Phys 561 Classical Electrodynamics
Midterm
Taner Akgün
Department of Astronomy and Space Sciences
Cornell University
October 23, 2000
Problem 1
An electric dipole of dipole moment p, fixed in direction, is located at a position r0 (t) with
respect to the origin. Its velocity, v = dr0 /dt is non-relativistic.
a. Show that the charge and current densities associated with the dipole can be expressed
as
ρ(x, t) = −(p · ∇)δ(x − r0 (t)); J(x, t) = −v(p · ∇)δ(x − r0 (t))
b. Show that the dipole gives rise to a magnetic dipole moment and an electric quadrupole
moment. Find them. (There are, of course, still higher moments.)
(a)
The averaged charge density of a medium can be written as:
< η(x, t) > = ρf ree (x, t) + ρbound (x, t) − ∇ · P(x, t) + ...
(1)
The first term accounts for the presence of free charges. The second is an average over all
bound charges, i.e. atoms, ions or molecules. Since the averaging procedure is usually taken
to be over a volume of the order of a molecule, or a few molecules, this second term vanishes
for a medium which is otherwise neutral (apart from the possible presence of free charges).
It is possible to combine the first two terms under the name external charge density, ρext
as well. The third term comes about from the presence of dipole moments (and sure, there
are higher order terms corresponding to higher moments). P is the polarization field and by
definition: 1
P(x, t) =
∑
pi δ(x − xi )
i
1
The interested reader is referred to Chapter VI of Jackson [1].
1
(2)
Phys 561 Classical Electrodynamics
2
Assume that we have no external charges and only a continuous polarization field. The dipole
moment of a small cell of volume Ωi can then be calculated to be (we choose to denote the
total charge density by ρ):
∫
pi =
Ωi
3 ′ ′
′
d x x ρ(x , t) = −
∫
Ωi
3 ′ ′
′
′
d x x ∇ · P(x , t) =
∫
d3 x′ P(x′ , t)
(3)
Ωi
Let’s now return to the problem. Since we have no external charges, and effectively a
single dipole located at r0 (t), we can write the polarization field as:
P(x, t) = pδ(x − r0 (t))
(4)
We are also given that the dipole moment is not a function of position. Hence:
ρ = −∇ · P = −∇ · (pδ) = −δ∇ · p − p · ∇δ = −(p · ∇)δ(x − r0 (t))
(5)
For the current density we simply receive:
J = vρ = −v(p · ∇)δ(x − r0 (t))
(6)
(b)
We are to calculate the magnetic dipole moment and the electric quadrupole moment. These
are defined through the following equations, respectively:
m(x, t) =
Qij (x, t) =
1 ∫ 3 ′ ′
d x x × J(x′ , t)
2c
∫
d3 x′ (3x′i x′j − δij x′2 )ρ(x′ , t)
(7)
(8)
We will begin with the magnetic dipole. However, first let’s consider the following integral:
∫
d3 x G f · ∇δ(x − r0 )
(9)
where whether G is a scalar, vector or tensor function is immaterial. We rewrite the integral
as:
∫
d3 x G fi
∂
δ(x − r0 )
∂xi
Note that there are three terms in the integration. Consider just the ith term (i.e. no assumed
summation), and for convention let δi denote δ(xi − r0i ):
∫
∂
δi δj δk =
∂xi
(
)
∫
∫
∂
=
dxj dxk Gfi δi δj δk − dxi
(G fi )δi δj δk
∂xi
∂
= −
(G fi )
∂xi
x=r0
dxi dxj dxk G fi
