EDlarranhaga

Relativistic Classical Electrodynamics

Alexis Larrañaga

June 18, 2013

2

Contents

I The Relativistic Description of Classical Particles 9

1 Special Relativity 111.1 The Minkowskian Space . . . . . . . . . . . . . . . . . . . . . . . . . . 111.2 Lorentz Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . 121.3 Covariant and Contravariant Components . . . . . . . . . . . . . . . . 131.4 Light Cones and Causality Relations . . . . . . . . . . . . . . . . . . . 131.5 Scalars, Vectors and Tensors . . . . . . . . . . . . . . . . . . . . . . . . 14

1.5.1 Scalars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.5.2 Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.5.3 Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.6 Relativistic Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.6.1 Addition of Velocities . . . . . . . . . . . . . . . . . . . . . . . 161.6.2 4-Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.6.3 Doppler Effect and Aberration . . . . . . . . . . . . . . . . . . 17

1.7 The Principle of Relativity . . . . . . . . . . . . . . . . . . . . . . . . . 181.8 Scalar, Vector and Tensor Fields . . . . . . . . . . . . . . . . . . . . . 18

1.8.1 Scalar Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.8.2 Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.8.3 Tensor Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

1.9 Some Useful Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.9.1 Differential Relations . . . . . . . . . . . . . . . . . . . . . . . . 191.9.2 Integral Relations . . . . . . . . . . . . . . . . . . . . . . . . . 20

1.10 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2 Relativistic Dynamics 232.1 The 4-Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.1.1 Particles of Zero Rest Mass . . . . . . . . . . . . . . . . . . . . 242.2 The Equation of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . 252.3 Explicit Forms of the 4-Force . . . . . . . . . . . . . . . . . . . . . . . 26

2.3.1 Constant Newtonian Force . . . . . . . . . . . . . . . . . . . . 262.3.2 4-Force Linear in the Velocity . . . . . . . . . . . . . . . . . . . 272.3.3 The Lorentz Force . . . . . . . . . . . . . . . . . . . . . . . . . 272.3.4 4-Force Quadratic in the Velocity . . . . . . . . . . . . . . . . . 29

3

4 CONTENTS

2.4 Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

II Relativistic Electromagnetic Field Theory 33

3 Maxwell Equations 353.1 The problem and the solution . . . . . . . . . . . . . . . . . . . . . . . 37

3.1.1 Magnetic Monopoles . . . . . . . . . . . . . . . . . . . . . . . . 383.2 Maxwell’s Equations in Matter . . . . . . . . . . . . . . . . . . . . . . 383.3 Electromagnetic Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

3.3.1 Monochromatic Plane Waves . . . . . . . . . . . . . . . . . . . 413.3.1.1 Polarization of Plane Waves . . . . . . . . . . . . . . 42

3.3.2 Propagation in Linear Media . . . . . . . . . . . . . . . . . . . 433.3.2.1 Reflection and Transmission at Normal Incidence . . . 443.3.2.2 Reflection and Transmission at Oblique Incidence . . 45

3.4 The Electromagnetic Field Equations . . . . . . . . . . . . . . . . . . . 463.5 Electromagnetic Potentials . . . . . . . . . . . . . . . . . . . . . . . . 473.6 Covariant Form of Maxwell-Lorentz Equations . . . . . . . . . . . . . 483.7 Gauge Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . 50

3.7.1 Lorentz Gauge . . . . . . . . . . . . . . . . . . . . . . . . . . . 513.7.2 The Coulomb or Transverse Gauge . . . . . . . . . . . . . . . . 51

3.8 Lorentz Invariants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 523.9 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

4 Lagrangian and Hamiltonian Formulation 554.1 Equation of Motion in Lagrangian Form . . . . . . . . . . . . . . . . . 554.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

4.2.1 Free Particle . . . . . . . . . . . . . . . . . . . . . . . . . . . . 574.2.1.1 Non-Relativistic Limit . . . . . . . . . . . . . . . . . . 574.2.1.2 Equation of Motion . . . . . . . . . . . . . . . . . . . 58

4.2.2 Charged Particle in Electromagnetic Fields . . . . . . . . . . . 594.3 Hamiltonian Equations of Motion . . . . . . . . . . . . . . . . . . . . . 60

4.3.1 Charged Particle in Electromagnetic Fields . . . . . . . . . . . 604.3.2 Other Forms of Hamiltonian . . . . . . . . . . . . . . . . . . . 61

4.3.2.1 Hamiltonian Tensor . . . . . . . . . . . . . . . . . . . 614.3.2.2 Hamiltonian as Energy . . . . . . . . . . . . . . . . . 624.3.2.3 Two Different Hamiltonians . . . . . . . . . . . . . . . 62

4.4 Classical Spin. Electric and Magnetic Moments . . . . . . . . . . . . . 634.4.1 Motion of the Particle in the Presence of Moments . . . . . . . 644.4.2 The Spin Vector . . . . . . . . . . . . . . . . . . . . . . . . . . 65

4.4.2.1 Spin Precession . . . . . . . . . . . . . . . . . . . . . 664.4.3 Equation of Motion for the Spin in the Presence of Electromag-

netic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 694.5 General Description of the Classical Spin . . . . . . . . . . . . . . . . 71

CONTENTS 5

4.5.1 Spacetime Translations . . . . . . . . . . . . . . . . . . . . . . 724.5.2 Lorentz Transformations . . . . . . . . . . . . . . . . . . . . . . 73

4.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

5 Lagrangian Description of the Fields 775.1 Symmetries and Conservation Laws. Noether’s Theorem . . . . . . . . 78

5.1.1 Noether Current and Conserved Charge . . . . . . . . . . . . . 795.2 The Lagrangian for the Real Scalar Field . . . . . . . . . . . . . . . . 81

5.2.1 Conserved Quantities . . . . . . . . . . . . . . . . . . . . . . . 815.2.1.1 Infinitesimal Translation. Energy-Momentum Tensor 815.2.1.2 Infinitesimal Lorentz Transformation . . . . . . . . . 82

5.3 The Lagrangian for the Complex Scalar Field . . . . . . . . . . . . . . 845.3.1 Conserved Quantities . . . . . . . . . . . . . . . . . . . . . . . 84

5.3.1.1 Infinitesimal Translation. Energy-Momentum Tensor 845.3.1.2 Infinitesimal Lorentz Transformation . . . . . . . . . 855.3.1.3 Internal Symmetries . . . . . . . . . . . . . . . . . . . 85

5.4 The Lagrangian for the Electromagnetic Field . . . . . . . . . . . . . . 865.4.1 Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . 875.4.2 Alternative Lagrangians for the Free Electromagnetic Field . . 88

5.5 Poincaré Transformations . . . . . . . . . . . . . . . . . . . . . . . . . 895.5.1 Infinitesimal Translation. The Energy-Momentum tensor . . . 89

5.5.1.1 Conserved Charge . . . . . . . . . . . . . . . . . . . . 915.5.2 Infinitesimal Lorentz Transformation . . . . . . . . . . . . . . . 94

5.6 Internal Symmetries . . . . . . . . . . . . . . . . . . . . . . . . . . . . 955.7 Canonical Form of the Field Equations . . . . . . . . . . . . . . . . . . 96

5.7.1 Hamiltonian for the Electromagnetic Field . . . . . . . . . . . . 975.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

III Interaction of Particles and Fields 99

6 Interacting Fields 1016.1 Interaction Field - External Current . . . . . . . . . . . . . . . . . . . 101

6.1.1 The Electromagnetic Field with Sources . . . . . . . . . . . . . 1016.2 Interaction Field - Particle . . . . . . . . . . . . . . . . . . . . . . . . . 104

6.2.1 A Charged Particle in an Electromagnetic Field . . . . . . . . . 1046.3 Interaction between Fields . . . . . . . . . . . . . . . . . . . . . . . . . 1086.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

7 Solution of the Equations of Motion 1117.1 Green’s Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1117.2 Electrostatic Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1137.3 Non-static Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114

7.3.0.1 The Residue Theorem in Bref . . . . . . . . . . . . . 1157.3.1 Retarded Green’s Functions . . . . . . . . . . . . . . . . . . . . 115

6 CONTENTS

7.3.2 Advanced Green’s Functions . . . . . . . . . . . . . . . . . . . 1197.3.3 Other Green’s Functions . . . . . . . . . . . . . . . . . . . . . . 120

7.4 Solving the Inhomogeneous Equation . . . . . . . . . . . . . . . . . . . 1227.4.1 Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

7.5 Gauge Condition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1257.6 Jefimenko’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 1267.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

8 Radiation 1298.1 Lienard-Wiechert Potentials . . . . . . . . . . . . . . . . . . . . . . . . 129

8.1.1 Coulomb Field . . . . . . . . . . . . . . . . . . . . . . . . . . . 1318.2 The Electromagnetic Field . . . . . . . . . . . . . . . . . . . . . . . . . 132

8.2.0.1 Coulomb Field . . . . . . . . . . . . . . . . . . . . . . 1338.2.0.2 Electromagnetic Field of a Uniformly Moving Charged

Particle . . . . . . . . . . . . . . . . . . . . . . . . . . 1348.3 The Radiation Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

8.3.1 Poynting Vector of Radiation . . . . . . . . . . . . . . . . . . . 1458.3.1.1 Small Velocities Limit . . . . . . . . . . . . . . . . . . 1458.3.1.2 General Case . . . . . . . . . . . . . . . . . . . . . . . 146

8.3.2 Larmor Formula . . . . . . . . . . . . . . . . . . . . . . . . . . 1478.3.2.1 Particle in Linear Motion (Bremsstrahlung) . . . . . . 148

8.4 General Properties of the Radiation Field . . . . . . . . . . . . . . . . 1528.4.1 Electromagnetic Plane Waves . . . . . . . . . . . . . . . . . . . 153

8.4.1.1 Polarization . . . . . . . . . . . . . . . . . . . . . . . 1558.4.2 Plane Wave Decomposition of the General Solution . . . . . . . 1568.4.3 Energy-Momentum Tensor for the Radiation Field . . . . . . . 157

8.5 Radiation Reaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1588.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160

9 Special Topics 1619.1 Time-Independent Multipolar Expansion . . . . . . . . . . . . . . . . . 161

9.1.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1629.1.2 General Time-Independent Multipolar Expansion . . . . . . . 162

9.2 Time-Dependent Multipolar Expansion . . . . . . . . . . . . . . . . . . 1639.2.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164

9.2.1.1 Electric Dipole Radiation . . . . . . . . . . . . . . . . 164

Introduction

7

Part I

The Relativistic Description ofClassical Particles

9

Chapter 1

Special Relativity

1.1 The Minkowskian Space

In order to describe the special theory of relativity, we introduce a 4-dimensionalspace

M :=

�

x =

�

x0, x1, x2, x3

�

(1.1)

whose elements x are called 4-vectors. The scalar product (Minkowskian product)between two 4-vectors x and y is defined by

xy := x0y0 � x1y1 � x2y2 � x3y3. (1.2)

Denoting the components of the 4-vector x by xµ (µ = 0, 1, 2, 3) and similarly fory, the Minkowskian product can be written as

xy = ⌘µ⌫

xµy⌫ (1.3)

where

⌘µ⌫

=

0

B

B

@

1 0 0 0

0 �1 0 0

0 0 �1 0

0 0 0 �1

1

C

C

A

(1.4)

is known as the Minkowskian metric tensor. Two vectors x and y are called orthogonal

if xy = 0.The scalar product induces a real norm (but not positive definite) as

x2

:= xx =

�

x0

�

2 ��

x1

�

2 ��

x2

�

2 ��

x3

�

2

= ⌘µ⌫

xµx⌫ . (1.5)

11

12 CHAPTER 1. SPECIAL RELATIVITY

1.2 Lorentz Transformations

A Lorentz Transformation is defined as a linear transformation

L : M �! M (1.6)x �! x0

= Lx (1.7)

that leaves invariant the Minkowskian product, i.e.

x0y0 = (Lx) (Ly) = xy (1.8)

The more general transformation of this kind is known as the inhomogeneusLorentz transformation, and it can be represented by

x0µ= ⇤

µ

↵

x↵ + aµ. (1.9)

This transformation reprents the Poincaré group and therefore it has 10 param-eters: aµ (1 time traslation + 3 space traslation) and ⇤µ

↵

(3 space rotations + 3Lorentz boost). The Poincaré group is the fundamental symmetry of physics becauseit represents the undamental properties of spacetime, i.e. homogeneity and isotropyof 3-dimensional space and time.

As a special case, the Lorentz homogeneus transformations between a referenceframes ⌃ and a frame ⌃0 that moves with velocity ~v along the axis x1 with respectoto ⌃, can be represented by the 4x4 symmetric matrix (Lorentz boost)

⇤

µ

↵

=

0

B

B

@

� �� 0 0

�� 0 0

0 0 1 0

0 0 0 1

1

C

C

A

=

@x0µ

@x↵(1.10)

where

� = � (v) =1

p

1� �2

(1.11)

and � =

v

c

. It is straightforward to note that this is a proper orthogonal matrix, i.e.

det⇤

µ

↵

= +1. (1.12)

From the definition of a Lorentz transformation, it is clear that the norm of a4-vector is an invariant, i.e.

x02= x0x0

= xx = x2. (1.13)

1.3. COVARIANT AND CONTRAVARIANT COMPONENTS 13

1.3 Covariant and Contravariant Components

The inverse ⌘µ⌫ of the Minkowskian metric tensor is defined by the relation

⌘µ�

⌘�⌫ = �⌫µ

(1.14)

where �⌫µ

is the Kronecker’s delta function. It is said that ⌘µ⌫

are the covariant

components of the metric while ⌘µ⌫ are its contravariant componets.Using the inverse of the metric, the scalar product between vectors x and y can

be written as

xy = ⌘µ⌫

xµy⌫ = ⌘µ⌫xµ

y⌫

(1.15)

where xµ are the contravariant components of vector x while xµ

are its covariant

components that can be obtained by

xµ

= ⌘µ⌫

x⌫ . (1.16)

From this relation we conclude that

xµ

=

�

x0,�x1,�x2,�x3

�

. (1.17)

1.4 Light Cones and Causality Relations

Since the norm of a 4-vector is invariant under Lorentz transformations and it is notpositive defined, we divide the vectors according to their norm into three groups:

1. If x2 > 0, x is a time-like vector

2. If x2

= 0, x is a light-like or null vector

3. If x2 < 0, x is a space-like vector.A time-like vector x with x0 > 0 is called future directed, while if x0 < 0 it is said

past directed.All the light-like vectors begining at some point p of spacetime form the surface

of a 3-dimensional cone called the light cone. This gives the surface on which lightrays move away from or into point p, of course with speed c. All time-like vectors areinside the light cone while all space-like ones are outside.

Consider two events (or points) in spacetime labeled by the coordinates x1

=

�

x0

1

, x1

1

, x2

1

, x3

1

�

and x2

=

�

x0

2

, x1

2

, x2

2

, x3

2

�

. The spacetime interval or spacetime lenghtbetween these two events is

�s2 = (x2

� x1

)

2

=

�

x0

2

� x0

1

�

2 ��

x1

2

� x1

1

�

2 ��

x2

2

� x2

1

�

2 ��

x3

2

� x3

1

�

2

. (1.18)

If the two points are infinitesimaly close, we define the infinitesima line element(or infinitesimal spacetime interval)

ds2 =

�

dx0

�

2 ��

dx1

�

2 ��

dx2

�

2 ��

dx3

�

2

. (1.19)


Because of the definitions, the line element ds2 is an invariant under Lorentztransformations. If the two events x

1

and x2

occur at the same spatial locaton, i.e.xi

1

= xi

2

for i = 1, 2, 3, the line element is just a time interval, called the proper time

interval,

ds2 = c2d⌧2, (1.20)

and it is also an invariant under Lorentz transformations.Given two any points, the slope of the line connecting them indicates the velocity

of the signals between the points. This fact, establishes that space-like points ( anypoint outside the light cone), can not be reached from other by signals with velocitiesequal or less than the speed of ligth c. Therefore, the light cone establishes thecausality relations between points in spacetime.

1.5 Scalars, Vectors and Tensors

Depending on the properties of transformation when we make a coordinates changex ! x0, the fundamental mathematical objects are classified as scalars, vectors ortensors.

1.5.1 ScalarsA Lorentz Scalar (tensor of rank zero) is a quantity that is invariant under Lorentztransformations. Some examples of this kind of quantities are the spacetime intervalds2, the norm of any 4-vector, the minkowskian product of any two 4-vectors, theproper time interval, the proper mass of a particle, etc.

1.5.2 VectorsA 4-vector (tensor of rank one), or simply vector, V is a quantity that transform as

V0= LV (1.21)

under a Lorentz transformation L. The contravariant components of the vector,denoted by V µ, transform as

V0µ

=

@x0µ

@x⌫V ⌫ . (1.22)

As we already know, the 4-vector can also be represented in covariant components,denoted by V

µ

, that are related with the contavariant ones by means of the metrictensor,

Vµ

= ⌘µ⌫

V ⌫ . (1.23)

These components transforms as

1.6. RELATIVISTIC KINEMATICS 15

V0

µ

=

@x⌫

@x0µ

V⌫

. (1.24)

Some examples of vectors are the 4-velocity Uµ, the 4-acceleration Aµ and the4-force fµ.

1.5.3 TensorsA second order tensor T is a set of 16 components that transform as

T 0= L†TL (1.25)

where L† is the hermitian conjugate of the Lorentz transformation L. For real trans-formations the hermitian conjugate reduces to the transpose, LT . The contravariantcomponents of a second order tensor are denoted by Tµ⌫ and satisfy

T0µ⌫

=

@x0µ

@x↵@x

0⌫

@x�T↵� . (1.26)

As in the case of vectors, one can define the covariant components of the tensor,denoted by T

µ⌫

and related with the contravariant components by

Tµ⌫

= ⌘µ↵

⌘⌫�

T↵� . (1.27)

However, for tensors one can define also the mixed components Tµ

⌫

by the relation

Tµ

⌫

= ⌘⌫�

Tµ�

= ⌘µ↵T↵⌫

. (1.28)

The notion of tensor can be generalized to the concept of r-order tensor as anobject with 4

r components which, in contravariant form Tµ1µ2...µr , transform as

T0µ1µ2...µr

=

@x0µ1

@x↵1

@x0µ2

@x↵2...@x

0µ

r

@x↵r

T↵1↵2...↵r . (1.29)

Some examples of tensors are the metric ⌘µ⌫

, the electromagnetic field Fµ⌫

andthe energy-momentum tensor T

µ⌫

. In general relativity, one of the most important isthe 4-order Riemann tensor R

↵��

.

1.6 Relativistic Kinematics

Consider an inertial frame ⌃ in which a particle moves. Given the 3-vector positionof the particle, its physical (Newtonian) velocity is the 3-vector

~u = (ux

, uy

, uz

) =

d~r

dt. (1.30)

In special relativity, we generalize the concept of velocity by means of the velocity4-vector


U :=

dx

d⌧(1.31)

where x (⌧) is the position 4-vector of the particle and ⌧ is its proper time. Thecomponents of the 4-velocity can be written

Uµ

=

�

U0, U1U2, U3

�

= � (u) (c, ~u) (1.32)

where u = |~u| is the norm of the physical velocity and

� (u) =dt

d⌧=

1

q

1� u

2

c

2

. (1.33)

Since the norm of any 4-vector is an invariant, it can be calculated in any referencesystem. In particular, if we calculate the components of the 4-velocity in the referenceframe moving with the particle, we have Uµ

=

⇣

c,~0⌘

and its norm is

U2

= ⌘µ⌫

UµU⌫

= c2. (1.34)

This relation shows that every 4-velocity is a time-like vector.

1.6.1 Addition of VelocitiesNow consider another system of reference ⌃0 that moves with constant velocity withrespecto to ⌃ and both systems have parallel axes. The components of the 4-velocityof the particle in frame ⌃0 are related with the components of the 4-velocity in ⌃ bya Lorentz transformation, U 0

= LU , or in components

U 0↵=

@x0↵

@xµ

Uµ. (1.35)

In the special case in which reference frame ⌃0 moves with velocity ~v along theaxis x1 with respecto to ⌃, this relation gives the well known addition of velocitiesrule

8

>

>

>

>

>

<

>

>

>

>

>

:

ux

0=

u

x

�v

(

1� vu

x

c

2 )

uy

0=

u

y

q1� v

2

c

2

(

1� vu

x

c

2 )

uz

0=

u

z

q1� v

2

c

2

(

1� vu

x

c

2 )

(1.36)

1.6.2 4-AccelerationThe acceleration 4-vector is defined by

A =

�

A0, A1, A2, A3

�

:=

dU

d⌧. (1.37)

1.6. RELATIVISTIC KINEMATICS 17

Its components can be expressed as

A↵ = � (u)

✓

d� (u)

dt,d� (u)

dt~u+ � (u)~a

◆

(1.38)

where~a =

d~u

dt(1.39)

is the physical acceleration 3-vector. In order to calculate the norm of the 4-acceleration,we consider a reference system that moves with the particle (~u = 0), where the com-ponents reduce to A↵ = (0,~a) and the norm is cleraly

A2

= � |~a|2 . (1.40)

This relation shows that the 4-acceleration is always a spacelike vector. Evenmore,it is easy to probe that the 4-velocity and the 4-acceleration are orthogonal vectors,i.e.

UA = ⌘µ⌫

UµA⌫ = 0. (1.41)

1.6.3 Doppler Effect and AberrationConsider an electromagnetic plane wave function described by the function

� (x) = Ae±i

(

!t�~k·~r) (1.42)

where � represents an electric or magnetic field or the electromagnetic potential,! is the frequency, ~k is the propagation vector and the sign ± gives the direction ofpropagation. The quantity

' = !t� ~k · ~r (1.43)

is known as the phase of the wave and must be a relativistic invariant (i.e. invariantunder Lorentz transformations). We define the wave 4-vector as

k :=

⇣!

c,~k⌘

(1.44)

and with the help of x = (ct,~r), the phase can be written as the scalar product

' = kx (1.45)

which is obiously invariant. From its definition it is easy to see that k is a light-likevector (in vaccum),

k2 =

!2

c2��

�

�

~k�

�

�

= 0. (1.46)

The wave 4-vector in another system of reference ⌃0 is obtained by a Lorentztransformation, k0 = Lk, or in components


k0↵ =

@x0↵

@xµ

kµ. (1.47)

Considering again the special case in which reference frame ⌃0 moves with velocity~v along the axis x1 with respecto to ⌃ and we define as ↵ the angle between ~k and thex1 directions and ↵0 as the angle between ~k0 and the x1 direction, the transformationrelation gives the equations

!0= !

1� � cos↵p

1� �2

(1.48)

cos↵0=

cos↵� �

1� � cos↵. (1.49)

The first equation gives the change in frequency (Doppler effect), while the secondgives the change in direction of the wave as observed from a moving coordinate frame(aberration).

1.7 The Principle of RelativityThe Lorentz transformations permit different observers to compare their coordinates.As stated by Einstein, the principle of relativity says that physical laws must beindependent of the coordinate frame in which they are written. This is equivalentto say that it is not possible to distinguish a particular inertial reference frame fromphysical experiments. In mathematical terms, the priciple of relativity states thatphysical laws must be invariant under (inhomogeneus) Lorentz transformations.

Under this principle, we can say that fundamental equations of physics must bewritten in terms of scalars, vectors or tensors in such a way that the form of theequations is the same for all observers. Then, we say that the equations are covariant.However, it is important to note that a particular equation may not be manifestlycovariant because it has been reduced to a particular frame. Therefore, relativisticinvariance does not neccessarily imply covariance.

1.8 Scalar, Vector and Tensor FieldsParticle properties are described by scalars, vectors or tensors with a unique value(s)as noted above. However, Fields are defined to be objects that are continuous func-tions of spacetime or momenta of the particles. Therefore we classify these accordingto their properties of transformation.

1.8.1 Scalar FieldsA scalar field � (x) has the same numerical value at each physical point P for all

observers. If two observers ⌃ and ⌃0 denote the point P by x and x0 respectively, thevalue of the scalar field for both of them satisfy

1.9. SOME USEFUL RELATIONS 19

�0 (x0) = � (x) . (1.50)

1.8.2 Vector FieldsThe four values of a vector field Aµ

(x) at some physical point P viewed by twodifferent observers are related by a Lorentz transformation,

A0µ

(x0) = ⇤

µ

⌫

A⌫ (x) . (1.51)

Similarly, the covariant components of the vector field satisfy

A0

µ

(x0) = ⇤

⌫

µ

A⌫

(x) . (1.52)

It is interesting to note that the derivatives of a scalar field, denoted by

@�

@xµ

:= �,µ

(1.53)

transform like a vector field,

�0

,µ

(x0) =

@�0 (x0)

@x0µ

=

@x⌫

@x0µ

@� (x)

@x⌫= ⇤

⌫

µ

�,⌫

(x) . (1.54)

Similarly, it is easy to probe that �0,µ

(x0) = ⇤

µ

⌫

�,⌫ (x).

1.8.3 Tensor FieldsIn this case, the physical quantity at each spacetime point has an array of compo-nents. The transformation law for the contravariant components of an r�order tensorFµ1µ2...µr

(x) is

F0µ1µ2...µr

(x0) = ⇤

µ1↵1⇤

µ2↵2...⇤µ

r

↵

r

F↵1↵2...↵r

(x) . (1.55)

Here is interesting to note that the derivative of a vector field is a tensor field ofrank two

A0µ

,⌫

(x0) = ⇤

µ

↵

⇤

�

⌫

A↵,�

(x) (1.56)

as can be verfied in the same manner as for �,µ

.

1.9 Some Useful Relations

1.9.1 Differential RelationsIn last section we introduced the partial derivatives of fields. The 4-dimensionalgradient of a scalar field is denoted by

@� = @µ

� =

@�

@xµ

= �,µ

. (1.57)


The 4-dimensional divergence of a vector field is

@µ

Aµ

= @µAµ

=

@Aµ

@xµ

= Aµ

,µ

. (1.58)

The gradient vector of a scalar field is orthogonal to the 3-dimensional surfacewhere � is constant,

@µ

�dxµ

= d� = 0, (1.59)

where dxµ lies on the surface.The equation of continuity for a 4-vector field jµ is written in covariant notation

as

@µ

jµ = @0

j0 + ~r ·~j = 0. (1.60)

The covariant generalization of the curl of a vector field is the antisymmetric tensor

Fµ⌫

(x) = @µ

A⌫

(x)� @⌫

Aµ

(x) = �F⌫µ

(1.61)

which is often written as

Fµ⌫

(x) = @[µ

A⌫]

. (1.62)

The D’Alambertian is the operator defined as

⇤ = @2 = @µ@µ

=

@

@x0

@

@x0

�r2. (1.63)

From its definition as a “norm”, it is clear that this is an invariant differential operator.The wave equation for the scalar is written simply as

⇤ = 0. (1.64)

The curl of a gradient is always zero,

@µ

�,⌫

� @⌫

�,µ

= @µ

@⌫

�� @⌫

@µ

� = 0. (1.65)

The divergence of the curl is not always zero,

@⌫Fµ⌫

= @⌫ (@µ

A⌫

� @⌫

Aµ

) = @µ

@⌫

A⌫ �⇤Aµ

. (1.66)

1.9.2 Integral RelationsThe vanishing of the closed line integral

˛A

µ

dxµ

= 0 (1.67)

implies Aµ

= @µ

�. This can be proben by noting that

1.10. PROBLEMS 21

˛�,µ

dxµ

=

˛d� = 0. (1.68)

The 4-dimensional volume element is denoted

d4x = dx0dx1dx2dx3. (1.69)

Consider a 4-dimensional volume ⌃ with 3-dimensional boundary @⌃ describedby three parameters u1, u2, u3. The surface element on @⌃ can be represented by the4-vector

d�µ

:=

�

dx1dx2dx3, dx0dx2dx3, dx0dx1dx3, dx0dx1dx2

�

. (1.70)

Note that in the particular case in which the boundary @⌃ is the ordinary 3-dimensionalspace with dx0

= 0 (or x0

= constant), the surface element d�µ

is proportional to thedirection n

µ

= (1, 0, 0, 0).The Gauss’ theorem can be written, using the parameters u1, u2, u3, as

ˆ⌃

@µ

Aµd4x =

ˆ@⌃

�

�

�

�

�

�

�

�

�

A0 A1 A2 A3

@x

0

@u

!@x

1

@u

!@x

2

@u

!@x

3

@u

!

@x

0

@u

2@x

1

@u

2@x

2

@u

2@x

3

@u

2

@x

0

@u

3@x

1

@u

3@x

2

@u

3@x

3

@u

3

�

�

�

�

�

�

�

�

�

du1du2du3

=

ˆ@⌃

Aµd�µ

, (1.71)

where

d�0

=

@�

x1x2x3

�

@ (u1u2u3

)

du1du2du3

= dx1dx2dx3 (1.72)

and similarly for d�1

, d�2

and d�3

.

1.10 Problems

1. Write out the matrix that describesa.) a Galilean transformationb.) a Lorentz transformation along the y axisc.) a Lorentz transformation with velocity ~v along the x axis followed by aLorentz transformation with velocity ~u along the y axis. Does it matter in whatorder the transformations are carried out?

2. Consider two particles moving with velocities ~v1

and ~v2

with respecto to aninertial observer ⌃. Find the magnitude of the relative velocity of one of theparticles with respect to the other.


3. Consider a particle in hyperbolic motion described by the cartesian coordinates

x (t) =

q

b2 + (ct)2 (1.73)

y (t) = 0 (1.74)z (t) = 0 (1.75)

where b is a constant.a.) Find the physical velocity of the particleb.) Find the 4-velocity and the 4-acceleration of the particle

4. The Levi-Civita symbol "↵�� is completely antisymmetric with "0123 = +1.a. Show that this symbol has the same components in any reference system.b. Find the covariant components "

↵��

.c. Find the product "↵��"

↵��

.

5. Show that the symmetry (or antisymmetry) property of a tensor is preservedunder arbitrary transformations of coordinates.

6. Consider an antisymmetric tensor F↵�

= �F�↵

.a. Find the components of the dual tensor F ⇤↵� defined by

F ⇤↵�=

1

2

"↵��F��

. (1.76)

b. Show that the covariant components of the dual tensor are

F ⇤↵�

=

1

2

"↵��

F ��. (1.77)

c. Show that(F ⇤

)

⇤= �F. (1.78)

7. From geometrical optics we know that the incidence angle of a light ray in arest flat mirror is equal to the reflection angle.a. Consider a flat mirror moving with velocity ~v normal to its plane and a lightray with frequency !

i

moving with an incidence angle ✓i

. Find the frequencyand the reflection angle of the reflected light ray.b. Repeat the analysis when the mirror is moving along its plane.

Chapter 2

Relativistic Dynamics

2.1 The 4-MomentumIn order to introduce the relativistic momentum we need the concept of proper mass

of a particle m0

. It corresponds to the inertial mass of the particle measured in theparticle’s rest reference frame and it is a relativistic invariant. The 4-momentum isdefined as

P = m0

U (2.1)

where U is the 4-velocity of the particle. Note that this definition implies that thenorm of the momentum 4-vector is

P 2

= m2

0

c2 (2.2)

which is obiously invariant and shows that the 4-momentum is a timelike vector. Fromequation (1.32), the components of the 4-momentum are

Pµ

= m0

� (u) (c, ~u) . (2.3)

Hence, defining the relativistic mass of the particle as

m = m0

� (u) =m

0

p

1� �2

(2.4)

with � =

u

c

, we can write the components of the 4-momentum as

Pµ

= m (c, ~u) = (mc, ~p) (2.5)

where the physical momentum is ~p = m~u. In the limit of small velocities u ⌧ c wecan expand

� (u) = 1 +

1

2

⇣u

c

⌘

2

+O

⇣u

c

⌘

4

�

(2.6)

23

24 CHAPTER 2. RELATIVISTIC DYNAMICS

and therefore

~p = m0

� (u) ~u = m0

~u

1 +

1

2

⇣u

c

⌘

2

+O

⇣u

c

⌘

4

��

⇡ m0

~u (2.7)

which corresponds to the Newtonian momentum. On the other hand, the first com-ponent of the 4-momentum can be expanded as

P 0

= mc = m0

c

1 +

1

2

⇣u

c

⌘

2

+O

⇣u

c

⌘

4

��

⇡ m0

c+1

2

m0

u2

c. (2.8)

Thus, noting that

P 0c = mc2 =⇡ m0

c2 +1

2

m0

u2 (2.9)

and identifing the second term in the right hand side as the kinetic energy, we define

E0

= m0

c2 (2.10)

as the proper (or rest) energy of the particle and

E = mc2 (2.11)

as its relativistic energy. Therefore, the components of the 4-momentum can bewritten

Pµ

=

✓

E

c, ~p

◆

(2.12)

and its norm as

P 2

= m2

0

c2 =

E2

c2� |~p|2 (2.13)

or

E2

= m2

0

c4 + |~p|2 c2. (2.14)

2.1.1 Particles of Zero Rest MassAs is well known, in Newtonian mechanics there are no particles with zero rest mass.However, the description of special relativity makes sense when m

0

= 0. Whentaking zero rest mass, the 4-momentum becomes a null vector, P 2

= 0 (as well as the4-velocity), and thus the velocity of the particle must be equal to c. The relativisticenergy of the massless particle is

E = |~p| c. (2.15)

The most important realization of a zero rest mass particle is the photon and aswe have seen in a previous chapter, a plane electromagnetic wave is characterized bythe null 4-vector k =

⇣

!

c

,~k⌘

, which satisfies

2.2. THE EQUATION OF MOTION 25

k2 =

!2

c2��

�

�

~k�

�

�

= 0. (2.16)

As is known, many experiments involving blackbody radiation, photoelectric effect,x-rays, etc. show that the energy of a plane wave is proportional to its frequency !and that its momentum is proportional to the propagation vector ~k. In both cases,the proportionality constant has the same dimensionas and numerical value, and iscalled the Planck’s constant, ~ =

h

2⇡

,

E = ~! (2.17)~p = ~~k. (2.18)

This shows clearly that the 4-momentum associated with the plane wave is simply

P = ~k. (2.19)

2.2 The Equation of MotionThe relativistic equation of motion of a particle with proper mass m

0

is given by

f =

dP

d⌧= m

0

A (2.20)

where ⌧ is the particle’s proper time, A is the 4-acceleration and f is the force 4-vector.In components we have

fµ

=

dPµ

d⌧=

dt

d⌧

dPµ

dt= � (u)

dPµ

dt(2.21)

which can be writen

fµ

= � (u)

✓

1

c

dE

dt,d~p

dt

◆

. (2.22)

The spatial component are identified with the physical force,

~F =

d~p

dt, (2.23)

and therefore

fµ

= � (u)

✓

1

c

dE

dt, ~F

◆

. (2.24)

From the equation of motion it is straightforward that the 4-force is always or-thogonal to the 4-velocity,

fU = m0

UA = 0. (2.25)


In components this relation is

fµUµ

= ⌘µ⌫

fµU⌫

= �2 (u)

dE

dt� ~F · ~u

�

= 0 (2.26)

which gives the well known relation

dE

dt=

~F · ~u (2.27)

or in terms of the classical work

dE =

~F · d~r = dW. (2.28)

2.3 Explicit Forms of the 4-Force

From equations (2.24) and (2.27) we conclude that the 4-force components are

fµ

=

⇣

f0, ~f⌘

= � (u)

✓

1

c~F · ~u, ~F

◆

. (2.29)

This gives the condition

f0

=

1

c~f · ~u (2.30)

that does not determine the 4-force uniquely but tell us that some (or all) componentsof f are velocity dependent. Now we will consider some simple examples of relativisticforces.

