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Tobia Carozzi Anders Eriksson Bengt Lundborg Bo Thidé Mattias Waldenvik

ELECTROMAGNETIC FIELD THEORY

EXERCISES



Companion volume to

ELECTROMAGNETIC FIELD THEORY

by

Bo Thidé



ELECTROMAGNETIC

FIELD THEORY

Exercises

Tobia Carozzi Anders Eriksson

Bengt Lundborg Bo Thidé

Mattias Waldenvik Department of Space and Plasma Physics

Uppsala University

and

Swedish Institute of Space Physics

Uppsala Division

Sweden

ΣIpsum



This book was typeset in LATEX 2 D

on an HP9000/700 series workstation

and printed on an HP LaserJet 5000GN printer.

Copyright cE

1998 by

Bo Thidé

Uppsala, Sweden

All rights reserved.

Electromagnetic Field Theory Exercises

ISBN X-XXX-XXXXX-X



CONTENTS

Preface vii

1 Maxwell’s Equations 1

1.1 Coverage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Formulae used . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.3 Solved examples . . . . . . . . . . . . . . . . . . . . . . . . . . 1Example 1.1 Macroscopic Maxwell equations . . . . . . . . . 1

Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

Example 1.2 Maxwell’s equations in component form . . . . . 4

Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

Example 1.3 The charge continuity equation . . . . . . . . . 5

Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Electromagnetic Potentials and Waves 9

2.1 Coverage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.2 Formulae used . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.3 Solved examples . . . . . . . . . . . . . . . . . . . . . . . . . . 9Example 2.1 The Aharonov-Bohm effect . . . . . . . . . . . 9

Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

Example 2.2 Invent your own gauge . . . . . . . . . . . . . 11

Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

Example 2.3 Fourier transform of Maxwell’s equations . . . . 13

Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

Example 2.4 Simple dispersion relation . . . . . . . . . . . . 15

Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3 Relativistic Electrodynamics 19

3.1 Coverage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

i



ii

3.2 Formulae used . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3.3 Solved examples . . . . . . . . . . . . . . . . . . . . . . . . . . 20

Example 3.1 Covariance of Maxwell’s equations . . . . . . . 20

Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

Example 3.2 Invariant quantities constructed from the field tensor 22

Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 22Example 3.3 Covariant formulation of common electrodynam-

ics formulas . . . . . . . . . . . . . . . . . . . . 23

Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

Example 3.4 Fields from uniformly moving charge via Lorentz

transformation . . . . . . . . . . . . . . . . . . . 25

Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

4 Lagrangian and Hamiltonian Electrodynamics 29

4.1 Coverage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

4.2 Formulae used . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

4.3 Solved examples . . . . . . . . . . . . . . . . . . . . . . . . . . 30

Example 4.1 Canonical quantities for a particle in an EM field . 30

Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

Example 4.2 Gauge invariance of the Lagrangian density . . . 31

Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

5 Electromagnetic Energy, Momentum and Stress 33

5.1 Coverage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

5.2 Formulae used . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

5.3 Solved examples . . . . . . . . . . . . . . . . . . . . . . . . . . 34

Example 5.1 EM quantities potpourri . . . . . . . . . . . . . 34

Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

Example 5.2 Classical electron radius . . . . . . . . . . . . 37Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

Example 5.3 Solar sailing . . . . . . . . . . . . . . . . . . 39

Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

Example 5.4 Magnetic pressure on the earth . . . . . . . . . 41

Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

6 Radiation from Extended Sources 43

6.1 Coverage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

6.2 Formulae used . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

6.3 Solved examples . . . . . . . . . . . . . . . . . . . . . . . . . . 44

Example 6.1 Instantaneous current in an infinitely long conductor 44



iii

Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

Example 6.2 Multiple half-wave antenna . . . . . . . . . . . 50

Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

Example 6.3 Travelling wave antenna . . . . . . . . . . . . . 53

Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

Example 6.4 Microwave link design . . . . . . . . . . . . . 54Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

7 Multipole Radiation 57

7.1 Coverage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

7.2 Formulae used . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

7.3 Solved examples . . . . . . . . . . . . . . . . . . . . . . . . . . 58

Example 7.1 Rotating Electric Dipole . . . . . . . . . . . . 58

Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

Example 7.2 Rotating multipole . . . . . . . . . . . . . . . 60

Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

Example 7.3 Atomic radiation . . . . . . . . . . . . . . . . 62Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

Example 7.4 Classical Positronium . . . . . . . . . . . . . . 63

Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

8 Radiation from Moving Point Charges 67

8.1 Coverage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

8.2 Formulae used . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

8.3 Solved examples . . . . . . . . . . . . . . . . . . . . . . . . . . 68

Example 8.1 Poynting vector from a charge in uniform motion 68

Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

Example 8.2 Synchrotron radiation perpendicular to the accel-eration . . . . . . . . . . . . . . . . . . . . . . 70

Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

Example 8.3 The Larmor formula . . . . . . . . . . . . . . 71

Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

Example 8.4 Vavilov-Cerenkov emission . . . . . . . . . . . 73

Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

9 Radiation from Accelerated Particles 75

9.1 Coverage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

9.2 Formulae used . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

9.3 Solved examples . . . . . . . . . . . . . . . . . . . . . . . . . . 76



Example 9.1 Motion of charged particles in homogeneous static

EM fields . . . . . . . . . . . . . . . . . . . . . 76

Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

Example 9.2 Radiative reaction force from conservation of energy 79

Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

Example 9.3 Radiation and particle energy in a synchrotron . . 81

Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

Example 9.4 Radiation loss of an accelerated charged particle . 83

Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

iv



LIST OF FIGURES

6.1 The turn-on of a linear current at FG I

0 . . . . . . . . . . . . . . 46

6.2 Snapshots of the field . . . . . . . . . . . . . . . . . . . . . . . . 47

6.3 Multiple half-wave antenna standing current . . . . . . . . . . . . 50

9.1 Motion of a charge in an electric and a magnetic field . . . . . . . 78

v



vi



PREFACE

This is a companion volume to the book Electromagnetic Field Theory by Bo Thidé.

Uppsala, Sweden B. T.

November, 1998

vii



viii PREFACE



LESSON 1

Maxwell’s Equations

1.1 Coverage

In this lesson we examine Maxwell’s equations, the cornerstone of electrodynam-

ics. We start by practising our math skill, refreshing our knowledge of vector

analysis in vector form and in component form.

1.2 Formulae used

QR

E IU T V W 0 (1.1)

QR

B I 0 (1.2)

QX

E I`

b

b

F

B (1.3)

QX

B I c 0 j d

1e 2

b

b

F

E (1.4)

1.3 Solved examples

f

MACROSCOPIC MAXWELL EQUATIONS EXAMPLE 1.1

The most fundamental form of Maxwell’s equations is

gh

E pq s

D 0 (1.5)g

h

B p 0 (1.6)

gt

Ep v

w

wx B (1.7)

gt

B py 0 j

1 2

w

wx E (1.8)

(1.9)

1



2 LESSON 1. MAXWELL’S EQUATIONS

sometimes known as the microscopic Maxwell equations or the Maxwell-Lorentz equa-

tions. In the presence of a medium, these equations are still true, but it may sometimes

be convenient to separate the sources of the fields (the charge and current densities) into

an induced part, due to the response of the medium to the electromagnetic fields, and an

extraneous, due to “free” charges and currents not caused by the material properties. One

then writes

j p jind jext (1.10)

qp q ind

q ext (1.11)

The electric and magnetic properties of the material are often described by the electric

polarisation P (SI unit: C/m2) and the magnetisation M (SI unit: A/m). In terms of these,

the induced sources are described by

jind p

w

P s

wx

gt

M (1.12)

q ind p v

gh

P (1.13)

To fully describe a certain situation, one also needs constitutive relations telling how P

and M depends on E and B. These are generally empirical relations, different for differentmedia.

Show that by introducing the fields

D p

D 0E P (1.14)

H p B sy 0 v M (1.15)

the two Maxwell equations containing source terms (1.5) and (1.8) reduce to

gh

D pq ext (1.16)

gt

H p jext

w

wx D (1.17)

(1.18)

known as the macroscopic Maxwell equations.

Solution

If we insert

j p jind jext (1.19)

qp q ind

q ext (1.20)

and



1 .3 . SOLVED EXAMPLES 3

jind p

w

wx P

gt

M (1.21)

q ind p v

gh

P (1.22)

into

gt

B py 0 j

1

2

w

wx E (1.23)

gh

E pq s

D 0 (1.24)

(1.25)

we get

gt

B py 0

jext

w

wx P

gt

M

1 2

w

wx E (1.26)

gh

E p

1

D 0

q ext v

gh

P (1.27)

which can be rewritten as

gt B

y 0

v

M

p

jext

w

wx

P

D

0E

(1.28)g

h

D 0E P p q ext (1.29)

Now by introducing the D and the H fields such that

D p

D 0E P (1.30)

H p

B

y 0

v M (1.31)

the Maxwell equations become

gt

H p jext

w

wx D (1.32)

gh

D pq ext (1.33)

QED

The reason these equations are known as “macroscopic” are that the material properties

described by P and M generally are average quantities, not considering the atomic prop-

erties of matter. Thus E and D get the character of averages, not including details around

single atoms etc. However, there is nothing in principle preventing us from using large-

scale averages of E and B, or even to use atomic-scale calculated D and H although this is

a rather useless procedure, so the nomenclature “microscopic/macroscopic” is somewhat

misleading. The inherent difference lies in how a material is treated, not in the spatial

scales.

END OF EXAMPLE 1.1




f

MAXWELL’S EQUATIONS IN COMPONENT FORMEXAMPLE 1.2

Express Maxwell’s equations in component form.

Solution

Maxwell’s equations in vector form are written:g

h

E pq s

D 0 (1.34)g

h

B p 0 (1.35)

gt

Ep v

w

wx B (1.36)

gt

B py 0 j

1 2

w

wx E (1.37)

In these equations, E, B, and j are vectors, while q is a scalar. Even though all the equations

contain vectors, only the latter pair are true vector equations in the sense that the equations

themselves have several components.

When going to component notation, all scalar quantities are of course left as they are.

Vector quantities, for example E, can always be expanded as Ep

3

1

ˆ

p

ˆ

,where the last step assumes Einstein’s summation convention: if an index appears twice in

the same term, it is to be summed over. Such an index is called a summation index. Indices

which only appear once are known as free indices, and are not to be summed over. What

symbol is used for a summation index is immaterial: it is always true that p m m ,

since both these expressions mean 1 1 2 2

3 3 p ah

b. On the other hand, the

expression p m is in general not true or even meaningful, unless

p or if a is the

null vector.

The three

are the components of the vector E in the coordinate system set by the three

unit vectors ˆ

. The

are real numbers, while the ˆ

are vectors, i.e. geometrical objects.

Remember that though they are real numbers, the

are not scalars.

Vector equations are transformed into component form by forming the scalar product of

both sides with the same unit vector. Let us go into ridiculous detail in a very simple case:

G p H (1.38)

Gh

ˆ

mp

Hh

ˆ

m (1.39)

ˆ

h

ˆ

mp

ˆ

h

ˆ

m (1.40)

mp

m (1.41)

mp

m (1.42)

This is of course unnecessarily tedious algebra for an obvious result, but by using this

careful procedure, we are certain to get the correct answer: the free index in the resulting

equation necessarily comes out the same on both sides. Even if one does not follow this

complicated way always, one should to some extent at least think in those terms.

Nabla operations are translated into component form as follows:




z

p ˆ

w

w

Φ v ~

w

w

(1.43)

gh

V p

w

w

v ~

w

w

(1.44)

gt

V p

m ˆ

w

w

m v ~

m

w

m

w

(1.45)

where V is a vector field and

is a scalar field.

Remember that in vector valued equations such as Ampère’s and Faraday’s laws, one must

be careful to make sure that the free index on the left hand side of the equation is the same

as the free index on the right hand side of the equation. As said above, an equation of the

form p

is almost invariably in error!

With these things in mind we can now write Maxwell’s equations as

gh

E p

q

D 0

v ~

w

w

p

q

D 0

(1.46)

gh

B p 0 v ~

w

w

p 0 (1.47)

gt

Ep v

w

Bw

x

v ~

m

w

m

w

p v

w

wx

(1.48)

gt

B py 0 j

1 2

w

Ew

x

v ~

m

w

m

w

py 0

1 2

w

wx (1.49)

END OF EXAMPLE 1.2

f

THE CHARGE CONTINUITY EQUATION EXAMPLE 1.3

Derive the continuity equation for charge density q from Maxwell’s equations using (a)

vector notation and (b) component notation. Compare the usefulness of the two systems of notations. Also, discuss the physical meaning of the charge continuity equation.

Solution

Vector notation In vector notation, a derivation of the continuity equation for charge

looks like this:

Computeg

h

E in two ways:

1. Apply

to Gauss’s law:

w

wx

gh

E

p

1

D 0

w

wx

q

(1.50)




2. Take the divergence of the Ampère-Maxwell law:

gh

gt

B p y 0

gh

j

1 2

gh

w

wx E (1.51)

Useg

h

gt

0 and y 0 D 0 2

p 1:

p

gh

w

wx E

p v

1

D 0

gh

j (1.52)

Comparison shows that

w

wx

q

gh

j p 0 (1.53)

Component notation In component notation, a derivation of the continuity equation

for charge looks like this:

Compute

E in two ways:

1. Take

of Gauss’s law:

w

wx

w

w

p

1

D 0

w

wx

q (1.54)

2. Take the divergence of the Ampère-Maxwell law:

w

w

D

m

w

m

w

p y 0

w

w

1 2

w

w

w

wx (1.55)

Use that the relation D

m

0 is valid also if p

, and that y 0 D 0 2

p 1:

v ~

w

wx

w

w

p v

1

D 0

w

w

(1.56)

Comparison shows that

w

wx

q

w

w

p

0

(1.57)




Comparing the two notation systems We notice a few points in the derivations

above:

It is sometimes difficult to see what one is calculating in the component system.

The vector system with div, curl etc. may be closer to the physics, or at least to our

picture of it.

In the vector notation system, we sometimes need to keep some vector formulas inmemory or to consult a math handbook, while with the component system you need

only the definitions of D

m and

.

Although not seen here, the component system of notation is more explicit (read

unambiguous) when dealing with tensors of higher rank, for which vector notation

becomes cumbersome.

The vector notation system is independent of coordinate system, i.e.,z

isz

in

any coordinate system, while in the component notation, the components depend on

the unit vectors chosen.

Interpreting the continuity equation The equation

w

wx

q

gh

j p 0 (1.58)

is known as a continuity equation. Why? Well, integrate the continuity equation over some

volume

bounded by the surface . By using Gauss’s theorem, we find that

x

p

w

wx

q

3

p v

gh

j 3

p v j

h

S (1.59)

which says that the change in the total charge in the volume is due to the net inflow of

electric current through the boundary surface . Hence, the continuity equation is the field

theory formulation of the physical law of charge conservation.

END OF EXAMPLE 1.3



8



LESSON 2

ElectromagneticPotentials and Waves

2.1 Coverage

Here we study the vector and scalar potentials A and

and the concept of gauge

transformation.

One of the most important physical manifestation of Maxwell’s equations isthe EM wave. Seen as wave equations, the Maxwell equations can be reduced

to algebraic equations via the Fourier transform and the physics is contained in

so-called dispersion relations which set the kinematic restrictions on the fields.

2.2 Formulae used

E Ia ` `

b

Ab

F

B I

QX

A

2.3 Solved examples

f

THE AHARONOV-B OHM EFFECT EXAMPLE 2.1

Consider the magnetic field given in cylindrical coordinates,

B 0

p

ˆ (2.1)

B 0

p

0 (2.2)

9



10 LESSON 2. ELECTROMAGNETIC POTENTIALS AND WAVES

Determine the vector potential A that “generated” this magnetic field.

