Core Electrodynamics Copyright Sandra C Chapman

CHAPMAN C

Core ElectrodynamicsSANDRA

Sandra C. Chapman

June 29, 2010 COPYRIGHT

ELECTRODYNAMICS

CORE Preface CHAPMAN C

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

CORE 1 2

are given with graduate study in mind. As SI units are used throughout here, conversions for units and equations are also given. There have been many who have contributed to the existence of this book. Thanks in particular go to Nick Watkins for valuable discussions and to George Rowlands for his insightful reading of the final draft. David Betts, the editor of this series, has also shown remarkable patience and tenacity as the many deadlines have come and gone. The completion of the book was also much assisted by a PPARC personal fellowship. Finally, my thanks go to the students themselves; their lively receptionCHAPMAN of the original course and their insightful questions were theC inspiration for this book. If the reader finds that this book provides a shortcut to experience the beauty of electrodynamics, without sacrificing the rigour needed for further study, then I have suceeded. Sandra Chapman, June, 1999 SANDRA

ELECTRODYNAMICS

CORE Contents CHAPMAN C

3.5 Four Vectors and Four Vector Calculus...... 60 3.5.1 Some mechanics, Newton’s laws...... 60 3.5.2 Some four vector calculus...... 63 3.6 A Frame Invariant Electromagnetism ...... 64 3.6.1 Charge conservation...... SANDRA . . . . 64 3.6.2 A manifestly covariant electromagnetism ...... 65

4 The Field Tensors 67 4.1 Invariant Form for E and B: TheEMFieldTensor. . . . . 67 4.2 Maxwell’s Equations in Invariant Form...... 70 4.3 Conservation of Energy-Momentum...... 73 4.4 LorentzForce...... 74 4.4.1 Manifestly covariantelectrodynamics...... 75 4.5 Transformation of the Fields .COPYRIGHT ...... 76 4.6 Field from a Moving Point Charge...... 77 4.7 Retarded Potential...... 81 3.5 Four Vectors and Four Vector Calculus...... 60 3.5.1 Some mechanics, Newton’s laws...... 60 3.5.2 Some four vector calculus...... 63 3.6 A Frame Invariant Electromagnetism ...... 64 3.6.1 Charge conservation...... 64 3.6.2 A manifestly covariant electromagnetism ...... 65

CORE 3 4 CONTENTS

4.7 Retarded Potential...... 81

1 A Brief Tour of Electromagnetism. 9 1.1 The Building Blocks...... 9 1.2 MaxwellIandII ...... 11 1.2.1 Flux of a vector field ...... 13 1.2.2 Flux of E ...... 15 1.2.3 Flux of B ...... 18 1.2.4 Conservative and nonconservative fields...... CHAPMAN . 19 1.3 Maxwell III and IV ...... C . . . 21 1.3.1 Faraday’s law and Galilean invariance...... 21 1.3.2 Ampère’s law and conservation of charge...... 25 1.4 Electromagnetic Waves...... 26 1.5 Conservation Equations...... 28

2 Field Energy and Momentum. 29 2.1 Tensors and Conservation Equations.SANDRA ...... 30 2.1.1 Momentum flux density tensor...... 30 2.1.2 Momentum flux, gas pressure and fluid equations. . 33 2.1.3 Cartesian tensors, some definitions...... 34 2.2 Field Momentum and Maxwell Stress ...... 37 2.2.1 Energy conservation: Poynting’s theorem ...... 37 2.2.2 Momentum conservation: Maxwell stress ...... 39 2.3 Radiation Pressure...... 41 COPYRIGHT 3 A Frame Invariant Electromagnetism 45 3.1 TheLorentzTransformation...... 46 3.2 The Moving Charge and Wire Experiment...... 50 3.3 Maxwell in Terms of Potentials...... 55 3.4 Generalized Coordinates...... 56 3.5 Four Vectors and Four Vector Calculus...... 60 3.5.1 Some mechanics, Newton’s laws...... 60 3.5.2 Some four vector calculus...... 63 3.6 A Frame Invariant Electromagnetism ...... 64 3.6.1 Charge conservation...... 64 3.6.2 A manifestly covariant electromagnetism ...... 65

4 The Field Tensors 67 4.1 Invariant Form for E and B: TheEMFieldTensor. . . . . 67 ELECTRODYNAMICS4.2 Maxwell’s Equations in Invariant Form...... 70 4.3 Conservation of Energy-Momentum...... 73 4.4 LorentzForce...... 74 CORE CONTENTS 5

4.4.1 Manifestly covariantelectrodynamics...... 75 4.5 Transformation of the Fields ...... 76 4.6 Field from a Moving Point Charge...... 77 4.7 Retarded Potential...... 81

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

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

C Some Advanced Problems 111 C.1 Maxwell Stress Tensor...... 111 C.2 Liouville and Vlasov Theorems: a Conservation Equation for PhaseSpace...... 112 C.3 Newton’s Laws and the Wave Equation under Galilean Trans- formation...... 112 C.4 Transformation of the Fields...... 112 C.5 Metric for Flat Spacetime.1 ...... 113 ELECTRODYNAMICSC.6 Length of the EM ﬁeld Tensor in Spacetime...... 114 C.7 Alternative Form for the Maxwell Homogenous Equations. . 114 C.8 Lorentz Transformation of the EM Field Tensor...... 114 CORE 6 CONTENTS

D Solution to Advanced Problems 117 D.1 Maxwell Stress Tensor...... 117 D.2 Liouville and Vlasov Theorems: a Conservation Equation in PhaseSpace...... 118 D.3 Newton’s Laws and the Wave Equation under Galilean Trans- formation...... 119 D.4 Transformation of the Fields...... 121 D.5 Metric for Flat Spacetime...... 123 D.6 Length of the EM Field Tensor in Spacetime...... CHAPMAN . 126 D.7 Alternative Form for the Maxwell Homogenous Equations.C . 127 D.8 Lorentz Transformation of the EM Field Tensor...... 127

E Vector identities 129 E.1 Diﬀerential Relations ...... 129 E.2 Integral Relations ...... 130

F Tensors SANDRA 133 F.1 CartesianTensors ...... 133 F.2 Special Tensors ...... 135 F.3 Generalized Tensors ...... 135 F.3.1 General properties of spacetime ...... 136 F.3.2 Flat spacetime ...... 137

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

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

I Physical Constants (SI) 147

ELECTRODYNAMICS

CORE List of Figures CHAPMAN C

4.1 Charge q is at rest at the origin of the S! frame and is moving with velocity vˆx1 w.r.t. frame S. The origins of the S and S! frames coincide at t =0...... 78 4.2 Sketch of E and E at point P versus x! /γ = x vt. . . . 79 1 2 1 1 − SANDRA 4.3 Sketch of the E! field around a charge at rest, and the E field around a charge moving at relative velocity v, in the x1,x2 plane...... 80 4.4 A volume element is located at rest at the origin of the S! frame which is moving at +v in the x1 direction w.r.t. the S frame. In the S! frame the volume element contains charge ρ!dx1! dx2! dx3! . The origins of the two frames are coincident at t = t! =0...... 81 COPYRIGHT 4.1 Charge q is at rest at the origin of the S! frame and is moving with velocity vˆx1 w.r.t. frame S. The origins of the S and S! frames coincide at t =0...... 78 4.2 Sketch of E and E at point P versus x! /γ = x vt. . . . 79 1 2 1 1 − 4.3 Sketch of the E! field around a charge at rest, and the E field around a charge moving at relative velocity v, in the x1,x2 plane...... 80 4.4 A volume element is located at rest at the origin of the S! frame which is moving at +v in the x1 direction w.r.t. the S frame. In the S! frame the volume element contains charge ρ!dx1! dx2! dx3! . The origins of the two frames are coincident at t = t! =0...... 81

1.1 Surface element dS on surface S spanning curve C with line ELECTRODYNAMICSelement dl...... 10 1.2 Surface element dS on volume V ...... 10 1.3 The right hand rule ...... 11

CORE 7 8 LIST OF FIGURES

1.4 Location of charges q1 and q2 ...... 12 1.5 The electric field due to charge qi = ρ(r!)dV ! ...... 13 1.6 Flux across dS is maximal when v is parallel to dS, and zero when v is perpendicular...... 14 1.7 Volume vdt containing particles that cross dS in time dt . . 14 1.8 Surfaces enclosing single charge q. The surface element dS lies on the arbitrary surface S and forms the end of the cone. There is no flux of E across the sides of the cone...... CHAPMAN . 15 1.9 Side view of cone enclosing single charge q. The surface element vector dS and field E lie in the plane of the paper.C 16 1.10 Cone enclosing single charge q in cartesian and cylindrical polar coordinates...... 17 1.11 The closed surface S has zero nett flux through it...... 19 1.12 E dl is integrated around the closed curve C in the vicinity· of a point charge. The radial sections of the path give contributions of equal size and oppositeSANDRA sign and the sections along contours of the potential (dotted lines) give zero contribution. The integral around the closed curve is zero. . 21 1.13 The observer is at rest in frame 1 (top) and moves with the wireinframe2(bottom)...... 22 1.14 The observer is at rest and the wire moves on conducting rails. 23 1.15 Wire element dl sweeps out surface element dS in time dt. . 24 1.16 Charge flows out of volume V ...... 26 1.17 An ideal dischargingCOPYRIGHT capacitor...... 27 2.1 Forces acting on fluid element dV form the components of a pressuretensor ...... 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 of motion. 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 is carried by the protons, the test charge q is at rest...... 51 ELECTRODYNAMICS3.4 The (dashed) loop over which we integrate B dl is a circle of radius r centred on the wire, with ˆr transverse· to the direction of motion u. Also shown is the direction of u B. 53 ∧ CORE LIST OF FIGURES 9

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

A.1 Problem 1a: A straight wire carrying a steady current I. . . 92 A.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...... 122 D.2 Curve C enclosing the wire withCOPYRIGHT element along the curve dl. 122

ELECTRODYNAMICS

CORE 10 LIST OF FIGURES

CHAPMAN C

SANDRA

ELECTRODYNAMICS

CORE Chapter 1 CHAPMAN C A Brief Tour of Electromagnetism. SANDRA

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

1.1 The Building Blocks.

ELECTRODYNAMICSWe will use the experimentally determined:

Coulomb’s Law • CORE 11 12 CHAPTER 1. A BRIEF TOUR OF ELECTROMAGNETISM.

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

V SANDRA

Figure 1.2: Surface element dS on volume V

Faraday’s Law • Amp`ere’s Law • and some mathematics:COPYRIGHT Stokes’ Theorem (for any vector ﬁeld A): •

A dl = A dS (1.1) · ∇∧ · !C "S where surface S spans curve C as shown in ﬁgure 1.1 Gauss’ or Divergence Theorem: •

A dS = AdV (1.2) · ∇· !S "V where surface S encloses volume V as shown in ﬁgure 1.1. and the right hand rule ELECTRODYNAMICS•

ˆi ˆj = kˆ ∧ CORE 1.2. MAXWELL I AND II 13

F k

i j v B

Figure 1.3: The right hand rule CHAPMAN C which is built in to the way the cross product is deﬁned, so that in the Lorentz force law the magnetic force

F = qv B M ∧ is positive for protons (figure 1.1). If we had used a left handed cross product so that î ˆj SANDRA= kˆ then ∧ − Lorentz force law would be written FM = qv B for protons. We have also defined the electric (vector) field to point− away∧ from positive charge so that FE = qE for protons.

1.2 Maxwell I and II

Coulomb experimentally determined the force between two static charges to be: COPYRIGHT 1 q q F = 1 2 (r r ) (1.3) 21 4π$ r r 3 1 − 2 0 | 1 − 2 | where the force is in Newtons (N), charge q is in Coulombs (C), r in meters 12 2 1 2 and $0 =8.85.10− C N − m− is the permittivity of free space (see ﬁgure 1.2). Coulomb’s law embodies three empirically known properties of F:

1. the force is proportional to the charges F q ,q ∼ 1 2 2. the force points radially away from the positive charges

3. the force is inverse square F 1/r2 ∼ ELECTRODYNAMICSIn addition we can show that the force obeys the principle of superposition so that for a collection of n charges; the force on the jth charge qj is CORE 14 CHAPTER 1. A BRIEF TOUR OF ELECTROMAGNETISM.

q r-r 1 21 q 2

rr2 1 CHAPMAN C 0

Figure 1.4: Location of charges q1 and q2

n 1 qiqj SANDRA Fj = 3 (ri rj) (1.4) 4π$0 ri rj − i=j ## | − | We then deﬁne the electric ﬁeld at the position of the jth charge as F E = j (1.5) qj

1 that is, the Lorentz force law for electric field only, so that E is in NC− 1 (we will see that it is alsoCOPYRIGHT in Vm− from energy considerations). The vector field E(rj ) is then defined at the position r = rj of the test charge qj: 1 qi E(rj)= 3 (ri rj ) (1.6) 4π$0 ri rj − i=j ## | − | This can be expressed in terms of a scalar field, the charge density ρ(r) provided that the collection of point charges can be treated as a smoothly varying function, that is qi = ρ(r!)dV ! (1.7) This description will therefore be valid on length and timescales over which 1.4 to 1.7 hold. When we consider small length and timescales there are two distinct considerations. First, we require that the charge density 1.7, and all other quantities that will be described by scalar and vector fields here, such as electromagnetic fields, energy and momentum densities ELECTRODYNAMICSand so forth, are still describable by continuous functions. Second, we need to use the correct mechanics in the equations of motion for the charges and hence the Lorentz force. Here we will develop electrodynamics in terms CORE 1.2. MAXWELL I AND II 15

r-r’ dV’ E(r)

’ rr CHAPMAN C

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

of mechanics that is consistent with special (and general) relativity,SANDRA but is classical. At some length and timescale quantum mechanics is needed to replace classical mechanics. The ﬁeld equations may also need to be quantized (to give Quantum Electrodynamics)1. In this sense, the Maxwell equations, the Lorentz force and the mechanics of special relativity that we will discuss are classical . With these caveats, we can write the electric ﬁeld in terms of a volume integral over the charge density:

1 ρCOPYRIGHT(r!)(r r!) E(r)= − dV ! (1.8) 4π$ r r 3 0 "V | − ! | hence the electric ﬁeld retains the experimentally determined properties: it points radially away from an element of positive charge, its magnitude is inverse square with distance, it is proportional to the charge and it obeys the principle of superposition. We will now take these properties and phrase them in a more profound form, in terms of the ﬂux of E.

1.2.1 Flux of a vector ﬁeld Flux is mathematically and conceptually the same for any vector ﬁeld. We can explore the concept with a simple example; a cold gas2 comprised of

1The need for a field theory that permits both a continuous, classical limit, and discrete quanta (photons in the case of electromagnetic fields) was highlighted by the ELECTRODYNAMICSdiscovery of the photoelectric effect 2We will make use of the ”cold gas” model several times in this book, in all cases it is defined as here: a population of identical particles which all have the same velocity v(r,t) at any given position and time. CORE 16 CHAPTER 1. A BRIEF TOUR OF ELECTROMAGNETISM.

v(r) v(r) dS

CHAPMAN dS C

Figure 1.6: Flux across dS is maximal when v is parallel to dS, and zero when v is perpendicular. SANDRA v(r) v(r) ! l dS

vdt l cos ! COPYRIGHT 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 the same velocity v(r) at any position r. We then deﬁne:

Flux through surface element dS = number of particles crossing dS per second.

Notice that the flux depends upon the angle between dS and v(r) as in figure 1.2.1. To find the flux across dS we just need to identify the volume containing all particles that will cross the surface element per second. This volume is sketched in figure 1.2.1, where we have rotated our point of view such that both v and dS are in the plane of the paper (we can always do this, since the two vectors will define a plane). The projection of dS in this ELECTRODYNAMICSplane is l and perpendicular to the plane is a so that dS = la. The angle between v and dS is θ. If the surface element dS is su| fficiently| small that n and v are constant across it, then in time dt all particles in the cold gas CORE 1.2. MAXWELL I AND II 17

q S CHAPMAN C

Figure 1.8: Surfaces enclosing single charge q. The surface element dS lies on the arbitrary surface S and forms the end of the cone. There is no flux of E across the sides of the cone. SANDRA in volume la cos θvdt = dS cos θvdt = v dSdt | | · cross the surface element dS. The number of particles crossing dS in time dt is then nv dSdt, so that a flux of nv dS particles crosses dS per second. 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 = nv dS (1.9) COPYRIGHT· "S 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 field we will find the flux due to a single positive charge q which is located somewhere inside the closed surface S

Flux of E = E dS (1.10) · !S We will utilize the fact that we can choose any convenient S as long as q is located inside, and will ”build in” the three experimentally determined properties of E = F/q from Coulomb’s law. We start by taking a surface ELECTRODYNAMICSelement dS on an arbitrary surface enclosing q. We can then consider the cone shaped surface with sides formed by radius vectors from q and an end formed by dS as shown in ﬁgure 1.2.2. From Coulomb’s law: CORE 18 CHAPTER 1. A BRIEF TOUR OF ELECTROMAGNETISM.

" E(r)

q dS n CHAPMAN C Figure 1.9: Side view of cone enclosing single charge q. The surface element vector dS and ﬁeld E lie in the plane of the paper.

1. E points radially out from the charge: that is, E points in the r direction. There is then no flux of E out of the sides of the cone. The flux of E must emerge from the end of theSANDRA cone dS. We can again sketch the surface element dS choosing the plane of the paper to be the plane defined by the vectors dS and E and the cone is shown cut by this plane in the figure 1. The flux of E out of the cone is then E dS = E dS (1.11) · | | n where dSn is the projection of dS parallel to E (i.e. E is normal to the plane in which surface dSn lies) as shown in figure 1. The element dSn hasCOPYRIGHT a very useful property: it lies on the surface of a sphere centred on the charge q.

The element dSn is sketched in cylindrical polar coordinates with q at the origin in ﬁgure 1, where the element on the surface of the sphere at radius r is dS = rdθ r sin θdφ (1.12) n × We can now exploit 2. E is inverse square and 3. E is proportional to q so that

1 q E = 2 | | 4π$0 r ELECTRODYNAMICSthen the ﬂux of E through the cone 1 q E dS = E dSn = 2 rdθ r sin θdφ · × 4π$0 r × × CORE 1.2. MAXWELL I AND II 19

rd! ! d! CHAPMAN C # d y x #

r sin !d# SANDRA

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

q = dθ sin θdφ 4π$0 which is independent of r (E is inverse square and the surface area of the sphere is proportional to r2).COPYRIGHT Now the evaluation of the ﬂux of E over the entire closed surface is simply the integral over the solid angle 4π:

E dS = · !S q π 2π q sin θdθ dφ = 4π$ $ 0 "0 "0 0 For a collection of charges we just add the electric ﬁeld from each one, by principle of superposition. So the ﬂux through S from n charges will be

E dS = (E + E + E + ...) dS · 1 2 3 · !S !S ELECTRODYNAMICS= E1 dS + E2 dS + E3 dS + ... S · S · S · ! ! q ! q q = 1 + 2 + 3 + ... $0 $0 $0 CORE 20 CHAPTER 1. A BRIEF TOUR OF ELECTROMAGNETISM.

n Q = q = i $ i=1 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 integrated over the volume V enclosed by S CHAPMAN Q = ρdV (1.13) "V C then we have Gauss’ law in integral form 1 E dS = ρdV (1.14) · $ !S 0 "V The diﬀerential form follows immediately fromSANDRA the Divergence theorem 1.2 1 EdV = ρdV (1.15) ∇· $ "V 0 "V which we can write as:

MAXWELL I : E = ρ (1.16) ∇· #0 In going from 1.14 to 1.16 we are implicitly treating the fields as classical, and our field theory willCOPYRIGHT not hold on the quantum scale. It is in this sense that 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 define the magnetic flux through any surface S as

B dS =Φ (1.17) · B "S Again, experimentally, it can be shown that the magnetic ﬂux through any closed surface is zero so that

B dS = 0 (1.18) · !S ELECTRODYNAMICSwhich immediately from Divergence theorem 1.2 gives

MAXWELL II B = 0 (1.19) ∇· CORE 1.2. MAXWELL I AND II 21

B B E

S CHAPMAN C

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

Field lines are drawn passing through a closed surface S in figureSANDRA 1.2.3. The electric field has nonzero divergence so that closed surfaces can be found that yield zero or nonzero nett flux from 1.14 depending upon whether charge is enclosed in the surface. The magnetic field has zero divergence so that 1.18 always yields zero nett flux as far as we are able to determine experimentally. Lines of E end on electric charge whereas lines of B are continuous. No particle has yet been identified that is a source of magnetic field, but as we shall see in Chapter 4 section 2, the Maxwell equations themselves and special relativity does not preclude the existence of such particles, that is, of magnetic monopoles.COPYRIGHT

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

W = F dl = q E dl (1.20) · · "C "C from the Lorentz force law (this is minus the work done by the field on the 1 charge). The work done per unit charge W/q (in units JC− ) is defined as 1 ELECTRODYNAMICSthe potential of the electric field (in V ) giving units of E as Vm− . Coulomb’s law then reveals an interesting property of the electrostatic field via 1.20. Knowing that the field is radial and is any function of r, for CORE 22 CHAPTER 1. A BRIEF TOUR OF ELECTROMAGNETISM.

the ﬁeld from a single point charge we have

r2 E dl = E(r)dr (1.21) · "C "r1 Now we can choose C(r, θ,φ ) (in spherical polar coordinates) to have any θ,φ dependence, 1.21 will yield the same result which will depend only on the endpoints r1 and r2. It then follows that 1.21 is zero if r1 = r2, that is, if the path C is closed. The electrostatic field is conservative. ThisCHAPMAN will also hold for the field from any collection of charges, since by the principle of superposition we can write 1.20 as a sum of integrals due to theC electric field from each point charge. For the electrostatic field, taking 1.21 around a closed path and using Stokes’ theorem 1.1 then immediately gives

E = 0 (1.22) ∇∧ then since for any scalar ﬁeld φ SANDRA

( φ)=0 ∇∧ ∇ we can write the conservative electrostatic ﬁeld in terms of a potential:

E = φ (1.23) −∇

For magnetic fields, Maxwell II implied that magnetic field lines form closed loops. From figureCOPYRIGHT 1.2.3 one might expect the magnetic field to have curl whereas the electrostatic field does not; this is what is found experimentally and Ampère’s law of magnetostatics is:

B dl = µ I (1.24) · 0 !C where the current I is in Amperes (A C/s) from which we can now define the units of B as Tesla (T ) and≡ the permeability of free space 7 1 µ0 =4π.10− T msC− . Again, if the collection of (moving) charges can be treated as a smoothly varying function, the current that they carry I flowing across surface S can be expressed in terms of a vector field, the current density J: I = J dS (1.25) · "S ELECTRODYNAMICSThis, along with 1.1 immediately gives, for magnetostatics:

B = µ J (1.26) ∇∧ 0 CORE 1.3. MAXWELL III AND IV 23

E #

+q C CHAPMAN C

Figure 1.12: E dl is integrated around the closed curve C in the vicinity of a point charge. The· radial sections of the path give contributions of equal size and opposite sign and the sections along contours of the potentialSANDRA (dotted lines) give zero contribution. The integral around the closed curve is zero.

