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Electromagnetic Theory
Summary:
• Maxwell’s equations
• EM Potentials
• Equations of motion of particles in electromagnetic fields
• Green’s functions
• Lienard-Weichert potentials
• Spectral distribution of electromagnetic energy from an arbitrarily moving charge
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1 Maxwell’s equations
curlE = –
∂B
∂t
Faraday’s law
1 ∂E
curlB = µ 0 J + ----c2 ∂t
ρ
divE = ----ε0
Field diverges from
electric charges
divB = 0
No magnetic
monopoles
– 12
ε 0 = 8.854 ×10
1
ε 0 µ 0 = ----c2
Ampere’s law
Farads/metre
µ 0 = 4π × 10 – 7 Henrys/metre
8
c = 2.998 ×10 m/s ≈ 300, 000 km/s
Conservation of charge
∂ρ
+ div J = 0
∂t
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Conservation of energy
E×B
1 B 2
∂ 1
2
--- ε 0 E + --- ------ + div  -------------- = – ( J ⋅ E )
 µ 
2 µ 0
∂t 2
0
Electromagnetic
energy density
Poynting - Work done on
flux
particles by EM field
Poynting Flux
This is defined by
E×B
S = -------------µ0
ε ijk E j B k
S i = --------------------µ0
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Conservation of momentum
Momentum
density
Maxwell’s
stress tensor
∂  S i  ∂M ij
----- –
= – ρE i – ( J × B ) i
∂ t  c 2 ∂ x j
Rate of change of
momentum due to EM
field acting on matter
Electric part
Magnetic part
 Bi B j B 2

1 2 

M ij = ε 0 E i E j – --- E δ ij +  ----------- – --------- δ ij

  µ
2
2µ 0 
0
= – Flux of i cpt. of EM momentum in j direction
2 Equations of motion
Charges move under the influence of an electromagnetic field according to the (relativistically correct)
equation:
dp
p×B
------ = q ( E + v × B ) = q  E + --------------

dt
γm 
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Momentum and energy of the particle are given by:
p = γmv
E = γmc 2
v 2 – 1 / 2

γ = 1 – ----
c 2
E 2 = p2c2 + m2c4
3 Electromagnetic potentials
3.1 Derivation
divB = 0 ⇒ B = curl A
curlE = –
⇒E+
∂B
∂A
⇒ curl  E +  = 0

∂t
∂t 
∂A
= – gradφ
∂t
⇒ E = – gradφ –
∂A
∂t
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Summary:
E = – gradφ –
∂A
∂t
B = curl A
3.2 Potential equations
Equation for the vector potential A
Substitute into Ampere’s law:
1 ∂
∂A
curl curl A = µ 0 J + ----- – gradφ –
∂t
c2 ∂t
2
1 ∂
1∂ A
----– ∇2 A + ----- gradφ + grad div A = µ 0 J
c2 ∂t
c2 ∂t 2
Equation for the scalar potential φ
Exercise:
Show that
∇2φ + div
∂A
ρ
= – ----∂t
ε0
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3.3 Gauge transformations
The vector and scalar potentials are not unique. One can see that the same equations are satisfied if one
adds certain related terms to φ and A , specifically, the gauge transformations
A′ = A – gradψ
φ′ = φ +
∂ψ
∂t
leaves the relationship between E and B and the potentials intact. We therefore have some freedom to
specify the potentials. There are a number of gauges which are employed in electromagnetic theory.
Coulomb gauge
div A = 0
Lorentz gauge
1 ∂φ
----- + div A = 0
c2 ∂t
2
1∂ A
⇒ ----– ∇2 A = µ 0 J
2
2
c ∂t
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Temporal gauge
φ = 0
⇒
∂
ρ
div A = – ----∂t
ε0
2
1∂ A
----+ curl curl A = µ 0 J
2
2
c ∂t
The temporal gauge is the one most used when Fourier transforming the electromagnetic equations. For
other applications, the Lorentz gauge is often used.
4 Electromagnetic waves
For waves in free space, we take
E = E 0 exp [ i ( k ⋅ x – ωt ) ]
B = B 0 exp [ i ( k ⋅ x – ωt ) ]
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and substitute into the free-space form of Maxwell’s equations, viz.,
curlE = –
∂B
∂t
divE = 0
1 ∂E
curlB = ----c2 ∂t
div B = 0
This gives:
k × E0
ik × E 0 = iωB 0 ⇒ B 0 = ---------------ω
1
ω
ik × B 0 = – ----- iωE 0 ⇒ k × B 0 = – ----- E 0
c2
c2
ik ⋅ E 0 = 0 ⇒ k ⋅ E 0 = 0
ik ⋅ B 0 = 0 ⇒ k ⋅ B 0 = 0
We take the cross-product with k of the equation for B 0 :
( k × E0 )
( k ⋅ E 0 )k – k 2 E 0
ω
k × B 0 = k × --------------------- = ----------------------------------------- = – ----- E 0
ω
ω
c2
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and since k ⋅ E 0 = 0
ω
 -----– k 2 E 0 = 0 ⇒ ω = ± ck
 c2

2
the well-known dispersion equation for electromagnetic waves in free space. The - sign relates to waves
travelling in the opposite direction, i.e.
E = E 0 exp [ i ( k ⋅ x + ωt ) ]
We restrict ourselves here to the positive sign. The magnetic field is given by
κ × E0
k × E0
1 k

