Phys.

506 Electricity and Magnetism Winter 2004

Prof. G. Raithel
Problem Set 4
Total 40 Points
1. Problem 9.10 10 Points
a). In the long-wavelength limit, in the source and its immediate vicinity electro- and magnetostatic
equations apply. Thus, with Eq. 5.53 the magnetization density M is, using r z = sin

=
sin ( xsin + y cos ), and v
0
= c
M =
1
2
x J(x)
=
1
2
_
iv
0
2
_
rr
_
r
1
2
+ z
a
0
z
_
=
1
2
_
iv
0
2
_
rr
_
r
1
2
+ z
a
0
z
_
(r, )
=
1
2
_
iv
0
2
_
ra
0
z

sin (r, )
=
1
2
_
ia
0
v
0
2
_

tan (r, )
= ( xsin y cos )
_
ia
0
c
4
_
tan (r, ) = M , q.e.d.
(1)
In the calculation of multipole moments at frequency
0
, we may thus replace the current by the given
eective magnetization density and set J = 0. (Note that both M and J carry a time factor exp(i
0
t),
which is not shown.) Since M is of the form
M =

f(r, )
with a function f that doesnt depend on , it is
M = 0 .
In the long-wavelength limit, for the multipole moments Eqs. 9.169 to 9.172 apply. Thus, with J = 0 and
M = 0 both M
lm
= 0 and M

lm
= 0. There are no magnetic multipoles.
From the orthogonality of the spherical harmonics, the only non-vanishing Q
lm
is
Q
10
=
_

_

0
r
_
2e

6a
4
0
r exp(
3r
2a
0
)Y
00
Y
10
_
Y

10
r
2
drd
=
2e

24a
4
0
_

0
r
4
exp(
3r
2a
0
)dr
=
1

6
256
81
ea
0
(2)
To nd the Q

lm
, we note
(r M) =
ica
0
4
(r

)r tan (r, )
=
ica
0
4
(

r tan (r, )
=
ica
0
4
1
r sin

r sin tan (r, )

(3)
Since the angular dependence of is Y
10
cos , this is
(r M) =
ica
0
4
1
sin
2 sin (r, ) =
ica
0
2
(r, )
From Eq. 9.170 and the previous result on Q
lm
, it is seen that the only non-vanishing Q

lm
is
Q

10
=
ka
0
4
Q
10
=

2
a
0

Q
10
Since the factor on the rhs is of order
1
1000
1
137
and the radiated power behaves as the square of the multipole
moments, we can safely assume
Q

10
= 0
b: Using Eq. 9.169, it is a
E
(1, 0) =
ck
3
3i

2Q
10
. Also, the radiated power P =
Z
0
2k
2
|a
E
|
2
. Inserting the results
of part a), it is
P =
256
2
81
2
9 6
Z
0
c
2
k
4
a
2
0
e
2
1.02 10
9
W
This can be expressed in the required unit, yielding
P =
_
2
3
_
8
h
0
_

4
c
a
0
_
c: The transition rate is
=
P
h
0
=
_
2
3
_
8
_

4
c
a
0
_
Numerically,
= 6.27 10
8
s
1
= (1.59ns)
1
This equals the quantum mechanical decay rate of the hydrogen 2P level.
Note. The only non-zero multipole moment found in the classical calculation conforms with quantum
mechanical selection rules explained in Chapter 9.8. First, in a transition from an upper 2P level into a
lower 1S level the atomic angular momentum changes from 1 to 0 (with spin neglected). Thus, only l = 1
radiation can occur. Further, the transition from the 2P level into 1S reverses the parity of the atomic state,
requiring an emission eld mode with odd parity (that is, odd magnetic eld). This only leaves electric l = 1
decay modes. Finally, in the given example both the upper and lower states have zero zangular momentum.
Thus, the emitted eld cannot carry any zangular momentum. In summary, the only multipole eld allowed
by selection rules is the a
E
(l = 1, m = 0), as found above.
d: According to an earlier homework problem, for an elementary charge orbiting in the xy-plane at a radius
2a
0
, the only radiation multipole moment for dipole radiation is
Q
11
= 2