2.3.1 Constant Newtonian Force

Consider a constant Newtonian force, ~F = constant. This gives the 4-force components

~f = � (u) ~F =

~Fp

1� �2

(2.31)

f0

=

1

c

~F · ~up

1� �2

. (2.32)

It is obvious that all components are increasing function of the velocity. Note theeven if we consider ~f as a constant, the component f0 would be velocity-dependent(and vice-versa).

2.3. EXPLICIT FORMS OF THE 4-FORCE 27

2.3.2 4-Force Linear in the VelocityThe simplest form of the 4-force is a linear combination of the components of the4-velocity. This can be written in the form

fµ

= �Fµ⌫

(x)U⌫ (2.33)

where � is a constant and Fµ⌫

(x) is a second rank tensor field. The condition oforthogonality fU = 0 gives the condition

fµ

Uµ

= �Fµ⌫

(x)U⌫Uµ

= 0 (2.34)

which clearly states that the tensor field F is antisymmetric, i.e.

Fµ⌫

(x) = �F⌫µ

(x) . (2.35)

2.3.3 The Lorentz ForceA particular case of a linear force is the Electromagnetic interaction. Choosing � =

q

c

with q the electric charge and defining the antisymmetric tensor in terms of the electricfield ~E = (E

1

, E2

, E3

) and the magnetic field ~B = (B1

, B2

, B3

) as

Fµ⌫

=

0

B

B

@

0 E1

E2

E3

�E1

0 �B3

B2

�E2

B3

0 �B1

�E3

�B2

B1

0

1

C

C

A

, (2.36)

the 4-force is given by

fµ

=

q

cFµ⌫

U⌫ . (2.37)

The electromagnetic tensor can be expressed also in contravariant indices as

Fµ⌫

=

0

B

B

@

0 �E1

�E2

�E3

E1

0 �B3

B2

E2

B3

0 �B1

E3

�B2

B1

0

1

C

C

A

(2.38)

or in mixed indices

Fµ

⌫

=

0

B

B

@

0 E1

E2

E3

E1

0 B3

�B2

E2

�B3

0 B1

E3

B2

�B1

0

1

C

C

A

, (2.39)

which let us write the 4-force as

fµ

=

q

cFµ

⌫

U⌫ . (2.40)

The equation of motion (2.20) gives in this case


d

d⌧

✓

m0

dxµ

d⌧

◆

=

q

cFµ

⌫

U⌫ . (2.41)

The µ = 0 component of this equation is

d

d⌧

✓

m0

dx0

d⌧

◆

=

q

cF 0

⌫

U⌫ (2.42)

dt

d⌧

d

dt

✓

m0

dt

d⌧

dx0

dt

◆

=

q

cF 0

i

U i (2.43)

� (u)d

dt(m

0

� (u) c) =q

cF 0

i

ui� (u) (2.44)

d

dt

�

mc2�

= qEi

ui (2.45)

d

dt

�

mc2�

= q ~E · ~u (2.46)

which can be written as

mc2 = q

ˆ~E · d~r. (2.47)

This equation shows that only the electric field contributes to the total energy. Onthe other hand, the spatial components of the equation of motion give

d

d⌧

✓

m0

d~r

d⌧

◆

=

q

c

⇣

~EU0

+ � (u) ~u⇥ ~B⌘

(2.48)

dt

d⌧

d

dt

✓

m0

dt

d⌧

d~r

dt

◆

=

q

c

⇣

~Ec� (u) + � (u) ~u⇥ ~B⌘

(2.49)

d

dt(m~u) = q ~E +

q

c~u⇥ ~B, (2.50)

which is the equation of motion with the Lorentz force in the right hand side.Finally it is important to remember that F

µ⌫

is a tensor and therefore under aLorentz transformation it satisfies

F0µ⌫

= ⇤

µ

↵

⇤

⌫

�

F↵� . (2.51)

This equation give us the transformation laws for the electric and magnetic fields. Forexample, the first component of the electric field satisfies

E0

1

= �⇤0

↵

⇤

1

�

F↵� . (2.52)

2.4. ANGULAR MOMENTUM 29

2.3.4 4-Force Quadratic in the VelocityConsider now a 4-force quadratic in velocities,

fµ

= Hµ⌫�

(x)U⌫

U�

(2.53)

where Hµ⌫�

(x) is a tensor field of order three. In order to satisfy the conditionfU = 0, or in components

fµUµ

= Hµ⌫�

(x)Uµ

U⌫

U�

= 0, (2.54)

this tensor must be anisymmetric in any two indices and therefore it has only 24independent components. However, we can take as a particular case a tensor anti-symmetric in all its indices, i.e.

Hµ⌫�

= �H⌫µ�

= H⌫�µ

= �H�⌫µ

= H�µ⌫

= �Hµ�⌫ . (2.55)

Hence, it can be written in a compact form as

Hµ⌫�

= ✏µ⌫�⇢H⇢

(2.56)

where ✏µ⌫�⇢ is the completly antisymmetric symbol with ✏0123 = +1 and H⇢

is calledthe dual of Hµ⌫�.

2.4 Angular MomentumIn classical mechanics we can write the equations of motion in terms of the angularmomentum, defined as ~l = ~r ⇥ ~p. Here we have to use the generalization of the crossproduct defining the second order antisymmetric tensor lµ⌫ in terms of the 4-positionand the 4-momentum as

lµ⌫ = xµP ⌫ � x⌫Pµ. (2.57)

It is related with the components of the 3-dimensional angular momentum ~l =(lx

, ly

, lz

) through the spatial components

l12 = �l21 = lz

(2.58)l13 = �l31 = �l

y

(2.59)l23 = �l32 = l

x

. (2.60)

The rest of the components are

l0j = x0P j � xjP 0

= ctmuj � xjmc = mct

✓

uj � xj

t

◆

. (2.61)

These components do not have a simple physical meaning. However, note that l0jvanishes for linear motion

�

xj

= ujt�

. Therefore, we can write


xj

= ujt� l0j

mc(2.62)

to interpret l0j as the deviation from the linear motion of the particle.From the angular momentum lµ⌫ we can construct different scalars. In particular

we define

lµ⌫ lµ⌫

= l0j l0j

+ li0li0

+ lij lij

+ ljilji

(2.63)

or

lµ⌫ lµ⌫

= 2l0j l0j

+ 2lij lij

(2.64)

with i < j. This gives

lµ⌫ lµ⌫

= �2

�

l0j�

2

+ 2

�

�

�

~l�

�

�

2

= 2

h

x2P 2 � (xP )

2

i

. (2.65)

The relativistic torque is defined in a similar fashion using the 4-force,

Nµ⌫

= xµf⌫ � x⌫fµ. (2.66)

Thus, the equations of motion can be written as

Nµ⌫

=

dlµ⌫

d⌧. (2.67)

2.5 Problems1. Probe equation (2.48).

2. Probe equation (2.65).

3. Using equation (2.51), find the electric and magnetic fields for a charge Q thatmoves with constant velocity ~u along the positive x axis.

4. Find the components of the dual tensor F ⇤ of the electromagnetic field tensor(2.36) to show that it is obtained by replacing E by +B and B by �E. Writethem in matricial form.

5. From the definition of the electromagnetic field tensor (2.36), define two Lorentzinvariants. (Hint: consider also the dual tensor F ⇤).

6. In an inertial frame ⌃ the electric field ~E and the magnetic field ~B are neitherparallel nor perpendicular ar a particular spacetime point (supposse that theangle between the two vectors is ↵). Show that in a different inertial system ⌃

0,moving relative to ⌃ with a velocity ~v given by

~v

1 +

v

2

c

2

=

~E ⇥ ~B�

�

�

~B�

�

�

2

+

�

�

�

~E�

�

�

2

, (2.68)

2.5. PROBLEMS 31

the fields ~E0 and ~B0 are parallel at that point. Is there a frame in which thetwo vectors are perpendicular?.

7. A charge q is released from rest at the origin in the presence of a uniform electricfield ~E = E

0

z and a uniform magnetic field ~B = B0

x. Determine the trajectoryof the particle by transforming to a system in which ~E = 0, finding the pathin that system and then transforming back to the original system. AssumeE

0

< B0

. (Hint: a Lorentz boost in an adequate direction is enough!).


Part II

Relativistic ElectromagneticField Theory

33

Chapter 3

Maxwell Equations

Faraday and Maxwell propose the description of the interaction between chargedparticles by using the intermediary electromagnetic field. We will assume, as a startingpoint, that this field is produced by the charged particles and that its physical realitycan be probed by measuring the acceleration that it produces on charged particles.

It is well known that charged particles at rest are acted on by a force ~FE

= q ~Ewhere q is an attribute of the particle called charge, and ~E is the electric field. Thisequation gives the interpretation of the electric field as the electric force per unitcharge1.

If the charged particle is moving with velocity ~v, it feels a total force given by

~F = q⇣

~E + ~v ⇥ ~B⌘

(3.1)

where ~B is the magnetic field. This equation known as the Lorentz Force may betaken as the defintion of the electric and magnetic fields.

From the conservation of energy, it is known that the electric field is conservative,i.e.

~r⇥ ~E = 0 (3.2)

or equivalently, ~E = �~r� with � an scalar function called the electrostatic or scalar

potential. We also know that the magnetic field does not do any work (because themagnetic force is a cross product between velocity and field).

Empirically, charges in free space and at rest produce an electric field given by

~E =

1

4⇡"0

ˆ⇢

r3~rd3x, (3.3)

where "0

is called the permitivity of free space. From this relation and applying Gauss’law we obtain the differential relation

1This is similar to the interpretation of the gravitational field ~g as the gravitational force per unitmass.

35

36 CHAPTER 3. MAXWELL EQUATIONS

~r · ~E =

⇢

"0

. (3.4)

Similarly, stationary currents produce a magnetic field given by

~B =

µ0

4⇡

ˆ ~j ⇥ ~r

r3d3x, (3.5)

where µ0

is called the permeability of free space and ~j is the current density. Thisequation leads to the differential relation

~r⇥ ~B = µ0

~j. (3.6)

Finally, in agreement with observation, the magnetic field lines are always closed.This fact can be written in differential form as

~r · ~B = 0. (3.7)

However, our study of special relativity shows that ~E and ~B are different facesof the same thing, the electromagnetic field F

µ⌫

. Therefore, it is not surprising tofind some relations between these fields. In fact, in 1831 Michael Faraday reporteda series of experiments in which a changing magnetic field induces an electric field.Faraday concluded that his results can be summarized by considering a closed curveC and the enclosed surface S in the integral equation

˛C~E · d~l = � @

@t

ˆS~B.d~a (3.8)

that, applying Stoke’s theorem, gives the differential relation

~r⇥ ~E = �@~B

@t. (3.9)

Note that this equation reduces to the relation (3.2) when the magnetic field isstatic.

Considering now the current crossing a closed surface S, we introduce the currentdensity ~j by the relation

I =

˛S~j · d~a. (3.10)

Applying Gauss’ theorem we obtain˛S~j · d~a =

ˆV

⇣

~r ·~j⌘

dV (3.11)

where V is the volume enclosed by S. Because of the empirical conservation of the

charge, whatever flows out through the surface must come at the expense of theremaining inside,

3.1. THE PROBLEM AND THE SOLUTION 37

ˆV

⇣

~r ·~j⌘

dV = � d

dt

ˆV⇢dV = �

ˆV

@⇢

@tdV. (3.12)

Note that the minus sign reflects the fact that an outward flow decreases the chargeleft in V. Since this relation applies to any volume, we obtain the continuity equation

@⇢

@t+

~r ·~j = 0. (3.13)

3.1 The problem and the solutionSo far, we have encountered the equations that represent the state of electromagnetictheory when Maxwell began his work,

~r · ~E =

⇢

"0

(3.14)

~r · ~B = 0 (3.15)

~r⇥ ~E = �@~B

@t(3.16)

~r⇥ ~B = µ0

~j. (3.17)

Maxwell notice a fatal inconsistency in these equations that has to do with therule that the divergence of a curl is always zero. Applying the divergence to equation(3.16) works fine,

~r ·⇣

~r⇥ ~E⌘

= �~r · @~B

@t= � @

@t

⇣

~r · ~B⌘

= 0 (3.18)

which is zero by equation (5.10). However, when applying the divergence to equation(3.17) we get

~r ·⇣

~r⇥ ~B⌘

= µ0

~r ·~j 6= 0 (3.19)

and it is obvious that the right hand side is not neccesarily zero. In order to solvethis problem, Maxwell consider the continuity equation (3.13) to obtain

~r ·~j = �@⇢@t

= � @

@t

⇣

"0

~r · ~E⌘

= �~r ·

"0

@ ~E

@t

!

. (3.20)

Therefore, the Ampere’s law must be corrected to be

~r⇥ ~B = µ0

~j + µ0

"0

@ ~E

@t. (3.21)

The new term is hard to detect in ordinary electromagnetic experiments, whereit must compete for recognition with ~j, and that’s why Faraday and others neverdiscovered it in the laboratory. It can be stated by saying that a changing electricfield induces a magnetic field.


3.1.1 Magnetic MonopolesIn vacuum, Maxwell’s equations have a nice symmetry,

~r · ~E = 0 (3.22)~r · ~B = 0 (3.23)

~r⇥ ~E = �@~B

@t(3.24)

~r⇥ ~B = µ0

"0

@ ~E

@t. (3.25)

However, this symmetry is spoiled by the electric charge term in Gauss’s law andthe current term in Ampere’s law. In order to preserve the symmetry, some people hasstudied the possible existence of magnetic charge. Under this supposition, Maxwell’sequations become

~r · ~E =

⇢e

"0

(3.26)

~r · ~B = µ0

⇢m

(3.27)

~r⇥ ~E = �µ0

~jm

� @ ~B

@t(3.28)

~r⇥ ~B = �µ0

~je

+ µ0

"0

@ ~E

@t, (3.29)

where ⇢m

represents the density of magnetic charge and ~jm

would be the current ofmagnetic charge. In this situation, both charges, electric and magnetic, would beconserved

@⇢m

@t+

~r ·~jm

= 0 (3.30)

@⇢e

@t+

~r ·~je

= 0. (3.31)

However, there is no proof of the existence of magnetic charges (see [7] for abibliography on the subject). Experimental facts show that ⇢

m

and ~Jm

are zeroeverywhere. For example, magnetic multipole expansions have no monopole termand magnetic dipoles (magnets) consist of current loops and not by separated northand south poles.

3.2 Maxwell’s Equations in MatterWhen working with materials that present electric and/or magnetic polarization,Maxwell’s equations must be rewritten. For example, in the static case, inside polar-

3.2. MAXWELL’S EQUATIONS IN MATTER 39

ized matter there will be an accumulation of bound charge. The electric polarization~P is related to the bound charge through

⇢b

= �~r · ~P . (3.32)

Similarly, a magnetic polarization or magnetization

~M is related to the bund cur-

rent by

~jb

=

~r⇥ ~M. (3.33)

When considering the non-static case there is one more thing to take into account.Any change in the electric polarization ~P involves a flow of bound charge, which givesrise to a polarization current density that can be written as

~jp

=

@ ~P

@t. (3.34)

It is important to note that ~jp

has nothing to do with ~jb

because the first one isthe result of linear motion of bound charge while the second one is associated with themagnetization and therefore it involves the spin and the orbital motion of electrons.Note that equation () gives rise to a continuity equation,

~r ·~jp

=

~r · @~P

@t(3.35)

~r ·~jp

=

@⇣

~r · ~P⌘

@t(3.36)

~r ·~jp

= �@⇢b@t

. (3.37)

With these results in mind we will rewrite Maxwell’s equations. The first step isto write the total charge density as

⇢ = ⇢f

+ ⇢b

= ⇢f

� ~r · ~P (3.38)

where ⇢f

represents the free charge. Therefore, Gauss’s law gives

~r · ~E =

1

"0

h

⇢f

� ~r · ~Pi

(3.39)

or

~r · ~D = ⇢f

(3.40)

where we defined the electric displacement as

~D = "0

~E +

~P . (3.41)

Likewise, the total current density can be written


~j = ~jf

+

~jb

+

~jp

(3.42)

~j = ~jf

+

~r⇥ ~M +

@ ~P

@t(3.43)

and thus, Ampere’s law becomes

~r⇥ ~B = µ0

~jf

+

~r⇥ ~M +

@ ~P

@t

!

+ µ0

"0

@ ~E

@t(3.44)

or

~r⇥ ~H =

~jf

+

@ ~D

@t(3.45)

where we defined the vector2

~H =

1

µ0

~B � ~M. (3.46)

In conclusion, inside matter, Maxwell’s equations can be written as

~r · ~D = ⇢f

(3.47)~r · ~B = 0 (3.48)

~r⇥ ~E = �@~B

@t(3.49)

~r⇥ ~H =

~jf

+

@ ~D

@t. (3.50)

3.3 Electromagnetic Waves

In regions of space where there are no sources, Maxwell’s equations are

~r · ~E = 0 (3.51)~r · ~B = 0 (3.52)

~r⇥ ~E = �@~B

@t(3.53)

~r⇥ ~B = µ0

"0

@ ~E

@t. (3.54)

2Some authors call ~H the magnetic field and change the name of ~B to “flux induction” or “magneticinduction”. However, for us the fundamental quantity is ~B and therefore we will call it always themagnetic filed while quantity ~Hwill be called simply “H”.

3.3. ELECTROMAGNETIC WAVES 41

Taking the curl of the third equation gives

~r⇥⇣

~r⇥ ~E⌘

= �~r⇥ @ ~B

@t(3.55)

and considering the vector identity

~r⇥⇣

~r⇥ ~E⌘

=

~r⇣

~r · ~E⌘

�r2 ~E, (3.56)

we have

~r⇣

~r · ~E⌘

�r2 ~E = � @

@t

⇣

~r⇥ ~B⌘

. (3.57)

Using the first and fourth of the Maxwell’s equations we obtain the wave equationfor the electric field,

r2 ~E = µ0

"0

@2 ~E

@t2. (3.58)

Similarly, from Maxwell’s equations is straightforward to obtain th wave equationfor the magnetic field,

r2 ~B = µ0

"0

@2 ~B

@t2. (3.59)

In both equations it is clear that the propagation speed of the waves is given bythe paramters "

0

and µ0

by the relation

c =1

p"0

µ0

. (3.60)

3.3.1 Monochromatic Plane WavesSinusoidal waves with one frequency ! are called monochromatic. A general solutionof the wave equations for the electric and magnetic fields are the plane waves

~E (t,~r) =

~E0

ei(~

k·~r�!t) (3.61)

~B (t,~r) =

~B0

ei(~

k·~r�!t) (3.62)

where ~E0

and ~B0

are two constant vectors. Replacing into the wave equations weobtain the condition

�

�

�

~k�

�

�

2

=

!2

c2. (3.63)

However, the ~E and ~B fields are not independent because they must obey Maxwell’sequations. For example replacing the electric wave solution in Gauss’ law we obtain


~r · ~E = 0

~r · ~E0

ei(~

k·~r�!t)

= 0

~E0

· ~rei(~

k·~r�!t)

= 0

i ~E0

· ~kei(~k·~r�!t) = 0 (3.64)

which gives the condition

~E0

· ~k = 0. (3.65)

With a similar procedure we obtain for the magnetic field

~B0

· ~k = 0. (3.66)

These two equations mean that the electric and magnetic fields are both orthogonalto the direction of propagation of the wave ~k. We conclude that electromagnetic wavesare transverse waves.

Using Faraday’s law, the wave solution gives

~r⇥ ~E = �@~B

@t

~r⇥h

~E0

ei(~

k·~r�!t)

i

= � @

@t

h

~B0

ei(~

k·~r�!t)

i

i~k ⇥ ~E0

ei(~

k·~r�!t)

= i! ~B0

ei(~

k·~r�!t)

~k ⇥ ~E0

= ! ~B0

. (3.67)

This equation shows that the three vectors ~k, ~E and ~B are mutually orthogonal.Taking the norm of this equation we obtain the relation between the magnitudes

�

�

�

~k�

�

�

�

�

�

~E�

�

�

= !�

�

�

~B�

�

�

(3.68)

�

�

�

~E�

�

�

= c�

�

�

~B�

�

�

. (3.69)

3.3.1.1 Polarization of Plane Waves

Because of relations (3.67) and (3.69), the fields for a plane wave moving in thedirection n

3

can be expressed as

~E (t,~r) = (E01

n1

+ E02

n2

) ei(~

k·~r�!t) (3.70)

~B (t,~r) =

1

c

⇣

n3

⇥ ~E⌘

(3.71)


where E01

and E02

are, in general, complex amplitudes (because the fields indifferent dirctions may have different phases).

The Polarization of a plane wave describes the direction, magnitude and phase ofthe electric part of the wave. Thus, we have some special cases of polarization:

1. Linear Polarization. When E01

and E02

have the same phase (and maybedifferent magnitudes). We define the polarization vector with a magnitude E =

p

E2

01

+ E2

02

and making an angle ✓ = tan

�1

⇣

E02E01

⌘

with respect to the direction n1

.It is convenient to choose coordinates so that E

02

= 0.2. Elliptical Polarization. When E

01

and E02

have different phases and differ-ent magnitudes. The polarization vector traces out an ellipse in the plane defined byn1

and n2

.3. Circular Polarization. Corresponds to a special case of the elliptical po-

larization in which E01

and E02

have a difference of phase of exactly ⇡

2

and themagnitudes are equal. Since ei

⇡

2= i, the electric field takes the form

~E (t,~r) =

E01p2

(n1

± in2

) . (3.72)

3.3.2 Propagation in Linear MediaInside matter, in regions where there is no free charges or currents, the Maxwell’sequations can be written

~r · ~D = 0 (3.73)~r · ~B = 0 (3.74)

~r⇥ ~E = �@~B

@t(3.75)

~r⇥ ~H =

@ ~D

@t. (3.76)

If the media is linear the electric displacement is

~D = "0

~E +

~P = " ~E (3.77)

with ~P the polarization vector and " the electric permitivity of the material. Sim-ilarly

~H =

1

µ0

~B +

~M =

1

µ~B (3.78)

with µ the magnetic permeability of the material. If the medium is homogeneous

(i.e. the parameters " and µ that characterize the media do not vary from point topoint), Maxwell’s equations become


~r · ~E = 0 (3.79)~r · ~B = 0 (3.80)

~r⇥ ~E = �@~B

@t(3.81)

~r⇥ ~B = µ"@ ~E

@t. (3.82)

Note that these equations are completely similar to Maxwell’s vacuum equationswith the new parameters µ and " instead of µ

0

and "0

. Therefore, electromagneticwaves appear in a similar form as in vacuum but they propagate in the linear andhomogeneus medium with speed

v =

1

p"µ

=

c

n(3.83)

wheren =

p"µ

p"0

µ0

=

c

v(3.84)

is called the index of refraction of the material.

3.3.2.1 Reflection and Transmission at Normal Incidence

Consider a plane wave with frequency ! moving in a medium I in the z direction,polarized in the x direction and approaching the plane xy which forms the boundarywith medium II. The electric and magnetic fields are given by

~EI

= E0

e�i(!t�k

I

z)x (3.85)

~BI

= B0

e�i(!t�k

I

z)y =

E0

vI

e�i(!t�k

I

z)y (3.86)

where vI

is the propagation speed in medium I. When the wave arrives to theinterface, it gives rise to a reflected wave described by the fields

~ER

= E0R

e�i(!t+k

I

z)x (3.87)

~BR

= �B0R

e�i(!t+k

I

z)y = �E0R

vI

e�i(!t+k

I

z)y (3.88)

which travels back to the left in medium I (that is represented by the minus sign inthe argument of the exponential function). Note that the minus sign in ~B

R

is requiredby the ortoghonality relations between the fields and the direction of propagation.

However, the incident wave also gives rise to a transmitted wave into medium II,described by


~ET

= E0T

e�i(!t�k

II

z)x (3.89)

~BT

= �B0T

e�i(!t�k

II

z)y =

E0T

vII

e�i(!t�k

II

z)y (3.90)

where vII

is the propagation speed in medium II.

3.3.2.2 Reflection and Transmission at Oblique Incidence

Now we will consider the general case of oblique incidence, in which the incomingwave meets the boundary at an arbitrary angle of incidence ✓

I

. Consider the incidentplane wave with frequency ! moving in medium I and described by the fields

~EI

=

~E0

e�i

(

!t�~kI

·~r) (3.91)

~BI

=

1

v1

⇣

ˆkI

⇥ ~EI

⌘

. (3.92)

It gives rise to a reflected wave described by the fields

~ER

=

~E0R

e�i

(

!t�~kR

·~r) (3.93)

~BR

=

1

v1

⇣

ˆkR

⇥ ~ER

⌘

. (3.94)

and to a transmitted wave into medium II, described by

~ET

=

~E0T

e�i

(

!t�~kT

·~r) (3.95)

~BT

=

1

v2

⇣

ˆkT

⇥ ~ET

⌘

. (3.96)

Note that all three waves have the same frequency ! becuase it is determined bythe source of the wave. Thus the three wave numbers are related with the velocity ofpropagation and the frequency by

�

�

�

~k�

�

�

I

v1

=

�

�

�

~k�

�

�

R

v1

=

�

�

�

~k�

�

�

T

v2

= ! (3.97)

or equivalently�

�

�

~k�

�

�

I

=

�

�

�

~k�

�

�

R

=

v2

v1

�

�

�

~k�

�

�

T

=

n1

n2

�

�

�

~k�

�

�

T

(3.98)

with n1

and n2

the indices of refraction in each region. The phase of the plane wavemust be the same at every point in the interface (i.e. z = 0). Hence, we obtain theconditions

~kI

· ~r =

~kR

· ~r =

~kT

· ~r (3.99)


with z = 0, or explicitly in cartesian coordinates,

kIx

x+ kIy

y = kRx

x+ kRy

y = kTx

x+ kTy

y (3.100)

for all values of x and y. This equation holds for the components separately becausetaking x = 0 gives

kIy

= kRy

= kTy

(3.101)

while taking y = 0 gives

kIx

= kRx

= kTx

. (3.102)

These conditions mean that if we orient our axes so that ~kI

lies in the xz planes(i.e. k

Iy

= 0), the vectors ~kR

and ~kT

are in the same plane. This result can be statedas

“The incident, reflected and transmitted wave vectors form a plane (plane

of incidence) which also includes the normal to the surface”.

Equation (3.102) also implies that�

�

�

~kI

�

�

�

sin ✓I

=

�

�

�

~kR

�

�

�

sin ✓R

=

�

�

�

~kT

�

�

�

sin ✓T

(3.103)

where ✓I

is the angle of incidence, ✓R

is the angle of reflection, and ✓T

is the angle of

refraction (remember that all angles are measured with respect to the normal). Usingequation (3.98) we conclude the law of reflection,

“The angle of incidence is equal to the angle of reflection”,

✓I

= ✓R

. (3.104)

Similarly, we conclude the law of refraction, or Snell’s law,

sin ✓T

sin ✓I

=

n1

n2

. (3.105)

3.4 The Electromagnetic Field Equations

The complete theory of electromagnetism is condensed in Maxwell’s equations. Inorder to write them in a convariant language, we will choose some special units.First, we will use units in which the magnetic field has the same dimensions as theelectric field. This is acomplished via equation (3.69) i.e. we will rescale the magneticfield by a c factor. Secondly, we will put "

0

= 1 and therefore µ0

"0

= µ0

=

1

c

2 . Thisgives the Maxwell’s equations

3.5. ELECTROMAGNETIC POTENTIALS 47

~r · ~E = ⇢ (Gauss’s law) (3.106)~r · ~B = 0 (no magnetic monopoles) (3.107)

~r⇥ ~E = �1

c

@ ~B

@t(Faraday’s law) (3.108)

~r⇥ ~B =

1

c~j +

1

c

@ ~E

@t(Ampere’s law with Maxwell’s term) (3.109)

together with the Lorentz force,

~F = q

✓

~E +

1

c~v ⇥ ~B

◆

, (3.110)

and the continuity equation

@⇢

@t+

~r ·~j = 0. (3.111)

3.5 Electromagnetic PotentialsNow we will present the representation of the fields in terms of potentials. In electro-statics, the electric field is conservative and we write it in terms of the scalar potential.In electrodynamics this is no longer possible because the curl of ~E is nonzero. How-ever, equation (3.107) tell us that the magnetic potential is divergenceless, so we canwrite it in terms of a potential vector as

~B =

~r⇥ ~A (3.112)

becuase the divergence of a curl is always zero. Using this relation in Faraday’s lawwe have

~r⇥ ~E = �1

c

@

@t

⇣

~r⇥ ~A⌘

(3.113)

~r⇥ ~E = �1

c~r⇥ @ ~A

@t(3.114)

~r⇥

~E +

1

c

@ ~A

@t

!

= 0. (3.115)

Since the curl of the term in the parenthesis is always zero,we can write it in termsof the gradient of a scalar potential �,

~E +

1

c

@ ~A

@t= �~r� (3.116)

or


~E = �~r�� 1

c

@ ~A

@t. (3.117)

The potential representation of the fields automatically fulfills the two homoge-neous Maxwell equations (3.107) and (3.108). Putting expression (3.117) in Gauss’slaw we obtain

~r ·

~r�+

1

c

@ ~A

@t

!

= �⇢ (3.118)

r2�+

1

c

@

@t

⇣

~r · ~A⌘

= �⇢. (3.119)

Finally, replacing equations (3.112) and (3.117) in Ampere/Maxwell’s law we ob-tain the equation

~r⇥⇣

~r⇥ ~A⌘

=

1

c~j � 1

c

@

@t

~r�+

1

c

@ ~A

@t

!

(3.120)

~r⇥⇣

~r⇥ ~A⌘

=

1

c~j � 1

c~r✓

@�

@t

◆

� 1

c2@2 ~A

@t2. (3.121)

Using the vector identity ~r⇥⇣

~r⇥ ~A⌘

=

~r⇣

~r · ~A⌘

�r2 ~A gives

~r⇣

~r · ~A⌘

�r2 ~A =

1

c~j � 1

c~r✓

@�

@t

◆

� 1

c2@2 ~A

@t2. (3.122)

Note that the system of four equations to obtain the fields ~E (t, ~x) and ~B (t, ~x)given the sources ⇢ (t, ~x) and ~j (t, ~x), has been simplified to a system of two equationsto obtain the potentials � (t, ~x) and ~A (t, ~x). However, the simplification obtained bythe introduction of the potentials must be paid for by the fact that � and ~A are notunique for a given electromagnetic field.

3.6 Covariant Form of Maxwell-Lorentz Equations

We define the 4-vector potential in terms of � and ~A as

Aµ

=

⇣

�, ~A⌘

(3.123)

or equivalently

Aµ

=

⇣

�,� ~A⌘

(3.124)

so the electromagnetic field is given by

Fµ⌫

= @µ

A⌫

� @⌫

Aµ

. (3.125)

3.6. COVARIANT FORM OF MAXWELL-LORENTZ EQUATIONS 49

which is obviously an antisymmetric tensor. We will postulate that the total charge Q,or equivalently the equation of continuity, is invariant under Lorentz transformations.We define the 4-current density jµ by

jµ := ⇢0

Uµ (3.126)

with ⇢0

the rest charge density which is an invariant quantity. We can write also

jµ = ⇢0

dxµ

d⌧= ⇢

0

dt

d⌧

dxµ

dt= ⇢

dxµ

dt(3.127)

where

⇢ = ⇢0

dt

d⌧= ⇢

0

� (u) (3.128)

is the actual charge density. Then, the components of the 4-current density can bewritten

jµ = (⇢c, ⇢~v) =⇣

⇢c,~j⌘

. (3.129)

The norm of this 4-vector is the invariant

j2 = ⌘µ⌫

jµj⌫ = ⇢20

(3.130)

and the equation of continuity can now be written as

@µ

jµ =

@⇢

@t+

~r ·~j = 0, (3.131)

which is simply the 4-dimensional divergence of the current vector.The Maxwell equations with sources (Gauss’s law and Ampere/Maxwell’s law) are

written in a covariant form as

@⌫

Fµ⌫

= �1

cjµ. (3.132)

This relation also includes the continuity equation because any antisymmetrictensor satisfies @

µ

@⌫

Fµ⌫

= 0 and thus

@µ

@⌫

Fµ⌫

= 0 = �1

c@µ

jµ. (3.133)

The homogeneous Maxwell equations are automatically satisfied because any an-tisymmetric tensor F

µ⌫

satisfies the identity

Fµ⌫,⇢

+ F⌫⇢,µ

+ F⇢µ,⌫

= 0 (3.134)

or in terms of the dual tensor,

F ⇤µ⌫

=

1

2

✏µ⌫⇢�

F ⇢�, (3.135)


they can be written in the simple form

@µ

F ⇤µ⌫= 0. (3.136)

To show that this identity is true note that

@µ

F ⇤µ⌫=

1

2

@µ

(✏µ⌫⇢�F⇢�

) (3.137)

@µ

F ⇤µ⌫=

1

2

@µ

(✏µ⌫⇢�@⇢

A�

� ✏µ⌫⇢�@�

A⇢

) (3.138)

@µ

F ⇤µ⌫= @

µ

(✏µ⌫⇢�@⇢

A�

) (3.139)

@µ

F ⇤µ⌫= ✏µ⌫⇢�@

µ

@⇢

A�

= 0, (3.140)where in the last term we use the fact that @

µ

@⇢

A�

is symmetric in the indices µ⇢while ✏µ⌫⇢� is completely antisymmetric.

From its definition, the dual tensor is obtained by replacing Ei

by Bi

and Bi

by�E

i

in the field tensor. thus, in matricial notation this tensor is

F ⇤µ⌫

=

0

B

B

@

0 B1

B2

B3

�B1

0 E3

�E2

�B2

�E3

0 E1

�B3

E2

�E1

0

1

C

C

A

(3.141)

or equivalently

F ⇤µ⌫=

0

B

B

@

0 �B1

�B2

�B3

B1

0 E3

�E2

B2

�E3

0 E1

B3

E2

�E1

0

1

C

C

A

. (3.142)

3.7 Gauge TransformationsAs we have mentioned, potentials do not determine uniquely the fields because atransformation of the potential 4-vector of the form

Aµ

! Ãµ

= Aµ

+

@⇤

@xµ

(3.143)

with ⇤ (x) an arbitrary scalar function, does not affect the components of the fieldtensor F

µ⌫

,

˜Fµ⌫

= @µ

Ã⌫

� @⌫

Ãµ

(3.144)˜Fµ⌫

= @µ

A⌫

+ @µ

@⌫

⇤� @⌫

Aµ

� @⌫

@µ

⇤ (3.145)˜Fµ⌫

= @µ

A⌫

� @⌫

Aµ

(3.146)˜Fµ⌫

= Fµ⌫

. (3.147)

3.7. GAUGE TRANSFORMATIONS 51

Equation (3.143) is called a gauge transformation and the quantities invariantunder this kind of transformation are called gauge invariants, and only those willhave a direct experimental significance.