Solution

A interesting question in electrodynamics is whether the EM potentials

and A are more

than mathematical tools, and alternatives to the Maxwell equations, from which we can

derive the EM fields. Could it be that the potentials and not Maxwell’s equations are more

fundamental? Although the ultimate answer to these questions is somewhat metaphysical,

it is exactly these questions that make the Aharonov-Bohm effect. Before we discuss this

effect let us calculate the vector field from the given magnetic field.

The equations connecting the potentials with the fields are

E p v

g

v

w

Aw

x (2.3)

B p

gt

A (2.4)

In this problem we see that we have no boundary conditions for the potentials. Also, let us

use the gauge

p 0.

This problem naturally divides into two parts: the part within the magnetic field and thepart outside the magnetic field. Let us start with the interior part:

w

Aw

x

p 0 (2.5a)

1

w

w

v

w

w

p 0 (2.5b)w

w

v

w

w

p 0 (2.5c)

1

w

w

v

w

w

p

(2.5d)

The first equation tells us that A is time independent so A p A

. Examining the

other three we find that there is no dependence on or either so Ap

A

. All thatremains is

1

w

w

p

(2.6)

Integrating this equation we find that

p

2(2.7)

Moving to the outer problem, we see that the only difference compared with the inner

problem is that B p 0 so that we must consider

1

w

w

p

0 (2.8)




This time integration leads to

p «

(2.9)

If we demand continuity for the function

over all space we find by comparing with (2.7)

the arbitrary constant«

and can write in outer solution as

p

20

2 -

p 0 ! (2.10)

Now in electrodynamics (read: in this course) the only measurable quantities are the fields.

So the situation above, where we have a region in which the magnetic field is zero but

the potential is non-zero has no measurable consequence in classical electrodynamics. In

quantum mechanics however, the Aharonov-Bohm effect shows that this situation does

have a measurable consequence. Namely, when letting charged particles go around this

magnetic field (the particles are do not enter the magnetic field region because of a im-

penetrable wall) the energy spectrum of the particles after passing the cylinder will have

changed, even though there is no magnetic field along their path. The interpretation is that

the potential is a more fundamental quantity than the field.

END OF EXAMPLE 2.1

f

INVENT YOUR OWN GAUGE EXAMPLE 2.2

Name some common gauge conditions and discuss the usefulness of each one. Then invent

your own gauge and verify that it is indeed a gauge condition.

Solution

Background The Maxwell equations that do not contain source terms can be “solved”

by using the vector potential A and the scalar potential

, defined through the relations

B p

gt

A (2.11)

E p v

z

v

w

wx

(2.12)

Assuming linear, isotropic and homogeneous media, we can use the constitutive relations

D p

D E H p B sy

, and jp ®

E j ¯ (where j ¯ is the free current density arising from

other sources than conductivity) and definitions of the scalar and vector potentials in the

remaining two Maxwell equations and find that

z 2

gh

w

Aw

x

p v

q

D

(2.13)

z 2Av y ®

w

wx A

v y

D

w

2Aw

x

2v

z

g h

A y

D

w

wx

y ®

p v y

j ¯ (2.14)




These equations are used to determine A and

from the source terms. And once we have

found A and

it is straight forward to derive the E and B fields from (2.11) and (2.12).

The definitions of the scalar and vector potentials are not enough to make A and

unique,

i.e. , if one is given A and

then (2.11) and (2.12) determine B and E, but if one is given

B and E there many ways of choosing A and

. This can be seen through the fact that A

and

can be transformed according to

A ¯

p A

z ±

(2.15)

¯

p

v

w

wx

±

(2.16)

where±

is an arbitrary scalar field, but the B and E fields do not change. This kind of

transformation is called a gauge transformation and the fact that gauge transformations do

not affect the physically observable fields is known as gauge invariance.

Gauge conditions The ambiguity in the definitions of A and

can be used to introduce

a gauge condition. In other words, since the definitions (2.11) and (2.12) do not completely

define A and

we are free to add certain conditions. Some common gauge conditions are

Coulomb gaugeg

h

A p 0

Lorentz gaugeg

h

A y

D

w

s

wx

y ®

p 0

Temporal gauge

p 0

The Coulomb gauge is most useful when dealing with static fields. Usingg

h

A p 0 then

(2.13) and (2.14, for static fields, reduces to

z 2

p v

q

D

(2.17)

z 2Ap v y

j (2.18)

The Lorentz gauge is the most commonly used gauge for time-varying fields. In this case

(2.13) and (2.14) reduce to

z 2

v y ®

w

wx

v y

D

w

2

wx

2

p v

q

D

(2.19)

z 2

v y ®

w

wx

v y

D

w

2

wx

2

Ap v y

j (2.20)

So the Lorentz transform decouples (2.13) and (2.14) and puts

and A on equal footing.

Furthermore, the resulting equations are manifestly covariant.

In the temporal gauge one “discards” the scalar potential by setting

p 0. In this gauge

(2.13) and (2.14) reduce to

1 2

w

2Aw

x

2

gt g t

Ap

y

j (2.21)




Thus the single vector A describes both E and B in the temporal gauge.

How to invent your own gauge Gauges other than Coulomb, Lorentz and the tem-

poral mentioned above are rarely used in introductory literature in Electrodynamics, but

it is instructive to consider what constitutes a gauge condition by constructing ones own

gauge.

Of course, a gauge condition is at least a scalar equation containing at least one of the

components of A or

. Once you have an equation that you think might be a gauge, it must

be verified that it is a gauge. To verify that a condition is a gauge condition it is sufficient

to show that any given set of A and

can be made to satisfy your condition. This is done

through gauge transformations. So given a A and a

which satisfy the physical conditions

through (2.13) and (2.14) we try to see if it is possible (at least in principle) to find a gauge

transformation to some new potential A ¯ and

¯ , which satisfy your condition.

END OF EXAMPLE 2.2

f

FOURIER TRANSFORM OF MAXWELL’S EQUATIONS EXAMPLE 2.3

Fourier transform Maxwell’s Equation. Use the Fourier version of Maxwell’s equations

to investigate the possibility of waves that do not propagate energy; such waves are called

static waves.

Solution

Maxwell’s equations contain only linear operators in time and space. This makes it easy to

Fourier transform them. By transforming them we get simple algebraic equations instead of

differential equations. Furthermore, the Fourier transformed Maxwell equations are useful

when working with waves or time-varying fields, especially since the response function,

i.e. the dielectric function, is in many case more fundamentally described as a function of

angular frequency ³ than length x.

To perform this derivation we need formulas on how to translate the operatorsg

h

,g

t

andw

s

wx

in Maxwell’s equations.

The Fourier transform in time, is defined by

˜µ

³

¶

¶

x¹ º »

µ

x

(2.22)

and the Fourier transform in space, is defined analogously

˜

k

¶

¶

3¹

º k¼x

x (2.23)

and so a combined spatial and time Fourier transform becomes




˜½

³

k p

¶

¶

x

3¹

º¿ k ¼ x »

Á

½

x

x (2.24)

If we apply the last formula ong

h

E we get

gh

E ~

¶

¶

x

3¹

º¿ k ¼ x »

Á

gh

E

p

¶

¶

x

3¹

º¿ k ¼ x »

Á

w

x

x

w

p

xÄ Å

x

x

¹ º¿ k ¼ x »

Á Æ

x È

¶

É Ê Ë Ì

0

v

x

3

v Í

x

x

pÍ

x

3

x

x

pÍ

˜

³

k (2.25)

where we have used partial integration. Forg

t

E we get (note: no summation over !)

gt

E ~

x

3¹

º¿ k ¼ x » Á

gt

E

x

³

p

x

3

¹º

¿ k ¼ x »

Á

m

w

x

x

w

m

e

p

xÐ

¹º

¿ k ¼ x »

Á Ñ

É Ê Ë Ì

0

v

x

3Ó

v Í

m

m

x

x

¹º

¿ k ¼ x »

Á

e Õ

pÍ

m

˜

m

³

k e (2.26)

where we have once again used partial integration. One may proceed analogously forw

s

wx

E

x

x . Trivially, one gets similar equations for the transformation of the D, H and B

fields. Thus we have found that

gh

V

x

x ~ Í k

h

V

³

k (2.27a)g

t

V

x

x ~ Í k

t

V

³

k (2.27b)w

V

x

x

wx

~ v Í ³V

³

k (2.27c)

where V

x

x is an arbitrary field and ~ denotes here “Fourier transform”. These trans-

formation rules are easy to remember by saying that roughly the Fourier transform of z

is

Í k and the Fourier transform of w

s

wx

isv Í ³

.

Now we can use (2.27a), (2.27b) and (2.27c) on Maxwell’s equations. We then get, after

some simple trimming




kt

E

³

k p ³

B

³

k (2.28a)

kt

H

³

k p

j

³

k v Í ³

D

³

k (2.28b)

kh

D

³

k Ö p v Í q

³

k (2.28c)

kh

B

³

k p 0 (2.28d)

where we have dropped the ˜ notation. These are the Fourier version of Maxwell’s equa-

tion.

As an example of the use of the Fourier transformed Maxwell’s equations let us derive

static waves. Static waves are one possible oscillation mode for the E and H fields. Let’s

say that we have a mode Ø such that the E p E Ù field is oscillating at ³p ³ Ù

-

p 0 and

that it has a k p k Ù

-

p 0 which is parallel to the electric field, so k ÙÛ E Ù . From (2.28a)

this implies that

³Ù

B Ùp

k Ù

t

E Ùp

0 (2.29)

B Ùp 0 (2.30)

So, we see that Sp

E

t

Hp

0 trivially. The lesson here is that you can time-varying fieldsthat do not transmit energy! These waves are also called longitudinal waves for obvious

reasons.

END OF EXAMPLE 2.3

f

SIMPLE DISPERSION RELATION EXAMPLE 2.4

If a progressive wave is travelling in a linear, isotropic, homogenous, nonconducting

dielectric medium with dielectricity parameter D and permeability y , what is the disper-

sion relation? And what is the group velocity in this case? Also, what is the dispersion

relation in a conducting medium?

Solution

A dispersion relation is a relation between ³ and k, usually something like

Ý

³

k p

0 (2.31)

From this one can solve for ³ which is then a function of

1

2

3 . For isotropic media

then ³ will be a function of Þ k Þp only. A dispersion relation determines what modes (i.e.

what combinations of k and ³ ) are possible. The dispersion relation is derivable in prin-

cipal once one has explicit knowledge of the dielectricity function (or response function)

for the medium in question.

The two vector equations in Maxwell’s equations are




gt

E p v

w

Bw

x (2.32)

gt

H p

w

Dw

x (2.33)

so for a progressive wave characterized by ³ and k propagating in a l.i.h. medium with

® p0 (so

p ®E p 0), then these equations give

kt

E p³ B (2.34)

kt B

y

p v ³

D E (2.35)

Operating on (2.34) with kt

and then using (2.35) we get a single vector equation:

kt

kt

E p ³ k

t

B (2.36)

kh

E

É Ê Ë Ì

0 Progressive wave!

kv

2Ep v ³

2D

y E (2.37)

2v ³

2D

y E p 0 (2.38)

Since E is not assumed to be zero then

2v ³

2D

yp 0 (2.39)

³

2p

2

D

y

(2.40)

³ p â

ã

D

y

pâ ä

(2.41)

where we have identified the phase velocity äp 1 s

ã

D

y .

The group velocity in this case is

åæ

w

³

w

p

w

w

ä p ä

(2.42)

so in this simple case the group velocity is the same as the phase velocity.

For the case of a conducting medium, in which jp ®

E, the two vector equations applied

on a wave which at first resembles the progressive wave we used above gives

kt

E p³

B (2.43)

kt B

y

p v Í ®E

v ³

D E (2.44)

Combining these two equation as done previously, we get

v

2Ep v Í ® y ³

Ev ³

2D

y E (2.45)

Ó

³

2

Í

®

D

³ v ä

2

2Õ

Ep

0 (2.46)




So that

³p

Í ®

2 D

â

1

2 ç

4 ä

2

2v

®

2

D

2(2.47)

and the group velocity is

w

³

w

pâ

2 ä

2

è

4 ä

2

2v

®

2s

D

2(2.48)

If ® p

0 then we are back again to the previous problem as can be verified.

END OF EXAMPLE 2.4



18



LESSON 3

RelativisticElectrodynamics

3.1 Coverage

We examine the covariant formulation of electrodynamics. We take up the concept

of 4-tensors and give examples of these. Also, show how 4-tensors are manipu-

lated. We discuss the group of Lorentz transformations in the context of electro-dynamics.

3.2 Formulae used

A Lorentz boost in the 3-direction

éê ë

ê

I ìí

íî ï

0 0`

ïð

0 1 0 0

0 0 1 0

`

ïð

0 0

ï

ò ó

ó

ô

(3.1)

The field tensor (components are 0, 1, 2, 3)

õ ê ö

I ìí

íî

0`

÷ 1 `÷ 2 `

÷ 3

÷ 1 0 `

eø

3e

ø

2

÷ 2e

ø

3 0`

eø

1

÷ 3 `

eø

2e

ø

1 0

ò ó

ó

ô

(3.2)

19




and so

2

2

1

p

2

2

1

; 2

2

2

p

2

2

2

; 2

2

3

p

2

2

3(3.11)

butw

wx

p

wx

¯

wx

w

wx

¯

w

¯1w

x

w

w

¯

1

w

¯2w

x

w

w

¯

2

w

¯3w

x

w

w

¯

3

p (3.12)

p

w

wx

¯

v

å

w

w

¯3

(3.13)

and sow

2

wx

2p

w

wx

¯

v

å

w

w

¯3

w

wx

¯

v

å

w

w

¯3

p

w

2

wx

¯

2

å 2

w

2

w

¯

23

v 2 å

w

2

wx

¯

w

¯3

(3.14)

where we have used the fact that the operatorsw

s

wx

¯ andw

s

w

¯3 commute. Thus we have

found that

ù 2~

ù 2¯

p

z

¯

2Ψ v

1 2

w

2Ψ

wx

¯

2v

å 2

2

w

2Ψ

w

¯

23

2 å

2

w

2Ψ

wx

¯

w

¯3

p 0 (3.15)

Which obviously does not have the same form as the d’Alembertian in the unprimed sys-tem!

Let us do the same calculations for the case of a Lorentz transformation; more specifically

we consider a boost along the 3 axis which is given by

þ

p

¢

þ

þ

þ

(3.16)

where

¢

þ

þ

p

£¤

¤

¥

¦ 0 0 v

¦§

0 1 0 0

0 0 1 0

v

¦§ 0 0 ¦

©

(3.17)

(remember that y runs over 0, 1, 2, 3). Since¦

and§

depend onå

The 4-gradient,w

þ

w

s

w

þ

transforms as

wþ

p

w

þ

w

þ

wþ

p

w

w

þ

Ó

¢

þ

ÿ

ÿ

Õ

wþ

p

ÿ

w

¢

þ

ÿ

w

þ

¢

þ

ÿ

wþ

p

¢

þ

þ

wþ

(3.18)

so

ù 2p

wþ w

þ

p

þ

ÿ ¢

þ

þ

¢ ÿ

ÿ

wþ

w

ÿ

p

¦ 2v

¦ 2 § 2

w

20 v

w

21 v

w

22 v

¦ 2v

¦ 2 § 2

w

23 p

þ

ÿ

wþ

w

ÿ

p

wþ

w

þ

(3.19)

In other words we have found that the d’Alembertian is invariant under Lorentz boosts.

END OF EXAMPLE 3.1



22 LESSON 3. RELATIVISTIC ELECTRODYNAMICS

f

INVARIANT QUANTITIES CONSTRUCTED FROM THE FIELD TENSOREXAMPLE 3.2

Construct the dual tensor

Ù

p

12

D

Ù! " #

½

" #of the field tensor

½

þ

ÿ . What quantities

constructed solely with the field tensor and it’s dual tensor, are invariant under Lorentz

transformations? Having found these quantities you should be able to answer the questions:

can a purely electric field in one inertial system be seen as a purely magnetic field

in another?

and, can a progressive wave be seen as a purely electric or a purely magnetic field

in an inertial system?