Then B is nonconservative unless there are no currents. The special case of current free systems (or regions) can be treated by defining a magnetostatic scalar potential in analogy to 1.23. Generally we define a magnetic vector potential B = A (1.27) ∇∧COPYRIGHT Since for any vector field ( A) = 0 this satisfies B = 0. The potentials φ and A are an equivalent∇· ∇∧ representation for the electrostatic∇· and magnetostatic fields. In section 1.3 we will complete this representation by explicitly considering the general case where the fields vary with time.

1.3 Maxwell III and IV

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

ELECTRODYNAMICS1.3.1 Faraday’s law and Galilean invariance. It is found experimentally that if the magnetic ﬂux passing through a loop of wire changes for any reason, a voltage is induced across the wire. This CORE 24 CHAPTER 1. A BRIEF TOUR OF ELECTROMAGNETISM.

B v^ B v

CHAPMAN C B

SANDRA v

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

is Faraday’s law:

COPYRIGHTdΦ d E dl = B = B dS (1.28) · − dt −dt · !C "S where the magnetic flux is through surface S spanned by the wire loop forming curve C. For this to be true, it has to hold in all frames of reference. To check Faraday’s law, we will consider a wire moving with respect to the magnetic field, so that the magnetic flux changes, and we will look at what happens to the charges in the wire in two frames 3. The two frames are shown in 1.3.1.

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

2. We transform frames to move with the wire, so that the magnetic ﬁeld depends on time. ELECTRODYNAMICS 3Note that this is a non relativistic, or Galilean, frame transformation. The relativistic treatment is in Chapter 3 CORE 1.3. MAXWELL III AND IV 25

B c dl b

d a CHAPMAN C 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 the wire so that Fe = ev B (when a steady state is reached the ends of the wire become charged,− ∧ and a back e.m.f is induced, that is, an electric field that acts opposite to Fe). When we transform to frameSANDRA (2) we (the observer) see a stationary wire. The electrons still respond to the Lorentz force however, so we now conclude that Fe = eE. So for the Lorentz force law to work in both frames, v B in one frame− is just equivalent to E in another. E and B are frame dependent∧ and form a single quantity, the electromagnetic field; a form of the Maxwell Equations must therefore also exist that is frame invariant, and we will derive it later. This electric field that is implied by the frame transformation modifies the conservative, curl free electrostatic field. We will now calculate its curl for the moving wire. COPYRIGHT To stop the charges ”piling up” at the ends of the wire we will complete the circuit by running the wire on conducting rails as sketched in figure 1.3.1; the rails and the rest of the circuit are at rest w.r.t. the observer. The wire and rails then form a closed loop shown in figure 1.3.1. The work done on the electrons around the loop is:

b c d a F dl = F dl + F dl + F dl + F dl (1.29) · · · · · ! "a "b "c "b Now in this frame only the wire is moving, so that all the terms in the path integral 1.29 are zero except between a and b (note that in the presence of an additional electrostatic ﬁeld, the contribution to the integral around ELECTRODYNAMICSthe closed loop would still be zero). This leaves b b F dl = F dl = e (v B) dl (1.30) · · − ∧ · ! "a "a CORE 26 CHAPTER 1. A BRIEF TOUR OF ELECTROMAGNETISM.

v dt CHAPMAN Figure 1.15: Wire element dl sweeps out surface element dS inC time dt. in the observer’s rest frame. To evaluate 1.30 consider the small element of wire dl shown in 1.3.1. The area element swept out by wire element dl in time dt is dS = vdt dl. Rearranging the r.h.s. of 1.30 gives ∧ b b dS d b dΦ F dl = e B (v dl)=e B = e B dS = e B (1.31) · · ∧ · dt SANDRAdt · dt ! "a "a "a since in the observer rest frame the magnetic field is time independent, so that work is done by the rate of change of magnetic flux. Now we can make the following assertion: the Lorentz force law yields the same force on the electrons in both frames. Hence the work done on the electrons in the moving wire in frame 1 (equation 1.31) must be equivalent to that done by an electric field in frame 2 where the wire is at rest: d COPYRIGHTe E dl e B dS (1.32) − · ≡ dt · ! "S 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 different frames. To obtain a relationship between E and B we need to work in a single frame: so let us consider a loop of wire a rest w.r.t the observer so that dS is fixed, and B changes with time. In this frame dS is fixed and v = dr/dt = 0 so from chain rule:

dB ∂B ∂B = + v (1.33) dt ∂t ∂r · then 1.32 becomes ∂B E dl = dS (1.34) · − ∂t · ! "S ELECTRODYNAMICSUsing Stokes’ theorem 1.1 this is just Maxwell III in diﬀerential form:

MAXWELL III E = ∂B (1.35) ∇∧ − ∂t CORE 1.3. MAXWELL III AND IV 27

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

E = E + v B (1.36) 2 1 ∧ 1 where the subscripts refer to frames 1 and 2 and v is the transformation velocity. Using the principle of superposition we have added the (arbitrary) CHAPMAN electric field in frame 1, E1, which to simplify the above discussion we assumed to be zero. The nonrelativistic 1.36 was needed to make the Lorentz C force law Galilean frame invariant; in section 4.5 equation 1.36 will be generalized for Lorentz frame invariance. Since E is no longer zero we cannot describe this field as the gradient of a scalar∇∧ potential. To retain B = A we can use ∇∧ ∂A E = φ SANDRA(1.37) −∇ − ∂t which is consistent with 1.35 and B = A. ∇∧ 1.3.2 Ampère’slaw and conservation of charge. Recall from section 1.2.4 that when the fields and currents are steady we have Ampère’s law 1.26, B = µ J (1.38) ∇∧ COPYRIGHT0 which will now be amended to apply to time dependent situations. To work out what happens to the fields when charge and current densities change with time we will make one important assumption, that nett charge is conserved that is, single charges are neither created or destroyed. Experimentally this has been verified to high precision, but we will assume that it is exactly true and later, that it is true in all frames of reference. Charge and current densities can then be related if we consider an arbitrary finite volume containing some total charge that is varying with time Q(t), as charges flow out of the volume as in figure 1.3.2. There is a total current flowing out of volume V : ∂Q ∂ρ I = = (1.39) − ∂t − ∂t "V ELECTRODYNAMICSand as I flows across the surface S which encloses V we can also write from 1.25: ∂ρ I = J dS = (1.40) · − ∂t !S "V CORE 28 CHAPTER 1. A BRIEF TOUR OF ELECTROMAGNETISM.

q dS

V q

q Q(t)

q q S CHAPMAN C Figure 1.16: Charge ﬂows out of volume V .

This can be written in differential form using 1.2: ∂ρ J = (1.41) ∇· − ∂t SANDRA Equation 1.41 immediately shows the problem with 1.26 in time dependent situations; the divergence of 1.26 gives J = 0, that is, currents must close in steady state. When we obtained∇ 1.41· we allowed nett current to flow out of a closed surface, so that 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 ∂ρ ∂E J + = J + $ =0 ∇· ∂t ∇· 0 ∂t $ % equating this with COPYRIGHT( B) = 0 then gives: ∇· ∇∧ ∂E MAXWELL IV B = µ J + µ $ (1.42) ∇∧ 0 0 0 ∂t which is Ampère corrected for time dependent fields. We have added a displacement current to the r.h.s. of 1.26. How does this work in practice? A simple example of the fields around an ideal capacitor that is discharging are sketched in figure 1.3.2, where B is given by the conduction current flowing in the circuit outside of∇∧ the capacitor plates, and by the displacement current due to the time dependent electric field between the plates.

1.4 Electromagnetic Waves. ELECTRODYNAMICSIt is now straightforward to show that Maxwell’s equations support free space waves. We use the vector relation: ( A)= ( A) 2A (1.43) ∇∧ ∇∧ ∇ ∇· −∇ CORE 1.4. ELECTROMAGNETIC WAVES. 29

-Q(t) +Q(t) - E(t) + - + - - + I(t) + I(t) - + - + CHAPMAN - + C B(t) - + B(t) B(t)

Figure 1.17: An ideal discharging capacitor.

and take the curl of Maxwell IV 1.42 to give SANDRA ∂ ( B) 2B = µ J + µ $ E (1.44) ∇ ∇· −∇ 0∇∧ 0 0 ∂t∇∧ which, with Maxwell II 1.19 and III 1.35 and in free space where there are no currents J = 0 gives 1 ∂2B 2B = (1.45) ∇ c2 ∂t2 1 where µ0$0 = c2 . If instead we take theCOPYRIGHT curl of Maxwell III 1.35 a similar procedure gives (with no charges in free space ρ = 0)

1 ∂2E 2E = (1.46) ∇ c2 ∂t2 Equations 1.45 and 1.46 are wave equations for E and B, and are linear. This means that any wave with frequency ω and wavenumber k of the form:

B, E f(ωt k r) (1.47) ∼ − · is a solution to 1.45 and 1.46. These waves will propagate at phase speed ω/k = c. Crucially, we identify these with light waves in free space. Since 1.45 and 1.46 are linear, we can superpose any solutions of the form 1.47 into a wave group or packet, and this will be nondispersive. The ELECTRODYNAMICSwavegroup will have speed c and will carry energy and momentum through free space4.

4The properties of these waves are explored in the revision problems CORE 30 CHAPTER 1. A BRIEF TOUR OF ELECTROMAGNETISM.

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 as the momentum flux density tensor) rather than the motions of individual particles. Electromagnetic fields can be quantized, that is, treated as a collection of photons which carry energy and momentum. We would then expect the free space electromagnetic fields to have equivalent energy and momentum flux (the latter given by the Maxwell stress tensor). CHAPMAN 1.5 Conservation Equations. C Equation 1.41 is a conservation equation and perhaps not surprisingly, all conservation equations are of this form; they embody the premise that the particles and the quantity that they carry (in this case, charge) is neither created nor destroyed. If we recall the cold gas from section 1.2.1 of number density n(r,t) with each particle moving with the same velocity v(r,t) carrying charge q, then ρ = nq and J =SANDRAnqv and 1.41 will become (cancelling q from both sides):

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

CORE Chapter 2 CHAPMAN C Field Energy and Momentum. SANDRA

So far we have discussed the free space macroscopic electromagnetic fields and the field equations: the Maxwell equations that describe how the fields evolve in space and time. Charges are included in this description as macroscopic charge density and current density. However point charges carry energy and momentum, the Maxwell equations have wave solutions and we might expect this to imply that the fields carry energy and momentum also. Indeed, the energy and the radiation pressure of waves, and of photons has been measured experimentally. COPYRIGHT In this chapter we will formalize the concept of the energy and momentum of the electromagnetic fields. This is most easily achieved through conservation equations, and we will use the cold gas model from Chapter 1 in this context. In a gas, individual particles carry momentum which is a vector quantity, the gas as a whole, when described by macroscopic or fluid variables has a corresponding tensor pressure. We will first use the simpler cold gas model to introduce Cartesian tensors, and their role in equations for conservation of momentum flux. We will then find the Maxwell Stress Tensor that is, the ”ram pressure tensor” for the electromagnetic field. The cold gas model will allow us to treat a system with free charges and electromagnetic fields (ie E and B). This is readily generalized to linear media1.

ELECTRODYNAMICS1The Maxwell equations and conservation equations for free space and charges discussed here are linear. Our approach then generalizes to media in which these equations remain linear. This is the case if the ﬁelds induced in the medium are linearly proportional to those in the surrounding free space. See the revision problems for examples.

CORE 31 32 CHAPTER 2. FIELD ENERGY AND MOMENTUM.

Pyz Pxz CHAPMAN Pyy C P xx P y Pxy yx x SANDRA Figure 2.1: Forces acting on ﬂuid element dV form the components of a pressure tensor

2.1 Tensors and Conservation Equations.

2.1.1 Momentum flux density tensor. Tensors arise in macroscopicCOPYRIGHT or bulk descriptions such as gases, fluids, and solids where we have vector fields describing distributed bulk properties rather than properties at a vanishingly small point (such as at a particle). A cube shaped elemental volume dV in a gas is sketched in figure 2.1.1 and the 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 forces normal to the surface, in the ˆx direction (Pxx), we can also twist the cube by exerting shear forces tangential± to the surface, in the ˆy direction ± (Pxy), and in the ˆz direction (Pxz). The same is true of the y and z faces, so that in our three± dimensional gas we have nine numbers describing the forces on the gas. These nine elements constitute the pressure tensor for the gas and we can write them as a matrix:

Pxx Pxy Pxz P = P P P (2.1) ij  yx yy yz  ELECTRODYNAMICS Pzx Pzy Pzz   If there were no shear forces the tensor has three independent elements CORE 2.1. TENSORS AND CONSERVATION EQUATIONS. 33

and becomes: Pxx 00 P = 0 P 0 (2.2) ij  yy  00Pzz   and if these normal forces on all sides of the cube are equal, the pressure is isotropic: P 00 Pij = 0 P 0 (2.3) CHAPMAN  00P    C We will now obtain the momentum ﬂux tensor for the cold gas in which all particles move with the same velocity v(r) at a given position r. In a given direction ˆr the momentum of one particle is

p = mv = mv ˆr (2.4) r r · and so momentum flowing across surface element dS in time dtSANDRAdue to this component is ρ(v ˆr)(v dSdt)=ρv v dsdt (2.5) · · r n where the term in the first bracket of the l.h.s. of 2.5 is the momentum density due to the ˆr component, and the term in the second bracket is the volume containing the particles that flow across dS in time dt. The subscripts r and n denote components along ˆr, and the normal to dS, ˆn respectively. The momentum flow per unit areaCOPYRIGHT per unit time from equation 2.5 is the scalar Prn = ρvrvn (2.6) We could write the momentum flux flowing in ˆn due to all components of 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 = ρ(vxˆx + vy ˆy + vzˆz)vn = ρvvn (2.7) If we then sum 2.7 over n =1, 3 (again, the x, y, z directions), we have the momentum flux flowing in all directions due to all components as the tensor: = ρ(v ˆx + v ˆy + v ˆz)(v ˆx + v ˆy + v ˆz)=ρvv (2.8) P x y z x y z Tensors written as vector outer products as in equation 2.8 are known as ELECTRODYNAMICSdyadics. 2Throughout this book we will use the index notation i, j to mean all values between i and j inclusive. CORE 34 CHAPTER 2. FIELD ENERGY AND MOMENTUM.

The dyadic vv can be written out in full: ˆx vxvxxˆˆx +vxvyxˆˆy +vxvzxˆˆz vxvx vxvy vxvz vv = +v v yˆˆx +v v yˆˆy +v v yˆˆz =(xˆˆyˆz) v v v v v v  ˆy  y x y y y z  y x y y y z  +vzvxˆzˆx +vzvyˆzˆy +vzvzˆzˆz vzvx vzvy vzvz        ˆz    (2.9)  For any orthogonal coordinate system we can use the following shorthand:CHAPMAN vxvx vxvy vxvz C vv = v v v v v v (2.10)  y x y y y z  vzvx vzvy vzvz For most purposes we don’t really need to write out all the elements of the matrix 2.10. Instead, we can simply write vivj where it is assumed in this three dimensional, Cartesian world i =1, 3 and j =1, 3. The momentum ﬂux density tensor is then written as: SANDRA

Pij = ρvivj (2.11) The notation in 2.11 is useful, for example we can immediately deduce from 2.11 that vv and hence is symmetric as vivj = vjvi (there are only 6 independent components forP the cold gas) without writing out all nine elements of the matrix 2.10. We can now calculate the force on the cube due to the momentum flux of the cold gas. The gas particles can deliver momentum to the cube either by flowing out of theCOPYRIGHT cube (the rocket effect), or by slowing down in the cube. The force on the cube of arbitrary volume V due to flow out of particles is minus the rate at which momentum flows out. The force due to particles slowing down in the cube is again minus the momentum inflow rate as dS points outwards on the surface. If we consider one component of v in the ˆr direction, the rate at which momentum is delivered through dS is given by dF = ρ(v ˆr)(v ds) (2.12) r − · · where the first bracket is the momentum density of the ˆr component. So we have from all three components: dF = ρvv ds = ds (2.13) − · −P· Thus the force on the whole volume is

ELECTRODYNAMICSFv = dF = ds = ρvv ds = dv = ρvvdv − S P· − S · − V ∇·P − V ∇· " " " " " (2.14)

CORE 2.1. TENSORS AND CONSERVATION EQUATIONS. 35

2.1.2 Momentum flux, gas pressure and fluid equations. The above then implies that the momentum flux density corresponds to a force per unit volume: F 1 v = dF = (2.15) V V −∇ ·P "V We can now obtain a conservation equation for momentum in the gas. CHAPMAN To do this we need to relate the l.h.s. of 2.15 to the rate of change of C momentum in the gas. For this we will make an important restriction: we will consider the force over a small volume dV in which the total number of particles is constant. This means we are considering the rate of change of momentum due to a local ensemble of particles slowing down, or speeding up, and we will choose dV accordingly; if this dV contains a constant mass M of particles then it experiences a force SANDRA dMv dv F = = M (2.16) V dt dt Then the force per unit volume F dv V = ρ (2.17) dV m dt One possibility is to choose dV to contain the same particles by moving with the fluid at velocity v, then 2.17 becomesCOPYRIGHT (using chain rule)

dv ∂v ρ = ρ +(v. )v (2.18) m dt m ∂t ∇ 0 1 This is equivalent to Liouville’s theorem3 The momentum flux density tensor that we have derived is not the same as the pressure tensor th. The pressure of a warm gas is the momentum flux due to the thermalP or random motions of the gas particles in the rest frame of the gas. In the warm gas, a particle will have total velocity u = v+c where c is the random velocity relative to the average or bulk velocity v(r,t) with which the gas as a whole moves, so that the average < u >= v and < c >= 0. 3Liouville’s theorem expresses conservation of probability density along a trajectory ELECTRODYNAMICSin phase space. In a system with no sources or sinks of particles, we can follow the phase space trajectory r(t), v(t) of any particle and along that trajectory the probability of finding the particle in elemental phase space volume drdv is constant. See advanced problem 2. CORE 36 CHAPTER 2. FIELD ENERGY AND MOMENTUM.

In our cold gas, the velocity due to random motions is zero so that for each particle u = v. From equation 2.8 above, we can find the nett momentum flux density in the warm gas due to an average over an ensemble of particles in small volume element dV at position r at time t. = ρ< uu >= ρvv + ρ< cc > (2.19) P In the rest frame of the gas v = 0 and the warm gas has thermal pressure = ρ< cc >. Generally then, Pth CHAPMAN = ρvv + (2.20) P Pth C The momentum flux density due to gas-fluid motion as a whole, ρvv is the ram pressure of the gas, and is just the momentum flux density of the cold gas. If we move with the gas (so that in our frame v = 0), equation 2.18 can be written as a fluid equation for the gas: dv ∂v ρ = ρ +(v )v SANDRA= (2.21) m dt m ∂t ·∇ −∇ ·Pth 0 1 and in this frame there is no ram pressure. If we are at rest w.r.t. the gas bulk flow then we experience both ram and thermal pressure.

2.1.3 Cartesian tensors, some deﬁnitions.

We can write or Pij for a tensor of the second rank where in the discussion so far i =1,P3 and j =1, 3 i.e. there are nine possible indices (or nine components). A tensorCOPYRIGHT of the ﬁrst rank (vector) would be written a or ai where i =1, 3, just three possibilities or three components. A scalar is thus a tensor of rank zero and has one component. We can extend this to as many indices as we wish, for example Pijkl is a tensor of rank four. A rank two dyadic would be ab which can also be written aibj. Note that all the indices are ”down” (written as subscripts) in this Cartesian system so that a dyadic is always of the form aibjck..., later, when we introduce generalized coordinates as opposed to Cartesian coordinates we j will come across a aj which has a diﬀerent meaning - we will cover this in Chapter 3 section 4. A key property of the tensor formalism is that it embodies coordinate transformations. Figure 2.1.3 shows a vector r w.r.t two orthogonal coordinate systems: x, y, z and x!,y!,z!. The primed axes are simply the unprimed axes rotated in space (x, y, z x!,y!,z!). A vector in the unprimed frame: → ELECTRODYNAMICS x r = y (2.22)  z    CORE 2.1. TENSORS AND CONSERVATION EQUATIONS. 37

z z’

r y’ CHAPMAN C

y x x’ SANDRA 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 +a13z ! y = a21x +a22y +a23z (2.23) ! z = a31x +a32y +a33z

The tensor a contains the directionCOPYRIGHT cosines of the rotation x, y, z ij → x!,y!,z!. This can be written as

r! = r (2.24) A· where ! x a11 a12 a13 r! = y! and = a a a (2.25)   A  21 22 23  z! a31 a32 a33     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

! rj = ajiri (2.26) ELECTRODYNAMICSwhich is shorthand (the Einstein summation convention) for ! rj = ajiri (2.27) i=1,3 # CORE 38 CHAPTER 2. FIELD ENERGY AND MOMENTUM.

