123

which expands to:0#!2 ! 1

c2

'

't2! 1

c2(v ·#!)2 +

2c2

(v ·#!)'

't!

1& = 0 (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

&(x, t) $ ei(%t"k·x) (D.34)

which transformed to the primed frame becomes:

&(x!, t!) $ ei([%"k·x]t!"k·x!) (D.35)

which is consistent with the doppler shift of frequency:

(! = ( ! k · x (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 experimentdiscussed in 3.2.

To in the rest frame calculate the electric field from one of the wires weenclose an elemental length dl of wire in surface S as shown in figure D.4.S is a cylinder of radius l. Gauss’ law in integral form then gives

!

SE · dS =

"

Sc

E · dS = E2#ldl ="

V

"

$0dV =

"

$0Adl (D.37)

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

E ="A

2#l(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 andthe force in this frame in which the charges are at rest is just dF = "AE(l)dlper length element dl of wire.

In the moving frame we can calculate the magnetic field by enclosingthe wire with a curve as shown in D.4 and by using Ampere:

!

SB · dS = µ0

"

SJ · dS = µ0

"

S"!v · dS (D.39)

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ldl

A

S

+

+++

+

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

+

+

+

+

+

ldl

A

C

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

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D.5. METRIC FOR FLAT SPACETIME. 125

but taking into account the Lorentz contraction along the direction of thewire in order to calculate "! in the moving frame as in 3.2. The electricfield can also be calculated in the moving frame using charge density "!.

Alternatively, the transformation of the fields can be used directly giventhe electric field in the unprimed (rest) frame where the magnetic field iszero:

B! = !!v " Ec2

(D.40)

andE! = !E (D.41)

where we have exploited the geometry, that is, in the unprimed frame theelectric field is perpendicular to the wires.

In each frame the Lorentz force is then just given directly from theLorentz force law.

D.5 Metric for Flat Spacetime.

a To showg)&g

&* = D*) = )*) (D.42)

where we will use the notation D*) for the l.h.s. and

g)* =

&

22'

1 0 0 00 !1 0 00 0 !1 00 0 0 !1

(

33) = g)* (D.43)

Now all o! axis , (= - terms of g)* and g)* are zero, hence all of theo! axis terms of the inner product D*

) will be zero. For example

D21 = g1&g

&2 = 0 (D.44)

for any !.

The remaining terms are: , = - = 0 giving

D00 = g0&g

&0 = g00g00+g01g

10+g02g20+g03g

30 = 1+0+0+0 (D.45)

, = - = 1:

D11 = g1&g

&1 = g10g01+g11g

11+g12g21+g13g

31 = 0+1+0+0 (D.46)

similarly, D22 = 1 and D3

3 = 1

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Then

D*) =

&

22'

1 0 0 00 1 0 00 0 1 00 0 0 1

(

33) = )*) (D.47)

as required.

b To showg)*x

!* = x!) (D.48)

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

g)&%&*x* (D.49)

this is covariant by inspection (the only remaining index after con-traction will be ”down”).Explicitly this can be calculated as follows: First we can calculate

g)&%&* = %*) (D.50)

by exploiting the fact that , (= - terms will be zero (from the defini-tion of g)*). The remaining nonzero terms are:

%*0 = g00%0* = %0*

%*1 = g11%1* = !%1*

%*2 = g22%2* = !%2*

%*3 = g33%3* = !%3*

(D.51)

theng)&%&*x* = %*

)x* = x!) (D.52)

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

%*0x* = %0*x* = x!

0

%*1x* = !%1*x* = x!

1

%*2x* = !%2*x* = x!

2

%*3x* = !%3*x* = x!

3

(D.53)

so that if covariant four vector

x* = (ct,!x) =

&

22'

ct!x!y!z

(

33) (D.54)

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and

%)*x* =

&

22'

ct!

x!

y!

z!

(

33) (D.55)

then we have

%*)x* =

&

22'

ct!

!x!

!y!

!z!