Q
11
where

Q
11
is a usual spherical multipole evaluated in the rotating frame. Here,

Q
11
= 2ea
0
_
3
8
exp(i
0
)
with a phase
0
that we may set to zero. Thus,
Q
11
= 4
_
3
8
ea
0
leading to a radiated power of
P
cl
=
_
3
2
2
6
_
h
0
_

4
c
a
0
_
The ratio of this classical power and the quantum power of part b) is
P
cl
P
qm
=
3
10
2
14
= 3.60
2. Problem 9.16 10 Points
a): In this problem, a calculation in cartesian coordinates is the most straightforward. The current density
is
J(x) = zI(x)(y) sin(kz)
for |z| /2. The radiation pattern is only relevant in the radiation zone. Thus, we calculate
A(x) =

0
4
exp(ikr)
r
_
source
exp(ik

n

)J(x

)dxdydz
= z

0
4
I
2i
exp(ikr)
r
_
z=+/2
z=/2
exp(ikz cos ) (exp(ikz) exp(ikz)) dz
= z

0
4
I
2i
exp(ikr)
r
_
1
ik(1 cos )
exp(ikz(1 cos )) +
1
ik(1 + cos )
exp(ikz(1 + cos ))
_
z=+/2
z=/2
= z

0
4
I
2k
2i exp(ikr)
r
_
sin((1 + cos ))
(1 + cos )

sin((1 cos ))
(1 cos )
_
= z

0
I
2
exp(ikr)
ikr
_
sin( cos )
sin
2

_
In the radiation zone, H =
ik

0
n A, and with r z = sin

H =

I
2
exp(ikr)
r
_
sin( cos )
sin
_
The radiation pattern is
dP
d
= r
2 1
2Z
0
E E

= r
2 Z
0
2
H H

, yielding
dP
d
=
|I|
2
Z
0
8
2
_
sin
2
( cos )
sin
2

_
The result is exact in the radiation-zone limit, kr 1. For the plot, see Problem 9.17.
b): The radiated power P
P =
|I|
2
Z
0
8
2
_
1
1
sin
2
( cos )
sin
2

2d cos
=
|I|
2
Z
0
4
_
1
1
1 cos
2
(x)
1 x
2
dx
=
|I|
2
Z
0
4
1.55718
Sine the radiation resistance is dened via P =
1
2
R
rad
|I|
2
, it is
R
rad
=
Z
0
2
1.55718 = 93.36
3. Problem 9.17 10 Points
a): We use Eqs. 9.167 and 9.168 to obtain multipole moments that are NOT in the small-source approxi-
mation. Since Eqs. 9.167f are processed most eciently in spherical coordinates, we use
J(x) = r
I(r)
2r
2
[(cos 1) + (cos + 1)]
and
(x) =
1
2ir
2
dI(r)
dr
[(cos 1) + (cos + 1)]
with I(r) = I sin(kr) for 0 < r < /2 and zero otherwise. It is easily veried that the continuity equation,
J = i, holds. Since there is no intrinsic magnetization M and since at all locations r where there is
current owing it is rJ = 0, the magnetic moments all vanish. From Eq. 9.167 we nd the electric-multipole
amplitudes
a
E
(l, m) =
k
2
i
_
l(l + 1)
_
Y

l,m
_
c
_
d
dr
rj
l
(kr)
_
+ ik(r J)j
l
(kr)
_
r
2
drd
=
k
2
i
_
l(l + 1)
1
2
__
Y

l,m
[(cos 1) + (cos + 1)] d
_

__
c
i
_
dI
dr
__
d
dr
rj
l
(kr)
_
+ ikrI(r)j
l
(kr)dr
_
=
k
2
_
l(l + 1)
{Y
l,0
( = 0) + Y
l,0
( = )}
m,0
__
1
k
_
dI
dr
__
d
dr
rj
l
(kr)
_
krI(r)j
l
(kr)dr
_
=
m,0