Solution

The dual tensor of ½

Ù

is given by

Ù

p

1

2D

Ù " #

½

" # p

1

2D

Ù " #

"

þ

#

ÿ

½

þ

ÿ

(3.20)

where

½

þ

ÿ

p

£¤

¤

¥

0 v

1 v

2 v

3

1 0 v

3

2

2

3 0 v

1

3 v

2

1 0

©

(3.21)

and

þ

ÿ

p

£¤

¤

¥

1 0 0 0

0 v 1 0 0

0 0 v 1 0

0 0 0 v 1

©

(3.22)

We determine first the field tensor with two covariant indices through the formula

½

þ

ÿ

p

þ

Ù

ÿ

½

Ù

p

£¤

¤

¥

0

1

2

3

v

1 0 v

3

2

v

2

3 0 v

1

3 v

2

1 0

©

(3.23)

so

Ù

p

£¤

¤

¥

0 v

1 v

2 v

3

1 0

1 v

2

2 v

3 0

1

3

2 v

1 0

©

(3.24)




We see that the dual tensor can be obtained from½

þ

ÿ

by putting E ~

B and B~ v

E s

.

From the formula for the dual tensor

þ

ÿ

we see that it is a 4-tensor since½

þ

ÿ

is a four

tensor and D is easily shown to be an invariant under orthogonal transforms for which the

Lorentz transform is a special case. What can we create from

þ

ÿ

and½

þ

ÿ

which is

invariant under Lorentz transformations? We consider the obviously invariant quantities

þ

ÿ

½

þ

ÿ and½

þ

ÿ

½

þ

ÿ .

þ

ÿ½

þ

ÿ

p v4 E

h

B (3.25)

½

þ

ÿ½

þ

ÿ

p 2

2

2v

2 (3.26)

This means that Eh

B and

2v

2

2 are Lorentz invariant scalars.

Relation of EM fields in different inertial systems Now that we know that Eh

B

and

2v

2

2 are Lorentz invariant scalars, let see what they say about EM fields in

different inertial systems. Let us say that % Eh

B and &

2 v

2 2. All inertialsystems must have the same value for % and & . A purely electric field in one inertial

system means that B p 0, so % p 0 and & ' 0. A purely magnetic field would mean that

E¯ p 0, so % p 0 but & ( 0. In other words it does not seem that a purely electric field

can be a purely magnetic field in any inertial system.

For a progressive wave E ) B so % p 0 and in a purely electric or a purely magnetic field

% p 0 also, but for a progressive wave

p

so & p 0 and if the other system has

E ¯p

0 or B ¯p

0 then &p

0 force both the fields to be zero. So this is not possible.

END OF EXAMPLE 3.2

f

COVARIANT FORMULATION OF COMMON ELECTRODYNAMICS FORMULAS EXAMPLE 3.3

Put the following well know formulas into a manifestly covariant form

The continuity equation

Lorentz force

The inhomogeneous Maxwell equations

The homogenous Maxwell equations




Solution

The Methodology To construct manifestly covariant formulas we have at our disposal

the following “building blocks”:

an event

þ

p

x

x ,

4-velocityä

þ

p

¦

¦

v

,4-momentum 0

þ

p

s

p ,

wave 4-vector

þ

p

³s

k ,

4-current density 1

þ

p

q

j ,

4-potential

þ

p

s

A ,

4-force½

þ

p

¦ vh

F s

¦ F

Also we have the 4-gradient

4-gradientw

þ

w

s

w

þ

p

w

s

w

x

w

s

w

x

as our operator building block and also the second rank 4-tensor

field tensor½

þ

ÿ

p

£¤

¤

¥

0 v

1 v

2 v

3

1 0 v

3

2

2

3 0 v

1

3 v

2

1 0

©

Observe that we use indices which run 0, 1, 2, 3 where the 0-component is time-like

component, there is also the system where indices run 1, 2, 3, 4 and the 4-component

is the time-like component. Beware!

A sufficient condition to formulate covariant electrodynamic formulas is that we make

our formulas by combine the above 4-vectors. To make sure we have a covariant form

we take outer product (i.e. simply combine the tensors so that all the indices are free)

and then perform zero or more contractions, i.e. equate two indices and sum over this

index (notationally this means we create a repeated index). In the notation we use here

contractions must be between a contravariant (upper) index and a covariant (lower) index.

One can always raise or lower a index by including a metric tensor Ù

. On top of this

sufficient condition, we will need to use our knowledge of the formulas we will try to

make covariant, to accomplish our goal.

The continuity equation We know that the continuity equation is a differential equa-

tion which includes the charge density and the current density and that it is a scalar equa-

tion. This leads us to calculate the contraction of the outer product between the 4-gradientw

þ

and the 4-current 1

ÿ

wþ

1

þ

p

0 (3.27)




This is covariant version of the continuity equation, thus in space-time the continuity equa-

tion is simply stated as the 4-current density is divergence-free!

Lorentz force We know that the left hand side of Lorentz force equation is a 3-force.

Obviously we should use the covariant 4-vector force instead. And on the right hand side

of th Lorentz equation is a 3-vector quantity involving charge density and current density

and the E and B fields. The EM fields are of course contained in the field tensor½

þ

ÿ

. To

get a vector quantity from½

þ

ÿ

and 1

þ

we contract these so our guess is

½

þ

p

½

þ

ÿ

1

ÿ (3.28)

Inhomogeneous Maxwell equations The inhomogeneous Maxwell may be written

as the 4-divergence of the field tensor

w

Ù

½

Ù

p1

(3.29)

Homogeneous Maxwell equations The homogenous Maxwell equations are written

most compactly using the dual tensor of the field tensor. Using the dual tensor we have

w

Ù

Ù

p 0 (3.30)

END OF EXAMPLE 3.3

f

FIELDS FROM UNIFORMLY MOVING CHARGE VIA LORENTZ TRANSFORMATION EXAMPLE 3.4

In the relativistic formulation of classical electrodynamics the E and B field vector form

the antisymmetic electrodynamic field tensor

p

£¤

¥

0 v 4 v

s

v 0

v

4s

4 v

0 v

s

s

4s

s

0

©

Show that the fields from a charge 6 in uniform, rectilinear motion

E p

6

4 7

D 0 8

3

1 v

å 2s

2 R0

B p vt

E s

2

are obtained via a Lorentz transformation of the corresponding fields in the rest system of

the charge.

Solution

We wish to transform the EM fields. The EM fields in a covariant formulation of electro-

dynamics is given by the electromagnetic field tensor




½

þ

ÿ

p

£¤

¤

¥

0 v

3

2

1

3 0 v

1

2

v

2

1 0

3

v

1 v

2 v

3 0

©

(3.31)

where we are using components running as 1, 2, 3, 4. To transform the EM fields is to

transform the field tensor. A Lorentz transformation of the field tensor can be written

½9 A

p

¢9

þ

¢A

ÿ

½

ÿ

þ

(3.32)

F p ~LF0L (3.33)

where

¢ÿ

þ

p

£¤

¤

¥

¦ 0 0 v

§¦

0 1 0 0

0 0 1 0

v

§¦ 0 0 ¦

©

(3.34)

where ¦

p 1 s

è

1 v

§ 2 and §

p

å

s

where v p

å

1 ˆ

. The fields in the rest system

0 are

E0p

6

4 7

D 0

0 ˆ

F

0 ˆG

0 ˆ

è

0 2 F

0 2

0 23

(3.35)

B0p 0 (3.36)

so that

½

þ

ÿ

p

£¤

¤

¥

0 0 0

01

0 0 0

02

0 0 0

03

v

01 v

02 v

03 0

©

(3.37)

A little matrix algebra gives

~LFL p ~L

£¤

¤

¥

0 0 0

01

0 0 0

02

0 0 0

03

v

01 v

02 v

03 0

©

£¤

¤

¥

¦ 0 0 v

§¦

0 1 0 0

0 0 1 0

v

§¦ 0 0 ¦

©

p (3.38)

p

£¤

¤

¥

¦ 0 0 v

§¦

0 1 0 0

0 0 1 0

v

§¦ 0 0

¦

©

£¤

¤

¥

v

§¦

01 0 0 ¦

01

v

§¦

02 0 0

¦

02

v

§¦

03 0 0 ¦

03

v

¦

01 v

02 v

03

§¦

01

©

p (3.39)

p

£¤

¤

¥

v

§¦ 2

01

§¦ 2

01

¦§

02

¦§

03

¦ 2

1 v

§ 2

01

v

¦§

02 0 0 ¦

02

v

¦§

03 0 0 ¦

03

v

¦

2 1 v

§

2

01 v

¦

02 v

¦

03 v

§¦

2

01

§¦

2

01

©

(3.40)




1 p

01 (3.41)

2 p

¦

02 (3.42)

3 p

¦

03 (3.43)

1 p 0 (3.44)

2 p

¦§

03 (3.45)

3 p v

¦§

02 (3.46)

£¤

¤

¥

0

F

0

0

x

0

©

p

¢

£¤

¤

¥

F

x

©

p

£¤

¤

¥

¦

v

8

F

¦

x

§

©

(3.47)

where8

p

å

x

E p

6

47

D

0

¦

v

8

ˆ

F ˆG

ˆ

è

¦

2

v

8

2 F

2

23

(3.48)

I 20 p

8

v

2 F

2

2 (3.49)

8

v

p

I

0 sin

v

7

2

p

I

0 cos

(3.50)

E p

6

4 7

D 0¦ 2

R0Q

v

8

2

4

2 R

2

"

2

3(3.51)

v

8

2

F

2

2

¦ 2p

I

20 v

F

2v

2

F

2

2

¦ 2p (3.52)

p

I 20

F

2

2

1¦ 2

v 1 p (3.53)

p

I 20

1 v

§ 2 sin2

(3.54)

E p

6

4 7

D 0

1 v

§ 2

R0

I

30

è

1 v

§ 2 sin2

3(3.55)

END OF EXAMPLE 3.4



28



LESSON 4

Lagrangian andHamiltonian

Electrodynamics

4.1 Coverage

We briefly touch the Lagrangian formulation of electrodynamics. We look at both

the point Lagrangian for charges in EM fields and the Lagrangian density of the

EM fields.

4.2 Formulae used

The Lagrangian for a charged particle in EM fields is

é

I` T

e 2

ï

`V d V v

R

A (4.1)

A useful Lagrangian density for EM field and its interaction with charged particles

is given by

Y

I`

1

2W 0 `

e 2 ø 2`

÷

2 a

` AR

j dT

(4.2)

29



30 LESSON 4. LAGRANGIAN AND HAMILTONIAN ELECTRODYNAMICS

4.3 Solved examples

f

CANONICAL QUANTITIES FOR A PARTICLE IN AN EM FIELDEXAMPLE 4.1

Derive the canonical momentum and the generalised force for the case of a charged particle

in EM field given by

and A. The Lagrangian is¢

p vb

c

2

"

v 6

6

vh

A.

Solution

We know from analytical mechanics that the canonical momentum P is found through

f

p

w

¢

w

å

(4.3)

so with

¢

p v g

2

¦

v 6

6

vh

A (4.4)

we find that

f

p

w

¢

w

å

p

w

w

å

v

g

2

ç

1 v

å

å

2v 6

6

å

p

g

2å

2

Q

1 v pq p q

c

2

6

p

g

¦å

6 (4.5)

P p p 6

A (4.6)

And on the other hand the generalised force is

p v

wu

w

x

wu

w

˙

(4.7)

whereu

is the generalised velocity dependent potential 6

v 6 vh

A, which gives us,

p v

w

w

6

v 6

å

x

w

w

å

6

v 6

å

p v6

w

w

6

w

å

w

v 6

x (4.8)

Qp v

6

g

6

g

vh

A v

6

A

x

p v6

g

6

vh

g

A 6

vt

g t

A v

6

w

Aw

x

v 6

vh

g

A

p v6

g

6 v

tg t

A 6

w

Aw

x

p6

E

6

v

t

B (4.9)




What does the canonical momentum P p p 6

A represent physically? Consider a charge

moving in a static magnetic field. This charge will perform gyro-harmonic motion. Ob-

viously the momentum is not conserved but on the other hand we did not expect it to be

conserved since there is a force on the charge. However we now from analytical mechan-

ics, that it is the conservation of canonical momentum that is more general. Conservation

of canonical momentum is found when the problem is translational invariant, which is true

in the case we have here since the potentials do not depend on spatial coordinates. So weexpect P p p

6 A to be a constant of the motion, but what is it?

END OF EXAMPLE 4.1

f

GAUGE INVARIANCE OF THE LAGRANGIAN DENSITY EXAMPLE 4.2

Consider the Lagrangian density of the EM fields in the form

x

p v

1

2 0

2

2v

2 v

Ah

j q

(4.10)

We know that EM field are invariant under gauge transformations

A p A ¯

g±

(4.11)

p

¯

v

w

±

wx (4.12)

Determine the Lagrangian density under a gauge transformation. Is it invariant? If not,

discuss the consequences this would have.

Solution

Let us insert the gauge transformation relations into the Lagrangian density. Remembering

that E and B are invariant under gauge transformations we find that

x

~

x

¯

p v

1

2 0

2

2v

2 v

A ¯

g ±

h

j q

¯

v

w

±

wx

(4.13)

v

1

2 0

2

2v

2 v

A ¯

h

j q

¯

v

g±

h

jv q

w

±

wx

p (4.14)

p v

1

2 0

2

2v

2 v

A ¯

h

j q

¯

v

gh

±

j

±

gh

j v

w

q

±

wx

±

w

q

wx

p(4.15)

p v

1

2 0

2

2v

2 v

A ¯

h

j q

¯

±

w

q

wx

gh

j v

gh

±

j v

w

q

±

wx

p(4.16)

p

x

0v

gh

±

j v

w

q

±

wx

(4.17)



32 LESSON 4. LAGRANGIAN AND HAMILTONIAN ELECTRODYNAMICS

where we have used the continuity equation. We see thatx

0 obviously has the same form

asx

, but what about the other two terms?

Some thought reveals that the neccessary condition for a Lagrangian to be physically ac-

ceptable is not the Lagrangian itself is invariant but rather that the variation of the action

integral p

yy

x

3

x

is invariant. So now we would like to check that the gauge

transformations indeed do not affect any the variation of the action. Now it is possible

to proceed in two different ways to do this: one is simply carry out the integration in thedefinition of the action integral and check that its variation is zero, or two, remembering

that the variation of the action is equivalent to the Euler-Lagrange equations, one could

plug in the Lagrangian density (4.17) into Euler-Lagrange equations to check the resulting

equations differ from the Maxwell equations.

Let us use the first alternative. Since the action is linear inx

it is sufficient to examine

ext p

g h

±

j

w

q

±

wx

3

x

p

±

jh

S

x

q

± 3

1

0

v

±

w

q

wx

gh

j

3

x

p

±

jh

S

x

q

±

3

1

0

(4.18)

where we have used the continuity equation. Furthermore, if we assume no flux source/sink

at infinity then we can write

ext p

q

± 3

1

0

(4.19)

Now when taking the variation, we realize that we must holdx

0 andx

1 the end point of the

particle path fixed, and thus

ext p 0 (4.20)

As one would expect, gauge transformations of the potentials do not effect the physics of

the problem!

END OF EXAMPLE 4.2



LESSON 5

ElectromagneticEnergy, Momentum

and Stress

5.1 Coverage

Here we study the force, energy, momentum and stress in an electromagnetic field.

5.2 Formulae used

Poynting’s vector

S I EX

H (5.1)

Maxwell’s stress tensor

I÷

d

ø

`

1

2

`

÷

d

ø

a (5.2)

33



34 LESSON 5. ELECTROMAGNETIC ENERGY, MOMENTUM AND STRESS

Table 5.1. The following table gives the relevant quantities. The field vectors in

this table are assumed to be real. If the given fields are complex, use the real part

in the formulas.