(

33) = x!) (D.56)

which is covariant in the primed frame.

c To show

F)* = g)&F&,g,* (D.57)

perform the contraction in two steps. First, the nonzero terms ofF &,g,* = F &

* are:

) = 0,- = 0 F &0 = +F &0

) = 1,- = 1 F &1 = !F &1

) = 2,- = 2 F &2 = !F &2

) = 3,- = 3 F &3 = !F &3

(D.58)

for ! = 0, 3. Then the nonzero terms of g)&F&* are:

, = ! = 0 F0* = +F 0*

, = ! = 1 F1* = !F 1*

, = ! = 2 F2* = !F 2*

, = ! = 3 F3* = !F 3*

(D.59)

for - = 0, 3. Here we have exploited the fact that all - (= ! terms ofg*& are zero.

We now just consider the signs of the terms of F)* for all ,,- .

row 0: , = 0 all change sign except F00 = 0row 1: , = 1 F10 changes sign onlyrow 2: , = 2 F20 changes sign onlyrow 3: , = 3 F30 changes sign only

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D.6 Length of the EM Field Tensor in Space-time.

We have the electromagnetic field tensor:

F)* =

&

22'

0 !Exc !Ey

c !Ezc

Exc 0 !Bz By

Ey


(

33) (D.60)

from Chapter 4, and also F)* is obtained via the transformation Ei ' !Ei

and Bi ' Bi in F)* .The symmetries of this tensor are then by inspection:

F *) = !F)*

F*) = !F)*(D.61)

and the trace is zero:F)) = F)) = 0 (D.62)

F)*F *) then has the following properties for ! = 0, 3:

i) The on axis terms are zero

F &&F&& = 0 (D.63)

ii) The timelike terms, that is, those involving E are:

F0&F &0 = !F0&F&0 = !(!F0&F0&)F&0F 0& = !F&0F0& = !(!F&0F&0)

(D.64)

contracting these terms over ! gives:

#

&

[F&0]2 =E2

c2=#

&

[F0& ]2 (D.65)

so that the total contribution of the timelike terms is 2E2/c2.

iii) The spacelike terms, that is, terms involving B then have the followingproperties:

F)&F&) = !F)&F

)& = !F)&F)& (D.66)

for , (= 0, ! (= 0. Contracting these terms gives:#

)=1,3

#

&=1,3

!(F)&)2 = !2B2 (D.67)

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D.7. ALTERNATIVE FORM FOR THE MAXWELL HOMOGENOUS EQUATIONS.129

The result of the contraction is then:

F)*F *) = 2$

E2

c2! B2

%(D.68)

In free space (photons only) we have E = Bc so that F)*F *) vanishes.For the dual tensor F)* we use the fact that it is obtained from F)* via

the duality transformation Ei ' Bic2, Bi ' !Ei (see 4.2). Transformingthe result gives

F)*F*) = 2$

B2c4

c2! E2

%= ! 2

c2

$E2

c2! B2

%= !c2F)*F *) (D.69)

as required.

D.7 Alternative Form for the Maxwell Ho-mogenous Equations.

This left as an exercise for the student.

D.8 Lorentz Transformation of the EM FieldTensor.

a In 4.1 we obtainF)* = ')A* ! '*A) (D.70)

then the Lorentz transformation of this is just the transformation ofeach of the component four vectors.

b To evaluate this ie:

F !)* = '!)A!* ! '!*A!) (D.71)

we apply the Lorentz transformation and the transpose rule:

F !)* = (%),',)(%*&A&) ! (%*&'&)(%),A,)= %),',A&%&* ! %*&'&A,%,)

= %),',A&%&* ! %),'&A,%&*

= %),(',A& ! '&A,)%&*

= %),F,&%&*

(D.72)

as required.