Let’s now switch the summation on, and also assume that the vector f is fixed (i.e. no space
variation, while time variation is still allowed):
∂fi ∂
∂G
∂G (G fi )
−G
−
= − fi
= − fi
= − f · ∇G|x=r0
∂xi
∂x
∂x
∂x
i
i
i
x=r0
x=r0
x=r0
(10)
Phys 561 Classical Electrodynamics
3
We have found an important result, which can immediately be tested on the electric dipole
moment. Recall that the electric dipole is defined in terms of the charge density in the
following way:
∫
p(x, t) =
d3 x′ x′ ρ(x′ , t)
(11)
Using the result for the charge density obtained in part-a, and the above result for the
intergral with G replaced by x′ and f replaced by −p we get:
p(x, t) = −
∫
d3 x′ x′ p · ∇δ(x − r0 (t))
= p · ∇x|x=r0 = pi ∇i x|x=r0 = pi x̂i |x=r0 = p
Confident in what we have so far been doing, we can now address the problem of calculating the magnetic dipole moment:
m(x, t) =
=
=
=
=
=
=
1 ∫ 3 ′ ′
d x x × J(x′ , t)
2c ∫
1
−
d3 x′ x′ × v(p · ∇)δ(x − r0 (t))
2c
1
p · ∇(x × v)
2c
x=r0
1
pi ∇i (x × v)
2c
x=r0
1
1
pi (∇i x) × v + pi x × (∇i v)
2c
2c
x=r0
1
1
pi x̂i × v + x × (pi ∇i v)
2c
2c
x=r0
1
1
p × v + x × [(p · ∇)v]
2c
2c
x=r0
(12)
That the second term of this equation vanishes can be best seen by noting that at any instant
in time r0 is fixed in space (or has no gradient). This is true for the present case since r0 is
a function of time only, and not a function of position. Thus:
∂
∂ ∂r0 (t)
∂ ∂r0 (t)
v(t) =
=
= 0
∂xi
∂xi ∂t
∂t ∂xi
We therefore obtain the following result for the magnetic dipole:
m(t) =
1
p × v(t)
2c
(13)
The magnetic dipole is only time dependent through the velocity.
We can see that this result is true by performing the calculations in completely different
way. We can treat a dipole as two particles of charge q and −q, respectively, separated in
space by a vector d, pointing from the minus charge to the plus charge. Specifically, suppose
that the positive charge is located at R + d/2 and that the negative charge is located at
R − d/2, where R is the position of center of charge (whereby we do not require the particles
to posses equal amounts of mass). Then the charge density of the system can be conveniently
written as a sum of Dirac delta functions:
ρ = qδ(x − R − d/2) − qδ(x − R + d/2)
(14)
Phys 561 Classical Electrodynamics
4
The electric dipole moment is simply:
∫
p =
d3 x′ x′ ρ(x′ , t)
= q(R + d/2) − q(R − d/2) = qd
(15)
as one would expect. The magnetic dipole moment can be similarly calculated:
∫
m(x, t) =
=
=
=
1
d3 x′ x′ × J(x′ , t)
2c ∫
1
d3 x′ x′ × vρ(x′ , t)
2c
1
(q(R + d/2) × v − q(R − d/2) × v)
2c
1
p×v
2c
(16)
We have only assumed that the particles are moving with the same velocity, an assumption
which is very proper for the present problem. We have made no specific requirements for
the separation distance, d between the charges; in fact it could be that the separation is
infinitesimally small.
We have thus shown the validity of our previous calculations, and are now in a position
to attack the problem of the electric quadrupole moment. Let’s rewrite the equation in the
form:
Q =
∫
d3 x′ G ρ
(17)
where Q and G are 3 × 3 tensors, and:
G = 3xxT − I|x|2


