EM_Theory.pdf

Electromagnetic Theory 1 /56
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

1 Maxwell’s equations
Conservation of charge
curlE
t
∂
∂B
–
=
curlB µ0J
1
c2
----
-
t
∂
∂E
+
=
divE
ρ
ε0
----
-
=
divB 0
=
Faraday’s law
Ampere’s law
Field diverges from
electric charges
No magnetic
monopoles
ε0 8.854
12
–
×10 Farads/metre
= µ0 4π 10 7
–
× Henrys/metre
=
ε0µ0
1
c2
----
-
= c 2.998
8
×10 m/s 300 000 km/s
,
≈
=
t
∂
∂ρ
divJ
+ 0
=

Conservation of energy
Poynting Flux
This is defined by
t
∂
∂ 1
2
--
-ε0E2 1
2
--
-
B2
µ0
-----
-
+
 
  div
E B
×
µ0
-------------
-
 
 
+ J E
⋅
⋅
⋅
⋅
( )
–
=
Electromagnetic
energy density
Poynting
flux
- Work done on
particles by EM field
S
E B
×
µ0
-------------
-
= Si
εijkE jBk
µ0
--------------------
-
=

Conservation of momentum
2 Equations of motion
Charges move under the influence of an electromagnetic field according to the (relativistically correct)
equation:
t
∂
∂ Si
c2
----
-
 
 
xj
∂
∂Mij
– ρEi
– J B
×
( )i
–
=
Momentum
density
Maxwell’s
stress tensor
Rate of change of
momentum due to EM
field acting on matter
Mij ε0 EiE j
1
2
--
-E2δij
–
 
 
BiBj
µ0
----------
-
B2
2µ0
--------
-δij
–
 
 
 
+
=
Flux of i cpt. of EM momentum in j direction
–
=
Electric part Magnetic part
dp
dt
-----
- q E v B
×
+
( ) q E
p B
×
γm
-------------
-
+
 
 
= =

Momentum and energy of the particle are given by:
3 Electromagnetic potentials
3.1 Derivation
p γmv
= γ 1
v2
c2
----
-
–
 
 
1 2
/
–
=
E γmc2
= E2 p2c2 m2c4
+
=
divB 0
= B
⇒ curlA
=
curlE
t
∂
∂B
–
= curl E
t
∂
∂A
+
 
 
⇒ 0
=
E
t
∂
∂A
+
⇒ gradφ
–
=
E
⇒ gradφ
–
t
∂
∂A
–
=

Summary:
3.2 Potential equations
Equation for the vector potential
Substitute into Ampere’s law:
Equation for the scalar potential
Exercise:
Show that
E gradφ
–
t
∂
∂A
–
= B curlA
=
A
curl curlA µ0J
1
c2
----
-
t
∂
∂
gradφ
–
t
∂
∂A
–
+
=
1
c2
----
-
t2
2
∂
∂ A
A
∇2
–
1
c2
----
-
t
∂
∂
gradφ grad divA
+ + µ0J
=
φ
φ
∇2 div
t
∂
∂A
+
ρ
ε0
----
-
–
=

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 and , specifically, the gauge transformations
leaves the relationship between and 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
Lorentz gauge
φ A
A′ A gradψ
–
= φ′ φ
t
∂
∂ψ
+
=
E B
divA 0
=
1
c2
----
-
t
∂
∂φ
divA
+ 0
=
1
c2
----
-
t2
2
∂
∂ A
A
∇2
– µ0J
=
⇒

Temporal gauge
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
φ 0
=
t
∂
∂
divA
⇒
ρ
ε0
----
-
–
=
1
c2
----
-
t2
2
∂
∂ A
curl curlA
+ µ0J
=
E E0 i k x
⋅ ωt
–
( )
[ ]
exp
=
B B0 i k x
⋅ ωt
–
( )
[ ]
exp
=

and substitute into the free-space form of Maxwell’s equations, viz.,
This gives:
We take the cross-product with of the equation for :
curlE
t
∂
∂B
–
= curlB
1
c2
----
-
t
∂
∂E
=
divE 0
= div B 0
=
ik E0
× iωB0
= B0
⇒
k E0
×
ω
---------------
-
=
ik B0
×
1
c2
----
-iωE0 k B0
×
⇒
–
ω
c2
----
-E0
–
= =
ik E0
⋅ 0
= k E0
⋅
⇒ 0
=
ik B0
⋅ 0
= k B0
⋅
⇒ 0
=
k B0
k B0
× k
k E0
×
( )
ω
--------------------
-
×
k E0
⋅
( )k k2E0
–
ω
----------------------------------------
-
ω
c2
----
-E0
–
= = =

and since
the well-known dispersion equation for electromagnetic waves in free space. The - sign relates to waves
travelling in the opposite direction, i.e.
We restrict ourselves here to the positive sign. The magnetic field is given by
The Poynting flux is given by
and we now take the real components of and :
k E0
⋅ 0
=
ω2
c2
------ k2
–
 
  E0 0
= ω
⇒ ck
±
=
E E0 i k x
⋅ ωt
+
( )
[ ]
exp
=
B0
k E0
×
ω
---------------
-
1
c
--
-
k
k
--
- E0
×
 
 
κ E0
×
c
---------------
-
= = =
S
E B
×
µ0
-------------
-
=
E B
E E0 k x
⋅ ωt
–
( )
cos
= B
κ
κ
κ
κ E0
×
c
---------------
- k x
⋅ ωt
–
( )
cos
=

where is now real, then
The average of over a period ( ) is so that the time-averaged value of
the Poynting flux is given by:
5 Equations of motion of particles in a uniform magnetic field
An important special case of particle motion in electromagnetic fields occurs for and
. 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:
E0
S
E0 κ
κ
κ
κ E0
×
( )
×
µ0c
---------------------------------
- k x
⋅ ωt
–
( )
cos2
=
cε0 E0
2κ
κ
κ
κ κ
κ
κ
κ E0
⋅
( )E0
–
( ) k x
⋅ ωt
–
( )
cos2
=
cε0E0
2κ
κ
κ
κ k x
⋅ ωt
–
( )
cos2
=
k x
⋅ ωt
–
( )
cos2 T 2π ω
⁄
= 1 2
⁄
S
〈 〉
cε0
2
-------
-E0
2κ
κ
κ
κ
=
E 0
=
B constant
=
dp
dt
-----
- q v B
×
( )
q
γm
------
- p B
×
( )
= =