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Appendix E

Vector identities

Assuming the right hand rule relates the orthogonal unit vectors i, j, andk:

i " j = k

E.1 Di!erential Relations

a · b = b · a = a1b1 + a2b2 + a3b3 (E.1)

a " b = !b " a =i j ka1 a2 a3

b1 b2 b3

(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)

#(+&) = &#+ + +#& (E.7)

# · (&a) = &# · a + a ·#& (E.8)

# " (&a) = & "#a + (#&) " a (E.9)

# · (a " b) = b · (# " a) ! a · (# " b) (E.10)

#(a · b) = (a ·#)b + (b ·#)a + a " (# " b) + b " (# " a) (E.11)

#" (a " b) = a(# · b) + (b ·#)a ! b(# · a) ! (a ·#)b (E.12)

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132 APPENDIX E. VECTOR IDENTITIES

a " (# " b) = (#B) · a ! (a ·#)b (E.13)

# " (# " a) = #(# · a) !#2a (E.14)

# · (# " a) = 0 (E.15)

#" (#&) = 0 (E.16)

E.2 Integral Relations

For the following integral relations vector surface element dS is directedalong the normal n to the surface S, line element dl is directed along curveC.

Flux of a vector field a:

Flux =!

Sa · dS (E.17)

If surface S spans curve C:

• !

C&dl =

"

SdS " (#&) (E.18)

• Stokes Theorem: !

Ca · dl =

"

S# " a · dS (E.19)

• !

Cdl " a =

"

s(dS "#) " a (E.20)

• "

SdS · (#& "#+) =

!

C&d+ = !

!

C+d& (E.21)

If surface S encloses volume V , dS points outwards:

• !

S&dS =

"

v(#&)dV (E.22)

• Gauss or Divergence Theorem:!

Sa · dS =

"

V# · adV (E.23)

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E.2. INTEGRAL RELATIONS 133

• !

S(dS " a) =

"

V(#" a)dV (E.24)

• Green I:!

S&(#+) · dS =

"

V

?&#2+ + (#&) · (#+)

@dV (E.25)

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

S(&#+ ! +#&) · dS

"

V(&#2+ ! +#2&)dV (E.26)

• Green II in vector form:!

S[b " (# " a) ! a " (# " b)] · dS = (E.27)

"

V(a · [# " (# " b)] ! b · [# " (# " a)]) dV

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134 APPENDIX E. VECTOR IDENTITIES

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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 toas many indices as we wish, for example Pijkl is a tensor of rank four.

• A Dyadic is formed of two vectors a, b for any orthogonal coordinatesystem:

ab =

&

'axbx axby axbz

aybx ayby aybz

azbx azby azbz

(

) (F.1)

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

Aij = aibj (F.2)

• Coordinate rotationA vector in the unprimed frame:

r =

&

'xyz

(

) (F.3)

will have coordinates in the primed frame:

x!

= a11x +a12y +a13zy

!= a21x +a22y +a23z

z!

= a31x +a32y +a33z

(F.4)

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136 APPENDIX F. TENSORS

The tensor aij contains the direction cosines of the rotation x, y, z 'x!, y!, z!. This can be written as

r! = A · r (F.5)

where

r!=

&

'x

!

y!

z!

(

) and A =

&

'a11 a12 a13

a21 a22 a23

a31 a32 a33

(

) (F.6)

• Einstein conventionSince x = r1, y = r2 and z = r3, equation 2.24 is

r!

j = ajiri (F.7)

which is shorthand (the Einstein summation convention) for

r!

j =#

i=1,3

ajiri (F.8)

The sum over index i on the r.h.s. of 2.27 contracts the number ofindices by one (the l.h.s. has single index j).

• Inner product.Equations 2.26 and 2.27 are an example of a tensor dot (or inner)product. The switch to index notation drops all reference to the basisvectors (or axes) x, y, z. Hence equation 2.28 is tensor dot product:

a · T = a1T11 + a2T21 + a3T31 + a1T12 + a2T22 + ...= xjaiTij

& aiTij

(F.9)dropping all reference to the basis vectors xj.

• Then # · T in Cartesian coordinates is

xj'

'xiTij = # · T& '

'xiTij (F.10)

It follows that

# · (&T ) = &(# ·T ) + (#&) · T (F.11)

# · (ab) = (# · a)b + (a ·#)b (F.12)

• Divergence theorem for tensors of arbitrary rank:!

