Homework 8 Answers, 95.657 Fall 2009, Electromagnetic Theory I
Dr. Christopher S. Baird, UMass Lowell
Problem 1
Jackson 4.1
Calculate the multipole moments qlm of the charge distributions shown as parts a) and b). Try to obtain
results for the nonvanishing moments valid for all l, but in each case find the first two sets of
nonvanishing moments at the very least.
z
a)
-q
-q
a
a
z
b)
q
a
a
-2 q
y
q
a
y
a
q
q
x
x
SOLUTION
a) The charge density is written down in spherical coordinates as:
=
q
 r −a cos  [ − −3/2− − −/2 ]
a2
Plug this into the multipole moments definition and evaluate:
q lm =∫ Y *lm  ' , 'r 'l x 'd x '
q lm =
q
Y * ' , 'r 'l  r '−a cos  ' [ '−  '−3 /2− '− '−/2 ] d x '
2 ∫ lm
a
2 
q lm =q al ∫ ∫ Y *lm ' , ' cos ' [ '− '−3/2−  '−'−/ 2 ] sin ' d  ' d '
0
l
q lm=q a
q lm =q al
l
q lm =q a
0
 
 
 
2
2 l1 l−m ! m
P l 0 ∫ e−i m  ' [ '−'−3 / 2−'−  '−/2 ] d '
4  lm !
0
2 l1 l−m ! m
P l 0 [1−i m−−1m−1m i m ]
4  lm !
2 l1 l−m ! m
m
m
P 01−−1 1−i 
4  lm ! l
q lm=2 q al
 
2 l1 l−m! m
P  01−i m  if m is odd, qlm = 0 if m is even
4  lm! l
This is the solution valid for all l.
Let us write our the first few multipoles explicitly to see what this means.
The monopole moment is:
q 00=0 because m is even
This makes sense because the total charge is zero.
The dipole moments are:
q 1,−1=q a

3
1i
2
q 1,0=0
q 11 =q a

3
−1i
2
Put these together into the Cartesian dipole moment vector:
p= p x i  p y j  p z k
p=

2
 q1,−1−q11  i−i q 1,−1q 11 j 2 q 10 k 
3
p=2 q a  i j 
The quadrupole moments are:
q 2,−2 =0
q 2,−1=0
q 2,0=0
q 21=0
q 2,2=0
The problem asks us for the first two sets of non-vanishing moments, so we have to keep going.
The non-zero octapole moments are:
q 3,−3=q a 3

35
1−i
16



q 3,−1=q a3
q 31 =q a3
q 33 =q a3
21
−1−i
16 
21
1−i
16 
35
−1−i
16 
b) The charge density is written down in spherical coordinates as:
=
q
2q
r −a [ cos −1 cos 1 ]−
 r 
2
2 a
4  r2
Plug this into the multipole moments definition and evaluate:
q lm =∫ Y *lm  ' , 'r 'l x 'd x '
*
l
q lm =∫ Y lm  ' , ' r ' [
q
2q
 r '−a [  cos '−1cos  '1 ]−
 r '] d x '
2
2 a
4  r '2
2 
q
q lm =
∫ ∫ Y * ' , 'a l [ cos '−1 cos '1 ] sin ' d ' d '
2  0 0 lm
2 
∞
2q
−
Y *lm ' , ' r 'l  r 'sin ' d r ' d  ' d '
∫
∫
∫
4 0 0 0
Expand the spherical harmonics and try to do the integral over the azimuthal coordinate. We find that it
vanishes except for when m = 0. This makes sense because the problem has azimuthal symmetry. In the
second integral, all terms disappear except for l = 0.
[ 
q lm =m ,0 q al
]
2 l1
[1−1l ]− l ,0 q 1
4

This becomes more clear if we break this into different cases, including the simplest monopoles first:
q 00=0
q 10 =q 1,−1=q11=0
q 20 =