2x2 − y 2 − z 2
3xy
3xz

3xy
−x2 + 2y 2 − z 2
3yz
= 


2
2
2
3xz
3yz
−x − y + 2z
(18)
Note the following in the above equation. First, I denotes the identity matrix. Second, the
first term appearing in the definition of G involves a dyadic vector multiplication, where the
superscript T stands for transpose.
By using the usual results, we receive:
Q = −
∫
d3 x′ G(p · ∇)δ(x − r0 )
= (p · ∇)G|x=r0
Now consider the elements of Q separately:
Qij =
=
=
=
=
(p · ∇)Gij
pk ∇k (3xi xj − δij xl xl )
pk (3xi ∇k xj + 3xj ∇k xi − 2δij xl ∇k xl )
pk (3xi δjk + 3xj δik − 2δij xl δkl )
3xi pj + 3xj pi − 2δij xl pl
(19)
Phys 561 Classical Electrodynamics
5
Calculating the terms and writing them in matrix form we get:
Q(t) = 3xpT + 3pxT − 2IxT p


4xpx − 2ypy − 2zpz
3xpy + 3ypx
3zpx + 3xpz


3xpy + 3ypx
−2xpx + 4ypy − 2zpz
3ypz + 3zpy
= 
 (20)
3zpx + 3xpz
3ypz + 3zpy
−2xpx − 2ypy + 4zpz
where it is to be understood that the vector x is now actually r0 (t).
Problem 2
A particle of charge e and mass m moves with speed v, v/c ≪ 1 in a uniform magnetic field
B. Suppose the motion is confined to the plane perpendicular to B.
a. Calculate the power radiated, P in terms of B and v, and show that:
P =−
dE
= γE,
dt
where E is the energy of the particle, and find γ. This gives
E(t) = E(0)e−γt ,
1/γ is the mean lifetime of the motion. For an electron, find 1/γ in seconds for a magnetic
field of 104 Gauss.
b. Describe qualitatively the trajectory of the particle as a function of time.
(a)
We shall make use of the Liénard result (1898)2 :
[
P =
2e2
1
1
2
(
v̇)
−
(v̇ × v)2
3c3 (1 − β 2 )3
c2
]
(21)
Since we can expand the acceleration into parallel and perpendicular components with respect to the velocity, we can write:
(v̇)2 −
1
1
(v̇ × v)2 = (v̇|| )2 + (v̇⊥ )2 − 2 (v̇⊥ )2 (v)2 = (v̇|| )2 + (v̇⊥ )2 (1 − β 2 )
2
c
c
(22)
whence the Liénard result becomes:
P =−
[
]
2e2
1
dE
2
2
2
= 3
(
v̇
)
+
(
v̇
)
(1
−
β
)
⊥
||
dt
3c (1 − β 2 )3
(23)
The minus sign appears since we choose to define E as the energy of the particle, and P
is just the rate of decrease of this energy. In the non-relativistic limit (β ≪ 1) and for
acceleration perpendicular to velocity, as in the present case, we receive:
P =
2
See for instance Jackson [1], eq.14.26.
2e2
(v̇⊥ )2
3
3c
(24)
Phys 561 Classical Electrodynamics
6
The acceleration can be found to be:
a = v̇⊥ =
F
ev × B
=
m
mc
(25)
Since we assume that the motion is confined to a plane perpendicular to the magnetic
field, the magnitude of acceleration is just:
(
)
eB
a=
v
mc
(26)
It is interesting to note that the proportionality constant between speed and radius of trajectory is also the same as that between acceleration and speed:
(
)
eB
v=
R
mc
(27)
2e4 B 2 2
v
3m2 c5
(28)
The result for the power is then:
P =
while the energy of the particle is given through:
1
E = mv 2
2
(29)
We thus receive:
P =−
dE
4e4 B 2
=
E = γE
dt
3m3 c5
(30)
with an obvious definition for the constant γ. A quick dimensional check will certify that γ
indeed has the proper dimensions of inverse time. The solution of this differential equation
is trivial:
E(t) = E(0)e−γt
(31)
For the specific values of:
e
B
m
c
=
=
=
=
4.8 × 10−10 esu
104 Gauss
9 × 10−28 g
3 × 1010 cm/s
we get:
3m3 c5
1
= 4 2 = 2.50sec
γ
4e B
(32)
Phys 561 Classical Electrodynamics
7
1
0.8
0.6
0.4
0.2
0
1
0.2
0.4
0.6
0.8
1
Figure 1: The spiral trajectory for highly dissipative motion (λ = 0.07rad−1 ). This can be
achieved, for instance, by increasing the magnetic field strength, B, increasing the charge, e
or decreasing the mass, m of the particle.
(b)
The particle will slow down since:
1
E(t) = mv 2 (t) = E(0)e−γt
2
i.e.
v(t) = v(0)e−γt/2
(33)
As it slows down the radius of the trajectory will decrease (see eq.27) and the particle
will spiral down to the center and will eventually come to rest at the center. In fact since
there is a constant ratio between the speed and the radius, we have:
R(t) = R(0)e−γt/2
(34)
In terms of the polar angle:
θ = ωt =
vt
eB
=
t
R
mc
(35)
we get the following equation in polar coordinates:
R(θ) = R(0)e−λθ
(36)
where:
λ=
2e3 B
1 4e4 B 2 mc
=
≈ 1.1 × 10−12
2 3m3 c5 eB
3m2 c4
(37)
Phys 561 Classical Electrodynamics
8
We have calculated λ for the above sample values. Note that this is an extremely slow decay
(fig.1) in polar angle, while it is rather fast in time (i.e. it takes many revolutions in a very
short time to decay to the center).
Problem 3
Consider an antenna described by the current density
Jx = Jy = 0
(
)
1
l
Jz = I sin kl − k|z| δ(x)δ(y) sin ωt, |z| <
2
2
ω
2π
k= =
c
λ
a. Compute the time-averaged power radiated into a given solid angle dP/dΩ.
b. Show that the power radiated vanishes whenever
(
4πn
kl
cos θ = ± 1 −
)
n = 1, 2, ...
4πn and 1 −
≤1
kl
where θ is the angle between n̂, the direction of observation and ẑ.
(a)
We will make use of the following equations for the radiated power:


cos ϕ sin θ
dP
1


2
=
(n̂ × c) , with n̂ =  sin ϕ sin θ 
3
dΩ
4πc
cos θ
(
∫
R n̂ · x′
d x J̇ x , t − +
c
c
3 ′
and c =
)
′
(38)
We begin by taking the time derivative of the current density and substituting it into the
integral for c. We also perform the trivial x and y integrations (i.e. replace x and y by zero):
∫
l/2
(
)
(
ωR zω cos θ
1
c = ẑIω
dz sin kl − k|z| cos ωt −
+
2
c
c
−l/2
)
We now invoke an old friend from high school:
cos(A + B) = cos A cos B − sin A sin B
(39)
We thus obtain:
∫
l/2
(
)
(
)
(
)
ωR
1
zω cos θ
c = ẑIω
dz sin kl − k|z| cos ωt −
cos
2
c
c
−l/2
(
)
(
)
(
)
∫ l/2
ωR
1
zω cos θ
− ẑIω
dz sin kl − k|z| sin ωt −
sin
2
c
c
−l/2
Phys 561 Classical Electrodynamics
9
Now, the second integral vanishes since it is the integral of an odd function (or rather the
integral of the multiplication of an odd function with an even function) over a symmetric
interval around the origin. Hence, reducing the interval of integration in half by making use
of the symmetry we get:
∫
l/2
c = 2ẑIω
0
(
)
(
)
(
1
ωR
zω cos θ
dz sin kl − kz cos ωt −
cos
2
c
c
)
We now invoke a second good old friend (from the times when mathematics was simple, and
so was the universe):
sin A cos B =
1
[sin(A + B) + sin(A − B)]
2
(40)
Using this equation in the integrand, we receive:
[
(
)
(
)]
)
ωR ∫ l/2
1
zω cos θ
1
zω cos θ
c = ẑIω cos ωt −
dz sin
kl − kz +
+ sin
kl − kz −
c
2
c
2
c
0
(
We can now safely carry out the integration:
(
)
(
) l/2

1
zω cos θ
) cos 1 kl − kz + zω cos θ
cos
kl
−
kz
−
ωR 
2
c
2
c

ẑIω cos ωt −
+
ω cos θ
ω cos θ
(
c =
c
(
)

(
(
){
(
k−
lk cos θ
2
)
c
(
k+
( )
c
(
lk
2
(0 )

cos
cos
cos lk2
ωR  cos

+
−
−
= ẑIω cos ωt −
c
k(1 − cos θ) k(1 + cos θ) k(1 − cos θ) k(1 + cos θ)
)
lk cos θ
2
)
)} [
ωR
lk cos θ
lk
1
1
= ẑIω cos ωt −
cos
− cos
+
c
2
2
k(1 − cos θ) k(1 + cos θ)
)
(
( )}
){
(
2Iω
lk cos θ
lk
ωR
= ẑ
cos
− cos
2 cos ωt −
c
2
2
k sin θ
]
Lastly, let’s remind the reader of a third old -and again good- equation:
(
)
(
A+B
A−B
cos A − cos B = −2 sin
sin
2
2
)
(41)
This equation allows us to finalize our result for c:
(
c = ẑ
)
4Ic
ωR
lk
lk
sin (1 + cos θ) sin (1 − cos θ)
2 cos ωt −
c
4
4
sin θ
(42)
Since c has only z component, the product n̂ × c yields:


sin ϕ sin θ
−
cos ϕ sin θ 
n̂ × c = |c| 


0
(43)
whence the radiated power per solid angle becomes:
dP
1
=
(n̂ × c)2
dΩ
4πc3
1
=
|c|2 sin2 θ
4πc3
(44)
Phys 561 Classical Electrodynamics
10
Let’s take the time averge of the power over a full cycle. The only time dependence comes
about from the term:
(
2
cos
ωR
ωt −
c
)
whose time average is obviously 1/2. We thus obtain the time-averaged power radiated per
solid angle:
dP
2I 2 /πc 2 lk
lk
=
(1 + cos θ) sin2 (1 − cos θ)
sin
2
dΩ
4
4
sin θ
(45)
(b)
Note that the above expression for the power per solid angle vanishes whenever one of the
sine functions goes to zero. This, on the other hand, happens whenever:
lk
(1 ± cos θ) = πn, n = 1, 2, ...
4
This we can rewrite as:
± cos θ =
4πn
−1
kl
or, equivalently:
(
4πn
cos θ = ± 1 −
kl
)
(46)
which is the desired result. By definition the right hand side may not exceed the interval
[−1, 1]. That is:
4πn 1 −
≤1
(47)
kl
This completes the solution.
Problem 4
A particle of charge e moves in a circular path of radius R in the x − y plane with a constant
angular velocity ω0 .
a. Show that the exact expression for the angular distribution of power radiated into the
mth multiple of ω0 is
[
e2 ω04 R2 2  dJm (mβ sin θ)
dPm
=
m
 d(mβ sin θ)
dΩ
2πc3
]2


cot2 θ 2
+
J
(mβ
sin
θ)

β2 m
where β = ω0 R/c, and Jm (x) is the Bessel function of order m.
b. Assume non-relativistic motion and obtain an approximate result for dPm /dΩ for the
value of m that is most important.
c. Assume extreme relativistic motion and obtain the results of Jackson [1], eq.14.79 for
a relativistic particle in instantaneously circular motion.3
3
This is problem 14.15 of Jackson [1].
Phys 561 Classical Electrodynamics
11
(a)
We will make use of an important formula derived in problem 14.13 of Jackson [1]. Here we
quote the result: “... if the motion of a radiating particle repeats itself with periodicity T ,
the continuous frequency spectrum becomes a discrete spectrum containing frequencies that
are integral multiples of the fundamental ... a general expression for the time-averaged power
radiated per unit solid angle in each multiple m of the fundamental frequency ω0 = 2π/T
is:” 4
e2 ω04 m2
dPm
=
dΩ
(2πc)3
∫
[
(
)] 2
2π/ω0
n̂ × x(t)
v(t) × n̂ exp imω0 t −
dt
0
c
(48)
We begin by writing the components of the vectors involved in the above integral:


cos ω0 t


x(t) = R  sin ω0 t  ,
0


− sin ω0 t


v(t) = Rω0  cos ω0 t  ,
0


cos φ sin θ


n̂ =  sin φ sin θ 
cos θ
(49)
The dot and cross products can easily be evaluated:
n̂ · x = R(cos ω0 t cos φ sin θ + sin ω0 t sin φ sin θ) = R cos(φ − ω0 t) sin θ


cos ω0 t cos θ
î
l̂
k̂ 

sin ω0 t cos θ
v × n̂ = Rω0 − sin ω0 t
 (50)
cos ω0 t
0 = Rω0 
cos φ sin θ sin φ sin θ cos θ − cos(φ − ω0 t) sin θ
Since the power will be symmetrically distributed over the azimuthal angle we set, without
loss of generality, φ = 0. Then we have an integral of the following form:


∫
)] 2
[
(
cos ω0 t cos θ
2π/ω0
R cos ω0 t sin θ


∆=
dt
Rω0  sin ω0 t cos θ  exp imω0 t −
0
c
− cos ω0 t sin θ
(51)
corresponding to the magnitude of a vector with three independent components. Switch to
variable ϕ = ω0 t, and set β = ω0 R/c, and x = mβ sin θ. Also define:
∫
I1 =
dϕ cos ϕeimϕ−ix cos ϕ
0
∫
I2 =
2π
2π
dϕ sin ϕeimϕ−ix cos ϕ
(52)
0
Then the above expression (eq.51), written as the sum of the absolute squares of the individual components, looks like:
∆ = R2 |I1 |2 cos2 θ + R2 |I2 |2 cos2 θ + R2 |I1 |2 sin2 θ = R2 |I1 |2 + R2 |I2 |2 cos2 θ
(53)
We therefore need to evaluate the integrals I1 and I2 alone. That we now do. But before
that we remind the reader of a couple of equations regarding Bessel functions. The Bessel
function of order m is given through:
Jm (x) =
4
Jackson [1], p.702.
1 ∫ 2π
dϕeix cos ϕ−imϕ
2πim 0
(54)
Phys 561 Classical Electrodynamics
12
Since Bessel functions are real, this is equivalent to:
Jm (x) =
∗
Jm
(x)
1
=
2π(−i)m
∫
2π
dϕeimϕ−ix cos ϕ
(55)
0
Therefore:
∫
I1 =
2π
∫
2π
dϕ cos ϕeimϕ−ix cos ϕ =
0
dϕ
0
1 d imϕ−ix cos ϕ 2π(−i)m d
e
=
Jm (x)
−i dx
−i dx
(56)
The second integral, I2 is a little more involved but still relatively easy. Define:
u = eimϕ
dv = dϕ sin ϕe−ix cos ϕ
(57)
and use partial integration:
∫
I2 =
udv = uv| −
= eimϕ
∫
vdu
∫
1 −ix cos ϕ 2π im
e
−
ix
ix
0
2π
0
dϕeimϕ−ix cos ϕ = −
2πm(−i)m
Jm (x)
x
(58)
Eq.53 now becomes:
∆ = R2 |I1 |2 + R2 |I2 |2 cos2 θ
(
)2
d
= (2πR)2 
Jm (x)
dx
(
)2 
m cos θ
+
Jm (x)
x

