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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 (x sin + y cos ), and v0 = c

M =1

2x J(x)

=1

2

iv02

rr

r

1

2+ z

a0z

= 12

iv02

rr

r 1

2+ z a

0

z

(r, )

=1

2

iv02

ra0

z sin (r, )

=1

2

ia0v0

2

tan (r, )

= (x sin y cos )

ia0c

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 giveneffective magnetization density and set J = 0. (Note that both M and J carry a time factor exp(i0t),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 Mlm = 0 and Mlm = 0. There are no magnetic multipoles.

From the orthogonality of the spherical harmonics, the only non-vanishing Qlm is

Q10 =

0

r

2e6a40

r exp( 3r2a0

)Y00Y10

Y10r

2drd

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=2e

24a40

0

r4 exp( 3r2a0

)dr

=16

256

81ea0

(2)

To find the Q

lm, we note

(r M) = ica04

(r )r tan (r, )

= ica04

(r tan (r, )

= ica04

1

r sin

r sin tan (r, )

(3)

Since the angular dependence of is Y10 cos , this is

(r M) = ica04

1

sin 2sin (r, ) = ica0

2(r, )

From Eq. 9.170 and the previous result on Qlm, it is seen that the only non-vanishing Q

lm is

Q10 = ka0

4Q10 =

2

a0

Q10

Since the factor on the rhs is of order 11000

1137

and the radiated power behaves as the square of the multipole

moments, we can safely assume

Q10 = 0

b: Using Eq. 9.169, it is aE(1, 0) =ck3

3i

2Q10. Also, the radiated power P =

Z02k2

|aE|2. Inserting the resultsof part a), it is

P =2562

812 9 6 Z0c2k4a20e2 1.02 10

9W

This can be expressed in the required unit, yielding

P =

2

3

8h0

4c

a0

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c: The transition rate is

=P

h0=

2

3

84c

a0

Numerically,

= 6.27 108s1 = (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 field mode with odd parity (that is, odd magnetic field). 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 field cannot carry any zangular momentum. In summary, the only multipole field allowedby selection rules is the aE(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

2a0, the only radiation multipole moment for dipole radiation is

Q11 = 2Q11

where Q11 is a usual spherical multipole evaluated in the rotating frame. Here, Q11 = 2ea0

38

exp(i0)with a phase 0 that we may set to zero. Thus,

Q11 = 4

3

8ea0

leading to a radiated power of

Pcl = 32

26 h04c

a0 The ratio of this classical power and the quantum power of part b) is

PclPqm

=310

214= 3.60

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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) =04

exp(ikr)

r

source

exp(ikn x)J(x)dxdydz

= z04

I

2i

exp(ikr)

r

z=+/2z=/2

exp(ikz cos ) (exp(ikz) exp(ikz)) dz

= z04

I

2i

exp(ikr)

r

1

ik(1 cos ) exp(ikz(1 cos )) +1

ik(1 + cos )exp(ikz(1 + cos ))

z=+/2z=/2

= z 04

I2k

2i exp(ikr)r

sin((1 + cos ))

(1 + cos ) sin((1 cos ))

(1 cos )

= z0I

2

exp(ikr)

ikr

sin( cos )

sin2

In the radiation zone, H = ik0 n A, and with r z = sin

H = I2

exp(ikr)

r

sin( cos )

sin

The radiation pattern isdP

d = r2 1

2Z0 E E

= r2 Z0

2 H H

, yielding

dP

d=|I|2Z0

82

sin2( cos )

sin2

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|2Z0

82

11

sin2( cos )

sin2 2d cos

=|I|2Z0

4

11

1 cos2(x)1 x2 dx

=|I|2Z0

4 1.55718

Sine the radiation resistance is defined via P = 12Rrad|I|2, it is

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Rrad =Z02

1.55718 = 93.36

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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 efficiently in spherical coordinates, we use

J(x) = rI(r)

2r2[(cos 1) + (cos + 1)]

and

(x) =1

2ir2dI(r)

dr[(cos 1) + (cos + 1)]

with I(r) = Isin(kr) for 0 < r < /2 and zero otherwise. It is easily verified that the continuity equation,

J = i, holds. Since there is no intrinsic magnetization M and since at all locations r where there iscurrent flowing it is rJ = 0, the magnetic moments all vanish. From Eq. 9.167 we find the electric-multipoleamplitudes

aE(l, m) =k2

i

l(l + 1)

Yl,m

c

d

drrjl(kr)

+ ik(r J)jl(kr)

r2drd

=k2

i

l(l + 1)