B 0 = ---------------- = --- --- × E 0 = ---------------
ck
ω
c
The Poynting flux is given by
E×B
S = -------------µ0
and we now take the real components of E and B :
E = E 0 cos ( k ⋅ x – ωt )
κ × E0
B = ---------------- cos ( k ⋅ x – ωt )
c
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where E 0 is now real, then
E0 × ( κ × E0 )
S = ---------------------------------- cos2 ( k ⋅ x – ωt )
µ0 c
= cε 0 ( E 02 κ – ( κ ⋅ E 0 )E 0 ) cos2 ( k ⋅ x – ωt )
= cε 0 E 02 κ cos2 ( k ⋅ x – ωt )
The average of cos2 ( k ⋅ x – ωt ) over a period ( T = 2π ⁄ ω ) is 1 ⁄ 2 so that the time-averaged value of
the Poynting flux is given by:
cε 0
------- E 02 κ
〈 S〉 =
2
5 Equations of motion of particles in a uniform magnetic field
An important special case of particle motion in electromagnetic fields occurs for E = 0 and
B = constant . This is the basic configuration for the calculation of cyclotron and synchrotron emission.
In this case the motion of a relativistic particle is given by:
q
dp
------ = q ( v × B ) = ------- ( p × B )
γm
dt
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Conservation of energy
There are a number of constants of the motion. First, the energy:
dp
p ⋅ ------ = 0
dt
and since
E 2 = p2c2 + m2c4
then
dp
dp
dE
E ------- = c 2 p ------ = c 2  p ⋅ ------ = 0 here.

dt
dt 
dt
There for E = γmc 2 is conserved and γ is constant - our first constant of motion.
Parallel component of momentum
The component of momentum along the direction of B is also conserved:
q B
B
dp B
d
-----  p ⋅ ---- = ------ ⋅ ---- = ------- ---- ⋅ ( p × B ) = 0
γm B
B
dt B
dt 
⇒ p || = γmv || is conserved
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where p || is the component of momentum parallel to the magnetic field.
We write the total magnitude of the velocity
v = cβ
and since γ is constant, so is v and we put
v || = v cos α
where α is the pitch angle of the motion, which we ultimately show is a helix.
Perpendicular components
Take the z –axis along the direction of the field, then the equations of motion are:
e1 e2 e3
q
dp
------ = ------- p p p
γm x y ||
dt
0 0 B
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In component form:
d px
q
--------- = ------- p y B = ηΩ B p y
γm
dt
d py
q
--------- = – ------- p x B = – ηΩ B p x
γm
dt
qB
Ω B = ---------- = Gyrofrequency
γm
q
η = ----- = Sign of charge
q
A quick way of solving these equations is to take the second plus i times the first:
d
----- ( p x + i p y ) = – iηΩ B ( p x + i p y )
dt
This has the solution
p x + i p y = A exp ( iφ 0 ) exp [ – iηΩ B t ]
⇒ p x = A cos ( ηΩ B t + φ 0 )
p y = – A sin ( ηΩ B t + φ 0 )
The parameter φ 0 is an arbitrary phase.
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Positively charged particles:
p x = A cos ( Ω B t + φ 0 )
p y = – A sin ( Ω B t + φ 0 )
Negatively charged particles (in particular, electrons):
p x = A cos ( Ω B t + φ 0 )
p y = A sin ( Ω B t + φ 0 )
We have another constant of the motion:
p x2 + p y2 = A 2 = p 2 sin2 α = p ⊥2
where p ⊥ is the component of momentum perpendicular to the magnetic field.
Velocity
The velocity components are given by:
vx
px
sin α cos ( Ω B t + φ 0 )
1
v y = ------- p y = cβ – η sin α sin ( Ω t + φ )
B
0
γm
vz
pz
cos α
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Position
Integrate the above velocity components:
cβ sin α
------------------ sin ( Ω B t + φ 0 )
ΩB
x0
x
+ y0
cβ sin α
y =
η ------------------ cos ( Ω B t + φ 0 )
ΩB
z
z0
cβt cos α
This represents motion in a helix with
cβ sin α
Gyroradius = r G = -----------------ΩB
The motion is clockwise for η > 0 and anticlockwise for η < 0 .
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r G sin ( Ω B t )
y
r G cos ( Ω B t ) Ω B t
y
η>0
η<0
x
– r G cos ( Ω B t ) Ω B t
x
r G sin ( Ω B t )
In vector form, we write:
x = x 0 + r G sin ( Ω B t + φ 0 ) η cos ( Ω B t + φ 0 ) 0
+ cβt cos α 0 0 1
The parameter x 0 represents the location of the guiding centre of the motion.
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6 Green’s functions
Green’s functions are widely used in electromagnetic and other field theories. Qualitatively, the idea behind Green’s functions is that they provide the solution for a given differential equation corresponding
to a point source. A solution corresponding to a given source distribution is then constructed by adding
up a number of point sources, i.e. by integration of the point source response over the entire distribution.
6.1 Green’s function for Poisson’s equation
A good example of the use of Green’s functions comes from Poisson’s equation, which appears in electrostatics and gravitational potential theory. For electrostatics:
ρe ( x )
∇2φ ( x ) = – -------------ε0
where ρ e is the electric charge density.
In gravitational potential theory:
∇2φ ( x ) = 4πGρ m ( x )
where ρ m ( x ) is the mass density.
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The Green’s function for the electrostatic case is prescribed by:
δ ( x – x′′ )
∇2G ( x, x′′ ) = – ---------------------ε0
where δ ( x – x′ ) is the three dimensional delta function. When there are no boundaries, this equation has
the solution
G ( x, x′′ ) = G ( x – x′′ )
1
G ( x ) = --------------4πε 0 r
The general solution of the electrostatic Poisson equation is then
φ( x) =
∫
G ( x – x′′ )ρ e ( x′ )d 3 x′
space
1
= -----------4πε 0
∫
space
ρ e ( x′ )
----------------- d 3 x′
x – x′
For completeness, the solution of the potential for a gravitating mass distribution is:
φ ( x ) = –G
∫
space
ρ m ( x′ )
----------------- d 3 x′
x – x′
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6.2 Green’s function for the wave equation
In the Lorentz gauge the equation for the vector potential is:
2
1∂ A
----– ∇2 A = µ 0 J
2
2
c ∂t
and the equation for the electrostatic (scalar) potential is
2
1 ∂
ρ
----φ ( t, x ) – ∇2φ ( t, x ) = ----ε0
c2 ∂t 2
These equations are both examples of the wave equation
2
1 ∂
----ψ ( t, x ) – ∇2ψ ( t, x ) = S ( t, x )
2
2
c ∂t
When time is involved a “point source” consists of a source which is concentrated at a point for an instant
of time, i.e.
S ( x, t ) = Aδ ( t – t′ )δ 3 ( x – x′ )
where A is the strength of the source, corresponds to a source at the point x = x′ which is switched on
at t = t′ .
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In the case of no boundaries, the Green’s function for the wave equation satisfies:
 1 ∂2