3.7.1 Lorentz GaugeThe freedom given by gauge transformations can be used to simplify the field equa-tions by imposing some specific condition. For example, imposing the Lorentz gauge

condition

@⌫

A⌫ = 0, (3.148)

the field equations (3.132) take the simple and interesting form

@⌫

Fµ⌫

= @⌫

(@µA⌫ � @⌫Aµ

) = �1

cjµ (3.149)

@⌫

@⌫Aµ

=

1

cjµ (3.150)

⇤Aµ

=

1

cjµ (3.151)

which is just a wave equation with sources for the vector field Aµ

. Note that thechoice of this gauge is made by adding to A

µ

a term ⇤ such that @µ

⇤

µ

= �@µ

Aµ.However, it is important to note that even with the Lorentz gauge imposed, the fieldis not uniquely defined because you can add to the potential a divergence @

µ

' of anyfunction ' solution of the equation ⇤' = 0, and the fields do not change.

In the Lorentz gauge, both the potential and the field satisfy a wave equation.This result is independent of coordinates and exhibits the causal propagation of thefields or potentials at the speed of light.

3.7.2 The Coulomb or Transverse GaugeOther important gauge is the Coulomb gauge, also known as transverse gauge, thatis defined by the condition

~r · ~A = 0. (3.152)

The field equations in terms of the potentials (3.119) and (3.122) become

r2� = �⇢ (3.153)

and

r2 ~A� 1

c2@2 ~A

@t2= �1

c~j +

1

c~r✓

@�

@t

◆

. (3.154)

The potential � is the well known solution


� (~r) =1

4⇡

ˆ⇢ (~r

0

)

|~r � ~r0

|d3x

0

(3.155)

which comes from the Green’s function

G (~r,~r0

) = � 1

4⇡ |~r � ~r0

| (3.156)

that solves the point-source differential equation

r2G (~r,~r0

) = � (~r � ~r0

) . (3.157)

However, function � (~r) represents an instantaneous potential. This means that achange in the charge distribution modifies instantly the electric potnetial throughoutall space. a potential that

3.8 Lorentz InvariantsUsing the Minkowskian product, we can define many Lorentz invariants from the 4-vector A

µ

and the tensor Fµ⌫

. Two important invariant quantities that contain onlythe fields (and therefore are also gauge invariants) are

Fµ⌫Fµ⌫

= 2

�

�

�

~B�

�

�

2

��

�

�

~E�

�

�

2

�

(3.158)

and✏µ⌫⇢�F

µ⌫

F⇢�

= 2F ⇤⇢�F⇢�

= �8

~E · ~B. (3.159)

These invariant will be important in following chapters.

3.9 Problems1. Obtain equation (3.4) using Gauss’ law.

2. Obtain equation (3.6).

3. Probe that the divergence of the curl of any vector is always zero, i.e. ~r ·⇣

~r⇥ ~C⌘

= 0 for any vector ~C.

4. Assume that there is a Coulomb law for magnetic charges in the form

~F =

µ0

4⇡

Qm

qm

r2r. (3.160)

Find the force that the magnetic charge qm

feels when moving with velocity ~vthrough electric and magnetic fields ~E and ~B. (Hint: See [12])

5. Consider a magnetic charge qm

passing through a loop of wire without resistanceand self inductance L. Find the current induced in the loop. (Hint: See [4])

3.9. PROBLEMS 53

6. Show that equation (3.125) gives the same expression as equation (2.36).

7. Show that equations (3.132) and (3.136) correspond to Maxwell’s equations.

8. Using the definition of the field tensor and its dual, probe equations (3.158) and(3.159).


Chapter 4

Lagrangian and HamiltonianFormulation

4.1 Equation of Motion in Lagrangian FormIn order to write a covariant Hamilton’s principle for a single particle we define theaction integral using the proper time as

S =

ˆ⌧2

⌧1

L [xµ

(⌧) , xµ

(⌧) , ⌧ ] d⌧ (4.1)

where xµ

=

dx

µ

d⌧

= Uµ and L is a functional such that the true trajectory of theparticle xµ

(⌧) gives an extremum of the action,

�S = 0. (4.2)

As always, the variation � is defined so that the end points xµ

(⌧1

) and xµ

(⌧2

) arefixed. However, this time we also have the extra condition

U2

(⌧) = c2. (4.3)

The lagrangian is a Lorentz scalar and it is invariant under reparameterizations.Hence, in order to obtain the equations of motion and include the condition (4.3),we will use another invariant parameter � instead of the proper time. This gives thevariational principle

�

ˆ�2

�1

L [xµ

(�) , vµ (�) ,�] d� = 0 (4.4)

where

vµ =

dxµ

d�=

d⌧

d�

dxµ

d⌧=

d⌧

d�Uµ. (4.5)

55

56 CHAPTER 4. LAGRANGIAN AND HAMILTONIAN FORMULATION

The norm of this 4-vector is

v2 =

✓

d⌧

d�

◆

2

U2

=

✓

d⌧

d�

◆

2

c2 (4.6)

which gives✓

d⌧

d�

◆

2

=

⇣v

c

⌘

2

(4.7)

and therefore we have

vµ =

�

�

�

v

c

�

�

�

Uµ. (4.8)

In order to calculate the variation of the action let us write

�L =

@L

@xµ

�xµ

+

@L

@vµ�vµ. (4.9)

Using the relation

d

d�(�xµ

) = �

✓

dxµ

d�

◆

= �vµ, (4.10)

the variation of the lagrangian can be written as

�L =

@L

@xµ

�xµ

+

@L

@vµd

d�(�xµ

) (4.11)

�L =

@L

@xµ

� d

d�

✓

@L

@vµ

◆�

�xµ

+

d

d�

✓

@L

@vµ�xµ

◆

. (4.12)

Because of the condition of fixed end points, the second term in the right handside does not contribute to the total variation of the action. Thus the variation gives

�S =

ˆ�2

�1

@L

@xµ

� d

d�

✓

@L

@vµ

◆�

�xµd� = 0 (4.13)

and since �xµ is arbitrary inside the boundaries, we obtain the equations of motion

d

d�

✓

@L

@vµ

◆

� @L

@xµ

= 0. (4.14)

Defining the generalized momenta

pµ

=

@L

@vµ, (4.15)

the equations of motion may be written

dpµ

d�=

@L

@xµ

. (4.16)

4.2. EXAMPLES 57

4.2 Examples

4.2.1 Free ParticleIn order to obtain the lagrangian of a free particle, we will consider first its action. Inspecial relativity the trajectory of a particle in spacetime is known as the worldline,

that can be parameterized by the particle’s proper time ⌧ . The action of the particlemust be a scalar and since ⌧ is relativistic invariant we will propose

S0

/ˆ

d⌧. (4.17)

The action has the units

[S] = [Energy] [Time] (4.18)

while

[⌧ ] = [Time] . (4.19)

Therefore, the proportionality constant between action an proper time must haveenergy units. For a particle with proper mass m

0

we proose that this constant is therest energy E

0

= m0

c2, giving the action

S0

= �m0

c2ˆ

d⌧ (4.20)

where we introduce a minus sign to obtain the correct non-relativistic limit as will beseen below. From the definition of line element we have

d⌧ =

ds

c(4.21)

and then the action can be written

S0

= �m0

c

ˆds. (4.22)

4.2.1.1 Non-Relativistic Limit

We can use the coordinate time t as the variable of integration by noting that the lineelement can be written as

ds = cdt

s

1� |~u|2

c2(4.23)

where we have defined the physical velocity as

|~u|2 =

✓

dx1

dt

◆

2

+

✓

dx2

dt

◆

2

+

✓

dx3

dt

◆

2

. (4.24)


Thus, the action is

S0

= �m0

c2ˆ

t2

t1

s

1� |~u|2

c2dt (4.25)

from which we have the lagrangian

L0

= �m0

c2

s

1� |~u|2

c2. (4.26)

In the limit of small speeds, |~u|c

⌧ 1, the lagrangian can be expanded as

L0

' �m0

c2

1� 1

2

|~u|2

c2

!

= �m0

c2 +1

2

m0

|~u|2 . (4.27)

The first term in the right hand side is a constant, so it does not affect theequation of motion while the second term is just the non-relativistic lagrangian for afree particle (Note that the minus sign introduced in the action is neccesary to obtainthe correct non-relativistic lagrangian).

4.2.1.2 Equation of Motion

From the equation (4.7) we can change the proper time for a new parameter � in theaction (4.20) by making

S0

= �m0

c2ˆ

d⌧

d�d� = �m

0

c

ˆ pv2d� (4.28)

and therefore the free particle is described by the Lagrangian

L0

= m0

cpv2 (4.29)

where we have ommited this time the irrelevant minus sign. The generalized momen-tum is

pµ

=

@L0

@vµ=

@

@vµ⇥

m0

cp

vµvµ

⇤

(4.30)

pµ

=

m0

cpv2

vµ

(4.31)

but using (4.8),

pµ

= m0

Uµ

, (4.32)

that corresponds to the definition of the 4-momentum in equation (2.1). The equationof motion is

dpµ

d�=

d

d�(m

0

Uµ

) = 0 (4.33)

4.2. EXAMPLES 59

d⌧

d�

d

d⌧(m

0

Uµ

) = 0 (4.34)

d

d⌧(m

o

Uµ

) = 0 (4.35)

which is the Minkowskian equation of motion for a free particle.

4.2.2 Charged Particle in Electromagnetic FieldsThe Lagrangian of a particle in the presence of a vector field is given by

L = L0

+ L1

= m0

cpv2 +

q

cA

µ

(x) vµ (4.36)

where the interaction part couples the particle velocity to the external vector fieldA

µ

(x). The generalized momentum is

pµ

=

@L

@vµ=

m0

cpv2

vµ

+

q

cA

µ

(4.37)

pµ

= m0

Uµ

+

q

cA

µ

(4.38)

or equivalently

pµ = m0

Uµ

+

q

cAµ. (4.39)

Note that in this case it is clear the difference between the generalized or canonicalmomentum and the 4-momentum used in the Minkowskian equation of motion (2.20).The Lagrangian equation of motion (4.14) is in this case

d

d�

h

m0

Uµ

+

q

cA

µ

(x)i

=

@L

@xµ

(4.40)

d⌧

d�

d

d⌧[m

0

Uµ

] +

d⌧

d�

d

d⌧

hq

cA

µ

(x)i

=

q

cv⌫@A

⌫

(x)

@xµ

(4.41)

d⌧

d�

d

d⌧[m

0

Uµ

] +

d⌧

d�

dx⌫

d⌧

d

dx⌫

hq

cA

µ

(x)i

=

q

cv⌫@A

⌫

(x)

@xµ

(4.42)

d⌧

d�

d

d⌧[m

0

Uµ

] +

d⌧

d�

q

cU⌫

@Aµ

(x)

@x⌫=

q

c

d⌧

d�U⌫

@A⌫

(x)

@xµ

(4.43)

d

d⌧[m

0

Uµ

] =

q

cU⌫

@A⌫

(x)

@xµ

� q

cU⌫

@Aµ

(x)

@x⌫(4.44)

d

d⌧[m

0

Uµ

] =

q

cU⌫

@A⌫

(x)

@xµ

� @Aµ

(x)

@x⌫

�

. (4.45)


Identifying the external vector field Aµ

(x) with the electromagnetic potential theterm in parenthesis corresponds to the antisymmetric electromagnetic field tensor F

µ⌫

and this equation becomes the Lorentz force found in (2.41),

d

d⌧[m

0

Uµ

] =

q

cU⌫F

µ⌫

. (4.46)

4.3 Hamiltonian Equations of MotionNow we will introduce formally a Hamiltonian from the Lagrangian L = L [xµ, vµ].Note that

dL

d�=

@L

@xµ

vµ +

@L

@vµdvµ

d�(4.47)

and using the equations of motion, this gives

dL

d�=

d

d�

✓

@L

@vµ

◆

vµ +

@L

@vµdvµ

d�(4.48)

dL

d�=

d

d�

✓

@L

@vµvµ◆

(4.49)

which can be written

d

d�

✓

@L

@vµvµ � L

◆

= 0. (4.50)

This relation let us define the Hamiltonian as the term in parenthesis,

H = pµ

vµ � L, (4.51)

even though this quantity is not the energy (remember that the relativistic energycoresponds to the zero component of the 4-momentum) and its numerical value iszero as we will see below. The canonical equations of motion are

@H

@xµ

= �dpµ

d�(4.52)

@H

@pµ=

dxµ

d�. (4.53)

4.3.1 Charged Particle in Electromagnetic FieldsAs an example of this formulation consider again the Lagrangian of a particle in thepresence of a vector field,

L = m0

cpv2 +

q

cA

µ

(x) vµ (4.54)

whose generalized momentum is

4.3. HAMILTONIAN EQUATIONS OF MOTION 61

pµ

=

@L

@vµ=

m0

cpv2

vµ

+

q

cA

µ

(x) . (4.55)

The Hamiltonian definition gives

H = pµ

vµ �h

m0

cpv2 +

q

cA

µ

(x) vµi

(4.56)

H =

m0

cpv2

vµ

vµ +

q

cA

µ

(x) vµ �h

m0

cpv2 +

q

cA

µ

(x) vµi

= 0 (4.57)

which is identically zero. However the equations of motion (4.52) and (4.53) can becalculated from its functional dependence,

@H

@xµ

=

@

@xµ

h

pµ

vµ �m0

cpv2 � q

cA⌫

(x) v⌫i

= �dpµ

d�(4.58)

q

cv⌫@A

⌫

(x)

@xµ

=

dpµ

d�(4.59)

and

@H

@pµ=

@

@pµ

h

pµ

vµ �m0

cpv2 � q

cA⌫

(x) v⌫i

=

dxµ

d�(4.60)

vµ

=

dxµ

d�. (4.61)

4.3.2 Other Forms of HamiltonianAlthough the above definition of the Hamiltonian function is correct, it has two sin-gular characteristics: it is identically vanishing and it does not corresponds to theenergy. Therefore, in order to avoid these unwanted characteristics we can definedifferent kinds of hamiltonians.

4.3.2.1 Hamiltonian Tensor

Instead of the function H some authors use the non-vanishing symmetric tensor

Hµ⌫

= pµv⌫ � ⌘µ⌫L (4.62)

and a set of canonical equations,

@Hµ⌫

@xµ

= �dp⌫

d�(4.63)

@Hµ⌫

@pµ=

dx⌫

d�. (4.64)

However, this tensor does not have the meaning of energy.


4.3.2.2 Hamiltonian as Energy

As we have seen above, the generalized or canonical momentum differ from the 4-momentum used in the Minkowskian equation of motion (2.20). In general, tehcanonical momentum pµ consist of the kinetic term m

0

Uµ plus a term that comesfrom the interaction with external fields (thus being independent of the proper massm

0

) as can be seen from equation (4.39). This can be written as

pµ = m0

Uµ

+ fµ

( , U) (4.65)

where f is a function of the fields and eventually of the 4-velocity U . From thisrelation we can write the generalization of the norm of the 4-momentum,

(p� f)2

= m2

0

U2

= m2

0

c2 (4.66)

Assuming that the function f is independent of the velocities we have

�

p0 � f0

�

2 ��

�

�

~p� ~f�

�

�

2

= m2

0

c2 (4.67)

and hence, the energy of the particle is the component ˆH = cp0,

ˆH = cp0 = c

r

�

�

�

~p� ~f�

�

�

2

+m2

0

c2. (4.68)

Note that this quantity is not a scalar (nor a vector and nor a tensor) but it is acomponent of a 4-vector. Therefore, ˆH transforms as the zero component of a fourvector under a change of coordinates. This energy can be used as the Hamiltonianfunction and the corresponding canonical equations of motion are

@ ˆH

@~r= �d~p

d⌧(4.69)

@ ˆH

@~p=

d~r

d⌧. (4.70)

These equations are clearly non-covariant but they are still relativistic invariant.

For a particle in an electromagnetic field, described by the canonical momentum(4.39), we have fµ

=

q

c

Aµ. Therefore, the hamiltonian function ˆH is

ˆH = c

r

�

�

�

~p� q

c~A�

�

�

2

+m2

0

c2. (4.71)

4.3.2.3 Two Different Hamiltonians

It is also possible to define two different scalar Hamiltonians with the meaning ofenergy, supossing that f is independent of the velocity, through the relations

4.4. CLASSICAL SPIN. ELECTRIC AND MAGNETIC MOMENTS 63

h =

1

2m0

(p� f)2

=

1

2

m0

c2 (4.72)

and

h0=

q

(p� f)2

= m0

c. (4.73)Note that function h satisfies the canoncial equations of motion,

@h

@pµ=

1

m0

(pµ

� fµ

) = Uµ

=

dxµ

d⌧(4.74)

@h

@xµ

= � 1

m0

(p⌫ � f⌫) @µ

f⌫

= �U⌫@µ

f⌫

= �dpµ

d⌧. (4.75)

It is easy to show that function h0 satisfies a similar set of equations.

4.4 Classical Spin. Electric and Magnetic MomentsOne of the greatest advantages of the Lagrangian and Hamiltonian formulations isthat new intrinsec properties of classical relativistic particles can be easily consideredby adding the adequate terms to the lagrangian.

Consider an external electromagnetic field and a particle with an intrinsec mag-netic moment ~µ = (µ

1

, µ2

, µ3

) (not the orbital magnetic moment!) and intrinsecdipole moment ~d = (d

1

, d2

, d3

). This particle has, in an external magnetic field ~B, anadditional energy ~µ · ~B and in the presence of an external electric field ~E, it has anadditional energy ~d · ~E (actually, these moments are defined phenomenologically bythe extra terms in the energy). These two extra terms of energy can be written incovariant notation as

1

2

�µ⌫Fµ⌫

(4.76)

where �µ⌫ is the antisymmetric tensor

�µ⌫

=

0

B

B

@

0 d1

d2

d3

�d1

0 µ3

�µ2

�d2

�µ3

0 µ1

�d3

µ2

�µ1

0

1

C

C

A

. (4.77)

The fact that this is a tensor means that electric and magnetic moments transformas

�0µ⌫

=

@x0µ

@x↵@x

0⌫

@x��↵� (4.78)

under a general coordinates transformation. Since the moments are not scalars butvectors in the rest frame of the particle, their orientation can change along the trajec-tory. Therefore, we have to solve not only the motion of the particle in the presenceof the moments but also the precessions of the moments.


4.4.1 Motion of the Particle in the Presence of MomentsTo obtain the new lagrangian consider the new term in the action of the particle dueto the presence of the moments,

Sm

=

ˆ1

2

�µ⌫Fµ⌫

d⌧ =

ˆ1

2

�µ⌫Fµ⌫

d⌧

d�d� (4.79)

or

Sm

=

ˆ1

2

�µ⌫Fµ⌫

pv2

cd�. (4.80)

The complete lagrangian that describes the particle is obtained by adding thisnew term to the lagrangian (4.36),

L = m0

cpv2 +

1

2c�↵�F

↵�

pv2 +

q

cA

µ

vµ (4.81)

and for simplicity, we will consider initially a constant �µ⌫ . In order to obtain theequation of motion note that the new term in L has

@

@xµ

1

2c�↵�F

↵�

pv2�

=

1

2c�↵�

pv2@

µ

F↵�

=

1

2

�↵�d⌧

d�@µ

F↵�

(4.82)

and

d

d�

@

@vµ

✓

1

2c�↵�F

↵�

pv2◆�

=

d

d�

1

2c�↵�F

↵�

vµpv2

�

(4.83)

d

d�

@

@vµ

✓

1

2c�↵�F

↵�

pv2◆�

=

d

d�

1

2c�↵�F

↵�

Uµ

c

�

(4.84)

d

d�

@

@vµ

✓

1

2c�↵�F

↵�

pv2◆�

=

d⌧

d�

d

d⌧

1

2c2�↵�F

↵�

Uµ

�

(4.85)

Hence, the lagrangian equation of motion is

d

d⌧[m

0

Uµ

] =

q

cU⌫F

µ⌫

+

1

2

�↵�@µ

F↵�

� d

d⌧

1

2c2�↵�F

↵�

Uµ

�

. (4.86)

In order to interpret physically this equation we rewrite the las term in the righthand side as

d

d⌧

1

2c2�↵�F

↵�

Uµ

�

=

1

2c2�↵�F

↵�

dUµ

d⌧+

1

2c2�↵�U

µ

U⌫

@⌫F↵�

(4.87)

which let us group some terms in the equation (4.86),

d

d⌧[m

0

Uµ

] =

q

cU⌫F

µ⌫

+

1

2

�↵�@µ

F↵�

� 1

2c2�↵�F

↵�

dUµ

d⌧� 1

2c2�↵�U

µ

U⌫

@⌫F↵�

(4.88)


✓

1 +

1

2m0

c2�↵�F

↵�

◆

d

d⌧[m

0

Uµ

] =

q

cU⌫F

µ⌫

+

1

2

�↵�@µ

F↵�

� 1

2c2�↵�U

µ

U⌫

@⌫F↵�

(4.89)

d

d⌧[m

0

Uµ

] =

1

1 +�

q

cU⌫F

µ⌫

+

1

2

�↵�@µ

F↵�

� 1

2c2�↵�U

µ

U⌫

@⌫F↵�

�

(4.90)

where we defined

� =

1

2m0

c2�↵�F

↵�

. (4.91)

The equation of motion shows that the effect of the moments is twofold: first,they change effectively the mass or the charge of the particle (because of the term1/ (1 +�)); and second, the moments add forces which depend on the gradient of theexternal field (i.e. @

µ

F↵�

).In the particular case of a homogeneus field, @

µ

F↵�

= 0, the equation of motionbecomes

d

d⌧[m

0

Uµ

] =

1

1 +�

hq

cU⌫F

µ⌫

i

(4.92)

which shows a deviation from the Lorentz force. However, if the energy of the momentsis small with respect to the proper energy, 1

2

�↵�F↵�

⌧ m0

c2 i.e. � ⇡ 0, the Lorentzequation is recovered.

4.4.2 The Spin VectorThe above analysis is just a first approximation because �µ⌫ is, in general, a functionof ⌧ and therefore we must consider the precession of the moments. To describe it wecan use the tensor �µ⌫ or its dual vector S⇢ which is defined through the relation

�µ⌫

=

1

2

✏µ⌫↵⇢

p↵S⇢ (4.93)

or equivalently

S⇢ =1

2

✏⇢↵µ⌫p↵

�µ⌫

(4.94)

where ✏⇢↵µ⌫ is the totally antisymmetric symbol with ✏0123 = +1 and p↵ is themomentum vector.

From these relations we obtain easily the properties

p⇢

S⇢ = 0 (4.95)pµ

�µ⌫

= 0 (4.96)Sµ

�µ⌫

= 0. (4.97)


Hence, the spin of the particle may be described by the 4-vector Sµ that is orthog-onal to the momentum. For example, in the rest frame of the particle its momentumis p = (m

0

c, 0, 0, 0) and the spin is the spacelike vector S =

⇣

0, ~S⌘

.

4.4.2.1 Spin Precession

Let S� =

⇣

S0, ~S⌘

be the spin 4-vector of a particle in an arbitrary system of referenceand Uµ its 4-velocity. We can write

U↵@↵

S� = U↵@↵

⇣

S0, ~S⌘

(4.98)

or using U↵@↵

=

dx

↵

d⌧

@

@x

↵

=

d

d⌧

,

U↵@↵

S� =

dS0

d⌧,d~S

d⌧

!

. (4.99)

If there are no forces interacting with the spin of the particle, the equation ofmotion for the spin vector in the rest frame must be d

~

S

d⌧

= 0, hence

U↵@↵

S� =

✓

dS0

d⌧,~0

◆

. (4.100)

Note that equation (4.95) gives

U↵@↵

�

S�

U�

�

= 0 (4.101)

U�

U↵@↵

S� + S�

U↵@↵

U�

= 0 (4.102)

and using U↵@↵

U�

=

dx

↵

d⌧

@

@x

↵

U�

=

dU

�

d⌧

= A� , with A� the 4-acceleration of theparticle, we have

U�

U↵@↵

S� = �S�

A� . (4.103)

Now we will consider the rest frame of the particle where its 4-velocity has com-ponents U�

=

⇣

c,~0⌘

and therefore equation (4.100) gives

U�

U↵@↵

S� = cdS0

d⌧= �S

⇢

A⇢ (4.104)

from which we have

dS0

d⌧= �S

�

A�

c. (4.105)

Hence, we can write

U↵@↵

S� =

✓

�S�

A�

c,~0

◆

(4.106)


or in terms of the 4-velocity,

U↵@↵

S = �S⇢

A⇢

c2U (4.107)

dS

d⌧= �S

⇢

A⇢

c2U. (4.108)

A general Lorentz boost (i.e. one with ~v in any direction but without rotations)is given by

ct0 = � (�)h

ct� ~� · ~ri

(4.109)

~r0 = ~r +

"

(� � 1)

~� · ~r�2

� �t

#

~� (4.110)

while the inverse transformation is obtained by replacing ~� ! �~�. In particular, thespin 4-vector obeys this kind of transformation. Thus, if the primed components areevaluated in the rest frame, S0

=

⇣

0, ~S⌘

, the components in the laboratory systemare given by

S =

�⇣

~� · ~S⌘

, ~S +

~� (� � 1)

~� · ~S�2

!

. (4.111)

However, we have that

� � 1

�2

=

�2

1 + �(4.112)

and therefore

S =

✓

�⇣

~� · ~S⌘

, ~S +

~��2

1 + �

⇣

~� · ~S⌘

◆

. (4.113)

In addition we know that the 4-velocity and the 4-acceleration in the laboratoryframe are

U =

⇣

�c, �~�c⌘

(4.114)

and

A =

cd�

d⌧,d�

d⌧~�c+ �

d~�

d⌧c

!

. (4.115)

�

�2

1 + �

⇣

~� · ~S⌘ d�

d⌧c2�

�"

~S�d~�

d⌧c+ ~�

�2

1 + �

⇣

~� · ~S⌘

�d~�

d⌧c

#

These relations give


SA = ��c"

~S · d~�

d⌧+

�2

1 + �

⇣

~� · ~S⌘

~� · d~�

d⌧

#

(4.116)

and therefore equation (4.108) becomes

dS

d⌧=

�

c

"

~S · d~�

d⌧+

�2

1 + �

⇣

~� · ~S⌘

~� · d~�

d⌧

#

U. (4.117)

Using equations (6.47) and (4.114) we can write the temporal and spatial compo-nents of this equation as

d

d⌧

h

�⇣

~� · ~S⌘i

= �2

"

~S · d~�

d⌧+

�2

1 + �

⇣

~� · ~S⌘

~� · d~�

d⌧

#

(4.118)

d

d⌧

~S +

~��2

1 + �

⇣

~� · ~S⌘

�

= �2

"

~S · d~�

d⌧+

�2

1 + �

⇣

~� · ~S⌘

~� · d~�

d⌧

#

~�. (4.119)

These two equations can be written as one relation

d

d⌧

~S +

~��2

1 + �

⇣

~� · ~S⌘

�

=

~�d

d⌧

h

�⇣

~� · ~S⌘i

(4.120)

which can be rewritten as

d~S

d⌧=

~�d

d⌧

�

1 + �

⇣

~� · ~S⌘

�

� d~�

d⌧

�2

1 + �

⇣

~� · ~S⌘

(4.121)

Finally, equation (4.121) may be rewritten (with some algebraic work and usingthe vectorial identity ~A⇥

⇣

~B ⇥ ~C⌘

=

⇣

~A · ~C⌘

~B �⇣

~A · ~B⌘

~C) as

d~S

d⌧=

✓

�2

1 + �

◆

"

~S ⇥

~� ⇥ d~�

d⌧

!#

(4.122)

or better as

d~S

d⌧=

~S ⇥ ~!T

(4.123)

where we have defined

~!T

=

✓

�2

1 + �

◆

~� ⇥ d~�

d⌧

!

. (4.124)

Equation (4.123) represents the Thomas Precession and correspondingly, ~!T

isknown as Thomas Precession Frequency. Note that in this case, the precession of thespin is produced only by the state of movement of the particle (there are no forces!)and is clearly a relativistic effect.


4.4.3 Equation of Motion for the Spin in the Presence of Elec-tromagnetic Fields

The magnetic moment of a particle can be expressed as proportional to its spin vector.The proportionality constant depends on a combination of fundamental constantswhich have the dimensions of a magnetic moment and is known as the Bohr magneton,µB

=

q~2m0c

. This relation is

~µ = gµB

~S

~ =

gq

2m0

c~S (4.125)

where g is a dimensionless number called the gyromagnetic ratio or Landé factor. Forexample, the works of Uhlenbeck and Goudsmit (1926) shown that the electron hasg = 2. In the presence of an external magnetic field ~B, the torque on the magneticmoment gives the equation of motion for the spin

~NB

= ~µ⇥ ~B =

d~S

d⌧(4.126)

with ⌧ the proper time that is equivalent to the coordinate time t in the non-relativisticlimit. Similarly, if the particle has an intrinsec electric dipole moment ~d, it can bewritten in terms of the spin as

~d =

fq

2m0

c~S (4.127)

where f is the electric analog of g. If there is an external electric field ~E, the torqueon the electric moment gives the equation of motion for the spin

~NE

=

~d⇥ ~E =

d~S

d⌧. (4.128)

If the particle has both moments and there are external electric and magneticfields, the complete equation of motion for the spin is

d~S

d⌧=

gq

2m0

c~S ⇥ ~B +

fq

2m0

c~S ⇥ ~E. (4.129)

From the condition pµ

Sµ

= 0 we also have

Uµ

Sµ

= 0. (4.130)

In an arbitrary frame of reference Uµ

= � (u) (c, ~u) and Sµ

=

⇣

S0, ~S⌘

. Thus

Uµ

Sµ

= � (u)h

cS0 � ~u · ~Si

= 0 (4.131)

from which

S0

=

~u · ~Sc

. (4.132)


While S0 vanishes in the instantaneous rest frame, its variation dS

0

d⌧

needs not,

dS0

d⌧=

1

c

d⇣

~u · ~S⌘

d⌧=

1

c

d~u

d⌧· ~S +

1

c~u · d

~S

d⌧. (4.133)

The equations of motion for the spin (4.129) and (4.133) can be generalized co-variantly by writting

dSµ

d⌧=

gq

2m0

c2[Fµ⌫S

⌫

+ S�

F�⇢U⇢

Uµ

]

+

fq

2m0

c2[Fµ⌫⇤S

⌫

+ S�

F�⇢⇤U⇢

Uµ

]� 1

c

dU⌫

d⌧S⌫

Uµ, (4.134)

where F ⇤µ⌫

is the dual of Fµ⌫

. To probe that (4.134) reproduces equations (4.129) and(4.133) you just have to evaluate the expression in the rest frame of the particle.

In order to complete the description of the motion of the spin vector there is justone more thing to consider. The factor dU

µ

d⌧

in the last term has to be replaced usingthe eqution of motion of the particle (4.90). It gives

1

c

dU⌫

d⌧S⌫

Uµ

=

1

c (1 +�m)

q

m0

cU⇢

F ⌫⇢ +1

2m0

�↵�@⌫F↵�

� 1

2m0

c2�↵�U⌫U

�

@�F↵�

�

S⌫

Uµ

and using the orthogonality S⌫

U⌫

= 0, the last term dissapears,

1

c

dU⌫

d⌧S⌫

Uµ

=

1

(1 +�m)

q

m0

c2U⇢

F ⌫⇢ +1

2m0

c�↵�@⌫F

↵�

�

S⌫

Uµ. (4.135)

Replacing in the equation of motion of the spin vector we obtain finally

dSµ

d⌧=

q

m0

c2

hg

2

Fµ⌫S⌫

+

✓

g

2

� 1

1 +�m

◆

S�

F�⇢U⇢

Uµ

+

f

2

(Fµ⌫⇤S⌫

+ S�

F�⇢⇤U⇢

Uµ

)

�

� �↵�@⌫F↵�

S⌫

Uµ

2m0

c (1 +�m)

. (4.136)

If we consider only magnetic moments (f = 0), homogeneus fields (@⌫F↵�

= 0)and that the energy of the moments is small with respect to the proper energy,1

2

�↵�F↵�

⌧ m0

c2, i.e. �m ⇡ 0, this equation reduces to

dSµ

d⌧=

q

m0

c2

hg

2

Fµ⌫S⌫

+

⇣g

2

� 1

⌘

S�

F�⇢U⇢

Uµ

i

that is the well known result reported by Bargmann, Michel and Telegdi [2].

4.5. GENERAL DESCRIPTION OF THE CLASSICAL SPIN 71

4.5 General Description of the Classical Spin

The description of classical spin described above is correct but it is most satisfactoryto introduce a Lagrangian that gives both, the equation of motion for the trajectoryof the particle and the equation for the spin. This new Lagrangian must containspacetime coordinates that give rise to the equation of motion and also a certainnumber of internal degrees of freedom describing the spin and that give rise to thespin equation. Therefore, we write the action of the particle with spin in the presenceof the vector field A

µ

as

S =

ˆLd⌧ (4.137)

where the Lagrangian is

L = Lh

xµ, xµ, qµ(↵)

, qµ(↵)

, Aµ

(x) , @⌫

Aµ

(x)i

(4.138)

where the internal coordinates are represented by the four 4-vectors qµ(↵)

labeled by↵ = 0, 1, 2, 3. Without loosing generality, we choose the four vector q

(0)

in the direc-tion of the 4-velocity U and the other three 4-vectors, q

(1)

, q(2)

and q(3)

, orthogonalto this direction, i.e. they are spacelike.

Variation of this action with respect to spacetime coordinates gives the equationsof motion,

�S

�xµ

= 0 ) d

d⌧

✓

@L

@xµ

◆

� @L

@xµ

= 0 (4.139)

while the variation of this action with respect to internal coordinates gives the equa-tions

�S

�qµ(↵)

= 0 ) d

d⌧

@L

@qµ(↵)

!