Conservation of energy
There are a number of constants of the motion. First, the energy:
and since
then
There for is conserved and is constant - our first constant of motion.
Parallel component of momentum
The component of momentum along the direction of is also conserved:
p
dp
dt
-----
-
⋅ 0
=
E2 p2c2 m2c4
+
=
E
dE
dt
------
- c2 p
dp
dt
-----
- c2 p
dp
dt
-----
-
⋅
 
  0 here.
= = =
E γmc2
= γ
B
d
dt
----
- p
B
B
---
-
⋅
 
  dp
dt
-----
-
B
B
---
-
⋅
q
γm
------
-
B
B
---
- p B
×
( )
⋅ 0
= = =
p||
⇒ γmv||
= is conserved

where is the component of momentum parallel to the magnetic field.
We write the total magnitude of the velocity
and since is constant, so is and we put
where is the pitch angle of the motion, which we ultimately show is a helix.
Perpendicular components
Take the –axis along the direction of the field, then the equations of motion are:
p||
v cβ
=
γ v
v|| v α
cos
=
α
z
dp
dt
-----
-
q
γm
------
-
e1 e2 e3
px py p||
0 0 B
=

In component form:
A quick way of solving these equations is to take the second plus times the first:
This has the solution
The parameter is an arbitrary phase.
d px
dt
--------
-
q
γm
------
- pyB ηΩB py
= =
d py
dt
--------
-
q
γm
------
- pxB
– ηΩB px
–
= =
ΩB
q B
γm
---------
- Gyrofrequency
= =
η
q
q
----- Sign of charge
= =
i
d
dt
----
- px ipy
+
( ) iηΩB px ipy
+
( )
–
=
px ipy
+ A iφ0
( )
exp iηΩBt
–
[ ]
exp
=
px
⇒ A ηΩBt φ0
+
( )
cos
= py A
– ηΩBt φ0
+
( )
sin
=
φ0

Positively charged particles:
Negatively charged particles (in particular, electrons):
We have another constant of the motion:
where is the component of momentum perpendicular to the magnetic field.
Velocity
The velocity components are given by:
px A ΩBt φ0
+
( )
cos
= py A
– ΩBt φ0
+
( )
sin
=
px A ΩBt φ0
+
( )
cos
= py A ΩBt φ0
+
( )
sin
=
px
2 py
2
+ A2 p2 α
sin2 p⊥
2
= = =
p⊥
vx
vy
vz
1
γm
------
-
px
py
pz
cβ
α ΩBt φ0
+
( )
cos
sin
η α ΩBt φ0
+
( )
sin
sin
–
α
cos
= =

Position
Integrate the above velocity components:
This represents motion in a helix with
The motion is clockwise for and anticlockwise for .
x
y
z
cβ α
sin
ΩB
-----------------
- ΩBt φ0
+
( )
sin
η
cβ α
sin
ΩB
-----------------
- ΩBt φ0
+
( )
cos
cβt α
cos
x0
y0
z0
+
=
Gyroradius rG
cβ α
sin
ΩB
-----------------
-
= =
η 0
> η 0
<

In vector form, we write:
The parameter represents the location of the guiding centre of the motion.
ΩBt
rG ΩBt
( )
sin
rG ΩBt
( )
cos
x
y
ΩBt
r
– G ΩBt
( )
cos
rG ΩBt
( )
sin
x
y
η 0
> η 0
<
x x0 rG ΩBt φ0
+
( )
sin η ΩBt φ0
+
( )
cos 0
+
=
cβt α 0 0 1
cos
+
x0

6 Green’s functions
Green’s functions are widely used in electromagnetic and other field theories. Qualitatively, the idea be-
hind 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 elec-
trostatics and gravitational potential theory. For electrostatics:
where is the electric charge density.
In gravitational potential theory:
where is the mass density.
φ x
( )
∇2
ρe x
( )
ε0
-------------
-
–
=
ρe
φ x
( )
∇2 4πGρm x
( )
=
ρm x
( )

The Green’s function for the electrostatic case is prescribed by:
where is the three dimensional delta function. When there are no boundaries, this equation has
the solution
The general solution of the electrostatic Poisson equation is then
For completeness, the solution of the potential for a gravitating mass distribution is:
G x x′
′
′
′
,
,
,
,
( )
∇2 δ x x′
′
′
′
–
( )
ε0
---------------------
-
–
=
δ x x′
–
( )
G x x′
′
′
′
,
,
,
,
( ) G x x′
′
′
′
–
( )
=
G x
( )
1
4πε0r
--------------
-
=
φ x
( ) G x x′
′
′
′
–
( )ρe x′
( )d3x′
space
∫
=
1
4πε0
-----------
-
ρe x′
( )
x x′
–
----------------
-d3x′
space
∫
=
φ x
( ) G
ρm x′
( )
x x′
–
----------------
-d3x′
space
∫
–
=