 ----2- 2 – ∇2  G ( t – t′, x – x′′ ) = δ ( t – t′ )δ 3 ( x – x′′ )
 c ∂t

The relevant solution (the retarded Green’s function) is:
1
r
G ( t, x′ ) = --------- δ  t – --
4πr  c
so that
1
x – x′
G ( t – t′, x – x′ ) = ----------------------- δ  t – t′ – ----------------
4π x – x′ 
c 
The significance of the delta function in this expression is that a point source at ( t′, x′ ) will only contribute to the field at the point ( t, x ) when
x – x′
t = t′ + ---------------c
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x – x′
i.e. at a later time corresponding to the finite travel time ---------------- of a pulse from the point x′ . Equivac
lently, a disturbance which arrives at the point t, x had to have been emitted at a time
x – x′
t′ = t – ---------------c
x – x′
The time t – ---------------- is known as the retarded time.
c
The general solution of the wave equation is
ψ ( t, x ) =
∞
∫–∞
dt′
∫
G ( t – t′, x – x′ )S ( t′, x′ )d 3 x′
space
1 ∞
= ------ ∫
dt′
4π – ∞
∫
space
S ( t′, x′ ) 
x – x′
-------------------δ t – ---------------- d 3 x′
x – x′ 
c 
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6.3 The vector and scalar potential
Using the above Green’s function, the vector and scalar potential for an arbitrary charge and current distribution are:
µ0 ∞
δ ( t – t′ – x – x′′ ⁄ c ) J ( t′, x′′ )
---------------------------------------------------------------------- d 3 x′
----A ( t, x ) = - ∫
dt′ ∫
x – x′′
4π – ∞
space
δ ( t – t′ – x – x′′ ⁄ c )ρ e ( t′, x′′ )
1 ∞
------------------------------------------------------------------------ d 3 x′
----------dt′ ∫
φ ( t, x ) =
∫
4πε 0 – ∞
x – x′′
space
7 Radiation from a moving charge – the Lienard-Weichert potentials
7.1 Deduction from the potential of an arbitrary charge distribution
The current and charge distributions for a moving charge are:
ρ ( t, x ) = qδ 3 ( x – X ( t ) )
J ( t, x ) = qvδ 3 ( x – X ( t ) )
where v is the velocity of the charge, q , and X ( t ) is the position of the charge at time t .The charge q
is the relevant parameter in front of the delta function since
∫
space
ρ ( t, x )d 3 x = q
∫
δ 3 ( x – X ( t ) )d 3 x = q
space
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Also, the velocity of the charge
dX ( t )
v ( t ) = -------------- = Ẋ ( t )
dt
so that
ρ ( t, x ) = qδ 3 ( x – X ( t ) )
J ( t, x ) = qX ˙( t ) δ 3 ( x – X ( t ) )
With the current and charge expressed in terms of spatial delta functions it is best to do the space integration first.
We have
∫
space
δ ( t – t′ – x – x′′ ⁄ c )ρ e ( t′, x′′ )
------------------------------------------------------------------------ d 3 x′ =
x – x′′
∫
space
δ ( t – t′ – x – x′′ ⁄ c )δ 3 ( x′ – X ( t′ ) ) 3
------------------------------------------------------------------------------------d x′
x – x′′
x – X ( t′ )
qδ  t – t′ – -------------------------


c
= -----------------------------------------------------x – X ( t′ )
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One important consequence of the motion of the charge is that the delta function resulting from the space
integration is now a more complicated function of t′ , because it depends directly upon t′ and indirectly
though the dependence on X ( t′ ) . The delta-function will now only contribute to the time integral when
x – X ( t′ )
t′ = t – ------------------------c
The retarded time is now an implicit function of ( t, x ) , through X ( t′ ) . However, the interpretation of t′
is still the same, it represents the time at which a pulse leaves the source point, X ( t′ ) to arrive at the field
point ( t, x ) .
We can now complete the solution for φ ( t, x ) by performing the integration over time:
x – X ( t′ )
qδ  t – t′ – -------------------------