Name Symbol Formula SI unit

Energy density

u

p

12 E

h

D

12 H

h

B J/m3

Intensity S Et

H W/m2

Momentum density PEM D

y Et

H kg/s m2

Stress T

Ý

v

12

m

Ý

m

m m

Pa

5.3 Solved examples

f

EM QUANTITIES POTPOURRIEXAMPLE 5.1

Determine the instantaneous values of the energy density, momentum density, intensity

and stress associated with the electromagnetic fields for the following cases:

a)

E p

0e2

¹

º¿

m

1

1» Á

H p

mþ

»

ˆ

t

Efor 1 p

ã

y ³ , 2 p 3 p 0.

b) same as in a) but for 1 pÍ Ø

c)

p

þ

0b

2

cos

3

p

þ

0b

4

sin

3

Also, identify these cases. Assume that D p

E and B py

H.

Solution

Background

(a) Progressive wave This case is an example of a progressive or propagating wave.

Since E and H are complex we must first take their real parts:

Re E p

0 cos

1

1 v ³

x

e2 (5.3)

Re Hp v

1

y³

0 cos

1

1 v ³

x

e3 (5.4)

The energy density is




u

p

p

D

2

20 cos2

1

1 v ³

x

21

2 y³

2

20 cos2

1

1 v ³

x

p

(5.5)

p

D

20 cos2

³

ã

D

y

1 v ³

x

(5.6)

The intensity or power density is

S p

0 cos

1

1 v ³

x

e2

t

v

1

y³

0 cos

1

1 v ³

x

e3 p

(5.7)

p v

ç

D

y

20 cos2

³

ã

D

y

1 v ³

x

e2

t

e3 p (5.8)

p v

ç

D

y

20 cos2

³

ã

D

y

1 v ³

x

e1 (5.9)

The momentum is

PEM p

D

y S p

D

ã

D

y

20 cos2

³

ã

D

y

1 v ³

x

(5.10)

The stress is

11 p v

D

2

22 v

y

2

23

p v

D

2

20 cos2

1

1 v ³

x

v

21

y³

2

20 cos2

1

1 v ³

x

p v

D

20 cos2

1

1 v ³

x

(5.11)

21 p

31 p

12 p 0 (5.12)

22 p

D

2

22 v

y

2

23

p

D

2

20 cos2

1

1 v ³

x

v

21

y³

2

20 cos2

1

1 v ³

x

(5.13)

32 p

13 p

23 p 0 (5.14)

33 p

y

2

23 v

D

2

22 p v

22 p 0 (5.15)

(b) Evanescent wave This case is a example of an evanescent wave. We take real part

of the fields keeping in mind the fact that 1 pÍ Ø is imaginary:

Re E p

0

¹

Ù

1 sin ³

x

e2 (5.16)

Re H p v

Ø

y³

e1

t

Im E p v

Ø

y³

e1

t

0

¹

Ù

1 cos ³

x

e2

p v

Ø

0

y³

¹

Ù

1

cos³

x

e3 (5.17)





u

p

p

D

2

20

¹

2 Ù

1 sin2³

x

Ø

2

20

2 y³

2

¹

2 Ù

1 cos2³

x

p (5.18)

p

20

¹

2 Ù

1

2

D sin2³

x

Ø

2

y³

2cos2

³

x

(5.19)

The intensity is

Sp v

Ø

20

y³

¹ 2 Ù

1 sin ³

x

cos ³

x

e1 (5.20)

The momentum is

P p

D

y Sp v

Ø

20

D

³

¹ 2 Ù

1 sin ³

x

cos ³

x

e1 (5.21)

The stress is

11 p v

D

2

22 v

y

2

23 p

20

¹

2 Ù

1

2

D sin2³

x

Ø

2

y³

2cos2

³

x

(5.22)

21 p

31 p

12 p 0 (5.23)

22 p

D

2

22 v

y

2

23 p

20

¹

2 Ù

1

2

D sin2³

x

v

Ø

2

y³

2cos2

³

x

(5.24)

32 p

13 p

23 p 0 (5.25)

33 p

y

2

23 v

D

2

22 p v

22 p v

20

¹

2 Ù

1

2

D sin2³

x

v

Ø

2

y³

2cos2

³

x

(5.26)

(c) Magnetic dipole This case is a magnetic dipole. The fields are real and in spherical

coordinates.


u

p

p

1

2 y 0

2

2

p

1

2 y 0

y

20

g

2

16 7

2

4cos2

6

y

20

g

2

16 7

2

sin2

6

p

y 0g

2

32 7

2

4cos2

sin2

1

6

p

y 0g

2

32 7

2

6

1 3cos2

(5.27)

The intensity is S p 0 since E p 0, and likewise for the momentum. The stress tensor

components are




p

1

y 0

2

v

u

p

p

y 0g

2

4 7

2

cos2

6v

y 0g

2

32 7

2

6

1 3cos2

p

y 0g

2

32 7

2

6

8cos2

v 1 v 3cos2

p v

y 0g

2

32 7

2

6

1 v 5cos2

(5.28)

p

1

y 0

p

y 0g

2

8 7

sin

cos

6(5.29)

p

0 (5.30)

p

(5.31)

p

1

y 0

2

v

u

p

p

y 0g

2

16 7

2

sin2

6v

y 0g

2

32 7

2

6

1 3cos2

p

y 0g

2

32 7

2

6

2sin2

v 1 v 3cos2

p v

y 0g

2

32 7

2

6

1 v 3cos2

2sin2

(5.32)

p 0 (5.33)

p

0 (5.34)

p 0 (5.35)

p 0 (5.36)

END OF EXAMPLE 5.1

f

CLASSICAL ELECTRON RADIUS EXAMPLE 5.2

Calculate the classical radius of the electron by assuming that the mass of an electron is the

mass of it’s electric field and that the electron is a homogenous spherical charge distribution

of radius k

and total charge Þ 6Þ p

¹

. Mass and energy are related through the equation

p

g

2.

Solution

One the “failures” of Maxwell’s equations or classical electrodynamics is on question of

mass of point particles. From relativity one can show that fundamental particles should be

point-like. However, if one calculates the electromagnetic mass from Maxwell’s equation

one gets an infinite result due to the singularity in Gauss lawg h

E pq s

D 0. This points

to the fact that Maxwell’s equations has a minimum length scale validity where quantum

mechanics takes over.

We will calculate this problem as follows: determine the electric field in all of space and

then integrate the formula for the energy density of the electric field. This integral will

contain the radius of the electron since it partitions the integration. We then relate this field

energy to the mass of the electron, which is a known quantity. We use this relation to solve




for the “electron radius”.

We start by determining the electric field. Consider a homogeneous spherical charge distri-

bution in a vacuum. Introduce spherical coordinates. We divide the whole space into two

parts: the region outside of the sphere, i.e.

k

, is given by Coulomb’s law:

k

p

¹

4 7

D 0 2

(5.37)

and the region inside the sphere is given by Gauss law

gh

E

k

p q s

D 0 (5.38)

which in this case gives

1

2

w

w

2

p

¹

4 s 3 7

3k

D 0

(5.39)

If we integrate this equation, i.e. (5.39), we find that

k

p

¹

4 7

3k

D 0 (5.40)

We can easily verify that (5.40) is indeed a solution to (5.39) and furthermore it is continu-

ous with (5.37), thus making the solution unique.

We may now determine the energy density of the electric field due to the electron. It is

simply

u

p

p

1

2D 0

2 (5.41)

l

u

p

k

p

k

2

32

2 n

0

1

4

u

p

k

p

k

2

32

2 n

0

6e

2(5.42)

Now we integrateu

p

over all space. That is

u

p

all space

u

p

k

u

p

k

3

p

4

0

e

0

¹

2

32 7

2D 0

6e

2

2

Ω

4

0

¶

e

¹

2

32 7

2D 0

1

4

2

Ω

p

¹

2

8 7

D 0

6e

5

5

e

0

¹

2

8 7

D 0

v

1

¶

e

p

3¹

2

20 7

D 0

1

e

(5.43)

Finally, we relate the total electric field energy to the rest mass of the electron and solve

for the electron radius,




u

p

g

k

2 (5.44)

3¹

2

20 7

D 0

1

e

p

g

k

2

e p

3

5

¹

2

4 7

D 0g

k

2(5.45)

This last result can be compared with the de facto classical electron radius which is defined

as

k

p

¹

2

4 7

D 0g

k

2(5.46)

and is found by calculating the scattering cross section of the electron.

END OF EXAMPLE 5.2

f

SOLAR SAILING EXAMPLE 5.3

Investigate the feasibility of sailing in our solar system using the solar wind. One technicalproposal uses kapton, 2 mm in thickness, with a 0.1 mm thick aluminium coating as ma-

terial for a sail. It would weigh 1 g/m2. At 1 AU (= astronomical unit o distance between

sun and earth) the intensity of the electromagnetic radiation from the sun is of the order

1 4t

103 W/m2. If we assume the sail to be a perfect reflector, what would the acceleration

be for different incidence angles of the sun’s EM radiation on the sail?

Solution

In this problem we will use the principle of momentum conservation. If the continuity

equation for electromagnetic momentumw

PEMs

wx

gh

T p

w

Pmechs

wx

is integrated over

all space the stress term disappears and what is left says that a change in electromagnetic

momentum is balanced by mechanical force.Imagine a localised pulse of a progressive electromagnetic wave incident on a plane. That

the wave is progressive means simply that it is “purely radiation” (more about progressive

waves in the next lesson). Let us characterize this wave by a Poynting vector S of duration

∆x

. It travels in space with velocity , it “lights up” the area ∆ on the surface on the

sail, and, furthermore, S makes an angle 180 v

with the normal of the sail surface. The

momentum carried by such a wave is

pbeforeEM p Pbefore

EM

∆

p

S s

2

∆

∆x

cos

(5.47)

so the momentum along the direction of the surface normal ˆ

is

pbefore

EM

h

ˆ

p v

Þ S Þ

∆

∆

x

cos2

(5.48)




After hitting the sail, the pulse will be characterized with all the same quantities as before

the impact, except that the component along the surface normal will be opposite in sign.

This is because the sail is assumed to be perfectly reflecting. Thus,

pafterEM

h

np v

pbeforeEM

h

n p

Þ S Þ

∆ ∆x

cos2

(5.49)

Now the continuity equation for EM momentum says thatw

PEM s

wx

p

w

Pmech p F so for

the component along the sail surface normal this implies that

Fh

n p

pafterEM v pbefore

EM

∆x

h

n p 2Þ S Þ

∆ cos2

(5.50)

The other force components are zero by symmetry.

So finally, the pressure exerted by the pulse on the sail is simply

p

Fh

n

∆

p 2Þ S Þ

cos2

(5.51)

We now have the pressure exerted by a pulse charcterized by S incident on a surface with

an angle . The solar wind can be seen as a multitude of such pulses radiating radiallyoutwards from the sun. At 1 AU, the solar constant is 1.3 kW s m2. The solar constant is

the intensity of EM radiation or in other words, the magnitude of the Poynting vector Þ S Þ .

From equation (5.51), we find that

p

21 3

t

03 W/m2

3t

108 m/scos2

(5.52)

Newton’s second law gives the acceleration of the sail (which, as we recall, weighs

3t

10 3 kg/m2) due to the solar wind can at most be

p

0 9t

10 5 N/m2

3h

10 3 kg/m2p 3 mm/s2 (5.53)

which is half of the acceleration due to the sun’s gravitational field.

END OF EXAMPLE 5.3




f

MAGNETIC PRESSURE ON THE EARTH EXAMPLE 5.4

Determine the magnetic pressure due to the Earth’s magnetic field at the magnetic

poles (take p

6t

10 5 T) and compare this with the Earth’s atmospheric pressure

(1atm p 1 01t

105 Pa). Now assume that the magnetic dipole moment is proportional to

the angular velocity of the earth, how much faster would the angular velocity need to be

for the magnetic pressure to be comparable to the atmospheric pressure?

Solution

In this exercise we see that EMF can exert pressure not only via radiation pressure, but also

through static fields.

The static magnetic pressure is quantified in the Maxwellian stress tensor, which for this

case is

T p

2

sy 0 0 0

0 0 0

0 0 0

v

u

p

0 0

0u

p

0

0 0u

p

(5.54)

So by taking the inner product of T

and the unit vector pointing in the direction the northpole, namely 1 0 0 z , we find the pressure in the radial direction to be

p

2

2 y 0

p

6h

10 5

2

1 3h

10 62

h

1 3h

10 6p 1 4

h

10 3 Pa (5.55)

Now we assume, according to the hypothesis in problem, that

1

period

g

ã

(5.56)

so that |

1|

0

p

ç

f

0f

1

(5.57)

where we denote the current values of the pressure and rotational period time, 0 and

0

respectively and the hypothetical values 1 and

1. Solving for

1 we arrive at

1 p

0

0

1

(5.58)

So with 1 p 105 Pa, we find that

1 p 24h

3600h

ç

1 4h

10 3

105p 10 s (5.59)

END OF EXAMPLE 5.4



42



LESSON 6

Radiation fromExtended Sources

6.1 Coverage

We will study the important general problem of how to calculate the EM fields

induced by spatially extended, time-varying sources. This problem is solved in

different ways depending on the explicit form of the source distribution. For trulyextended bodies with non-monochromatic time dependence, we use the general

expressions for the retarded potentials. And for monochromatic, one-dimensional

current distributions, which we will call antennas here, we use the formulas given

below.

6.2 Formulae used

General expressions for retarded potentials

A `

F~ x

a

I

c

04

3

G

j`

x G~

F

G

a

x ` xG

`

F~ x a

I

1

4 0

3

G

T

`

x G~

F

G

a

x ` xG

For “antennas”, we use

BI

`

c 0

sin

4

x x ë

x ` xG

ˆ

EI

`

sin

4

0

e

x x ë

x`

xG

ˆ

43



44 LESSON 6. RADIATION FROM EXTENDED SOURCES

S I

2 sin2

c 0e

32

2

x ` xG

2

2 ˆ

where

I

`

G

a

ë

cos

G

6.3 Solved examples

f

INSTANTANEOUS CURRENT IN AN INFINITELY LONG CONDUCTOREXAMPLE 6.1

Consider the following idealised situation with an infinitely long, thin, conducting wire

along the

axis. For

x

0, it is current free, but at time

x

p

0 a constant current1

isapplied simultaneously over the entire length of the wire. Consequently, the wire carries

the current

p

0

x

0

1

x

' 0

It is assumed that the conductor can be kept uncharged, i.e., qp 0 . Determine B, E and S

in whole space.

Hint: Calculate first the vector potential A .

Solution

This problem belongs to the most general category of problems of the kind where given

a source distribution one wants to find the EM fields. This is because the source is not

monochromatic, so it is not an antenna, and furthermore it is an extended distribution, so

multipole expansion analysis is not possible. So we must use the most general formula for

calculating fields induced by time-varying sources, which in the Lorentz gauge take the

form

A

x

x p

y 0

4 7

3

j

x

x

Þ x v x

Þ

(6.1)

x

x p

1

4 7

D 0

3

q

x

x

Þ x v x

Þ

(6.2)




where the source timex

is to be replaced byx

p

x

v Þx v x

Þs

sox

is seen as a function

of x

, x and x

. The functionx

v Þx v x

Þs

is known as the retarded timex

ret, an expression

which is meaningful only relative the field point

x

x .

The algorithm of solution now that we have decided to use (6.1) is to find the explicit

form for

x

x

and q

x

x

, perform the integrations to obtain the potentials A

x

x and

x

x and, finally, derive the E and B fields from the potentials in the normal fashion and

the Poynting vector from the fields.

Let us find the explicit expression for the current density

x

x

, which is illustrated in

Fig. 6.1. In many problems, the expressions for the sources consist of a time-dependent

part times a space-dependent part. This is one such case. The “switching on” atx

p 0 can

be written as a step function

x

. And if we orient the wire along the

axis, the space

dependent part can be written

F

1

ˆ

.

On the other hand, the charge distribution q

x

x p

0 as given in the problem formulation.

This can be seen as and v charges flowing in opposite directions such as to keep the

total charge density q

x

x p 0 but this could still have a total current density j

x

x

-

p 0.