5
q a2

q 2,−2 =q 2,−1=q2,1 =q 2,2=0
q 3,−3=q3,−2=q 3,−1=q3,0 =q 3,1=q3,2=q3,3 =0
q 40=

9 4
qa

and all higher terms can be expressed as:
q lm=

2 l1 l
q a if m = 0 and l = even, qlm = 0 otherwise

c) For the charge distribution of the second set b) write down the multipole expansion for the potential.
Keeping only the lowest-order term in the expansion, plot the potential in the x-y plane as a function of
distance from the origin for distances greater than a.
SOLUTION:
The multipole expansion of the potential is:
l
1 ∞
4
=
q lm r −l−1 Y lm  , 
∑
∑
4 0 l=0 m =−l 2 l1
=
[
∞
Y  ,
1 4
1
4
1
4
q20 3 Y 20  ,
q 40 5 Y 40  ,  ∑
ql , 0 l0 l 1
4 0 5
9
r
r
r
l =6,even 2 l1
q
=
4 0 a
[ 
3
5

a
a
3 cos2 −1
r
r
]
l 1

∞
1
a
35cos 4 −20 cos 2 32 ∑
4
l =6, even r
]
P l  ,
Now if r is much greater than a, then (a/r) is much less than one and (a/r) raised to higher powers is
even smaller, so that we only need to keep the first term.
3
=

q
a
3 cos 2 −1
4 0 a r
In the x-y plane this becomes:
=
3

−q
a
4 0 a r
1.0
0.9
0.8
We can plot the potential
in units of (-q/4πε0a) and
the distance in unit of a:
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
1
1.5
2
2.5
3
3.5
4
4.5
5
d) Calculate directly from Coulomb’s law the exact potential for b) in the x-y plane. Plot it as a function
of distance and compare with the result found in part c).
SOLUTION
We have three points charge and can write out an exact solution for the point charges:
=
[
]
q
1
1
2

−
4 0 ∣ra z∣ ∣r−a z∣ ∣r∣
In the x-y plane this becomes:
[
[
1
2q
4 0
=
−q
2
2
−
4 0 a r /a r /a21
2
r a
2
−
1
r
]
=
]
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
It becomes apparent now that the first term in the multipole expansion is a good approximation to as
close as r = 2a, but becomes inaccurate closer than that.
Dividing out the assymptotic form (a3/r3) lets us compare them for clearly:
1.2
1.0
0.8
0.6
0.4
0.2
0.0
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
Problem 2
Consider the localized charge distribution defined by ρ = r e-r cos2θ. Find the monopole, dipole, and
quadrupole moment of this charge distribution. Write down the potential far away due to this charge
distribution in terms of these moments. Also find the exact potential anywhere due to this charge
distribution by applying Columb's Law and doing the integral. Write down an expression for the error
we would encounter if we tried to use the multipole approach close to the charge distribution as the
exact potential minus the multipole-expanded potential. What does this error expression reduce to if we
only care about points on the z-axis? Plot the error along the z-axis versus distance from the origin.
SOLUTION
The monopole moment is defined as:
q 00 =

1
∫  x 'd x '
4
Plugging in the charge distribution we have here:
2 


∞
1
q 00 =
r ' e −r ' cos 2 ' r ' 2 sin  ' d r ' d ' d  '
∫
∫
∫
4 0 0 0

∞
1
q 00 =
2  ∫ r '3 e −r ' d r ' ∫ cos 2 ' sin ' d  '
4
0
0
Make the substitution x = cos θ'


1
∞
1
q 00 =
2  ∫ r '3 e −r ' d r ' ∫ x 2 dx
4
0
−1
q 00 =
∞
1 4
∫ r '3 e−r ' d r '
4 3 0
Use integration by parts to solve the last integral:
q 00 =

1 4
[−e−r ' r '33 r '2 6 r '6]∞0
4 3
q 00=4 
The dipole moment is:
p=∫ x '  x ' d x '
2  ∞
 r ' e−r ' cos 2 ' r ' 2 sin ' d r ' d ' d '
p=∫ ∫ ∫ r ' cos ' sin ' ir 'sin 'sin ' jr ' cos ' k
0
0 0
∞