� @L

@qµ(↵)

= 0. (4.140)

Now consider the general infinitesimal spacetime and internal transformations

xµ ! x0µ= xµ

+ �xµ, (4.141)

qµ(↵)

! q0µ

(↵)

= qµ(↵)

+ �qµ(↵)

. (4.142)

Since the field A↵ (x) depends on spacetime coordinates, the infinitesimal trans-formation change it as

A↵ (x) ! A0↵

(x0) = A↵ (x) + �A↵ (x) . (4.143)

The variation of the field can be written at first order as

�A↵ (x) = A0↵

(x0)�A↵ (x) = A

0↵

(x) + �xµ

(@µ

A↵)�A↵ (x) (4.144)


�A↵ (x) = �0

A↵ + �xµ

(@µ

A↵) . (4.145)

The variation of the action under these transformations is

�S =

ˆ�Ld⌧, (4.146)

where the variation of the lagrangian is

�L =

@L

@xµ

�xµ

+

@L

@xµ

�xµ

+

@L

@qµ(↵)

�qµ(↵)

+

@L

@qµ(↵)

�qµ(↵)

++

@L

@A⇢

�0

A⇢

+

@L

@ (@⌫

A⇢

)

�0

(@⌫

A⇢

) ,

(4.147)which can be written as

�L =

@L

@xµ

� d

d⌧

✓

@L

@xµ

◆�

�xµ

+

"

@L

@qµ(↵)

� d

d⌧

@L

@qµ(↵)

!#

�qµ(↵)

+

d

d⌧

"

@L

@xµ

�xµ

+

@L

@qµ(↵)

�qµ(↵)

#

+

@L

@A⇢

[�A⇢

� @µ

A⇢

�xµ

] +

@L

@ (@⌫

A⇢

)

[� (@⌫

A⇢

)� @µ

@⌫

A⇢

�xµ

] . (4.148)

In this exprression the first two brackets vanish because they correspond to theEuler-Lagrange equations for coordinates x and q, respectively. In order to analyzethe conserved quantities we will consider the following transformations:

�L =

d

d⌧

"

@L

@xµ

�xµ

+

@L

@qµ(↵)

�qµ(↵)

#

+

@L

@A⇢

[�A⇢

� @µ

A⇢

�xµ

] +

@L

@ (@⌫

A⇢

)

[� (@⌫

A⇢

)� @µ

@⌫

A⇢

�xµ

] . (4.149)

4.5.1 Spacetime TranslationsIf the spacetime translations defined by the relations

�xµ

= aµ, d

d⌧

(�xµ

) = 0, �qµ(↵)

= 0, �Aµ

= 0, (4.150)

let invariant the lagrangian, i.e. �L = 0 in Eq. (4.149), they give the equations ofmotion for the particle

d

d⌧

✓

@L

@xµ

◆

= pµ

=

@L

@A⇢

@µ

A⇢

+

@L

@ (@⌫

A⇢

)

@µ

(@⌫

A⇢

) (4.151)

which is a generalization of the equation of motion (4.90).

4.5. GENERAL DESCRIPTION OF THE CLASSICAL SPIN 73

4.5.2 Lorentz Transformations

Now consider the infinitesimal homogeneous Lorentz transformations described by

�xµ

= ✏µ⌫

x⌫ , �qµ(↵)

= ✏µ⌫

q⌫(↵)

. (4.152)

The 4-vector Aµ

and its derivative ,wich is a (0, 2) tensor, transform as

�Aµ

= ✏ ⌫

µ

A⌫

, � (@⌫

Aµ

) = ✏ ⇢

µ

@⌫

A⇢

+ ✏ ⇢

⌫

@⇢

Aµ

, (4.153)

Replacing in equation (4.149), this transformation gives the variation

�L =

d

d⌧

"

@L

@xµ

�xµ

+

@L

@qµ(↵)

�qµ(↵)

#

+

@L

@Aµ

�Aµ

� @L

@A⇢

@µ

A⇢

�xµ

+

@L

@ (@⌫

Aµ

)

� (@⌫

Aµ

)� @L

@ (@�

A⇢

)

@µ

@�

A⇢

�xµ

�L =

d

d⌧

@L

@xµ

✏µ⌫x⌫

�

+

d

d⌧

"

@L

@qµ(↵)

✏µ⌫q⌫(↵)

#

+

@L

@Aµ

✏µ⌫A⌫

� @L

@A⇢

@µ

A⇢

✏µ⌫x⌫

+

@L

@ (@⌫

Aµ

)

✏µ⇢@⌫

A⇢

+

@L

@ (@⌫Aµ

)

✏⌫⇢@⇢

Aµ � @L

@ (@�

A⇢

)

@µ

@�

A⇢

✏µ⌫x⌫

.

Renaming some mute indices,

�L =

d

d⌧[p

µ

x⌫

] ✏µ⌫ +d

d⌧

"

@L

@qµ(↵)

q⌫(↵)

#

✏µ⌫ +@L

@Aµ

✏µ⌫A⌫

� @L

@A⇢

@µ

A⇢

✏µ⌫x⌫

+

@L

@ (@⇢

Aµ

)

✏µ⌫@⇢

A⌫

+

@L

@ (@µA⇢)✏µ⌫@

⌫

A⇢ � @L

@ (@�

A⇢

)

@µ

@�

A⇢

✏µ⌫x⌫

,

and using the antisymmetry of ✏µ⌫

, we obtain

2�L =

d

d⌧

⇥

p[µ

x⌫]

⇤

✏µ⌫ +d

d⌧

2

4

@L

@q[µ

(↵)

q⌫](↵)

3

5 ✏µ⌫ +@L

@A[µ

A⌫]

✏µ⌫ � @L

@A⇢

A⇢,[µ

x⌫]

✏µ⌫

+

@L

@�

@⇢

A[µ

�A⌫],⇢

✏µ⌫ +@L

@�

A⇢,[µ�@

⌫]

A⇢✏µ⌫ � @L

@ (@�

A⇢

)

A⇢,�[µ

x⌫]

✏µ⌫ .

If the lagrangian (and correspondingly the action) is invariant under the consideredtransformation, i.e. �L = 0, this equation can be written


d

d⌧

2

4

@L

@q[µ

(↵)

q⌫](↵)

3

5

+ p[µ

x⌫]

+

n

p[µ

x⌫]

� @L

@A

⇢

A⇢,[µ

x⌫]

� @L

@A

⇢,�

A⇢,�[µ

x⌫]

o

=

� @L

@A

[µ A⌫]

� @L

@A

[µ,⇢

A⌫],⇢

� @L

@

(

A

⇢,[µ)

@⌫]

A⇢.

However, note that the term in brackets in the left hand side is zero because ofthe equation of motion (4.151). Thus

d

d⌧

2

4

@L

@q[µ

(↵)

q⌫](↵)

3

5

+ p[µ

x⌫]

= � @L

@A[µ

A⌫]

� @L

@A[µ

,⇢

A⌫],⇢

� @L

@�

A⇢,[µ�@

⌫]

A⇢. (4.154)

Finally, we can write eq. (4.154) as

˙Sµ⌫

+ p[µ

U⌫]

= �j[µ

A⌫]

�M ⇢

[µ

A⌫],⇢

�M⇢[µ

@⌫]

A⇢ (4.155)

where we have defined

Sµ⌫

=

@L

@q[µ

(↵)

q⌫](↵)

(4.156)

jµ

=

@L

@Aµ

(4.157)

andM⇢µ

= �Mµ⇢

=

@L

@A⇢,µ

. (4.158)

The last two term in the right hand side may be expanded to write

M ⇢

[µ

A⌫],⇢

+M⇢,[µ

@⌫]

A⇢ = M ⇢

µ

A⌫,⇢

�M ⇢

⌫

Aµ,⇢

+M⇢µ

@⌫

A⇢ �M⇢⌫

@µ

A⇢

M ⇢

[µ

A⌫],⇢

+M⇢,[µ

@⌫]

A⇢ = M ⇢

µ

A⌫,⇢

�M ⇢

⌫

Aµ,⇢

+M⇢

µ

@⌫

A⇢

�M⇢

⌫

@µ

A⇢

M ⇢

[µ

A⌫],⇢

+M⇢,[µ

@⌫]

A⇢ = M ⇢

µ

A⌫,⇢

�M ⇢

⌫

Aµ,⇢

�M ⇢

µ

@⌫

A⇢

+M ⇢

⌫

@µ

A⇢

M ⇢

[µ

A⌫],⇢

+M⇢,[µ

@⌫]

A⇢ = M ⇢

µ

[A⌫,⇢

� @⌫

A⇢

]�M ⇢

⌫

[Aµ,⇢

� @µ

A⇢

]

and defining the antisymmetric tensor Fµ⌫

= A⌫,µ

�Aµ,⌫

,

4.6. PROBLEMS 75

M ⇢

[µ

A⌫],⇢

+M⇢,[µ

@⌫]

A⇢ = M ⇢

µ

F⇢⌫

�M ⇢

⌫

F⇢µ

M ⇢

[µ

A⌫],⇢

+M⇢,[µ

@⌫]

A⇢ = �M ⇢

µ

F⌫⇢

+M ⇢

⌫

Fµ⇢

M ⇢

[µ

A⌫],⇢

+M⇢,[µ

@⌫]

A⇢ = �M ⇢

[µ

F⌫]⇢

Therefore the equation for the spin finally becomes

˙Sµ⌫

+ p[µ

U⌫]

= �j[µ

A⌫]

+M ⇢

[µ

F⌫]⇢

. (4.159)

4.6 Problems1. Show that the canonical equations of motion are given by equations (4.52) and

(4.53).

2. Write the Hamiltonian tensor Hµ⌫ given by equation (4.62) for a charged particlein an electromagnetic field. Obtain the corresponding canonical equations ofmotion (4.63) and (4.64).

3. Write the scalar Hamiltonian ˆH given by equation (4.68) for a charged particlein an electromagnetic field. Obtain the corresponding canonical equations ofmotion (4.69) and (4.70).

4. Show that the Hamiltonian h0 defined by equation (4.73) satisfies a set of canon-ical equations of motion.

5. Write the scalar Hamiltonians h and h0 given by equations (4.72) and (4.73)for a charged particle in an electromagnetic field. Obtain the correspondingcanonical equations of motion.

6. Probe properties (4.95), (4.96) and (4.97).

7. Show that a general Lorentz boost is given by equations (4.109) and (4.110).


9. Derive equation (4.116).

10. Find equations (4.118) and (4.119).

11. Probe that equation (4.121) is obtained from (4.118) and (4.119).

12. Find (4.122) from (4.121).

13. Probe that (4.134) reproduces equations (4.129) and (4.133) by evaluating theexpression in the rest frame of the particle.


Chapter 5

Lagrangian Description of theFields

Fields as physical systems will be described by a set of functions ↵ (x) with ↵ =

1, 2, ..., N , that satisfy certain partial differential equations called the field equations.Insetad of initial conditions, this set of field equations have boundary conditions toensure a unique solution. In this section we will obtain the field equations startingfrom an action principle, which guarantees the relativistic invariance of the theory.The action is written as

S =

ˆ⌃

Ld4x (5.1)

where ⌃ is a 4-dimensional spacetime region with a 3-dimensional boundary @⌃ and

L = L [ ↵ (x) , @µ

↵ (x) , xµ

] (5.2)

is called the Lagrangian density and it is assumed to be a Lorentz scalar. We proposethe variational principle

�S = �

ˆ⌃

Ld4x = 0 (5.3)

and consider a fixed boundary @⌃. Then, we have

@L =

@L@ ↵

� ↵ +

@L@ (@

µ

↵)� (@

µ

↵) . (5.4)

Note that the variation is taken by changing the fields but the coordinates xµ arenot varied. hence, we write

@L =

@L@ ↵

� @µ

@L@ (@

µ

↵)

�

� ↵ + @µ

@L@ (@

µ

↵)� ↵

�

(5.5)

and then

77

78 CHAPTER 5. LAGRANGIAN DESCRIPTION OF THE FIELDS

�S =

ˆ⌃

@L@ ↵

� @µ

@L@ (@

µ

↵)

�

� ↵d4x+

ˆ⌃

@µ

@L@ (@

µ

↵)� ↵

�

d4x = 0. (5.6)

As always, the last divergence term can be transfromed by Gauss’s theorem into a3-dimensional integral at @⌃ and vanishes because of the condition of fixed boundaryfor the � ↵.

Because the � ↵ are arbitrary in ⌃, the action principle implies the field equations

@L@ ↵

� @µ

@L@ (@

µ

↵)= 0. (5.7)

It is important to note that this deduction is valid for scalar, vectorial and tensorialfields and that the Lagrangian for a specific field is not unique, we can always add toL a term with the form @

µ

�

µ with an arbitrary �µ = �

µ

( , x), and the field equationsdo not change.

5.1 Symmetries and Conservation Laws. Noether’sTheorem

The invariance of the action under some transformation gives a conserved quantity.One of the greatest advantages of lagrangian formulation is that these conservedquantities can be easily identificable. Consider a field ↵ (x) and some infinitesimaltransformation that affects the coordinates,

xµ ! x0µ= xµ

+ �xµ, (5.8)

which induces a change in the field as

↵ (x) ! 0↵

(x0) = ↵ (x) + � ↵ (x) . (5.9)

The variation of the field can be written at first order as

� ↵ (x) = 0↵

(x0)� ↵ (x) =

0↵

(x) + �xµ

(@µ

↵)� ↵ (x) (5.10)

� ↵ (x) = �0

↵ + �xµ

(@µ

↵) . (5.11)

The variation of the action under this coordinate transformation has two contri-butions,

�S =

ˆ��

d4x�

L+

ˆd4x�L. (5.12)

To obtain the variation of the integration measure we use the Jacobian,

d4x0=

�

�

�

�

det

dx0µ

dx⌫

�

�

�

�

�

d4x, (5.13)

5.1. SYMMETRIES AND CONSERVATION LAWS. NOETHER’S THEOREM 79

which using the coordinate transformation becomes

d4x0= |det [�µ

⌫

+ @⌫

(�xµ

)]| d4x = [1 + @µ

(�xµ

)] d4x (5.14)

to obtain finally

��

d4x�

= d4x0 � d4x = @µ

(�xµ

) d4x. (5.15)

On the other hand, the variation of the lagrangian density comes from �x and�0

↵,

�L =

@L@xµ

�xµ

+

@L@ ↵

�0

↵ +

@L@ (@

µ

↵)@µ

(�0

↵) (5.16)

�L =

@L@xµ

�xµ

+

@L@ ↵

�0

↵ + @µ

@L@ (@

µ

↵)�0

↵�

� @µ

@L@ (@

µ

↵)

�

�0

↵, (5.17)

and using the field equations we get,

�L =

@L@xµ

�xµ

+ @µ

@L@ (@

µ

↵)�0

↵�

. (5.18)

Replacing these results in the action gives

�S =

ˆd4x

L@µ

(�xµ

) +

@L@xµ

�xµ

+ @µ

@L@ (@

µ

↵)�0

↵��

(5.19)

or grouping some terms

�S =

ˆd4x@

µ

L�xµ

+

@L@ (@

µ

↵)�0

↵�

. (5.20)

5.1.1 Noether Current and Conserved ChargeGiven a particular transformation, the variation in the coordinates and in the fieldcan be expressed using a certain number of independent infinitesimal parameters �!r,

�xµ

= Xµ

r

(x) �!r (5.21)

� ↵ (x) = �↵r

(x) �!r. (5.22)

Replacing these expressions in equation (5.11) gives

�0

↵ = [�

↵

r

(x)�X⌫

r

(x) (@⌫

↵)] �!r (5.23)

and the variation of the action (5.20) can be written as


�S = �ˆ

d4x@µ

[jµr

�!r

] , (5.24)

where we have defined the Noether current jµr

,

jµr

=

@L@ (@

µ

↵)(@⌫

↵)� L�µ⌫

�

X⌫

r

(x)� @L@ (@

µ

↵)�

↵

r

(x) . (5.25)

Since the parameters �!r are independent, the variation �S is

�S = �ˆ

d4x [(@µ

jµr

) �!r

+ jµr

@µ

(�!r

)] . (5.26)

Suppose that S is invariant under a global (constant) transformation, i.e. that for�!r

= constant, we have �S = 0. Then, it implies thatˆ

d4x (@µ

jµr

) �!r

= 0, (5.27)

from which it follows that the Noether current is conserved,

@µ

jµr

= 0. (5.28)

Associated with this current we define a charge Q in a certain 3-dimensionalvolume V as

Qr

=

ˆVd3xj0

r

, (5.29)

which is a constant (independent of t) if there is no flux of Noether’s current thorughthe boundary of V. In order to probe this statement consider the derivative

dQr

dt=

ˆVd3x@

0

j0r

(5.30)

dQr

dt=

ˆVd3x

⇥

@µ

jµr

� @i

jir

⇤

(5.31)

dQr

dt= �

ˆVd3x@

i

jir

(5.32)

and using the theorem of divergence, the integral can be taken over the 2-dimensionalboundary S

dQr

dt= �

˛S~jr

· d~a = 0, (5.33)

which vanishes if there is no flux of Noether’s current thorugh S.

5.2. THE LAGRANGIAN FOR THE REAL SCALAR FIELD 81

5.2 The Lagrangian for the Real Scalar FieldIt is easy to choose the Lagrangian densities that give us the known field equations.For example, for a massive real scalar field we can use

L =

1

2

(@µ

�) (@µ�)� 1

2

m2�2 (5.34)

The conjugate field is defined as

⇡µ

=

@L@ (@

µ

�)= @µ� (5.35)

and the field equations are

@L@�

= @µ

⇡µ (5.36)

�m2� = @µ

@µ� (5.37)

�

⇤+m2

�

� = 0, (5.38)

which is the Klein-Gordon equation.

5.2.1 Conserved Quantities5.2.1.1 Infinitesimal Translation. Energy-Momentum Tensor

Consider the infinitesimal transformation

xµ ! x0µ= xµ

+ ✏µ (5.39)

that corresponds to an infinitesimal translation with constant parameters �!⌫ = ✏⌫

and

Xµ

⌫

(x) = �µ⌫

(5.40)�

⌫

(x) = 0. (5.41)

The action for the real scalar field is invariant under this kind of transformation,since it is just a shift of the origin of coordinates. Thus, the Noether current (5.25) is

jµ⌫

=

@L@ (@

µ

�)(@⇢

�)� L�µ⇢

�

�⇢⌫

(5.42)

jµ⌫

=

@L@ (@

µ

�)(@⌫

�)� L�µ⌫

(5.43)

jµ⌫

= (@µ�) (@⌫

�)� 1

2

(@�

�) (@��) �µ⌫

+

1

2

m2�2�µ⌫

(5.44)


or raising the ⌫ index,

jµ⌫ = (@µ�) (@⌫�)� 1

2

(@�

�) (@��) ⌘µ⌫ +1

2

m2�2⌘µ⌫ . (5.45)

The Noether current is called in this case the energy-momentum tensor and isclearly a symmetric tensor. The associated charge is called in this case the energy-

momentum vector,

P⌫

=

ˆd3xP

⌫

=

ˆd3xj0

⌫

(5.46)

where P⌫

is called the density of momentum vector. The zero component of P⌫

is

P0

=

ˆd3x

@L@ (@

0

�)(@

0

�)� L�

, (5.47)

or using the definition of the conjugate momenta,

P0

=

ˆd3x

h

⇡0

˙�� Li

, (5.48)

where ˙� = @0

�. It is clear now that P0

corresponds to the Hamiltonian.

5.2.1.2 Infinitesimal Lorentz Transformation

Consider now the infinitesimal Lorentz transformation,

x0µ= xµ

+ ✏µ⌫x⌫

, (5.49)

where the coeficients ✏µ⌫ are antisymmetric, i.e.

✏µ⌫ = �✏⌫µ. (5.50)

The action is invariant now if we consider in addition that the fields are subjectedto the corresponding spin transformation,

�0 (x) = � (x) +i

2

X

✏µ⌫sµ⌫

� (x) , (5.51)

where sµ⌫

is the spin tensor of the theory,

sµ⌫

=

8

>

>

<

>

>

:

0 (spin 0)1

2

⌃

µ⌫

(spin 1

2

)Sµ⌫

(spin 1)Sµ⌫

+

1

2

⌃

µ⌫

(spin 3

2

)

In the case of spinless fields the spin term is obviously abscent. This time, we take

Xµ

⇢

�!⇢ = ✏µ⌫x⌫

(5.52)�

µ

⇢

�!⇢ = 0. (5.53)

5.2. THE LAGRANGIAN FOR THE REAL SCALAR FIELD 83

From equation (5.24),

�S = �ˆ

d4x@µ

⇥

jµ⇢

�!⇢⇤

= 0, (5.54)

the invariance of the action under this transformation implies the condition

@µ

⇥

jµ⇢

�!⇢⇤

= @µ

✓

@L@ (@

µ

�)@⌫

�� L�µ⌫

◆

X⌫

⇢

�!⇢�

= 0 (5.55)

@µ

✓

@L@ (@

µ

�)@⌫

�� L�µ⌫

◆

✏⌫�x�

�

= 0 (5.56)

or in the terms of the energy-momentum tensor,

@µ

[jµ⌫

✏⌫�x�

] = 0 (5.57)

@µ

1

2

(jµ⌫

✏⌫�x�

+ jµ⌫

✏⌫�x�

)

�

= 0 (5.58)

@µ

1

2

(jµ⌫

✏⌫�x�

� jµ⌫

✏�⌫x�

)

�

= 0 (5.59)

@µ

1

2

✏⌫� (jµ⌫

x�

� jµ�

x⌫

)

�

= 0 (5.60)

1

2

✏⌫�@µ

[jµ⌫

x�

� jµ�

x⌫

] = 0. (5.61)

Defining the antisymmetric object

Mµ

⌫�

= jµ⌫

x�

� jµ�

x⌫

= �Mµ

�⌫

, (5.62)the six conserved currents satisfy

@µ

Mµ

�⌫

= 0 (5.63)and consequently, there are six constants of motion defined as

M�⌫

=

ˆM0

�⌫

d3x (5.64)

M�⌫

=

ˆ⇥

j0⌫

x�

� j0�

x⌫

⇤

d3x. (5.65)

In terms of the density momentum vector P⌫

this is

M�⌫

=

ˆd3x [x

�

P⌫

� x⌫

P�

] . (5.66)

M�⌫

is known as the total angular momentum tensor of the field. In this caseit corresponds only to the orbital term because the scalar field is spinless. If youconsider a non-spinless field (e.g Dirac field), there appear a spin term in the totalangular momentum tensor.


5.3 The Lagrangian for the Complex Scalar FieldA massive complex scalar field is described by the lagrangian

L =

1

2

(@µ

�⇤) (@µ�)� 1

2

m2��⇤. (5.67)

The conjugate fields are defined as

⇡µ

=

@L@ (@

µ

�)=

1

2

@µ�⇤ (5.68)

and

⇡⇤µ=

@L@ (@

µ

�⇤)=

1

2

@µ�. (5.69)

This time we have two field equations given by

@L@�

= @µ

⇡µ (5.70)

and

@L@�⇤

= @µ

⇡⇤µ. (5.71)

These give the Klein-Gordon equations

�

⇤+m2

�

� = 0 (5.72)�

⇤+m2

�

�⇤ = 0. (5.73)

5.3.1 Conserved Quantities5.3.1.1 Infinitesimal Translation. Energy-Momentum Tensor

The infinitesimal translation

xµ ! x0µ= xµ

+ ✏µ (5.74)

leaves the action for the complex scalar field invariant. Thus, the Noether current(5.25) is called again the energy-momentum tensor and is given by

jµ⌫

= (@µ�⇤) (@⌫

�) + (@µ�) (@⌫

�⇤)� �µ⌫

L (5.75)

or

jµ⌫ = (@µ�⇤) (@⌫�) + (@µ�) (@⌫�⇤)� ⌘µ⌫L. (5.76)

The associated charge is the energy-momentum vector,

5.3. THE LAGRANGIAN FOR THE COMPLEX SCALAR FIELD 85

P⌫

=

ˆd3xP

⌫

=

ˆd3xj0

⌫

(5.77)

and the zero component of P⌫

is the Hamiltonian

P0

=

ˆd3x

@L@ (@

0

�)(@

0

�) +@L

@ (@0

�⇤)(@

0

�⇤)� L�

, (5.78)

or using the definition of the conjugate momenta,

P0

=

ˆd3x

h

⇡0

˙�+ ⇡⇤0˙�⇤ � L

i

. (5.79)

5.3.1.2 Infinitesimal Lorentz Transformation

Following the same procedure as for the real scalar field, the infinitesimal Lorentztransformation,

x0µ= xµ

+ ✏µ⌫x⌫

, (5.80)

gives the continuity equation

@µ

[jµ⌫

x�

� jµ�

x⌫

] = 0. (5.81)

We define again the antisymmetric object

Mµ

⌫�

= jµ⌫

x�

� jµ�

x⌫

= �Mµ

�⌫

, (5.82)

from which

@µ

Mµ

�⌫

= 0 (5.83)

and consequently, there are again six constants of motion identified with the angularmomentum components,

M�⌫

=

ˆd3x [x

�

P⌫

� x⌫

P�

] . (5.84)

5.3.1.3 Internal Symmetries

An internal symmetry is one in which

�xµ

= 0. (5.85)

For example, consider the field � (x) changing by an overall constant phase factor,

�0 (x) = eiq�� (x) (5.86)

and the complex conjugate field changing by


�0⇤ (x) = e�iq��⇤ (x) . (5.87)

For an infinitesimal such transformation we have

�� = iq�� (x) (5.88)��⇤ = �iq��⇤ (x) . (5.89)

where we have only one parameter �� that controls de transformation. Evidently theaction is invariant under such transformation because � always appears multiplied by�⇤. This time we have

Xµ

(x) = 0 (5.90)� (x) = iq� (x) (5.91)�

⇤(x) = �iq�⇤ (x) (5.92)

Then, we will define the Noether current, eq. (5.25), as1

jµ

c=

@L@ (@

µ

�)�+

@L@ (@

µ

�⇤)�

⇤ (5.93)

jµ

c=

1

2

(@µ�⇤) iq�� 1

2

(@µ�) iq�⇤ (5.94)

jµ =

1

2

iqc [� (@µ�⇤)� �⇤ (@µ�)] . (5.95)

This result will be important later when we study the interaction between fields.

5.4 The Lagrangian for the Electromagnetic Field

We already know that Maxwell’s equations involve first derivatives of Fµ⌫

(or secondderivatives of A

µ

) and the lagrangian must be both Lorentz and gauge invariant.Thus, the only possibility is a lagrangian quadratic in the tensor field F

µ⌫

but wemake the variations with respect to the vector field A

µ

. The usual choice is fordescribing the free electromagnetic field is

L = �1

4

Fµ⌫Fµ⌫

(5.96)

which is equivalent to L =

1

2

✓

�

�

�

~E�

�

�

2

��

�

�

~B�

�

�

2

◆

.

1Note that this definition differs from the Noether current in eq. (5.25) by a factor of �c.

5.4. THE LAGRANGIAN FOR THE ELECTROMAGNETIC FIELD 87

However, if we want to obtain Maxwell’s equations with sources we need to in-troduce a term proportional to the 4-vector j (which breaks the guage invariance).Hence, the complete lagrangian for the field is

L = �1

4

Fµ⌫Fµ⌫

� 1

cjµ

Aµ. (5.97)

This lagrangian is not invariant under the gauge transformation (3.143) becauseit becomes

˜L = �1

4

˜Fµ⌫

˜Fµ⌫

� 1

cjµ

Ãµ (5.98)

˜L = �1

4

Fµ⌫Fµ⌫

� 1

cjµ

Aµ � 1

cjµ

@µ⇤ (5.99)

or in term of the original lagrangian,

˜L� L = �1

cjµ

@µ⇤ (5.100)


˜L� L = �1

c@µ (j

µ

⇤) +

1

c⇤@µj

µ

. (5.101)

Note that the first term is a 4-divergence and therefore it does not affect theequation of motion. The second term permit us to conclude that the physical lawscoming from this lagrangian are invariant under gauge transformations if @µj

µ

= 0,i.e. as long as charge is absolutely conserved.

5.4.1 Equations of MotionThe variation of the corresponding action gives

�S = �ˆ

d4x�

1

4

Fµ⌫Fµ⌫

+

1

cjµ

Aµ

�

(5.102)

�S = �ˆ

d4x

1

2

Fµ⌫�Fµ⌫

+

1

cjµ

�Aµ

�

. (5.103)

The definition of the electromagnetic tensor gives

�Fµ⌫

= � (@µ

A⌫

� @⌫

Aµ

) = @µ

�A⌫

� @⌫

�Aµ

(5.104)

and using the antisymmetry of Fµ⌫

,

Fµ⌫�Fµ⌫

= Fµ⌫@µ

�A⌫

� Fµ⌫@⌫

�Aµ

(5.105)Fµ⌫�F

µ⌫

= Fµ⌫@µ

�A⌫

+ Fµ⌫@µ

�A⌫

(5.106)Fµ⌫�F

µ⌫

= 2Fµ⌫@µ

�A⌫

. (5.107)


Hence,

�S = �ˆ

d4x

Fµ⌫@µ

�A⌫

+

1

cjµ

�Aµ

�

(5.108)

�S = �ˆ

d4x

@µ

(Fµ⌫�A⌫

)� @µ

Fµ⌫�A⌫

+

1

cjµ

�Aµ

�

(5.109)

The first integral vanishes because of the condition of fixed �A⌫

at the boundaryand as we assume that this variation is arbitrary in the region of integration we obtainthe field equations

@⌫

Fµ⌫

+

1

cjµ = 0. (5.110)

As we have seen, the two homogeneus Maxwell equations correspond to an identityfulfilled because F

µ⌫

is derived from the potential Aµ

, therefore they are not deducedfrom this action.

5.4.2 Alternative Lagrangians for the Free ElectromagneticField

The standard lagrangian for the free electromagnetic field, i.e. without sources, is theone in equation (5.96). However, we already know that the definition of the lagrangianis not unique, and hence, there are severals forms which differ from each other by adivergence term.

For example consider the non gauge-invariant Fermi lagrangian,

LI

= �1

4

Fµ⌫Fµ⌫

� 1

2

(@µ

Aµ

)

2

, (5.111)

which give, variating the field Aµ

, the field equations

⇤Aµ

= 0, (5.112)

i.e. Maxwell’s equations in the Lorentz gauge. The same field equations are obtainedfrom the lagrangian

LII

= �1

2

@⌫

Aµ

@⌫Aµ (5.113)

whch differ from L by the divergence term 1

2

@µ

(A⌫

@⌫Aµ

).Another form is the Schwinger lagrangian

LIII

=

1

4

Fµ⌫Fµ⌫

+

1

2

[A⌫

@µ

Fµ⌫ � @µ

A⌫

Fµ⌫

] (5.114)

which give the field equations

⇤Aµ

= 0 (5.115)

5.5. POINCARÉ TRANSFORMATIONS 89

and

@⌫

Fµ⌫

= 0 (5.116)

when varying independently both Aµ

and Fµ⌫ .

5.5 Poincaré TransformationsThe Lagrangian density for the free electromagnetics field,

L = �1

4

Fµ⌫Fµ⌫

, (5.117)

is invariant under Poincaré transformations,

xµ ! x0µ= ⇤

µ

⌫

x⌫ + aµ, (5.118)

where ⇤µ

⌫

represents a Lorentz transformation and aµ corresponds to spacetime trans-lations. Therefore, there are some conserved quantities (in fact there are in general10 conserved quantities associated with the ten parameters of the transformation).We will consider some special cases in order to interpret physically these quantities.

5.5.1 Infinitesimal Translation. The Energy-Momentum ten-sor

Consider the infinitesimal transformation

xµ ! x0µ= xµ

+ ✏µ (5.119)

that corresponds to an infinitesimal translation with constant parameters �!⌫ = ✏⌫

and

Xµ

⌫

(x) = �µ⌫

(5.120)�

⌫

(x) = 0. (5.121)

The action for the free electromagnetic field is invariant under this kind of trans-formation, since it is just a shift of the origin of coordinates. Considering the vectorfield Aµ, the Noether current (5.25) is

jµ⌫

=

@L@ (@

µ

A�)(@⇢

A�)� L�µ⇢

�

�⇢⌫

(5.122)

or

jµ⌫

=

@L@ (@

µ

A�

)

(@⌫

A�

)� L�µ⌫

. (5.123)

From (5.117) we have


@L@ (@

µ

A�

)

= �Fµ� (5.124)

and then

jµ⌫

= �Fµ�@⌫

A�

+

1

4

�µ⌫

F �⇢F�⇢

. (5.125)

We rise the ⌫ index to obtain

jµ⌫ = �Fµ

�

@⌫A� +

1

4

⌘µ⌫F �⇢F�⇢

(5.126)

which is clearly non-symmetric in the indices µ, ⌫. However, adding the term ˜jµ⌫ =

Fµ

�

@�A⌫ , we make it symmetric and it becomes

Tµ⌫

= jµ⌫ + ˜jµ⌫ = Fµ

�

F�⌫ +1

4

⌘µ⌫F�⇢F�⇢

(5.127)

which is called the energy-momentum tensor. Now we will show that the added termdoes not violate the conservation law, so the tensor Tµ⌫ will be conserved. The newterm has

@µ

˜jµ⌫

= @µ

(Fµ�@�

A⌫

) = @µ

Fµ�@�

A⌫

+ Fµ�@µ

@�

A⌫

. (5.128)

Note that the last term vanishes because it is the product of an antisymmetrictensor Fµ� and the symmetric @

µ

@�

A⌫

. Meanwhile the first term vanishes because ofMaxwell’s equations in empty space @

µ

Fµ�

= 0. Hence we have

@µ

˜jµ⌫

= 0. (5.129)

On the other hand, for jµ⌫

we have

@µ

jµ⌫

= �@µ

(Fµ�@⌫

A�

) +

1

4

�µ⌫

@µ

(F�⇢F�⇢

) (5.130)

@µ

jµ⌫

= �@µ

(Fµ�@⌫

A�

) +

1

4

@⌫

(F�⇢F�⇢

) (5.131)

@µ

jµ⌫

= �@µ

Fµ�@⌫

A�

� Fµ�@µ

@⌫

A�

+

1

4

@⌫

(F�⇢F�⇢

) (5.132)

which, using again the Maxwell’s equations @µ

Fµ�

= 0, gives

@µ

jµ⌫

= �Fµ�@µ

@⌫

A�

+

1

4

@⌫

(F �⇢F�⇢

) (5.133)

@µ

jµ⌫

= �Fµ�@µ

@⌫

A�

+

1

2

F�⇢@⌫

F�⇢

. (5.134)

Writting

@µ

A�

=

1

2

(@µ

A�

� @�

Aµ

) +

1

2

(@µ

A�

+ @�

Aµ

) (5.135)


@µ

A�

=

1

2

Fµ�

+

1

2

(@µ

A�

+ @�

Aµ

) (5.136)

we have the first term in equation (5.134) as

Fµ�@µ

@⌫

A�

=

1

2

Fµ�@⌫

Fµ�

+

1

2

Fµ�@⌫

(@µ

A�

+ @�

Aµ

) (5.137)

Fµ�@µ

@⌫

A�

=

1

2

Fµ�@⌫

Fµ�

(5.138)

where the second term vanishes because of the antisymmetry of Fµ� and the symmetryof the term in parenthesis. Thus, equation (5.134) becomes

@µ

jµ⌫

= �1

2

Fµ�@⌫

Fµ�

+

1

2

F�⇢@⌫

F�⇢

= 0, (5.139)

and therefore it is clear that the energy-momentum tensor satisties the conservationlaw

@µ

Tµ⌫

= 0. (5.140)

5.5.1.1 Conserved Charge

In this case, the associated charge is called the energy-momentum vector and is definedby

P ⌫ =

ˆP⌫d3x =

ˆT o⌫d3x =

ˆd3x

F 0

�

F�⌫ +1

4

⌘0⌫F �⇢F�⇢

�

(5.141)

P ⌫ =

ˆd3x

�F 0

�

F ⌫� +

1

4

⌘0⌫F�⇢F�⇢

�

. (5.142)

The zero component of this vector is

P 0

=

ˆd3x

�F 0

�

F 0�

+

1

4

F�⇢F�⇢

�

(5.143)

P 0

=

ˆd3x

�F 0

i

F 0i

+

1

4

F�⇢F�⇢

�

. (5.144)

Remembering that

F 0i

= @0Ai � @iA0

=

1

c

@Ai

@t+r

i

� = �Ei

(5.145)

andF 0

i

= @0Ai

� @i

A0

= �1

c

@Ai

@t�r

i

� = Ei

, (5.146)

we have


F 0

i

F 0i

= ��

�

�

~E�

�

�

2

. (5.147)

Similarly, the other term gives

1

4

F�⇢F�⇢

= �1

2

✓

�

�

�

~E�

�

�

2

��

�

�

~B�

�

�

2

◆

. (5.148)

Therefore the component P 0 is

P 0

=

ˆd3x

�

�

�

~E�

�

�

2

� 1

2

✓

�

�

�

~E�

�

�

2

��

�

�

~B�

�

�

2

◆�

(5.149)

P 0

=

ˆd3x

1

2

✓

�

�

�

~E�

�

�

2

+

�

�

�

~B�

�

�

2

◆

(5.150)

and we identify

P0

=

1

2

✓

�

�

�

~E�

�

�

2

+

�

�

�

~B�

�

�

2

◆

= U (5.151)

with the energy density of the electromagnetic field.The spatial components of P ⌫ are

P i

=

ˆd3x

⇥

�F 0

�

F i�

⇤

. (5.152)

To interpret these quantities consider, for example, the component P 3,

P 3

=

ˆd3x

⇥

�F 0

�

F 3�

⇤

=

ˆd3x

⇥

�F 0

1

F 31 � F 0

2

F 32

⇤

. (5.153)

Using

F 31

= @3A1 � @1A3

= �@Ax

@z+

@Az

@x= �B

y

(5.154)

F 32

= @3A2 � @2A3

= �@Ay

@z+

@Az

@y= B

x

(5.155)

we obtain

P 3

=

ˆd3x [E

x

By

� Ey

Bx

] =

ˆd3x

⇣

~E ⇥ ~B⌘

z

. (5.156)

Thus, the spatial components of the conserved quantity P ⌫ form the 3-dimensionalvector

~P =

ˆd3x

⇣

~E ⇥ ~B⌘

=

ˆd3x

~S

c(5.157)


where ~S = c⇣

~E ⇥ ~B⌘

is called the Poynting Vector and represents the energy per

unit time and per unit area transported by the fields. Thus we can write

P⌫ =

U ,~S

c

!