l,even
2k
_
l(l + 1)
_
2l + 1
4
__ _
dI
dr
__
d
dr
rj
l
(kr)
_
k
2
rI(r)j
l
(kr)dr
_
=
m,0

l,even
2k
_
l(l + 1)
_
2l + 1
4
__ _
d
dr
_
rj
l
(kr)
dI
dr
__
rj
l
(kr)
_
d
2
I
dr
2
_
k
2
rI(r)j
l
(kr)dr
_
=
m,0

l,even
2k
_
l(l + 1)
_
2l + 1
4
_
_
rj
l
(kr)
dI
dr
_
L
0

_
L
0
rj
l
(kr)
_
d
2
I
dr
2
+ k
2
I(r)
_
dr
_
where the antenna half-length L = /2. We also use the denition
l,even
= 1 for even l and
l,even
= 0 for
odd l. For the given I(r) = I sin(kr) it is
d
2
I
dr
2
+ k
2
I(r) = 0, and
a
E
(l, m) =
m,0

l,even
2Ik
_
l(l + 1)
_
2l + 1
4
[rj
l
(kr)k cos(kr)]
/2
0
=
m,0

l,even
2Ik
_
l(l + 1)
_
2l + 1
4

2
j
l
()k
=
m,0

l,even
Ikj
l
()

(2l + 1)
l(l + 1)
(4)
In the long-wavelength approximation, we use Eq. 9.169-9.172. We already note that the long-wavelength
approximation cannot be expected to be tremendously accurate in the given case, because the antenna length
is not small compared with the wavelength.
As before, all moments vanish except the Q
l,0
with even l. It is
Q
l,m
=
_
Y

l,m
r
l
d
3
x
=
1
2i
__
r
l
_
dI(r)
dr
_
dr
___
Y

l,m
[(cos 1) + (cos + 1)] d
_
=
Ik
2i
_
_
L
0
r
l
cos(kr)dr
__
4
_
2l + 1
4

m,0

l,even
_
=
m,0

l,even
2Ik
i
_
2l + 1
4
_
_
L
0
r
l
cos(kr)dr
_
With a
E
(l, m) =
ck
l+2
i(2l+1)!!
_
l+1
l
Q
lm
, the electric-multipole amplitudes are, in the long-wavelength limit,
a
E
(l, m) =
m,0

l,even
2Ik
l+2
(2l + 1)!!
_
l + 1
l
_
2l + 1
4
_
_
L
0
r
l
cos(kr)dr
_
b): The exact lowest non-vanishing amplitude is
a
E
(2, 0) = Ikj
2
()
_
5
6
Using only this moment, the radiated power is
P =
Z
0
2k
2
|a
E
(2, 0)|
2
=
1
2
_
Z
0
5
6
|j
2
()|
2
_
|I|
2
=
1
2
_
Z
0
5
6
9

4
_
|I|
2
=
1
2
_
15Z
0
2
3
_
|I|
2
The numerical value for the radiation resistance (term in rectangular brackets) is
R
rad
=
_
15Z
0
2
3
_
= 91.12
The radiation pattern follows from Eq. 9.151 and Table 9.1,
dP
d
=
Z
0
2k
2
|a
E
(2, 0)|
2
|X
2,0
|
2
=
1
2
R
rad
|I|
2
15
8
sin
2
cos
2

The lowest non-vanishing amplitude in the long-wavelength approximation is

a
E
(2, 0) =
2Ik
4
15
_
15
8
_
_
/2
0
r
2
cos(kr)dr
_
=
Ik
4

30
_
1
k
3
_

0
x
2
cos(x)dx
_
=
2Ik

30
= Ik
_
2
15
Using only this moment, the radiated power is
P =
Z
0
2k
2
|a
E
(2, 0)|
2
=
1
2
_
2Z
0

15
_
|I|
2
The numerical value for the radiation resistance (term in rectangular brackets) is
R
rad
=
_
2Z
0

15
_
= 157.8
The radiation pattern follows from Eq. 9.151 and Table 9.1,
dP
d
=
Z
0
2k
2
|a
E
(2, 0)|
2
|X
2,0
|
2
=
1
2
R
rad
|I|
2
15
8
sin
2
cos
2

Discussion of 9.16 and 9.17.