c
1 ∞
φ ( t, x ) = ------------ ∫ ------------------------------------------------------ dt′
x – X ( t′ )
4πε 0 – ∞
This equation is not as easy to integrate as might appear because of the complicated dependence of the
delta-function on t′ .
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7.2 Aside on the properties of the delta function
The following lemma is required.
We define the delta-function by
∞
∫–∞ f ( t )δ ( t – a ) dt
Some care is required in calculating
= f (a)
∞
∫–∞ f ( t )δ ( g ( t ) – a ) dt .
Consider
∞
∫–∞
f ( t )δ ( g ( t ) – a ) dt =
=
∞
dt
----- dg
f
(
t
)δ
(
g
(
t
)
–
a
)
∫–∞
dg
f (t)
--------∫–∞ ġ ( t )- δ ( g – a ) dg
∞
f ( g –1 ( a ) )
= ------------------------ġ ( g – 1 ( a ) )
where g – 1 ( a ) is the value of t satisfying g ( t ) = a .
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7.3 Derivation of the Lienard-Wierchert potentials
In the above integral we have the delta function δ ( t – t′ – x – X ( t′ ) ⁄ c ) = δ ( t′ + x – X ( t′ ) ⁄ c – t ) so
that
g ( t′ ) = t′ + x – X ( t′ ) ⁄ c – t
Differentiating this with respect to t′ :
dg ( t′ )
∂ x – X ( t′ )
--------------- = ġ ( t′ ) = 1 + ------------------------dt′
∂ t′
c
To do the partial derivative on the right, express x – X ( t′ ) 2 in tensor notation:
x – X ( t′ ) 2 = x i x i – 2x i X i ( t′ ) + X i ( t′ )X i ( t′ )
Now,
∂
∂
x – X ( t′ ) 2 = 2 x – X ( t′ ) × x – X ( t′ )
∂ t′
∂t
Differentiating the tensor expression for x – X ( t′ ) 2 gives:
∂
x – X ( t′ ) 2 = – 2x i Ẋ i ( t′ ) + 2X i ( t′ )Ẋ i ( t′ ) = – 2Ẋ i ( t′ ) ( x i – X i ( t′ ) )
∂ t′
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Hence,
2 x – X ( t′ ) ×
∂
x – X ( t′ ) = – 2Ẋ i ( t′ ) ( x i – X i ( t′ ) )
∂t
Ẋ i ( t′ ) ( x i – X i ( t′ ) )
Ẋ ( t′ ) ⋅ ( x – X ( t′ ) )
∂
⇒
x – X ( t′ ) = – -------------------------------------------- = – -------------------------------------------x – X ( t′ )
x – X ( t′ )
∂ t′
The derivative of g ( t′ ) is therefore:
ġ ( t′ ) = 1 +
Ẋ ( t′ ) ⋅ ( x – X ( t′ ) ) ⁄ c
∂ x – X ( t′ )
------------------------- = 1 – --------------------------------------------------∂ t′
c
x – X ( t′ )
x – X ( t′ ) – Ẋ ( t′ ) ⋅ ( x – X ( t′ ) ) ⁄ c
= ---------------------------------------------------------------------------------x – X ( t′ )
Hence the quantity 1 ⁄ ( ġ ( t′ ) ) which appears in the value of the integral is
1
x – X ( t′ )
----------- = ---------------------------------------------------------------------------------ġ ( t′ )
x – X ( t′ ) – Ẋ ( t′ ) ⋅ ( x – X ( t′ ) ) ⁄ c
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Thus, our integral for the scalar potential:
x – X ( t′ )
qδ  t – t′ – -------------------------