. (5.158)

The conservation law for the µ = 0 component of the energy-momentum tensorsays

@⌫

T 0⌫

= @⌫

P⌫ = @0

P0

+

1

c~r · ~S = 0 (5.159)

@U@t

+

~r · ~S = 0. (5.160)

This result, called Poynting’s Theorem, confirms that the Poynting vector is theenergy flux density or the momentum radiation flowing into or out of a volume whichleads to an increase or decrease in the energy.

However, for the electromagnetic field not only is the energy conserved but mo-mentum is itself conserved. The rate of change of total momentum in a volume isdue to momentum flux flowing in and out of the system. Just as T 0i

= Pi

=

S

i

c

isthe energy flux, the components T 1i, T 2i and T 3i are the momentum flux. To probeit, consider the µ = i components of the conservation law for the energy-momentumtensor,

@⌫

T ⌫i =1

c

@T 0i

@t+r

j

T ji

= 0 (5.161)

or in terms of the components of the Poynting vector,

1

c2@Si

@t+r

j

T ji

= 0. (5.162)

Defining the 3-dimensional vector ~T i

=

�

T 1i, T 2i, T 3i

�

, this equation can be writ-ten

1

c2@Si

@t+

~r · ~T i

= 0 (5.163)

which shows that ~T i can be interpreted as the flux of the energy flux.We conclude that for the electromagnetics field not only the total energy in a a

volumeÚd3x is conserved but also the total momentum of the radiation

´~Sd3x is

conserved. The above analysis permit us to interpret the different components of theenergy-momentum tensor as

Tµ⌫

=

✓

U [Energy Density] ~S [Energy Flux]~S [Energy Flux] T ij

[Momentum Flux]

◆

. (5.164)


5.5.2 Infinitesimal Lorentz TransformationConsider now the infinitesimal Lorentz transformation,

x0µ= ⇤

µ

⌫

x⌫ ⇡ xµ

+ ✏µ⌫x⌫

, (5.165)where the coeficients ✏µ⌫ are antisymmetric,

✏µ⌫ = �✏⌫µ. (5.166)There are six independent parameters in this Lorentz trasnformation. As well

known, three of them correspond to spatial rotations while the other three charac-terize the Lorentz boosts. In general, the relativistic fields are chosen to belong to arepresentation of the Lorentz group. This means that under a Lorentz transforma-tion, the components of the field mix together, as, for instance, a vector field doesunder rotations. Therefore, we will require the transformation of the fields under theinfinitesimal Lorentz transformation to be of first order in the parameters ✏µ⌫ in theform

A0µ

= Aµ � 1

2

✏↵�

�

⌃

↵�

�

µ

⌫

A⌫ , (5.167)

where the coefficients ⌃↵� (antisymmetric in its indices ↵,�) is the spin tensor of thetheory and define a matrix in the indices (µ, ⌫) which can be shown to be the represen-tative of the infinitesimal generators of the Lorentz group in the field representation2.This time, the infinitesimal parameters �!⇢ are given by the relations

Xµ

⇢

�!⇢ = ✏µ⌫x⌫

(5.168)

�

µ

⇢

�!⇢ = �1

2

✏↵�

�

⌃

↵�

�

µ

⌫

A⌫ . (5.169)

From equation (5.24),

�S = �ˆ

d4x@µ

⇥

jµ⇢

�!⇢⇤

= 0, (5.170)

the invariance of the action under this transformation implies the conservation

@µ

⇥

jµ⇢

�!⇢⇤

= @µ

✓

@L@ (@

µ

A�)@⌫

A� � L�µ⌫

◆

X⌫

⇢

�!⇢ � @L@ (@

µ

A�)�

�

⇢

�!⇢�

= 0 (5.171)

@µ

✓

@L@ (@

µ

A�)@⌫

A� � L�µ⌫

◆

✏⌫�x�

+

@L@ (@

µ

A�)

1

2

✏↵�

�

⌃

↵�

�

�

⇢

A⇢�

= 0 (5.172)

2The general form of the transformation of a field is 12 ✏↵�

�

⌃↵�

�

µ

⌫

where the coefficient ⌃ takesthe values

�

⌃↵�

�

µ

⌫

= 0 for scalar fields,�

⌃↵�

�

µ

⌫

= gµ↵g�⌫

� gµ�g↵⌫

for scalar fields,�

⌃↵�

�

µ

⌫

= 14

�

�↵�� ↵

�

µ

⌫

for scalar fields, etc.

5.6. INTERNAL SYMMETRIES 95

or in the terms of the energy-momentum tensor,

@µ

Tµ

⌫

✏⌫�x�

+

@L@ (@

µ

A�)

1

2

✏↵�

�

⌃

↵�

�

�

⇢

A⇢�

= 0 (5.173)

@µ

1

2

(Tµ

⌫

✏⌫�x�

+ Tµ

⌫

✏⌫�x�

) +

@L@ (@

µ

A�)

1

2

✏↵�

�

⌃

↵�

�

�

⇢

A⇢�

= 0 (5.174)

@µ

1

2

(Tµ

⌫

✏⌫�x�

� Tµ

⌫

✏�⌫x�

) +

@L@ (@

µ

A�)

1

2

✏↵�

�

⌃

↵�

�

�

⇢

A⇢�

= 0 (5.175)

@µ

1

2

✏⌫� (Tµ

⌫

x�

� Tµ

�

x⌫

) +

@L@ (@

µ

A�)

1

2

✏↵�

�

⌃

↵�

�

�

⇢

A⇢�

= 0 (5.176)

1

2

✏⌫�@µ

Tµ

⌫

x�

� Tµ

�

x⌫

+

@L@ (@

µ

A↵)(⌃

⌫�

)

↵

⇢

A⇢�

= 0. (5.177)

Defining the antisymmetric object

Mµ

⌫�

= Tµ

⌫

x�

� Tµ

�

x⌫

+

@L@ (@

µ

A↵)(⌃

⌫�

)

↵

⇢

A⇢ = �Mµ

�⌫

, (5.178)

the six conserved currents satisfy

@µ

Mµ

�⌫

= 0 (5.179)

and consequently, there are six constants of motion defined as

M�⌫

=

ˆM0

�⌫

d3x. (5.180)

Three of these constants (the ones with spatial values of the indices � and ⌫)correspond to the components of the angular momentum of the field.

5.6 Internal SymmetriesAn internal symmetry is one in which

�xµ

= 0. (5.181)

For example, consider the lagrangian given in Eq. (5.117) and the gauge transfor-mation

A0µ

= Aµ

+ @µ⇤ (5.182)

which evidently lets the action invariant. This time we have

�xµ

= Xµ�! = 0 (5.183)�0

Aµ

= �

µ�! = @µ⇤ (5.184)


and from equation (5.24),

�S = �ˆ

d4x@µ

[jµ�!] = 0, (5.185)

the invariance of the action under this transformation implies the conservation law

@µ

⇥

jµ⇢

�!⇢⇤

= @µ

✓

@L@ (@

µ

A�)@⌫

A� � L�µ⌫

◆

X⌫�! � @L@ (@

µ

A�)�

��!

�

= 0 (5.186)

@µ

@L@ (@

µ

A�)@�⇤

�

= 0 (5.187)

@µ

[Fµ�@�

⇤] = 0. (5.188)

5.7 Canonical Form of the Field Equations

Now we will discuss the Hamiltonian form of the field equations. In general, thelagrangian density is a function of the field variables µ and their derivatives,

L = L [ ↵ (x) , @µ

↵ (x) , xµ

] , (5.189)

and we define the conjugate field variables (conjugated momenta) by

⇡µ

↵

=

@L@ (@

µ

↵). (5.190)

The formal definition of the Hamiltonian density is the quantity

Hµ

⌫

= ⇡µ

↵

@⌫

↵ � �µ⌫

L, (5.191)

which is known as the canonical energy-momentum tensor and is a function of thefield and its conjugate,

Hµ

⌫

= Hµ

⌫

( ↵,⇡⇢↵

, x⇢) . (5.192)

From this definition it is clear that the formal covariant Hamilton equations of thefield theory are

@Hµ

⌫

@⇡µ

↵

= @⌫

↵ (5.193)

@Hµ

⌫

@ ↵= ��µ

⌫

@L@ ↵

= ��µ⌫

@�

⇡�↵

(5.194)

where, in the last step, we have made use of the field equations @L@

↵

= @�

h

@L@(@

�

↵

)

i

.We may also define a scalar function

5.8. PROBLEMS 97

ˆH = ⇡µ

↵

@µ

↵ � L, (5.195)

which gives the Hamilton equations as

@ ˆH@⇡µ

↵

= @µ

↵ (5.196)

@ ˆH@ ↵

= � @L@ ↵

= �@µ

⇡µ

↵

. (5.197)

5.7.1 Hamiltonian for the Electromagnetic FieldThe free electromagnetic field in equation (5.96) gives the conjugate field

⇡↵� =

@L@ (@

↵

A�

)

=

@

@ (@↵

A�

)

�1

4

Fµ⌫Fµ⌫

�

= �F↵� (5.198)

as shown in equation (5.124). Therefore, the hamiltonian for the free electromagneticfield is

H = �F↵�@↵

A�

+

1

4

Fµ⌫Fµ⌫

. (5.199)

Note that this hamiltonian is equivalent to the contraction of the Noether’s current(energy-momentum tensor) given in equation (5.126).

Tµ⌫

= Fµ

�

F �⌫ +1

4

⌘µ⌫F�⇢F�⇢

(5.200)

jµ⌫

= �Fµ�@⌫

A�

+

1

4

�µ⌫

F �⇢F�⇢

(5.201)

j = jµµ

= �Fµ�@µ

A�

+ F�⇢F�⇢

(5.202)

5.8 Problems

1. Obtain the field equations (5.112) from the Fermi lagrangian (5.111).

2. Obtain the field equations (5.115) and (5.116) from the Schwinger lagrangian(5.114).


4. Show that the energy-momentum tensor for the free electromagnetic field givenin equation (5.127) is traceless.

5. Probe relation (5.148).


6. Show that the hamiltonian for the free electromagnetic field is given by equation(5.199).

7. Show that the hamiltonian for the free electromagnetic field can be written asH = T �F↵�@

�

A↵

= �F↵�@�

A↵

, wher T is the trace of the energy-momentumtensor, T = T ⇢

⇢

= 0.

8. Find the conjugate fields and hamiltonian functions for the Fermi and Schwingerlagrangians.

Part III

Interaction of Particles andFields

99

Chapter 6

Interacting Fields

As is well known, an accelerated charged particle produces a field which changes theexternal field in which the particle moves. The new field, in turn, affects the motion ofthe particle. This kind of interaction between particles and fields will be the subjectof study along this chapter using, of course, the lagrangian formalism.

6.1 Interaction Field - External CurrentConsider a field ↵ (x) described by the free field lagrangian L

0

. If there is a variableexternal current j (x) acting as source of the field, it will modify the lagrangianand hence the equations of motion. The interaction between field and current isrepresented by an additional term to the lagrangian,

L = L0

[ ↵ (x) , @µ

↵ (x) , xµ

] + L1

[j (x) , ↵ (x)] . (6.1)

6.1.1 The Electromagnetic Field with SourcesConsider the electromagnetic field Aµ described by the free field lagrangian

L0

= �1

4

Fµ⌫Fµ⌫

. (6.2)

In the presence of a current 4-vector jµ (x) we can add the simplest scalar interactionterm j

µ

Aµ to obtain the lagrangian

L = �1

4

Fµ⌫Fµ⌫

� 1

cjµ

Aµ (6.3)

which gives, as we have seen, Maxwell’s field equations with sources,

@v

Fµ⌫

= �1

cjµ. (6.4)

Making the gauge transformation

101

102 CHAPTER 6. INTERACTING FIELDS

Aµ

! A0

µ

= Aµ

+ @µ

⇤ (6.5)

the lagrangian change as L ! L0where

L0= �1

4

F0µ⌫F

0

µ⌫

� 1

cjµA

0

µ

(6.6)

L0= �1

4

Fµ⌫Fµ⌫

� 1

cjµ (A

µ

+ @µ

⇤) (6.7)

L0= �1

4

Fµ⌫Fµ⌫

� 1

cjµA

µ

� 1

cjµ@

µ

⇤ (6.8)

L0= L� 1

cjµ@

µ

⇤. (6.9)

However, the last term can be rewritten as

L0= L� 1

c@µ

(jµ⇤) +1

c⇤@

µ

jµ (6.10)

and because the second term in the right hand side is a divergence it can be neglected,giving

L0= L+

1

c@µ

(jµ⇤) ⌘ L (6.11)

assuming that the continuity equation @µ

jµ = 0 holds.The energy-momentum tensor for the lagrangian (6.3) is

⇥

µ⌫

= Tµ⌫ � 1

c⌘µ⌫A

�

j� (6.12)

where Tµ⌫ is the free field energy-momentum tensor (5.127),

Tµ⌫

= Fµ

�

F�⌫ +1

4

⌘µ⌫F �⇢F�⇢

. (6.13)

Since the lagrangian (6.3) is not invariant under translations, the energy-momentumtensor ⇥µ⌫ does not satisfy an equation of continuity, in fact

@µ

⇥

µ⌫

= @µ

Tµ⌫ � 1

c⌘µ⌫@

µ

(A�

j�) . (6.14)

Note that

@µ

Tµ⌫

= @µ

�Fµ

�

F ⌫� +

1

4

⌘µ⌫F�⇢F�⇢

�

(6.15)

@µ

Tµ⌫

= � (@µ

Fµ

�

)F ⌫� � Fµ

�

@µ

F ⌫� +

1

2

(@⌫F �⇢)F�⇢

(6.16)

and by Maxwell’s equations,

6.1. INTERACTION FIELD - EXTERNAL CURRENT 103

@µ

Tµ⌫

= �1

cj�

F ⌫� +

�@µF ⌫� +

1

2

@⌫Fµ�

�

Fµ�

(6.17)

@µ

Tµ⌫

= �1

cj�F ⌫

�

+

1

2

[�@µF ⌫� � @µF ⌫� + @⌫Fµ�

]Fµ�

(6.18)

@µ

Tµ⌫

= �1

cj�F ⌫

�

+

1

2

[�@µF ⌫� + @µF �⌫ + @⌫Fµ�

]Fµ�

. (6.19)

Applying the Maxwell’s equations @µF ⌫� + @⌫F�µ + @�Fµ⌫

= 0 we have

@µ

Tµ⌫

= �1

cj�F ⌫

�

+

1

2

[@⌫F�µ + @�Fµ⌫

+ @µF�⌫ + @⌫Fµ�

]Fµ�

(6.20)

@µ

Tµ⌫

= �1

cj�F ⌫

�

+

1

2

[�@⌫Fµ�

+ @�Fµ⌫

+ @µF�⌫ + @⌫Fµ�

]Fµ�

(6.21)

@µ

Tµ⌫

= �1

cj�F ⌫

�

+

1

2

[@�Fµ⌫

+ @µF�⌫ ]Fµ�

. (6.22)

Since the term in parenthesis is symmetric in the indices µ,� and is multiplied bythe antisymmetic tensor F

µ�

, the second term in the right hand side vanishes and weobtain finally

@µ

Tµ⌫

= �1

cj�F ⌫

�

= �k⌫ (6.23)

where we introduced the vector k⌫ called the force density. Equation (6.23) is veryinteresting because it tell us that when the field is driven by a current j (x), theenergy momentum tensor Tµ⌫ no longer satisfies a continuity equation (as in the caseof free fields). Therefore, the energy and momentum of the field change as a functionof time, and the rate of change are given in terms of the force density 4-vector k⌫ .

Replacing this result in the divergence of ⇥µ⌫ we obtain

@µ

⇥

µ⌫

= �1

cj�F ⌫

�

� 1

c⌘µ⌫@

µ

(A�

j�) (6.24)

@µ

⇥

µ⌫

= �1

cj�

F ⌫� � 1

c@⌫ (A�j

�

) (6.25)

that tell us that this tensor does not satisfy a continuity equation neither. However,introducing an energy-momentum tensor ✓µ⌫ for the current, i.e. the matter tensor,thorugh the equation

@µ

✓µ⌫ = k⌫ =

1

cj�F ⌫

�

, (6.26)

the relation (6.23) takes the form of an equation of continuity


@µ

(Tµ⌫

+ ✓µ⌫) = 0. (6.27)

In conclusion, we say that the total energy and momentum of the field plus matter

are conserved, although they are not conserved separately. This means that there isan exchange of energy and momentum between field and matter systems.

6.2 Interaction Field - ParticleConsider now a system composed by the field ↵ and a set of particles when thereis an energy exchange between them. We may consider that there is a current j (x)that is a function of the particle’s coordinates yµ (⌧) and their derivatives. Hence, thetotal lagrangain can be written as

L = L0

[ ↵ (x) , @µ

↵ (x) , xµ

] + L(P )

0

[yµ (⌧) , yµ (⌧)]

+L1

[ ↵ (x) , @µ

↵ (x) , yµ (⌧) , yµ (⌧)] , (6.28)

where L0

is the lagrangian of the free field, L(P )

0

is the lagrangian of the freeparticle, and L

1

is the lagrangian of interaction. Varying the total lagrangian withrespect to the particle’s coordinates gives the equations of motion

m0

yµ = fµ

( ↵) (6.29)

where the force depends now on the field variables. Similarly, the variation of L withrespect to the variables ↵ gives the field equations,

�L� ↵

= K↵

(y) , (6.30)

where the inhomogeneus source term is due to the presence of particles. The free fieldequations are recovered when k

↵

= 0. The coupled equations (6.29) and (6.30) forma complete system to describe the interaction between particles and fields.

6.2.1 A Charged Particle in an Electromagnetic FieldAs an example of this interaction, consider a single charged particle with world lineyµ = yµ (⌧). This particle defines a current jµ (x) that, as a function of xµ, is zeroeverywhere except along the world line. The corresponding charge density is expressedin terms of a � function,

⇢ (x) = q� [x� y (⌧)] , (6.31)

and the current density is defined as jµ = qtot

yµ =

´⇢yµds, which is written in this

case as

jµ (x) = q

ˆdsyµ (⌧) � (x� y (⌧)) (6.32)

6.2. INTERACTION FIELD - PARTICLE 105

where the integral is taken along the world line of the particle. In terms of its propertime, this is

jµ (x) = qc

ˆ+1

�1d⌧ yµ (⌧) � (x� y (⌧)) (6.33)

where the 4-velocity satisfies yµ

yµ = c2. The current density jµ satisfies the equationof continuity. To probe it, note that

@µ

jµ = qc

ˆ+1

�1d⌧ yµ (⌧)

@

@xµ

[� (x� y (⌧))] . (6.34)

Using the property of the Dirac’s delta function

@

@xµ

[� (x� y (⌧))] = � @

@yµ[� (x� y (⌧))] (6.35)

and the relation

@f

@yµ@yµ

@⌧=

df

d⌧(6.36)

we have

@µ

jµ = �qc

ˆ+1

�1d⌧

d

d⌧[� (x� y (⌧))] = �qc � (x� y (⌧))|⌧=+1

⌧=�1 (6.37)

@µ

jµ = 0. (6.38)

The complete lagrangian describing the field ands the particle is

L = L0

+ L(P )

0

+ L1

(6.39)

donde

L0

=

ˆd3xL

0

= �ˆ

d3x1

4

Fµ⌫Fµ⌫

(6.40)

L1

=

ˆd3xL

1

= �ˆ

d3x1

cjµ

Aµ (6.41)

y

L(P )

0

= �m0

c

q

[yµ (⌧)]2

. (6.42)

In order to describe the field, we consider the terms in the lagrangian involvingAµ, i.e.

L0

+ L1

= �1

4

Fµ⌫Fµ⌫

� 1

cjµ

Aµ (6.43)


which give the field equations (6.4),

@⌫

Fµ⌫

= �q

ˆ+1

�1d⌧ yµ (⌧) � (x� y (⌧)) . (6.44)

This time, the divergence of Tµ⌫ (6.23) gives

@µ

Tµ⌫

= qF ⌫�ˆ

+1

�1d⌧ y

�

(⌧) � (x� y (⌧)) (6.45)

@µ

Tµ⌫

= qF ⌫� y�

(⌧)|y(⌧)=x

. (6.46)

This equation shows that the divergence of the energy-momentum tensor of theelectromagnetic field at point x is essentially the Lorentz force on the particle at pointyµ (⌧) = x.

In order to obtain the equation of motion for the particle, we consider the termsin the lagrangian involving the particle’s coordinates, i.e.

L(P )

0

+ L1

= �m0

c

q

[yµ (⌧)]2 � 1

c

ˆd3xj

µ

Aµ (6.47)

from which the equation of motion is

d

d⌧[m

0

yµ

] =

q

cy⌫F

µ⌫

. (6.48)

Now, we will consider only the free particle lagrangian

L(P )

0

= m0

c

q

[yµ (⌧)]2 (6.49)

which gives the variation

�L(P )

0

=

@L(P )

0

@yµ�yµ +

@L(P )

0

@yµ�yµ =

@L(P )

0

@yµ�yµ +

@L(P )

0

@yµ@ (�yµ)

@⌧(6.50)

�L(P )

0

=

@L(P )

0

@yµ�yµ +

@L(P )

0

@yµ@ (�yµ)

@⌧(6.51)

and applying Euler-Lagrange equations,

�L(P )

0

=

@

@⌧

"

@L(P )

0

@yµ

#

�yµ +

@L(P )

0

@yµ@ (�yµ)

@⌧. (6.52)

Under the particular case of a traslation, �yµ = ✏µ = cte. and thus

�L(P )

0

=

@

@⌧

"

@L(P )

0

@yµ

#

✏µ (6.53)

6.2. INTERACTION FIELD - PARTICLE 107

�L(P )

0

= y⌫@

@y⌫

"

@L(P )

0

@yµ

#

✏µ. (6.54)

Assuming invariance under translations, �L(P )

0

= 0, we obtain the equation

y⌫@

@y⌫

"

@L(P )

0

@yµ

#

= 0 (6.55)

y⌫@

@y⌫[m

0

cyµ

] = 0. (6.56)

Since the 4-position and the 4-velocity are assumed to be independent we can write

@

@y⌫[m

0

yµy⌫ ] = 0. (6.57)

This equation shows that the energy-momentum tensor (conserved quantity associatedwith translational invariance) can be defined as

✓µ⌫ = m0

ˆd⌧ yµy⌫� (x� y (⌧)) (6.58)

which obviously has

✓ = ✓µµ

= m0

c2. (6.59)

The divergence of this tensor is

@µ

✓µ⌫ = m0

ˆd⌧ yµy⌫@

µ

[� (x� y (⌧))] (6.60)

@µ

✓µ⌫ = �m0

ˆd⌧ yµy⌫

@

@yµ[� (x� y (⌧))] (6.61)

@µ

✓µ⌫ = �m0

ˆd⌧ y⌫

d

d⌧[� (x� y (⌧))] (6.62)

@µ

✓µ⌫ = �m0

ˆd⌧

d

d⌧[y⌫� (x� y (⌧))] +m

0

ˆd⌧ y⌫� (x� y (⌧)) (6.63)

but the first term vanishes and thus

@µ

✓µ⌫ = m0

ˆd⌧ y⌫� (x� y (⌧)) . (6.64)

Equations (6.45) and (6.64) give

@µ

(Tµ⌫

+ ✓µ⌫) =

ˆ+1

�1[�qF ⌫� y

�

(⌧) +m0

cy⌫ ] � (x� y (⌧)) d⌧ (6.65)


which vanishes by using the equation of motion (6.48),

@µ

(Tµ⌫

+ ✓µ⌫) = 0. (6.66)

6.3 Interaction between FieldsFinally, we will consider the interaction between fields. Remember that quantum fieldtheory considers particles and fields as the same entity, hence it is natural to considernow the interaction between the electromagnetic field with another kind of field.

Consider the vector field A↵ and let the charged particles be described by a com-plex scalar field � (x). The total lagrangian will be written as

L = L0

[A↵ (x)] + L(�)

0

[� (x)] + L1

[A↵ (x)� (x)] (6.67)

where L0

and L(�)

0

are the lagrangian of the free electromagnetic and scalar fieldsgiven by equations (5.96) and (5.67), respectively. The interaction lagrangian, L

1

, iswritten by taking into account the requirement of Lorentz and gauge invariance asfollows.

As we have seen, the complex scalar field is invariant under gauge transformationsof the first kind,

�! �0 = �eiq� (6.68)

with � a constant, while the electromagnetic field is invariant under gauge transfor-mations

Aµ

! A0µ

= Aµ

+ @µ

⇤ (x) . (6.69)

In order to build the interaction term, we will require the whole lagrangian tobe invariant simultaneously under (6.69) and under the gauge transformation of thesecond kind,

�! �0 = �eiq⇤(x). (6.70)

Note that the expression

Dµ

� = (@µ

� iqAµ

)� (6.71)

changes under the desired transformations as

D0µ

�0 =�

@µ

� iqA0µ

�

�0 = [@µ

� iq (Aµ

+ @µ

⇤ (x))]⇣

�eiq⇤(x)

⌘

(6.72)

D0µ

�0 = eiq⇤(x)

[@µ

� iqAµ

]�+ � [@µ

� iq@µ

⇤ (x)] eiq⇤(x) (6.73)

D0µ

�0 = eiq⇤(x)Dµ

�+ � [iq@µ

⇤ (x)� iq@µ

⇤ (x)] eiq⇤(x) (6.74)

6.3. INTERACTION BETWEEN FIELDS 109

D0µ

�0 = e�iq⇤(x)Dµ

�. (6.75)

Taking the complex conjugate of this equation gives

D0⇤µ

�0⇤

= e�iq⇤(x)D⇤µ

�⇤ (6.76)

and therefore we obtain the invariance of the product,

Dµ

�0Dµ⇤�0⇤

= Dµ

�Dµ⇤�⇤. (6.77)

This property invites us to replace the partial derivative operator as

@µ

! Dµ

= @µ

� iqAµ

(6.78)

in the free fields lagrangian . This gives

L = �1

4

Fµ⌫Fµ⌫

+

1

2

[@µ

� iqAµ

]� [@µ + iqAµ

]�⇤ � 1

2

m2��⇤ (6.79)

L = �1

4

Fµ⌫Fµ⌫

+

1

2

(@µ

�) (@µ�⇤)+1

2

q2Aµ

Aµ��⇤� 1

2

iqAµ

[�@µ

�⇤ � �⇤@µ

�]� 1

2

m2��⇤

(6.80)

L = L0

[A↵ (x)] + L(�)

0

[� (x)] +1

2

q2Aµ

Aµ��⇤ � 1

2

iqAµ

[�@µ

�⇤ � �⇤@µ

�] . (6.81)

Defining the current density as in equation (5.95),

jµ

=

1

2

iqc [�@µ

�⇤ � �⇤@µ

�] (6.82)

we write the lagrangian as

L = L0

[A↵ (x)] + L(�)

0

[� (x)]� 1

cjµ

Aµ

+

1

2

q2Aµ

Aµ��⇤ (6.83)

from which we identify the interaction lagrangian as

L1

[A↵ (x)� (x)] = �1

cjµ

Aµ

+

1

2

q2Aµ

Aµ��⇤. (6.84)

The variation of the lagrangian with respect to the fields Aµ and � gives thecoupled field equations

@v

Fµ⌫

= �1

cjµ + q2Aµ��⇤ (6.85)

andh

(@µ

� iqAµ

)

2 �m2

i

� = 0. (6.86)


These equations show that one field acts as the source of the other. However, itis not obvious the way in which the coupling between fields is made.

Since the obtained lagrangian is invariant under inhomogeneous Lorentz trans-formations, there are several conserved quantities. In particular, the total energy-momentum tensor ⇥µ⌫ is conserved,

@µ

⇥

µ⌫

= 0, (6.87)

showing that there is an exchange of energy and momentum between the fields.

6.4 Problems1. Using the definition of the Noether current, show that the energy momentum

tensor associated with the lagrangian (6.3) is given by equation (6.12).

2. Derive particle’s equation of motion (6.48) from the lagrangian (6.47).

3. Show that the lagrangian (6.83) gives the coupled field equations (6.85) and(6.86).

4. Deduce the energy-momentum tensor for the lagrangian (6.83) and show thatit satisfies a continuity equation.

Chapter 7

Solution of the Equations ofMotion

In this section we will begin the solution and interpretation of the coupled equationsof motion of fields and particles. In this chapter we will illustrate the method ofGreen’s Functions to solve one of the coupled equations.

7.1 Green’s Functions

Consider the free electromagnetic field described by the equation

@v

Fµ⌫

= �1

cjµ. (7.1)

Choosing the gauge

@⌫

A⌫ = 0 (7.2)

the field equation becomes

⇤Aµ

=

1

cjµ. (7.3)

If G (x� x0) is a solution of the equation

⇤G (x� x0) = �4 (x� x0

) , (7.4)

then the most general solution of the field equation (7.3) is

Aµ

(x) = Aµ

0

(x) +1

c

ˆd4x0G (x� x0

) jµ (x0) , (7.5)

where Aµ

0

(x) is any solution of the homogeneous equation

111

112 CHAPTER 7. SOLUTION OF THE EQUATIONS OF MOTION

⇤Aµ

0

(x) = 0. (7.6)

Clearly, solution (7.5) is not unique, unless specified by boundary conditions. TheGreen’s Function G (x� x0

) represents the field at the point x due to a unit currentdensity at the point x0. In the solution (7.5) this function is integrated over all pointsx0 with a weight factor jµ (x) to obtain the total field due to currents. In order toprobe that the potential in (7.5) is indeed a solution of the non-homogeneous equation(7.3), note that by applying the D’Alambertian operator we obtain

⇤Aµ

(x) = ⇤Aµ

0

(x) +1

c⇤ˆ

d4x0G (x� x0) jµ (x0

) (7.7)

⇤Aµ

(x) =1

c

ˆd4x0⇤G (x� x0

) jµ (x0) (7.8)

⇤Aµ

(x) =1

c

ˆd4x0�4 (x� x0

) jµ (x0) =

1

cjµ. (7.9)

To obtain the Green’s function we will take the Fourier transform of equation(7.4). Remember that a function f (x) and its the Fourier transform f (k) are relatedby

f (x) =1

(2⇡)4

ˆd4kf (k) e�ikx. (7.10)

For example, to obtain the Fourier transform of the � function, consider its integraldefinition

�4 (x� x0) =

1

(2⇡)4

ˆd4ke�ik

(

x�x

0), (7.11)

i.e. the Fourier transform of the � function is one. Now consider the correspondingrelation for the Green’s function,

G (x� x0) =

1

(2⇡)4

ˆd4kG (k) e�ik

(

x�x

0), (7.12)

where G (k) is the Foureir transform of G (x� x0). Inserting these two equations into

(7.4) we get

1

(2⇡)4

⇤ˆ

dkG (k) e�ik

(

x�x

0)

=

1

(2⇡)4

ˆdke�ik

(

x�x

0) (7.13)

ˆdkG (k)⇤e�ik

(

x�x

0)

=

ˆdke�ik

(

x�x

0) (7.14)

�ˆ

dkG (k) e�ik

(

x�x

0)k2 =

ˆdke�ik

(

x�x

0) (7.15)

from which we conclude that

7.2. ELECTROSTATIC FIELD 113

G (k) = � 1

k2(7.16)

which is known as the propagator (in momentum space). Replacing this function intothe Fourier transform of the Green’s function we have

G (x� x0) = � 1

(2⇡)4

ˆd4k

e�ik

(

x�x

0)

k2(7.17)

and the general solution of the field equation becomes

Aµ

(x) = Aµ

0

(x)� 1

(2⇡)4

c

ˆ ˆd4kd4x0jµ (x0

)

e�ik

(

x�x

0)

k2. (7.18)

7.2 Electrostatic FieldConsider the simple case of a static electric field, described by the 4-potential

Aµ

= (�, 0, 0, 0) (7.19)

where � = � (~r). The field equation (7.3) reduces to the Poisson equation

r2� = �1

cj0 = �⇢. (7.20)

The solution of this equation will be given in terms of the 3-dimensional versionof the Green’s function (7.17)

G (~x� ~x0) =

1

(2⇡)3

ˆd3k

ei~

k·(

~x�~x0)

�

�

�

~k�

�

�

2

(7.21)

where the measure in the integral can be written as d3k = K2dKd⌦ and the argumentof the exponential as ~x� ~x0

=

~R so the inner product is ~k · ~R = KR cos ✓. Note thatthe argument of the integral has a positive sign (because of our sign convention forthe spatial part) as well as the measure of the integral. Then it becomes

G (~x� ~x0) =

1

(2⇡)3

ˆ 1

0

ˆ⇡

0

eiKR cos ✓dK sin ✓d✓

ˆ2⇡

0

d� (7.22)

G (~x� ~x0) =

2⇡

(2⇡)3

ˆ 1

0

�eiKR cos ✓

iKR

�

⇡

0

dK (7.23)

G (~x� ~x0) =

1

iR (2⇡)2

ˆ 1

0

⇥

eiKR � e�iKR

⇤

KdK (7.24)

G (~x� ~x0) =

1

iR (2⇡)2

ˆ 1

0

2i sin (KR)

KdK (7.25)


G (~x� ~x0) =

2

R (2⇡)2

ˆ 1

0

sin (KR)

KdK (7.26)

G (~x� ~x0) =

2

R (2⇡)2

⇡

2

(7.27)

G (~x� ~x0) =

1

4⇡R=

1

4⇡ |~x� ~x0| (7.28)

This result represents the static field at a distance R from a unit source and as iswell known, it depends only on the magnitude R = |~x� ~x0|. The general solution ofthe Poisson equation is then

� = �0

�ˆ

⇢ (~x0) d3x0

4⇡ |~x� ~x0| (7.29)

where �0

is a solution of the Laplace equation, r2�0

= 0. This part of the solution isdetermined by the boundary conditions. For example, if we require that �! 0 when|~x| ! 1, we need �

0

= 0.