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

a = a T + a T + a T + a T + a T + ... ·T 1 11 2 21 3 31 1 12 2 22 (2.28) = ˆxj aiTij CHAPMAN which is aiTij if we again drop all reference to the basis vectorsCˆxj . So we now can write down in Cartesian coordinates which is ∇ ·T ∂ ˆxj Tij = (2.29) ∂xi ∇ ·T

There are other special tensors. One is the trace which is δij = 1 when i = j and 0 when i = j: ( SANDRA 1 0 0 δij = 0 1 0 (2.30)  0 0 1    Contracting δij over i and j with two vectors extracts the dot product:

δ a b = a b + a b + a b = a b = a b (2.31) ij i j 1 1 2 2 3 3 · i i

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

$ijkajbk = $i11a1b1 +$i12b1a2 +$i13b1a3 +$i21a2b1 +$i22b2a2 +$i23b3a2 (2.32) +$i31a3b1 +$i32b2a3 +$i33b3a3

If we consider the i = 1 terms we ﬁnd that all but $123a2b3 and $132a3b2 are zero. These last two give a2b3 a3b2 which we recognise as the x component of the vector cross product.− Similarly, i =2, 3 gives the y, z components respectively, so that contracting with $ijk extracts the cross product. In this way, vector operations, vector calculus, and coordinate transformations can all be performed using index notation. This Cartesian formal- ELECTRODYNAMICSism works ﬁne for the three space dimensions, for six phase space dimensions, etc. Later, when we consider special relativity we will work in spacetime coordinates, and will need to generalize this Cartesian formalism. CORE 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 and momentum. Again, we will consider a volume element dV that contains a fixed number of charges, and field energy and momentum that can flow in and out of this volume element. In this sense, the charges act as a source or sink of electromagnetic energy and momentum. At a microscopic level, photons move in and out of dV and carry energy and momentum. Here in our field theory, we can only discuss the ”smoothed” behaviour in CHAPMAN element dV , and all quantities are obtained from fluid variables so that the C number of charges is n(r,t)dV , the energy in the electromagnetic fields is U(r,t)dV and in the particles is ε(r,t)dV where n(r,t) is number density and U(r,t) and ε(r,t) are the energy densities carried by the fields and particles respectively. Strictly speaking, we would derive the conservation equations in integral form, by considering integrals over some volume V , in a similar manner to the derivation of Maxwells’ equations in Chapter 1. Here we will take the short cut and work with the differentialSANDRA forms directly, by assuming that all bulk quantities are well defined per unit area or per unit volume. For field energy density for example this means that the energy density: 1 U(r,t)dV = U(r,t) V "V is well defined for arbitrary (small) V . COPYRIGHT 2.2.1 Energy conservation: Poynting’s theorem Energy is lost from the electromagnetic fields if work is done on the particles. From the Lorentz force law, a single positive charge moving in the fields gains energy dv d 1 v.m = mv2 = qv E (2.33) dt dt 2 · $ % then n charges per unit volume gain energy density ε at a rate dε = J E (2.34) dt · This energy density gain by the charges has to be balanced by an energy density decrease in the fields and an energy flux into the (unit) volume ELECTRODYNAMICScontaining the particles. Using 1.42 to substitute for J: B ∂ E2 J E = E ∇∧ $0 (2.35) · · µ0 − ∂t 2 CORE 40 CHAPTER 2. FIELD ENERGY AND MOMENTUM.

Now from the form of the conservation equation in Chapter 1 (equation 1.48) we expect the r.h.s. to be the divergence of energy ﬂux (a vector) minus rate of change of energy density. Using the vector identity:

(E B)=B ( E) E ( B) (2.36) ∇· ∧ · ∇∧ − · ∇∧ 2.35 can, using 1.35 be rearranged as

E B ∂ E2 1 B2 CHAPMAN J E = ∧ $ + (2.37) · −∇· µ − ∂t 0 2 µ 2 C $ 0 % 0 0 1 equation 2.37 is Poynting’s theorem for the case of an ensemble of free charges in an electromagnetic ﬁeld, where the ﬁeld energy density

E2 1 B2 U = $0 + (2.38) 2 µ0 2 SANDRA

3 in Jm− and the Poynting ﬂux

E B S = ∧ (2.39) µ0

2 1 which has units of energy per unit area per unit time (Jm− s− ). For some arbitrary volume V , enclosed by surface S the divergence theorem 1.2 gives: COPYRIGHT SdV = S dS (2.40) ∇· · "V !S so that the surface integral of the Poynting flux S gives the energy carried out of the volume by the electromagnetic fields per unit time (dS always points outwards); energy gain by the particles is balanced by minus the Poynting flux. Rearranging Poyntings theorem:

dε ∂U = + S = J E (2.41) − dt ∂t ∇· − ·

Poyntings theorem 2.41 expresses conservation of energy between the ensemble of charges and the electromagnetic ﬁelds4.

ELECTRODYNAMICS4In linear media, that is, where the response from the bound charges in the medium is linearly proportional to the applied ﬁeld, B = µrµ0H, D = !r!0E and S = E H and U = 1 [E D + B H] then conservation of energy is still given by 2.41. In nonlinear∧ 2 · · media 2.41 is no longer valid. See revision problem 8. CORE 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 exerted on the particles, again given by the Lorentz force law. Again we assume that the number of charges in volume element dV is fixed, and that we can always integrate over V to find the value of any quantity per unit volume. The Lorentz force law then implies a rate of change of momentum to the particles from the fields. For one particle with momentum pq dp CHAPMAN q = q(E + v B) (2.42) dt ∧ C

then for a collection of charges with momentum density Pq = npq dP q = ρ E + J B (2.43) dt q ∧ This expresses the rate of change of momentum density of the charges. From an understanding of electromagnetism based on experimentSANDRA we might also expect a conservation equation to contain a term describing the rate of change of momentum density of the fields; either because waves carry momentum and Maxwell’s equations support wave solutions, or because of the particle- like behaviour of the electromagnetic wavefields. On that basis let’s make a guess as to what this might be, from the following argument. If we consider the cold gas, all particles will have rest energy ε = mc2 and momentum p = mv. Then the energy flux Sp is just

2 Sp = nvε =COPYRIGHTnvmc (2.44)

(the flux across surface element dS will be Sp dS see section 1.2.1). The momentum density is just · P = nmv (2.45) so that 2 Sp = Pc (2.46) This is true for any particles, including photons. In the case of photons, the energy flux is just the Poynting flux 2.39 that we obtained from conservation of energy. Our ”guess” for the momentum density of the fields is then: S P = = $ E B (2.47) f c2 0 ∧ To obtain the conservation equation we then just rearrange 2.43 to look ELECTRODYNAMICSlike a conservation equation for momentum: dP ∂P p = f (2.48) − dt ∂t − ∇ ·T CORE 42 CHAPTER 2. FIELD ENERGY AND MOMENTUM.

This is in (almost) the same form as 2.41 except that the terms are now vectors; the rate of change of momentum density to the charges and fields are balanced by the divergence of a momentum flux density tensor . To find , Maxwell I and III ( 1.16, 1.35) are used to substituteT for ρ and J in 2.43T to give B ( B) ∂E ρE + J B = $0E( E) ∧ ∇∧ + $0B (2.49) ∧ ∇· − µ0 ∧ ∂t then since by expanding the l.h.s. and using Maxwell III 1.35: CHAPMAN ∂ S ∂E C = $ B + E ( E) (2.50) ∂t c2 − 0 ∧ ∂t ∧ ∇∧ $ % 0 1 equation 2.49 is ∂ S B ( B) (ρE + J B)= 2 $0E( E)+$0E ( E)+ ∧ ∇∧ (2.51) − ∧ ∂t c − ∇· ∧ ∇∧SANDRAµ0 Now the remaining terms on the r.h.s. of 2.51 must give the divergence of a tensor. Looking at the terms in E only, we can write E2 E( E) E ( E)=E( E) + (E )E (2.52) ∇· − ∧ ∇∧ ∇· ·∇ −∇ 2 by using vector identity E.11. In index notation the r.h.s. of 2.52 is ∂ ∂ ∂ E2 ∂ ∂ E2 Ej Ei + Ei Ej δij = EiEj δij (2.53) ∂xi ∂xCOPYRIGHTi − ∂xi 2 ∂xi − ∂xi 2 which is just the divergence of a tensor: E2 T E = E E δ (2.54) ij i j − ij 2 we can use the same manipulation for the B terms in 2.51, by simply adding the (zero) term (from Maxwell II 1.19) B( B) ∇· (2.55) − µ0 The Maxwell stress tensor in the conservation of momentum 2.48 is then

1 1 2 1 2 Tij = $0EiEj + BiBj δij ($0E + B ) (2.56) µ0 − 2 µ0 Notice that we wrote 2.48 with a term where Poynting’s theorem ELECTRODYNAMICShas a term + S. This is a matter of convention:−∇ ·T a Poynting ﬂux into a volume ( S∇negative)· will correspond to a radiation pressure acting upon it ( ∇positive).· ∇ ·T CORE 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 solution to the free space wave equations introduced in section 1.4. We can show that the operators ik ∇ ≡ − (2.57) ∂ iω ∂t ≡ CHAPMAN apply in the case of the plane wave solution C i(ωt k r) E = E0 e − · i(ωt k r) (2.58) B = B0 e − · The operators 2.57 can now be used to save some algebra. The relationship between E, B and k follows from the Maxwell equations directly. Maxwell III 1.35 for the plane wave becomes: SANDRA ik E = iωB (2.59) − ∧ − which rearranges to: 1 B = kˆ E (2.60) c ∧ Similarly, Maxwell I and II ( 1.16 1.19) give k E = 0 (2.61) · k B = 0 (2.62) · COPYRIGHT Maxwells equations then reveal that the free space plane wave E, B, k form an orthogonal set of vectors. Again for simplicity, lets choose the direction of k to be along the ˆz axis. The wave propagates in the z direction i.e. k = kˆz and E =(E ,E , 0) x y (2.63) B =(Bx,By, 0) We will now write down the Maxwell stress tensor for free space plane waves. Generally E2 1 B2 Tij = $0(EiEj δij )+ (BiBj δij ) (2.64) − 2 µ0 − 2 and for the ˆz propagating plane wave 2.63 any term with a z index including Txz,Tzx,Tzy is zero. This gives

B2 $ E2 + x P $ E E + BxBy 0 0 x µ0 EM 0 x y µ0 ELECTRODYNAMICS− 2 T = By Bx 2 By (2.65) ij  $0EyEx + $0E + PEM 0  µ0 y µ0 − 00 P  EM   −  CORE 44 CHAPTER 2. FIELD ENERGY AND MOMENTUM.

where E2 B2 PEM =+$0 + (2.66) 2 2µ0 1 ω from 2.60 and with c = = we have Bx = Ey µ0$0 and By = √µ0#0 k √ Ex√µ0$0 so that ByBx = µ0$0EyEx, and the terms Txy and Tyz are zero. These relationships also simplify− 2.66 to

B2 PEM = CHAPMAN(2.67) µ0 C As a result the Maxwell stress tensor for the free space plane wave becomes diagonal with terms

2 2 Bx Txx = $0E + PEM x µ0 2 − 2 By (2.68) Tyy = $0E + PEM y µ0 − T = P SANDRA zz − EM We can now calculate the force due to the radiation pressure, that is, the rate of change of momentum delivered by the free space wave per unit area, from the momentum flux conservation equation 2.48 this is just ∂ = ˆxj Tij (2.69) ∇ ·T ∂xi For the ˆz propagating wave,COPYRIGHT 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 off axis) terms are zero. We are left with

∂ ∂ ∂ B2 = ˆz T = ˆz [ P ]= ˆz (2.70) ∇ ·T ∂z zz ∂z − EM − ∂z µ $ 0 % for the plane wave solution, using 2.57 the radiation pressure is just: 1 = 2ikB2 (2.71) ∇ ·T µ0 Lets compare this with the rate of change of momentum ﬂux in the ﬁelds:

∂Pf ∂ E B = ∧ 2 (2.72) ∂t ∂t µ0c ELECTRODYNAMICSMaxwell IV in free space for the plane wave solution gives, using 2.57

k B = µ $ ωE (2.73) − ∧ 0 0 CORE 2.3. RADIATION PRESSURE. 45

hence ( kˆ B) B B2 Pf = − ∧ ∧ = kˆ (2.74) µ0c µ0c where we have used 2.62. We then have ∂P B2 B2 f =2iω kˆ =2ik (2.75) ∂t µ0c µ0 so that in the absence of free charges, the divergence in radiation pressure CHAPMAN balances the rate of increase in momentum density in the ﬁelds. C

SANDRA

ELECTRODYNAMICS

CORE 46 CHAPTER 2. FIELD ENERGY AND MOMENTUM.

CHAPMAN C

SANDRA

ELECTRODYNAMICS

CORE Chapter 3 CHAPMAN C A Frame Invariant Electromagnetism SANDRA

So far we have the field equations, the Maxwell equations, and using the Lorentz force law have constructed equations describing the conservation of energy and momentum between the fields and charges. Our next task is to cast these equations in a form that is consistent with special relativity. Special relativity has two fundamental predictions: Frame Invariance, that is, the laws of physics are the same in all • inertial frames of reference. COPYRIGHT The speed of light is the same in all inertial frames. • The second of these yields the Lorentz transformation between one frame moving at constant velocity w.r.t. another. Our new formalism for electromagnetism will tie up a number of loose ends that have already appeared so far. In 1.3.1 Galilean invariance was imposed on Faraday’s Law to give Maxwell III ( 1.35) and this showed that E and B are equivalent: B in one frame can look like E in another( 1.36). We will in Chapter 4 obtain a single object, the Electromagnetic Field Tensor that describes the fields E and B and which gives a Lorentz transformation of the fields valid for relativistic frame transformations, using the formalism developed in this Chapter. In 1.4 we found that the Maxwell equations predict light waves in free space. Since the speed of light is constant under Lorentz transformation we must be able to find a form for the ELECTRODYNAMICSwave equations, and the Maxwell equations, that is invariant under Lorentz transformation. So far we have used the Lorentz force law to give the force on individual charges but have not concerned ourselves with the equations

CORE 47 48 CHAPTER 3. A FRAME INVARIANT ELECTROMAGNETISM

specifying the particle motion under that force; we will now incorporate relativistic rather than Newtonian mechanics. Finally, we have found conservation equations for charge, mass, energy and momentum, between col- lections of charges and the electromagnetic fields; a Lorentz invariant form must also exist for these. Some individual quantities (such as energy) will be different between one frame and another, but other properties, such as charge and particle number, will be (inertial) frame invariant and this leads us to form conservation equations that are invariant under Lorentz transformation. CHAPMAN To construct an electromagnetic field theory that is invariantC under the Lorentz frame transformation will require a generalization of Cartesian tensors that were introduced in Chapter 2. We have seen that coordinate transformations in x, y, z space can be written as tensor operators (or 3 3 rotation matrices). The Lorentz transformation can be written as a tensor× operator in space-time (a 4 4 rotation matrix), and the key to unifying electromagnetism with special× relativity is toSANDRA construct four vectors and four 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 formulation of general relativity where space-time is curved by the gravitational field. Here for special relativity we neglect gravity (and therefore do not treat accelerating frames), so we consider inertial frame transformation only and space-time is flat. COPYRIGHT 3.1 The Lorentz Transformation

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

2 2 2 (c∆t!) =(u∆t) +∆y (3.1) ELECTRODYNAMICS 1This experiment was actually performed by Michelson and Morley in 1887 to attempt to measure our relative velocity w.r.t. the aether which was believed to ﬁll the vacuum to allow the propagation of electromagnetic waves. CORE 3.1. THE LORENTZ TRANSFORMATION 49

c$t’

u$t’ u CHAPMAN C

u SANDRA Figure 3.1: The light clock is oriented perpendicular 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.

In the clock’s rest frame the time taken for the light to travel from one mirror 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.COPYRIGHT If the distance perpendicular to the direction of motion∆ y is the same in both frames then

2 2 2 2 2 ∆t! (c u )=c ∆t (3.3) − so that there is time dilation

∆t! = γ∆t (3.4) where 1 γ = (3.5) √ 1 u2 − c2 in other words ”moving clocks go slow”.4 5 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 from ELECTRODYNAMICSthe forward going light pulse. When the light moves in the direction of the clock motion it travels a longer distance

c∆tf! =∆x! + u∆tf! (3.6) CORE 50 CHAPTER 3. A FRAME INVARIANT ELECTROMAGNETISM

$x’ u$tf’

u$t’ b $x’ u CHAPMAN C

$x u SANDRA 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 rest frame ie we move with the clock.

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

c∆t! =∆x! u∆t! (3.7) COPYRIGHTb − b the total round trip time of the light pulse in the moving clock is then

2c∆x! ∆t! =∆t! +∆t! = = γ∆t (3.8) f b c2 u2 − 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 length of the clock ∆x ∆x! = (3.10) ELECTRODYNAMICS γ in the moving frame. In terms of distances x, x! and times t, t! measured from origins in the CORE 3.1. THE LORENTZ TRANSFORMATION 51

two (observer) frames, the Lorentz transformation is:

vx t! = γ t 2 − c x! = γ(x vt) 4 − 5 (3.11) y! = y z! = z

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

x2 + y2 + z2 = c2t2 (3.12)

at time t. If we use the Lorentz transformation 3.11 on this expressionSANDRA 3.12, we obtain: 2 2 2 2 2 x! + y! + z! = c t! (3.13) Whatever frame we are in, the distance propagated by the wavefront always obeys an expression of the form of 3.12. This looks almost like the length squared of a vector in x, y, z, ct space. We can formalize this idea by writing 3.12 as s2 = c2t2 x2 y2 z2 (3.14) − COPYRIGHT− − 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, Lorentz transforming from one moving frame to another simply corresponds to a rotation of the axes x, y, z, ct. We have already developed notation for coordinate rotations in Cartesian x, y, z space. Lets see what happens if we stick to Cartesian rules. If our four vector is sj and

ct x s =(ct, x)= (3.15)  y   z      then the Lorentz transformation will be of the form:

ELECTRODYNAMICSsj! =Λ jksk (3.16) 2This is known as a spacelike interval, if s is imaginary then it describes a timelike interval, that is, an object travelling faster than c. CORE 52 CHAPTER 3. A FRAME INVARIANT ELECTROMAGNETISM

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

v γ c γ 00 v γγ− 00 Λ = c (3.17) jk  −0 0 1 0   0 0 0 1      then if we work out the coordinate rotation we have: CHAPMAN v xv γ c γ 00 ct cγ(t c2 ) C ct! v − − γγ00 x γ(x vt) x! s! =Λ s =  − c    =  −  =   · 0 0 1 0 y y y!  0 0 0 1   z   z   z!                 (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.SANDRA Lets calculate the length of our Cartesian four vector:

ct x 2 2 2 2 2 2 s s sjsj =[ct, x, y, z]   = x + y + z + c t = s (3.19) · ≡ y (  z      so we have a problem! The Cartesian system isn’t consistent when we try to extend our coordinateCOPYRIGHT system from three space dimesions to four spacetime dimensions. The solution is to generalize the coordinate system, and we will do this in section 3.4. First, we will look at a simple example to illustrate just why we need generalized coordinates and four vectors in a complete description of electromagnetism.

3.2 The Moving Charge and Wire Experi- ment.

The following ”thought experiment” is a neat way to understand the consequences of special relativity in electromagnetism. An infinitely long wire is oriented along the direction of motion of the observer (the ˆx direction); a section of the wire is sketched in figure 3.2. In the S1 frame the observer is at rest and the current in the wire is carried by the electrons moving to ELECTRODYNAMICSthe right with velocity uˆx. In the S2 frame the observer moves to the right with velocity uˆx, now the electrons in the wire are at rest and the current is carried by the protons moving to the left at uˆx. We can find out what − CORE 3.2. THE MOVING CHARGE AND WIRE EXPERIMENT. 53

q u S 1 + J v=0 1 v-=u z

y x CHAPMAN q

S2 C

J + 2 v=-u

v-=0 u SANDRA 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 the protons, the test charge q is at rest.

happens to the electromagnetic fields in both frames by releasing a test charge q located outside of the wire (q doesn’t contribute to the fields). The test charge starts at rest in the S2 frame, so that in the S1 frame it starts by moving to the right with velocityCOPYRIGHTuˆx. Now here is the dilemma: if in the S1 frame, we have zero charge density in the wire (ρ1 = 0) then the only force acting on the test charge is due to the magnetic field, ie F = qu B (3.20) 1 ∧ 1 and E1 = 0. This will cause the charge to move transverse to the wire in the y, z plane, that is, transverse to the ˆx direction of the frame transformation velocity. In the S2 frame, the test charge is initially at rest, so there can be no v B force. But any motion of the charge in y or z, that is, transverse to the direction∧ of the frame transformation velocity will appear to be exactly the same in both frames, from the Lorentz frame transformation 3.11. So what causes the force in the S2 frame? To resolve this, we need to calculate the electric and magnetic fields in the two frames. We only know the total charge density in S1. To calculate it in S2 we will assume that charge is invariant under Lorentz ELECTRODYNAMICStransformation, that is, that charges are not created or destroyed by moving from one inertial frame to the other. Then all we need to do to find the charge density is to consider a bunch of charge Q and find the volume that CORE 54 CHAPTER 3. A FRAME INVARIANT ELECTROMAGNETISM

it occupies in the two frames. Lets assume that the charge Q occupies a box with sides of lengths∆ x, ∆y, and∆ z in a frame where it is at rest w.r.t. the observer. The observer measures a charge density: Q ρ = (3.21) ∆x∆y∆z If the observer now moves in the ˆx direction w.r.t. the charge, distances will Lorentz contract in the x direction from 3.10 and the observerCHAPMAN will measure a charge density C Q Q ρ! = = γ = γρ (3.22) ∆x!∆y∆z ∆x∆y∆z If charges carry a current, then the corresponding charge density is always larger than if the charges are at rest3 from 3.22. In our moving wire experiment: + + SANDRA ρ2 = γρ1 (3.23) ρ1− ρ2− = γ then the total charge density in the two frames is:

+ ρ1 = ρ1 + ρ1− =0 + + 1 ρ2 = ρ + ρ− = ρ γ (3.24) 2 2 1 − γ + u2 = ρ1 γ c2 =0 6 7 COPYRIGHT( 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 the force on the test charge q:

F = qu B 1 ∧ 1 (3.25) F2 = qE2

Let’s calculate the forces in the two frames by using Maxwell’s equations to calculate the fields. In S1 the force is due to the magnetic field which can be obtained from the integral form of 1.42; we perform a line integral around the curve C enclosing the wire as shown in figure 3.2. Then since

B dl = µ J dS = µ I (3.26) · 0 · 0 ELECTRODYNAMICS !C " 3In everyday experience this eﬀect is small, the electrons in a copper wire of 1mm2 4 1 cross sectional area carrying 1 A have an average drift speed of 10− ms− which is why household wiring doesn’t charge up. ∼ CORE 3.2. THE MOVING CHARGE AND WIRE EXPERIMENT. 55

A CHAPMAN J Bv B C

r u

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

r S dl

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

CORE 56 CHAPTER 3. A FRAME INVARIANT ELECTROMAGNETISM

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

B dl =2πrB (3.27) · !C Since the current I is carried by the electrons moving at speed u through the wire which has cross section A, I = ρ−Au. The ”u B” force 3.25 in 1 ∧ CHAPMAN frame 1 then has magnitude C 2 qρ1−A u F1 = 2 (3.28) 2π$0r c 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 integral form of 1.16; we perform a surface integral over the closed cylindrical surface shown in figure 3.2: SANDRA ρ dQ E dS = EdV = 2 dV = (3.29) · ∇· $ $ !S "V "V 0 0 From 1.35 E is directed radially out from the wire so there is no contribution to the surface integral from the ends of the cylinder. From the geometry E = E(r) so that E dS =2πrEdl (3.30) · COPYRIGHT!S and excluding the test charge, the charge enclosed in the cylinder dQ = ρ2Adl. The force due to the electric field in frame S2 then has magnitude

+ 2 qρ2A qρ1 A u F2 = = 2 γ (3.31) 2π$0r 2π$0r c where we have used 3.24, this again acts in the +ˆr direction to repel the test charge. Comparing 3.28 and 3.31 we find that the forces acts in the same (ˆr) direction in both frames but the magnitude differs by a factor of γ; F2 = γF1 (3.32) Now recall that any motion transverse to the direction of motion u must be the same in both frames. What is happening here? Lets look at the equation of motion of the test charge. During the first (infinitesimally ELECTRODYNAMICSsmall) time interval∆ t after it is released from rest it will gain transverse momentum ∆p = F1∆t1 (3.33) CORE 3.3. MAXWELL IN TERMS OF POTENTIALS. 57

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

∆t1 = γ∆t2 (3.34)

then F1∆t1 = F1γ∆t2 = F2∆t2 (3.35) so that the change in transverse momentum∆ p = F ∆t in any frame, and CHAPMAN the transverse motion will be frame independent, as we would expect. Special relativity implies that length, time, force etc. change as we C Lorentz transform from one frame to another. This example has shown how the electromagnetic ﬁelds, current and charge densities are also not invariant under Lorentz transformation. We have anticipated that space and time must form a single coordinate system, space-time, and in this coordinate system, the (3) space coordinates and and (1) time coordinate of a point combines to form a single four-vector which hasSANDRA length that is invariant under Lorentz transformation. In the same way, current and charge density are in themselves not invariant, but can be combined to form a four vector that has invariant length. This leads to a frame invariant electromagnetism that we will introduce next.