The radiation resistances found are
R
1
= R
rad,exact
= 93.36
R
2
= R
rad,a20,exact
= 91.12
R
3
= R
rad,a20,approx
= 157.8
It is R
2
< R
1
. This is to be expected, because the total radiated powers of multipoles add incoherently.
Thus, by neglecting higher exact multipoles we will slightly underestimate the radiated power, which is
equivalent to underestimating the radiation resistance. In the given case, from R
1
and R
2
it follows that by
neglecting higher-order exact multipoles we underestimate the radiated power by 2.4% (this is not so bad).
It is R
3
>> R
1
. This is not unexpected, because by making the small-source approximation we essentially
neglect destructive interference of radiation originating from dierent portions of the source. The destructive
interference reduces the radiation eciency of sources that are not much smaller than the wavelength. In the
case of large sources, neglecting this destructive interference can lead to gross overestimates of the radiated
power, as in our case.
Figure 1: Radiation patterns for the indicated cases. Bold and solid: exact calculation. Solid: Lowest
exact multipole term (this term is due to a
E
(2, 0)). Dashed: Same multipole term in the long-wavelength
approximation.
4. Problem 9.22 10 Points
a): Electric-multipole modes = TM modes. We use Eq. 9.122 as starting point. Since the elds must be
regular at r = 0, we choose j
l
(kr) for all radial functions. The generic form of the eld of a TM
lm
-mode,
with amplitude a
E
(l, m) set to 1, then is
H = j
l
(kr)X
lm
E =
iZ
0
k
H =
iZ
0
k
j
l
(kr)X
lm
The boundary conditions are that at r = a the electric eld must only have a radial component and the mag-
netic eld must be transverse. The second condition is automatically satised because of the transversality
of the X
lm
. To match the rst, we use
X
l,m
=
1
_
l(l + 1)

LY
l,m
=
1
_
l(l + 1)
1
i
_

1
sin

_
Y
l,m
to rst write out the Held components,
H
r
= 0
H

=
m
_
l(l + 1)
j
l
(kr)
1
sin
Y
l,m
H

=
1
i
_
l(l + 1)
j
l
(kr)

Y
l,m
Then, the electric-eld components follow from E =
iZ
0
k

_

+

H

_
,
E
r
=
iZ
0
k
1
r sin
[

sin H

]
=
iZ
0
k
1
i
_
l(l + 1)
j
l
(kr)
r sin
_

sin

+ im

1
sin
_
Y
l,m
=
Z
0
k
1
_
l(l + 1)
j
l
(kr)
r
_
1
sin

sin

m
2
sin
2

_
Y
l,m
= Z
0
_
l(l + 1)
j
l
(kr)
kr
Y
l,m
E

=
iZ
0
k
1
r

r
rH

=
Z
0
_
l(l + 1)
1
kr
_
d
dr
rj
l
(kr)
_
[

Y
l,m
]
E

=
iZ
0
k
1
r

r
rH

=
iZ
0
m
_
l(l + 1)
1
kr
_
d
dr
rj
l
(kr)
_ _
1
sin
Y
l,m
_
The cavity frequencies follow from the requirement E