7.3 Non-static Field

In th general 4-dimensional non-static case, we need to solve the integral in Green’sfunction (7.17). The measure can be written as d4k = dk0d3k and then we can writethe integral

G (x� x0) = � 1

(2⇡)4

ˆd3k

ˆ+1

�1dk0

e�ik

0⇣x

0�x

00⌘

ei~

k·(

~x�~x0)

(k0)2 �

�

�

�

~k�

�

�

2

(7.30)

where the integrand is singular because it is a function of k0 with two poles at

k0 = ±r

�

�

�

~k�

�

�

2

= ±K. (7.31)

A straightforward technique for solving this integral employs integration in thecomplex plane (treating k0 as the complex variable). In the lower or upper half k0

plane, depending on the sign of⇣

x0 � x00

⌘

, the integrand goes to zero at infinityexponentially. Therefore the k0 integration transforms into a contour integration. Bychoosing different contours, we get different solutions which differ from each otherby the solution of the homogeneous equations, i.e. different contours correspond todifferent boundary conditions.

7.3. NON-STATIC FIELD 115

7.3.0.1 The Residue Theorem in Bref

Before integrating the different contours, we will review shortly the residue theorem.Let f (z) be an analytic function inside and on a closed curve C, except for isolatedsingular points z

1

, z2

, ..., zN

lying inside C. Then the contour integration of f (z) gives

˛C

f (z) dz = 2⇡i

N

X

k=1

Resz=z

k

[f (z)] (7.32)

where the contour C is evaluated counterclockwise and Resz=z

k

[f (z)] stands for theresidue of f (z) at z = z

k

which can be evaluated as the coefficient c�1

in the Laurentexpansion of f (z),

f (z) =

1X

n=�1cn

(z � zk

)

n

. (7.33)

If the pole zk

is of order m, the residue can be calculated as

Resz=z

k

[f (z)] = lim

z!z

k

1

(m� 1)!

dm�1

dzm�1

[(z � zk

)

m

f (z)] . (7.34)

In Figure 7.1, we can see three different contours representing three importantphysical situations. Lifting the integration segment along the real axis by the amount±i� in the direction of the imaginary axis Im (k

0

), we can close the integration loopwith a semicircle at infinity in either the positive or negative imaginary half plane.To choose the correct loop we need to ensure that the integral from the semicircle is

zero. When⇣

x0 � x00

⌘

> 0, we need that the integrand e�ik0

⇣x

0�x

00⌘

! 0 in thelimit |Im (k

0

)| ! 1, therefore we need that Im (k0

) < 0 and hence the loop mustbe closed in the negative half-plane. On the other hand, when

⇣

x0 � x00

⌘

< 0 weneed that Im (k

0

) > 0 and hence the loop must be closed in the positive half-plane.Depending on the choosen trajectory, the loop encloses two singularities or none.

7.3.1 Retarded Green’s FunctionsConsider the contour C

R

for which the corresponding Green’s function is denotedDret

(x� x0). Obviously we need to close the contour an we will make it in the upper

half plane for⇣

x0 � x00

⌘

< 0 or in the lower half plane for⇣

x0 � x00

⌘

> 0. In theformer case, the contour does not include any poles, so the integral is zero. In thelatter case, the k0 integration gives a residue of the integrand at the two poles. Thisgives

˛C

R

dk0e�ik

0⇣x

0�x

00⌘

(k0)2 �

�

�

�

~k�

�

�

2

= �2⇡iResk0=±|~k|

2

6

4

e�ik

0⇣x

0�x

00⌘

(k0)2 �

�

�

�

~k�

�

�

2

3

7

5

(7.35)


Figure 7.1: Contours for the k0 integration in Green’s function.

where the minus sign comes from the clockwise sense of CR

when closing in the lowerhalf-plane. In order to evaluate the residue, we write the denominator as

�

k0�

2��

�

�

~k�

�

�

2

=

⇣

k0 ��

�

�

~k�

�

�

⌘ ⇣

k0 +�

�

�

~k�

�

�

⌘

and use relation (7.34) with m = 1. Thus we write

˛C

R

dk0e�ik

0⇣x

0�x

00⌘

(k0)2 �

�

�

�

~k�

�

�

2

= �2⇡i

2

4

lim

k

0!|~k|e�ik

0⇣x

0�x

00⌘

k0 +�

�

�

~k�

�

�

+ lim

k

0!�|~k|e�ik

0⇣x

0�x

00⌘

k0 ��

�

�

~k�

�

�

3

5

(7.36)which gives

˛C

R

dk0e�ik

0⇣x

0�x

00⌘

(k0)2 �

�

�

�

~k�

�

�

2

= �2⇡i

2

4

e�i|~k|

⇣x

0�x

00⌘

2

�

�

�

~k�

�

�

� ei|~k|

⇣x

0�x

00⌘

2

�

�

�

~k�

�

�

3

5 . (7.37)

Therefore, Green’s function is

Dret

(x� x0) =

2⇡i

(2⇡)4

ˆd3k

1

2

�

�

�

~k�

�

�

e�i|~k|

⇣x

0�x

00⌘

� ei|~k|

⇣x

0�x

00⌘�

⇥

⇣

x0 � x00

⌘

ei~

k·(

~x�~x0)

(7.38)where we introduce the Heaviside function ⇥ (t),

⇥ (t) =

(

0 for t < 0

1 for t > 0

(7.39)


in order to account for both possible loops in the contour. Replacing R = |~x� ~x0|and d3k = K2dKd⌦ we have

Dret

(x� x0) =

i⇥⇣

x0 � x00

⌘

(2⇡)3

ˆdKd✓

2⇡

2

K sin ✓

e�iK

⇣x

0�x

00⌘

� eiK

⇣x

0�x

00⌘�

eiKR cos ✓

Dret

(x� x0) =

i⇡⇥⇣

x0 � x00

⌘

(2⇡)3

ˆ 1

0

dKK

ˆ⇡

0

eiKR cos ✓

sin ✓d✓

e�iK

⇣x

0�x

00⌘

� eiK

⇣x

0�x

00⌘�

Dret

(x� x0) =

i⇡⇥⇣

x0 � x00

⌘

(2⇡)3

ˆ 1

0

dKK

�eiKR cos ✓

iKR

�

⇡

0

e�iK

⇣x

0�x

00⌘

� eiK

⇣x

0�x

00⌘�

Dret

(x� x0) =

⇡⇥⇣

x0 � x00

⌘

(2⇡)3

R

ˆ 1

0

dK⇥

eiKR � e�iKR

⇤

e�iK

⇣x

0�x

00⌘

� eiK

⇣x

0�x

00⌘�

Dret

(x� x0) =

⇡⇥⇣

x0 � x00

⌘

(2⇡)3

R

ˆ 1

0

dK

eiK

hR�

⇣x

0�x

00⌘i

+ e�iK

hR�

⇣x

0�x

00⌘i

�eiK

hR+

⇣x

0�x

00⌘i

� e�iK

hR+

⇣x

0�x

00⌘i�

.

By using the limits of integration, we can write this integrals as

Dret

(x� x0) =

⇡⇥⇣

x0 � x00

⌘

(2⇡)3

R

ˆ 1

�1dK

eiK

hR�

⇣x

0�x

00⌘i

� eiK

hR+

⇣x

0�x

00⌘i�

.

In virtue of 2⇡� (u) =´dkeiku and the property � (�u) = � (u), we get [1]

Dret

(x� x0) =

⇡⇥⇣

x0 � x00

⌘

(2⇡)2

R

h

�⇣

x0 � x00 �R

⌘

� �⇣

x0 � x00

+R⌘i

Dret

(x� x0) =

⇥

⇣

x0 � x00

⌘

4⇡R

h

�⇣

x0 � x00 �R

⌘

� �⇣

x0 � x00

+R⌘i

.(7.40)

Due to the Heaviside function, we have⇣

x0 � x00

⌘

> 0 and since we need R > 0

the second � function does not contribute, leaving us with


Dret

(x� x0) =

1

4⇡R�⇣

x0 � x00 �R

⌘

(7.41)

or better

Dret

(x� x0) =

1

4⇡ |~x� ~x0|�⇣

x0 � x00 � |~x� ~x0|

⌘

. (7.42)

Equation (7.40) can be rewritten covariantly by using the property of the � func-tion1

��

t2 � a2�

=

1

2a[� (t� a) + � (t+ a)] . (7.43)

Specifically we have

�

⇣

x0 � x00

⌘

2

�R2

�

=

1

2R

h

�⇣

x0 � x00 �R

⌘

+ �⇣

x0 � x00

+R⌘i

(7.44)

and multiplying by the Heavyside function,

⇥

⇣

x0 � x00

⌘

�

⇣

x0 � x00

⌘

2

�R2

�

=

1

2R

h

⇥

⇣

x0 � x00

⌘

�⇣

x0 � x00 �R

⌘

+⇥

⇣

x0 � x00

⌘

�⇣

x0 � x00

+R⌘i

⇥

⇣

x0 � x00

⌘

�

⇣

x0 � x00

⌘

2

�R2

�

=

1

2R⇥

⇣

x0 � x00

⌘

�⇣

x0 � x00 �R

⌘

,(7.45)

where the second term in the right hand side vanishes in virtue of the arguments ofthe � and ⇥ functions. Therefore, equation (7.40) can be written as

Dret

(x� x0) =

1

2⇡�

⇣

x� x0⌘

2

�

⇥

⇣

x0 � x00

⌘

. (7.46)

This is known as the retarded function because of its dependence in the � function.Note that the field at the point x due to the unit source at point x0 is different fromzero only if the time at the source point ~x0 is earlier than the time at which oneevaluates the field at ~x. Moreover, the retarded Green’s function shows that onlysignals travelling with a velocity of light from ~x0 to ~x contribute to the field becausethe � function is zero unless its argument is zero. In Figure 7.2 we draw the light conecentered at x and the green’s function tell us that x0 must lie on the backward lightcone.

1This is a particular case of the composition property of the � function, see for example [1] and[6], � [g (x)] =

P

k

�(x�x

k

)|g0(x

k

)| , where xk

are the roots of function g (x).


Figure 7.2: Field at point (t, ~x) produced by a unit source at point (t0, ~x0).

7.3.2 Advanced Green’s FunctionsNow we will consider the contour C

A

, which produces the Advanced Green’s Function

Dadv

(x� x0). This time we have

Dadv

(x� x0) =

(

= 0 for x0 > x00

6= 0 for x0 < x00

, (7.47)

i.e. that the advanced function will represents a field at the point ~x at time t due tounit charge at ~x0 at a later time t0. Of course this kind of solution is contrary to thephysical idea of propagation with increasing time, but due to the form of the fieldequations, it is a valid solution. Following the same procedure as for the retardedfunction, we obtain

Dadv

(x� x0) =

1

4⇡ |~x� ~x0|�⇣

x0 � x00

+ |~x� ~x0|⌘

(7.48)

or

Dadv

(x� x0) =

1

2⇡�

⇣

x� x0⌘

2

�

⇥

⇣

x00 � x0

⌘

. (7.49)

The advanced Green’s function shows that only signals travelling with a velocityof light from ~x to ~x0 contribute to the field because the � function is zero unless itsargument (x� x0

) is a light-like vector. Hence, this time x0 must lie on the forwardlight cone of point x and we can write

Dret

(�x) = Dadv

(x) . (7.50)


7.3.3 Other Green’s FunctionsNow consider the contour C

F

in Figure 7.1. Depending on the sign of⇣

x0 � x00

⌘

we get the contribution of one or the other pole only. Thus we write the completeGreen’s function as

DF

(x� x0) = ⇥

⇣

x0 � x00

⌘

D+

(x� x0)�⇥

⇣

x00 � x0

⌘

D�(x� x0

) (7.51)

where D+ and D� are the contributions of the poles +K and �K, respectively. The+ function is

˛C

+F

dk0e�ik

0⇣x

0�x

00⌘

(k0)2 �

�

�

�

~k�

�

�

2

= �2⇡iResk0=+|~k|

2

6

4

e�ik

0⇣x

0�x

00⌘

(k0)2 �

�

�

�

~k�

�

�

2

3

7

5

(7.52)

where the minus sign comes from the clockwise sense of CF

when closing in the lowerhalf-plane. In order to evaluate the residue, we write the denominator as

�

k0�

2��

�

�

~k�

�

�

2

=

⇣

k0 ��

�

�

~k�

�

�

⌘ ⇣

k0 +�

�

�

~k�

�

�

⌘

and use relation (7.34) with m = 1,

˛C

+F

dk0e�ik

0⇣x

0�x

00⌘

(k0)2 �

�

�

�

~k�

�

�

2

= �2⇡i

2

4

lim

k

0!|~k|e�ik

0⇣x

0�x

00⌘

k0 +�

�

�

~k�

�

�

3

5 (7.53)

˛C

+F

dk0e�ik

0⇣x

0�x

00⌘

(k0)2 �

�

�

�

~k�

�

�

2

= �2⇡ie�i|~k|

⇣x

0�x

00⌘

2

�

�

�

~k�

�

�

. (7.54)

Thus, the Green’s function is

D+

(x� x0) =

i

(2⇡)3

ˆd3k

1

2

�

�

�

~k�

�

�

e�i|~k|

⇣x

0�x

00⌘

ei~

k·(

~x�~x0). (7.55)

The first exponential in in the integral can be written as

e�i|~k|

⇣x

0�x

00⌘

=

ˆdk0�

⇣

k0 ��

�

�

~k�

�

�

⌘

e�ik

0⇣x

0�x

00⌘

(7.56)

and therefore

D+

(x� x0) =

i

(2⇡)3

ˆd4k

�⇣

k0 ��

�

�

~k�

�

�

⌘

2

�

�

�

~k�

�

�

e�ik

0⇣x

0�x

00⌘

ei~

k·(

~x�~x0) (7.57)

D+

(x� x0) =

i

(2⇡)3

ˆd4k

�⇣

k0 ��

�

�

~k�

�

�

⌘

2

�

�

�

~k�

�

�

e�ik

(

x�x

0). (7.58)


Now we will introduce the � function

�

�

k0�

2 ��

�

�

~k�

�

�

2

�

=

1

2⇡

ˆd!e

i!

h(

k

0)

2�|~k|2i

(7.59)

or

��

k2�

=

1

2⇡

ˆd!ei!k

2

, (7.60)

which, following eq. (7.43), satisfies

��

k2�

=

1

2

�

�

�

~k�

�

�

h

�⇣

k0 ��

�

�

~k�

�

�

⌘

+ �⇣

k0 +�

�

�

~k�

�

�

⌘i

. (7.61)

Multiplying by the Heavyside function,

⇥

�

k0�

��

k2�

=

1

2

�

�

�

~k�

�

�

h

⇥

�

k0�

�⇣

k0 ��

�

�

~k�

�

�

⌘

+⇥

�

k0�

�⇣

k0 +�

�

�

~k�

�

�

⌘i

⇥

�

k0�

��

k2�

=

1

2

�

�

�

~k�

�

�

⇥

�

k0�

�⇣

k0 ��

�

�

~k�

�

�

⌘

, (7.62)

where the second term in the right hand side vanishes in virtue of the arguments ofthe � and ⇥ functions if k0 > 0. Thus, for the D+ function we obtain finally

D+

(x� x0) =

i

(2⇡)3

ˆd4ke�ik

(

x�x

0)��

k2�

⇥

�

k0�

. (7.63)

Similarly, the D� function can be written as

D�(x� x0

) = � i

(2⇡)3

ˆd4ke�ik

(

x�x

0)��

k2�

⇥

�

�k0�

. (7.64)

It is straightforward to show the additional properties

D+

[� (x� x0)] = �D�

[x� x0] (7.65)

and⇥

D+

(x� x0)

⇤⇤= D�

(x� x0) (7.66)

where ⇤ denotes complex conjugation.Finally, other Green’s function is obtained as the principal value of the k0 inte-

gration. The resulting function can be written as2

2Remember that the principal value of a singular integral is equal to 1/2 the residue of the poleterm.


¯D (x) =1

2

⇥

Dret

(x) +Dadv

(x)⇤

(7.67)

and is an even and real function, i.e.

¯D (x) = ¯D (�x) =⇥

¯D (x)⇤⇤

. (7.68)

7.4 Solving the Inhomogeneous EquationThe solution of the electrodynamics equation (7.3),

⇤Aµ

=

1

cjµ, (7.69)

depends on the boundary conditions (Cauchy data). Let us choose these conditionson the spacelike surface t = �1. The corresponding solutions of the homogeneousequation will be denoted as A

in

(x) and we must use the retarded Green’s functionto write the complete solution

Aµ

(x) = Aµ

in

(x) +1

c

ˆd4x0Dret

(x� x0) jµ (x0

) (7.70)

Aµ

(x) = Aµ

in

(x) +1

4⇡c

ˆ �⇣

x0 � x00 � |~x� ~x0|

⌘

|~x� ~x0| jµ (x0) d4x0. (7.71)

where we have determined the field at point x starting with the initial value att = �1. Now, let us choose the Cauchy data at t = +1. The solutions of thehomogeneous equation will be denoted as A

out

(x) and we will use the advancedGreen’s function,

Aµ

(x) = Aµ

out

(x) +1

c

ˆd4x0Dadv

(x� x0) jµ (x0

) (7.72)

Aµ

(x) = Aµ

out

(x) +1

4⇡c

ˆ �⇣

x0 � x00

+ |~x� ~x0|⌘

|~x� ~x0| jµ (x0) d4x0. (7.73)

In this case we determine the field at point x from the future backwards startingwith its value at t = +1.

Dirac noted that the above solutions, i.e. choosing the retarded Green’s func-tion (ingoing wave) or an advanced function (outgoing wave), give a misleading pic-ture of the physical system because it must have a complete time-reversal symmetry.Therefore, he introduced the radiation field as the difference between the ingoing andoutgoing waves,

Aµ

radiation

(x) = Aµ

out

(x)�Aµ

in

(x) =1

c

ˆd4x0D (x� x0

) jµ (x0) (7.74)

7.4. SOLVING THE INHOMOGENEOUS EQUATION 123

where

D (x� x0) = Dret

(x� x0)�Dadv

(x� x0) (7.75)

is a solution of the homogeneous equation. Some people interpret the radiation fieldas the only physically important one because it represents the change in the vectorpotential at an event produced symmetrically by any given 4-current due to its pastand future motion (i.e. radiation reaction is produced by transfer of momentum bothto a charge from other charges in its past and from a charge to those same charges inits future!).

The radiation Green’s function is

D (x� x0) =

1

2⇡�

⇣

x� x0⌘

2

�

h

⇥

⇣

x0 � x00

⌘

�⇥⇣

x00 � x0

⌘i

(7.76)

D (x� x0) =

1

2⇡�

⇣

x� x0⌘

2

�

sgn⇣

x0 � x00

⌘

, (7.77)

where the sign function sgn (t) is defined as

sgn (t) = ⇥ (t)�⇥ (�t) =

(

�1 for t < 0

+1 for t > 0.(7.78)

Since

Dret

[� (x� x0)] = Dadv

(x� x0) , (7.79)

the homogeneous Green’s function is and odd function,

D (x0 � x) = �D (x� x0) . (7.80)

Using the retarded and advanced Green’s function together with property (7.43)of the � function, we get

D (x� x0) =

i

(2⇡)3

ˆd4k�

�

k2�

sgn⇣

x0 � x00

⌘

e�ik

(

x�x

0). (7.81)

This expression can be rewritten as

D (x� x0) =

i

(2⇡)3

ˆd4k�

�

k2�

sgn�

k0�

e�ik

(

x�x

0) (7.82)

where we have changed the argument in the sign function because the 4-vector k

must be future directed (retarded) or past directed (advanced) when⇣

x0 � x00

⌘

> 0

or⇣

x0 � x00

⌘

< 0, respectively. From this equation we calculate the time derivativeof the Green’s function as


@D (x� x0)

@x0

�

�

�

�

x

0=x

00=

i

(2⇡)3

ˆd4k�

�

k2�

sgn�

k0� �

�ik0�

e�ik

0⇣x

0�x

00⌘

ei~

k·(

~x�~x0)

�

�

�

�

�

x

0=x

00

(7.83)

@D (x� x0)

@x0

�

�

�

�

x

0=x

00=

1

(2⇡)3

ˆd3kei

~

k·(

~x�~x0)

ˆ+1

�1dk0k0�

�

k2�

sgn�

k0�

. (7.84)

The integral over k0 gives one because k0sgn�

k0�

=

�

�k0�

� and thus, using equation(7.61), we have

ˆ+1

�1dk0

�

�k0�

� ��

k2�

=

ˆ+1

�1dk0

�

�k0�

�

2

�

�

�

~k�

�

�

h

�⇣

k0 +�

�

�

~k�

�

�

⌘

+ �⇣

k0 ��

�

�

~k�

�

�

⌘i

(7.85)

ˆ+1

�1dk0

�

�k0�

� ��

k2�

=

1

2

�

�

�

~k�

�

�

h

�

�

�

��

�

�

~k�

�

�

�

�

�

+

�

�

�

�

�

�

~k�

�

�

�

�

�

i

= 1. (7.86)

Hence the time derivative gives

@D (x� x0)

@x0

�

�

�

�

x

0=x

00=

1

(2⇡)3

ˆd3kei

~

k·(

~x�~x0) (7.87)

@D (x� x0)

@x0

�

�

�

�

x

0=x

00= �3 (~x� ~x0

) . (7.88)

This equation can be written covariantly in the form

ˆ�

d�µ

@D (x� x0)

@xµ

f (x) = f (x0) (7.89)

where � is a spacelike surface on which x and x0 lie. Introducing a covariant spatialfunction �µ by

@D

@xµ

= �µ (x� x0) = nµ� (~x� ~x0

) (7.90)

with nµ the normal vector to �. Thus we write

ˆ�

d�µ

�µ (x� x0) f (x) = f (x0

) . (7.91)

7.5. GAUGE CONDITION 125

7.4.1 ScatteringFrom equation (7.74) we relate the incident radiation field with the radiated field by

Aµ

out

(x) = Aµ

in

(x) +1

c

ˆD (x� x0

) jµ (x0) d4x0. (7.92)

In an Scattering Experiment we measure the undisturbed field at distant past ordistant future and then we introduce a disturbance at finite time. Hence, equation(7.92) determines the net effect of the interaction and just as in quantum mechanics,we can introduce a classical scattering matrix S as the ratio between A

out

and Ain

,i.e.

Aµ

out

(x) = Sµ

⌫

A⌫in

(x) , (7.93)

which can be given in terms of D (x� x0).

7.5 Gauge Condition

Along this chapter we have worked in the gauge @µ

Aµ

= 0. Now we will show thathis condition is satisfied by the presented solutions. First we will assume that thehomogeneous part of the solutions satisfy the condition,

@µ

Aµ

in

= @µ

Aµ

out

= 0. (7.94)

Then, equations (7.71) and (7.73) give

@µ

Aµ

(x) =1

c

ˆ@

@xµ

⇥

Dadv,ret

(x� x0)

⇤

jµ (x0) d4x0 (7.95)

or equivalently

@µ

Aµ

(x) = �1

c

ˆ@

@x0µ

⇥

Dadv,ret

(x� x0)

⇤

jµ (x0) d4x0. (7.96)

The integrand can be rewritten to obtain

@µ

Aµ

(x) = �1

c

ˆ@

@x0µ

⇥

Dadv,ret

(x� x0) jµ (x0

)

⇤

d4x0

+

1

c

ˆDadv,ret

(x� x0)

@jµ

@x0µ

d4x0. (7.97)

where the second term vanishes by the equation of continuity @j

µ

@x

0µ

= 0 and the firstintegral can be transformed into a surface integral

@µ

Aµ

(x) = �1

c

ˆ�

d�0µ

Dadv,ret

(x� x0) jµ (x0

) , (7.98)


with � a spacelike surface containing x and x0. Finally, the right hand side integralvanishes because Dadv,ret

(x� x0) = 0 for space-like separated x and x0. Hence we

obtain the desired condition

@µ

Aµ

= 0. (7.99)

7.6 Jefimenko’s EquationsConsidering a region where the external free field satisfies Aµ

in

(x) = 0, the retardedpotential is written in terms of the retarded Green’s function as

Aµ

(x) =1

c

ˆd4x0Dret

(x� x0) jµ (x0

) (7.100)

Aµ

(x) =1

4⇡c

ˆ �⇣

x0 � x00 � |~x� ~x0|

⌘

|~x� ~x0| jµ (x0) d4x0. (7.101)

Performing the integration in the retarded time x00 we obtain

Aµ

(x) =1

4⇡c

ˆjµ�

x0

R

, ~x0�

|~x� ~x0| d3x0, (7.102)

where the retarded time is x0

R

= x0 � |~x� ~x0|. Since jµ =

⇣

⇢c,~j⌘

, the components ofthis equation are

� (x) =1

4⇡

ˆ⇢�

x0

R

, ~x0�

|~x� ~x0| d3x0 (7.103)

~A (x) =1

4⇡c

ˆ ~j�

x0

R

, ~x0�

|~x� ~x0| d3x0. (7.104)

The electromagnetic field is written in term of the retarded potential as

Fµ⌫

(x) = @µ

A⌫

(x)� @⌫

Aµ

(x) (7.105)

or separating its components as

~E = �~r�� 1

c

@ ~A

@t(7.106)

and~B =

~r⇥ ~A. (7.107)

Using the retarded potentials and the relation

@~j�

x0

R

, ~x0�

@x0

=

@~j�

x0

R

, ~x0�

@x0

R

@x0

R

@x0

=

@~j�

x0

R

, ~x0�

@x0

R

(7.108)

7.7. PROBLEMS 127

we obtain the electric field

~E (x) =1

4⇡

ˆ "⇢�

x0

R

, ~x0� (~x� ~x0)

|~x� ~x0|3+

@⇢�

x0

R

, ~x0�

@x0

R

(~x� ~x0)

|~x� ~x0|2� 1

c |~x� ~x0|@~j�

x0

R

, ~x0�

@x0

R

#

d3x0

(7.109)which is the time-dependent generalization of Coulomb’s law (note that when thecharge and current densities do not depend on time this expresion recovers theCoulomb field obtaine from the potential given in equation (7.29)).

On the other hand, using the relation

~r⇥~j�

x0

R

, ~x0�=

@~j�

x0

R

, ~x0�

@x0

R

⇥ (~x� ~x0)

|~x� ~x0| (7.110)

the magnetic field can be written as

~B (x) =1

4⇡c

ˆ "~j �x0

R

, ~x0�

|~x� ~x0|2+

1

|~x� ~x0|@~j�

x0

R

, ~x0�

@x0

R

#

⇥ (~x� ~x0)

|~x� ~x0| d3x0 (7.111)

which corresponds to the generalization of the well-known Biot-Savart law.Equations (7.109) and (7.111) are the causal solutions of Maxwell’s equations (i.e.

retarded solutions) and are called Jefimenko’s Equations. It is interesting to note thatthis equations where presented in the literature jus some decades ago (1966) [8, 9].

7.7 Problems1. Derive explicitly the advanced Green’s function (7.49).

2. Obtain explicitly the D�(x� x0

) function given by equation (7.64).

3. Probe properties (7.65) and (7.66).

4. Show that the radiation Green’s function can be written as equation (7.81).

5. Derive explicitly equation (7.110).


Chapter 8

Radiation

8.1 Lienard-Wiechert Potentials

In this section we will discuss the field of a moving charged particle in a region wherethe external free field satisfies Aµ

in

(x) = Aµ

out

(x) = 0. The current density associatedto the charged particle traveling along the worldline y (⌧) is given by equation (6.33),

jµ (x0) = qc

ˆ+1

�1yµ (⌧) � [x0 � y (⌧)] d⌧, (8.1)

where y (⌧) is the 4-velocity and ⌧ is its proper time. The generated potential at pointx is described by the retarded function (7.71),

Aµ

(x) =1

c

ˆDret

(x� x0) jµ (x0

) d4x0 (8.2)

Aµ

(x) = q

ˆ ˆ+1

�1Dret

(x� x0) yµ (⌧) � [x0 � y (⌧)] d⌧d4x0. (8.3)

Using equation (7.46) for Dret,

Aµ

(x) =q

2⇡

ˆ ˆ+1

�1�

⇣

x� x0⌘

2

�

⇥

⇣

x0 � x00

⌘

yµ (⌧) � [x0 � y (⌧)] d⌧d4x0, (8.4)

we can perform the x0 integration to obtain

Aµ

(x) =q

2⇡

ˆ+1

�1�h

(x� y (⌧))2

i

⇥

�

x0 � y0�

yµ (⌧) d⌧. (8.5)

The argument of the function �h

(x� y (⌧))2

i

has two roots but the ⇥ functionpicks up the root with x0 > y0 (⌧). Note that

129

130 CHAPTER 8. RADIATION

d

d⌧

h

(x� y (⌧))2

i

= �2 (x� y (⌧))�

y�

, (8.6)

and from the property of the � function,

� [g (x)] =X

k

� (x� xk

)

|g0 (xk

)| , (8.7)

where xk

are the roots of function g (x), we have

�h

(x� y (⌧))2

i

=

� (x� y (⌧))

|�2 (x� y (⌧))�

y�

|⌧=⌧0

, (8.8)

where the expression is evaluated at the retarded proper time ⌧0

such that x� y (⌧0

)

is light-like and x0 � y0 (⌧0

) = R > 0. Thus Aµ becomes

Aµ

(x) =q

4⇡

ˆ+1

�1

� (x� y (⌧))

[(x� y (⌧))�

y�

]

⌧=⌧0

yµ (⌧) d⌧. (8.9)

which gives the Lienard-Wiechert potentials,

Aµ

(x) =q

4⇡

yµ

(x� y)�

y�

�

⌧=⌧0

. (8.10)

The conditons defining ⌧0

show that it corresponds to the point where the trajectoryof the particle cuts the past light cone with vertex x. From Figure 8.1, it is clearthat only from the point y (⌧

0

), a light signal can reach the point x. The particle andthe field point are separated by a 3-dimensional distance R =

�

�

�

~R�

�

�

= |~x� ~y (⌧0

)| and

a time interval R

c

=

|x0�y

0(⌧0)|

c

= t � tR

. With a similar procedure, but using theadvanced Green’s function, we obtain the point y (⌧

1

) shown in Figure 8.1 which isthe only point of the trajectory which contributes to the field at point x.

With ~R = ~x� ~y (⌧0

) and writting the 4-velocity as

yµ = (c, ~u)dt

d⌧, (8.11)

we have

(x� y)�

y�

=

⇣

Rc� ~R · ~u⌘ dt

d⌧= Rc

1�~R · ~uRc

!

dt

d⌧. (8.12)

Introducing the normalized vector

n =

~R

R(8.13)

this expression becomes

(x� y)�

y�

= Rc⇣

1� n · ~�⌘ dt

d⌧(8.14)

8.1. LIENARD-WIECHERT POTENTIALS 131

Figure 8.1: Field at point x produced by a point particle moving along the worldliney (⌧). The points at ⌧

0

and ⌧1

contribute to the retarded and adanced potential,respectively.

where ~� =

~u

c

. Hence, the Lienard-Wiechert potential can be written as

Aµ

(x) =q

4⇡cR

2

4

yµ⇣

1� n · ~�⌘

dt

d⌧

3

5

⌧=⌧0

(8.15)

or its components as

� =

q

4⇡R

1

⇣

1� n · ~�⌘

�

�

�

�

�

�

⌧=⌧0

(8.16)

~A =

q

4⇡R

~�⇣

1� n · ~�⌘

�

�

�

�

�

�

⌧=⌧0

= �~�. (8.17)

8.1.1 Coulomb FieldAs an example, consider a particle at rest described by

yµ =

⇣

c⌧,~0⌘

(8.18)

yµ =

⇣

c,~0⌘

. (8.19)

Therefore, the components of the Lienard-Wiechert potential become

� =

q

4⇡R(8.20)


~A = 0, (8.21)

which correspond to the usual Coulomb field.

8.2 The Electromagnetic FieldNow we will evaluate the field tensor Fµ⌫ using equation (8.3). Performing the x0

integration we get

Aµ

(x) = q

ˆ+1

�1Dret

(x� y (⌧)) yµ (⌧) d⌧. (8.22)

By differentiation we have

@⌫Aµ

(x) = q

ˆ+1

�1d⌧

@

@x⌫

⇥

Dret

(x� y (⌧))⇤

yµ (⌧) (8.23)

and since the retarded Green’s function depends on (x� y)2, it can be written as

@⌫Aµ

(x) = q

ˆ+1

�1d⌧

@Dret

@h

(x� y)2

i

@h

(x� y)2

i

@x⌫

yµ (8.24)

@⌫Aµ

(x) = 2q

ˆ+1

�1d⌧

@Dret

@h

(x� y)2

i

(x� y)⌫

yµ (8.25)

@⌫Aµ

(x) = 2q

ˆ+1

�1d⌧

dDret

d⌧

d⌧

dh

(x� y)2

i

(x� y)⌫

yµ. (8.26)

Using equation (8.6) we obtain

@⌫Aµ

(x) = �q

ˆ+1

�1d⌧

dDret

d⌧

(x� y)⌫

yµ

(x� y)�

y�

(8.27)

and integrating by parts

@⌫Aµ

(x) = �q

Dret

(x� y)⌫

yµ

(x� y)�

y�

�

+1

�1+ q

ˆ+1

�1d⌧

d

d⌧

(x� y)⌫

yµ

(x� y)�

y�

�

Dret

(x� y) .