So we have for the current that

j

x

x

p

F

1

x

ˆ (6.3)

and for the charge density

q

x

x

p

0 (6.4)

We insert (6.3) into (6.1), remembering to replace the source timex

withx

v Þ x v x

Þ s

,

and perform the integration. The integrations over

and F

are trivial:

A

x

x p

ˆ

y 0 1

4 7

x

v Þx v

ˆ

Þs

Þ x v

ˆ

Þ

(6.5)

For the remaining

integration we use cylindrical coordinates (see Figure BLP1.2cyl) so

we can write

x p

ˆ

(6.6)

Þ x v x

Þp

è

2

2 (6.7)

The step function in the integrand is zero when its argument is less than zero. This means

that when integrating over

, only those

contribute which satisfy

x

v Þ

ˆ

v

ˆ

Þs

0

x

è

2

2 (6.8)

which can be written, if we assume

x

0

(6.9)

as




r

z

Figure 6.1. The current density distribution j is along the axis and is turned on

at

0. We use cylindrical coordinates.

2 2 2

2 (6.10)

ª 2

2 ¬

2 (6.11)

or

¬

ª 2

2 ¬

2

ª 2

2 ¬

2 -® (6.12)

where we have introduced ® simply as a shorthand. These limits can be understood as

follows (cf. Fig. 6.2): after the current is switched on, EM fields are sent from each point

along the wire and travel at the speed of light. Since the information (i.e., the current turn-

on) carried by the fields travel in the “line of sight”, or in other words in a straight line,

each field point only sees those parts of the current which are close enough. This illustrates

the concept of retarded time, which is only meaningful relative the field point, and also the

information gathering sphere.

So we have now that

A

ˆ°±

0 ³

4 ´µ ·

·

1¹

2 2 º

ˆ°±

0 ³

4 ´

ln »¼

1 ½ 1 ¬ ¾

2

¿ 2 À 2

1 ¬

½ 1 ¬ ¾

2

¿ 2 À 2 ÁÂ

(6.13)

On the other hand, the scalar potential we may set so ÃÄ Æ

x È

0, since ÉÄ Ê Æ

x ÊÈ

0.

Now that we have the vector potential we can derive the E and B fields. The B field is

B

Ì Î

A (6.14)

and since A only has its ˆ° -component different from zero in the cylindrical system, we

have




r

P P

ÐÑ Ó Ô Õ Ð Ñ Ö × 2

Ø

Ó

2Ô

Õ

×

2×

P

ÐÑ Ù Ú

2×

Û

2 Ø

Ó

2Ô

2

Figure 6.2. This series of snapshots shows what the part of the current is seen at

the field point P at different field times Ü .

BÝ Þ

ˆßà â ã

à å

(6.15)

Thus we only need æç è

æé

. Introducing êë í 1 Þ

å

2 ïð ò 2

Ü

2 ô , which satisfies

à

ê

à å

Ý Þ

å

ò 2Ü

2

1

ê

(6.16)

we then can write

àâ ã

à å

Ýõ

0 ÷

4 ø

à

à å ù

lnù

1 úê

1Þ ê ü ü

Ý Þ õ

0 ÷

4 ø

2

å

ê

(6.17)

þ

B Ý ˆß

õ

0 ÷

2ø

å ÿ

1Þ

é

2

2¡

2

(6.18)

Observe that for large Ü we have the “static” case: B Ý ˆß

õ

0 ÷

ïð 2 ø

å

ô

The electric field is derived from

E Ý Þ

à

A

à

Ü

Ý Þ ˆ¢

õ

0 ÷

4 ø

à

à

Ü

ù

lnù

1 úê

1Þ ê ü ü

(6.19)

But

à

ê

à

Ü

Ý

1

2 ê

2å

2

ò 2Ü

3Ý

1

ê

å

2

ò 2Ü

3 (6.20)




and so

£

£

¤

ln

¤

1 ¥

1 ¬

¥ § §

1 ¬

¥

1 ¥

2

À

Ä 1 ¬

¥

È

2

2¥

(6.21)

and

E

¬ ˆ° ±

0 ³

2 ´

1

½ 1 ¬ ¾

2

¿ 2 À 2

(6.22)

Notice that as

then E

0!

All we have left is to determine the Poynting vector S

S 1

±

0

EÎ

B 1

±

0

±

0 ³

2 ´

½ 1 ¬ ¾

2

¿ 2 À 2

±

0 ³

2 ´

½ 1 ¬ ¾

2

¿ 2 À 2

Ä ˆ°

Î

ˆ

È (6.23)

±

0

Ä 2 ´ È

2

³

2

Ä1 ¬ ¾

2

¿ 2À

2 È

ˆ (6.24)

From this expression we can for example calculate the radiated power per unit length byintegrating over a cylindrical surface C enclosing the wire:

S"

º

C

±

0

2 ´

³

2

Ä 1 ¬ ¾

2

¿ 2À

2 È

(6.25)

We see that an infinite power is transmitted starting at

0 and

0 which travels out

to infinite . So in practice it is impossible to produce this physical setup. This is due to

the quick “turn on”. In the physical world only gradual turn ons are possible. This many

be seen as a consequence of what is known as Gibb’s phenomenom.

In the above we tacitly employed the retarded potential without discussing the possibil-ity of using the advanced potential. Let us see what happens if we apply the advanced

potential to this problem. The only thing that changes from the outset is that the source

time Ê is replaced by the advanced time

·

-

x ¬ x Ê

# instead of the retarded time

¾

-

¬

x ¬ x Ê

# . With

·

as the argument to the step function, the contribution to the

integral comes only from those Ê which satisfy

ˆ

¬

Ê ˆ°

# $ 0

¬

ˆ

¬

Ê ˆ°

# (6.26)

If we assume that

0 Æ (6.27)

then we may write this as




Ê

2 2 $ 2

2 (6.28)

Ê

$ % ª 2

2 ¬

2 (6.29)

or

Ê

¬

ª 2

2 ¬

2Æ

ª 2

2 ¬

2

Ê (6.30)

One can proceed further and calculate the resulting integral. But what is interesting is that

now we see that the relation (6.9) seems to say that we have no information about what

happened before turn on, while the relation (6.27) says we have no information about what

happened after turn on. Physics seems to be conspiring on us in such a way that we cannot

compare the advanced and the retarded potential at the same time!

END OF EXAMPLE 6.1 &




'

MULTIPLE HALF-WAVE ANTEN NAEXAMPLE 6.2

A wire antenna with a length of (Ä

) 1 È half wavelengths is assumed to have a current dis-

tribution in the form of a standing wave with current nodes at the its endpoints. Determine

the angular distribution of the radiated electromagnetic power from the antenna.

Hint: It can me convenient to treat even(

and odd(

separately.

Solution

One realises that the setup in this problem is an antenna, since we have a monochromatic

source and the current is an extended one-dimensional distribution. Thus, we may use the

“antenna formulae,” but since only the radiated effect is ask for all we need is

3

S4

5

2 sin2 7

±

0

32 ´

2 2

8 2 ˆ (6.31)

where

8

µ9

9

A B

Ä

Ê

ÈC

E F H P cos R S

Ê (6.32)

so we seek a form for the current distributionA B

Ä Ê È

.

It simplifies matters if we consider the cases for(

an even and(

an odd multiple of

half wavelengths separately. As shown in Figure 6.3, for the case of ( even, we will use

sin Ä5 Ê È

and for(

odd we use cos Ä5

Ê È

.

Ê

a) even

sin Ä5 Ê È b) odd

cos Ä5

Ê È

¬V V

Ê

¬V V

Figure 6.3. Depending on whether the length of the wire is an even (as seen in

a) or odd (as seen in b) multiple of half the wave length, the current distribution

is sin Ä5

Ê È

or cos Ä5

Ê È

.

We perform both integrations over Ê

from ¬V to V which is the total length

W, which is

a multiple(

of X

# 2. These facts give us a relation between the integration limits and5

,

namely




W

2 V

( X

# 2

V

´

2

(

5 Y

(6.33)

Let us consider the odd case. The current distribution is written

A B

Ä

Ê

È

³ 0 cos5

Ê (6.34)

and so

8

µ9

9

³ 0 cos 5

Ê

C

E F H P cos R S

Ê

a c

5

Ê

S

Ê

1 #

5

S

c d

³ 0

25

µf h p

2

f h p

2 r

C

E s

C

E s t

C

E s cosR S

c

³ 0

25

aC

E s v 1 cosR x

y

Ä 1 ¬ cos 7

È

¬

C

E s v 1 cosR x

y

Ä 1 ¬ cos 7

È

d

f h p

2

f h p

2

³ 0

5sin2 7

sin(

´

2Ä 1 ¬ cos 7

È

sin(

´

2Ä 1 cos 7

È

cos 7 sin´

2Ä 1 ¬ cos 7

È

¬ sin´

2Ä 1 cos 7

È

³ 0

5sin2 7

2sin Ä(

´

# 2 È cos Ä(

´

# 2cos 7

È

2cos 7 cos Ä(

´

# 2 È sin Ä(

´

# 2cos 7

È

2³ 0

5sin2 7

cos Ä(

´

# 2cos 7

È (6.35)

Inserting this into (6.32) gives us

3

S4

³

20 ±

0 cos2

Ä(

´

# 2cos 7

È

8 ´

2 2 sin2 7

ˆ (6.36)

Let us consider the case when ( is even so the current distribution is written

A B

Ä

Ê

È

³ 0 sin 5

Ê

Y

(6.37)

We remember the relation (6.33) which is still valid but now ( is an even number. So we

have that

8

³ 0µ

2

2

C

E F H P cosR sin 5

Ê

S

Ê

a c

5

Ê

S

Ê

1s

S

cd

³ 0

2y

5 µf h p

2

f h p

2

C

E s cos R

r

C

E s

¬

C

E st

S

c

³ 0

2y

5 µ f h p

2

f h p

2

C

E s v 1 cos R x

¬

C

E s v 1 cos R xS

c

³ 0

2

y

5

a C

E v 1 cos R x

2 ¬

C

E v 1 cos R x

2

y

Ä

1¬

cos7

È

¬

C

E v 1 cos R x

2 ¬

C

E v 1 cos R x

2

y

Ä

1

cos7

È

d

f h p

2

f h p

2




³ 0y

5 sin2 7

Ä 1 cos 7

È sin

Ä 1 ¬ cos 7

È

(´

2

¬

Ä 1 ¬ cos 7

È sin Ä

1 cos 7

È

(´

2

³ 0y

5sin2 7

sin Ä 1 ¬ cos 7

È

(´

2

¬ sin Ä 1 cos 7

È

(´

2

cos 7 sin Ä 1 ¬ cos 7

È

(´

2

sin Ä 1 cos 7

È

(´

2

2³

0y

5sin2 7

cos

(´

2 sin

(´

2cos

7

cos7

sin

(´

2 cos

(´

2cos

7

¬

y2

³ 0

5 sin2 7

sin

(´

2cos 7

(6.38)

8

2 4³

20

5

2 sin4 7

sin2

(´

2cos 7

(6.39)

3

S4

³

20 ±

0

8 ´

2 2

sin2

r

f h

2cos 7

t

sin2 7

ˆ (6.40)

END OF EXAMPLE 6.2&




'

TRAVELLING WAVE ANTENNA EXAMPLE 6.3

A wire antenna of length

is fed at one of its endpoints by a transmitter signal and is at

its other end terminated with a resistance to ground. The termination is adjusted such that

no current is reflected back on the wire. This means that the current distribution comprises

travelling waves emanating from the feed point so one can assume thatA

Ä Ê È

³

exp Ä

y

5

Ê È

along the wire. Determine the angular distribution of the electromagnetic radiation from

this antenna.

Solution

We need the formula

3

S4

5

2 sin2 7

32 ´

2 2

±

0j

0

8

2 ˆ (6.41)

where

8

µk p

2

k p

2º

Ê

C

E F cos R

H P

A

Ä

Ê

È (6.42)

In this case the distribution is given:A

Ä Ê È

³ 0 C

E F H P

. If we insert this distribution into

(6.41), we get

8

³ 0µ

k p

2

k p

2º

Ê

C

E F v 1 cos R x

H P

³ 0 m

C

E F v 1 cos R x

H P

y

5Ä 1 ¬ cos 7

Èn

k p

2

k p

2

a

sin o

1

2y

r

C

E

¬

C

E t

d

³ 0

2

5 Ä 1 ¬ cos 7

È

sin

¤

5 W

2Ä 1 ¬ cos 7

È

§

(6.43)

8 2 4

³

20

5

2Ä 1 ¬ cos 7

È

2sin2

¤

5 W

2Ä 1 ¬ cos 7

È

§

(6.44)

Finally we have

3

S4

³

20 sin2 7

8 ´

2 2

±

0j

0

sin2

r

F

k

2Ä 1 ¬ cos 7

È

t

Ä 1 ¬ cos 7

È

2ˆ (6.45)





'

MICROWAVE LINK DESIGNEXAMPLE 6.4

Microwave links are based on direct waves, i.e., propagation along the line of sight between

the transmitter and receiver antennas. Reflections from the ground or a water surface may

in unfavourable cases distort the transmission. Study this phenomenon using the following

simple model:

The transmitter antenna T is a horizontal half-wave dipole placed a distance 1 above the

ground level. The receiving antenna R is in the main lobe from T, at a horizontal distance

from T, and at height 2. The signal at R is considered to be composed of the direct

wave and a wave reflected from the ground. The reflection is assumed to cause a phase

shift ´ in the wave, but no loss of power. The ground is considered flat over the distance

, and 1 Æ

2

.

1. Calculate the electric field E (magnitude and direction) in at R if the transmitter

antenna is fed with a currentA

.

2. Discuss the meaning of the result.

3. Simplify the result for the case 1 2

X

.

z

z

z

|

|

|

|

|

|

|

|

|

|

|

|

|

|

|

|

|

|

|

|

2

~

~

~

~

~

~

~

z

R

D

2

1

1

T

1

Solution

The Fourier transform of the E field in the far zone (radiation field) from a half-wave dipole

antenna, i.e., a linearly extended current distribution with lengthX

# 2 in the direction,

EB

rad Ä r È

y

4 ´ 0

C

E F

¾

ˆ

Î

µ p

4

p

4

S

Ê

C

E kx P

jB

Î

k

Since k

5ˆ and we study the radiation in the maximum direction, i.e., perpendicular to

j

B

Ä Ê È

³ cosÄ

5

Ê È

ˆ°

so that k"

xÊ

0, this expressions simplifies to




EB

rad Ä r È

y

5

4 ´ 0

C

E F

¾

µ p

4

p

4

S

Ê

³

cos Ä5

Ê

È ˆ

Î

Ä ˆ°

Î

ˆ

È

y

5

4 ´ 0

C

E F

¾

³

5

sin Ä5

Ê

È

p

4

p

4

2

ˆ°

y

³

2 ´ 0

C

E F

¾

ˆ°

A superposition of the direct and reflected contributions (with the distance from the trans-

mitter

to the receiver

equal to

1 and

2, respectively), with due regard to the phase

shift ´ (corresponding to a change of sign in the current), gives the Fourier transform of

the total E field at the far zone point

:

EtotB

rad Ä r È

EdirB

rad Ä r È

EreflB

rad Ä r È

y

³

2 ´ 0

¤

C

E F

¾ 1

1

¬

C

E F

¾ 2

2§

ˆ°

(6.46)

Since 1 Æ 2

, Pythagoras’ theorem gives

1

ª

2

Ä 1

¬

2 È

2

Ä 1

¬

2 È

2

2

21

22

2

¬

1 2

¬

∆

2

2

ª

2

Ä 1

2 È 2

Ä 1

2 È

2

2

2

1

2

22

1 2

∆

2

where

21

22

2

(6.47)

and

∆

2 1 2 (6.48)

is the difference in path distance. Insertion of (6.47) and (6.48) into (6.46), with

5

#

2 ´

#

X, gives the Fourier transform of the field at

EtotB

rad Ä r È

y

³

2 ´ 0

C

E F

¾

C

E F ∆¾

p

2¬

C

E F ∆¾

p

2 ˆ

°

³

´ 0

C

E F

¾

sin 2 ´ 1 2

X

ˆ°

from which we obtain the physical E field

Etotrad Ä r Æ

È

Re EtotB

rad C

E

B

À

³

´ 0

sin2 ´

1 2

X

cosa 2 ´

X

¤

21

22

2

§

¬

d

ˆ°

We see that the received signal at will be extinct if

2 ´ 1 2

X

´




56



LESSON 7

Multipole Radiation

7.1 Coverage

We look at electric dipole, magnetic dipole and electric quadrupole radiation. Mul-

tipole radiation analysis is important since it simplifies the calculation of radiation

fields from time-varying field and since EM multipoles exist in many fields of

physics such as astrophysics, plasma physics, atomic physics, and nuclear physics.