6.2 Green’s function for the wave equation
In the Lorentz gauge the equation for the vector potential is:
and the equation for the electrostatic (scalar) potential is
These equations are both examples of the wave equation
When time is involved a “point source” consists of a source which is concentrated at a point for an instant
of time, i.e.
where is the strength of the source, corresponds to a source at the point which is switched on
at .
1
c2
----
-
t2
2
∂
∂ A
A
∇2
– µ0J
=
1
c2
----
-
t2
2
∂
∂
φ t x
,
( ) φ t x
,
( )
∇2
–
ρ
ε0
----
-
=
1
c2
----
-
t2
2
∂
∂
ψ t x
,
( ) ψ t x
,
( )
∇2
– S t x
,
( )
=
S x t
,
( ) Aδ t t′
–
( )δ3 x x′
–
( )
=
A x x′
=
t t′
=

In the case of no boundaries, the Green’s function for the wave equation satisfies:
The relevant solution (the retarded Green’s function) is:
so that
The significance of the delta function in this expression is that a point source at will only con-
tribute to the field at the point when
1
c2
----
-
t2
2
∂
∂
∇2
–
 
 
 
G t t′ x x′
′
′
′
–
,
–
( ) δ t t′
–
( )δ3 x x′
′
′
′
–
( )
=
G t x′
,
( )
1
4πr
--------
-δ t
r
c
-
-
–
 
 
=
G t t′ x x′
–
,
–
( )
1
4π x x′
–
----------------------
-δ t t′
–
x x′
–
c
---------------
-
–
 
 
=
t′ x′
,
( )
t x
,
( )
t t′
x x′
–
c
---------------
-
+
=

i.e. at a later time corresponding to the finite travel time of a pulse from the point . Equiva-
lently, a disturbance which arrives at the point had to have been emitted at a time
The time is known as the retarded time.
The general solution of the wave equation is
x x′
–
c
---------------
- x′
t x
,
t′ t
x x′
–
c
---------------
-
–
=
t
x x′
–
c
---------------
-
–
ψ t x
,
( ) t′ G t t′ x x′
–
,
–
( )S t′ x′
,
( )d3x′
space
∫
d
∞
–
∞
∫
=
1
4π
-----
- t′
S t′ x′
,
( )
x x′
–
-------------------δ t
x x′
–
c
---------------
-
–
 
  d3x′
space
∫
d
∞
–
∞
∫
=

6.3 The vector and scalar potential
Using the above Green’s function, the vector and scalar potential for an arbitrary charge and current dis-
tribution are:
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:
where is the velocity of the charge, , and is the position of the charge at time .The charge
is the relevant parameter in front of the delta function since
A t x
,
( )
µ0
4π
-----
- t′
δ t t′
– x x′
′
′
′
– c
⁄
–
( )J t′ x′
′
′
′
,
( )
x x′
′
′
′
–
---------------------------------------------------------------------
-d3x′
space
∫
d
∞
–
∞
∫
=
φ t x
,
( )
1
4πε0
-----------
- t′
δ t t′
– x x′
′
′
′
– c
⁄
–
( )ρe t′ x′
′
′
′
,
( )
x x′
′
′
′
–
-----------------------------------------------------------------------
-d3x′
space
∫
d
∞
–
∞
∫
=
ρ t x
,
( ) qδ3 x X t
( )
–
( )
=
J t x
,
( ) qvδ3 x X t
( )
–
( )
=
v q X t
( ) t q
ρ t x
,
( )d3x
space
∫ q δ3 x X t
( )
–
( )d3x
space
∫ q
= =

Also, the velocity of the charge
so that
With the current and charge expressed in terms of spatial delta functions it is best to do the space inte-
gration first.
We have
v t
( )
dX t
( )
dt
-------------
- Ẋ t
( )
= =
ρ t x
,
( ) qδ3 x X t
( )
–
( )
=
J t x
,
( ) qX t
( )
˙ δ3
x X t
( )
–
( )
=
δ t t′
– x x′
′
′
′
– c
⁄
–
( )ρe t′ x′
′
′
′
,
( )
x x′
′
′
′
–
-----------------------------------------------------------------------
-d3x′
space
∫
δ t t′
– x x′
′
′
′
– c
⁄
–
( )δ3 x′ X t′
( )
–
( )
x x′
′
′
′
–
------------------------------------------------------------------------------------d3x′
space
∫
=
qδ t t′
–
x X t′
( )
–
c
------------------------
-
–
 
 
x X t′
( )
–
------------------------------------------------------
=

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 , because it depends directly upon and indirectly
though the dependence on . The delta-function will now only contribute to the time integral when
The retarded time is now an implicit function of , through . However, the interpretation of
is still the same, it represents the time at which a pulse leaves the source point, to arrive at the field
point .
We can now complete the solution for by performing the integration over time:
This equation is not as easy to integrate as might appear because of the complicated dependence of the
delta-function on .
t′ t′
X t′
( )
t′ t
x X t′
( )
–
c
------------------------
-
–
=
t x
,
( ) X t′
( ) t′
X t′
( )
t x
,
( )
φ t x
,
( )
φ t x
,
( )
1
4πε0
-----------
-
qδ t t′
–
x X t′
( )
–
c
------------------------
-
–
 
 
x X t′
( )
–
------------------------------------------------------ t′
d
∞
–
∞
∫
=
t′

7.2 Aside on the properties of the delta function
The following lemma is required.
We define the delta-function by
Some care is required in calculating .
Consider
where is the value of satisfying .
f t
( )δ t a
–
( ) t
d
∞
–
∞
∫ f a
( )
=
f t
( )δ g t
( ) a
–
( ) t
d
∞
–
∞
∫
f t
( )δ g t
( ) a
–
( ) t
d
∞
–
∞
∫ f t
( )δ g t
( ) a
–
( )
dt
dg
-----
- g
d
∞
–
∞
∫
=
f t
( )
ġ t
( )
---------
-δ g a
–
( ) g
d
∞
–
∞
∫
=
f g 1
– a
( )
( )
ġ g 1
– a
( )
( )
------------------------
-
=
g 1
– a
( ) t g t
( ) a
=