c
1 ∞
φ ( t, x ) = ------------ ∫ ------------------------------------------------------ dt′
x – X ( t′ )
4πε 0 – ∞
q
x – X ( t′ )
1
= ------------ ---------------------------------------------------------------------------- × ------------------------4πε 0
Ẋ ( t′ ) ⋅ ( x – X ( t′ ) ) x – X ( t′ )
x – X ( t′ ) – -------------------------------------------c
1
q
= ------------ ---------------------------------------------------------------------------4πε 0
Ẋ ( t′ ) ⋅ ( x – X ( t′ ) )
x – X ( t′ ) – -------------------------------------------c
where, it needs to be understood that the value of t′ involved in this solution satisfies, the equation for
retarded time:
x – X ( t′ )
t′ = t – ------------------------c
We also often use this equation in the form:
x – X ( t′ )
t′ + ------------------------- = t
c
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7.4 Nomenclature and symbols
We define the retarded position vector:
r′ = x – X ( t′ )
and the retarded distance
r′ = x – X ( t′ ) .
The unit vector in the direction of the retarded position vector is:
r′
n′ ( t′ ) = ---r′
The relativistic β of the particle is
Ẋ ( t′ )
β ( t′ ) = ------------c
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7.5 Scalar potential
In terms of these quantities, therefore, the scalar potential is:
q
1
φ ( t, x ) = ------------ ---------------------------------------------------------------------------4πε 0
Ẋ ( t′ ) ⋅ ( x – X ( t′ ) )
x – X ( t′ ) – -------------------------------------------c
1
q
= ------------ -----------------------------4πε 0 r′ – β ( t′ ) ⋅ r′
1
q
=  ---------------- ---------------------------------- 4πε r′ [ 1 – β ( t′ ) ⋅ n′ ]
0
This potential shows a Coulomb-like factor times a factor ( 1 – β ( t′ ) ⋅ n′ ) – 1 which becomes extremely
important in the case of relativistic motion.
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7.6 Vector potential
The evaluation of the integral for the vector potential proceeds in an analogous way. The major difference is the velocity Ẋ ( t′ ) in the numerator.
µ0 q
Ẋ ( t′ )
A ( t, x ) = --------- ---------------------------------------------------------------------------4π
Ẋ ( t′ ) ⋅ ( x – X ( t′ ) )
x – X ( t′ ) – -------------------------------------------c
µ0 q
Ẋ ( t′ )
= ----------- ----------------------------------4πr′ [ 1 – β ( t′ ) ⋅ n′ ]
Hence we can write
q
Ẋ ( t′ )
1 q
Ẋ ( t′ )
A ( t, x ) = µ 0 ε 0 × ---------------- ----------------------------------- = ----- ---------------- ----------------------------------4πε 0 r′ [ 1 – β ( t′ ) ⋅ n′ ]
c 2 4πε 0 r′ [ 1 – β ( t′ ) ⋅ n′ ]
= c – 1 β ( t′ )φ ( t, x )
This is useful when for expressing the magnetic field in terms of the electric field.
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7.7 Determination of the electromagnetic field from the Lienard-Wierchert potentials
To determine the electric and magnetic fields we need to determine
E = – gradφ –
∂A
∂t
B = curl A
The potentials depend directly upon x and indirectly upon x, t through the dependence upon t′ . Hence
we need to work out the derivatives of t′ with respect to both t and x .
∂t′
Expression for -----∂t
Since,
x – X ( t′ )
t′ + ------------------------- = t
c
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∂t′
We can determine ------ by differentiation of this implicit equation.
∂t
∂t′ 1 ∂
------ + --- ( x – X ( t′ ) ) = 1
∂t c ∂ t
Ẋ ( t′ ) ∂t′
∂t′ ( x – X ( t′ ) )
⇒ ------ + --------------------------- ⋅  – ------------- ------ = 1
x – X ( t′ )  c  ∂t
∂t
Ẋ ( t′ ) ⋅ ( x – X ( t′ ) ) ∂t′
1 – -------------------------------------------- ------ = 1
∂t
c x – X ( t′ )
Solving for ∂t′ ⁄ ∂t :
∂t′
x – X ( t′ )
------ = --------------------------------------------------------------------------------------∂t
x – X ( t′ ) – ( Ẋ ( t′ ) ⁄ c ) ⋅ ( x – X ( t′ ) )
r′
= -------------------------------------r′ – Ẋ ( t′ ) ⋅ r′ ⁄ c
∂t′
1
------ = -----------------------------∂t 1 – β ( t′ ) ⋅ n′
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∂t′
Expression for ------- = ∇t′
∂x i
Again differentiate the implicit function for t′ :
∂t′ ( x i – X i ( t′ ) ) ( x j – X j ( t′ ) )Ẋ j ( t′ ) ⁄ c ∂t′
------- + ----------------------------- – ----------------------------------------------------- × ------- = 0
x – X ( t′ )
∂x i
∂x i c x – X ( t′ )
xi – X i
β j ( x j – X j ( t′ ) )
∂t′
------- 1 – -----------------------------------+ ---------------------------- = 0
∂x i
x – X ( t′ )
c x – X ( t′ )
xi – X i
∂t′ x – X ( t′ ) – β ( t′ ) ⋅ ( x – X ( t′ ) )
⇒ ------- -------------------------------------------------------------------------- = – ---------------------------∂x i
x – X ( t′ )
c x – X ( t′ )
– ( x i – X i ( t′ ) ) ⁄ c
∂t′
-----= -------------------------------------------------------------------------⇒
∂x i
x – X ( t′ ) – β ( t′ ) ⋅ ( x – X ( t′ ) )
i.e.
or
xi ′
c –1 ni ′
∂t′
------- = – -------------------------------------- = – -----------------------------∂x i
1 – β ( t′ ) ⋅ n′
r′ – Ẋ ( t′ ) ⋅ r′ ⁄ c
c – 1 n′
∇t′ = – -----------------------------1 – β ( t′ ) ⋅ n′