(8.28)The first term is zero, while the second term can be integrated by replacing the

retarded Green’s function as the � function in equation (7.46),

@⌫Aµ

(x) =q

2⇡

ˆ+1

�1d⌧

d

d⌧

(x� y)⌫

yµ

(x� y)�

y�

�

�h

(x� y)2

i

⇥

�

x0 � y0�

. (8.29)

8.2. THE ELECTROMAGNETIC FIELD 133

Using again equation (8.6) we get

@⌫Aµ

(x) =q

4⇡

ˆ+1

�1d⌧

d

d⌧

(x� y)⌫

yµ

(x� y)�

y�

�

� (x� y)

(x� y)⇢

y⇢

(8.30)

@⌫Aµ

(x) =q

4⇡

1

(x� y)⇢

y⇢

d

d⌧

(x� y)⌫

yµ

(x� y)�

y�

�

. (8.31)

Introducing

Y = (x� y)�

y�

(8.32)Z = (x� y)

�

y�

(8.33)

we have

@⌫Aµ

(x) =q

4⇡

1

(x� y)⇢

y⇢

"

�y⌫ yµ + (x� y)⌫

yµ

(x� y)�

y�

� (x� y)⌫

yµ (�y� y�

+ (x� y)�

y�

)

((x� y)�

y�

)

2

#

@⌫Aµ

(x) =q

4⇡

1

Y

"

�y⌫ yµ + (x� y)⌫

yµ

Y�

(x� y)⌫

yµ�

�c2 + Z�

Y 2

#

. (8.34)

Hence, the electromagnetic field is

Fµ⌫

=

q

4⇡Y

3

⇥

(x� y)µ

y⌫Y � (x� y)µ

y⌫�

Z � c2�

� (x� y)⌫

yµY + (x� y)⌫

yµ�

Z � c2�⇤

. (8.35)

8.2.0.1 Coulomb Field

A particle at rest is characterized by equations (8.18) and (8.19). This gives Z = 0

and the field

Fµ⌫

=

q

4⇡Y

3

⇥

��

xµ � c⌧�µ0�

c�⌫0�

�c2�

+

�

x⌫ � c⌧�⌫0�

c�µ0�

�c2�⇤

(8.36)

Fµ⌫

=

qc

3

4⇡Y

3

⇥�

xµ � c⌧�µ0�

�⌫0 ��

x⌫ � c⌧�⌫0�

�µ0⇤

. (8.37)

The electric field has components

E1

= F 10

=

qc3

4⇡Y 3

x1 (8.38)

E2

= F 20

=

qc3

4⇡Y 3

x2 (8.39)

E3

= F 30

=

qc3

4⇡Y 3

x3. (8.40)


Note thatY = (x� y)

�

y�

=

�

x0 � y0�

c = Rc (8.41)

and ~R = (x1

, x2

, x3

). Thus

~E =

q

4⇡R3

~R. (8.42)

Similarly, the magnetic field has components

B1

= F 32

= 0 (8.43)B

2

= F 13

= 0 (8.44)B

3

= F 21

= 0 (8.45)

and hence

~B = 0. (8.46)

It is clear that equations (8.42) and (8.46) represent the Coulomb field.

8.2.0.2 Electromagnetic Field of a Uniformly Moving Charged Particle

Consider now a uniformly moving particle with a velocity ~u = (u, 0, 0). Its 4-acceleration and 4-velocity are (see equation (1.38))

yµ = 0 (8.47)

yµ = (c, u, 0, 0)dt

d⌧= (c, u, 0, 0) � (u) (8.48)

yµ =

�

y0, y1, 0, 0�

. (8.49)

Then

Y = (x� y)�

y�

=

�

x0 � y0�

c� ��

x1 � y1�

u� = Rc� �Ru� (8.50)Y = (c� u)R� (8.51)

and

Z = 0. (8.52)

The field is

Fµ⌫

=

q

4⇡ (c� u)3

R3�3

⇥

� (x� y)µ

y⌫�

�c2�

+ (x� y)⌫

yµ�

�c2�⇤

(8.53)

Fµ⌫

=

qc2

4⇡ (c� u)3

R3�3[(x� y)

µ

y⌫ � (x� y)⌫

yµ] (8.54)

8.3. THE RADIATION FIELD 135

Fµ⌫

=

q

4⇡c (1� �)3

R3�3[(x� y)

µ

y⌫ � (x� y)⌫

yµ] (8.55)

with � =

u

c

. This can be written as

Fµ⌫

=

q

4⇡c (1� �)3

R3�3

⇥

(x� y)µ

c��⌫0 + (x� y)µ

u��⌫1 � (x� y)⌫

c��µ0 � (x� y)⌫

u��µ!⇤

Fµ⌫

=

q

4⇡c (1� �)3

R3�2

⇥

(x� y)µ

c�⌫0 + (x� y)µ

u�⌫1 � (x� y)⌫

c�µ0 � (x� y)⌫

u�µ1⇤

.

The electric field has components

E1

= F 10

=

q

4⇡c (1� �)3

R3�2

⇥�

x1 � y1�

c��

x0 � y0�

u⇤

.

E1

=

q

4⇡c (1� �)3

R3�2[Rc�Ru]

E1

=

q

4⇡ (1� �)2

R2�2

E1

=

q

4⇡R2

(1 + �)

(1� �)(8.56)

and

E2

= F 20

= 0 (8.57)E

3

= F 30

= 0. (8.58)

8.3 The Radiation FieldFrom the two examples above we can interpret the general electromagnetic field givenin equation (8.35). It has two contributions: the first one is the near field whichdepends only on the velocities y and is porportional to 1

R

2 . The second contirbutionis the far field or radiation which depends on the acceleration y and is proportionalto 1

R

.Using the Lienard-Wiechert potentials (8.16) and (8.17), we will obtain the electric

and magnetic fields by the relations

~E = �~r�� 1

c

@ ~A

@t(8.59)

and~B =

~r⇥ ~A. (8.60)

Note that


~r� =

q

4⇡~r

2

4

1

⇣

R� ~R · ~�⌘

3

5 (8.61)

~r� = � q

4⇡

1

⇣

R� ~R · ~�⌘

2

~r⇣

R� ~R · ~�⌘

(8.62)

where all quantities are evaluated at the retarded time ⌧ = ⌧0

. Since R = x0�y0 (⌧0

) =

ct� ctR

, the first term in the gradient is

~rR = �c~rtR

. (8.63)

The second term in the gradient can be rewritten as1

~r⇣

~R · ~�⌘

=

⇣

~R · ~r⌘

~� +

⇣

~� · ~r⌘

~R+

~R⇥⇣

~r⇥ ~�⌘

+

~� ⇥⇣

~r⇥ ~R⌘

. (8.65)

Now we will evaluate the four terms in the right hand side,

⇣

~R · ~r⌘

~� = (Rx

@x

+Ry

@y

+Rz

@z

)

~u

c

=

✓

Rx

d~u

dtR

@tR

@x+R

y

d~u

dtR

@tR

@y+R

z

d~u

dtR

@tR

@z

◆

1

c

=

˙~�⇣

~R · ~rtR

⌘

(8.66)

where ˙~� =

1

c

d~u

dt

R

is the acceleration of the particle at the retarded time. The secondterm is

⇣

~� · ~r⌘

~R = (�x

@x

+ �y

@y

+ �z

@z

)

~R (8.67)

and using ~R = ~x� ~y (⌧0

) we write

⇣

~� · ~r⌘

~R =

~� � (�x

@x

+ �y

@y

+ �z

@z

) ~y

=

~� �✓

�x

d~y

dtR

@tR

@x+ �

y

d~y

dtR

@tR

@y+ �

z

d~y

dtR

@tR

@z

◆

=

~� � c~�⇣

~� · ~rtR

⌘

. (8.68)

1Here we use the vector identity

~r⇣

~A · ~B⌘

=⇣

~A · ~r⌘

~B +⇣

~B · ~r⌘

~A+ ~A⇥⇣

~r⇥ ~B⌘

+ ~B ⇥⇣

~r⇥ ~A⌘

(8.64)

.


In the third term we have the curl

~r⇥ ~� =

✓

@�z

@y� @�

y

@z

◆

x+

✓

@�x

@z� @�

z

@x

◆

y +

✓

@�y

@x� @�

x

@y

◆

z (8.69)

~r⇥~� =

✓

d�z

dtR

@tR

@y� d�

y

dtR

@tR

@z

◆

x+

✓

d�x

dtR

@tR

@z� d�

z

dtR

@tR

@x

◆

y+

✓

d�y

dtR

@tR

@x� d�

x

dtR

@tR

@y

◆

z

(8.70)

~r⇥ ~� = � ˙~� ⇥ ~rtR

. (8.71)

Finally, in the fourth term we have the curl

~r⇥ ~R =

~r⇥ [~x� ~y (⌧0

)] (8.72)

~r⇥ ~R =

~r⇥ ~x� ~r⇥ ~y (⌧0

) (8.73)

~r⇥ ~R = �~r⇥ ~y (⌧0

) (8.74)

~r⇥ ~R = c~� ⇥ ~rtR

. (8.75)

Replacing equations (8.66), (8.68), (8.71) and (8.75) in (8.65) we obtain

~r⇣

~R · ~�⌘

=

˙~�⇣

~R · ~rtR

⌘

+

~� � c~�⇣

~� · ~rtR

⌘

� ~R⇥⇣

˙~� ⇥ ~rtR

⌘

+

~� ⇥⇣

c~� ⇥ ~rtR

⌘

.

(8.76)Using the vector identity ~A ⇥ ~B ⇥ ~C =

~B⇣

~A · ~C⌘

� ~C⇣

~A · ~B⌘

in the last twoterms we have

~r⇣

~R · ~�⌘

=

˙~�⇣

~R · ~rtR

⌘

+

~� � c~�⇣

~� · ~rtR

⌘

� ˙~�⇣

~R · ~rtR

⌘

+

⇣

˙~� · ~R⌘

~rtR

+ c~�⇣

~� · ~rtR

⌘

� c⇣

~� · ~�⌘

~rtR

~r⇣

~R · ~�⌘

=

~� +

⇣

˙~� · ~R⌘

~rtR

� c�2~rtR

(8.77)

~r⇣

~R · ~�⌘

=

~� +

h

˙~� · ~R� c�2

i

~rtR

. (8.78)

Equations (8.63) and (8.78) replaced in (8.62) gives the gradient of the electricpotential as


~r� =

q

4⇡

1

⇣

R� ~R · ~�⌘

2

h

~� +

⇣

˙~� · ~R+ c� c�2

⌘

~rtR

i

(8.79)

~r� =

q

4⇡R2

1

⇣

1� n · ~�⌘

2

h

~� +

⇣

˙~� · ~R+ c� c�2

⌘

~rtR

i

. (8.80)

From the relation R = x0 � y0 (⌧0

) = x0 � ctR

we can write

c~rtR

=

~r�

x0 �R�

= �~rR = �~r⇣

p

~R · ~R⌘

(8.81)

c~rtR

= � 1

2R~r⇣

~R · ~R⌘

(8.82)

which using the vector identity (8.64) becomes

c~rtR

= � 1

2R

h

2

⇣

~R · ~r⌘

~R+ 2

~R⇥⇣

~r⇥ ~R⌘i

(8.83)

c~rtR

= � 1

R

h⇣

~R · ~r⌘

~R+

~R⇥⇣

~r⇥ ~R⌘i

. (8.84)

Replacing ~R = ~x� ~y (⌧0

) we have⇣

~R · ~r⌘

~R =

~R� ~u⇣

~R · ~rtR

⌘

(8.85)

and

~r⇥ ~R = ~u⇥ ~rtR

. (8.86)

Hence,

c~rtR

= � 1

R

h

~R� ~u⇣

~R · ~rtR

⌘

+

~R⇥⇣

~u⇥ ~rtR

⌘i

(8.87)

and using the vector identity ~A⇥ ~B ⇥ ~C =

~B⇣

~A · ~C⌘

� ~C⇣

~A · ~B⌘

,

c~rtR

= � 1

R

h

~R� ~u⇣

~R · ~rtR

⌘

+ ~u⇣

~R · ~rtR

⌘

�⇣

~R · ~u⌘

~rtR

i

(8.88)

c~rtR

= � 1

R

h

~R�⇣

~R · ~u⌘

~rtR

i

. (8.89)

This gives

c~rtR

= �~R

R+

~R

R· ~u!

~rtR

(8.90)

c~rtR

� (n · ~u) ~rtR

= �n (8.91)


c~rtR

h

1� n · ~�i

= �n (8.92)

~rtR

= � n

c⇣

1� n · ~�⌘ . (8.93)

Therefore, the gradient of the potential becomes

~r� =

q

4⇡R2

1

⇣

1� n · ~�⌘

2

2

4

~� �⇣

˙~� · ~R+ c� c�2

⌘ n

c⇣

1� n · ~�⌘

3

5 (8.94)

~r� =

q

4⇡R2

1

⇣

1� n · ~�⌘

3

"

⇣

1� n · ~�⌘

~� �

˙~�

c· ~R+ 1� �2

!

n

#

. (8.95)

In order to complete the deduction of the electric field we need to compute thetime derivative

1

c

@ ~A

@t=

q

4⇡c

@

@t

2

4

~�⇣

R� ~R · ~�⌘

3

5 (8.96)

1

c

@ ~A

@t=

q

4⇡c

2

6

4

1

⇣

R� ~R · ~�⌘

@~�

@t�

~�⇣

R� ~R · ~�⌘

2

@⇣

R� ~R · ~�⌘

@t

3

7

5

(8.97)

1

c

@ ~A

@t=

q

4⇡c

8

>

<

>

:

1

⇣

R� ~R · ~�⌘

˙~�@t

R

@t�

~�⇣

R� ~R · ~�⌘

2

dR

dt� d

dtR

⇣

~R · ~�⌘ @t

R

@t

�

9

>

=

>

;

.

(8.98)The relation R = x0 � y0 (⌧

0

) = x0 � ctR

gives

@R

@t=

@

@t

�

x0 � ctR

�

= c� c@t

R

@t(8.99)

and hence

1

c

@ ~A

@t=

q

4⇡c

8

>

<

>

:

1

⇣

R� ~R · ~�⌘

˙~�@t

R

@t�

~�⇣

R� ~R · ~�⌘

2

c� c@t

R

@t� d

dtR

⇣

~R · ~�⌘ @t

R

@t

�

9

>

=

>

;

(8.100)


1

c

@ ~A

@t=

q

4⇡R2c

1

⇣

1� n · ~�⌘

2

⇢

˙~�R⇣

1� n · ~�⌘ @t

R

@t�

c� c@t

R

@t� d

dtR

⇣

~R · ~�⌘ @t

R

@t

�

~�

�

(8.101)

1

c

@ ~A

@t=

q

4⇡R2c

1

⇣

1� n · ~�⌘

2

8

<

:

�c~� +

2

4

˙~�R⇣

1� n · ~�⌘

+

0

@c+d⇣

~R · ~�⌘

dtR

1

A

~�

3

5

@tR

@t

9

=

;

(8.102)

1

c

@ ~A

@t=

q

4⇡R2c

1

⇣

1� n · ~�⌘

2

(

�c~� +

"

˙~�R⇣

1� n · ~�⌘

+

c+ ~R · d~�

dtR

+

d~R

dtR

· ~�!

~�

#

@tR

@t

)

.

(8.103)Using ~R = ~x� ~y (⌧

0

), we obtain

d~R

dtR

=

d

dtR

(~x� ~y) = �~u (8.104)

and therefore,

1

c

@ ~A

@t=

q

4⇡R2c

1

⇣

1� n · ~�⌘

2

⇢

�c~� +

h

˙~�R⇣

1� n · ~�⌘

+

⇣

c+ ~R · ˙~� � ~u · ~�⌘

~�i @t

R

@t

�

(8.105)

1

c

@ ~A

@t=

q

4⇡R2c

1

⇣

1� n · ~�⌘

2

⇢

�c~� +

h

˙~�R⇣

1� n · ~�⌘

+

⇣

c+ ~R · ˙~� � c~� · ~�⌘

~�i @t

R

@t

�

(8.106)

1

c

@ ~A

@t=

q

4⇡R2c

1

⇣

1� n · ~�⌘

2

⇢

�c~� +

h

˙~�R⇣

1� n · ~�⌘

+

⇣

c� c�2

+

~R · ˙~�⌘

~�i @t

R

@t

�

(8.107)On the other hand, the relation R = x0 � y0 (⌧

0

) = x0 � ctR

let us write

c@t

R

@t=

@

@t

�

x0 �R�

= c� @

@tR = c� @

@t

⇣

p

~R · ~R⌘

(8.108)

c@t

R

@t= c� 1

R~R · @

~R

@t(8.109)


c@t

R

@t= c� n · @

~R

@t. (8.110)

Replacing ~R = ~x� ~y (⌧0

) we have

c@t

R

@t= c� n · @

@t(~x� ~y) (8.111)

c@t

R

@t= c+ n · @~y

@tR

@tR

@t(8.112)

c@t

R

@t= c+ cn · ~� @tR

@t(8.113)

@tR

@t=

1

1� n · ~�. (8.114)

Hence,

1

c

@ ~A

@t=

q

4⇡R2c

1

⇣

1� n · ~�⌘

3

n

�c~�⇣

1� n · ~�⌘

+

˙~�R⇣

1� n · ~�⌘

+

⇣

c� c�2

+

~R · ˙~�⌘

~�o

(8.115)

1

c

@ ~A

@t=

q

4⇡R2

1

⇣

1� n · ~�⌘

3

(

�~�⇣

1� n · ~�⌘

+

˙~�

cR⇣

1� n · ~�⌘

+

1� �2

+

~R ·˙~�

c

!

~�

)

(8.116)

1

c

@ ~A

@t=

q

4⇡R2

1

⇣

1� n · ~�⌘

3

"

�~� +

˙~�

cR

!

⇣

1� n · ~�⌘

+

1� �2

+

~R ·˙~�

c

!

~�

#

.

(8.117)Combining equations (8.95) and (8.117), the electric field (8.59) is

~E = � q

4⇡R2

1

⇣

1� n · ~�⌘

3

"

⇣

1� n · ~�⌘

~� �

˙~�

c· ~R+ 1� �2

!

n

#

� q

4⇡R2

1

⇣

1� n · ~�⌘

3

"

�~� +

˙~�

cR

!

⇣

1� n · ~�⌘

+

1� �2

+

~R ·˙~�

c

!

~�

#


~E = � q

4⇡R2

1

⇣

1� n · ~�⌘

3

"

�

˙~�

c· ~R+ 1� �2

!

n+

˙~�

cR⇣

1� n · ~�⌘

+

1� �2

+

~R ·˙~�

c

!

~�

#

(8.118)

~E = � q

4⇡R2

1

⇣

1� n · ~�⌘

3

"

�

1� �2

�

⇣

~� � n⌘

�

˙~�

c· ~R!

n+

˙~�

cR⇣

1� n · ~�⌘

+

~R ·˙~�

c

!

~�

#

(8.119)

~E = � q

4⇡R2

1

⇣

1� n · ~�⌘

3

"

�

1� �2

�

⇣

~� � n⌘

+

˙~�

cR⇣

1� n · ~�⌘

�

~R ·˙~�

c

!

⇣

n� ~�⌘

#

(8.120)

~E = � q

4⇡R2

1

⇣

1� n · ~�⌘

3

"

�

1� �2

�

⇣

~� � n⌘

+

˙~�

c

⇣

R� ~R · ~�⌘

�

~R ·˙~�

c

!

⇣

n� ~�⌘

#

.

(8.121)Now note that using ~A⇥ ~B ⇥ ~C =

~B⇣

~A · ~C⌘

� ~C⇣

~A · ~B⌘

we have

~R⇥h⇣

n� ~�⌘

⇥ ˙~�i

=

⇣

n� ~�⌘⇣

~R · ˙~�⌘

�h

~R ·⇣

n� ~�⌘i

˙~� (8.122)

~R⇥h⇣

n� ~�⌘

⇥ ˙~�i

=

⇣

n� ~�⌘⇣

~R · ˙~�⌘

�⇣

R� ~R · ~�⌘

˙~� (8.123)

and therefore the electric field can be written

~E = � q

4⇡R2

1

⇣

1� n · ~�⌘

3

�

1� �2

�

⇣

~� � n⌘

� 1

c~R⇥

h⇣

n� ~�⌘

⇥ ˙~�i

�

(8.124)

~E =

q

4⇡R2

1

⇣

1� n · ~�⌘

3

�

1� �2

�

⇣

n� ~�⌘

+

1

c~R⇥

h⇣

n� ~�⌘

⇥ ˙~�i

�

. (8.125)

In order to obtain the magnetic field we need to evaluate, using equation (8.17),the curl

~B =

~r⇥ ~A =

~r⇥h

�~�i

. (8.126)


The vector identity ~r⇥⇣

f ~A⌘

= f⇣

~r⇥ ~A⌘

� ~A⇥⇣

~rf⌘

gives

~B = �⇣

~r⇥ ~�⌘

� ~� ⇥⇣

~r�⌘

(8.127)

which using equations (8.71) and (8.95) becomes

~B = �� ˙~� ⇥ ~rtR

� q

4⇡R2

~� ⇥

2

6

4

1

⇣

1� n · ~�⌘

3

"

⇣

1� n · ~�⌘

~� �

˙~�

c· ~R+ 1� �2

!

n

#

3

7

5

(8.128)

~B = � q

4⇡R

1

⇣

1� n · ~�⌘

˙~�⇥ ~rtR

+

q

4⇡R2

1

⇣

1� n · ~�⌘

3

˙~�

c· ~R+ 1� �2

!

~�⇥n (8.129)

~B = � q

4⇡R

1

⇣

1� n · ~�⌘

2

6

4

˙~� ⇥ ~rtR

� 1

R

1

⇣

1� n · ~�⌘

2

˙~�

c· ~R+ 1� �2

!

~� ⇥ n

3

7

5

.

(8.130)The gradient of the retarded time is written using equation (8.93), giving

~B = � q

4⇡R

1

⇣

1� n · ~�⌘

2

2

4� ˙~� ⇥ n

c� 1

R

1

⇣

1� n · ~�⌘

˙~�

c· ~R+ 1� �2

!

~� ⇥ n

3

5

(8.131)

~B = � q

4⇡R

1

⇣

1� n · ~�⌘

2

n⇥

2

4

˙~�

c+

1

R

1

⇣

1� n · ~�⌘

˙~�

c· ~R+ 1� �2

!

~�

3

5 (8.132)

~B = � q

4⇡R2

1

⇣

1� n · ~�⌘

3

n⇥"

R⇣

1� n · ~�⌘

˙~�

c+

˙~�

c· ~R+ 1� �2

!

~�

#

. (8.133)

Comparing this equation with (8.118) and noting that n⇥ n = 0 we write finally themagnetic field as the simple expression

~B = n⇥ ~E. (8.134)

Now we are in position to interpret the different term in the electric and magneticfields. The first term in the electric field (8.125) is called the generalized Coulomb


field (or velocity field or near field) because it does not depend on the accelerationand decays with 1

R

2 . Togheter with the corresponding term in the magnetic field,they form the electromagnetic near field:

~Enear

=

q

4⇡R2

�

1� �2

�

⇣

n� ~�⌘

⇣

1� n · ~�⌘

3

(8.135)

~Bnear

=

q

4⇡R2

�

1� �2

�

⇣

~� ⇥ n⌘

⇣

1� n · ~�⌘

3

. (8.136)

Note that for for a static particle (i.e. ~� = 0) the electric field reduces to theCoulomb field,

~Enear

=

q

4⇡R2

n. (8.137)

On the other hand, the second term in the electric field (8.125) depends on theacceleration and falls with 1

R

, so it is dominant at large distances. This is the respon-isble term for electromagnetic radiation and togheter with the corresponding term inthe magnetic field, they formthe radiation field (or far field),

~Erad

=

q

4⇡cR

n⇥h⇣

n� ~�⌘

⇥ ˙~�i

⇣

1� n · ~�⌘

3

(8.138)

~Brad

=

q

4⇡cR

n⇥n

n⇥h⇣

n� ~�⌘

⇥ ˙~�io

⇣

1� n · ~�⌘

3

. (8.139)

Now we will concentrate on the radiation field. Note that for slowly movingparticles, � ⌧ 1, the far fields reduce to2

2Here we used the vector identity ~A⇥ ~B ⇥ ~C = ~B⇣

~A · ~C⌘

� ~C⇣

~A · ~B⌘

to write

n⇥h⇣

n� ~�⌘

⇥ ~�i

=⇣

n · ~�⌘⇣

n� ~�⌘

�h

n ·⇣

n� ~�⌘i

~� =⇣

n · ~�⌘⇣

n� ~�⌘

�⇣

1� n · ~�⌘

~� (8.140)

and hence

n⇥n

n⇥h⇣

n� ~�⌘

⇥ ~�io

= �⇣

n · ~�⌘⇣

n⇥ ~�⌘

�⇣

1� n · ~�⌘⇣

n⇥ ~�⌘

(8.141)

and for small � we obtain

n⇥n

n⇥h⇣

n� ~�⌘

⇥ ~�io

' �⇣

1� n · ~�⌘⇣

n⇥ ~�⌘

. (8.142)


~Erad

' q

4⇡cRn⇥

⇣

n⇥ ˙~�⌘

(8.143)

~Brad

' � q

4⇡cRn⇥ ˙~�. (8.144)

8.3.1 Poynting Vector of Radiation8.3.1.1 Small Velocities Limit

The pointing vector for the radiation field in the � ⌧ 1 limit is

~S = c⇣

~E ⇥ ~B⌘

= � q2

(4⇡)2

cR2

h

n⇥⇣

n⇥ ˙~�⌘i

⇥h

n⇥ ˙~�i

. (8.145)

This expression can be written as using the vector identity ~A⇥ ~B⇥ ~C =

~B⇣

~A · ~C⌘

�~C⇣

~A · ~B⌘

again. Note that

n⇥⇣

n⇥ ˙~�⌘

=

⇣

n · ˙~�⌘

n� (n · n) ˙~� =

⇣

n · ˙~�⌘

n� ˙~� (8.146)

and henceh

n⇥⇣

n⇥ ˙~�⌘i

⇥h

n⇥ ˙~�i

=

h⇣

n · ˙~�⌘

n� ˙~�i

⇥h

n⇥ ˙~�i

(8.147)

h

n⇥⇣

n⇥ ˙~�⌘i

⇥h

n⇥ ˙~�i

=

⇣

n · ˙~�⌘ h

n⇥⇣

n⇥ ˙~�⌘i

� ˙~� ⇥⇣

n⇥ ˙~�⌘

(8.148)

h

n⇥⇣

n⇥ ˙~�⌘i

⇥h

n⇥ ˙~�i

=

⇣

n · ˙~�⌘ h

n⇣

n · ˙~�⌘

� ˙~�i

� ˙�2n+

˙~�⇣

n · ˙~�⌘

(8.149)

h

n⇥⇣

n⇥ ˙~�⌘i

⇥h

n⇥ ˙~�i

=

⇣

n · ˙~�⌘

2

n� ˙�2n. (8.150)

If ✓ is the angle between ˙~� and n we haveh

n⇥⇣

n⇥ ˙~�⌘i

⇥h

n⇥ ˙~�i

=

˙�2

cos

2 ✓n� ˙�2n (8.151)

h

n⇥⇣

n⇥ ˙~�⌘i

⇥h

n⇥ ˙~�i

=

�

cos

2 ✓ � 1

�

˙�2n (8.152)

h

n⇥⇣

n⇥ ˙~�⌘i

⇥h

n⇥ ˙~�i

= � sin

2 ✓ ˙�2n (8.153)

or


h

n⇥⇣

n⇥ ˙~�⌘i

⇥h

n⇥ ˙~�i

= �⇣

n⇥ ˙~�⌘

2

n (8.154)

and therefore,

~S =

q2

(4⇡)2

cR2

⇣

n⇥ ˙~�⌘

2

n. (8.155)

This expression shows explicitly that the energy flux is radial in the direction of n,i.e. in the direction of ~R. Note that the Poynting vector has magnitude

�

�

�

~S�

�

�

=

q2u2

sin

2 ✓

(4⇡)2

c3R2

(8.156)

which represents the characteristic radiation pattern (i.e. / sin

2 ✓) in the dipoleapproximation.

8.3.1.2 General Case

In general, the radiation field is described by equations (8.138) and (8.139),

~Erad

=

q

4⇡cR

n⇥h⇣

n� ~�⌘

⇥ ˙~�i

⇣

1� n · ~�⌘

3

(8.157)

~Brad

= n⇥ ~Erad

. (8.158)

The Poynting vector is then

~S = c⇣

~E ⇥ ~B⌘

= c ~Erad

⇥⇣

n⇥ ~Erad

⌘

(8.159)

and from the figure it is easy to see that its norm is�

�

�

~S�

�

�

= c�

�

�

~Erad

�

�

�

2

. (8.160)

or

�

�

�

~S�

�

�

=

q2

(4⇡)2

cR2

�

�

�

�

�

�

�

n⇥h⇣

n� ~�⌘

⇥ ˙~�i

⇣

1� n · ~�⌘

3

�

�

�

�

�

�

�

2

. (8.161)

The angular distribution of the emitted radiation can be analyzed using this re-lation. Defining the solid angle subtended by an element of area �An a distance Rfrom the source as �⌦ =

�A

R

2 , the power radiated into �⌦ is

�P = �⌦R2

�

�

�

~S · n�

�

�

@t

@tR

, (8.162)


where we ahve taken into account that the interval of time �t for the observer is notthe same as the interval �t

R

for the emitter. In other words, in order to obtain theradiated power we need to take into account that the radiated energy per unit timemust be calculated in terms of the retarded time t

R

, i.e.

dW

dtR

=

@t

@tR

dW

dt. (8.163)

Therefore, we obtained that the radiated power per unit solid angle is

dP

d⌦= R2

�

�

�

~S · n�

�

�

@t

@tR

. (8.164)

Using equations (8.114) we have

@t

@tR

= 1� n · ~� = 1� � cos ✓ (8.165)

and from (8.161) we obtain

dP

d⌦=

q2

(4⇡)2

cR2

�

�

�

�

�

�

�

n⇥h⇣

n� ~�⌘

⇥ ˙~�i

⇣

1� n · ~�⌘

3

�

�

�

�

�

�

�

2

⇣

1� n · ~�⌘

(8.166)

dP

d⌦=

q2

(4⇡)2

cR2

�

�

�

n⇥h⇣

n� ~�⌘

⇥ ˙~�i

�

�

�

2

⇣

1� n · ~�⌘

5

. (8.167)

8.3.2 Larmor FormulaConsider a non-relativistiv particle, � ⌧ 1, whose Poynting vector is given by equation(8.155). The energy radiated per unit time is obtained by integrating ~S over the sphereof radius R. This gives

P =

ˆ~S · d~⌃ =

ˆ~S · nR2d⌦ =

q2

(4⇡)2

c

ˆ⇣

n⇥ ˙~�⌘

2

d⌦. (8.168)

Using⇣

n⇥ ˙~�⌘

2

=

�

u

c

sin ✓�

2 and d⌦ = sin✓d✓d' we write

P =

q2u2

(4⇡)2

c32⇡

ˆ⇡

0

sin

2 ✓ sin ✓d✓ (8.169)

P =

q2u2

8⇡c3

ˆ⇡

0

�

1� cos

2 ✓�

sin ✓d✓ (8.170)

P =

q2u2

8⇡c3

� cos ✓ +cos

3 ✓

3

�

⇡

0

(8.171)


P =

q2u2

8⇡c3

1 + 1 +

�1� 1

3

�

=

q2u2

8⇡c3

2� 2

3

�

=

q2u2

8⇡c34

3

(8.172)

P =

q2

6⇡

u2

c3. (8.173)

This expression is known as the Larmor Formula for the power radiated from anonrelativistic accelerated point charge. The covariant version of this equation canbe written3

P = � q2

6⇡

1

m2

0

c3dp↵

d⌧

dp↵

d⌧. (8.179)

By substituting E = �m0

c2 and ~p = �m0

~u in the components of the momentum 4-vector and using some vectors identities it is possible to write the relativistic radiatedpower (due to Liénard) as

P = � q2

6⇡

�6

c3

˙�2 �⇣

~� ⇥ ˙~�⌘

2

�

. (8.180)

Note the �6 factor which increases the radiated power enormously as the particleapproaches the speed of light. The procedure is extremely large and therefore we willnot deduce this equation completly. However, we include the following example tovisualize the apparition of the �6 factor.

8.3.2.1 Particle in Linear Motion (Bremsstrahlung)

Suppose that a particle is moving instantaneously in a straight line, so at time t = tR

it has ~� and ˙~� colinear. The complete radiation fields (8.138) and (8.139) give4

3Note that the momentum 4-vector p↵ =⇣

E

c

, ~p⌘

has

(8.174)dp↵

d⌧=

✓

1

c

dE

d⌧,d~p

d⌧

◆

(8.175)

and thus

dp↵

d⌧

dp↵

d⌧=

1

c2

✓

dE

d⌧

◆2

��

�

�

�

d~p

d⌧

�

�

�

�

2

. (8.176)

In the non-relativistic limit this quantity reduces to

dp↵

d⌧

dp↵

d⌧= �m2

0

�

�

�

�

d~u

dt

�

�

�

�

2

. (8.177)

Hence, the coavariant generalization of equation (8.173) is proposed to be

P = �q2

6⇡

1

m20c

3

dp↵

d⌧

dp↵

d⌧. (8.178)

4In the magnetic field we used the vector identity ~A⇥ ~B ⇥ ~C = ~B⇣

~A · ~C⌘

� ~C⇣

~A · ~B⌘

to write


~Erad

=

q

4⇡cR

n⇥⇣

n⇥ ˙~�⌘

⇣

1� n · ~�⌘

3

(8.183)

~Brad

=

q

4⇡cR

n⇥h

n⇥⇣

n⇥ ˙~�⌘i

⇣

1� n · ~�⌘

3

= � q

4⇡cR

⇣

n⇥ ˙~�⌘

⇣

1� n · ~�⌘

3

(8.184)

and therefore the corresponding Poynting vector is

~S = c⇣

~E ⇥ ~B⌘

= � q2

(4⇡)2

cR2

1

⇣

1� n · ~�⌘

6

h

n⇥⇣

n⇥ ˙~�⌘i

⇥h

n⇥ ˙~�i

. (8.185)

As it was shown before, the numerator can be rewritten to obtain

~S =

q2

(4⇡)2

cR2

⇣

n⇥ ˙~�⌘

2

n⇣

1� n · ~�⌘

6

. (8.186)

If ✓ is the angle between n and ~� we have

~S =

q2

(4⇡)2

cR2

˙�2

sin

2 ✓n

(1� � cos ✓)6

. (8.187)

Therefore, the radiated power per unit solid angle gives, see eq. (8.164),

dP

d⌦=

@t

@tR

~S · nR2

=

q2 ˙�2

(4⇡)2

c

sin

2 ✓

(1� � cos ✓)5

(8.188)

which integrating becomes

P =

q2 ˙�2

2⇡

(4⇡)2

c

ˆ⇡

0

@t

@tR

sin

2 ✓

(1� � cos ✓)6

sin ✓d✓. (8.189)

Using equation (8.165) we have

P =

q2 ˙�2

8⇡c

ˆ⇡

0

sin

2 ✓

(1� � cos ✓)5

sin ✓d✓. (8.190)

n⇥⇣

n⇥ ~�⌘

=⇣

n · ~�⌘

n� (n · n) ~� =⇣

n · ~�⌘

n� ~� (8.181)

and hence

n⇥h

n⇥⇣

n⇥ ~�⌘i

= �⇣

n⇥ ~�⌘

(8.182)


In order to integrate we perform the sustitution x = cos ✓ obtaining

P =

q2 ˙�2

8⇡c

ˆ1

�1

1� x2

(1� �x)5

dx (8.191)

which can be intergrated by parts as

P =

q2 ˙�2

8⇡c

"

4

3 (1 + �)3

(1� �)3

#

(8.192)

P =

q2 ˙�2

6⇡c

1

(1� �2

)

3

(8.193)

P =

q2 ˙�2

6⇡c�6. (8.194)

Note the obtained �6 factor and the whole expression for the radiated power asexpected from equation (8.180) when ~� and ˙~� are colinear. It is also interesting to notethat the angular distribution of this radiation has the same form whether the particleis accelerating ˙� > 0 or decelerating ˙� < 0 because it depends on the square of ˙�. Ineither case, the the radiation is concentrated in the forward direction (with respect tothe velocity). This example describes the classical bremsstrahlung or braking radiation(e.g. when a high speed electron hits a metal target and decelerates).