7.3 Derivation of the Lienard-Wierchert potentials
In the above integral we have the delta function so
that
Differentiating this with respect to :
To do the partial derivative on the right, express in tensor notation:
Now,
Differentiating the tensor expression for gives:
δ t t′
– x X t′
( )
– c
⁄
–
( ) δ t′ x X t′
( )
– c
⁄ t
–
+
( )
=
g t′
( ) t′ x X t′
( )
– c
⁄ t
–
+
=
t′
dg t′
( )
dt′
--------------
- ġ t′
( )
= 1
t′
∂
∂ x X t′
( )
–
c
------------------------
-
+
=
x X t′
( )
– 2
x X t′
( )
– 2 xixi 2xiXi t′
( )
– Xi t′
( )Xi t′
( )
+
=
t′
∂
∂
x X t′
( )
– 2 2 x X t′
( )
–
t
∂
∂
x X t′
( )
–
×
=
x X t′
( )
– 2
t′
∂
∂
x X t′
( )
– 2 2xiẊi t′
( )
– 2Xi t′
( )Ẋi t′
( )
+ 2Ẋi t′
( ) xi Xi t′
( )
–
( )
–
= =

Hence,
The derivative of is therefore:
Hence the quantity which appears in the value of the integral is
2 x X t′
( )
–
t
∂
∂
x X t′
( )
–
× 2Ẋi t′
( ) xi Xi t′
( )
–
( )
–
=
t′
∂
∂
x X t′
( )
–
Ẋi t′
( ) xi Xi t′
( )
–
( )
x X t′
( )
–
-------------------------------------------
-
–
Ẋ t′
( ) x X t′
( )
–
( )
⋅
x X t′
( )
–
-------------------------------------------
-
–
= =
⇒
g t′
( )
ġ t′
( ) 1
t′
∂
∂ x X t′
( )
–
c
------------------------
-
+ 1
Ẋ t′
( ) x X t′
( )
–
( )
⋅ c
⁄
x X t′
( )
–
--------------------------------------------------
-
–
= =
x X t′
( )
– Ẋ t′
( ) x X t′
( )
–
( )
⋅ c
⁄
–
x X t′
( )
–
---------------------------------------------------------------------------------
-
=
1 ġ t′
( )
( )
⁄
1
ġ t′
( )
----------
-
x X t′
( )
–
x X t′
( )
– Ẋ t′
( ) x X t′
( )
–
( )
⋅ c
⁄
–
---------------------------------------------------------------------------------
-
=

Thus, our integral for the scalar potential:
where, it needs to be understood that the value of involved in this solution satisfies, the equation for
retarded time:
We also often use this equation in the form:
φ t x
,
( )
1
4πε0
-----------
-
qδ t t′
–
x X t′
( )
–
c
------------------------
-
–
 
 
x X t′
( )
–
------------------------------------------------------ t′
d
∞
–
∞
∫
=
q
4πε0
-----------
-
x X t′
( )
–
x X t′
( )
–
Ẋ t′
( ) x X t′
( )
–
( )
⋅
c
-------------------------------------------
-
–
---------------------------------------------------------------------------
-
1
x X t′
( )
–
------------------------
-
×
=
q
4πε0
-----------
-
1
x X t′
( )
–
Ẋ t′
( ) x X t′
( )
–
( )
⋅
c
-------------------------------------------
-
–
---------------------------------------------------------------------------
-
=
t′
t′ t
x X t′
( )
–
c
------------------------
-
–
=
t′
x X t′
( )
–
c
------------------------
-
+ t
=

7.4 Nomenclature and symbols
We define the retarded position vector:
and the retarded distance
.
The unit vector in the direction of the retarded position vector is:
The relativistic of the particle is
r′ x X t′
( )
–
=
r′ x X t′
( )
–
=
n′ t′
( )
r′
r′
---
-
=
β
β t′
( )
Ẋ t′
( )
c
------------
-
=

7.5 Scalar potential
In terms of these quantities, therefore, the scalar potential is:
This potential shows a Coulomb-like factor times a factor which becomes extremely
important in the case of relativistic motion.
φ t x
,
( )
q
4πε0
-----------
-
1
x X t′
( )
–
Ẋ t′
( ) x X t′
( )
–
( )
⋅
c
-------------------------------------------
-
–
---------------------------------------------------------------------------
-
=
q
4πε0
-----------
-
1
r′ β t′
( ) r′
⋅
–
------------------------------
=
q
4πε0r′
----------------
 
  1
1 β t′
( ) n′
⋅
–
[ ]
----------------------------------
-
=
1 β t′
( ) n′
⋅
–
( ) 1
–

7.6 Vector potential
The evaluation of the integral for the vector potential proceeds in an analogous way. The major differ-
ence is the velocity in the numerator.
Hence we can write
This is useful when for expressing the magnetic field in terms of the electric field.
Ẋ t′
( )
A t x
,
( )
µ0q
4π
--------
-
Ẋ t′
( )
x X t′
( )
–
Ẋ t′
( ) x X t′
( )
–
( )
⋅
c
-------------------------------------------
-
–
---------------------------------------------------------------------------
-
=
µ0q
4πr′
----------
-
Ẋ t′
( )
1 β t′
( ) n′
⋅
–
[ ]
----------------------------------
-
=
A t x
,
( ) µ0ε0
q
4πε0r′
----------------
×
Ẋ t′
( )
1 β t′
( ) n′
⋅
–
[ ]
----------------------------------
-
1
c2
----
-
q
4πε0r′
----------------
Ẋ t′
( )
1 β t′
( ) n′
⋅
–
[ ]
----------------------------------
-
= =
c 1
– β t′
( )φ t x
,
( )
=