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The potentials include explicit dependencies upon the spatial coordinates of the field point and implicit
dependencies on ( t, x ) via the dependence on t′
The derivatives of the potentials can be determined from:
∂t′
∂φ = ∂φ + ∂φ -----x
∂
∂ xi t
∂ x i t′ ∂ t′ x i i
∂ A i ∂t′
∂ Ai
= -------- ------------∂t′ ∂t
x
∂t i
∂ A i ∂t′
∂ Ai
∂ Ai
-------- -------=
+
--------------∂x j t
∂x j t′ ∂t′ ∂x j
∂ Ak
∂ Ak
∂t′ ∂ A k
ε ijk --------- = ε ijk --------- + ε ijk -------- --------∂x j ∂t′
∂x j t
∂x j t′
In dyadic form:
∇φ t = ∇φ t′ + ∂φ ∇t′
∂ t′
∂A =
------∂t x
x
A
curl A t = curl A t′ + ∇t′ × ∂-----∂t′
Electromagnetic Theory
t′
∂ A ∂----------- ∂t∂t′ x
x
36 /56
Electric field
The calculation of the electric field goes as follows. Some qualifiers on the partial derivatives are omitted
since they should be fairly obvious
∂φ
∂[ c – 1 β ( t′ )φ ] ∂t′
∂A
E = – ∇φ – ------- = – ∇φ t′ – ∇t′ – -------------------------------  ------
 ∂t 
∂ t′
∂t′
∂t
∂t′
∂φ
β ∂t′ φ
= – ∇φ t′ –
∇t′ + --- ------ – --- β̇ ( t′ ) -----∂t
c
∂ t′
c ∂t
The terms
1 ( n′ – β )
β ∂t′
∇t′ + --- ------ = – --- ---------------------c 1 – β ⋅ n′
c ∂t
∂t′
1
------ = -------------------------∂t
( 1 – β ⋅ n′ )
Other useful formulae to derive beforehand are:
∂r′
------- = – c – 1 β
∂t′
∂r′
------- = – c – 1 β ⋅ n′
∂t′
1
1
In differentiating ------------------------------- it is best to express it in the form ---------------------- .
r′ ( 1 – β ⋅ n′ )
r′ – β ⋅ r′
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With a little bit of algebra, it can be shown that
q
n′ – β
∇φ t′ = – ------------------- ----------------------------4πε 0 r′ 2 ( 1 – β ⋅ n′ ) 2
˙ ⋅ n′ ]
qc [ β ⋅ n′ – β 2 + c – 1 r′β
∂φ
= ------------------- ------------------------------------------------------------∂ t′
( 1 – β ⋅ n )2
4πε 0 r′ 2
Combining all terms:
β̇
q [ ( n′ – β ) ( 1 – β′ 2 + c – 1 r′β̇ ⋅ n′ ) – c – 1 r′β
β( 1 – β ⋅ n′ ) ]
------------------ -------------------------------------------------------------------------------------------------------------------------------E =
4πε 0 r′ 2
( 1 – β ⋅ n′ ) 3
The immediate point to note here is that many of the terms in this expression decrease as r′ – 2 . However,
the terms proportional to the acceleration only decrease as r′ – 1 . These are the radiation terms:
q [ ( n′ – β )β̇ ⋅ n′ – β̇ ( 1 – β̇ ⋅ n′ ) ]
E rad = ------------------ -------------------------------------------------------------------------4πcε 0 r
( 1 – β ⋅ n′ ) 3
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Magnetic field
We can evaluate the magnetic field without going through more tedious algebra. The magnetic field is
given by:
∂A
B = curl A = curl A t′ + ∇t′ × ------∂t′
∂A
= curl ( c – 1 φβ ) t′ + ∇t′ × ------∂t′
where
∂t′
– c – 1 n′
∇t′ = ---------------------- = – c – 1 n′ × -----∂t
1 – β ⋅ n′
Now the first term is given by:
curl ( c – 1 φβ ) = c – 1 ∇φ t′ × β
and we know from calculating the electric field that
n′ – β
q
∇φ t′ = – ------------------- ----------------------------- = ξ ( n′ – β )
4πε 0 r′ 2 ( 1 – β ⋅ n′ ) 2
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Therefore,
c – 1 ∇φ t′ × β = c – 1 ξ ( n′ – β ) × β = c – 1 ξ ( n′ – β ) × ( β – n + n ) = c – 1 ξ ( n′ – β ) × n′
= c – 1 ∇φ t′ × n′
Hence, we can write the magnetic field as:
∂ A ∂t′
∂A
B = c – 1 ∇φ t′ × n′ – c – 1 n′ × ------- ------ = c – 1 n′ × – ∇φ t′ – ------∂t′ ∂t
∂t
Compare the term in brackets with
∂A
E = – ∇φ t′ + ∂φ ∇t′ – ------∂t
∂ t′
x
Since ∇t′ ∝ n′ , then
B = c – 1 ( n′ × E )
This equation holds for both radiative and non-radiative terms.
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40 /56
Poynting flux
The Poynting flux is given by:
E×B
E × ( n′ × E )
S = -------------- = ------------------------------- = cε 0 [ E 2 n′ – ( E ⋅ n′ )E ]
µ0
cµ 0
We restrict attention to the radiative terms in which E rad ∝ r′ – 1
For the radiative terms,
q [ ( 1 – β̇ ⋅ n′ ) ( n′ ⋅ β̇ ) – n′ ⋅ β̇ ( 1 – β̇ ⋅ n′ ) ]
E rad ⋅ n′ = ------------------ ------------------------------------------------------------------------------------------------ = 0
4πcε 0 r
( 1 – β ⋅ n′ ) 3
so that the Poynting flux,
S = cε 0 E 2 n′
This can be understood in terms of equal amounts of electric and magnetic energy density ( ( ε 0 ⁄ 2 )E 2 )
moving at the speed of light in the direction of n′ . This is a very important expression when it comes to
calculating the spectrum of radiation emitted by an accelerating charge.
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8 Radiation from relativistically moving charges
Trajectory of
particle
v
2⁄γ
Illustration of the beaming of radiation from a relativistically moving
particle.
Note the factor ( 1 – β ⋅ n′ ) – 3 in the expression
for the electric field. When 1 – β′ ⋅ n′ ≈ 0 the
contribution to the electric field is large; this occurs when β′ ⋅ n′ ≈ 1 , i.e. when the angle between the velocity and the unit vector from the
retarded point to the field point is approximately
zero.
We can quantify this as follows: Let θ be the angle between β ( t′ ) and n′ , then
1
1
1 – β′ ⋅ n′ = 1 – β′ cos θ ≈ 1 –  1 – ---------  1 – --- θ 2