7.2 Formulae used

Fields in far regions from an electric dipole

B rad

x

0

4b

x

x p

1

k

E rad

x

1

4 0

x

x

p 1

k

k

Fields in the far zone from a magnetic dipole

B rad

x

0

4ª

x

x

m

k

k

E rad

x ¬

4 0

-

x

x m

k

Field in the far zone from an electric quadrupole

B rad ®

0

8°

x

x

k ± Q

k

E rad

®

8

0

x

x

²

k ± Q

k

k

57



58 LESSON 7. MULTIPOLE RADIATION

7.3 Solved examples

'

ROTATING ELECTRIC DIPOLEEXAMPLE 7.1

An electric dipole with constant electric dipole moment magnitude is located at a point in

theµ ¶

plane and rotates with constant angular frequency.

(a) Determine the time-dependent electromagnetic fields at large distances from the di-

pole.

(b) Determine the radiated average power angular distribution and the total radiated

power.

Solution

(a) We can write the time-varying dipole momentum relative the location of the dipoleas

p Ä È

·

0 Ä cos

ˆ¹

sin

ˆº

È (7.1)

which represents a constant dipole moment·

0 times unit vector rotating with angular fre-

quency

. This can also be rewritten in complex form

p Ä È

Re

·

0 C

E

B

À

ˆ¹

y ·

0 C

E

B

À

ˆº

(7.2)

or

p Ä È

Ä

·

0 C

E

B

À

ˆ¹

y ·

0 C

E

B

À

ˆº

È

cY

cY

(7.3)

where c.c., stands for the complex conjugate of the term opposite the sign. In what

follows we use the convention that we write the dipole expression as a complex quantity

but we drop the c.c. term, which is commonplace when discussing harmonic oscillation. It

is easy to identify the Fourier component of the dipole moment in this case

pB

Ä È

·

0 ˆ¹

y ·

0 ˆº

Y

(7.4)

We notice, in this complex space variable space, that the ˆº component has the phase factory

C

E

h p

2 relative the ˆ¹ component, which is due the circular rotation.

We would like to express (7.4) in spherical components rather than Cartesian components

since the expressions for dipole fields in spherical components are simpler, so we transform

the base vectors ˆ¹

and ˆº

in the conventional fashion and get




pB

Ä È

·

0 ÄÄsin 7 cos

»ˆ

cos 7 cos»

ˆ¼

¬ sin»

ˆ

È (7.5)

y

Ä sin 7 sin»

ˆ

cos 7 sin»

ˆ¼

cos»

ˆ

ÈÈ

(7.6)

·

0 Ä ˆ sin 7

Ä cos »

y

sin »È

(7.7) ˆ¼

cos 7

Ä cos »

y

sin » È

(7.8) ˆ

Ä

¬ sin 7

y

cos »È

È

½

cos »

y

sin »

C

E ¿ À

(7.9)

·

0 C

E ¿

Ä ˆ

sin7

ˆ¼

cos7

y

ˆ

È (7.10)

Now that we have the Fourier component of the dipole moment expressed in spherical

components we insert this into the dipole radiation fomulae:

BB

¬

±

0

4 ´

C Á

F

¾

pB

Î

k (7.11)

EB

¬

1

4 ´

j

0

C Á

F

¾

Ä pB Î

k È

Î

k (7.12)

First we calculate

pÎ

k

·

0 5 C

Á

¿ Â

Â

Â

Â

Â

Â

ˆ

ˆ¼

ˆ

sin 7 cos 7 Ä

1 0 0

Â

Â

Â

Â

Â

Â

(7.13)

·

0 5 C

Á

¿

Ä

Ä ˆ¼

¬ cos 7 ˆ

È (7.14)

from which we get

Ä pÎ

k È

Î

k

·

0 5 C

Á

¿Â

Â

Â

Â

Â

Â

ˆ ˆ¼

ˆ

0 Ä

¬ cos 7

1 0 0

Â

Â

Â

Â

Â

Â

(7.15)

·

0 5

2C

Á

¿

Ä

¬ cos 7 ˆ¼

¬

Ä ˆ

È (7.16)

so finally we can write the field in space and time coordinates (remember:B Ä Æ x È

Re Æ

B

Ä x ÈC

E

B

À

),

B Ä Æ x È

¬

±

0

4 ´

Re Ç

CÁ

v F

¾

B

À

x

·

0 5 C

Á

¿

Ä

Ä ˆ¼

¬ cos 7

»È

È

(7.17)

¬

±

0

·

0 5

4 ´

Re

C

Á

v F

¾

B

À

¿

x

C

Á

¿

Ä

Ä ˆ¼

¬ cos 7

» È (7.18)

So our final expression is

B

±

0

2 ·

0

4 ´

Ä sin Ä5

¬

» È

ˆ¼

cos 7 cos Ä5

¬

» È ˆ

È (7.19)

E

2 ·

0

4 ´

j

0

2

Ä

cos

7

cosÄ

5

¬

»È

ˆ¼

¬

sinÄ

5

¬

»È

ˆ

È

(7.20)




(b) We use the formula3

S4

12 Ê 0

EÎ

BË

where the E and B fields are complex and

monochromatic so

3

S4

1

2±

0

EÎ

B Ë

1

2±

0

±

0

4 ´

1

4 ´

j

0

1 2

ÄÄp

Î

k È

Î

k È

Î

Ä p Ë

Î

k È

(7.21)

32 ´

2 j

0 2

Ä k Ä pÎ

k È" Ä

p Ë

Î

k È

¬

Ä pÎ

k È k" Ä

pÎ

k ÈÈ

(7.22)

32 ´

2 j

0 2

pÎ

k 2 ˆ

(7.23)

±

0

· 20

4

32 ´

2 2Ä 1 cos2 7

È ˆ (7.24)

The total power is then

µ Ω

3 Í

¾

4

2

º

Ω

(7.25)

±

0

· 20

4

32 ´ 2

2 ´

µh0

S

7 sin

Ä 1 cos2 7

È

(7.26)

±

0

· 20

4

32 ´

2

2 ´

µ

1

1º

µ Ä 1

µ

2È

(7.27)

±

0

· 20

4

6 ´

(7.28)

END OF EXAMPLE 7.1&

'

ROTATING MULTIPOLEEXAMPLE 7.2

Two point charges of equal chargeÏ

are located in theµ ¶

plane at either end of the diameter

of a circle of radius®

. The particles rotate with a constant angular speed

in the plane of the circle.

Determine

(a) The Fourier components of p1, m, and Q

and

(b) The radiation diagram when

®

Solution




r Ê1

®

Ä cos

0 ˆ¹

sin

0 ˆº

È (7.29)

r Ê2

¬ r Ê1 (7.30)

v Ê1

0®

Ä

¬ sin

ˆ¹

cos

0 ˆº

È (7.31)

v Ê2

¬ v Ê1 (7.32)

p

Ñ Ò

Ï

Ò

rÊ

Ò

0 (7.33)

m 1

2

Ñ Ò

r Ê

Ò

Î

Ï v Ê

Ò

1

2Ä r Ê1

Î

Ï v Ê1

Ä

¬ r Ê1 È

Î

ÏÄ

¬ v Ê1 ÈÈ

Ï

0®

2 ˆ°

(7.34)

mB

ˆ°

Ï

0®

2 1

2´ µ Ó

Ó

C

E

B

À

º

Ï

0®

2 ˆ°

Ô

Ä

È (7.35)

Ö

Á ×

Ñ Ò

Ï

Ò

µ

Ê

Á

Ò

µ

Ê

×

Ò

(7.36)

Ö Ù Ù

ÏÄ

µ 21

µ 22 È

2 Ï

®

2 cos2

0

Ï

®

2 Ä 1

cos 2

0 È (7.37)

Ö Û Û

2Ï

®

2 sin2

0

Ï

®

2Ä 1 ¬ cos2

0 È

(7.38)Ö Ù Û

Ö Û Ù

Ï

®

2 sin2

0 (7.39)Ö

Á

H

Ö

H

Á

0 (7.40)

Ö Ù Ù

Ï

®

2Ä 1

C

Á

2

B

0À

È (7.41)Ö Û Û

Ï

®

2Ä 1 ¬

C

Á

2

B

0À

È (7.42)Ö Ù Û

Ö Û Ù

Ï

®

2 Ä

C

Á

2

B

0À

(7.43)

Fourier transform

Ö Ù Ù

B 1

2´ µ Ó

Ó

Ï

®

2Ä 1 cos 2

0 È C

E

B

À

º

(7.44)

Ï

®

2Ä

Ô

Ä

È

1 # 2 Ô

Ä

2

0 È

1 # 2 Ô

Ä

¬ 2

0 ÈÈ

(7.45)

Ö Û Û

B

Ï

®

2 1

2´ µ Ó

Ó

Ä 1 ¬ 1 # 2 C

Á

2

B

0À

¬ 1 # 2 C

Á

2

B

0À

ÈC

E

B

À

º

(7.46)

Ï

®

2 1

2 ´

Ä

Ô

Ä

È

¬

Ô

Ä

¬ 2

0 ÈÈ (7.47)

Ö Ù Û

B

Ï

®

2 1

2´ µ Ó

Ó

C

Á

v 2

B

0

B

x

À

º

Ï

®

2 y

Ô

Ä

¬ 2

0 È (7.48)




Ö

B

Ï

®

2 Ô

Ä

È»¼

1 0 0

0 1 0

0 0 0ÁÂ

Ï

®

2 Ô

Ä

¬ 2

0 È»¼

1 0 0

0 1 0

0 0 0ÁÂ

(7.49)

B

y

±

0

8´

C Á

F

¾

Ä k"Q È

Î

k (7.50)

E

Ä

8 ´

j

0

CÁ

F

¾

ÄÄ k " Q È

Î

k È

Î

k (7.51)

3 Í

4

1

2±

0

Ä EÎ

B Ë

È (7.52)


'

ATOMIC RADIATIONEXAMPLE 7.3

A transition in an atom is described as quantum matrix element of a radiation operator

between the un-normalised eigenstates

Ç

Ψ

C

¾ exp Ä

¬

y ß

1à

È

cos 7

Ψ á

C

¾ exp Ä

¬

y ß

0à

È

(7.53)

At a certain moment, the atom is therefore described by

Ψ

Ψ

áΨ

á(7.54)

where

f and

e can be viewed as given constants, chosen such that Ψ becomes normalised.

According to semiclassical theory, one can interpret the magnitude squared of the wave

function as a particle density function. Determine, according to this semiclassical inter-

pretation, the power emitted by the atom via the dipole radiation which appears due to the

transition between the two states.

The power from an electric dipole is given by

±

0

4 p 2

12

(7.55)

Solution

The charge density is

É

Ï ΨΨ Ë

Ï

r

Ψ

2

Ë

á

Ψ Ë

á

Ψ

Ë

á Ψ á Ψ Ë

á Ψ á

2 t

(7.56)

Ï

r

2 2 exp Ä

¬ 2

È cos2 7

á

2 exp Ä

¬ 2

È

(7.57)

½

Ë

á expÄ

¬

Ä

Äâ

1¬

â

0È È

Ë

á

expÄ

Ä

Äâ

1¬

â

0È È

À

cos7

expÄ

¬

2

È

t

(7.58)




Only the two last terms contribute because they are non-static. Via inspection we find the

Fourier components for

Äâ 1

¬

â 0 È

# ä

É

B

Ä

Æ

7

È

Ï

Ë

á

cos 7 exp Ä

¬ 2

È (7.59)

Consequently, the corresponding Fourier component of the dipole moment p

å

É

B

r S 3µ

is

p

µ

r

Ï

Ë

á

cos 7 exp Ä

¬ 2

È

ˆ

t

2

º

º

Ω

(7.60)

Ï

Ë

á

¤

µ

4 exp Ä

¬ 2

È

S

§

µ

2

h0 µh0

cos 7 sin 7 ˆ

S

7

S

Ã (7.61)

Ï

Ë

á

1

24!

µ

2

h0 µh0

cos 7 sin 7

Ä sin 7 cos ÃÆ sin 7 sin Ã

Æ cos 7

È

S

7

S

Ã

(7.62)

12 Ï

Ë

áÄ 0 Æ 0 Æ

µ

2

h0 µh0

cos2 7 sin 7

S

7

S

ÃÈ

(7.63)

12Ï

Ë

áÄ 0 Æ 0 Æ 2 ´

µh0

1 # 2 Ä sin 7 sin 7 cos2 7

È

S

7

È

(7.64)

12 ´Ï

Ë

áÄ 0 Æ 0 Æ 1 1 # 2

µh0

Ä sin3 7

¬ sin 7

È

S

7

È

(7.65)

12 ´Ï

Ë

áÄ 0 Æ 0 Æ 1 1 # 2 Ä 1 # 3 ¬ 1 È

È

(7.66)

8 ´Ï

Ë

áÄ 0 Æ 0 Æ 1 È (7.67)

8 ´Ï

Ë

áˆ° (7.68)

and a similar integral for the complex conjugate term. The power from the electric dipole

radiation due to the transition between the two states is given by

±

0

4 p 2

12 ´

16 ´Ï

2±

0 Äâ 1

¬

â 0 È

4

3 ä 4

Ä

Ë

á È

2

Y

(7.69)

END OF EXAMPLE 7.3&

'

CLASSICAL POSITRONIUM EXAMPLE 7.4

Calculate the radiation from a positron-electron head-on collision and subsequent annihil-

ation, classically by assuming that the particles travel at a constant velocity ç 0

up until

the time the annihilate.

Solution

Background Consider a system of localised charges in motion. If we assume that we

are observing at a distance x

much greater than extension x Ê

of the charge system

and further thatç

, it can be argued that the source time Ê

is approximately

¬

#

(where

x

) instead of

¬

x¬

xÊ

#

. This is because, due to the first assumption




x ¬ x Ê

¬ x Ê

"

x

(7.70)

Since the timescale for the system is of the order è

x Ê

#

çand since

ç

we have that

x Ê

(7.71)

and this allows us to write

Ê

¬

# (7.72)

The vector potential is this case

A

±

0

4 ´ µ

j Ä

¬

#

È

S 3µ

(7.73)

since the denominator in the integrand is now not dependent on the source coordinates.

Substituting j

É v, we rewrite the vector potential as

A

±

0

4´

Ä Σ Ïv È (7.74)

Observe that the summation can be written as

ΣÏ

v

º

º

ΣÏ

x Ê

p (7.75)

where p is the electric dipole. Thus,

A

±

0

4 ´

p (7.76)

Deriving the EM field in the usual manner we get

E 1

4´

0

2Ä

Ä pÎ

R

È

Î

R

È (7.77)

B

±

0

4 ´

2Ä p

Î

RÈ (7.78)

It can be shown that the angular and spectral distribution of energy is

dê

d

dΩ

Ä Ä

p È

BÎ

R #

È

2

2 ´

3(7.79)

The Calculation The dipole electric moment of the positronium is

p

Ç

2Ï á ç 0 ˆ°

0

ë

0

$

0(7.80)




We take the second time derivative of this

£

p£

2Ï á ç 0

7

Ä

¬

È

ˆ° (7.81)

£

2p£

2

¬ 2 Ï á ç 0Ô

Ä È

ˆ°

(7.82)

For the spectral and angular distributions of the radiation we have

d ê

d

dΩ

Ä p

BÎ

R #

È

2

2 ´

3(7.83)

where subscript

denotes the Fourier transform of the dipole. Now the Fourier transform

of the dipole is simply pB

¬ 2 Ï ç 0 ˆ°

#

Ä 2 ´ È

1

p

2, so that

d ê

d

dΩ

Ï

2á

ç

20 Ä ˆ°

Î

R È

2#

´

2 3 (7.84)

In the final result we notice that there is no dependence on

so the spectral density is white

noise.




