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Tobia Carozzi Anders Eriksson Bengt Lundborg Bo Thidé Mattias Waldenvik
ELECTROMAGNETIC FIELD THEORY
EXERCISES
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Companion volume to
ELECTROMAGNETIC FIELD THEORY
by
Bo Thidé
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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
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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
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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
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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
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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
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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
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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
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vi
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PREFACE
This is a companion volume to the book Electromagnetic Field Theory by Bo Thidé.
Uppsala, Sweden B. T.
November, 1998
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viii PREFACE
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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
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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
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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
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4 LESSON 1. MAXWELL’S EQUATIONS
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:
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1 .3 . SOLVED EXAMPLES 5
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)
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6 LESSON 1. MAXWELL’S EQUATIONS
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)
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1 .3 . SOLVED EXAMPLES 7
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
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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)
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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)
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2 .3 . SOLVED EXAMPLES 11
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)
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12 LESSON 2. ELECTROMAGNETIC POTENTIALS AND WAVES
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)
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2 .3 . SOLVED EXAMPLES 13
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
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14 LESSON 2. ELECTROMAGNETIC POTENTIALS AND WAVES
˜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
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2 .3 . SOLVED EXAMPLES 15
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
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16 LESSON 2. ELECTROMAGNETIC POTENTIALS AND WAVES
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)
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2 .3 . SOLVED EXAMPLES 17
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
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18
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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
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3 .3 . SOLVED EXAMPLES 21
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
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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)
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3 .3 . SOLVED EXAMPLES 23
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
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24 LESSON 3. RELATIVISTIC ELECTRODYNAMICS
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)
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3 .3 . SOLVED EXAMPLES 25
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
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26 LESSON 3. RELATIVISTIC ELECTRODYNAMICS
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)
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3 .3 . SOLVED EXAMPLES 27
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
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28
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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
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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)
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4 .3 . SOLVED EXAMPLES 31
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)
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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
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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
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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
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5 .3 . SOLVED EXAMPLES 35
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)
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36 LESSON 5. ELECTROMAGNETIC ENERGY, MOMENTUM AND STRESS
The energy density is
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.
The energy density is
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
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5 .3 . SOLVED EXAMPLES 37
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
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38 LESSON 5. ELECTROMAGNETIC ENERGY, MOMENTUM AND STRESS
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,
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5 .3 . SOLVED EXAMPLES 39
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)
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40 LESSON 5. ELECTROMAGNETIC ENERGY, MOMENTUM AND STRESS
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
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5 .3 . SOLVED EXAMPLES 41
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
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42
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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
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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)
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6 .3 . SOLVED EXAMPLES 45
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
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46 LESSON 6. RADIATION FROM EXTENDED SOURCES
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
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6 .3 . SOLVED EXAMPLES 47
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)
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48 LESSON 6. RADIATION FROM EXTENDED SOURCES
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
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6 .3 . SOLVED EXAMPLES 49
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 &
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50 LESSON 6. RADIATION FROM EXTENDED SOURCES
'
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
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6 .3 . SOLVED EXAMPLES 51
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
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52 LESSON 6. RADIATION FROM EXTENDED SOURCES
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&
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6 .3 . SOLVED EXAMPLES 53
'
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)
END OF EXAMPLE 6.3 &
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54 LESSON 6. RADIATION FROM EXTENDED SOURCES
'
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
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6 .3 . SOLVED EXAMPLES 55
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
´
END OF EXAMPLE 6.4 &
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56
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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
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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
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7 .3 . SOLVED EXAMPLES 59
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)
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60 LESSON 7. MULTIPOLE RADIATION
(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
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7 .3 . SOLVED EXAMPLES 61
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)
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62 LESSON 7. MULTIPOLE RADIATION
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)
END OF EXAMPLE 7.2 &
'
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)
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7 .3 . SOLVED EXAMPLES 63
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
à (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
ሰ (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
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64 LESSON 7. MULTIPOLE RADIATION
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)
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7 .3 . SOLVED EXAMPLES 65
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.
END OF EXAMPLE 7.4 &
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66
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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
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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)
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8 .3 . SOLVED EXAMPLES 69
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
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70 LESSON 8. RADIATION FROM MOVING POINT CHARGES
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.
END OF EXAMPLE 8.1 &
'
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)
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8 .3 . SOLVED EXAMPLES 71
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)
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72 LESSON 8. RADIATION FROM MOVING POINT CHARGES
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)
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8 .3 . SOLVED EXAMPLES 73
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.
END OF EXAMPLE 8.3 &
'
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
ĵ
¬
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.
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74 LESSON 8. RADIATION FROM MOVING POINT CHARGES
Ä
¬
Ê
È
¬ 2ç µ
%
ª 4ç
2µ
2 ¬ 4 ĵ
2
¶
2È
Ä ç
2 ¬
2 #
2È
2 Äç
2 ¬
2 #
2È
¬
ç µ
%
½
¿ 2
Ò
2 ĵ
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
ĵ
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&
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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
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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)
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9 .3 . SOLVED EXAMPLES 77
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)
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78 LESSON 9. RADIATION FROM ACCELERATED PARTICLES
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.
END OF EXAMPLE 9.1 &
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9 .3 . SOLVED EXAMPLES 79
'
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)
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80 LESSON 9. RADIATION FROM ACCELERATED PARTICLES
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&
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9 .3 . SOLVED EXAMPLES 81
'
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)
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82 LESSON 9. RADIATION FROM ACCELERATED PARTICLES
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)
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9 .3 . SOLVED EXAMPLES 83
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
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84 LESSON 9. RADIATION FROM ACCELERATED PARTICLES
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)
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9 .3 . SOLVED EXAMPLES 85
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.
END OF EXAMPLE 9.4 &
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86 LESSON 9. RADIATION FROM ACCELERATED PARTICLES