7.7 Determination of the electromagnetic field from the Lienard-Wierchert potentials
To determine the electric and magnetic fields we need to determine
The potentials depend directly upon and indirectly upon through the dependence upon . Hence
we need to work out the derivatives of with respect to both and .
Expression for
Since,
E gradφ
–
t
∂
∂A
–
=
B curlA
=
x x t
, t′
t′ t x
t′
∂
t
∂
-----
-
t′
x X t′
( )
–
c
------------------------
-
+ t
=

We can determine by differentiation of this implicit equation.
Solving for :
t′
∂
t
∂
-----
-
t′
∂
t
∂
-----
-
1
c
--
-
t
∂
∂
x X t′
( )
–
( )
+ 1
=
t′
∂
t
∂
-----
-
x X t′
( )
–
( )
x X t′
( )
–
--------------------------
-
Ẋ t′
( )
c
------------
-
–
 
  t′
∂
t
∂
-----
-
⋅
+
⇒ 1
=
1
Ẋ t′
( ) x X t′
( )
–
( )
⋅
c x X t′
( )
–
-------------------------------------------
-
–
t′
∂
t
∂
-----
- 1
=
t′
∂ t
∂
⁄
t′
∂
t
∂
-----
-
x X t′
( )
–
x X t′
( )
– Ẋ t′
( ) c
⁄
( ) x X t′
( )
–
( )
⋅
–
--------------------------------------------------------------------------------------
-
=
r′
r′ Ẋ t′
( ) r′
⋅ c
⁄
–
-------------------------------------
-
=
t′
∂
t
∂
-----
-
1
1 β
β
β
β t′
( ) n′
⋅
–
-----------------------------
-
=

Expression for
Again differentiate the implicit function for :
t′
∂
xi
∂
------
- t′
∇
=
t′
t′
∂
xi
∂
------
-
xi Xi t′
( )
–
( )
c x X t′
( )
–
----------------------------
-
xj X j t′
( )
–
( )Ẋ j t′
( ) c
⁄
x X t′
( )
–
----------------------------------------------------
-
t′
∂
xi
∂
------
-
×
–
+ 0
=
t′
∂
xi
∂
------
- 1
βj xj X j t′
( )
–
( )
x X t′
( )
–
------------------------------------
–
xi Xi
–
c x X t′
( )
–
---------------------------
-
+ 0
=
t′
∂
xi
∂
------
-
x X t′
( )
– β t′
( ) x X t′
( )
–
( )
⋅
–
x X t′
( )
–
--------------------------------------------------------------------------
⇒
xi Xi
–
c x X t′
( )
–
---------------------------
-
–
=
t′
∂
xi
∂
------
-
⇒
xi Xi t′
( )
–
( )
– c
⁄
x X t′
( )
– β t′
( ) x X t′
( )
–
( )
⋅
–
--------------------------------------------------------------------------
=
i.e.
t′
∂
xi
∂
------
-
xi′
r′ Ẋ t′
( ) r′
⋅ c
⁄
–
-------------------------------------
-
–
c 1
– ni′
1 β
β
β
β t′
( ) n′
⋅
–
-----------------------------
-
–
= =
or t′
∇
c 1
– n′
1 β
β
β
β t′
( ) n′
⋅
–
-----------------------------
-
–
=

The potentials include explicit dependencies upon the spatial coordinates of the field point and implicit
dependencies on via the dependence on
The derivatives of the potentials can be determined from:
In dyadic form:
t x
,
( ) t′
xi
∂
∂φ
t
xi
∂
∂φ
t′
t′
∂
∂φ
xi
t′
∂
xi
∂
------
-
+
=
Ai
∂
t
∂
-------
-
xi
Ai
∂
t′
∂
-------
-
t′
∂
t
∂
-----
-
=
Ai
∂
xj
∂
-------
-
t
Ai
∂
xj
∂
-------
-
t′
Ai
∂
t′
∂
-------
-
t′
∂
xj
∂
-------
-
+
=
εijk
Ak
∂
xj
∂
--------
-
t
εijk
Ak
∂
xj
∂
--------
-
t′
εijk
t′
∂
xj
∂
-------
-
Ak
∂
t′
∂
--------
-
+
=
φ
∇ t φ t′
∇
t′
∂
∂φ
x
t′
∇
+
= A
∂
t
∂
------
-
x
A
∂
t′
∂
------
-
x
t′
∂
t
∂
-----
-
=
curl A t curl A t′ t′
∇ A
∂
t′
∂
------
-
x
×
+
=

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
The terms
Other useful formulae to derive beforehand are:
In differentiating it is best to express it in the form .
E φ
∇
–
A
∂
t
∂
------
-
– φ t′
∇
–
t′
∂
∂φ
t′
c 1
– β t′
( )φ
[ ]
∂
t′
∂
------------------------------
-
t′
∂
t
∂
-----
-
 
 
–
∇
–
= =
φ t′ t′
∂
∂φ
t′
∇
β
c
--
-
t′
∂
t
∂
-----
-
+
–
φ
c
--
-β̇ t′
( )
t′
∂
t
∂
-----
-
–
∇
–
=
t′
∇
β
c
--
-
t′
∂
t
∂
-----
-
+
1
c
--
-
n′ β
–
( )
1 β n′
⋅
–
---------------------
-
–
=
t′
∂
t
∂
-----
-
1
1 β n′
⋅
–
( )
--------------------------
=
r′
∂
t′
∂
------
- c 1
– β
–
=
r′
∂
t′
∂
------
- c 1
– β n′
⋅
–
=
1
r′ 1 β n′
⋅
–
( )
------------------------------
-
1
r′ β r′
⋅
–
---------------------
-