66



LESSON 8

Radiation fromMoving Point

Charges

8.1 Coverage

In our series on deriving fields from given sources we have come to the most fun-

damental case: the moving point charge. The fields are derived from the Liénard-

Wiechert potentials. In what follows we will assume that the motion xì

í

ì

is known

“in advance”. We find that accelerating charges radiate. We also look other mech-

anisms for a point charge to radiate such as Cerenkov emission.

8.2 Formulae used

According to the Formulae (F.23–26), the fields from a charge in arbitrary motion

are given by

E í î

x ï

4 ð 0 ò

3ó

Rô õ

1 ö

2

-

2ø ù

x

x ì

Rô

v-

2ú

(8.1)

B í î

x

x

x ì

E í î

x

-

x

xì

(8.2)

ò

x

x ì

x

x ì

±

v

-

(8.3)

Rô

x x ì

ü

ü

x x ì

ü

ü

v

-

(8.4)

õ ý

í

ì

ý

í

ø

x

x

xì

ò

(8.5)

67



68 LESSON 8. RADIATION FROM MOVING POINT CHARGES

Alternative formulae

x x ì

r (8.6)

v

-

þ

î

þ

ö

-

(8.7)

Rô

x

x0 r

þ r

ô r0 (8.8)

ò

þ ± r

cos¡

1

cos¡

(8.9)

õ ý

í

ý

í

ì

ø

x

ò

1

cos¡

(8.10)

8.3 Solved examples

'

POYNTING VECTOR FROM A CHARGE IN UNIFORM MOTIONEXAMPLE 8.1

Determine the Poynting vector for the fields from a chargeÏ

which moves with constant

velocity v. Show that no power is emitted from the charge during the motion.

Solution

In general the fields due to a single point charge may be written as

E

E¢

Erad (8.11)

B

B ¢

Brad (8.12)

E¢

and B¢

are known as the velocity fields and Erad and Brad are known as accelerationfields. These can be derived from the Liénard-Wiechert potentials and result in

E ¢

a Ï

4 ´ 0 £

3Ä r ¬

¤

È Ä 1 ¬¥

2È

d

(8.13)

Erad

a Ï

4 ´ 0

£

3r

rÎ

½

Ä r ¬

¤

È

Î ˙¤

À t

d

(8.14)

and

B¢

a Ï

4 ´ 0

£

3

¤

Î

rr

1 ¬¥

2 t

d

(8.15)

Brad

a Ï

4 ´ 0

2£

3ˆ

Î

r

Î ½

Ä r ¬

¤

È

Î ˙¤

À

d

(8.16)




where ¤

- v # and

£

¬

Ä

¤

"r È (8.17)

The big outer square brackets

indicate that one should evaluate their content at the

retarded time Ê

¬

x ¬ x ÊÄ Ê È

# , where x Ê

Ä Ê È is the given motion of the charged

particle. It is this that makes the equations for the fields difficult to evaluate in general. Itmay not be difficult to get an expression for x

Ê Ä Ê È , but then to solve the equation for the

retarded time Ê

¬

x ¬ x ÊÄ Ê È

# for

Ê

Ê Ä Æ

x È . In this case it is not necessary to

perform this transformation of variables since we are not interested in the time evolution,

so we drop the brackets.

It is easy to verify that for uniform motion of the charge Ï , or in other words ˙¤

- 0, that

Erad

Brad- 0 and that

B 1

¤

Î

E (8.18)

so the Poynting vector in this case is

S

1±

0

EÎ

Ä

¤

Î

E È

0

±

0E

Î

Ä

¤

Î

E È (8.19)

Furthermore, it can be shown that

£

¬

Ä

¤

"r È

½

20

¬

Ä r0

Î

¤

È

2 (8.20)

where we have introduced the virtual position vector r0- r ¬

¤ . With these last relations

we may write

E

Ï

4 ´ 0

Ä 1 ¬¥

2È

1 3

0 Ä 1 ¬¥ 2 sin2 7

È

3

p

2r0 (8.21)

where

7

is the angle between r0 and

¤

. Inserting this relation for E into the relation for Swe obtain

S

0

±

0

Ï

2

16 ´

2

20

Ä 1 ¬¥

2È

2 1 6

0 Ä 1 ¬¥ 2 sin2 7

È

3r0

Î

Ä

¤

Î

r0 È (8.22)

So

¤

Î

r0

¥

0 sin 7 ˆ (8.23)

and

r0

Î

Ä

¤

Î

r0 È

¬¥

20 sin 7 ˆ

¼

(8.24)

So finally




S

¬

Ï

2

16 ´

2 0

çÄ 1 ¬

¥

2È

2 sin 7

40 Ä 1 ¬

¥ 2 sin2 7

È

3 ˆ

¼

(8.25)

And now when we integrate the Poynting vector over a spherical surface A with radius

which encloses the moving charge which results in

S "

S

A

Í

Ä ˆ¼

" ˆ

È

º

Ω

0 (8.26)

since the Poynting vector is not radial, so a charge in uniform motion in vacuo, does not

radiate energy.


'

SYNCHROTRON RADIATION PERPENDICULAR TO THE ACCELERATIONEXAMPLE 8.2

Determine the angular distribution of synchrotron radiation in the plane perpendicular to

the acceleration v for a charged particle moving with velocity v.

Solution

We consider only the formulas for the radiation fields, for which the denominator is the

cube of the retarded relative distance

£

¬ r "

¤

¬

¥ cos 7

Ä 1 ¬¥ cos 7

È (8.27)

Now, we have that v"

r

0, so

rÎ

Ä r ¢

Î

v È

r ¢ Ä r " v È

¦ 0

¬ v Ä r " r ¢ È

¬ vr

" Är ¬

¤

È

(8.28a)

¬ v Ä

2¬

2¥ cos 7

È (8.28b)

¬ v 2Ä 1 ¬

¥ cos 7

È (8.28c)

¬ v

£

(8.28d)

where 7 is the angle between the velocity and r. So that

Erad

¬

±

0 Ï

4 ´

£

3v

£

¬

±

0 Ï

4 ´

v

£

2(8.29)

The Poynting vector is given by

S 1

±

0

E

2 ˆ

±

20 Ï

2

16 ´ 2±

0

˙ç

2 ˆ

2 Ä 1¬

¥

cos7

È 4(8.30)




Remember that the Poynting vector represents energy flux per unit time at the field point

at the time . The charge’s energy loss must be related to the time Ê, i.e., the time when

the energy was emitted!

£ §

£

Í

¾

2 (8.31)

but£ §

£

Ê

£

£

Ê

£ §

£

Ä 1 ¬¥ cos 7

È

±

0 Ï

2

16 ´

2

˙ç

2

2Ä 1 ¬

¥ cos 7

È

4

2

±

0 Ï

2 ˙ç

2

16 ´

2

1

Ä 1 ¬¥ cos 7

È

3(8.32)

END OF EXAMPLE 8.2&

'

THE LARMOR FORMULA EXAMPLE 8.3

Derive the Larmor formula by calculating the radiated power of an accelerating point

charge due to electric dipole emission. Apply the Larmor formula to linear harmonic mo-

tion and circular motion.

Solution

The Larmor formula is a very useful equation for deriving the power of emission from

non-relativistic accelerating charged particles. It can be derived from the radiation fields

of a non-relativistic ( ç

) charged particle, but it can also be seen as an electric dipole

relative a co-moving coordinate system. We shall investigate the latter.

First derive the power emitted by an electric dipole. Using a time domain (non-Fourier)

version of the dipole fields:

E 1

4 ´

0

2Ä p

Î

ˆ

È

Î

ˆ (8.33)

B

±

0

4 ´

Ä pÎ

ˆ

È (8.34)

we find the Poynting vector to be

S 1

±

0

EÎ

B

1

16 ´

2 0

2 3

Ä pÎ

ˆ

È

Î

ˆ

Î

Ä pÎ

ˆ

È

1

16 ´

2 0

2 3

p

Î

ˆ

2 ˆ

¬ ˆ

" Ä pÎ

ˆ

È Ä pÎ

ˆ

È

1

16 ´

2 0

2

3

p

Î

ˆ

2

ˆ

(8.35)




So integrating for the power

Ä È

Ä È

µh0 µ

2

h0

S"

ˆ

2 sin 7

S

Ã

S

7

1

16´

2

0

3µ

h

0µ

2

h

0

pÎ

ˆ

2sin 7

S

Ã

S

7

1

16 ´

2 0

3µ

h0 µ

2

h0

p

2sin3 7

S

Ã

S

7

p

2

8 ´ 0

3µ

h0

sin37

S

7

p

2

8 ´ 0

3

4

3

p

2

6 ´ 0

3(8.36)

p Ä È

Ï x Ä È (8.37)

Ä È

Ï

2 x

2

6 ´ 0

3

Ï

2 a

2

6 ´ 0

3(8.38)

where we have identified the acceleration a Ä È

- x Ä È .

Linear harmonic motion

µ Ä È

µ 0 cos Ä

0 È Æ

®

Ä È

¨µ Ä È

¬

20 µ 0 cos Ä

0 È

Y

(8.39)

Ä È

Ï

2 4

0µ

2

0 cos

2Ä

0

È

6 ´ 0

3 (8.40)

3

4

Ï

2 40 µ

20

12 ´ 0

3(8.41)

Circular motion

µ Ä È

0 cos Ä

0 È Æ ¶

Ä È

0 sin Ä

0 È Æ (8.42)

Ä È

Ï

2 40

20

6 ´ 0

3 (8.43)




Validity of the Larmor formula The Larmor formula although not covariant in form

can indeed be extended such as to be valid for all inertial frames.

One point that should definitely be raised is power radiated for more than one accelerating

charge. It is not so simple that one may assume that power is proportional to the number of

sources © . What must be understood is whether or not the sources are radiating coherently

or incoherently. For example, consider the above case of circular motion. If we had a

large collection of particles such as the case with in a particle storage ring or in circular

wire, the radiation is not automatically proportional to©

. If they are bunched the power is

proportional to©

2, this is coherent radiation. If the charges are distributed homogeneously

the radiated power is 0. And if the charges are distributed evenly but fluctuate thermally

then power is proportional to©

, this is incoherent radiation.


'

VAVILOV-CERENKOV EMISSION EXAMPLE 8.4

Show that the potentials at time at a point inside the Vavilov-Cerenkov cone receive

contributions from exactly two positions of the charged particle.

Solution

The motion of the charge is given by µ Ê Ä Ê È

çÄ

Ê

¬

È so that

2

Äµ

¬

µ

Ê

È

2

¶

2

µ

¬

çÄ

Ê

¬

È

2

¶

2

µ

2

ç

2Ä

¬

Ê

È

2 2 ç µ Ä

¬

Ê

È

¶

2 (8.44)

but

¿

Ò

Ä

¬

Ê

È is the retarded distance so that

2 ¬

2

2Ä

¬

Ê

È

2

0 (8.45)

or, from the expression for 2 above:

µ

2

ç

2Ä

¬

Ê

È

2 2ç µ Ä

¬

Ê

È

¶

2¬

2

2Ä

¬

Ê

È

2

0 (8.46)

¤

ç

2¬

2

2 §

Ä

¬

Ê

È

2 2 ç µ Ä

¬

Ê

È

¶

2

µ

2

0 (8.47)

This is a quadratic equation in Ä

¬

Ê È

. For a fixed we would in other words have two

values of

Ê

. It remains to show that the result is physically reasonable.




Ä

¬

Ê

È

¬ 2ç µ

%

ª 4ç

2µ

2 ¬ 4 Äµ

2

¶

2È

Ä ç

2 ¬

2 #

2È

2 Äç

2 ¬

2 #

2È

¬

ç µ

%

½

¿ 2

Ò

2 Äµ

2

¶

2È

¬

ç

2¶

2

ç

2 ¬

¿ 2

Ò

2

(8.48)

Ä

¬

Ê

È)

2

2Ä

µ

2

¶

2È

$

ç

2¶

2 (8.49)

2

2ç

2

$

¶

2

µ

2

¶

2(8.50)

sin2o c

$ sin2o

(8.51)

o

o c (8.52)

where o c is the critical angle of the Cerenkov radiation; it is half of the opening angle

of the shock wave of the radiation. So that, in other words, Ä µÆ

¶ È is inside the cone! It

remains to be shown that µ

0

Ä

¬

Ê

È

¤

ç

2 ¬

2

2 §

%

2

2Ä

µ

2

¶

2È

¬

ç

2¶

2 ¬

ç µ

%

ç µ

2

2ç

2

Äµ

2

¶

2È

µ

2¬

¶

2

µ

2¬

ç µ

ç µ

%

sin2o c

1

cos2o

¬ tan2o

¬ 1

ç µ

%

Ä 1 ¬ cos2o c È

¬

Ä 1 ¬ cos2o È

cos2o

¬ 1

ç µ

%

1 ¬

cos2o c

cos2o

¬ 1

(8.53)

Ä

¬

Ê

È

¤

ç

2¬

2

2 §

0

ç µ

%

1 ¬

cos2o c

cos2o

¬ 1

0

(8.54)

END OF EXAMPLE 8.4&



LESSON 9

Radiation fromAccelerated Particles

9.1 Coverage

The generation of EM fields via Liénard-Wiechert potentials are considered sim-

ultaneously with the Lorentz force to give a self-consistent treatment of radiation

problems. In the previous lesson we solve radiation problems from given expres-sions for the motion. Now we consider how charges actually move in the EM fields

and thus present the Lorentz force. We also discuss the effect of radiation on the

motion of the radiating body itself known as radiative reaction

9.2 Formulae used

The covariant Lorentz force

v

í

ï

Eù

v

B (9.1)

75



76 LESSON 9. RADIATION FROM ACCELERATED PARTICLES

9.3 Solved examples

'

MOTION OF CHARGED PARTICLES IN HOMOGENEOUS STATIC EM FIELDSEXAMPLE 9.1

Solve the equations of motion for a charged particle in a static homogenous electric field

E and magnetic field B.

Hint : Separate the motion v into motion parallel and perpendicular to the magnetic field

respectively.

Solution

Background on equations of motion for charged particles As we know, the mo-

tion of a charged particle in electric and magnetic fields is given by the Lorentz force

F

ÏÄ E v

Î

B È (9.2)

This can be seen as a definition of the E and B fields and also the fundamental equation

for measuring the fields. The E field gives the force parallel to direction of motion and B

gives the force perpendicular to the direction of motion. In this certain sense the Lorentz

force is trivial: it is simply a definition of the EM fields.

The equation for the Lorentz force is relativistically correct as it stands, as long as one

interpretes

F

º

Ä( v È

º

(9.3)

where(

is the mass of the particle,

- 1 #

ª 1 ¬

ç

2 # 2, v is the three-velocity, and is

the time. The fact that the Lorentz force in this form

º

Ä(

v È

º

ÏÄ E v

Î

B È (9.4)

is Lorentz invariant is not immediately clear but can easily be shown. On the other hand

this equation is difficult to solve for v Ä È

because

contains v.

In many cases one has conditions which are non-relativistic and under such conditions it

is possible to simplify (9.3) and thus also (9.4). One simply uses the fact that as ç

#

0

then 1. In this case F

(

º

v #

º

, (i.e. the Newtonian force definition), so the Lorentz

force becomes

º

v

º

Ï

(

Ä

E

v

Î

BÈ

(9.5)




This equation is the equation of motion for non-relativistic charge particles in EM fields.

