Date post: | 04-Apr-2018 |
Category: |
Documents |
Upload: | razarizvi1 |
View: | 236 times |
Download: | 1 times |
of 13
7/31/2019 Jackson 9.10, 9.16
1/13
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
7/31/2019 Jackson 9.10, 9.16
2/13
=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
7/31/2019 Jackson 9.10, 9.16
3/13
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
7/31/2019 Jackson 9.10, 9.16
4/13
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
7/31/2019 Jackson 9.10, 9.16
5/13
Rrad =Z02
1.55718 = 93.36
7/31/2019 Jackson 9.10, 9.16
6/13
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)
7/31/2019 Jackson 9.10, 9.16
7/13
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
7/31/2019 Jackson 9.10, 9.16
8/13
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.
7/31/2019 Jackson 9.10, 9.16
9/13
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.
7/31/2019 Jackson 9.10, 9.16
10/13
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
7/31/2019 Jackson 9.10, 9.16
11/13
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
7/31/2019 Jackson 9.10, 9.16
12/13
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):
7/31/2019 Jackson 9.10, 9.16
13/13
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)