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

B = A (3.36) ∇∧ ∂A E = φ (3.37) −∇ − ∂t which we showed always satisfy the homogenous Maxwell equations II ( 1.19) and III ( 1.35), that is, the Maxwell equations which do not refer to the currents or charges. It is easy to see that the potentials are not unique: since ψ is zero for any ψ we can add any ψ to A where ψ is a scalar ∇∧∇ ∇ ELECTRODYNAMICSfunction and still satisfy 3.36. This will then give new potentials: A! = A + ψ ∂ψ∇ (3.38) φ! = φ + ∂t CORE 58 CHAPTER 3. A FRAME INVARIANT ELECTROMAGNETISM

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 in terms of A and φ. Using 3.37 to substitute for E in Maxwell I gives:

∂ ρ 2φ A = (3.39) −∇ − ∂t∇· $0 CHAPMAN and using 3.37 and 3.36 to substitute for E and B in Maxwell IVC gives: 1 ∂ ∂2A 2A + ( A)+ φ + = µ J (3.40) −∇ ∇ ∇· c2 ∂t∇ ∂t2 0 0 1 We can now choose a gauge that reveals a nice symmetry in 3.39 and 3.40. This is the Lorentz gauge4: SANDRA 1 ∂φ A = (3.41) ∇· −c2 ∂t and substituting 3.41 into 3.39 and 3.40 gives:

2 1 ∂2φ ρ φ 2 2 = c ∂t #0 ∇ − 2 − (3.42) 2A 1 ∂ A = µ J ∇ − c2 ∂t2 − 0 in free space (ρ =0, JCOPYRIGHT= 0) equations 3.42 are simply wave equations for A and φ and predict electromagnetic waves with speed c. They contain the same information as the Maxwell equations, all we need is to put 3.42 and the Lorentz gauge 3.41 in Lorentz invariant form and we will have a description of electromagnetism that incorporates special relativity.

3.4 Generalized Coordinates.

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

4In electrostatics this reduces to the Coulomb gauge A = 0 which gives Poisson equations for both φ and A ∇· ρ 2φ = −∇ !0 ELECTRODYNAMICSand 2 A = µ0J −∇

CORE 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) · CHAPMAN 3. The Lorentz transformation corresponds to a rotation of the four C vector s in space-time.

The system that we devise will work for any four tensor (four vectors being the special case of four tensors of rank 1). We will start by deﬁning two ”versions” of our four vector5 and for the vector s these look like: SANDRA Covariant: ct x xα =(ct, x)= −  (3.47) − y −  z   −    and Contravariant: ct x xα =(ct, +x)= (3.48) COPYRIGHT y   z      Covariant rank 1 tensors (vectors) such as 3.47 always have a single ”down” index (subscript) and contravariant vectors such as 3.48 always have an ”up” index (superscript). Rank 2 or more tensors can be covariant (all indices down), contravariant (all indices up) or mixed (indices up and down). We are also going to use the summation convention introduced for Cartesian tensors 2.27. Now, if we simply deﬁne the scalar (dot, or inner) product of

5 α α 0 1 2 3 In general if we have a well deﬁned transformation that gives x" = x" (x ,x ,x ,x ) a covariant vector transforms as: ∂x0 ∂x1 ∂x2 ∂x3 ∂xβ a = a + a a a = a (3.45) α" α 0 α 1 α 2 α 3 α β ∂x" ∂x" ∂x" ∂x" ∂x" and a contravariant vector transforms as α ELECTRODYNAMICSα ∂ x" β a" = a (3.46) ∂xβ See the appendix on tensors for details. CORE 60 CHAPTER 3. A FRAME INVARIANT ELECTROMAGNETISM

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

ct 2 α x 2 2 2 2 s = xαx = x x =˜xx =[ct, x, y, z]   =(ct) x y z · − − − y − − −  z      (3.49) as required. Then we just need to know how to turn a covariant tensorCHAPMAN into a contravariant one, and vice versa, and from inspection of 3.47C and 3.48 we simply need an operation that changes all the signs on the spacelike components 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 or inner product or contraction over a pair of contravariant indices, this time between a rank 1 tensor (vector) and a rank 2 tensor:

β SANDRA xα = gαβx (3.50) and

α αβ x = g xβ (3.51) where 1000 0 1 0 0 g = = gαβ (3.52) αβ  00− 10 COPYRIGHT−  0001   −    gαβ is known as the metric of the spacetime. The spacetime interval is then written as 2 αβ α β s = xαg xβ = gαβx 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 interval s2, via 3.53. This particular metric then is just what is needed to embody the ﬂat spacetime of special relativity, in which the length of a spacetime interval (the length of s) has the same constant value anywhere in spacetime, i.e. spacetime itself is uniform. The formalism that we are de- ELECTRODYNAMICSveloping here can just as easily be applied where spacetime is nonuniform or curved by gravity. The metric for curved spacetime embodies general relativity. CORE 3.4. GENERALIZED COORDINATES. 61

The dot product operation that we deﬁned to obtain the desired behaviour for s2 has to be the same as the dot product between two four tensors 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...α (3.55) · ...α ... αβ that is, a contraction over the index α. Contraction with gαβ or g changes a index from contravariant to covariant (up to down) or vice versa (down to up). It works like this: CHAPMAN Ccontravariant to covariant (up to down): C ...... β a...α = gαβa... (3.56) and covariant to contravariant (down to up): ...α αβ ... a... = g a...β (3.57) Now we can write down the Lorentz transformation -rotationSANDRA matrix as a 4- tensor. A rotation of four vector s will be a operation of the form

! β xα =Λαβx (3.58) on the contravariant form and

α! αβ x =Λ xβ (3.59) on the covariant form. The contractions 3.58 and 3.59 change a contravariant four vector into a covariant one and vice versa. The rotation matrix6 is (almost) the same as that discussedCOPYRIGHT in Cartesian formalism and is: v γ c γ 00 + v γ −γ 00 Λ = c (3.60) αβ  00− 10  −  0 0 0 1   −  αβ   whereΛ is the transpose ofΛ αβ. The rotation of coordinates in spacetime written out in full is: ct! v vx γ c γ 00 ct cγ(t c2 ) ! v − − ! β x + c γ γ 00 x γ(x vt) x =Λαβx =  −  =  −    =  − −  α y! 00 10 y y − − −  z!   0 0 0 1   z   z     −     −   −       (3.61)  which just give the Lorentz transformation 3.11 as required (the inverse ELECTRODYNAMICStransformation 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. CORE 62 CHAPTER 3. A FRAME INVARIANT ELECTROMAGNETISM

3.5 Four Vectors and Four Vector Calculus.

The rules of ﬂat spacetime were deﬁned from the Lorentz transformation of the four vector xα but will apply to any four vector. If we can write the laws of mechanics and electromagnetism in four vector form, then we will have a description of mechanics and electromagnetism that is Lorentz invariant (i.e. manifestly covariant). The required four vectors will always have the same length in spacetime in all frames, and will Lorentz transform simply by coordinate rotation in spacetime. CHAPMAN C 3.5.1 Some mechanics, Newton’s laws. As we have shown, length and time form the four-vector:

ct x x = (3.62) α  −y  SANDRA −  z   −    Another example is the energy- momentum four vector:

ε c px ε pα =  −  =( , p) (3.63) py c − −  pz   −  COPYRIGHT  The length of pα should be constant. In full it is:

ε c 2 α ε px ε 2 pαp = , px, py, pz   = p (3.64) c − − − py c2 − 8 9  pz      Now recall some expressions from special relativity:

2 2 2 2 4 ε = p c + m0c (3.65)

The energy-momentum four vector then has length

α 2 2 pαp = m0c (3.66)

α ELECTRODYNAMICSThis just gives the rest energy of a particle (pαp /m0); if we can ﬁnd a frame where the particle is at rest (ie it is an electron, proton but not a photon) then since the length of pα is Lorentz invariant, its length in CORE 3.5. FOUR VECTORS AND FOUR VECTOR CALCULUS. 63

all frames must be just that found in the rest frame. If the particle is at rest the spacelike components of pα are zero and the timelike component is ε p0 = c = m0c. Now from special relativity the relativistic mass m = γm0 of a moving particle is larger than the rest mass m0, so we have in the moving frame, the timelike component given by:

0 2 2 p c = ε = mc = γm0c (3.67) CHAPMAN and the spacelike components p1,3 are given by: C

p = mu = γm0u (3.68)

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

α α p = m0u SANDRA(3.69) where we have introduced another four vector, the four velocity:

cγ pα γu uα = =  x  (3.70) m0 γuy  γuz      The four velocity uα must transform in the same way as the energy- mo- mentum four vector as we have simplyCOPYRIGHT divided by m0 which is invariant under Lorentz transformation. The expression 3.69 is manifestly covariant, that is, under Lorentz transformation it will retain the same form and the component four vectors will Lorentz transform, with invariant lengths. What about Newton’s laws? In a nonrelativistic world, frame transformations are Galilean and correspond to rotations of Cartesian position vector x in x, y, z plus translations x! = x + ut, where u is the constant velocity between the unprimed and the primed frames. The nonrelativistic (Newton) laws of motion are:

1. An object is at rest or moves in a straight line at constant velocity unless subject to some force.

2. Momentum is conserved (”for every action there is an equal and op- ELECTRODYNAMICSposite reaction”) under Galilean transformation. dp dx 3. The equations of motion F = dt and u = dt are invariant under Galilean transformation. CORE 64 CHAPTER 3. A FRAME INVARIANT ELECTROMAGNETISM

From (1) and (2) there will be a preferred rest frame in the sense that x =(x, y, z) is constant and the total p =(px,py,pz) = 0. The four vectors we have already written down for xα and pα will allow us to rewrite laws (1) and (2). We just need to look at diﬀerentiation w.r.t. time in (3) before we can rewrite all of Newton’s laws. Intuitively, we might expect that since time is dilated in the moving frame 3.4, an invariant interval to replace dt in the derivative would be the proper time dt/γ. We can formalize this by considering the coordinates of two particles that are moving apart at constant velocity in spacetime.CHAPMAN At time t, the particles are both at the same position (x, y, z)C and have spacetime coordinates (ct, x, y, z). As the particles move apart at constant velocity u =(ux,uy,uz), at time t+∆t, they will be separated by a distance:

∆x = ux∆t ∆y = uy∆t (3.71) ∆z = uz∆t SANDRA the spacetime interval between the two particles is then

∆s2 = c2∆t2 ∆x2 ∆y2 ∆z2 (3.72) =(c2 u2)∆t−2 − − − so that an invariant interval is ∆s ∆t = (3.73) COPYRIGHTc γ The time derivative in Newton’s laws is then to be replaced by the invariant spacetime derivative d d = γ (3.74) ds dt The spacetime derivative of (four) position in spacetime is then

ct cγ α dx dx d x γ dt α = γ   =  dy  = u (3.75) ds dt y γ dt  z   γ dz     dt      i.e. the four- velocity as we would expect. ELECTRODYNAMICSNewton’s laws of motion then become: 1. The spacetime interval xα =(ct, x, y, z) has constant length under Lorentz transformation. CORE 3.5. FOUR VECTORS AND FOUR VECTOR CALCULUS. 65

α ε 2. The energy- momentum four- vector p =(c ,px,py,pz) has constant length under Lorentz transformation.

α dpα α dxα 3. The equations of motion f = ds and u = ds are invariant under Lorentz transformation.

Then f α, the ”four - force”, must also be a four vector; in Chapter 4 we will find the four vector force on a charged particle in an electromagnetic field, that is, the covariant version of the Lorentz force law. CHAPMAN C 3.5.2 Some four vector calculus. Since we know how to define intervals in space time we can do some calculus in terms of four vectors. First let’s consider the four gradient of some scalar field φ(t, x, y, z)= φ(xα) defined in spacetime. The scalar field itself is an invariant,SANDRA the coordinates transform as the components of a four vector should. Since we have defined φ as a function of contravariant position in spacetime xα we can use chain rule to obtain the change in scalar field dφ on some spacetime interval ds given the contravariant coordinates of the interval dxα:

∂φ ∂φ ∂φ ∂φ dφ dφ = dt + dx + dy + dz = ds (3.76) ∂t ∂x ∂y ∂z ds The quantity dφ is also invariant, so theCOPYRIGHT r.h.s. of 3.76 must simply be in the form of the dot product of two four vectors:

α dφ = ∂α(φ)dx (3.77)

This means that the covariant four gradient is

1 ∂ c ∂t ∂ ∂x 1 ∂ ∂α =  ∂  =( , ) (3.78) ∂y c ∂t ∇  ∂   ∂z    which is almost what we would expect, except that the spacelike part is positive. We could have instead written the scalar ﬁeld as a function of covariant position xα, ie φ = φ(ct, x, y, z); the invariant interval would ELECTRODYNAMICSthen need to be written as a four− vector − dot− product

α dφ = ∂ (φ)dxα (3.79) CORE 66 CHAPTER 3. A FRAME INVARIANT ELECTROMAGNETISM

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

1 ∂ c ∂t ∂ α ∂x 1 ∂ ∂ =  − ∂  =( , ) (3.80) c ∂t −∇ − ∂y  ∂   ∂z   −  which again has spacelike components with opposite signs than we would expect. CHAPMAN As a consistency check, if covariant ∂α is a four vector thenC the contravariant ∂α should also be deﬁned via the spacetime metric:

1 ∂ c ∂t ∂ α αβ ∂x 1 ∂ ∂ = g ∂β =  − ∂  =( , ) (3.81) c ∂t −∇ − ∂y  ∂   ∂z  SANDRA  −  Having found the ”four gradient” operator of a scalar field we can define ”four divergence” of a four vector field (and more generally, a four tensor). We would expect a ”four - divergence” of a four vector to be of the form of a dot product: 1 ∂ ∂ aα = ∂αa = a0 + a (3.82) α α c ∂t ∇· (where a is the spacelike part of four vector field aα). Finally, the lengthCOPYRIGHT of the four vector ∂α is the ”four 2” D’Alembertian: ∇ 1 ∂2 ∂ ∂α = 2 = 2 (3.83) α c2 ∂t2 −∇ 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 invariant electromagnetism in terms of scalar and vector∇ potentials.

3.6 A Frame Invariant Electromagnetism

3.6.1 Charge conservation. Charge conservation was introduced in Chapter 1 to obtain 1.42 and so is ELECTRODYNAMICS”built in” to Maxwell’s equations. This is expressed by: ∂ρ J + = 0 (3.84) ∇· ∂t CORE 3.6. A FRAME INVARIANT ELECTROMAGNETISM 67

which was derived in 1.3.2 by assuming that charge is conserved. If we now 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 the form of the four divergence of some four vector:

α ∂αJ = 0 (3.85)

and by inspection, knowing ∂α we have the four current: cρ CHAPMAN J J α =(cp, J)= x  (3.86) C Jy  Jz      The length of J α is invariant under Lorentz transformation so that its components, the charge and current density are not, as we found when we considered the moving charge and wire experiment in section 3.2. SANDRA 3.6.2 A manifestly covariant electromagnetism We can ﬁnally write down the laws of electromagnetism in four vector form, if we do so in terms of the scalar and vector potentials for the electromagnetic ﬁelds. Recall from section 3.3 that the Maxwell equations can be written in terms of scalar potential φ and vector potential A, and take on (what we will see is a very useful) symmetry if we also choose the Lorentz gauge: 1 ∂ A + COPYRIGHTφ = 0 (3.87) ∇· c2 ∂t from Maxwell. This has the same form as the charge conservation equation so again we can write α ∂αA = 0 (3.88) where the new four vector is the four vector potential

φ c φ A Aα =( , A)= x  (3.89) c Ay  A   z    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 φ using the Lorentz gauge 3.42. The equation in terms of φ (originally ELECTRODYNAMICSMaxwell I 1.16) has been divided by c to give: 2 2 φ 1 ∂ φ ρ + 2 2 = = cρµ0 (3.90) −∇ c c ∂t c $0c CORE 68 CHAPTER 3. A FRAME INVARIANT ELECTROMAGNETISM

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

1 ∂ 2A + A = µ J (3.91) −∇ c2 ∂t 0 and the l.h.s. of this looks like the ”four 2” of the spacelike part of Aα. Together they are a single 4 vector equation:∇ φ CHAPMAN 2( , A)=µ (cρ, J) (3.92) c 0 C or α α 2A = µ0J (3.93) which is our equation for electromagnetism in manifestly covariant form. Since the invariance of charge implies that J α is a four vector from 3.85, 3.93 shows that Aα must also be a four vector.SANDRA So, to Lorentz transform the electromagnetic fields, all we need to do is write the fields in terms of A and φ, transform (ie rotate the four vectors in spacetime) Aα and J α, then work out the electromagnetic fields again from the new A and φ using 3.36 and 3.37. In the next Chapter we will find the electromagnetic field tensors that lead to a direct transformation of the fields. For the moment, we have already shown that Maxwell’s equations are consistent with special relativity. The Maxwell equations can be written in terms of four vectors. In free space J α = 0 and we are left with the free space wave equations for A and φ in manifestlyCOPYRIGHT covariant form: 2Aα = 0 (3.94)

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

ELECTRODYNAMICS

CORE Chapter 4 CHAPMAN C The Field Tensors

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

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

Let’s begin by trying to write down an invariant form for E and B. In Cartesian space we have already deﬁned B in terms of a vector ﬁeld: B = A (4.1) ∇∧ Now in spacetime, the vector potential A and the scalar potential φ has ELECTRODYNAMICSbeen replaced with the four vector φ Aα =( ,A ,A ,A ) = (A0,A1,A2,A3) (4.2) c x y z CORE 69 70 CHAPTER 4. THE FIELD TENSORS

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

1 ∂ ∂ ∂ ∂ ∂α = , , , =(∂0,∂1,∂2,∂3) (4.3) c ∂t −∂x −∂y −∂z $ % to construct a ”four- curl” of Aα. Writing out the components of A gives: ∇∧ B = ∂Az ∂Ay = (∂2A3 ∂3A2)= F 23 x ∂y − ∂z − − − CHAPMAN B = ∂Az ∂Ax = ∂1A3 ∂3A1 =+F 13 (4.4) y ∂x − ∂z − C B = ∂Ay ∂Ax = (∂1A2 ∂2A1)= F 12 z ∂x − ∂y − − − We can arrange the components of A to look like combinations of the components of the four vectors if we∇∧ use terms such as ∂2A3. From our notation these are not a contraction over an index as in the case of ”four 2 div” which had terms such as ∂2A , instead, they look like components of a four tensor of rank 2. All the three componentsSANDRA of 4.4 have the same pattern, that is, they look like they are part of a single rank 2 tensor F αβ. We might guess that more terms of F αβ originate in the electric ﬁeld, since it too is represented by components of Aα: ∂A E = φ (4.5) −∇ − ∂t Writing this in component form gives COPYRIGHT E = ∂φ ∂Ax = c(∂0A1 ∂1A0)= F 01c x − ∂x − ∂t − − − E = ∂φ ∂Ay = c(∂0A2 ∂2A0)= F 02c (4.6) y − ∂y − ∂t − − − E = ∂φ ∂Az = c(∂0A3 ∂3A0)= F 03c z − ∂z − ∂t − − − So, the three components of B in 4.4 and the three components of E in 4.6 give six terms of a rank 2 tensor

F αβ = ∂αAβ ∂βAα (4.7) − The other terms of the 4 4 ”four- tensor F αβ can be obtained from inspection of 4.7. The on axis× terms are when α = β:

F αα = ∂αAα ∂αAα = 0 (4.8) − that is, all four are zero (α =0, 1, 2, 3). The oﬀ axis terms are when α = β, ELECTRODYNAMICSand are for example when α = 0, β =1 (

F 01 = ∂0A1 ∂1A0 = (∂1A0 ∂0A1)= F 10 (4.9) − − − − CORE 4.1. INVARIANT FORM FOR E AND B: THE EM FIELD TENSOR.71

so that F αβ is antisymmeric: F αβ = F βα. The 4 4 = 16 terms in F αβ is comprised of the four on axis terms− which are zero,× plus the six different terms from the 3E ( 4.6) and 3B ( 4.4) components, each used twice: 0 Ex Ey Ez − c − c − c Ex 0 B B F αβ =  c − z y  (4.10) Ey B 0 B c z − x  Ez B B 0   c y x   −  CHAPMAN This is the electromagnetic field tensor F αβ which has ”packaged” E and B in in terms of four vectors from 4.7. This means that we can use F αβ C to describe laws of physics directly in terms of E and B in a form that is ”manifestly covariant”, that is, invariant in form under Lorentz transformation. We will in the next section write down the Maxwell equations in manifestly covariant form directly, instead of in terms of the scalar and vector potentials (the four vector Aα) as in the previous Chapter. What if we had used the covariant four vectors to form a ”fourSANDRA curl” of A? The starting point would have been: φ A =( , A , A , A ) = (A ,A ,A ,A ) (4.11) α c − x − y − z 0 1 2 3 and 1 ∂ ∂ ∂ ∂ ∂ = , , , =(∂ ,∂ ,∂ ,∂ ) (4.12) α c ∂t ∂x ∂y ∂z 0 1 2 3 $ % then A would look like: ∇∧ ∂A ∂A B = z y = ∂ A + ∂ A COPYRIGHT= (∂ A ∂ A )= F (4.13) x ∂y − ∂z − 2 3 3 2 − 2 3 − 3 2 − 23 23 13 so Bx = F23 = F , and similarly for By = F13 = F , Bz = F12 = F 12, so− the form− is not unlike the contravariant case 4.4. − − For the E field, we now have ∂φ ∂A E = x = c∂ A + c∂ A = c(∂ A ∂ A )=F c (4.14) x − ∂x − ∂t − 1 0 0 1 0 1 − 1 0 01 and E /c = F = F 01, that is, it has the opposite sign to the contravari- x 01 − ant case 4.6, and similarly for Ey and Ez. We can then write F = ∂ A ∂ A (4.15) αβ α β − β α with the covariant form of the electromagnetic field tensor:

Ex Ey Ez 0 c c c Ex ELECTRODYNAMICSc 0 Bz By Fαβ =  − −  (4.16) Ey B 0 B − c z − x  Ez B B 0   c y x   − −  CORE 72 CHAPTER 4. THE FIELD TENSORS

To get from the covariant to the contravariant forms of the electromagnetic ﬁeld tensor we just need to use the spacetime metric gαγ to change the signs on the timelike part of the tensor, that is, the F 0α and F α0 terms (in this case, the ”E” terms) γδ Fαβ = gαγF gδβ (4.17) which is equivalent to E E, B B. Finally, if F αβ is manifestly covariant we would expect→− it to have constant→ ”length”, this just turns out to be zero for electromagnetic waves in free space (see advanced problemCHAPMAN 6). C The usefulness of F αβ will now become clear as we use it to write down manifestly covariant forms of the Maxwell equations.