= E

= 0 at r = a. The frequencies can be obtained

from the transcendental equation
_
d
dr
rj
l
(kr)
_
r=a
=
_
d
dx
xj
l
(x)
_
x=ka
= 0
Denoting the n-th root of
d
dx
(xj
l
(x)) with x

ln
, it is ka =
lmn
a
c
= x

ln
. The resonance frequencies thus are

lmn
=
x

ln
c
a
Note that l = 0 does not exist, and that the frequencies are degenerate in m, i.e. for given l and n there are
2l + 1 TM-modes with the same frequency.
Magnetic-multipole modes = TE modes. We use Eq. 9.122 as starting point. The generic form of the eld
of a TE
lm
-mode, with amplitude a
M
(l, m) set to 1, then is
H =
i
k
j
l
(kr)X
lm
E = Z
0
j
l
(kr)X
lm
Comparison with the analogous equation for TM-modes shows that the elds of the TE-modes are obtained
by replacing the former H with E/Z
0
and the former E with Z
0
H. Thus, for TE-modes it is
E
r
= 0
E

=
Z
0
m
_
l(l + 1)
j
l
(kr)
1
sin
Y
l,m
E

=
Z
0
i
_
l(l + 1)
j
l
(kr)

Y
l,m
and
H
r
=
_
l(l + 1)
j
l
(kr)
kr
Y
l,m
H

=
1
_
l(l + 1)
1
kr
_
d
dr
rj
l
(kr)
_
[

Y
l,m
]
H

=
im
_
l(l + 1)
1
kr
_
d
dr
rj
l
(kr)
_ _
1
sin
Y
l,m
_
The conditions of vanishing transverse electric and vanishing normal magnetic eld at r = a are satised via
the transcendental equation
j
l
(ka) = 0
Denoting the n-th root of j
l
(x) with x
ln
, it is ka =
lmn
a
c
= x
ln
. The resonance frequencies thus are

lmn
=
x
ln
c
a
Again, l = 0-modes dont exist, and for given l and n there are 2l + 1 TE-modes with the same frequency.
b): (required for TE-modes only). From
lmn
=
2c

lmn
=
x
ln
c
a
we see that

lmn
a
=
2
x
ln
Numerically we nd the lowest roots of spherical Bessel functions to be x
11
= 4.493, x
21
= 5.763, x
31
= 6.988
and x
12
= 7.725. The lowest four TE-modes therefore are:
l n

lmn
a
1 1 1.398
2 1 1.090
3 1 0.899
1 2 0.813
Figure 2: Lowest spherical Bessel functions and their roots.
c):
The lowest TE-modes are the degenerate TE
l=1,m=1,n=1
, TE
l=1,m=0,n=1
and TE
l=1,m=1,n=1
-modes. To
obtain their elds, use the above general equations for the TE-elds to obtain:
l = 1, m = 1:
E
r
= 0
E

=
Z
0

2
_
3
8
j
1
(
x
11
a
r) exp(i)
E

=
Z
0
i

2
_
3
8
j
1
(
x
11
a
r) cos exp(i)
H
r
=

2
_
3
8
j
1
(
x
11
a
r)
x
11
a
r
sin exp(i)
H

=
1

2
_
3
8
a
x
11
r
_
d
dr
rj
1
(
x
11
a
r)
_
cos exp(i)
H

=
i

2
_
3
8
a
x
11
r
_
d
dr
rj
1
(
x
11
a
r)
_
exp(i)
l = 1, m = 0:
E
r
= 0
E

= 0
E

=
Z
0
i

2
_
3
4
j
1
(
x
11
a
r) sin
H
r
=

2
_
3
4
j
1
(
x
11
a
r)
x
11
a
r
cos
H

=
1

2
_
3
4
a
x
11
r
_
d
dr
rj
1
(
x
11
a
r)
_
sin
H

= 0
l = 1, m = 1:
E
r
= 0
E

=
Z
0

2
_
3
8
j
1
(
x
11
a
r) exp(i)
E

=
Z
0
i

2
_
3
8
j
1
(
x
11
a
r) cos exp(i)
H
r
=

2
_
3
8
j
1
(
x
11
a
r)
x
11
a
r
sin exp(i)
H

=
1

2
_
3
8
a
x
11
r
_
d
dr
rj
1
(
x
11
a
r)
_
cos exp(i)
H

=
i

2
_
3
8
a
x
11
r
_
d
dr
rj
1
(
x
11
a
r)
_
exp(i)