With a little bit of algebra, it can be shown that
Combining all terms:
The immediate point to note here is that many of the terms in this expression decrease as . However,
the terms proportional to the acceleration only decrease as . These are the radiation terms:
φ
∇ t′
q
4πε0r′2
------------------
-
n′ β
–
1 β n′
⋅
–
( )2
----------------------------
-
–
=
t′
∂
∂φ qc
4πε0r′2
------------------
-
β n′
⋅ β2
– c 1
– r′β
˙ n′
⋅
+
[ ]
1 β n
⋅
–
( )2
-------------------------------------------------------------
=
E
q
4πε0r′2
------------------
-
n′ β
β
β
β
–
( ) 1 β′2
– c 1
– r′β̇ n′
⋅
+
( ) c 1
– r′β
β
β
β̇ 1 β n′
⋅
–
( )
–
[ ]
1 β
β
β
β n′
⋅
–
( )3
-------------------------------------------------------------------------------------------------------------------------------
-
=
r′ 2
–
r′ 1
–
Erad
q
4πcε0r
-----------------
-
n′ β
–
( )β̇ n′
⋅ β̇ 1 β̇ n′
⋅
–
( )
–
[ ]
1 β n′
⋅
–
( )3
-------------------------------------------------------------------------
-
=

Magnetic field
We can evaluate the magnetic field without going through more tedious algebra. The magnetic field is
given by:
where
Now the first term is given by:
and we know from calculating the electric field that
B curl A curl A t′ t′
∇
A
∂
t′
∂
------
-
×
+
= =
curl c 1
– φβ
( ) t′ t′
∇
A
∂
t′
∂
------
-
×
+
=
t′
∇
c 1
– n′
–
1 β n′
⋅
–
---------------------
- c 1
– n′
t′
∂
t
∂
-----
-
×
–
= =
curl c 1
– φβ
( ) c 1
– φ
∇ t′ β
×
=
φ
∇ t′
q
4πε0r′2
------------------
-
n′ β
–
1 β n′
⋅
–
( )2
----------------------------
-
– ξ n′ β
–
( )
= =

Therefore,
Hence, we can write the magnetic field as:
Compare the term in brackets with
Since , then
This equation holds for both radiative and non-radiative terms.
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′
×
=
B c 1
– φ
∇ t′ n′
× c 1
– n′
A
∂
t′
∂
------
-
×
t′
∂
t
∂
-----
-
– c 1
– n′ φ
∇ t′
–
A
∂
t
∂
------
-
–
×
= =
E φ t′
∇
t′
∂
∂φ
x
t′
A
∂
t
∂
------
-
–
∇
+
–
=
t′
∇ n′
∝
B c 1
– n′ E
×
( )
=

Poynting flux
The Poynting flux is given by:
We restrict attention to the radiative terms in which
For the radiative terms,
so that the Poynting flux,
This can be understood in terms of equal amounts of electric and magnetic energy density ( )
moving at the speed of light in the direction of . This is a very important expression when it comes to
calculating the spectrum of radiation emitted by an accelerating charge.
S
E B
×
µ0
-------------
-
E n′ E
×
( )
×
cµ0
------------------------------
- cε0 E2n′ E n′
⋅
( )E
–
[ ]
= = =
Erad r′ 1
–
∝
Erad n′
⋅
q
4πcε0r
-----------------
-
1 β̇ n′
⋅
–
( ) n′ β̇
⋅
( ) n′ β̇ 1 β̇ n′
⋅
–
( )
⋅
–
[ ]
1 β n′
⋅
–
( )3
-----------------------------------------------------------------------------------------------
- 0
= =
S cε0E2n′
=
ε0 2
⁄
( )E2
n′

8 Radiation from relativistically moving charges
Note the factor in the expression
for the electric field. When the
contribution to the electric field is large; this oc-
curs when , i.e. when the angle be-
tween the velocity and the unit vector from the
retarded point to the field point is approximately
zero.
We can quantify this as follows: Let be the angle between and , then
2 γ
⁄
v
Illustration of the beaming of radi-
ation from a relativistically moving
particle.
Trajectory of
particle
1 β n′
⋅
–
( ) 3
–
1 β′ n′
⋅
– 0
≈
β′ n′
⋅ 1
≈
θ β t′
( ) n′
1 β′ n′
⋅
– 1 β′ θ 1 1
1
2γ2
--------
-
–
 
  1
1
2
--
-θ2
–
 
 
–
≈
cos
–
=
1 1
1
2γ2
--------
-
θ2
2
-----
–
–
 
 
–
=
1
2γ2
--------
-
θ2
2
-----
+
=
1
2γ2
--------
- 1 γ2θ2
+
( )
=

So you can see that the minimum value of is and that the value of this quantity only
remains near this for . This means that the radiation from a moving charge is beamed into a nar-
row cone of angular extent . This is particularly important in the case of synchrotron radiation for
which (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, , resulting from a moving charge, at a point in space and rep-
resent it in the form:
where and 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:
1 β′ n′
⋅
– 1 2γ2
( )
⁄
θ 1 γ
⁄
∼
1 γ
⁄
γ 104
∼
E t
( )
E t
( ) E1 t
( )e1 E2 t
( )e2
+
=
e1 e2
Eα ω
( ) eiωtEα t
( ) t
d
∞
–
∞
∫
= Eα t
( )
1
2π
-----
- e iωt
– Eα ω
( ) ω
d
∞
–
∞
∫
=

The condition that be real is that
Note: We do not use a different symbol for the Fourier transform, e.g. . The transformed variable
is indicated by its argument.
9.2 Spectral power in a pulse
Outline of the following calculation
• Consider a pulse of radiation
• Calculate total energy per unit area in the radia-
tion.
• 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:
Eα t
( )
Eα ω
–
( ) Eα
* ω
( )
=
Ẽα ω
( )
Eα t
( )
t
Diagrammatic representation of a
pulse of radiation with a duration
.
∆t
∆t
dW
dtdA
-----------
- Poynting Flux cε0
( )E2 t
( ) cε0
( ) E1
2 t
( ) E2
2 t
( )
+
[ ]
= = =

where and are the components of the electric field wrt (so far arbitrary) unit vectors and
in the plane of the wave.
The total energy per unit area in the –component of the pulse is
From Parseval’s theorem,
The integral from to can be converted into an integral from 0 to using the reality condition. For
the negative frequency components, we have
so that
E1 E2 e1 e2
α
dWαα
dA
--------------
- cε0
( ) Eα
2 t
( ) t
d
∞
–
∞
∫
=
Eα
2 t
( ) t
d
∞
–
∞
∫
1
2π
-----
- Eα ω
( ) 2 ω
d
∞
–
∞
∫
=
∞
– ∞ ∞
Eα ω
–
( ) Eα
* ω
–
( )
× Eα
* ω
( ) Eα ω
( )
× Eα ω
( ) 2
= =
Eα
2 t
( ) t
d
∞
–
∞
∫
1
π
--
- Eα ω
( ) 2 ω
d
0
∞
∫
=