1

2

Yl,m [(cos 1) + (cos + 1)] d

c

i

dI

dr

d

drrjl(kr)

+ ikrI(r)jl(kr)dr

= k2

l(l + 1)

{Yl,0( = 0) + Yl,0( = )} m,0

1

k

dI

dr

d

drrjl(kr)

krI(r)jl(kr)dr

= m,0l,even2k

l(l + 1)2l + 1

4dI

dr d

dr rjl(kr) k2rI(r)jl(kr)dr

= m,0l,even 2kl(l + 1)

2l + 1

4

d

dr

rjl(kr)

dI

dr

rjl(kr)

d2I

dr2

k2rI(r)jl(kr)dr

= m,0l,even 2kl(l + 1)

2l + 1

4

rjl(kr)

dI

dr

L0

L0

rjl(kr)

d2I

dr2+ k2I(r)

dr

where the antenna half-length L = /2. We also use the definition l,even = 1 for even l and l,even = 0 for

odd l. For the given I(r) = Isin(kr) it is d2I

dr2 + k2I(r) = 0, and

aE(l, m) = m,0l,even 2Ikl(l + 1)

2l + 1

4[rjl(kr)k cos(kr)]

/20

= m,0l,even2Ik

l(l + 1)

2l + 1

4

2jl()k

= m,0l,evenIkjl()

(2l + 1)

l(l + 1)

(4)

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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 Ql,0 with even l. It is

Ql,m =

Yl,mr

ld3x

=1

2i

rl

dI(r)

dr

dr

Yl,m [(cos 1) + (cos + 1)] d

=Ik

2i

L0

rl cos(kr)dr

4

2l + 1

4m,0l,even

= m,0l,even2Ik

i

2l + 1

4

L0

rl cos(kr)dr

With aE(l, m) =ckl+2

i(2l+1)!!

l+1

l Qlm, the electric-multipole amplitudes are, in the long-wavelength limit,

aE(l, m) = m,0l,even 2Ikl+2

(2l + 1)!!

l + 1

l

2l + 1

4

L0

rl cos(kr)dr

b): The exact lowest non-vanishing amplitude is

aE(2, 0) = Ikj2()

56

Using only this moment, the radiated power is

P =Z02k2

|aE(2, 0)|2 = 12

Z05

6|j2()|2

|I|2 = 1

2

Z05

6

9

4

|I|2 = 1

2

15Z023

|I|2

The numerical value for the radiation resistance (term in rectangular brackets) is

Rrad =15Z

023

= 91.12

The radiation pattern follows from Eq. 9.151 and Table 9.1,

dP

d=

Z02k2

|aE(2, 0)|2 |X2,0|2 = 12

Rrad |I|2 158

sin2 cos2

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

aE(2, 0) = 2Ik4

15

15

8

/20

r2 cos(kr)dr

=

Ik4

30 1

k3

0

x2 cos(x)dx=

2I k30

= Ik

2

15

Using only this moment, the radiated power is

P =Z02k2

|aE(2, 0)|2 = 12

2Z0

15

|I|2

The numerical value for the radiation resistance (term in rectangular brackets) is

Rrad =

2Z0

15

= 157.8

The radiation pattern follows from Eq. 9.151 and Table 9.1,

dP

d=

Z02k2

|aE(2, 0)|2 |X2,0|2 = 12

Rrad |I|2 158

sin2 cos2

Discussion of 9.16 and 9.17.

The radiation resistances found are

R1 = Rrad,exact = 93.36

R2 = Rrad,a20,exact = 91.12

R3 = Rrad,a20,approx = 157.8

It is R2 < R1. 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 isequivalent to underestimating the radiation resistance. In the given case, from R1 and R2 it follows that by

neglecting higher-order exact multipoles we underestimate the radiated power by 2.4% (this is not so bad).

It is R3 >> R1. This is not unexpected, because by making the small-source approximation we essentially

neglect destructive interference of radiation originating from different portions of the source. The destructive

interference reduces the radiation efficiency 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.

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Figure 1: Radiation patterns for the indicated cases. Bold and solid: exact calculation. Solid: Lowestexact multipole term (this term is due to aE(2, 0)). Dashed: Same multipole term in the long-wavelengthapproximation.