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

Ex Ey Ez E 0 c c c ∇c· Ex − − − 1 ∂Ex αβ 1 ∂ ∂ ∂ ∂ c 0 Bz By ( B)x c2 ∂t ∂αF = , , ,  Ey −  =  ∇∧ − 1 ∂Ey  c ∂t ∂x ∂y ∂z c Bz 0 Bx ( B)y c2 ∂t 0 1 Ez − ∇∧ − 1 ∂Ez  B B 0   ( B) 2   c y x   ∇∧ z − c ∂t  COPYRIGHT −   (4.18)  i.e a single four vector containing the (slightly rearranged) left hand side of the Maxwell inhomogeneous equations with timelike component giving

E ∇· = µ cρ (4.19) c 0 and with spacelike components giving:

1 ∂E B = µ J (4.20) ∇∧ − c2 ∂t 0

The r.h.s. of 4.19 and 4.20 taken together look like (µ0 times) the four- current J α =(cρ, J) (4.21)

ELECTRODYNAMICSso the inhomogeneous Maxwell equations are just the single equation

αβ β ∂αF = µ0J (4.22) CORE 4.2. MAXWELL’S EQUATIONS IN INVARIANT FORM. 73

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

B = 0 (4.23) ∇· and spacelike components

∂B E + = 0 (4.24) CHAPMAN ∇∧ ∂t C so we need some transformation that swaps the position of the E and B terms in F αβ so that the timelike components involve B and the spacelike components involve E. This is known as the duality transformation and is E Bc2 and B E; performing the transformation on F αβ gives the Dual→ ﬁeld tensor →− SANDRA 0 Bxc Byc Bzc B c − 0 − E − E ˜αβ =  x z − y  (4.25) F Byc Ez 0 Ex −  BzcEy Ex 0   −    The ”four div” of ˜αβ then yields the required homogenous Maxwell equations: F c B ∇· ∂Bx αβ ( E)x ∂t ∂α ˜ =  − ∇∧COPYRIGHT− ∂B  (4.26) F ( E) y − ∇∧ y − ∂t  ( E) ∂Bz   z ∂t   − ∇∧ −  so that the homogenous Maxwell equations are

∂ ˜αβ = 0 (4.27) αF But what is the signiﬁcance of ˜αβ? First, we need to relate it directly to F αβ to demonstrate that just likeF F αβ , ˜αβ is indeed a ”four tensor” and Lorentz transforms in the same way. ThisF is done via the ”four alternating tensor” (the 4 4 4 version of the Cartesian 3 3 alternating tensor that we ﬁrst met in× Section× 2.1.3): ×

c2 ˜αβ = $αβγδF (4.28) ELECTRODYNAMICSF 2 γδ Second, the dual tensor has an interesting physical signiﬁcance. If we apply the duality transformation to Maxwell’s equations we have: CORE 74 CHAPTER 4. THE FIELD TENSORS

Maxwell equations Dual equations ρ E = B = µ0ρ ∇· #0 ∇· B =0 E =0 (4.29) ∇· ∇· E = ∂B B = ∂E ∇∧ − ∂t ∇∧ ∂t B = µ J + 1 ∂E E = µ J ∂B ∇∧ 0 c2 ∂t ∇∧ − 0 − ∂t The Dual Maxwell equations 4.29 describe a world in which we would identify ρ as magnetic charge and J as magnetic current. In thisCHAPMAN dual world, E is nonconservative, that is, since E = 0 lines of E must close. In static situations B will be conservative as∇· B = 0 if E doesC not vary with time. ∇∧ Two properties of the Maxwell equations implied that they must embody Lorentz invariance if only we could write them in an invariant form. Both charge and the speed of light waves in free space c are invariant under Lorentz transformation. If we put ρ = 0 and J = 0 in 4.29 then the Maxwell and Dual equations are identical, ieSANDRA they will both predict light waves moving at invariant speed c in free space. Charge conservation is implicit in Maxwell IV, we can look at of conservation of magnetic charge using the dual equations. The Dual IV equation is ∂B E = µ J (4.30) ∇∧ − 0 − ∂t Taking the divergence of 4.30 gives ∂ ( E) = 0 = µ J ( B) (4.31) ∇· ∇∧COPYRIGHT− 0∇· − ∂t ∇· which from Dual I ( B = µ ρ) gives ∇· 0 ∂ρ J + = 0 (4.32) ∇· ∂t so (in this case magnetic) charge is again conserved, and we would from the Dual equations be able to write charge conservation in invariant form in the same way as in section 3.6.1. Writing the Maxwell equations in Lorentz invariant form then tells us that the existence of ”electric charge” or ”magnetic charge” is not excluded by special relativity. The only reason we wrote down the original Maxwell equations with electric charge and zero magnetic charge is that experimentally the magnetic charge is known to be zero to very high precision. The interesting question is whether all particles have the same ratio of electric ELECTRODYNAMICSto magnetic charge; if so, then we can always perform a duality transformation similar to that discussed here to make the Maxwell equations appear as if the magnetic charge is zero. CORE 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 conservation. We also found in section 3.5.1 that the energy and momentum of individual particles forms a single four vector, which will also have invariant length under Lorentz transformation. The energy- momentum of an ensemble of particles, a mixture of charged particles and photons, say, must also form a four vector. We might then guess that the conservation equations for ﬁeld energy (Poynting’s theorem 2.37, 2.41) and momentum CHAPMAN ( 2.48) could be combined to form an equation for the conservation of ﬁeld C and charged particle energy-momentum that is manifestly covariant. To demonstrate this we need to ”package” Energy conservation (Poynting’s Theorem)

1 ∂U S J E + = · (4.33) c ∂t ∇· c − c SANDRA and Momentum conservation 1 ∂ S = (ρE + J B) (4.34) c ∂t c − ∇ ·T − ∧ into one object. The r.h.s. of these equations should then form a four vector, with timelike component 4.33 and spacelike components 4.34. The l.h.s. of the desired expression must be the four divergence of a four tensor αβ ∂αT ; we can guess that this is the caseCOPYRIGHT since it involves ∂α and must be rank 1, that is, a four vector, to match the r.h.s. Looking at the l.h.s. first then, since 1 ∂ ∂ ∂ ∂ ∂ =( , , , ) (4.35) α c ∂t ∂x ∂y ∂z we need a 4 4 energy- momentum field tensor: × s1 s2 s3 U c c c s1 αβ c 11 12 13 T =  s2 −T −T −T  (4.36) c 21 22 23 s3 −T −T −T  31 32 33   c −T −T −T    where the Tij are the 9 components of the 3 3 Cartesian Maxwell stress αβ × tensor 2.56. Then ∂αT will give the l.h.s. of 4.33 and 4.34. ELECTRODYNAMICSWe have to deal with the r.h.s. in terms of E and B directly, which means that we need either the electromagnetic field tensor F αβ or its dual. We can guess which one by inspection of the r.h.s. terms; the spacelike part CORE 76 CHAPTER 4. THE FIELD TENSORS

needs a J B term, and we formed B by taking the ”four div” of F αβ ∧ ∇∧ αβ ( 4.22). So if we replace the ”four div” in ∂αF with a dot product with the four current Jα we have

Ex Ey Ez 1 0 c c c c J E Ex − − − − · αβ c 0 Bz By (ρE + J B)x JαF =(cρ, Jx, Jy, Jz)  −  =  − ∧  Ey (ρE + J B) − − − c Bz 0 Bx y E − − ∧  z B B 0   (ρE + J B)z   c − y x   − ∧     CHAPMAN(4.37)  which gives the r.h.s. as required. Our expression for energy-momentumC conservation is then αβ αβ ∂αT = JαF (4.38)

The r.h.s of equation 4.38 is constructed from four tensors and ∂α is a four vector, so the only remaining term T αβ must be a four tensor and the expression is manifestly covariant. SANDRA 4.4 Lorentz Force.

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

α dpα f = ds COPYRIGHTα dxα (4.39) u = ds We now just need to ﬁnd the four- force f α for a charged particle in an electromagnetic ﬁeld. Since the Lorentz force contains a +v B term, we αβ ∧ can again guess that since JαF gives J B terms we require something of the form − ∧

Ex Ey Ez 1 0 c c c cγ c γv E Ex − − − · αβ c 0 Bz By γvx γ(E + v B)x F uβ =  −   −  =  ∧  Ey γv γ(E + v B) c Bz 0 Bx y y E − − ∧  z B B 0   γvz   γ(E + v B)z   c y x   −   ∧   −     (4.40) αβ (we have swapped the order of the uα and the F around to get + instead of on the components of the r.h.s. in 4.40). The spacelike parts of 4.40 ELECTRODYNAMICSjust− look like the Cartesian components of the Lorentz force: d(mv) F = = e(E + v B) (4.41) dt ∧ CORE 4.4. LORENTZ FORCE. 77

multiplied by γ. The timelike part of 4.40 just looks like the Cartesian particle energy equation obtained from 4.41

d( 1 mv2) d$ F v = 2 = = ev E (4.42) · dt dt · multiplied by γ/c. Taking 4.41 and 4.42 together (and taking care of the e), we then have CHAPMAN d # d # γ dt c ds c C dmvx dmvx αβ γ dt ds eF uβ =  dmv4 y5  =  dmv4 y5  (4.43) γ dt ds  γ dmvz   dmvz   dt   ds      that is, α αβ dp eF uβ = SANDRA(4.44) ds the Lorentz force law in manifestly covariant form. We now have a complete system, the Maxwell equations, equations for conservation of charge and energy momentum, and a force law and equations of motion, all in manifestly covariant form. The rest of this chapter will be devoted to looking at the implications of what we have found so far.

4.4.1 Manifestly covariant electrodynamics. COPYRIGHT To summarize, we have: The inhomogenous Maxwell equations: • αβ β ∂αF = µ0J

The homogenous Maxwell equations: • ∂ ˜αβ =0 αF Charge -current conservation: •

α ∂αJ =0

ELECTRODYNAMICSEnergy-momentum conservation: • αβ αβ ∂αT = JαF CORE 78 CHAPTER 4. THE FIELD TENSORS

The equations of motion: • α dpα f = ds α dxα u = ds

The Lorentz force law: • dpα eF αβu = β ds CHAPMAN and with the Lorentz gauge: C •

α ∂αA =0

we have the Maxwell equations in terms of the four vector potential: • α α SANDRA 2A = µ0J

4.5 Transformation of the Fields

The electromagnetic field tensor F αβ allows us to ”package” the components of E and B in terms of four vectors as in 4.7. The Lorentz transformation of a four vector is simply a rotation of its spacetime coordinates as defined by 3.58 and 3.59. We can then find out how to Lorentz transform F αβ by applyingCOPYRIGHT the coordinate rotation to all of its’ component four vectors (see avanced problem 8):

αβ α β β α F ! = ∂! A! ∂! A! − = (Λαδ∂ )(ΛβγA ) (Λβγ∂ )(ΛαδA ) =Λ αδF Λγβ δ γ − γ δ δγ To obtain a simple formula to transform individual ﬁeld components we αβ αβ would then just need to compare terms in F and F ! . This (not diﬃcult, but rather long) operation gives:

E! = E & & B! = B & & (4.45) E! = γ(E + v B) ⊥ v∧E ⊥ B! = γ B c∧2 ⊥ − ⊥ ELECTRODYNAMICSwhere subscripts ” ” and ” 4” mean components5 parallel and perpendicular to the transformation* ⊥ velocity v. The inverse transform is just obtained by putting v v in 4.45. →− CORE 4.6. FIELD FROM A MOVING POINT CHARGE. 79

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

F 2 = e(E2 + v2 B2) = eE 2 (4.46) & ∧ & & CHAPMAN and we now know from 4.45 that the Lorentz force should be: C

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

as we chose E1 to be zero (by choosing ρ1 = 0). Also the Lorentz force law gives: F 2 = e(E2 + v2 B2) = eE 2 (4.48) ⊥ ∧ ⊥ ⊥ SANDRA which from 4.45 transforms to

F 2 = eγ(E1 + v1 B1) = eγv1 B1 (4.49) ⊥ ∧ ⊥ ∧

Hence F2 = γF1 which is just what we found in 3.32. Next, we will look at an application of 4.45. Before we do, there is one interesting point. Since the dual of F αβ also contains the electromagnetic fields in Lorentz transformable form we could have used it to find the transformation of the fields. We can find outCOPYRIGHT what this would yield by applying the duality transformation E Bc2 and B E to 4.45, and it is easy to verify that we obtain the same→ transformations→− for the fields.

4.6 Field from a Moving Point Charge.

In nonrelativistic situations (ie under Galilean frame transformation) we know that the field from a positive point charge just points radially outwards, and this was used in Chapter 1 to discuss Gauss’ law. Now that we know how to transform the fields we can look at what happens to the fields from charges moving at relativistic speeds. Lets consider a positive charge moving past us at speed v as shown in figure 4.6. If we are in the rest frame of the charge (the S! frame), there is ELECTRODYNAMICSno magnetic field and the electric field is radial giving (from Chapter 1): B! =0 q r! (4.50) E! = 3 4π#0 r! CORE 80 CHAPTER 4. THE FIELD TENSORS

x S x’2 S’ 2 P

r v a

x’ vt q x 1 CHAPMAN 1 C x x’ 3 3

Figure 4.1: Charge q is at rest at the origin of the S! frame and is moving with velocity vˆx1 w.r.t. frame S. The origins of the S and S! frames coincide at t = 0. SANDRA

We now transform to the S frame by moving in vˆx so that in the S frame − 1 the charge appears to be moving past us in the x1 direction with speed +v. The inverse transform gives us (with B! = 0 and v = v ) the ﬁelds in the S frame: &

E = E! & COPYRIGHTB =0& & (4.51) E = γE! ⊥ γ ⊥ v E B = 2 (v E!) = ∧2 ⊥ c ∧ ⊥ c

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

x! = γ(x vt) 1 1 − x2! = x2 = a (4.52) ELECTRODYNAMICS x3! = x3 =0

Substituting for x1! ,x2! ,x3! in 4.51 then gives the electric ﬁeld at P from the CORE 4.6. FIELD FROM A MOVING POINT CHARGE. 81

E2 E1 % >1

%=1

x-vt x-vt % >1 1 1 CHAPMAN C %=1

Figure 4.2: Sketch of E and E at point P versus x! /γ = x vt. 1 2 1 1 −

moving point charge: SANDRA q x1! q γ(x1 vt) E1 = E1! = 4π# 3 = 4π# − 3 0 (x 2+x 2+x 2) 2 0 (γ2(x vt)2+a2) 2 1! 2! 3! 1− γx! q 1 q γqa (4.53) E2 = γE2! = 4π# 3 = 4π# 3 0 (x 2+x 2+x 2) 2 0 (γ2(x vt)2+a2) 2 1! 2! 3! 1− E3 = γE3! =0 We can always obtain the magnetic field components at P from E via 4.51. To understand what is happening at relativistic speeds, lets sketch the electric field. Figure 4.6 shows the electric field components in the S frame E and E at point P , plotted versus x! /COPYRIGHTγ = x vt. As the relative speed of 1 2 1 1− the charge becomes relativistic, and γ> 1, the x1! /γ axis Lorentz contracts. The magnitude of E1 doesn’t change (as E1 = E1! ) instead the E1(x1! /γ) profile becomes compressed. The magnitude of E2 = γE2! increases, and again, the profile is compressed as γ> 1. The electric field occupies a more disc like region as γ is increased. This is sketched in figure 4.6 which shows the E field vectors in the x1,x2 plane. We expect the moving charge to have an accompanying magnetic field whether or not it is relativistic as it is carrying a current. In the nonrelativistic limit γ 1, 4.51 becomes: → v E µ dI r B = ∧ = 0 ∧ (4.54) c2 4πr3 where dI = qv is the current carried by the charge in its direction of relative ELECTRODYNAMICSmotion. This it just the Biot Savart law 1 in magnetostatics. Generally, the spatial 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 CORE 82 CHAPTER 4. THE FIELD TENSORS

E E’

v CHAPMAN C

Figure 4.3: Sketch of the E! ﬁeld around a charge at rest, and the E ﬁeld around a charge moving at relative velocity v, in the x1,x2 plane. SANDRA the charge moves at relativistic speeds, B will also be compressed into a more disc like region. As the charge moves closer to the speed of light v c and → ˆv E B ∧ (4.56) → c We can compare this to the free space electromagnetic wave solution to Maxwell’s equations 1.4. In section 2.4 we found that if we take a plane wave solution of the formCOPYRIGHT

i(ωt k r) E, B e − · (4.57) ∼ and substitute into Maxwell III we obtain 2.60:

kˆ E B = ∧ (4.58) c which is just 4.56, the magnetic ﬁeld from the (strongly) relativistic charge. So as v c, and both E and B become ”compressed” into a disc around the relativisitic→ charge, the ﬁelds start to look like a pulse of light centered on the charge.

situations is obtained by adding up the contributions due to all the current elements dI along the loop: µ0dI r ELECTRODYNAMICS B = ∧ (4.55) 4πr3 !C

CORE 4.7. RETARDED POTENTIAL. 83 P x2 x’2 S S’ r r’ v

x1 CHAPMAN x 3 x’ C x’ 1 3 ’dx’dx’dx’ & 1 2 3

Figure 4.4: A volume element is located at rest at the origin of the S! frame which is moving at +v in the x1 direction w.r.t. the S frame. In the S! frame the volume element contains charge ρ!dx1! dx2! dx3! . The originsSANDRA 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 dilation gave us a good idea of what happens to the fields around a single point charge as it moves at relativistic speeds. We will now obtain a general solution to Maxwell’s equations by findingCOPYRIGHT the fields from a collection of charges described in terms of the (frame dependent) charge and current density. To make life simple we can work in terms of the scalar and vector potentials rather than the field transformations directly. Everything needed for the calculation then forms four vectors; the scalar and vector potentials Aα, the charge and current densities J α, and spacetime xα, these then are straightforward to transform using the Lorentz transformation. The problem is posed in figure 4.7. We will first consider the scalar and vector potential due to the charge dQ contained in a small (elemental) volume element; since all our equations are still linear under the Lorentz transformation we can find the scalar and vector potential from charges distributed over a larger volume later by doing a volume integral. To simplify the algebra, the volume element containing dQ is located at the origin of the S! frame, and in this frame the charges are at rest. The S! frame then moves at speed +v along the x1! direction w.r.t. the S frame, ELECTRODYNAMICSand again for simplicity we will arrange for the origins of both frames to coincide at t = t! = 0. We now observe the scalar and vector potential dAα at some point P due to the charges when the frames coincided. Since CORE 84 CHAPTER 4. THE FIELD TENSORS

disturbances in the electromagnetic ﬁelds travel at speed c in all frames, in the S frame the ”signal” takes a time r t = (4.59) c to reach P from the S origin. By the time it has done so, the charges have moved; they are now located at the S! origin which is distance r! from P . α α We can ﬁnd dA from dA! in the S! frame where the charges are at rest, this is: CHAPMAN 0 dQ ρ!dx1! dx2! dx3! cdA! = dφ! = = 4π#0r! 4π#0r! C 1,2,3 (4.60) dA! =0 αβ The inverse Lorentz transformΛ (since S is moving in vˆx w.r.t. S!) (i) − 1 of 4.60 (see section 3.4) is

γ v γ 00 dφ! γ dφ! dφ c c c c v − vdφ! 1 α αβ + γ γ 00 SANDRA0 γ dA A =Λ A! =  c −    =  c2  =   (i) β 0010 0 0 dA2 −  0 0 0 1   0   0   dA3   −              (4.61)  so that dφ = γdφ! v (4.62) dA = ˆx1γ c2 dφ! The four-current J α will inverse transform in just the same way, from 2 charges at rest in the SCOPYRIGHT! frame to the charges moving in the S frame: ρ = γρ ! (4.63) J = γvρ! = vρ

Next, we can write dφ!(dx1! , dx2! , dx3! ,r!) as a function of coordinates in the S frame by using the Lorentz transformation of spacetime coordinates ie 4.52 to get expressions for the volume element dx1! , dx2! , dx3! and for r!. If we substitute in the time delay for t ( 4.59) then vr x! = γ x (4.64) 1 1 − c 6 7 This gives an expression for r!:

2 2 2 1 r! =(x1! + x2! + x3! ) 2 1 2 2 (4.65) = γ2 x vr + x2 + x2 ELECTRODYNAMICS 1 − c 2 3 8 9 2This just yields what we found4 for the moving5 charge and wire experiment in section 3.2. CORE 4.7. RETARDED POTENTIAL. 85

2 2 2 2 since r = x1 + x2 + x3 this can be written as:

vx1 r! = γ r (4.66) − c 8 9 The volume element as seen in S is obtained by implicit diﬀerentiation of 4.52: v dr v x1 dx! = γdx 1 = γdx 1 1 1 c dx1 1 c r − − (4.67) dx2! = dx2 6 7 4 5 CHAPMAN dx3! = dx3 C where we used the change in r across the volume element, dr/dx1, obtained by diﬀerentiating 2 2 2 2 r = x1 + x2 + x3 (4.68)

w.r.t. x1 to give dr x = 1 (4.69) dx1 r SANDRA Putting all this together, in the S frame we then have

ρ v x ρ dx dx dx γ dx1γ 1 c r dx2dx3 dφ = γ ! 1! 2! 3! = γ − (4.70) vx1 4π$0r! 6 7 4π$0γ4 r 5 − c which is just 4 5 ρdx dx dx dφ = 1 2 3 (4.71) 4π$COPYRIGHT0r This is the potential at P that we would expect from electrostatics if charge dQ were located at distance r instead of r!; that is, as if dQ were still at the origin of the S frame. So to get a solution of Maxwell’s equations, we have two approaches. One is to ”do the relativistic treatment” and Lorentz transform four vectors as above. The other is to insist that charge dQ is 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 reach point P , so we see a retarded potential consistent with the electrostatics of the charges at their previous location a time t = r/c earlier. The vector potential in the S frame is then v µ Jdx dx dx dA = ˆx dφ = 0 1 2 3 (4.72) 1 c2 4πr ELECTRODYNAMICSwhich, with B = A will again give the Biot Savart law, consistent with the magnetostatics∇∧ of the current due to the moving charges at their previous location a time t = r/c earlier. CORE 86 CHAPTER 4. THE FIELD TENSORS

dV’ r-r’ r’ P

r O CHAPMAN Figure 4.5: Volume element dV ! and point P located w.r.t. a single ﬁxed origin O. C

The algebra was simplified by choosing the charges to be at the origin of the S! frame and having the two frames coincident at t = 0. A more general arrangement is shown in figure 4.7 where all positions are known relative to a fixed origin. A volume element dVSANDRA! is at position r!, point P is then a distance r r! from the charges in dV !. It takes light a time | − | t = r r! /c to propagate from the charges to point P , so the potentials l | − | seen 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− and current densities in the S frame are:

r r! 1 ρ r!,t | −c | φ(r,t)= − dV ! (4.73) 4π$ 6 r r 7 COPYRIGHT0 "V | − ! | and r r! J r!,t | − | µ0 c A(r,t)= − dV ! (4.74) 4π 6 r r 7 "V | − ! | Why does this work? The whole ediﬁce of the Lorentz transformation and four vectors that we have developed amounts to three things: the speed of light c, the laws of physics, and certain quantities, like charge, are the same in all inertial frames. Hence we can use Gauss’ law of electrostatics, the invariance of charge, and the invarance of the speed of light, to obtain the (retarded) potential in any given frame. Provided that the resulting integrals for the retarded potentials are tractable, that’s all there is to it.