The total energy per unit area in the pulse, associated with the component, is
(The reason for the 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 . That factor results from the time integration of which comes
from, in effect, . This factor, of course, is not evaluated here since the pulse has an arbi-
trary spectrum.]
We identify the spectral components of the contributors to the Poynting flux by:
The quantity represents the energy per unit area per unit circular frequency in the entire pulse,
i.e. we have accomplished our aim and determined the spectrum of the pulse.
α
dWαα
dA
--------------
- cε0 Eα
2 t
( ) t
d
∞
–
∞
∫
cε0
π
-------
- Eα ω
( ) 2 ω
d
0
∞
∫
= =
αα
1 2
⁄ ωt
cos2
Eα ω
( ) 2 ω
d
0
∞
∫
dWαα
dωdA
--------------
-
cε0
π
-------
- Eα ω
( ) 2
=
dWαα
dωdA
--------------
-

We can use this expression to evaluate the power associated with the pulse. Suppose the pulse repeats
with period , then we define the power associated with component by:
This is equivalent to integrating the pulse over, say several periods and then dividing by the length of
time involved.
9.3 Emissivity
Consider the surface to be located a
long distance from the distance over
which the particle moves when emitting
the pulse of radiation. Then
and
T α
dWαα
dAdωdt
-------------------
-
1
T
---
dW
dAdω
--------------
cε0
πT
-------
- Eα ω
( ) 2
= =
dΩ
dA r2dΩ
=
r
Variables used to define the emis-
sivity in terms of emitted power.
Region in which particle moves
during pulse
dA
dA r2dΩ
=
dWαα
dAdωdt
-------------------
-
1
r2
----
-
dWαα
dΩdωdt
--------------------
-
=
dWαα
dΩdωdt
--------------------
-
⇒ r2
dWαα
dAdωdt
-------------------
-
=

The quantity
is the emissivity corresponding to the 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:
The definition of 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.
dWαα
dΩdωdt
--------------------
-
cε0r2
πT
------------- Eα ω
( ) 2
cε0r2
πT
-------------Eα ω
( )Eα
* ω
( )
= = (Summation not implied)
eα
Iω
cε0
πT
-------
- E1 ω
( )E1
* ω
( ) E2 ω
( )E2
* ω
( )
+
[ ]
=
Qω
cε0
πT
-------
- E1 ω
( )E1
* ω
( ) E2 ω
( )E2
* ω
( )
–
[ ]
=
Uω
cε0
πT
-------
- E1
* ω
( )E2 ω
( ) E1 ω
( )E2
* ω
( )
+
[ ]
=
Vω
1
i
--
-
cε0
πT
-------
- E1
* ω
( )E2 ω
( ) E1 ω
( )E2
* ω
( )
–
[ ]
=
Iω

We define a polarisation tensor by:
We have calculated above the emissivities,
corresponding to . More generally, we define:
and these are the emissivities related to the components of the polarisation tensor .
Iαβ ω
,
1
2
--
-
Iω Qω
+ Uω iVω
–
Uω iVω
+ Iω Qω
–
cε0
πT
-------
-Eα ω
( )Eβ
* ω
( )
= =
dWαα
dΩdωdt
--------------------
-
cε0r2
πT
-------------Eα ω
( )Eα
* ω
( )
= (Summation not implied)
EαEα
*
dWαβ
dΩdωdt
--------------------
-
cε0
πT
-------
-r2Eα ω
( )Eβ
* ω
( )
=
Iαβ

In general, therefore, we have
Consistent with what we have derived above, the total emissivity is
and the emissivity into the Stokes is
dW11
dΩdωdt
--------------------
- Emissivity for
1
2
--
- Iω Qω
+
( )
→
dW22
dΩdωdt
--------------------
- Emissivity for
1
2
--
- Iω Qω
–
( )
→
dW12
dΩdωdt
--------------------
- Emissivity for
1
2
--
- Uω iVω
–
( )
→
dW21
dΩdωdt
--------------------
-
dW12
*
dΩdωdt
--------------------
- Emissivity for
1
2
--
- Uω iVω
+
( )
→
=
εω
I
dW11
dΩdωdt
--------------------
-
dW22
dΩdωdt
--------------------
-
+
=
Q
εω
Q
dW11
dΩdωdt
--------------------
-
dW22
dΩdωdt
--------------------
-
–
=

Also, for Stokes and :
Note the factor of in the expression for . In the expression for the –vector of the
radiation field
. Hence and consequently are independent of , consistent with the above expres-
sions for emissivity.
The emissivity is determined by the solution for the electric field.
U V
εω
U
dW12
dΩdωdt
--------------------
-
dW12
*
dΩdωdt
--------------------
-
+
=
εω
V i
dW12
dΩdωdt
--------------------
-
dW12
*
dΩdωdt
--------------------
-
–
 
 
=
r2 dWαβ dΩdωdt
⁄ E
E
q
4πcε0r
-----------------
-
n′ β
–
( ) n′ β̇
⋅
( ) β̇ 1 β̇ n′
⋅
–
( )
–
[ ]
1 β n′
⋅
–
( )3
------------------------------------------------------------------------------
=
E 1 r
⁄
∝ rE r2EαEβ
* r