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4. Problem 9.22 10 Points

a): Electric-multipole modes = TM modes. We use Eq. 9.122 as starting point. Since the fields must be

regular at r = 0, we choose jl(kr) for all radial functions. The generic form of the field of a T Mlm-mode,

with amplitude aE(l, m) set to 1, then is

H = jl(kr)Xlm

E =iZ0

k H = iZ0

kjl(kr)Xlm

The boundary conditions are that at r = a the electric field must only have a radial component and the mag-

netic field must be transverse. The second condition is automatically satisfied because of the transversality

of the Xlm. To match the first, we use

Xl,m =1

l(l + 1) LYl,m =1

l(l + 1)1

i 1

sin Yl,m

to first write out the Hfield components,

Hr = 0

H = ml(l + 1)

jl(kr)1

sin Yl,m

H =1

i

l(l + 1)jl(kr)Yl,m

Then, the electric-field components follow from E = iZ0k

H + H

,

Er =iZ0

k

1

r sin [ sin H H]

=iZ0

k

1

i

l(l + 1)

jl(kr)

r sin

sin + im

1

sin

Yl,m

=Z0k

1l(l + 1)

jl(kr)

r

1

sin sin m

2

sin2

Yl,m

=

Z0l(l + 1)jl(kr)kr

Yl,m

E = iZ0k

1

rrrH

= Z0l(l + 1)

1

kr

d

drrjl(kr)

[Yl,m]

E =iZ0

k

1

rrrH

= iZ0ml(l + 1)

1

kr

d

drrjl(kr)

1

sin Yl,m

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The cavity frequencies follow from the requirement E = E = 0 at r = a. The frequencies can be obtained

from the transcendental equation

d

drrjl(kr)

r=a

=

d

dxxjl(x)

x=ka

= 0

Denoting the n-th root of ddx (xjl(x)) with x

ln, it is ka = lmn ac = x

ln. The resonance frequencies thus are

lmn =xlnc

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 field

of a T Elm-mode, with amplitude aM(l, m) set to 1, then is

H =ikjl(kr)Xlm

E = Z0jl(kr)Xlm

Comparison with the analogous equation for TM-modes shows that the fields of the TE-modes are obtained

by replacing the former H with E/Z0 and the former E with Z0H. Thus, for TE-modes it is

Er = 0

E = Z0ml(l + 1)

jl(kr)1

sin Yl,m

E =Z0

i

l(l + 1)jl(kr)Yl,m

and

Hr =

l(l + 1)

jl(kr)

kr Yl,m

H =1

l(l + 1)

1

kr

d

drrjl(kr)

[Yl,m]

H =im

l(l + 1)

1

kr

d

drrjl(kr)

1

sin Yl,m

The conditions of vanishing transverse electric and vanishing normal magnetic field at r = a are satisfied via

the transcendental equation

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jl(ka) = 0

Denoting the n-th root of jl(x) with xln, it is ka = lmnac = xln. The resonance frequencies thus are

lmn =xlnc

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= xlnca we see that

lmna

=2

xln

Numerically we find the lowest roots of spherical Bessel functions to be x11 = 4.493, x21 = 5.763, x31 = 6.988

and x12 = 7.725. The lowest four TE-modes therefore are:

l n lmna1 1 1.3982 1 1.0903 1 0.8991 2 0.813

Figure 2: Lowest spherical Bessel functions and their roots.

c):

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The lowest TE-modes are the degenerate T El=1,m=1,n=1, T El=1,m=0,n=1 and T El=1,m=1,n=1-modes. To

obtain their fields, use the above general equations for the TE-fields to obtain:

l = 1, m = 1:

Er = 0

E =Z0

23

8

j1(x11

a

r) exp(i)

E = Z0i

2

3

8j1(

x11a

r)cos exp(i)

Hr =

2

3

8

j1(x11

a r)x11

a rsin exp(i)

H = 12

3

8

a

x11r

d

drrj1(

x11a

r)

cos exp(i)

H = i2

3

8

a

x11r

d

drrj1(

x11a

r)

exp(i)

l = 1, m = 0:

Er = 0

E = 0

E = Z0i

2

3

4j1(

x11a

r)sin

Hr =

2

3

4

j1(x11

a r)x11

a rcos

H = 12

3

4

a

x11r

d

drrj1(

x11a

r)

sin

H = 0

l = 1, m = 1:Er = 0

E =Z0

2

3

8j1(

x11a

r)exp(i)

E =Z0

i

2

3

8j1(

x11a

r)cos exp(i)

Hr =

2

3

8

j1(x11

a r)x11

a rsin exp(i)

H = 12

3

8a

x11r

d

drrj1( x

11

ar)

cos exp(i)

H = i2

3

8

a

x11r

d

drrj1(

x11a

r)

exp(i)