ELECTRODYNAMICS

CORE Suggested Texts. CHAPMAN C

This is not an exhaustive list, they are the books that author has found the most useful in the context of electrodynamics, so if you enjoyed this book, read on...

The Feynman Lectures on Physics, Vol II, R. P. Feynman, R. B. • Leighton, M. Sands, Addison-Wesley, 1975: A lively andSANDRA insightful journey through electrodynamics, best read in one go. Training in the formalism should be sought elsewhere. Classical Electrodynamics, Second Edition, J. D. Jackson, Wiley, • 1975: A must for every working physicists bookshelves. Thorough and rigorous treatment of electromagnetism from first principles to electrodynamics. Includes all the material that I left out. The six core theories of modernCOPYRIGHT 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 electrodynamics. Mathematical Methods in the Physical Sciences Second Edition, M. • L. Boas, Wiley, 1983: A nice introduction to tensors, and to vector calculus. The Classical Theory of Fields, Fourth Edition, Course of Theoretical • Physics Vol. 2, L. D. Landau and E. M. Lifshitz, Pergamon, 1983: Extends the electrodynamics of flat spacetime treated here to curved ELECTRODYNAMICSspacetime, i.e. general relativity.

CORE 87 Index CHAPMAN alternating tensor, 36, 71 coordinate transformationC Ampère’s law, 18, 23 Cartesian, 34 differential form, 19 Lorentz, 49, 59 Coulomb basis vectors, 36 unit, 10 Biot Savart law, 79 Coulomb gauge, 56 Coulomb’sSANDRA law Cartesian experiment, 9 inconsistency with spacetime, 50 covariant, 57 charge conservation, 23, 75 cross product, 36 equation of, 24 current density, 18 charge density, 10 charge invariance, 51 D’Alembertian, 64 classical limit direction cosines, 35 fields, 11, 16 displacement current, 24 fluids, 37 COPYRIGHTin capacitors, 24 particles, 26 Divergence theorem cold gas, 12 definition of, 8 conservation equation, 26 dual tensor, 71 definition, 11 duality transformation, 71, 77 energy flux, 39 dyadic, 29, 34 momentum flux density, 30 conservation Einstein summation convention, 35 energy-momentum, 74, 75 electric field, 10 of current, 75 definition, 17 of energy density, 38 definition of, 10 of field-particle momentum, 39 flux, 13 of four-current, 65 in terms of charge density, 11 contraction in terms of potentials, 23, 55 in Cartesian tensors, 36 lines, continuous, 17 ELECTRODYNAMICScontravariant, 57 units, 10 coordinate rotation work done by, 17 in spacetime, 49 electric potential

CORE 88 INDEX 89

definition, 18 contravariant, 57 electromagnetic field tensor, 69, 70 covariant, 57 length, 70 energy-momentum, 60 electromagnetic waves, 41 force, 63, 75, 76 free space, 25 four vector potential, 65 from Maxwell equations, 25 four-current, 65, 82 general solution, 25 four-divergence, 64 group velocity, 25 four-gradient, 63 phase speed in vacuum, 25 four-velocity, 61, 76 CHAPMAN relativistic charge, 80 inner product, 57 C electrostatic field length-time, 60 conservative, 18 potential, 68 electrostatic potential, 18 frame transformation energy- momentum field tensor, 73 Galilean, 23 Energy-momentum, 75 nonrelativistic, 23 equivalence of electric and magnetic SANDRA fields, 21, 23, 51 Galilean frame transformation, 23 Galilean transformation, 61 Faraday’s law, 20, 22 gas from Lorentz force, 21 finite temperature, 26 field warm, 31 energy density, 37, 38 gauge transformation, 56 from moving point charge, 79 Gauss’ law, 16 momentum density, 39 integral form, 16 momentum flux density, 40 generalCOPYRIGHT relativity, 58 fields generalized coordinates classical limit, 16 inner product, 59 transformation of, 23 generalized coordinates, 56 flat spacetime, 58 fluid bulk variables, 37 inner product, 36, 59 fluid equation, 32 flux, 13 kinetic energy, 37 definition, 12 of electric field, 13 light clock, 46 of energy, 39 linear media, 38 force Liouville’s theorem, 31 from momentum flux tensor, 30, Lorentz contraction, 48 31 Lorentz force, 39 four vector Lorentz force law ELECTRODYNAMICSfour current, 75 invariant form, 75 four-curl, 68 Lorentz gauge, 56, 65, 76 four-vector, 55 in four-vector form, 65, 76 CORE 90 INDEX

Lorentz transformation, 49, 59 Poynting flux, 38, 39 inverse, 59 units, 38 of charge density, 52 Poynting’s theorem, 38 retarded potential, 82 pressure tensor, 28, 31 Lorentz transformation of fields, 76 isotropic, 29 shear free, 29 magnetic field principle of superposition, 9 lines, continuous, 17 proper time, 62 units, 18 CHAPMAN magnetic flux, 16 radiation pressure, 42 C magnetostatic scalar potential, 19 rest energy, 60 Maxwell equations retarded potential, 83 dual, 72 right hand rule homogenous, 55, 71, 75 in cross product, 8 inhomogenous, 56, 66, 71, 75 in Lorentz force law, 9 manifestly covariant, 66, 76 rocket effSANDRAect, 30 Maxwell homogenous equations rotation matrix, 35 invariant form, 71 rotation of four vector, 57 Maxwell I, 16 integral form, 16 spacelike interval, 49 Maxwell II, 17 spacetime integral form, 16 deridative, 62 Maxwell III, 23 interval, 49, 56, 62 integral form, 22 metric, 58 Maxwell inhomogenousCOPYRIGHT equations rotation in, 59 invariant form, 71 special relativity, 45 Maxwell IV, 24 speed of light, 66, 72 Maxwell stress from Maxwell’s equations, 25 for free space EM waves, 41 from potentials, 56 tensor, 40 speed of light in vacuum, 25 mechanics, invariant form, 74 Stokes theorem momentum flux density, 30 definition of, 8 momentum flux tensor, 29 superposition monopoles, 72 in Galilean frame transformation, moving charge and wire, 50, 77, 82 23 of electromagnetic waves, 25 Newton’s laws, 61 of electrostatic fields, 18 and special relativity, 62 tensor ELECTRODYNAMICSphase space, 31 alternating, 36, 71 photons, 26 Cartesian, 30 and relativistic charge, 80 coordinate transformation, 35 CORE INDEX 91

divergence of, 36 work dot product, 36 done by Lorentz force, 37 dual, 71 electromagnetic field, 69, 70 energy- momentum field, 73 ensor Cartesian, 34 field momentum flux density, 40 force from, 31 CHAPMAN inner product, 36 C Lorentz transformation, 59 Maxwell stress, 40, 73 momentum flux density, 29 pressure, 28, 31 rank, 34 trace, 36 SANDRA warm gas pressure, 31 Tesla, 18 time dilation, 47 timelike interval, 49 trace, 36 transformation as rotation, 35, 49, 57 Cartesian, 34 charge invariance, 51 COPYRIGHT duality, 71, 77 Galilean, 61 gauge, 56 Lorentz, 49, 59 of charge density, 52 of EM field tensor, 76 retarded potential, 82 rotation in spacetime, 59

vector potential, 55 from magnetostatics, 19

warm gas pressure tensor, 32 wave equation ELECTRODYNAMICSfor electrmagnetic ﬁelds, 25 in terms of potentials, 56 manifestly covariant, 66 CORE 92 INDEX

CHAPMAN C

SANDRA

ELECTRODYNAMICS

CORE Appendix A CHAPMAN C Revision Problems

A.1 Static Magnetic Fields. SANDRA

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 a circle of radius r as shown.≤ Use the integral form of .B = 0 to show that B is always perpendicular to r. From Amp`ere’s∇ law (the integral form of B = µ0J) calculate B(r) and hence using Stokes theorem (ie from∇∧B dl) calculate B. · ∇∧ COPYRIGHT Calculate: B directly. In a cylindrical coordinate system you can use the identity∇∧

1 ∂ B =( B) ˆz = (rB )ˆz ∇∧ ∇∧ z r ∂r θ (A.1)

when B = Bθθˆ.

b Outside the wire ie for r > a use Amp`ere’s law to calculate B(r) using curve C1 shown below (ie a circle of radius r centred on the wire). Use this expression for B to calculate B dl around curve C . Cal- · 2 culate B directly at a point within C2 and compare it with your value for∇∧B dl. Sketch the variation: of the magnitude of the cur- ELECTRODYNAMICSrent density J·(r) as a function of r and use Maxwell’s IVth equation : for steady ﬁelds B = µ0J to sketch the variation of B as a function of r. ∇∧ ∇∧

CORE 93 94 APPENDIX A. REVISION PROBLEMS

I CHAPMAN C r

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

C 1

C1 C2

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

ELECTRODYNAMICS

CORE A.2. STATIC ELECTRIC FIELDS. 95

A.2 Static Electric Fields.

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

1 ∂ C .E = (rE ) (A.2) ∇ r ∂r r

when E = Erˆr.

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

b i) Amp`ere’s law for steady currents is

B dl = µ I · 0 !c Write this in diﬀerential form. ii) Using conservation of charge obtain Maxwell IV for time dependent ﬁelds: B = µ J + µ $ ∂E ∇∧ 0 0 0 ∂t c In general (ρ = 0, J = 0) Poynting’s theorem in media is ELECTRODYNAMICSf ( f ( ∂ 1 (E H)= ( [E D + B H]) J E (A.3) ∇· ∧ −∂t 2 · · − f · CORE 96 APPENDIX A. REVISION PROBLEMS

Given that the energy density

1 U = [E D + B H] (A.4) 2 · ·

∂U starting with ∂t use Maxwell’s equations in media and the identity

(E H)=H ( E) E ( H) (A.5) ∇· ∧ · ∇∧ − · ∇∧ CHAPMAN to derive Poynting’s theorem. C

A.4 The Wave Equation: Linearity and Dis- persion. SANDRA A general form for the wave equation is:

∂2ψ = v2 2ψ (A.6) ∂t2 ∇ Show that

i(ω1t k1x) ψ1 = ψ01e − (A.7) is a solution of this equation.COPYRIGHT What is the phase speed and direction of propagation of this wave? If the wave

i(ω2t k2x) ψ2 = ψ02e − (A.8)

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

ψ3 = ψ1 + ψ2 (A.9)

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

∂2ψ = v2(ω) 2ψ + bψ (A.10) ELECTRODYNAMICS ∂t2 ∇ where b is a constant. Show that ψ3 is a solution. Give an example of a PDE for which ψ1 and ψ2 are solutions, but ψ3 is not. CORE 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: i(ωt k.r) E = E0e − i(ωt k.r) B = B0e − a Show that Maxwell’s equations in free space can be rewritten as: k E =0 k B =0 · 0 · 0 CHAPMAN k E = ωB k B = µ $ ωE ∧ 0 0 ∧ 0 − 0 0 0 C b Show that E, B and k (in that order) form a right-handed orthogonal system. c If, with an electric antenna, it is established that E only has a y component, what are the possible directions of propagation of the wave, and the directions of the associated B components.SANDRA d Describe the polarization of this wave. Write down E and B for a wave 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: 6 i(ωt+3x y z) B(r,t) = 10− [1, 2,BzCOPYRIGHT]e − − Determine: a The direction of propagation kˆ, wavelength λ, angular frequency ω. 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 (ρf = 0,Jf = 0). Use Maxwell’s equations in media and B = µµ0H, D = $$0E to derive wave equations for E and B. (Use the identity ( A)= ( A) 2A). ω ∇∧ ∇∧ ∇ ∇· −∇ What is the phase speed v = k of the waves in SI units? ELECTRODYNAMICSShow that 1 B = kˆ E (A.11) v ∧ CORE 98 APPENDIX A. REVISION PROBLEMS

A.8 Dielectrics and Polarization.

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

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

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

∂E 1 ∂2E 2E = µµ σ + (A.12) ∇ 0 ∂t v2 ∂t2 Derive the dispersion relation (that is, k as a function of ω) in the low frequency limit (terms O(ω2) terms O(ω)) by examining the properties of plane wave solutions. Hence- show that the skin depth (distance waves can penetrate into the conductor before they are appreciably attenuated) is 2 1 δ = 2 (A.13) ELECTRODYNAMICS {µµ0σω } What is the signiﬁcance of this result for i) radio propagation ii) microwave ovens iii) power transmission? CORE A.10. CAVITY RESONATOR. 99

A.10 Cavity Resonator.

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

CHAPMAN C

SANDRA

ELECTRODYNAMICS

CORE 100 APPENDIX A. REVISION PROBLEMS

CHAPMAN C

SANDRA

ELECTRODYNAMICS

CORE Appendix B CHAPMAN C Solutions to Revision Problems SANDRA B.1 Static Magnetic Fields.

a The current density I J = (B.1) | | πa2 is uniform within the wire. Current enclosed within circle, radius r: I Ir2 I = J dS =COPYRIGHTπr2 = (B.2) r · πa2 × a2 " Then since B dS = 0 (B.3) · ! over a surface deﬁned by a cylinder of radius r, B only has components parralel to the surface ie perpendicular to r. From Amp`ere B dl = µ I (B.4) · 0 ! and using cylindrical symmetry to infer that B is independent of r: | | Ir2 B dl = B 2πr = µ (B.5) · × 0 a2 ELECTRODYNAMICS! that is, µ Ir B = 0 (B.6) 2πa2 CORE 101 102 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 Ir2 µ = B πr2 (B.8) 0 a2 | ∇ ∧ | CHAPMAN giving C µ I B = 0 (B.9) | ∇ ∧ | πa2 directed in +J. Directly: 1 ∂ µ Ir mu I B = ˆz r 0 SANDRA= ˆz 0 (B.10) ∇∧ r ∂r 2πa2 πa2 0 1

b Outside the total current enclosed by C1 is I then

B dl = µ I (B.11) · 0 !C1

B2πr = µ0I (B.12) COPYRIGHT ie B = µ0I/2πr. Around C2

B dl = 0 (B.13) · !C2 as no current is enclosed. Evaluating this integral the contributions from the straight line segments of C2 along ˆr are zero and perpendicular to ˆr along the curved segments (at constant r1 and r2) give

µ I µ I B dl = 0 r dθ 0 r dθ = 0 (B.14) · 2πr × 1 − 2πr × 2 !C1 1 "1 2 "2

where the integral over θ just gives the angle subtended by curve C2 at r = 0. ELECTRODYNAMICSDirectly 1 ∂ rµ I B = ˆz 0 = 0 (B.15) ∇∧ r ∂r 2πr 0 1 CORE 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, cylintrical coordinates z, r,θ so that r dl = 0 we have either (i) E assymetric (ie · with a θ dependence) and E dl at some θ1 cancels that at some θ2, or (ii) E has cylindrical symmetry· and hence must be perpendicular to dl CHAPMAN everywhere. Since E due to a point charge is symmetric and the principle C of superposition holds we have case (ii). Thus this also admits a constant component along ˆz. To use Gauss in integral form over a cylinder of radius r < a, length L we have charge enclosed Q = ρV where

ρ L ρ ρ = l = l SANDRA(B.17) πa2L πa2 ie ρ Q = l πr2L (B.18) πa2 × The only contribution to E is on the curved surface of the cylinder on which E is constant. 1 E dS = E2πrL = ρ r2La2 (B.19) · $ l ! 0 hence COPYRIGHT ρlr E = 2 ˆr (B.20) 2$0a π Then 1 ∂ ρ r2 ρ rho E = l = l = (B.21) ∇· r ∂r 2$ a2π $ a2π $ 0 0 1 0 0 For cylinder radius r > a the enclosed Q = ρlL so that ρ L E dS = E2πrL = l (B.22) · $ ! 0 then ρ E = l ˆr (B.23) 2πr$0 ELECTRODYNAMICSand 1 ∂ ρ r E = l = 0 (B.24) ∇· r ∂r 2πr$ 0 0 1 CORE 104 APPENDIX B. SOLUTIONS TO REVISION PROBLEMS

B.3 Conservation and Poynting’s Theorem.

a i) Number through dS in time dt is nv dSdt since volume containing particles that cross surface in dt is v dS· dt. Then number crossing S in dt is · nv dSdt (B.25) · "S 1 Flux = No/sec/unit area = nv dS (B.26) S · CHAPMAN "S C ii) Number enclosed in S is

N = ndV (B.27) "V so number crossing S in unit time is ∂N ∂ SANDRA∂n = ndV = dV (B.28) ∂t ∂t ∂t "V "V Now using our expression for the ﬂux across S, the number ﬂowing out of V in time dt is ∂n nv dSdt = dV dt (B.29) · − ∂t !S "V From DivergenceCOPYRIGHT theorem then ∂n = (nv) (B.30) − ∂t ∇· For charges we have ρ = nq, J = nqv giving ∂ρ = J (B.31) − ∂t ∇·

b Given in the text. c Poynting theorem in media: from the energy density 1 U = [E.D + B.H] (B.32) 2 ELECTRODYNAMICSwe have ∂U 1 ∂D ∂E ∂H ∂B = E + D + B + H (B.33) ∂t 2 · ∂t · ∂t · ∂t · ∂t 0 1 CORE B.4. THE WAVE EQUATION: LINEARITY AND DISPERSION. 105

then since D = $$0E, B = µµ0H we have ∂U ∂D ∂B = E + H (B.34) ∂t · ∂t · ∂t which is ∂U = E ( H J )+H ( E)= J E H E+E H ∂t · ∇∧ − f · −∇∧ − f · − ·∇∧ ·∇∧ (B.35) Using the identity CHAPMAN .(E H)=H.( E) E.( H) (B.36) C ∇ ∧ ∇∧ − ∇∧ we obtain ∂U = J E .(E H) (B.37) ∂t − f · −∇ ∧ B.4 The Wave Equation: Linearity andSANDRA Dis- persion.

Substitute ψ1 into the wave equation: (iω )2ψ = v2( ik )2ψ (B.38) 1 1 − 1 1 ie if ω1 = vk1 then ψ1 is a solution. v is the phase speed, that is, the speed of a point of constant phase. Consider some point x at phase φ = ω1t k1x at time t. At time t + dt this point has moved to x + dx where dx =−vdt. So COPYRIGHT ω (t + dt) k (x + dx)=φ (B.39) 1 − 1 Hence ωdt = kdx and the point of constant phase moves at dx ω v = = 1 = v (B.40) dt k1

Now ψ2 is a solution. Subsitute ψ3 = ψ1 + ψ2 into the wave equation: ∂2 ∂2 ∂2 ψ = ψ + ψ = v2 2ψ = v2[ 2ψ + 2ψ ] (B.41) ∂t2 3 ∂t2 1 ∂t2 2 ∇ 3 ∇ 3 ∇ 2 and since ∂2 ψ = v2 2ψ (B.42) ∂t2 1,2 ∇ 1,2 ψ3 is also a solution. This will hold for any linear PDE, hence ψ3 can be ELECTRODYNAMICSshown to be a solution of the dispersive wave equation by the same method. Require nonlinear PDE (with terms in ψn,( ψ)n etc) to break the principle of superposition. ∇ CORE 106 APPENDIX B. SOLUTIONS TO REVISION PROBLEMS

B.5 Free Space EM waves I.