2 
2γ 2 
1 θ 2

= 1 – 1 – --------- – ---- 2γ 2 2 
1
1
θ2
= --------- + ----- = --------- ( 1 + γ 2 θ 2 )
2γ 2
2γ 2 2
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42 /56
So you can see that the minimum value of 1 – β′ ⋅ n′ is 1 ⁄ ( 2γ 2 ) and that the value of this quantity only
remains near this for θ ∼ 1 ⁄ γ . This means that the radiation from a moving charge is beamed into a narrow cone of angular extent 1 ⁄ γ . This is particularly important in the case of synchrotron radiation for
which γ ∼ 10 4 (and higher) is often the case.
9 The spectrum of a moving charge
9.1 Fourier representation of the field
Consider the transverse electric field, E ( t ) , resulting from a moving charge, at a point in space and represent it in the form:
E ( t ) = E 1 ( t )e 1 + E 2 ( t )e 2
where e 1 and e 2 are appropriate axes in the plane of the wave. (Note that in general we are not dealing
with a monochromatic wave, here.)
The Fourier transforms of the electric components are:
Eα(ω) =
∞
∫–∞
e iωt E α ( t ) dt
1 ∞
E α ( t ) = ------ ∫ e – iωt E α ( ω ) dω
2π – ∞
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43 /56
The condition that E α ( t ) be real is that
E α ( – ω ) = E α* ( ω )
Note: We do not use a different symbol for the Fourier transform, e.g. Ẽ α ( ω ) . The transformed variable
is indicated by its argument.
9.2 Spectral power in a pulse
Outline of the following calculation
Eα(t )
Diagrammatic representation of a
pulse of radiation with a duration
∆t .
∆t
t
• Consider a pulse of radiation
• Calculate total energy per unit area in the radiation.
• Use Fourier transform theory to calculate the
spectral distribution of energy.
• Show this can be used to calculate the spectral
power of the radiation.
The energy per unit time per unit area of a pulse of radiation is given by:
dW
------------ = Poynting Flux = ( cε 0 )E 2 ( t ) = ( cε 0 ) [ E 12 ( t ) + E 22 ( t ) ]
dtdA
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44 /56
where E 1 and E 2 are the components of the electric field wrt (so far arbitrary) unit vectors e 1 and e 2
in the plane of the wave.
The total energy per unit area in the α –component of the pulse is
dW αα
∞
--------------- = ( cε 0 ) ∫ E α2 ( t ) dt
dA
–∞
From Parseval’s theorem,
∞
1 ∞
2 ( t ) dt = ----- ∫ E α ( ω ) 2 dω
E
∫–∞ α
2π – ∞
The integral from – ∞ to ∞ can be converted into an integral from 0 to ∞ using the reality condition. For
the negative frequency components, we have
E α ( – ω ) × E α* ( – ω ) = E α* ( ω ) × E α ( ω ) = E α ( ω ) 2
so that
∞
1 ∞
2 ( t ) dt = -- ∫ E α ( ω ) 2 dω
E
∫–∞ α
π 0
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45 /56
The total energy per unit area in the pulse, associated with the α component, is
cε 0 ∞
dW αα
∞
2
--------------- = cε 0 ∫ E α ( t ) dt = -------- ∫ E α ( ω ) 2 dω
π 0
dA
–∞
(The reason for the αα subscript is evident below.)
[Note that there is a difference here from the Poynting flux for a pure monochromatic plane wave in
which we pick up a factor of 1 ⁄ 2 . That factor results from the time integration of cos2 ωt which comes
∞
from, in effect, ∫ E α ( ω ) 2 dω . This factor, of course, is not evaluated here since the pulse has an arbi0
trary spectrum.]
We identify the spectral components of the contributors to the Poynting flux by:
cε 0
dW αα
--------------- = -------- E α ( ω ) 2
π
dωdA
dW αα
The quantity --------------- represents the energy per unit area per unit circular frequency in the entire pulse,
dωdA
i.e. we have accomplished our aim and determined the spectrum of the pulse.
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46 /56
We can use this expression to evaluate the power associated with the pulse. Suppose the pulse repeats
with period T , then we define the power associated with component α by:
cε 0
dW αα
1 dW
-------------------- = --- -------------- = -------- E α ( ω ) 2
T dAdω
πT
dAdωdt
This is equivalent to integrating the pulse over, say several periods and then dividing by the length of
time involved.
9.3 Emissivity
dA = r 2 dΩ
Variables used to define the emissivity in terms of emitted power.
r
dΩ
Region in which particle moves
during pulse
Consider the surface dA to be located a
long distance from the distance over
which the particle moves when emitting
the pulse of radiation. Then dA = r 2 dΩ
and
dW αα
dW αα
dW αα
1 dW αα
2
-------------------- = ----- --------------------- ⇒ --------------------- = r -------------------dΩdωdt
dAdωdt
dAdωdt r 2 dΩdωdt
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47 /56
The quantity
cε 0 r 2
cε 0 r 2
dW αα
--------------------- = ------------- E α ( ω ) 2 = -------------E α ( ω )E α* ( ω )
πT
πT
dΩdωdt
(Summation not implied)
is the emissivity corresponding to the e α component of the pulse.
9.4 Relationship to the Stokes parameters
We generalise our earlier definition of the Stokes parameters for a plane wave to the following:
cε 0
I ω = -------- [ E 1 ( ω )E 1* ( ω ) + E 2 ( ω )E 2* ( ω ) ]
πT
cε 0
Q ω = -------- [ E 1 ( ω )E 1* ( ω ) – E 2 ( ω )E 2* ( ω ) ]
πT
cε 0
U ω = -------- [ E 1* ( ω )E 2 ( ω ) + E 1 ( ω )E 2* ( ω ) ]
πT
1 cε 0 *
V ω = --- -------- [ E 1 ( ω )E 2 ( ω ) – E 1 ( ω )E 2* ( ω ) ]
i πT
The definition of I ω is equivalent to the definition of specific intensity in the Radiation Field chapter.
Also note the appearance of circular frequency resulting from the use of the Fourier transform.
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48 /56
We define a polarisation tensor by:
cε 0
1 I ω + Q ω U ω – iV ω
I αβ, ω = --= -------- E α ( ω )E β* ( ω )
2 U + iV I – Q
πT
ω
ω ω
ω
We have calculated above the emissivities,
cε 0 r 2
dW αα
--------------------- = -------------E α ( ω )E α* ( ω )
πT
dΩdωdt
(Summation not implied)
corresponding to E α E α* . More generally, we define:
cε 0
dW αβ
--------------------- = -------- r 2 E α ( ω )E β* ( ω )
πT
dΩdωdt
and these are the emissivities related to the components of the polarisation tensor I αβ .
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49 /56
In general, therefore, we have
dW 11
--------------------- → Emissivity for
dΩdωdt
dW 22
--------------------- → Emissivity for
dΩdωdt
dW 12
--------------------- → Emissivity for
dΩdωdt
1
--- ( I ω + Q ω )
2
1
--- ( I ω – Q ω )
2
1
--- ( U ω – iV ω )
2
*
dW 12
dW 21
1
--------------------- = --------------------- → Emissivity for --- ( U ω + iV ω )
2
dΩdωdt
dΩdωdt
Consistent with what we have derived above, the total emissivity is
I
εω
dW 22
dW 11
= --------------------- + --------------------dΩdωdt dΩdωdt
and the emissivity into the Stokes Q is
Q
εω
dW 11
dW 22
= --------------------- – --------------------dΩdωdt dΩdωdt
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50 /56
Also, for Stokes U and V :
U
εω
*
dW 12
dW 12
= --------------------- + --------------------dΩdωdt dΩdωdt
*
dW 12
dW 12
V = i  --------------------- – ---------------------
εω
 dΩdωdt dΩdωdt
Note the factor of r 2 in the expression for dW αβ ⁄ dΩdωdt . In the expression for the E –vector of the
radiation field
q [ ( n′ – β ) ( n′ ⋅ β̇ ) – β̇ ( 1 – β̇ ⋅ n′ ) ]
E = ------------------ -----------------------------------------------------------------------------4πcε 0 r
( 1 – β ⋅ n′ ) 3
E ∝ 1 ⁄ r . Hence rE and consequently r 2 E α E β* are independent of r , consistent with the above expressions for emissivity.
The emissivity is determined by the solution for the electric field.
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10 Fourier transform of the Lienard-Wierchert radiation field
The emissivities for the Stokes parameters obviously depend upon the Fourier transform of
q n′ × [ ( n′ – β′ ) × β̇′ ]
rE ( t ) = --------------- ------------------------------------------------4πcε 0
( 1 – β′ ⋅ n′ ) 3
where the prime means evaluation at the retarded time t′ given by
r′
t′ = t – ---c
r′ = x – X ( t′ )
The Fourier transform involves an integration wrt t . We transform this to an integral over t′ as follows:
1
∂t
dt = ------ dt′ = --------------- dt′ = ( 1 – β′ ⋅ n′ )dt′
∂t′ ⁄ ∂t
∂t′
using the results we derived earlier for differentiation of the retarded time. Hence,
q ∞ n′ × [ ( n′ – β′ ) × β̇′ ]
rE ( ω ) = --------------- ∫ ------------------------------------------------- e iωt ( 1 – β′ ⋅ n′ ) dt′
4πcε 0 – ∞
( 1 – β′ ⋅ n′ ) 3
q ∞ n′ × [ ( n′ – β′ ) × β̇′ ]
= --------------- ∫ ------------------------------------------------- e iωt dt′
4πcε 0 – ∞
( 1 – β′ ⋅ n′ ) 2
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52 /56
The next part is
r′
e iωt = exp iω  t′ + ----