ST · dS =

"

V(# · T )0V (F.13)

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F.2. SPECIAL TENSORS 137

F.2 Special Tensors

• The trace is )ij = 1 when i = j and 0 when i (= j:

)ij =

&

'1 0 00 1 00 0 1

(

) (F.14)

Contracting )ij over i and j with two vectors extracts the dot product:

)ijaibj = a1b1 + a2b2 + a3b3 = a · b = aibi (F.15)

This generalizes to tensors of arbitrary rank:

)ijaiTijkl.. = a · T (F.16)

• The alternating tensor:

$ijk =1 if ijk = 123, 231, 312

!1 if ijk = 321, 132, 2130 any two indices alike

(F.17)

Contracting $ijk over j and k extracts the following vector from adyadic:

$ijkajbk = $i11a1b1 +$i12b1a2 +$i13b1a3

+$i21a2b1 +$i22b2a2 +$i23b3a2

+$i31a3b1 +$i32b2a3 +$i33b3a3

(F.18)

If we consider the i = 1 terms, all except $123a2b3 and $132a3b2 arezero. These last two give a2b3 ! a3b2 which are the x component ofthe vector cross product. Similarly, i = 2, 3 gives the y, z componentsrespectively, so that contracting with $ijk 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-timeinterval

s2 = c2t2 ! x2 ! y2 ! z2 (F.19)

The space-time interval is the length of the four vector:

s2 = s · s = x)x) (F.20)

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F.3.1 General properties of spacetime

• s has 2 forms:Covariant:

x) = (ct,!x) =

&

22'

ct!x!y!z

(

33) (F.21)

and Contravariant:

x) = (ct, +x) =

&

22'

ctxyz

(

33) (F.22)

• In general (for any spacetime geometry) if we have a well definedtransformation x!) = x!)(x0, x1, x2, x3) a covariant vector transformsas:

a!) =

'x0

'x!) a0 +'x1

'x!) a1 +'x2

'x!) a2 +'x3

'x!) a3 ='x*

'x!) a* (F.23)

and a contravariant vector transforms as

a!) ='x!)

'x0a0 +

'x!)

'x1a1 +

'x!)

'x2a2 +

'x!)

'x3a3 =

'x!)

'x*a* (F.24)

• For tensors of rank 2, a covariant tensor G)* transforms as

G!)* =

'x&

'x!)'x,

'x!* G&, (F.25)

a contravariant rank 2 tensor G)* as:

G!)* ='x!)

'x&'x!*

'x,G&, (F.26)

and a mixed tensor of rank 2 G)* transforms as:

G!)* =

'x!)

'x&'x,

'x!* G&, (F.27)

• The dot or inner product is:

a · b = a......)b...)

... (F.28)

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F.3. GENERALIZED TENSORS 139

that is, a contraction over the index ,.For general spacetime geometry the inner product is invariant undertransformation since:

a! · b! ='x*

'x!)'x!)

'x&a...

...*b...&... =

'x*

'x&a...

...*b...&... = )*&a...

...*b...&... = a · b

(F.29)

F.3.2 Flat spacetime

The Lorentz transformation corresponds to a rotation of the four vector sin flat space-time, under which s2 is constant.

• The norm or metric for flat spacetime is then the invariant di!erentiallength element:

(ds)2 = (dx0)2 ! (dx1)2 ! (dx2)2 ! (dx3)2 (F.30)

which is a special case of the general di!erential length element

(ds)2 = g)*dx)dx* (F.31)

• The spacetime metric for flat space is given by:

g)* =

&

22'

1 0 0 00 !1 0 00 0 !1 00 0 0 !1

(

33) = g)* (F.32)

• Covariant and contravariant vectors are related by

x) = g)*x* (F.33)

and

x) = g)*x* (F.34)

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

x!

) = %)*x* (F.35)

on the contravariant form and

x)!= %)*x* (F.36)

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on the covariant form.where the Lorentz transformation is:

%)* =

&

22'

! ! vc! 0 0

+ vc! !! 0 0

0 0 !1 00 0 0 !1

(

33) (F.37)

the inverse transformation matrix is then just obtained by v ' !vin 3.60.