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

ik ∇· ≡ − · ik ∇∧ ≡∂ − ∧ iω CHAPMAN ∂t ≡ C Then substituting the wave solutions into the free space Maxwell equations:

E k E =0 ∇· ⇒ · 0 B k B =0 ∇· ⇒ · 0 ∂ E = B k ESANDRA= ωB ∇∧ −∂t ⇒ ∧ 0 0 ∂ B = µ $ E k B = ωµ $ B ∇∧ 0 0 ∂t ⇒ ∧ 0 − 0 0 0

b Use result (a):

k E =0 k B =0 E, B k · 0 · 0 ⇒ ⊥ 1 B = kˆ E then handedness COPYRIGHTc ∧ ⇒

c k E = 0 so k = ˆxk + ˆyk , then since B = 1 kˆ E, B = ˆxB +ˆzB . · 0 x y c ∧ x z d Linearly polarized. i) Linear polarization:

i(ωt k.r) E = E0e − i(ωt k.r) B = B0e −

plane of polarization is E, k ii) Circular polarization:

i(ωt kz) i(ωt kz π ) ELECTRODYNAMICS E = E0ˆxe − + E0ˆye − ± 2 (B.43)

with B = 1 kˆ E. c ∧ CORE B.6. FREE SPACE EM WAVES II. 107

B.6 Free Space EM Waves II.

1 a k = 3ˆx + ˆy + ˆz, k = √11m− so λ =2π/k =1.89m. ω = ck = − 8 1 | | 9.93 10 s− . × 6 b Use k B = 0, k from (a) giving B = 10− T . · z c Use k B = ωµ $ B with above k, B to give ∧ 0 − 0 0 0 300 1 CHAPMAN E0 = (ˆx 4ˆy +7ˆz)Vm− √11 − C i(ωt+3x y z) E = E0e − −

d 1 2 2 S = E B = 432 cos (ωt k r)kˆWm− (B.44) µ0 ∧ − · SANDRA B.7 EM Waves in a Dielectric.

Using ∂D H = (B.45) ∇∧ ∂t we have using H = 0: ∇· ∂D ( H)= 2H = = (B.46) ∇∧ ∇∧ −∇COPYRIGHT∇∧ ∂t Then since Maxwell III ∂B E = (B.47) ∇∧ − ∂t can be written as ∂H D = $$ µµ (B.48) ∇∧ 0 0 − ∂t $ % we have ∂2H 2H = $$ µµ (B.49) ∇ 0 0 ∂t2 similarly taking ( D) yields ∇∧ ∇∧ ∂2D 2D = $$ µµ (B.50) ∇ 0 0 ∂t2 ELECTRODYNAMICSUsing B = µµ0H and D = $$0E gives wave equations in terms of E and B: ∂2E 2E = $$ µµ (B.51) ∇ 0 0 ∂t2 CORE 108 APPENDIX B. SOLUTIONS TO REVISION PROBLEMS

∂2B 2B = $$ µµ (B.52) ∇ 0 0 ∂t2 which by inspection have wave solutions with phase speed 1 ω v = = (B.53) ($$0µµ0) k As in free space plane waves, Maxwell; III yields: CHAPMAN k 1 B = kˆ E = kˆ E (B.54) ω ∧ v ∧ C 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. SANDRA a Gauss in integral form for the surface becomes: 1 E dS = σ dS (B.55) · $ f ! 0 " Then σ Q E = f = (B.56) $0 $0A Since Q = CV and V = Ed (from E = 0 and E directed normal to the plates) weCOPYRIGHT have ∇∧

Q Q $ A C = = = 0 (B.57) V Ed d

b From slab geometry E is uniform in the capacitor, hence no polarization charge is induced within the volume of dielectric. From the deﬁnition of polarization: 1 P dS = σ dS = P dS (B.58) · $ p ! 0 " " as P is then also normal to the (approximately inﬁnate) plates. This 2 gives P = σp in units of Coulomb m− . ELECTRODYNAMICSc In the presence of the dielectric 1 E dS = (σ σ )dS (B.59) · $ f − p ! 0 " CORE B.9. EM WAVES IN A CONDUCTOR: SKIN DEPTH. 109

which in this slab geometry is σ σ σ P E = f − p = f (B.60) $0 $0 − $0 then since for Linear media P =($ 1)$ E we have − 0 σ E = f (B.61) $$0 CHAPMAN This again yields the capacitance from Q = CV and V = Ed: C Q Q $ $A C = = = 0 (B.62) V Ed d B.9 EM Waves in a Conductor: Skin Depth. In media Maxwell IV is SANDRA ∂D H = J + (B.63) ∇∧ f ∂t which can be written as 1 ∂E B = Jf + $$0 (B.64) µµ0 ∇∧ ∂t

using Jf = σE this becomes: COPYRIGHT1 ∂E B = µµ σJ + (B.65) ∇∧ 0 f v2 ∂t We take ∂/∂t of this equation, and use

( E)= ( E) 2E (B.66) ∇∧ ∇∧ ∇ ∇· −∇ with E = 0 since D = ρ = 0 to obtain a modiﬁed wave equation: ∇· ∇· f ∂E 1 ∂2E 2E = µµ σ (B.67) ∇ 0 ∂t v2 ∂t2 Assume plane wave solutions of the form

i(ωt k r) E = E0e − · (B.68) ELECTRODYNAMICSthen substitute in to obtain the dispersion relation: w2 k2 = µµ σiω (B.69) v2 − 0 CORE 110 APPENDIX B. SOLUTIONS TO REVISION PROBLEMS

The two terms on the r.h.s. are due to the displacement and conduction currents respectively. Taking the low frequency limit of this expression is equivalent to neglecting the displacement currents in comparison to the conduction currents in the conductor and yields: k2 = µµ σiω (B.70) − 0 Clearly k has real and imaginary parts. Writing k = β iα gives plane wave solutions of the form: ± CHAPMAN i(ωt βkˆ r) αkˆ r E = E0e − · e± · C (B.71) which corresponds to waves that grow (+α) or decay ( α) as they propagate in +r. For a semi- infinate slab of conductor, the− waves must have finite amplitude as r which excludes growing solutions. The waves then decay and a measure→∞ of the distance that they can penetrate into the conductor before they are appreciably attenuated (the skin depth δ) is that required for the amplitude to fall by a factor 1SANDRA/e: 1 δ = (B.72) α α is then obtained by equating coefficients, with k2 = β2 α2 2iαβ (B.73) − − giving α = β,2αβ = µµ0σω and 2 COPYRIGHTδ = (B.74) µµ σω < 0 Implication: lower frequency waves propagate further into conductors. Some examples are: The earth’s ionosphere (a conductor) will not propagate radio waves • at all frequencies, hence longwave can be bounced off the ionosphere to be received at points on the earth’s surface below the horizon. Re-entering astronauts lose radio contact with the ground as their • spacecraft is engulfed in plasma. A few mm of conductor are sufficient to attenuate microwaves so that • it is safe to use microwave ovens. household power (50Hz) can be carried over short distances by copper ELECTRODYNAMICS• with limited power loss, whereas computer network signals require co axial cable or twisted pair ethernet (computer clock speeds are several 100Mhz). ×∼ CORE B.10. CAVITY RESONATOR. 111

your TV uses copper wire to carry the power, and coaxial cable to • carry the TV signal. over long distances copper wire would dissipate too much power to • be eﬃcient. Instead, high voltage, low current transmission lines are used.

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

SANDRA

ELECTRODYNAMICS

CORE 112 APPENDIX B. SOLUTIONS TO REVISION PROBLEMS

CHAPMAN C

SANDRA

ELECTRODYNAMICS

CORE Appendix C CHAPMAN C Some Advanced Problems

C.1 Maxwell Stress Tensor. SANDRA Momentum ﬂux conservation in an ideal Magnetohydrodynamic (MHD) plasma gives the following:

Du = P + J B (C.1) Dt −∇· ∧ For equilibrium the l.h.s. of this equation must be zero.

1. Show that J B can be writtenCOPYRIGHT as the divergence of a stress tensor ∧ TM .

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

BiBj Pij = P δij +(P P ) (C.2) ⊥ & − ⊥ B2 where the local magnetic ﬁeld direction is ˆz.

Use the equilibrium condition (TM P ) = 0 to obtain the following stability criteria: ∇· −

B2 P + = K (C.3) ⊥ 2µ0 B2 P P = + K! (C.4) ELECTRODYNAMICS& − ⊥ µ0

where K,K! are constants.

CORE 113 114 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 probability density f(x, v,t) in 6 dimensional phase space x, v, and is:

Df ∂f = + v f + a vf = 0 (C.5) Dt ∂t ·∇ ·∇ CHAPMAN a Show (by using Taylor expansion) that this corresponds toC the change in f along a particle orbit (Liouville’s theorem). b Show that for a charged particle moving under Lorentz force law that the Liouville Theorem is equivalent to a conservation equation in 6 dimensional phase space (hint, treat x and v as independent phase space coordinates). SANDRA C.3 Newton’s Laws and the Wave Equation under Galilean Transformation.

Determine whether or not the form of the following equations is invariant under the Galilean transformation x! = x vt, t! = t: − a Newton’s equations of motion for the ith particle in an ensemble where th th the force betweenCOPYRIGHT the i and j particle is F(xi xj). − dv m i = F(x x ) (C.6) i dt i − j j # b The wave equation for some ﬁeld Ψ(x,t)

1 ∂ 2 Ψ = 0 (C.7) ∇ − c2 ∂t2 $ %

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

ELECTRODYNAMICSC.4 Transformation of the Fields. Two charged straight wires carrying a charge density ρ per unit volume are a distance l apart and are at rest. The charges are not free to move within CORE C.5. METRIC FOR FLAT SPACETIME.1 115

+ + + + + + + + + + + +

CHAPMAN C + u + + + + + + + + + + + SANDRA Figure C.1: A pair of wires

the wires. Calculate the electric field at each wire due to the other from Gauss’ 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 the force on each wire due to this magneticCOPYRIGHT field from the Lorentz force law. The total Lorentz force must be the same in both frames of reference. What has happened to the electric field?

C.5 Metric for Flat Spacetime.1

γβ β a If gαβ is the metric for flat spacetime show that gαγ g = δα where δβ is a 4 4 rank 2 tensor with zero off axis terms and trace 1. α × b The Lorentz transformation of covariant four vector xα and contravariant four vector xα can be written as β xα! =Λαβx α αβ ELECTRODYNAMICSx! =Λ xβ show that the metric for flat spacetime gαβ transforms a contravariant vector into a covariant vector in all frames under Lorentz transforma- CORE 116 APPENDIX C. SOME ADVANCED PROBLEMS

tion, ie show β gαβx! = xα! (C.8)

γδ c Show that Fαβ = gαγF gδβ.

C.6 Length of the EM ﬁeld Tensor in Space- time. CHAPMAN C The tensor scalar product is

βα T : T = TαβT (C.9)

where we sum over both β and α. Show that the tensor scalar product of the electromagnetic ﬁeld tensor is SANDRA E2 F F βα =2 B2 (C.10) αβ c2 − $ % And that this is zero for free space electromagnetic waves (no charges present). Hence show that ˜ ˜βα = c2F F βα (C.11) FαβF − αβ where ˜ is the dual of F Fαβ COPYRIGHTαβ

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

Show that Maxwells’ homogenous equations can be written as

∂αF βγ + ∂βF γα + ∂γ F αβ = 0 (C.12)

C.8 Lorentz Transformation of the EM Field Tensor.

ELECTRODYNAMICSa Show that the Lorentz transformation of the ﬁeld tensor

αβ α β β α F ! = ∂! A! ∂! A! (C.13) − CORE C.8. LORENTZ TRANSFORMATION OF THE EM FIELD TENSOR.117

b The transpose rule is:

aαβ..xγδ.. = x...δγ a..βα (C.14)

for both contravariant and covariant tensors. Use the transpose rule and the Lorentz transformation of four vectors to show that the Lorentz transformation of the electromagnetic ﬁeld tensor is: αδ γβ F ! =Λ FδγΛ (C.15) CHAPMAN αβ C

SANDRA

ELECTRODYNAMICS

CORE 118 APPENDIX C. SOME ADVANCED PROBLEMS

CHAPMAN C

SANDRA

ELECTRODYNAMICS

CORE Appendix D CHAPMAN C Solution to Advanced Problems SANDRA D.1 Maxwell Stress Tensor.

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

1 1 2 Tij = BiBj δij B (D.1) µ0 − 2µ0 as shown in 2.2. COPYRIGHT 2. We can combine this with the pressure tensor as given:

BiBj Pij = P δij +(P P ) (D.2) ⊥ & − ⊥ B2 to obtain 1 (P P ) 1 & − ⊥ 2 (TM P )ij = 2 BiBj + B P δij − µ0 − B −2µ0 − ⊥ $ % $ % (D.3) This is divergence free if B2 (P P )=K! (D.4) µ0 − & − ⊥ and 1 ELECTRODYNAMICSB2 + P = K (D.5) 2µ0 ⊥

where K,K! are both constants.

CORE 119 120 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 f(x + δx, v + δv,t+ δt)=f(x, v,t)+δt + δx f + δv f + ... ∂t ·∇ ·∇v (D.6) so that the total change in f as we move from x x+δx, v CHAPMANv+δv, and t t + δt is → → → C f(x + δx, v + δv,t+ δt) f(x, v,t) Df − = (D.7) δt Dt and along a particle orbit such that:

δr v = SANDRA (D.8) δt δv a = (D.9) δt this becomes: Df ∂f = + v f + a f (D.10) Dt ∂t ·∇ ·∇v b If x and v are independent then COPYRIGHT ∂f ∂vjf (v f)j = vj = (D.11) ·∇ ∂xj ∂xj

that is, v f = (vf) (D.12) ·∇ ∇· Also, from the Lorentz force law q a = [E(x,t)+v B(x,t)] (D.13) m ∧ then again as x and v are independent

∂f ∂ E = (Ef) (D.14) · ∂v ∂v · ELECTRODYNAMICSand using an identity we can write ∂ ∂f ∂ (v Bf) = (v B) + f (v B) (D.15) ∂v · ∧ ∧ · ∂v ∂v · ∧ CORE D.3. NEWTON’S LAWS AND THE WAVE EQUATION UNDER GALILEAN TRANSFORMATION.121

by expanding into components the second term on the r.h.s. can be shown to be zero. Then ∂f ∂ (v B) = (v Bf) (D.16) ∧ · ∂v ∂v ∧ so that we can ﬁnally rewrite the Vlasov equation as ∂f + (vf)+ (af) = 0 (D.17) ∂t ∇· ∇v · CHAPMAN ie in the form of a conservation equation for f in 6 dimensional phase C space.

D.3 Newton’s Laws and the Wave Equation under Galilean Transformation. SANDRA Consider Galilean frame transformation x! = x vt, t! = t, with v constant. − th a For the i particle the velocity in the transformed frame is vi! = vi v so that the acceleration is − dv dv dv i! = i = i! (D.18) dt dt dt! then the force between the ith and jth particle dv m i = COPYRIGHTF(x x ) (D.19) i dt i − j j # transforms as

dvi! m = F(x! + vt x! vt)= F(x! x! ) (D.20) i dt i − j − i − j ! j j # # Hence Newton’s equations are Galilean invariant in form. b To transform the wave equation 1 ∂ 2 Ψ = 0 (D.21) ∇ − c2 ∂t2 $ % ∂ we need to transform and ∂t . These are deﬁned via the chain rule ELECTRODYNAMICSso that in the unprimed∇ frame: ∂ dΨ(x,t)= dx + dt Ψ(x,t) (D.22) ·∇ ∂t 0 1 CORE 122 APPENDIX D. SOLUTION TO ADVANCED PROBLEMS

so that dΨ(x,t) Ψ(x,t)= (D.23) ∇ dx |t const and ∂Ψ(x,t) dΨ(x,t) = (D.24) ∂t dt |x const we can compare this with the chain rule applied in the primed frame: ∂ dΨ(x!,t!)= dx! ! + dt! Ψ(x!,t!)CHAPMAN (D.25) ·∇ ∂t 0 ! 1 C then dΨ(x!,t!) !Ψ(x!,t!)= t! const (D.26) ∇ dx! | and ∂Ψ(x!,t!) dΨ(x!,t!) = x! const (D.27) ∂t! dt! SANDRA| Now t = t! so that t! constant is equivalent to t constant which implies that ! . However, x! = x, instead we have x! constant equivalent to x ∇v≡∇t constant when obtaining( the partial deridatives from chain rule.−

We can write dΨ(x!,t!) in terms of elements in the unprimed frame using dx! = dx vdt, dt! = dt: − ∂ ∂ dΨ(x!,t!)= dx! ! + dt! Ψ= (dx vdt) ! + dt Ψ COPYRIGHT·∇ ∂t! − ·∇ ∂t! 0 1 0 (D.28)1 Then, given that we can write Ψ as a function of variables in either frame: Ψ(x,t) Ψ(x!,t!) (D.29) ≡ this yields:

dΨ(x!,t!) ∂Ψ(x!,t!) ∂ = = v ! + Ψ(x!,t!) (D.30) dt |x const ∂t − ·∇ ∂t 0 ! 1 Then the wave equation in the unprimed frame is also

2 1 ∂ Ψ(x!,t!) = 0 (D.31) ∇ − c2 ∂t2 $ % ELECTRODYNAMICSwhich transforms to 2 2 1 ∂ ! v ! + Ψ(x!,t!) = 0 (D.32) ∇ − c2 − ·∇ ∂t = 0 ! 1 > CORE D.4. TRANSFORMATION OF THE FIELDS. 123

which expands to:

2 1 ∂ 1 2 2 ∂ ! (v !) + (v !) Ψ = 0 (D.33) ∇ − c2 ∂t2 − c2 ·∇ c2 ·∇ ∂t 0 ! 1 Hence the wave equation is not invariant in form.

Solutions to the wave equation in the unprimed frame will be of the form

i(ωt k x) Ψ(x,t) e − · (D.34) CHAPMAN ∼ which transformed to the primed frame becomes: C

i([ω k x]t! k x!) Ψ(x!,t!) e − · − · (D.35) ∼ which is consistent with the doppler shift of frequency:

ω! = ω k x (D.36) − · SANDRA We can verify that this is a solution of D.33 by direct substitution.

D.4 Transformation of the Fields.

This question is an extension of the moving charge and wire experiment discussed in 3.2. To in the rest frame calculate the electric field from one of the wires we enclose an elemental length dl of wire inCOPYRIGHT surface S as shown in figure D.4. S is a cylinder of radius l. Gauss’ law in integral form then gives ρ ρ E dS = E dS = E2πldl = dV = Adl (D.37) · · $ $ !S "Sc "V 0 0 where A is the c.s.a. of the wire. Then ρA E = (D.38) 2πl from each wire and is directed perpendicular to the wire from geometry. The total electric field is just the sum of the contribution from each wire and the force in this frame in which the charges are at rest is just dF = ρAE(l)dl per length element dl of wire. In the moving frame we can calculate the magnetic field by enclosing ELECTRODYNAMICSthe wire with a curve as shown in D.4 and by using Ampere:

B dS = µ J dS = µ ρ!v dS (D.39) · 0 · 0 · !S "S "S CORE 124 APPENDIX D. SOLUTION TO ADVANCED PROBLEMS

+ l S dl CHAPMAN C

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

A COPYRIGHT +

+ +

l + C dl +

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

CORE D.5. METRIC FOR FLAT SPACETIME. 125

but taking into account the Lorentz contraction along the direction of the wire in order to calculate ρ! in the moving frame as in 3.2. The electric field can also be calculated in the moving frame using charge density ρ!. Alternatively, the transformation of the fields can be used directly given the electric field in the unprimed (rest) frame where the magnetic field is zero: v E B! = γ ∧ (D.40) − c2 and CHAPMAN E! = γE (D.41) C where we have exploited the geometry, that is, in the unprimed frame the electric field is perpendicular to the wires. In each frame the Lorentz force is then just given directly from the Lorentz force law. SANDRA D.5 Metric for Flat Spacetime.

a To show γβ β β gαγg = Dα = δα (D.42) β where we will use the notation Dα for the l.h.s. and

1000 0 100 g = = gαβ (D.43) αβ  00− COPYRIGHT10 −  0001   −    αβ Now all oﬀ axis α = β terms of gαβ and g are zero, hence all of the ( β oﬀ axis terms of the inner product Dα will be zero. For example

2 γ2 D1 = g1γ g = 0 (D.44)

for any γ. The remaining terms are: α = β = 0 giving

0 γ0 00 10 20 30 D0 = g0γg = g00g +g01g +g02g +g03g = 1+0+0+0 (D.45)

α = β = 1:

ELECTRODYNAMICS1 γ1 01 11 21 31 D1 = g1γg = g10g +g11g +g12g +g13g = 0+1+0+0 (D.46)

2 3 similarly, D2 = 1 and D3 =1 CORE 126 APPENDIX D. SOLUTION TO ADVANCED PROBLEMS

Then 1 0 0 0 0 1 0 0 Dβ = = δβ (D.47) α  0 0 1 0  α  0 0 0 1    as required.   b To show β gαβx! = xα! CHAPMAN(D.48) we write the l.h.s. of this as C

γβ gαγΛ xβ (D.49)

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

γβ β gαγΛ =Λα (D.50)

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

β 0β 0β Λ0 = g00Λ =Λ Λβ = g Λ1β = Λ1β 1 11 − (D.51) Λβ = g Λ2β = Λ2β COPYRIGHT2 22 − Λβ = g Λ3β = Λ3β 3 33 − then γβ β gαγΛ xβ =Λαxβ = xα! (D.52) is a four vector with components α =0, 3:

β 0β Λ0 xβ =Λ xβ = x0! β 1β Λ1 xβ = Λ xβ = x1! β − 2β (D.53) Λ2 xβ = Λ xβ = x2! β − 3β Λ x = Λ x = x! 3 β − β 3 so that if covariant four vector ct ELECTRODYNAMICS x xβ =(ct, x)= −  (D.54) − y −  z   −    CORE D.5. METRIC FOR FLAT SPACETIME. 127

and

ct! αβ x! Λ xβ =   (D.55) y!  z!      then we have ct! CHAPMAN β x! Λαxβ =  −  = xα! (D.56) y! C −  z!   −    which is covariant in the primed frame.

c To show γδ Fαβ = gαγF gδβ SANDRA(D.57)

perform the contraction in two steps. First, the nonzero terms of γδ γ F gδβ = Fβ are:

γ γ0 δ =0,β =0 F0 =+F γ γ1 δ =1,β =1 F1 = F γ − γ2 (D.58) δ =2,β =2 F2 = F δ =3,β =3 F γ = −F γ3 COPYRIGHT3 − γ for γ =0, 3. Then the nonzero terms of gαγFβ are:

0 α = γ =0 F0β =+Fβ α = γ =1 F = F 1 1β − β (D.59) α = γ =2 F = F 2 2β − β α = γ =3 F = F 3 3β − β for β =0, 3. Here we have exploited the fact that all β = γ terms of ( gβγ are zero.