c
Since
r′ = x – X ( t′ ) ≈ x when x » X ( t′ )
then we expand r′ to first order in X . Thus,
r′ = x j – X j ( t′ )
xi
– ( x i – X i ( t′ ) )
∂r′
-------- = -------------------------------- = – ---- at X i = 0
r
r′
∂X i
xi
∂
r′
--r′ = r + -------× X i ( t′ ) = r – - X i ( t′ ) = r – n i X i = r – n ⋅ X ( t′ )
r′
∂X i X j = 0
r
Note that it is the unit vector n = - which enters here, rather than the retarded unit vector n′
r
Hence,
r n ⋅ X ( t′ )
exp ( iωt ) = exp iω  t′ + -- – --------------------


c
c
iωr
n ⋅ X ( t′ )
= = exp iω  t′ – -------------------- × exp -------

c
c
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53 /56
iωr
The factor exp -------- is common to all Fourier transforms rE α ( ω ) and when one multiplies by the comc
plex conjugate it gives unity. This also shows why we expand the argument of the exponential to first
order in X ( t′ ) since the leading term is eventually unimportant.
The remaining term to receive attention in the Fourier Transform is
n′ × [ ( n′ – β′ ) × β̇′ ]
------------------------------------------------( 1 – β′ ⋅ n′ ) 2
We first show that we can replace n′ by n by also expanding in powers of X i ( t′ ) .
x i – X i ( t′ )
xi
xi – X i
∂
× Xj
----------------------------n i ′ = ----------------------------- = ---- +
x j – X j ( t′ )
r
∂ X j x j – X j ( t′ ) X = 0
j
( xi – X i ) ( x j – X j )
xi
= ---- + – δ ij + ------------------------------------------× Xj
r
x j – X j ( t′ ) 3
Xj = 0
xi x j X j
= n i + – δ ij + --------- -----r2 r
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54 /56
So the difference between n and n′ is of order X ⁄ r , i.e. of order the ratio the dimensions of the distance
the particle moves when emitting a pulse to the distance to the source. This time however, the leading
term does not cancel out and we can safely neglect the terms of order X ⁄ r . Hence we put,
n′ × [ ( n′ – β′ ) × β̇′ ]
n × [ ( n – β′ ) × β̇′ ]
------------------------------------------------- = --------------------------------------------( 1 – β′ ⋅ n′ ) 2
( 1 – β′ ⋅ n ) 2
It is straightforward (exercise) to show that
d n × ( n × β′ )
n × [ ( n – β′ ) × β̇′ ]
------ ---------------------------- = --------------------------------------------dt′ 1 – β′ ⋅ n′
( 1 – β′ ⋅ n ) 2
Hence,
iωr ⁄ c ∞ d n × ( n × β′ )
q
n ⋅ X ( t′ )
 t′ – ---------------------------------------------------exp
iω
dt′
rE ( ω ) = --------------- e
∫–∞ dt′ 1 – β′ ⋅ n


4πcε 0
c
One can integrate this by parts. First note that
n × ( n × β′ ) ∞ = 0
----------------------------1 – β′ ⋅ n – ∞
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55 /56
since we are dealing with a pulse. Second, note that,
d
n ⋅ X ( t′ )
------ exp iω  t′ – --------------------


dt′
c
n ⋅ X ( t′ )
= exp iω  t′ – -------------------- × iω [ 1 – β′ ⋅ n ]


c
and that the factor of [ 1 – β′ ⋅ n′ ] cancels the remaining one in the denominator. Hence,
– iωq iωr ⁄ c ∞
n ⋅ X ( t′ )
 t′ – -------------------rE ( ω ) = --------------- e
n
×
(
n
×
β′
)
exp
iω
dt′
∫–∞


4πcε 0
c
In order to calculate the Stokes parameters, one selects a coordinate system ( e 1 and e 2 ) in which this is
as straightforward as possible. The motion of the charge enters through the terms involving β ( t′ ) and
X ( t′ ) in the integrand.
Remark
The feature associated with radiation from a relativistic particle, namely that the radiation is very strongly peaked in the direction of motion, shows up in the previous form of this integral via the factor
( 1 – β ⋅ n′ ) – 3 . This dependence is not evident here. However, when we proceed to evaluate the integral
in specific cases, this dependence resurfaces.
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