• The covariant four gradient is

') =

&

22'

1c$$t$$x$$y$$z

(

33) = (1c

'

't,#) (F.38)

• The contravariant four gradient is

') =

&

22'

1c$$t

! $$x

! $$y

! $$z

(

33) = (1c

'

't,!#) = g)*'* (F.39)

• Hence the ”four - divergence” of a four vector

')a) = ')a) =1c

'

'ta0 + # · a (F.40)

(where a is the spacelike part of four vector field a)).

• The ”4 curl” of four vector a is a rank 2 four tensor:

F)* = ')a* ! '*a) (F.41)

• The D’Alembertian invariant operator is then:

')') =

1c2

'2

't2!#2 = ! (F.42)

which is just the wave equation operator in vacuum.

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142 APPENDIX G. UNITS AND DIMENSIONS

Appendix G

Units and Dimensions

G.1 SI Nomenclature

Physical Quantity Name Symbollength metre mmass kilogram kgtime second scurrent Ampere Atemperature Kelvin Kamount of substance mole molluminous intensity candela cdplane angle radian radsolid angle steradian srfrequency Hertz Hzenergy Joule Jforce Newton Npressure Pascal Papower Watt Welectric charge Coulomb Celectric potential Volt Velectric resistance Ohm 'electric conductance Siemens Selectric capacitance Farad Fmagnetic flux Weber Wbmagnetic inductance Henry Hmagnetic intensity Tesla Tluminous flux lumen lmilluminance lux lxactivity (of radioactive source) Bequerel Bqabsorbed dose (of ionising radiation) Gray GyCORE E

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G.2. METRIC PREFIXES 143

G.2 Metric Prefixes

Multiple Prefix Symbol Multiple Prefix Symbol10"1 deci d 10 deca da10"2 centi c 102 hecto h10"3 milli m 103 kilo k10"6 micro µ 106 mega M10"9 nano n 109 giga G10"12 pico p 1012 tera T10"15 femto f 1015 peta P10"18 atto a 1018 exa E

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Appendix H

Dimensions and Units

H.1 Physical Quantities

SI has been used throughout this book. Gaussian units are still often usedhawever and conversion factors between SI and Gaussian units are givenhere.

To obtain the value of a quantity in Gaussian Units, multiply thevalue expressed in SI units by the conversion factor. Multiples of 3 inthe conversion factor result from approximating the speed of light c =2.99979% 108ms"1 1 3 % 108ms"1.

DimensionsPhysicalQuantity

Sym-

bol

SI Gaussian SIUnit

ConversionFactor

GaussianUnit

mass m m m kg 103 gram (g)length l l l m 102 centimetre

(cm)time t t t s 1 second (s)

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146 APPENDIX H. DIMENSIONS AND UNITS


Sym-

bol

SI Gaussian SIUnit

ConversionFactor

GaussianUnit

Capacitance C t2q2

ml2 l F 9 % 1011 cm

Charge q q m12 l

32

t C 3 % 109 stat-coulomb

ChargeDensity

" ql3

m12

l32 t

Cm"3 3 % 103 stat-coulombcm"3

Conductance tq2

ml21t S 9 % 1011 cms"1

Conductivity / tq2

ml31t S/m 9 % 109 s"1

Current I, i qt

m12 l

32

t2 A 3 % 109 statampere

Currentdensity

J ql2t

m12

l12 t2

Am"2 3 % 105 statamperecm"2

Density " ml3

ml3 kgm"3 10"3 gcm"3

Displacement D ql2

m12

l12 t

Cm"2 12# % 105 stat-coulombcm"2

Electric field E mlt2q

m12

l12 t

V/m 13 % 10"4 statvoltcm"1

Electro-motance

$,emf

ml2

t2qm

12 l

12

t V 13 % 10"2 statvolt

Energy W ml2

t2ml2

t2 J 107 ergEnergydensity

U mlt2

mlt2 J/m3 10 ergcm"3

Force F mlt2

mlt2 N 105 dyne

Frequency f,0 1t

1t Hz 1 Hz

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H.1. PHYSICAL QUANTITIES 147