The motion As is well known, charged particles are uniformly accelerated in a static

and homogenous electric field, and in a static and homogenous magnetic field the charged

particles perform circular motion. So what happens in a combined electric and magnetic

fields?

The equations of motion are given in (9.5) The first step in solving this equation is to

separate the motion into motion parallel with the B field which we will denote with v!

and

motion perpendicular to B, which we will denote v"

; so v

v!

v"

and thus

º

v!

º

º

v "

º

Ï

(

E!

Ï

(

Ä E"

v"

Î

B È (9.6)

and since the # and $ are mutually orthogonal the equation separates into

º

v!

º

Ï

(

E!

(9.7)

and for the perpendicular direction

º

v"

º

Ï

(

Ä E "

v "

Î

B È (9.8)

º

v "

º

¬

Ï

(

v "

Î

B Ï

(

E " (9.9)

This is a first order linear ordinary differential equation in v"

. It is also inhomogeneous

which means that we have both a solution to the homogenous equation and solution to the

inhomogeneous equation. Let us call the solution to the inhomogeneous equation %

and

the solution to the homogenous equation '

. For the inhomogeneous solution, we notice

that neither the inhomogeneity nor the coefficient of the zeroth order term depends on so

the solution %

itself cannot be time dependent. With this assumption, and that is left is¬

% Î

B

E" . This is easily solved, by taking the cross product of this equation with B

we find

¬

Ä

% Î

B È

Î

B

E"

Î

B (9.10)

¬

Ä

%

" B È

¦ 0

B

Æ

2 %

E "

Î

B (9.11)

% E

"

Î

B

Æ

2(9.12)

The solution to the homogenous equation

º

¾

º

Ï

(

¾

Î

B (9.13)




on the other hand, can be shown to have the form

¾

c" C

E

B

À

where c"

is a constant

vector perpendicular to B and satisfies c" "

y

c"

0 (where the"

denotes scalar product

defined as the inner product of the vectors), so for example c "

Ä 1 Æ

y

Æ 0 È if B is in the 3

direction.

x

y

z

Figure 9.1. Motion of a charge in an electric and a magnetic field. Here the

electric field is along the¶

axis and the magnetic field is along the axis. There

is a background velocity in the axis for reasons of clarity.

Interpretation of the Motion Having derived the solutions for the motion of the

charged particle we are now in a position to describe it in words. First of all we see that the

motion consists of three separate parts. First we have the motion along the B field whichsimply is not effected by the B field, so that part of the E field which is along the B field

accelerates the charge as if there were no B field. The motion perpendicular to the B field

can further be separated into two parts. One part, which here we have denoted

¾

, repres-

ents so called gyro-harmonic rotation, i.e. the particle moves in a circular orbit, with its

axis of rotation parallel with the B field with period of rotation 2 ´

#

¿ where

¿

Ï Æ

#

(

is known as the cyclotron frequency. This gyro-harmonic motion is the complete solution

to our problem if the E field was not considered. The second part of the motion in the

perpendicular plane however, which we here denoted %

involves both the B field and

the E field. It is known as the drift velocity since this is the velocity of the centre of the

gyro-harmonic motion.

The motion is illustrated in Figure 9.1.





'

RADIATIVE REACTION FORCE FROM CONSERVATION OF ENERGY EXAMPLE 9.2

Correct the equations of motion for a charged particle in EM fields to include the effects

of the energy loss due to radiation emitted when the particle accelerates. Assume non-

relativistic conditions, so radiated energy is given by Larmor formula.

Solution

Background Among other things, Maxwell’s introduction of the electric and magnetic

fields was a mathematical technique to divide the work of solving the motion of charge

particles in EM fields. Instead of action-at-a-distance, the fields naturally divide the prob-

lem into two parts: the generation of fields from moving charged particles and the motion

of charged particles in EM fields. In a certain sense this can be seen as a division into, on

the one side a cause and on the other side, an effect of EM interaction, but the formulas

governing these processes are not very symmetric with respect to each other in classical

electrodynamics: the generation of fields from moving point charges is determined via the

Liénard-Wiechert potentials and the motion of point charged particles is determined by the

Lorentz force.

Despite the lack of symmetry this division of seems successful at describing EM interactionexcept for one thing: it fails to describe the self-interaction of charged point particles . We

know that motion of charged particles is governed by the Lorentz force but at the same

time we know that acceleration of charged particles causes energy emission. These two

facts have been until now treated separately, but taken together we realize something is

missing. This can be seen for example in the case of a single charge under the influence

of a mechanical force in a region of space with no EM fields except for field from charge

itself. From the Liénard-Wiechert potentials we know that the mechanical force will cause

the charge to radiate and if energy is to be conserved the emission must take its energy from

the kinetic energy but since the Lorentz force is zero and there are no other electromagnetic

as of yet that we know of, we have no way of accounting for this radiative “friction.” This,

as of yet not mentioned, force is known as radiative reaction or radiative damping force.

One question that comes to mind after the above discussion is how can so many problems

be described by classical electrodynamics without considering radiative reaction? Obvi-

ously it should have a negligible effect in most cases but what are the limiting conditions?

Certainly, these should be determined by considering the conditions under which the en-

ergy emitted is of the same order as the kinetic energy of the charge. If we consider non-

relativistic motion, the energy emitted by an charge accelerating at the order of a, under a

period of duration of order , is given by the Larmor formula and is of the order

â rad è

Ï

2®

2

6 ´

j

0 3

(9.14)

On the other hand, the acceleration bestows the charge with kinetic energy on th order of

â

kinè

(Ä

®

È

2

2 (9.15)




So if we demand thatâ rad

â kin if we wish to neglect the radiation reaction, then this is

equivalent to

Ï

2®

2

6 ´

j

0 3

(Ä

®

È

2

2(9.16)

or

)

Ï

2

3 ´

j

0 3

(9.17)

and if we define the characteristic time as 1

-

Ï

2 # 3 ´

j

0 3 we can say that the effects of the

radiative reaction are negligible in measurements made over timescales on order of ) 1

.

Accounting for radiative reaction Having demonstrated the need for a force which

accounts for radiative effects of accelerating charges we set out to determine its form. From

the conservation of energy it is clear that the force we are looking for, which we will denote

Frad, must satisfy

µ

Frad "

vº

¬

µ

Ï

2v"

v

6 ´

j

0 3

º

(9.18)

where have integrated the Larmor formula over time on the right hand side. Partial integ-

ration yields

µ

Frad "v

º

µ

Ï

2v"

v

6 ´

j

0 3

º

¬

aÏ

2v"

v

6 ´

j

0 3

d

(9.19)

if assume periodic motion then we find that

µ

¤

Frad¬

Ï

2v

6 ´

j

0 3 §

"v

º

0 (9.20)

and so

Frad

Ï

2v

6 ´

j

0 3

( 1 v (9.21)

We now can correct the equation of motion to include the radiative reaction

(Ä v ¬

1v È

Fext (9.22)

This equation is know as Abraham-Lorentz equation of motion.

Unfortunately, the Abraham-Lorentz equation of motion is not without its own inherent

problems.

END OF EXAMPLE 9.2&




'

RADIATION AND PARTICLE ENERGY IN A SYNCHROTRON EXAMPLE 9.3

An electron moves in a circular orbit in a synchrotron under the action of the homogeneous

magnetic field B0.

(a) Calculate the energy which is lost in electromagnetic radiation per revolution if the

speedç

of the electron is assumed to be constant.

(b) At which particle energy does the radiated energy per revolution become equal to the

particle’s total energy, if Æ 0

1Y

5 T?

Hint:

º

Ä( 0

v È

º

Ê

F Æ

º

§

º

Ê

±

0 Ï

2v2

4

6 ´

Solution

(a) Since we wish to consider synchrotron motion we use the relativistically correct

equation of motion

º

p

º

Ê

Ïv

Î

B (9.23)

where the right hand side is the Lorentz force with E

0, and p is the space part of the

momentum 4-vector so

º

p

º

Ê

º

º

Ê

Ä( 0

v È

( 0 v

( 0v º

º

Ê

(9.24)

where

º

º

Ê

º

º

Ê

1

ª 1 ¬

ç

2 # 2

¬

1

2

3

¤

¬

º

º

Ê

v2

2 §

(9.25)

1

2 2

3º

º

Ä v " v È

3

2v " v (9.26)

But v"

v

0, since we know that the electron moves in a circular orbit , so3 4

3

À

0 and

thus

º

p

º

Ê

( 0 v (9.27)

This means that the power necessary to keep the particle in a circular orbit, disregarding

radiation losses, is

v"

º

p

º

Ê

(

0

¤

v"

v

2

2ç

2Ä v

"v È

§

0 (9.28)




which comes as no surprise. However, the circular motion does emit EM radiation. To find

the expression for the power loss due to radiation we need an expression for the accelera-

tion v . To this end we use what is left of (9.23), that is

( 0 v

Ïv

Î

B (9.29)

The acceleration is found by taking the norm of this last equation and since v#

B all wehave is the scalar equation

( 0

v

Ï

ç Æ 0 (9.30)

This equation is easily solved to give

˙ç

Ï

Æ 0

( 0

ç(9.31)

The factor 5 6 5 8

0

f

04

-

¿ is known as the synchrotron angular frequency and is the relativistic

value of the angular frequency for gyro-harmonic motion or cyclotron angular frequency.

Now we can insert this expression into the relativistic generalisation of the Larmor formula

º

§

º

Ê

±

0 Ï

2 ˙ç

2

4

6 ´

±

0 Ï

2 2¿

4

6 ´

ç

2 (9.32)

To find the energy per revolution, we need the period of revolution which is

2 ´

#

¿ ,

so

§

rev

º

§

º

Ê

2 ´

¿

±

0 Ï

2 2¿

4

6 ´

ç

2

±

0 Ï

2 2¿

4

6 ´

ç

2

(9.33)

±

0

Ï

3

3Æ 0 ç

2

3

( 0

(9.34)

(b) The task here is to equate the total energy and the radiated energy and solve for

velocity and then see what total energy that velocity is associated with.

The total energy isâ

( 0

2 and the radiated energy is§

rev

±

0

Ï

3

3ç

2Æ 0

#

Ä 3

( 0 È ,

which when equated gives

2

ç

2

1

2

±

0 Ï

3Æ 0

3

(

20

(9.35)

2

ç

2¬ 1

±

0 Ï

3Æ 0

3

(

20

(9.36)

ç

2

2

1

1

Ê 06

3

8

0

3¿

f

20

(9.37)




So, after a little algebra,

9

1

2

(

20

±

0

Ï

3Æ 0

(9.38)

Thus the particle energy for which the radiated energy is equal to the total particle energy

is

â

( 0 2 9

1

2

(

20

±

0

Ï

3Æ 0

0Y

16 TeV (9.39)

This is obviously the upper limit for which radiation effect can be neglected in the treatment

of synchrotron motion of the particle.

END OF EXAMPLE 9.3&

'

RADIATION LOSS OF AN ACCELERATED CHARGED PARTICLE EXAMPLE 9.4

A charged particle, initially at rest, is accelerated by a homogeneous electric field E0. Show

that the radiation loss is negligible, even at relativistic speeds, compared to the particle’s

own energy gain. Assume that under all circumstancesâ 0

108 V/m.

Solution

The relativistic equation of motion is

º

p

º

Ê

º

º

Ê

Ä( 0

v È

( 0

¤

v

2

2v Ä v

"v È

§

ÏE0 (9.40)

Since v#

v#

E0 then

( 0 ˙

çÄ 1

¥

2

2È

( 0 ˙

ç

¤

1

¥

2

1 ¬¥ 2 §

( 0 ˙ç

¤

1

1 ¬¥

2 §

( 0 ˙ç

2

Ï

â 0 (9.41)

( 0

3 ˙ç

Ï

â 0 (9.42)

˙ç

Ï

â 0

( 0

3(9.43)

Having found v we try to derive the radiation field generated by this motion; we have that




Erad

±

0 Ï

4 ´

£

3r

Î

Ä r ¢

Î

v È

±

0 Ï

4 ´

1 3

Ä 1 ¬¥ cos 7

È

3r

Î

r ¬

v

Î

v

±

0 Ï

4 ´

1 3

Ä 1 ¬¥ cos 7

È

3r

Î

»¼ r

Î

v ¬

vÎ

v

¦

0ÁÂ

±

0 Ï

4 ´

1 3

Ä 1 ¬¥ cos 7

È

3r

Î

Ä rÎ

v È

±

0 Ï

4 ´

1 3

Ä 1 ¬¥ cos 7

È

3

2 ˙ç sin 7 ˆ¼

±

0 Ï˙

ç

4 ´

sin 7

Ä 1 ¬¥ cos 7

È

3 ˆ¼

(9.44)

where have used the particular geometry of the vectors involved. Then we may determine

the Poynting vector, which is

S 1

±

0

Erad

2 ˆ

Í

¾

ˆ (9.45)

The radiated energy per unit area per unit time corresponding to this is

£ §

£

Í

¾

2

Â

Â

E2rad

Â

Â

±

0

2 (9.46)

but the energy radiated per unit area per unit time at the source point is

£ §

£

Ê

£ §

£

£

£

Ê

£ §

£

£

£

±

0

Erad

2

2Ä 1 ¬

¥ cos 7

È

±

0

±

20 Ï

2 ˙ç

2

16 ´

2

sin2 7

2Ä 1 ¬

¥ cos 7

È

6

±

0Ï

2

˙ç

2

16 ´

2

sin

2 7

Ä 1 ¬¥ cos 7

È

5 (9.47)

The total radiated energy per unit time is

£

˜§

£

Ê

µ

£ §

£

Ê

S

Ω

±

0 Ï

2 ˙ç

2

16 ´

2

2 ´

µ h0

sin3 7

Ä 1 ¬¥ cos 7

È

5

S

7

a

µ

¬ cos 7

S

7

C

Ù

sin R

d

2 ´

±

0 Ï

2 ˙ç

2

8 ´

2

µ

1

1

1 ¬

µ

2

Ä 1

¥

µ È

5

S

µ

2 ´

±

0 Ï

2 ˙ç

2

8 ´

2

µ

1

1

¤

1

Ä 1

¥

µ È

5¬

µ

2

Ä 1

¥

µ È

5 §

º

µ

2 ´

±

0 Ï

2 ˙ç

2

8 ´ 2

4

3

1

Ä 1¬

¥

2 È 3

2 ´

±

0 Ï

2 ˙ç

2

8 ´ 2

4

3

6

±

0 Ï

4â

20

6 ´( 20

(9.48)




We compare this expression for the radiated energy with the total energyâ

( 0

2, so

S

â

S

Ê

( 0 2

S

S

Ê

(9.49)

From the equation of motion we find that

º

º

Ê

Ä( 0 ç

È

( 0 ˙ç

( 0 ç

º

º

Ê

Ï

â 0 (9.50)

and, from the expression for ˙ç

,

( 0 ˙ç

Ï

â 0

2(9.51)

Hence

( 0 ç

º

º

Ê

Ï

â 0

¤

1 ¬

1

2 §

ç

2

2

(9.52)

º

º

Ê

Ï

â 0 ç

( 0 2

(9.53)

and

º

â

º

Ê

( 0 2

Ï

â 0 ç

( 0 2

Ï

â 0 ç(9.54)

Finally, for â 0

108 V/m, and an electron for which

Ï

C , we find that

S ˜§

S

ÊE

S

â

S

Ê

±

0 Ï

4â

20

6 ´

( 20

1

Ï

â 0 ç

±

0

C

3â 0

6 ´

( 20 ç

â 0

ç

Ä 4 ´

Î

10 7È

" Ä1

Y

6Î

10 19È

3

6 ´"

2Y

998Î

108" Ä

9Y

11Î

10 31È

2

1Y

1Î

10 12 â 0

ç

(9.55)

Which of course is a very small ratio for a relativistic electron (ç F

2Y

998Î

108 m/s)

even for â

108 V/m.


zemft thidé book exercises eletromagnetic field magnetic

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