We now just consider the signs of the terms of Fαβ for all α,β .

row 0: α = 0 all change sign except F00 =0 ELECTRODYNAMICSrow 1: α =1 F10 changes sign only row 2: α =2 F20 changes sign only row 3: α =3 F30 changes sign only CORE 128 APPENDIX D. SOLUTION TO ADVANCED PROBLEMS

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

We have the electromagnetic ﬁeld tensor:

0 Ex Ey Ez − c − c − c Ex 0 B B F αβ =  c − z y  (D.60) Ey B 0 B c z − x CHAPMAN  Ez B B 0   c y x   −  C from Chapter 4, and also Fαβ is obtained via the transformation Ei Ei αβ →− and Bi Bi in F . The→ symmetries of this tensor are then by inspection: F βα = F αβ (D.61) F = −F βα − αβ SANDRA and the trace is zero: αα F = Fαα = 0 (D.62) βα FαβF then has the following properties for γ =0, 3: i) The on axis terms are zero

γγ F Fγγ = 0 (D.63) ii) The timelike terms,COPYRIGHT that is, those involving E are: F F γ0 = F F = ( F F ) 0γ 0γ γ0 0γ 0γ (D.64) F F 0γ = −F F = −(−F F ) γ0 − γ0 0γ − − γ0 γ0 contracting these terms over γ gives: E2 [F ]2 = = [F ]2 (D.65) γ0 c2 0γ γ γ # # so that the total contribution of the timelike terms is 2E2/c2. iii) The spacelike terms, that is, terms involving B then have the following properties: F F γα = F F αγ = F F (D.66) αγ − αγ − αγ αγ for α = 0, γ = 0. Contracting these terms gives: ELECTRODYNAMICS( ( (F )2 = 2B2 (D.67) − αγ − α=1,3 γ=1,3 # # CORE D.7. ALTERNATIVE FORM FOR THE MAXWELL HOMOGENOUS EQUATIONS.129

The result of the contraction is then: E2 F F βα =2 B2 (D.68) αβ c2 − $ % βα In free space (photons only) we have E = Bc so that FαβF vanishes. For the dual tensor ˜αβ we use the fact that it is obtained from Fαβ via F 2 the duality transformation Ei Bic , Bi Ei (see 4.2). Transforming the result gives → →− CHAPMAN B2c4 2 E2 C ˜ ˜βα =2 E2 = B2 = c2F F βα (D.69) FαβF c2 − −c2 c2 − − αβ $ % $ % as required.

D.7 Alternative Form for the MaxwellSANDRA Ho- mogenous Equations.

This left as an exercise for the student.

D.8 Lorentz Transformation of the EM Field Tensor. a In 4.1 we obtain COPYRIGHT F αβ = ∂αAβ ∂βAα (D.70) − then the Lorentz transformation of this is just the transformation of each of the component four vectors. b To evaluate this ie:

αβ α β β α F ! = ∂! A! ∂! A! (D.71) − we apply the Lorentz transformation and the transpose rule:

αβ αδ βγ βγ αδ F ! = (Λ ∂δ)(Λ Aγ ) (Λ ∂γ )(Λ Aδ) αδ γβ −βγ δα =Λ ∂δAγ Λ Λ ∂γAδΛ αδ γβ − αδ γβ =Λ ∂δAγ Λ Λ ∂γ AδΛ (D.72) αδ − γβ =Λ (∂δAγ ∂γ Aδ)Λ αδ γβ− ELECTRODYNAMICS=Λ FδγΛ as required.

CORE 130 APPENDIX D. SOLUTION TO ADVANCED PROBLEMS

CHAPMAN C

SANDRA

ELECTRODYNAMICS

CORE Appendix E CHAPMAN C Vector identities

Assuming the right hand rule relates the orthogonal unit vectors ˆi, ˆj, and kˆ: SANDRA

ˆi ˆj = kˆ ∧ E.1 Diﬀerential Relations

a b = b a = a b + a b + a b (E.1) · · 1 1 2 2 3 3 COPYRIGHTˆi ˆj kˆ a b = b a = a a a (E.2) ∧ − ∧ 1 2 3 b1 b2 b3 a (b c)=(a b) c =(c a) b (E.3) · ∧ ∧ · ∧ · a (b c)=(a c)b (a b)c (E.4) ∧ ∧ · − · (a b) (c d) = (a c)(b d) (a d)(b c) (E.5) ∧ · ∧ · · − · · (a b) (c d) = [a (b d)] c [a (b c)] d (E.6) ∧ ∧ ∧ · ∧ − · ∧ (ψφ)=φ ψ + ψ φ (E.7) ∇ ∇ ∇ (φa)=φ a + a φ (E.8) ∇· ∇· ·∇ (φa)=φ a +( φ) a (E.9) ∇∧ ∧∇ ∇ ∧ (a b)=b ( a) a ( b) (E.10) ELECTRODYNAMICS∇· ∧ · ∇∧ − · ∇∧ (a b) = (a )b +(b )a + a ( b)+b ( a) (E.11) ∇ · ·∇ ·∇ ∧ ∇∧ ∧ ∇∧ (a b)=a( b)+(b )a b( a) (a )b (E.12) ∇∧ ∧ ∇· ·∇ − ∇· − ·∇ CORE 131 132 APPENDIX E. VECTOR IDENTITIES

a ( b) = ( B) a (a )b (E.13) ∧ ∇∧ ∇ · − ·∇ ( a)= ( a) 2a (E.14) ∇∧ ∇∧ ∇ ∇· −∇ ( a) = 0 (E.15) ∇· ∇∧ ( φ) = 0 (E.16) ∇∧ ∇ E.2 Integral Relations CHAPMAN For the following integral relations vector surface element dS Cis directed along the normal ˆn to the surface S, line element dl is directed along curve C. Flux of a vector ﬁeld a:

Flux = a dS (E.17) · !S SANDRA If surface S spans curve C:

• φdl = dS ( φ) (E.18) ∧ ∇ !C "S Stokes Theorem: • a dl = a dS (E.19) · ∇∧ · COPYRIGHT!C "S • dl a = (dS ) a (E.20) ∧ ∧∇ ∧ !C "s • dS ( φ ψ)= φdψ = ψdφ (E.21) · ∇ ∧∇ − "S !C !C If surface S encloses volume V , dS points outwards:

• φdS = ( φ)dV (E.22) ∇ !S "v Gauss or Divergence Theorem: ELECTRODYNAMICS• a dS = adV (E.23) · ∇· !S "V CORE E.2. INTEGRAL RELATIONS 133

• (dS a)= ( a)dV (E.24) ∧ ∇∧ !S "V Green I: • φ( ψ) dS = φ 2ψ +( φ) ( ψ) dV (E.25) ∇ · ∇ ∇ · ∇ !S "V ? @ Green II or Green’s Theorem in a plane: CHAPMAN • C (φ ψ ψ φ) dS (φ 2ψ ψ 2φ)dV (E.26) ∇ − ∇ · ∇ − ∇ !S "V Green II in vector form: • [b ( a) a ( b)] dS = (E.27) ∧ ∇∧ − ∧ ∇∧ · SANDRA !S (a [ ( b)] b [ ( a)]) dV · ∇∧ ∇∧ − · ∇∧ ∇∧ "V

ELECTRODYNAMICS

CORE 134 APPENDIX E. VECTOR IDENTITIES

CHAPMAN C

SANDRA

ELECTRODYNAMICS

CORE Appendix F CHAPMAN C Tensors

F.1 Cartesian Tensors SANDRA

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

axbx axby axbz ab = a b COPYRIGHTa b a b (F.1)  y x y y y z  azbx azby azbz   which can, where i =1, 3 and j =1, 3 be written as

Aij = aibj (F.2) Coordinate rotation • A vector in the unprimed frame: x r = y (F.3)  z  will have coordinates in the primed frame:

! ELECTRODYNAMICSx = a11x +a12y +a13z ! y = a21x +a22y +a23z (F.4) ! z = a31x +a32y +a33z

CORE 135 136 APPENDIX F. TENSORS

The tensor a contains the direction cosines of the rotation x, y, z ij → x!,y!,z!. This can be written as

r! = r (F.5) A· where ! x a11 a12 a13 ! ! r = y and = a21 a22 a23 (F.6)   A   z! a31 a32 a33     CHAPMAN Einstein convention • C Since x = r1, y = r2 and z = r3, equation 2.24 is

! rj = ajiri (F.7) which is shorthand (the Einstein summation convention) for

! rj = ajirSANDRAi (F.8) i=1,3 # The sum over index i on the r.h.s. of 2.27 contracts the number of indices 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 basis vectors (or axes)COPYRIGHTx, y, z. Hence equation 2.28 is tensor dot product: a = a1T11 + a2T21 + a3T31 + a1T12 + a2T22 + ... ·T aiTij = ˆxjaiTij ≡ (F.9) dropping all reference to the basis vectors ˆxj. Then in Cartesian coordinates is • ∇ ·T ∂ ∂ ˆxj Tij = Tij (F.10) ∂xi ∇ ·T≡ ∂xi It follows that (φ )=φ( )+( φ) (F.11) ∇· T ∇·T ∇ ·T (ab)=( a)b +(a )b (F.12) ∇· ∇· ·∇ Divergence theorem for tensors of arbitrary rank: ELECTRODYNAMICS• dS = ( ) (F.13) S T· ∇ ·T 0V ! "V CORE F.2. SPECIAL TENSORS 137

F.2 Special Tensors

The trace is δ = 1 when i = j and 0 when i = j: • ij ( 1 0 0 δij = 0 1 0 (F.14)  0 0 1    Contracting δij over i and j with two vectors extracts the dot product: CHAPMAN

δij aibj = a1b1 + a2b2 + a3b3 = a b = aibi (F.15) C · This generalizes to tensors of arbitrary rank:

δ a T = a (F.16) ij i ijkl.. ·T The alternating tensor: • SANDRA 1 if ijk = 123, 231, 312 $ijk = 1 if ijk = 321, 132, 213 (F.17) −0 any two indices alike

Contracting $ijk over j and k extracts the following vector from a dyadic:

$ijkajbk = $i11a1b1 +$i12b1a2 +$i13b1a3 +$i21a2b1 COPYRIGHT+$i22b2a2 +$i23b3a2 (F.18) +$i31a3b1 +$i32b2a3 +$i33b3a3

If we consider the i = 1 terms, all except $123a2b3 and $132a3b2 are zero. These last two give a2b3 a3b2 which are the x component of the vector cross product. Similarly,− i =2, 3 gives the y, z components respectively, 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-time interval s2 = c2t2 x2 y2 z2 (F.19) − − − ELECTRODYNAMICSThe space-time interval is the length of the four vector:

s2 = s s = x xα (F.20) · α CORE 138 APPENDIX F. TENSORS

F.3.1 General properties of spacetime s has 2 forms: • Covariant: ct x xα =(ct, x)= −  (F.21) − y −  z   −  and Contravariant:   CHAPMAN C ct x xα =(ct, +x)= (F.22)  y   z      In general (for any spacetime geometry) if we have a well deﬁned • α α 0 1 2 3 SANDRA transformation x! = x! (x ,x ,x ,x ) a covariant vector transforms as: ∂x0 ∂x1 ∂x2 ∂x3 ∂xβ aα! = α a0 + α a1 + α a2 + α a3 = α aβ (F.23) ∂x! ∂x! ∂x! ∂x! ∂x! and a contravariant vector transforms as α α α α α α ∂x! 0 ∂x! 1 ∂x! 2 ∂x! 3 ∂x! β a! = a + a + a + a = a (F.24) ∂x0 COPYRIGHT∂x1 ∂x2 ∂x3 ∂xβ For tensors of rank 2, a covariant tensor G transforms as • αβ ∂xγ ∂xδ Gαβ! = α β Gγδ (F.25) ∂x! ∂x! a contravariant rank 2 tensor Gαβ as:

α β αβ ∂x! ∂x! γδ G! = G (F.26) ∂xγ ∂xδ α and a mixed tensor of rank 2 Gβ transforms as:

α δ α ∂x! ∂x γ Gβ! = γ β Gδ (F.27) ∂x ∂x!

ELECTRODYNAMICSThe dot or inner product is: • a b = a... b...α (F.28) · ...α ... CORE F.3. GENERALIZED TENSORS 139

that is, a contraction over the index α. For general spacetime geometry the inner product is invariant under transformation since: β α β ∂x ∂x! ...... γ ∂x ...... γ β ...... γ a! b! = α γ a...βb... = γ a...βb... = δγ a...βb... = a b · ∂x! ∂x ∂x · (F.29)

F.3.2 Flat spacetime CHAPMAN The Lorentz transformation corresponds to a rotation of the four vector s C in flat space-time, under which s2 is constant. The norm or metric for flat spacetime is then the invariant differential • length element:

(ds)2 =(dx0)2 (dx1)2 (dx2)2 (dx3)2 (F.30) − − − SANDRA which is a special case of the general diﬀerential length element

2 α β (ds) = gαβdx dx (F.31)

The spacetime metric for ﬂat space is given by: • 1000 0 100 g = = gαβ (F.32) αβ  00− 10 COPYRIGHT−  0001   −    Covariant and contravariant vectors are related by • β xα = gαβx (F.33)

and

α αβ x = g xβ (F.34)

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

! β xα =Λαβx (F.35) ELECTRODYNAMICSon the contravariant form and

α! αβ x =Λ xβ (F.36) CORE 140 APPENDIX F. TENSORS

on the covariant form. where the Lorentz transformation is:

v γ c γ 00 + v γ −γ 00 Λ = c (F.37) αβ  00− 10  −  0 0 0 1   −    the inverse transformation matrix is then just obtained by vCHAPMANv in 3.60. C →− The covariant four gradient is • 1 ∂ c ∂t ∂ ∂x 1 ∂ ∂α =  ∂  =( , ) (F.38) ∂y c ∂t ∇  ∂   ∂z  SANDRA   The contravariant four gradient is • 1 ∂ c ∂t ∂ α ∂x 1 ∂ αβ ∂ =  − ∂  =( , )=g ∂β (F.39) c ∂t −∇ − ∂y  ∂   ∂z   −  Hence the ”fourCOPYRIGHT - divergence” of a four vector • 1 ∂ ∂ aα = ∂αa = a0 + a (F.40) α α c ∂t ∇· (where a is the spacelike part of four vector ﬁeld aα). The ”4 curl” of four vector a is a rank 2 four tensor: •

F αβ = ∂αaβ ∂βaα (F.41) − The D’Alembertian invariant operator is then: • 1 ∂2 ∂ ∂α = 2 = 2 (F.42) α c2 ∂t2 −∇ ELECTRODYNAMICSwhich is just the wave equation operator in vacuum.

CORE CHAPMAN C

SANDRA

ELECTRODYNAMICS

CORE 141 142 APPENDIX G. UNITS AND DIMENSIONS

Appendix G CHAPMAN C Units and Dimensions

G.1 SI Nomenclature SANDRA Physical Quantity Name Symbol length metre m mass kilogram kg time second s current Ampere A temperature Kelvin K amount of substance mole mol luminous intensity COPYRIGHTcandela cd plane angle radian rad solid angle steradian sr frequency Hertz Hz energy Joule J force Newton N pressure Pascal Pa power Watt W electric charge Coulomb C electric potential Volt V electric resistance Ohm Ω electric conductance Siemens S electric capacitance Farad F magnetic ﬂux Weber Wb magnetic inductance Henry H ELECTRODYNAMICSmagnetic intensity Tesla T luminous ﬂux lumen lm illuminance lux lx activity (of radioactive source) Bequerel Bq CORE absorbed dose (of ionising radiation) Gray Gy G.2. METRIC PREFIXES 143

G.2 Metric Prefixes Multiple Prefix Symbol Multiple Prefix Symbol 1 10− deci d 10 deca da 2 2 10− centi c 10 hecto h 3 3 10− milli m 10 kilo k 6 6 10− micro µ 10 mega M 9 9 10− nano n 10 giga G 12 12 10− pico p 10 tera T CHAPMAN 15 15 10− femto f 10 peta P 18 18 C 10− atto a 10 exa E

SANDRA

ELECTRODYNAMICS

CORE 144 APPENDIX G. UNITS AND DIMENSIONS

CHAPMAN C

SANDRA

ELECTRODYNAMICS

CORE Appendix H CHAPMAN C

Dimensions and Units SANDRA

H.1 Physical Quantities

SI has been used throughout this book.COPYRIGHT Gaussian units are still often used hawever and conversion factors between SI and Gaussian units are given here.

To obtain the value of a quantity in Gaussian Units, multiply the value expressed in SI units by the conversion factor. Multiples of 3 in the conversion factor result from approximating the speed of light c = 8 1 8 1 2.99979 10 ms− 3 10 ms− . × 1 ×

Dimensions Physical Sym-SI Gaussian SI Conversion Gaussian Quantity Unit Factor Unit bol mass m m m kg 103 gram (g) ELECTRODYNAMICSlength l l l m 102 centimetre (cm) time t t t s 1 second (s)

CORE 145 146 APPENDIX H. DIMENSIONS AND UNITS

Dimensions Physical Sym-SI Gaussian SI Conversion Gaussian Quantity Unit Factor Unit bol t2q2 11 Capacitance C ml2 l F 9 10 cm 1 3 × m 2 l 2 9 Charge q q t C 3 10 stat- × coulomb 1 q m 2 3 3 Charge ρ l3 3 Cm− 3 10 stat-CHAPMAN Density l 2 t × coulomb C 3 cm− tq2 1 11 1 Conductance ml2 t S 9 10 cms− tq2 1 × 9 1 Conductivity σ ml3 t S/m 9 10 s− 1 3 × q m 2 l 2 9 Current I, i t t2 A 3 10 statampere 1 × q m 2 2 5 Current J l2t 1 Am− 3 10 statampere l 2 t2 SANDRA× 2 density cm− m m 3 3 3 Density ρ l3 l3 kgm− 10− gcm− 1 q m 2 2 5 Displacement D l2 1 Cm− 12π 10 stat- l 2 t × coulomb 2 cm− 1 ml m 2 1 4 1 Electric ﬁeld E t2q 1 V/m 3 10− statvoltcm− l 2 t 1 1 × ml2 m 2 l 2 1 2 Electro- $, t2q t V 3 10− statvolt motance emf COPYRIGHT × ml2 ml2 7 Energy W t2 t2 J 10 erg m m 3 3 Energy U lt2 lt2 J/m 10 ergcm− density ml ml 5 Force F t2 t2 N 10 dyne 1 1 Frequency f,ν t t Hz 1 Hz

ELECTRODYNAMICS

CORE H.1. PHYSICAL QUANTITIES 147

Dimensions Physical Sym-SI Gaussian SI Conversion Gaussian Quantity Unit Factor Unit bol ml2 t 1 11 1 Impedance Z tq2 l Ω 9 10− scm− 2 2 × ml t 1 11 2 1 Inductance L q2 l H 9 10− s cm− 1 × q m 2 3 Magnetic H lt 1 A/m 4π 10− Oersted intensity l 2 t × CHAPMAN 1 3 ml2 m 2 l 2 8 Magnetic Φ tq t Wb 10 Maxwell C ﬂux 1 m m 2 4 Magnetic B tq 1 T 10 Gauss induction l 2 t 2 1 5 l q m 2 l 2 2 3 3 Magnetic m, µ t t Am 10 Oersted cm moment 1 q m 2 1 3 SANDRA Magnetisation M lt 1 Am− 10− Oersted l 2 t 1 1 q m 2 l 2 4π Magneto- t t2 A 10 Gilbert motance M ml ml 1 5 1 Momentum p t t kgms− 10 gcms− m m 2 1 2 1 Momentum l2t l2t kgm− 10− gcm− s− density ml 1 1 7 Permeability µ q2 1 Hm− 4π 10 - 2 2 × t q 1 9 Permittivity Φ$ ml3 1 COPYRIGHTFm− 36π 10 - 1 × q m 2 2 5 Polarization P l2 1 Cm− 3 10 stat- l 2 t × coulomb 2 cm− 1 1 ml2 m 2 l 2 1 2 Potential V,φ t2q t V 3 10− statvolt 2 2 × ml ml 7 1 Power P t3 t3 W 10 erg s− m m 3 3 1 Power lt3 lt3 Wm− 10 erg cm− s− density m m 2 Pressure P lt2 lt2 Pa 10 dyne cm−

ELECTRODYNAMICS

CORE 148 APPENDIX H. DIMENSIONS AND UNITS

Dimensions Physical Sym-SI Gaussian SI Conversion Gaussian Quantity Unit Factor Unit bol q2 1 1 9 1 Reluctance ml2 l AW b− 4π 10− cm− R ml2 t 1 × 11 1 Resistance R tq2 l Ω 9 10− scm− ml3 1 × 9 Resistivity η,ρ tq2 t Ωm 9 10− s ml ml 1 1×5 Thermal κ t3 t3 Wm− K−10 erg CHAPMAN 1 1 1 conductivity cm− s− K− 1 1 C ml m 2 l 2 1 6 Vector A tq t W bm− 10 Gauss cm potential l l 1 2 1 Velocity v t t ms− 10 cms− m m 1 1 Viscosity η, µ lt lt kgm− s−10 Poise 1 1 1 1 Vorticity ζ t t s− 1 s− ml2 ml2 7 Work W t2 t2 J SANDRA10 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, and the relative permittivity and permeability, $r and µr are dimensionless. SI Gaussian 7 2 µ0 4π 10− Hm− 1 × 1 $0 COPYRIGHT 1 c√µ0 D = $0E + P E +4πP H = B M B 4πM µ0 − − Maxwell I E = ρ ∇· #0 D = ρ D =4πρ Maxwell II ∇· B =0 ∇· B =0 ∇· ∂B ∇· 1 ∂B Maxwell III E = ∂t E = c ∂t ∇∧ − ∂E ∇∧ − Maxwell IV B = µ0J + µ0$0 ∂t ∇∧ ∂D 4π 1 ∂D H = J + ∂t H = c J + c ∂t ∇∧ ∇∧ v Lorentz force E + v B E + c B per unit ∧ ∧ charge

ELECTRODYNAMICS

CORE Appendix I CHAPMAN C Physical Constants (SI)

Physical quantity Symbol Value UnitsSANDRA 12 1 Free space permittivity $0 8.854 10− Fm− × 7 1 Free space permeability µ0 4π 10− Hm− 1 × 8 1 Free space speed of light c = 2.998 10 ms− √µ0#0 × 19 Elementary charge e 1.602 10− C × 24 2 Bohr magneton µB 9.274 10− Am− × 31 Electron mass me 9.109 10− kg × 27 Proton mass mp 1.673 10− kg × 27 Atomic mass unit mu 1.661 10− - × 34 Planck constant h COPYRIGHT6.626 10− Js × 23 1 Boltzmann constant k 1.381 10− JK− × 23 1 Avrogado number N 6.022 10 mol− 0 ×

ELECTRODYNAMICS

CORE 149