(59)
Substituting this into the equation for the radiated power, and recalling that x = mβ sin θ,
we get:
(
dPm
e2 ω04 R2 2  dJm (mβ sin θ)
=
m
dΩ
2πc3
d(mβ sin θ)
)2

cot2 θ 2
+
J (mβ sin θ)
β2 m
(60)
(b)
In the non-relativistic limit β ≈ 0, so that:
1 ∫ 2π
dϕeimβ sin θ cos ϕ−imϕ
2πim 0
1 ∫ 2π
≈
dϕe−imϕ (1 + imβ sin θ cos ϕ)
m
2πi 0
Jm (mβ sin θ) =
(61)
since ϕ ≫ β cos θ cos ϕ. The first term integrates to zero over the interval. The second term
is also zero as long as m ̸= 1 since exp(−imϕ) is a function of cos mϕ and sin mϕ, and cos mϕ
is orthogonal to cos ϕ unless m = 1, while sin mϕ is always orthogonal to cos ϕ. We therefore
conclude that Jm (x) is nonzero only for m = 1.
Actually, by far, the largest element of the Bessel sequence is J0 (x); though m = 0 is not
physical, and therefore is discarded. See for yourself that the radiated power is zero in that
case. On the other hand, we also assume that m cannot be infinitely large for a physical
system. And, in fact, should be relatively small in the present case.
Phys 561 Classical Electrodynamics
13
′
The derivative Jm
(x) can be analyzed by invoking the following relation, that can be
found in any standard mathematical handbook5 :
dJm (x)
m
m
= −Jm+1 (x) + Jm (x) = − Jm (x) + Jm−1 (x)
dx
x
x
(62)
We observe that Bessel functions can be written in terms of one another in recursion relations.
However, we proved above that only J1 (x) is of significant magnitude for nonrelativistic
speeds, which also implies that the argument x is very small, while we do not expect m to be
′
(x) comes from the mJm (x)/x term, which
too large. Then the greatest contribution to Jm
is largest again for m = 1.
We therefore find that in the nonrelativistic limit, only the fundamental mode counts.
Let’s now calculate the approximate values of the Bessel functions and subsitute them in the
power equation:
1 ∫ 2π
J1 (β sin θ) ≈
dϕe−iϕ iβ sin θ cos ϕ
2πi 0
1
1 ∫ 2π
dϕβ sin θ cos2 ϕ = β sin θ
=
2π 0
2
(63)
and:
J1′ (β sin θ) ≈
1
1
J1 (β sin θ) =
β sin θ
2
(64)
whence:
dP1
e2 ω04 R2
≈
dΩ
2πc3
[( )2
1
2
(
1
+
cos θ
2
)2 ]
=
)
e2 ω04 R2 (
2
1
+
cos
θ
8πc3
(65)
References
[1] Jackson, John David. “Classical Electrodynamics.” John Wiley & Sons, 3rd edition,
1998. (ISBN 0-471-30932-X)
[2] Lecture notes and homeworks.
[3] Abramowitz, M. and I. A. Stegun. “Handbook of Mathematical Functions.” Dover Publications. (ISBN 0-486-61272-4)
[4] Hildebrand, Francis B. “Advanced Calculus for Applications.” Prentice Hall, 2nd edition, 1976. (ISBN 0-13-011189-9)
[5] Watson, G.N. “Theory of Bessel Functions.” Cambridge University Press, 2nd edition,
1952.
5
See, for instance, Abramowitz & Stegun [3].