Expression (8.188) give us the angular distribution of the radiated power and itlaso tell us that the intesity reaches a maximum at the angle ✓

m

given by the condition

d

d✓

✓

dP

d⌦

◆

�

�

�

�

✓

m

= 0. (8.195)

This gives

d

d✓

sin

2 ✓

(1� � cos ✓)5

!

�

�

�

�

�

✓

m

= 0 (8.196)

2 sin ✓m

cos ✓m

(1� � cos ✓m

)

5 � 5� sin3 ✓m

(1� � cos ✓m

)

4

(1� � cos ✓m

)

10

= 0 (8.197)

2 cos ✓m

(1� � cos ✓m

)� 5� sin2 ✓m

= 0 (8.198)

2 cos ✓m

(1� � cos ✓m

)� 5��

1� cos

2 ✓m

�

= 0 (8.199)

2 cos ✓m

� 2� cos2 ✓m

� 5� + 5� cos2 ✓m

= 0 (8.200)

3� cos2 ✓m

+ 2 cos ✓m

� 5� = 0. (8.201)


This quadratic equation can be solved as

cos ✓m

=

�1 +

p

1 + 15�2

3�(8.202)

where the + sign has been choosen in order to keep the value of cos ✓m

between �1

and 1.To consider the ultra-relativistic case, � ! 1, we will expand the velocity as

� = 1� ✏ where ✏⌧ 1. Expanding the right hand side to first order in ✏ we obtain

cos ✓m

=

�1 +

q

1 + 15 (1� ✏)2

3 (1� ✏)⇡ 1

3

(1 + ✏)h

p

1 + 15 (1� 2✏)� 1

i

(8.203)

cos ✓m

⇡ 1

3

(1 + ✏)⇥

p16� 30✏� 1

⇤

(8.204)

cos ✓m

⇡ 1

3

(1 + ✏)

"

4

r

1� 15

8

✏� 1

#

(8.205)

cos ✓m

⇡ 1

3

(1 + ✏)

4

✓

1� 15

16

✏

◆

� 1

�

(8.206)

cos ✓m

⇡ 1

3

(1 + ✏)

4� 15

4

✏� 1

�

(8.207)

cos ✓m

⇡ 1

3

(1 + ✏)

3� 15

4

✏

�

(8.208)

cos ✓m

⇡ (1 + ✏)

1� 5

4

✏

�

(8.209)

cos ✓m

⇡ 1 + ✏� 5

4

✏ (8.210)

cos ✓m

⇡ 1� 1

4

✏. (8.211)

This expression shows in this limit the angles is ✓m

⇡ 0. Hence, we approximatethe Cosine function as cos ✓

m

⇡ 1� 1

2

✓2m

and therefore

1� 1

2

✓2m

⇡ 1� 1

4

✏ (8.212)

✓2m

⇡ 1

2

✏ (8.213)


✓m

⇡r

✏

2

=

r

1� �

2

⌧ 1. (8.214)

We conclude that most of the power is therefore emitted within a relatively narrowcone with opening angle ✓

m

⇡p

✏

2

with respect to the direction ˙~� / ~�.

8.4 General Properties of the Radiation FieldIn this section we will consider the general electromagnetic field produced by a movingparticle and given by equation (8.35),

Fµ⌫

=

q

4⇡Y

3

⇥

(x� y)µ

y⌫Y � (x� y)µ

y⌫�

Z � c2�

� (x� y)⌫

yµY + (x� y)⌫

yµ�

Z � c2�⇤

. (8.215)

This field satisfies the free Maxwell equations

@⌫

Fµ⌫

= 0 (8.216)

everyhwere except on the world-line of the particle and the corresponding electromag-netic energy-momentum tensor satisfies the conservation law

@⌫

Tµ⌫

= 0 (8.217)

everyhwere except on the world-line. The field Fµ⌫ also satisfies the properties

Fµ⌫

✏µ⌫⇢�F⇢�

= 0 (8.218)

and

Fµ⌫

✏⇢�µ⌫ (x� y)�

= 0 (8.219)

because ✏µ⌫⇢� is completely antisymmetric whereas the products Fµ⌫

F⇢�

and Fµ⌫

(x� y)�

are symmetric in the indices µ,� and ⌫,� respectively. Introducing again the dualtensor

F ⇤µ⌫= ✏µ⌫⇢�F

⇢�

, (8.220)

the first property can be written

Fµ⌫

F ⇤µ⌫= 0. (8.221)

This equation means that the vectors ~E and ~B are orthogonal. Similarly, thesecond property, equation (8.219), means that ~E and ~B are perpendicular to thevector (~x� ~y).

As stated before, the radiation field corresponds to the part of Fµ⌫ that dependson the acceleration y. From equation (8.35) this can be written

8.4. GENERAL PROPERTIES OF THE RADIATION FIELD 153

Fµ⌫

(rad)

=

q

4⇡Y 3

[(x� y)µ

y⌫Y � (x� y)µ

y⌫Z � (x� y)⌫

yµY + (x� y)⌫

yµZ]|⌧=⌧0

.

(8.222)Similarly to equations (8.223) and (8.33), it is easy to show that the radiation field

satisfies the properties

Fµ⌫(rad)

Fµ⌫

(rad)

= 0 (8.223)

and(x� y)

⌫

Fµ⌫(rad)

= 0. (8.224)

In order to probe these relations we also use the relation

[x� y (⌧0

)]

2

= 0, (8.225)

which expresses the fact that x and y (⌧0

) are separated by a null distance (i.e. theretarded time definition). Equation (8.223) gives the equality of the magnitudes ofthe fields ~E and ~B.

When an electromagnetic field has the two invariants Fµ⌫

Fµ⌫ and Fµ⌫

F ⇤µ⌫ iden-tically zero, it is said to be a pure radiation field or is also called a null field.

8.4.1 Electromagnetic Plane WavesConsider a plane wave moving in the direction x1. Hence, the potential Aµ that de-scribes this wave depends only on the time coordinate and just one spatial coodinate.The gauge choosing condition @

µ

Aµ

= 0 becomes

@0

A0

+ @1

A1

= 0. (8.226)

Consider a potential with

A0

= 0 (8.227)

and~r · ~A = 0 (8.228)

which is equivalent in this case to @1

A1

= 0. The field equations @µ

Fµ⌫

= 0 are

@µ

(@µA⌫ � @⌫Aµ

) = 0 (8.229)

@µ

@µA⌫ � @⌫@µ

Aµ

= 0 (8.230)

and by the gauge condition it reduces to

@µ

@µAi

= ⇤Ai

= 0 (8.231)

or in vector terms this is the one-dimensional wave equation


⇤ ~A = 0. (8.232)

The component A1 of this equation gives

@0

@0A1

= 0 (8.233)

or

@0

A1

= constant = C1

. (8.234)

If this constant is different from zero, C1

6= 0, this equation gives the constantelectric field E

1

= �@0

A1

= �C1

, which does not represent a wave. Thus, we choose

A1

= 0. (8.235)

For the component A2 the wave equation gives

@0

@0A2

+ @1

@1A2

= 0 (8.236)

and therefore A2 must be a function of the difference�

x0 � x1

�

, i.e.

@0

A2

= �@1

A2. (8.237)

Similarly, the A3 component satisfies

@0

A3

= �@1

A3. (8.238)

These relations give the non-vanishing components of the field,

Fµ⌫

=

0

B

B

@

0 0 �E2

�E3

0 0 �B3

B2

E2

B3

0 0

E3

�B2

0 0

1

C

C

A

(8.239)

and hence the dual tensor is

F ⇤µ⌫=

0

B

B

@

0 0 �B2

�B3

0 0 E3

�E2

B2

�E3

0 0

B3

E2

0 0

1

C

C

A

. (8.240)

From these tensors we calculate the field invariants. The first one, Fµ⌫

Fµ⌫

= 0,gives the well known relation

�

�

�

~E�

�

�

=

�

�

�

~B�

�

�

. (8.241)

The second invariant, Fµ⌫

F ⇤µ⌫= 0, gives the condition

~E · ~B = 0. (8.242)


8.4.1.1 Polarization

The relations between the electric and magnetic fields found above are not sufficientto fix completely ~E and ~B. For example, these vectors at a given point can stillchange as a function of time in a plane perpendicular to the direction of propagation~k. Thus, plane waves may be distinguished by the direction of one of these vectors,say the electric field, by introducing the so-called polarization degree of freedom. Since~E and ~B vary in a plane (perpendicular to ~k), two numbers are suficient to describethe polarization of radiation, i.e. there are two degrees of polarization. In order todescribe polarization covariantly, consider a monochromatic plane wave

Aµ

(x) = "µe�ikx (8.243)

where "µ is a constant vector. Replacing this solution in the wave equation ⇤Aµ

= 0

we obtain the condition

k2"µ = 0 (8.244)

or

k2 = 0. (8.245)

On the other hand, the gauge condition @µ

Aµ

= 0 implies

kµ"µ

= 0 (8.246)

which means that the amplitude "µ is orthogonal to kµ. We will choose a basisfor the Minkowkski space dependent on k. Denoting this basis vectors by e(�) with� = 0, 1, 2, 3, we impose the conditions

kµe(�)µ

=

�

k0, k0, 0, 0�

, (8.247)

i.e. in this basis the wave vector only has a time component and the first spatialcomponent. Since the amplitude "µ is expanded in the new basis as

"µ

=

X

�

"(�)

e(�)µ

, (8.248)

conditions (8.246) and (8.247) becomeX

�

"(�)

e(�)µ

kµ = k0"(0)

� k0"(1)

= k0⇥

"(0)

� "(1)

⇤

= 0 (8.249)

or

"(0)

= "(1)

. (8.250)

Therefore, "µ

in this basis is written as

"µe(�)µ

=

⇣

"(0), "(0), "(2)"(3)⌘

. (8.251)


The corresponding field calculated from the potential (8.243) is

Fµ⌫

= i ("µk⌫ � "⌫kµ) e�ikx. (8.252)

From here it is straightforward to writet the components of ~E and ~B along thedirection of ~k,

E1

= F 01

= i�

"0k1 � "1k0�

e�ikx (8.253)

and

B1

= F 23

= i�

"2k3 � "3k2�

e�ikx. (8.254)

Using the basis e(�)

µ

we have (see equations (8.247) and (8.250))

E1

= F 01

= i⇣

"(0)k0 � "(1)k0⌘

e�ikx

= 0 (8.255)

and

B1

= F 23

= i�

"2k3 � "3k2�

e�ikx

= 0. (8.256)

These results show clearly the transverse character of the electromagnetic wave.

8.4.2 Plane Wave Decomposition of the General SolutionAn arbitrary field satisfying the field equation

⇤Aµ

= 0 (8.257)

and the condition@µ

Aµ

= 0 (8.258)

can be expanded in a Fourier series as

Aµ

(x) =1

(2⇡)4

âµ (k) e�ikxdk. (8.259)

Since the field Aµ is real, the complex coefficients in the expansion must satisfy

a⇤µ

(k) = aµ

(�k) . (8.260)

Replacing the general solution (8.259) in the wave equation (8.257) gives the con-dition

k2 = 0 (8.261)

while replacing in the gauge condition (8.258) gives

aµ

(k) kµ = 0. (8.262)

The Fourier decomposition of Aµ gives the field tensor


Fµ⌫

(x) =�i

(2⇡)4

ˆ[kµa⌫ (k)� k⌫aµ (k)] e�ikxdk (8.263)

or

Fµ⌫

(x) =i

(2⇡)4

ˆ[k⌫aµ (k)� kµa⌫ (k)] e�ikxdk. (8.264)

8.4.3 Energy-Momentum Tensor for the Radiation FieldThe energy momentum tensor (6.12) is given by equation5

Tµ⌫

= Fµ

�

F�⌫ +1

4

⌘µ⌫F �⇢F�⇢

. (8.267)

From equation (8.239) we have

Fµ

�

F�⌫ =

0

B

B

@

E2 E2

B3

� E3

B2

0 0

E2

B3

� E3

B2

B2

0 0

0 0 E2

2

+B2

3

E2

E3

�B2

B3

0 0 E2

E3

�B2

B3

E2

3

+B2

2

1

C

C

A

(8.268)and

F�⇢F�⇢

= 2B2 � 2E2

= 2

�

B2 � E2

�

= 0 (8.269)

because the radiation field satisfies E2

= B2. Thus the energy-momentum tensor is

Tµ⌫

=

0

B

B

@

E2 E2

B3

� E3

B2

0 0

E2

B3

� E3

B2

E2

0 0

0 0 E2

2

+B2

3

E2

E3

�B2

B3

0 0 E2

E3

�B2

B3

E2

3

+B2

2

1

C

C

A

.

(8.270)It is also true for radiation fields that its transverse character gives

~E ⇥ ~B = (E2

B3

� E3

B2

)

ˆk (8.271)5A general formula that hold for tensors and its duals is the following: Any two tensors Fµ⌫ and

Gµ⌫ with dual tensors F ⇤µ⌫ and G⇤µ⌫ satisfy the general formula

1

2⌘µ⌫F ⇢�G

⇢�

= Fµ

�

G�⌫ �G⇤µ�

F ⇤�⌫ . (8.265)

Hence, the energy-momentum tensor can be written using the dual tensor F ⇤�⌫ and this propertyas

Tµ⌫ =1

2[Fµ

�

F�⌫ + F ⇤µ�

F ⇤⌫� ] . (8.266)


and

~E ⇥ ~B =

�

�

�

~E�

�

�

�

�

�

~B�

�

�

ˆk = E2

ˆk. (8.272)

Thus

E2

B3

� E3

B2

= E2 (8.273)

and the energy-momentum tensor becomes

Tµ⌫

=

0

B

B

@

E2 E2

0 0

E2 E2

0 0

0 0 E2

2

+B2

3

E2

E3

�B2

B3

0 0 E2

E3

�B2

B3

E2

3

+B2

2

1

C

C

A

. (8.274)

The Poynting vector for the monochormatic wave can be written simply as

~S = c⇣

~E ⇥ ~B⌘

= cE2

ˆk. (8.275)

For the specific potential (8.243) we can write the energy-momentum tensor as

Tµ⌫

= "2kµk⌫e�2ikx. (8.276)

Since kµ is a null 4-vector, the energy-momentum tensor satisfies the relations

Tµ

µ

= 0 (8.277)

Tµ⌫k⌫

= 0 (8.278)

and

Tµ⌫Tµ⌫

= 0. (8.279)

8.5 Radiation Reaction

As we have seen, an accelerated charge radiates. The radiation carries energy (andmomentum) which is supplied by the kinetic energy of the particle. This means thatthe radiation exerts a force back to the charge known as the radiation reaction. FromLarmor’s formula for a non-relativistic particle we know that the radiated power is

P =

q2

6⇡

u2

c3. (8.280)

Calling ~Frad

the radiation reaction force, the conservation of energy says that therate at which the particle loses energy under the influence of this force must be related

8.5. RADIATION REACTION 159

with P. Considering a time interval �t = t2

� t1

, the energy carried away by theradiation must be equal to the energy lost 6, i.e.

ˆt2

t1

~Frad

· ~udt = � q2

6⇡c3

ˆt2

t1

u2dt. (8.281)

Integrating by parts the right hand side gives

ˆt2

t1

u2dt =

ˆt2

t1

˙~u · ˙~udt =ˆ

t2

t1

d~u

dt· d~udt

dt = ~u · d~udt

�

�

�

�

t2

t1

�ˆ

t2

t1

d2~u

dt2· ~udt. (8.282)

The first term vanishes because the velocities and accelerations are the same at t1

and t2

(same state of the system),ˆ

t2

t1

u2dt = �ˆ

t2

t1

¨~u · ~udt. (8.283)

Hence,ˆ

t2

t1

~Frad

· ~udt = q2

6⇡c3

ˆt2

t1

¨~u · ~udt (8.284)

or

ˆt2

t1

"

~Frad

� q2¨~u

6⇡c3

#

· ~udt = 0. (8.285)

This equation gives finally the Abraham-Lorentz Formula for the radiation reactionforce,

~Frad

=

q2¨~u

6⇡c3. (8.286)

This equation has disturbing implications as we will see. For example, consider aparticle that is subject to no external forces. Its equation of motion is simply

~Frad

=

q2¨~u

6⇡c3= m ˙~u (8.287)

from which we write

d ˙~u

dt=

6⇡mc3

q2˙~u (8.288)

˙~u = a0

exp

6⇡mc3

q2t

�

(8.289)

6The time interval is choosen such that over �t the system returns to its initial state. For example,when considering periodic motion we must integrate over an integral number of full cycles. Whenthe motion is not periodic , we need to asure that the state (specifically the velocity fields) of thesystem at t1 is equal to the state at t2.


or

˙~u = a0

et

⌧ (8.290)

where ⌧ =

q

2

6⇡mc

3 . This equation tell us that the acceleration incresas

es exponentially with time (without external forces!!). For an electron we havethe mean time ⌧ = 6 ⇥ 10

�24s. which is roughly the timeit takes light to cross thaclassical electron radius r

e

=

e

2

m

e

c

2 .

8.6 Problems1. Using the Lienard-Wiechert potential, find the electromagnetic potentials pro-

duced by a point charge moving with constant velocity ~u = ux.

2. Explain why the first term in equation (8.28) vanishes.

3. Integrate equation (8.191) to obtain (8.192).

4. Probe equations (8.218) and (8.219) for the electromagnetic field (8.35).

5. Show explicitly that equation (8.221) implies that the vectors ~E and ~B areorthogonal and that equation (8.219) implies that ~E and ~B are perpendicularto the vector (~x� ~y).

6. Probe equations (8.223) and (8.224) for the radiation electromagnetic field(8.222).

7. Show that equation (8.223) implies the equality of the magnitudes of the fields~E and ~B.

8. Show explicitly that the energy-momentum tensor (8.276) satisfies equations(8.277),(8.278) and (8.279).

Chapter 9

Special Topics

9.1 Time-Independent Multipolar Expansion

The electric potential

� (~x) =

ˆ⇢ (~x0

)

|~x� ~x0|d3x0 (9.1)

includes an arbitrary charge distribution described by ⇢ (~x0). However, this equation

does not show that the corresponding field may have angular symmetries due to thecharge configuration. In order to show the different angular components we willconsider that the distribution ⇢ is time independent and we will decompose the fieldinto spherical waves with a common center. The distance modulus 1

|~x�~x0| can bewritten

1

|~x� ~x0| =1

q

|~x|2 + |~x0|2 � 2~x · ~x0(9.2)

1

|~x� ~x0| =1

|~x|

"

1 +

|~x0|2

|~x|2� 2~x · ~x0

|~x|2

#� 12

(9.3)

and expanding by using the binomial theorem for observation distances much largerthan the source size, i.e. for |~x| � |~x0|, gives

1

|~x� ~x0| ⇡1

|~x|

"

1 +

~x · ~x0

|~x|2� 1

2

|~x0|2

|~x|2+

3

2

(~x · ~x0)

2

|~x|4

#� 12

. (9.4)

We replace this expansion in the electric potential and integrate to obtain

� (~x) =q

r+

X

i

xi

di

r3+

1

2

X

i,j

Qij

xi

xj

r5+ ... (9.5)

161

162 CHAPTER 9. SPECIAL TOPICS

where

q =

ˆ⇢ (~x0

) d3x0 (9.6)

di

=

ˆx0i

⇢ (~x0) d3x0 (9.7)

Qij

=

ˆh

3x0i

x0j

� r02�

ij

i

⇢ (~x0) d3x0 (9.8)

are identified as the monopole moment q, the dipole moment vector ~d and the quadrupolemoment tensor Q

ij

, respectively.

9.1.1 ExamplesA) Consider the distribution ⇢ (~x0

) = q�3 (~x0) wich represents a point particle located

at ~x0= 0. This gives the potential

� (~x) =q

r+ 0 + 0 + ... (9.9)

where only the radial spherically symmetric contribution survives. It is clear that noangular dependence is present in this distribution

B) Consider the distribution ⇢ (~x0) = q�3 (~x0 � ~x

1

) � q�3 (~x0 � ~x2

) which describesto point particles with charges q and �q located at ~x0

= ~x1

and ~x0= ~x

2

, respectively.The electric potential gives this time

� (~x) = 0 +

X

i

xi

di

r3+ 0 + ... (9.10)


� (~x) = 0 +

~x · ~dr3

+ 0 + ... (9.11)

defining the dipole moment vector ~d = q~x1

� q~x2

. Note the starightforward angulardependence cos ✓, where ✓ is the angle between the observation point vector ~x andthe line passing through the charges (represented by the vector ~d).

9.1.2 General Time-Independent Multipolar ExpansionIn the general case, the angular dependence of the electric potential may be compli-cated. Therefore, it is usual to expand � in terms of spherical harmonics,

� (~x) =

1X

l=0

l

X

m=�l

qlm

Ylm

(✓,')

rl+1

4⇡

2l + 1

(9.12)

9.2. TIME-DEPENDENT MULTIPOLAR EXPANSION 163

where qlm

are the multipole moments. In order to identify this terms, consider theexpansion of the distance modulus

1

|~x� ~x0| = 4⇡

1X

l=0

l

X

m=�l

rl<

rl+1

>

1

2l + 1

Y ⇤lm

(✓0,'0)Y

lm

(✓,') (9.13)

where r<

(r>

) is the smaller (larger) between |~x| and |~x0|. The potential is written

� (~x) = 4⇡

ˆ⇢ (~x0

)

1X

l=0

l

X

m=�l

rl<

rl+1

>

1

2l + 1

Y ⇤lm

(✓0,'0)Y

lm

(✓,') d3x0 (9.14)

and the comparison term by term give us the identification of the coefficients

qlm

=

ˆ⇢ (~x0

) r0lY ⇤

lm

(✓0,'0) d3x0 (9.15)

which contain the information about the ✓-dependence through l and the '�dependencethrough m.

The electric field is calculated as

~E = �~r� (9.16)

which gives the components 1

Er

(l,m) =

4⇡ (l + 1)

2l + 1

qlm

Ylm

(✓,')

rl+2

(9.18)

E✓

(l,m) = � 4⇡

2l + 1

qlm

1

rl+2

@Ylm

(✓,')

@✓(9.19)

Er

(l,m) = � 4⇡

2l + 1

qlm

1

rl+2

im

sin ✓Ylm

(✓,') . (9.20)

9.2 Time-Dependent Multipolar ExpansionNow we will consider fields that are no longer static. In the free space, the potential4-vector must be a solution of the wave equation, so it can be written as a series oftime-harmonic terms (Fourier series) as

Aµ

(x) = Aµ

0

(~x) e�i!t (9.21)

where function Aµ

0

(~x) is a solution of the Helmholtz equation1Remember that the gradient in spherical coordinates is

~r� =@�

@rr +

1

r

@�

@✓✓ +

1

r sin ✓

@�

@''. (9.17)


⇥

r2

+ k2⇤

Aµ

0

(~x) = 0 (9.22)

where, as always,

k2 =

!2

c2. (9.23)

The method of solution is to separate out the angular and radial variables usingthe expansion

Aµ

0

(~x) =X

l,m

Rl

(r)Ylm

(✓,') (9.24)

for each µ. Hence 2, we obtain two equations, the angular component

�

1

sin ✓

@

@✓sin ✓

@

@✓+

1

sin

2 ✓

@2

@'2

�

Ylm

= l (l + 1)Ylm

(9.26)

and the radial equation

�1

r

d2

dr2r +

l (l + 1)

r2

�

Rl

= k2Rl

. (9.27)

9.2.1 Examples9.2.1.1 Electric Dipole Radiation

Consider two tiny metal spheres separated by a distance d and connected by a finewire along the z�direction. At time t the charge on the upper sphere is q (t) and thecharge on the lower sphere is �q (t). Suppose that we drive the charge back and forththrough the wire, from one end to the other, with a constant angular frequency ! inthe form

q (t) = q0

cos (!t) . (9.28)

The system is the oscillating electric dipole

~d (t) = d0

cos (!t) z (9.29)

where d0

= q0

d is the maximum value of the dipole moment. The retarded potential(7.103) in this example becomes

� (x) =1

4⇡

ˆ⇢�

x0

R

, ~x0�

|~x� ~x0| d3x0 (9.30)

2Remember that the laplacian in spherical coordinates is

r2� =1

r2@

@r

✓

r2@�

@r

◆

+1

r2 sin ✓

@

@✓

✓

sin ✓@�

@✓

◆

+1

r2 sin2 ✓

@2�

@'2. (9.25)


� (x) =1

4⇡

8

<

:

q0

cos

h

!⇣

t0R

� R+

c

⌘i

R+

�q0

cos

h

!⇣

t0R

� R�c

⌘i

R�

9

=

;

(9.31)

whereR± = |~x� ~x±| (9.32)

and by the law of cosines

R± =

s

r2 +

✓

d

2

◆

2

⌥ rd cos ✓. (9.33)

Now we will perform some approximations to describe in a simple way the dipole.The first approximation corresponds to consider a extremely small separation distancebetween the spheres, i.e. d ⌧ r. To first order in d we obtain

R± = r

s

1 +

✓

d

2r

◆

2

⌥ d

rcos ✓ (9.34)

R± ⇡ r

1⌥ d

2rcos ✓

�

(9.35)

and therefore

1

R±⇡ 1

r

1± d

2rcos ✓

�

. (9.36)

Similarly, to first order in d we write

cos

!

✓

t0R

� R±c

◆�

⇡ cos

!

✓

t0R

� r

c

1⌥ d

2rcos ✓

�◆�

(9.37)

cos

!

✓

t0R

� R±c

◆�

⇡ cos

!⇣

t0R

� r

c

⌘

⌥ !d

2ccos ✓

�

(9.38)

or using the well known identity,

cos

!

✓

t0R

� R±c

◆�

⇡ cos

h

!⇣

t0R

� r

c

⌘i

cos

!d

2ccos ✓

�

⌥sin

h

!⇣

t0R

� r

c

⌘i

sin

!d

2ccos ✓

�

.

(9.39)The second approximation corresponds to the perfect dipole limit, in which we

consider d ⌧ c

!

3. This gives

cos

!

✓

t0R

� R±c

◆�

⇡ cos

h

!⇣

t0R

� r

c

⌘i

⌥ !d

2ccos ✓ sin

h

!⇣

t0R

� r

c

⌘i

. (9.40)

3Note that this approximation corresponds to d ⌧ � where � = 2⇡c

!

is the wavelenght associatedwith the frequency ! (which will correspond to the frequency of the emitted radiation).


Replacing this quantity back into the potential we obtain

� (x) =

q0

4⇡

⇢

cos

h

!⇣

t0R

� r

c

⌘i

� !d

2ccos ✓ sin

h

!⇣

t0R

� r

c

⌘i

�

1

r

1 +

d

2rcos ✓

�

�

cos

h

!⇣

t0R

� r

c

⌘i

+

!d

2ccos ✓ sin

h

!⇣

t0R

� r

c

⌘i

�

1

r

1� d

2rcos ✓

��

(9.41)

� (x) =q0

4⇡r

⇢

d

rcos ✓ cos

h

!⇣

t0R

� r

c

⌘i

� !d

ccos ✓ sin

h

!⇣

t0R

� r

c

⌘i

�

(9.42)

� (x) =d0

cos ✓

4⇡r

⇢

1

rcos

h

!⇣

t0R

� r

c

⌘i

� !

csin

h

!⇣

t0R

� r

c

⌘i

�

. (9.43)

Note that this expression gives, in the static limit ! ! 0, the potential for astationary dipole (9.11),

� (x) =d0

cos ✓

4⇡r2. (9.44)

Finally, we will consider the radiation field (far field) and therefore we will makeour third approximation, r � c

!

4 or equivalently 1

r

⌧ !

c

. In this region the first termin the potential is negligible and then

�rad

(x) = �d0

! cos ✓

4⇡crsin

h

!⇣

t0R

� r

c

⌘i

. (9.45)

On the other hand, to obtain the vector potential ~A we start from the currentflowing in the wire,

~I =

dq

dtz = �q

0

! sin [!t] z. (9.46)

Thus, equation (7.104) gives the vector potential

~A (x) =1

4⇡c

ˆ ~j�

x0

R

, ~x0�

|~x� ~x0| d3x0 (9.47)

~A (x) =1

4⇡c

ˆ d

2

� d

2

�q0

! sin

⇥

!�

t0R

� R

c

�⇤

z

Rdz0. (9.48)

Applying the first and second approximations described above, we obtain to firstorder in d the value

~Arad

(x) = � 1

4⇡c

q0

!d

rsin

h

!⇣

t0R

� r

c

⌘i

z (9.49)

4This approximation corresponds to consider the far region, i.e. at a large distance comparedwith the wavelenght, r � �.


~Arad

(x) = � 1

4⇡c

!d0

rsin

h

!⇣

t0R

� r

c

⌘i

z (9.50)

or~Arad

(x) = � 1

4⇡c

!d0

rsin

h

!⇣

t0R

� r

c

⌘i⇣

cos ✓r � sin ✓ˆ✓⌘

. (9.51)

To obtain the fields we need to calculate the gradient

~r�rad

=

@�rad

@rr +

1

r

@�rad

@✓ˆ✓ (9.52)

~r�rad

= �d0

!

4⇡c

⇢

cos ✓

� 1

r2sin

h

!⇣

t0R

� r

c

⌘i

� !

rccos

h

!⇣

t0R

� r

c

⌘i

�

r � sin ✓

r2sin

h

!⇣

t0R

� r

c

⌘i

ˆ✓

�

(9.53)which can be written, using the third approximation, as

~r�rad

⇡ d0

!2

4⇡c2cos ✓

rcos

h

!⇣

t0R

� r

c

⌘i

r. (9.54)

Likewise,

@ ~Arad

@tR

= � 1

4⇡c

!2d0

rcos

h

!⇣

t0R

� r

c

⌘i

z (9.55)

@ ~Arad

@tR

= � 1

4⇡c

!2d0

rcos

h

!⇣

t0R

� r

c

⌘i⇣


. (9.56)

Thus, the electric field is

~Erad

= �~r�rad

� 1

c

@ ~Arad

@tR

(9.57)

~Erad

= �d0

!2

4⇡c2cos ✓

rcos

h

!⇣

t0R

� r

c

⌘i

r+1

4⇡c2!2d

0

rcos

h

!⇣

t0R

� r

c

⌘i⇣


(9.58)

~Erad

= �!2d

0

4⇡c2sin ✓

rcos

h

!⇣

t0R

� r

c

⌘i

ˆ✓. (9.59)

To obtain the magnetic field we calculate the curl

~r⇥ ~Arad

=

1

r

@

@r(rA

✓

)� @Ar

@✓

�

' (9.60)

~r⇥ ~Arad

= � 1

4⇡c

!d0

r

⇢

!

csin ✓ cos

h

!⇣

t0R

� r

c

⌘i

+

sin ✓

rsin

h

!⇣

t0R

� r

c

⌘i

�

'.

(9.61)


Using the third approximation, the second term is neglected and hence

~Brad

=

~r⇥ ~Arad

= �!2d

0

4⇡c2sin ✓

rcos

h

!⇣

t0R

� r

c

⌘i

'. (9.62)

The energy radiated by the oscillating electric dipole is calculated by the Poyntingvector,

~S = c⇣

~Erad

⇥ ~Brad

⌘

(9.63)

~S =

!4d20

(4⇡)2

c

sin

2 ✓

r2cos

2

h

!⇣

t0R

� r

c

⌘i

r (9.64)

and the intensity is obtained by averaging over a complete cycle in time,

D

~SE

=

!4d20

32⇡2c

sin

2 ✓

r2r. (9.65)

It is important to note that there is no radiation along the axis of the dipole (becausethis axis is defined by ✓ = 0). In fact, the intensity profile takes the form of a donut,with the maximum in the equatorial plane (✓ = ⇡

2

).Finally, the total radiated power corresponds to the integral

P =

!4d20

32⇡2c

ˆ2⇡

0

ˆ⇡

0

sin

2 ✓

r2r2 sin ✓d✓d' (9.66)

P =

!4d20

32⇡2c2⇡

4

3

(9.67)

P =

!4d20

12⇡c. (9.68)

Bibliography

[1] G. B.Arfken, H. J. Weber, F. E. Harris. Mathematical Methods for Physicists: A

Comprehensive Guide. Academic Press; 7 edition (2012)

[2] V. Bargmann, L. Michel and V. Telegdi. Phys. Rev. Lett. 2 , 435 (1959)

[3] R. G. Brown. Classical Electrodynamics. Part II. Lec-ture Notes. Duke Unversity Physics Department (2007)http://www.phy.duke.edu/~rgb/Class/Electrodynamics.php

[4] B. Cabrera, Phys. Rev. Lett. 48, 1378 (1982)

[5] R. Casalbuoni. Advanced Quantum Field Theory. Dipartimento di Fisica. Uni-verità di Firneze (2005)

[6] I. M. Gel’fand and G. E. Shilov, G.E. Generalized functions, V. 1–5, AcademicPress. (1968)

[7] A. S. Goldhaber and W. P. Trower, Am. J. Phys. 58,429 (1990).

[8] O. D. Jefimenko. Electricity and Magnetism: An Introduction to the Theory of

Electric and Magnetic Fields. Sect. 15.7. Electret Scientific Co. 2nd Ed. (1990)

[9] O. D. Jefimenko. Causality, Electromagnetic Induction, and Gravitation: A Dif-

ferent Approach to the Theory of Electromagnetic and Gravitational Fields. Elec-tret Scientific Co. 2nd Ed. (2000)

[10] F. Melia. Electrodynamics. The University of Chicago Press. (2001)

[11] M. J. Perry. Electrodynamics. Lecture Notes. DAMTP. University of Cambridge.(2000)

[12] W. Rindler, Am. J. Phys. 57,993 (1989)

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