10 Fourier transform of the Lienard-Wierchert radiation field
The emissivities for the Stokes parameters obviously depend upon the Fourier transform of
where the prime means evaluation at the retarded time given by
The Fourier transform involves an integration wrt . We transform this to an integral over as follows:
using the results we derived earlier for differentiation of the retarded time. Hence,
rE t
( )
q
4πcε0
--------------
-
n′ n′ β′
–
( ) β̇′
×
[ ]
×
1 β′ n′
⋅
–
( )3
------------------------------------------------
-
=
t′
t′ t
r′
c
---
-
–
= r′ x X t′
( )
–
=
t t′
dt
t
∂
t′
∂
-----
-dt′
1
t′
∂ t
∂
⁄
--------------
-dt′ 1 β′ n′
⋅
–
( )dt′
= = =
rE ω
( )
q
4πcε0
--------------
-
n′ n′ β′
–
( ) β̇′
×
[ ]
×
1 β′ n′
⋅
–
( )3
------------------------------------------------
-eiωt 1 β′ n′
⋅
–
( ) t′
d
∞
–
∞
∫
=
q
4πcε0
--------------
-
n′ n′ β′
–
( ) β̇′
×
[ ]
×
1 β′ n′
⋅
–
( )2
------------------------------------------------
-eiωt t′
d
∞
–
∞
∫
=

The next part is
Since
then we expand to first order in . Thus,
Note that it is the unit vector which enters here, rather than the retarded unit vector
Hence,
eiωt iω t′
r′
c
---
-
+
 
 
exp
=
r′ x X t′
( )
– x when x X t′
( )
»
≈
=
r′ X
r′ xj X j t′
( )
–
=
r′
∂
Xi
∂
-------
-
xi Xi t′
( )
–
( )
–
r′
--------------------------------
xi
r
---
- at Xi
– 0
= = =
r′ r r′
∂
Xi
∂
-------
-
X j 0
=
Xi t′
( )
×
+ r
xi
r′
---
-Xi t′
( )
– r niXi
– r n X t′
( )
⋅
–
= = = =
n
r
r
-
-
= n′
iωt
( )
exp iω t′
r
c
-
-
n X t′
( )
⋅
c
--------------------
–
+
 
 
exp iω t′
n X t′
( )
⋅
c
--------------------
–
 
  iωr
c
--------
exp
×
exp
=
= =

The factor is common to all Fourier transforms and when one multiplies by the com-
plex conjugate it gives unity. This also shows why we expand the argument of the exponential to first
order in since the leading term is eventually unimportant.
The remaining term to receive attention in the Fourier Transform is
We first show that we can replace by by also expanding in powers of .
iωr
c
--------
exp rEα ω
( )
X t′
( )
n′ n′ β′
–
( ) β̇′
×
[ ]
×
1 β′ n′
⋅
–
( )2
------------------------------------------------
-
n′ n Xi t′
( )
ni′
xi Xi t′
( )
–
xj X j t′
( )
–
----------------------------
-
xi
r
---
-
X j
∂
∂ xi Xi
–
xj X j t′
( )
–
----------------------------
-
X j 0
=
X j
×
+
= =
xi
r
---
- δij
–
xi Xi
–
( ) xj X j
–
( )
xj X j t′
( )
– 3
------------------------------------------
-
+
X j 0
=
X j
×
+
=
ni δij
–
xixj
r2
---------
+
X j
r
-----
-
+
=

So the difference between and is of order , 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 . Hence we put,
It is straightforward (exercise) to show that
Hence,
One can integrate this by parts. First note that
n n′ X r
⁄
X r
⁄
n′ n′ β′
–
( ) β̇′
×
[ ]
×
1 β′ n′
⋅
–
( )2
------------------------------------------------
-
n n β′
–
( ) β̇′
×
[ ]
×
1 β′ n
⋅
–
( )2
--------------------------------------------
-
=
d
dt′
------
n n β′
×
( )
×
1 β′ n′
⋅
–
----------------------------
n n β′
–
( ) β̇′
×
[ ]
×
1 β′ n
⋅
–
( )2
--------------------------------------------
-
=
rE ω
( )
q
4πcε0
--------------
-e
iωr c
⁄ d
dt′
------
n n β′
×
( )
×
1 β′ n
⋅
–
----------------------------
- iω t′
n X t′
( )
⋅
c
--------------------
–
 
 
exp t′
d
∞
–
∞
∫
=
n n β′
×
( )
×
1 β′ n
⋅
–
----------------------------
-
∞
–
∞
0
=

since we are dealing with a pulse. Second, note that,
and that the factor of cancels the remaining one in the denominator. Hence,
In order to calculate the Stokes parameters, one selects a coordinate system ( and ) in which this is
as straightforward as possible. The motion of the charge enters through the terms involving and
in the integrand.
Remark
The feature associated with radiation from a relativistic particle, namely that the radiation is very strong-
ly peaked in the direction of motion, shows up in the previous form of this integral via the factor
. This dependence is not evident here. However, when we proceed to evaluate the integral
in specific cases, this dependence resurfaces.
d
dt′
------ iω t′
n X t′
( )
⋅
c
--------------------
–
 
 
exp iω t′
n X t′
( )
⋅
c
--------------------
–
 
 
exp iω 1 β′ n
⋅
–
[ ]
×
=
1 β′ n′
⋅
–
[ ]
rE ω
( )
iωq
–
4πcε0
--------------
-e
iωr c
⁄
n n β′
×
( )
× iω t′
n X t′
( )
⋅
c
--------------------
–
 
 
exp t′
d
∞
–
∞
∫
=
e1 e2
β t′
( )
X t′
( )
1 β n′
⋅
–
( ) 3
–

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