Sym-

bol

SI Gaussian SIUnit

ConversionFactor

GaussianUnit

Impedance Z ml2

tq2tl ' 1

9 % 10"11 scm"1

Inductance L ml2

q2t2

l H 19 % 10"11 s2cm"1

Magneticintensity

H qlt

m12

l12 t

A/m 4#% 10"3 Oersted

Magneticflux

# ml2

tqm

12 l

32

t Wb 108 Maxwell

Magneticinduction

B mtq

m12

l12 t

T 104 Gauss

Magneticmoment

m, µ l2qt

m12 l

52

t Am2 103 Oersted cm3

Magnetisation M qlt

m12

l12 t

Am"1 10"3 Oersted

Magneto-motance

M qt

m12 l

12

t2 A 4!10 Gilbert

Momentum p mlt

mlt kgms"1 105 gcms"1

Momentumdensity

ml2t

ml2t kgm"2 10"1 gcm"2s"1

Permeability µ mlq2 1 Hm"1 1

4! % 107 -Permittivity #$ t2q2

ml3 1 Fm"1 36# % 109 -

Polarization P ql2

m12

l12 t

Cm"2 3 % 105 stat-coulombcm"2

Potential V,& ml2

t2qm

12 l

12

t V 13 % 10"2 statvolt

Power P ml2

t3ml2

t3 W 107 erg s"1

Powerdensity

mlt3

mlt3 Wm"3 10 erg cm"3s"1

Pressure P mlt2

mlt2 Pa 10 dyne cm"2

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148 APPENDIX H. DIMENSIONS AND UNITS


Sym-

bol

SI Gaussian SIUnit

ConversionFactor

GaussianUnit

Reluctance R q2

ml21l AWb"1 4#% 10"9 cm"1

Resistance R ml2

tq2tl ' 1

9 % 10"11 scm"1

Resistivity 1," ml3

tq2 t 'm 19 % 10"9 s

Thermalconductivity

2 mlt3

mlt3 Wm"1K"1105 erg

cm"1s"1K"1

Vectorpotential

A mltq

m12 l

12

t Wbm"1 106 Gauss cm

Velocity v lt

lt ms"1 102 cms"1

Viscosity 1, µ mlt

mlt kgm"1s"110 Poise

Vorticity 3 1t

1t s"1 1 s"1

Work W ml2

t2ml2

t2 J 107 erg

H.2 Equations

In SI, in media the permittivity $ = $0$r and the permeability µ = µ0µr

where $0 and µ0 are the permittivity and permeability in free space, andthe relative permittivity and permeability, $r and µr are dimensionless.

SI Gaussianµ0 4# % 10"7Hm"2 1$0

1c$

µ01

D = $0E + P E + 4#PH = B

µ0! M B! 4#M

Maxwell I # ·E = "#0

# · D = " # ·D = 4#"Maxwell II # ·B = 0 # ·B = 0Maxwell III #" E = !$B

$t #" E = ! 1c$B$t

Maxwell IV # " B = µ0J + µ0$0$E$t

#" H = J + $D$t # " H = 4!

c J + 1c$D$t

Lorentz forceper unitcharge

E + v " B E + vc "B

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Appendix I

Physical Constants (SI)

Physical quantity Symbol Value UnitsFree space permittivity $0 8.854 % 10"12 Fm"1

Free space permeability µ0 4# % 10"7 Hm"1

Free space speed of light c = 1$µ0#0

2.998 % 108 ms"1

Elementary charge e 1.602 % 10"19 CBohr magneton µB 9.274 % 10"24 Am"2

Electron mass me 9.109 % 10"31 kgProton mass mp 1.673 % 10"27 kgAtomic mass unit mu 1.661 % 10"27 -Planck constant h 6.626 % 10"34 JsBoltzmann constant k 1.381 % 10"23 JK"1

Avrogado number N0 6.022 % 1023 mol"1

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