0
0
∞
1
0
−1
p=2  k ∫ r ' 4 e−r ' d r ' ∫ cos 3  'sin ' d  '
p=2  k ∫ r ' 4 e−r ' d r ' ∫ x 3 dx
1
[ ]
∞
x4
p=2  k ∫ r ' 4 e−r ' d r '
4
0
−1
p=0
Because of azimuthal symmetry, only the m=0 quadrupole moment element will be non-zero:
q 20=




5
∫ 3 cos2 '−1r ' 2 x ' d x '
16
2 
∞
5
q 20 =
3 cos 2  '−1 r '2 r ' e−r ' cos 2  ' r '2 sin ' d r ' d  ' d '
∫
∫
∫
16 0 0 0
q 20 =
q 20 =

∞
5
2∫ 3 cos 2 '−1 cos 2 ' sin  ' d ' ∫ r '5 e−r ' d r '
16
0
0
1
∞
5
2 ∫ 3 x 4 −x 2  dx ∫ r '5 e−r ' d r '
16
−1
0
q 20=32 5 
The potential far away due to this charge distribution, if we only keep up to the quadrupole terms
becomes:
=
1
0 r
=


1
1
5
q00
q20 3 cos 2 −1
3
4
5 0 r 16
2
8

3 cos 2 −1
3
 0 r 0 r
The solution valid everywhere is found by using Coulomb's Law and expanding the denominator:
=
 x '
1
dx'
∫
∣
4 0 x−x '∣
=
2  ∞
l
∞
r l< *
1
1
−r '
2
2
r
'
e
cos
'
r
'
sin

'
∫ ∫∫
∑ ∑ 2 l1 r l 1 Y lm ' , 'Y lm  ,  d r ' d ' d  '
4 0 0 0 0
l =0 −l
>
2  ∞
l
∞
r l< *
1
1
−r '
2
2
=
∫ ∫∫ r ' e cos ' r ' sin  ' 4  ∑ ∑ 2 l1 r l 1 Y lm  ' , 'Y lm  ,d r ' d  ' d '
4 0 0 0 0
l =0 m=−l
>
Due to azimuthal symmetry, only the m=0 terms will be non-zero:
=

∞
∞
r l< 3 −r '
1
2  ∑ P l cos ∫ l 1
r ' e d r '∫ cos2 ' sin  ' P l  cos ' d '
4 0
l =0
0 r>
0
=
1
∞
∞
r l< 3 −r '
1
2  ∑ P l cos ∫ l 1
r ' e d r ' ∫ x 2 P l  x dx
4 0
l =0
0 r>
−1
Use the identity
2
1
x 2= P 2  x  P 0  x 
3
3

1
1
∞
∞
r l< 3 −r '
1
2
1
=
2  ∑ P l cos ∫ l 1 r ' e d r ' ∫ P 2  x  Pl  x  dx ∫ P 0  x P l  x  dx
4 0
3 −1
3 −1
l =0
0 r>

Because of orthogonality, every term in the sum vanishes except when l = 0 and l = 2:
∞
r 2< 3 −r '
1 ∞ 1 3 −r '
1
2
=
∫ r ' e d r ' 15  3 cos −1∫ r 3 r ' e d r '
3 0 0 r >
0
0
>
We must break the integrals into two cases, when r' < r and r' > r
=
[
]
r
[
r
∞
∞
1 1
1
1
3 −r '
2 −r '
2
5 −r '
2
r
'
e
d
r
'
r
'
e
d
r
'

3
cos
−1
r
'
e
d
r
'r
e −r ' d r '
∫
∫
∫
3∫
3 0 r 0
15
r 0
r
0
r
[
]
r
∞
1 1
−e−r ' r '33 r '2 6 r '6 ]0  [−e −r ' r ' 22 r '2 ]r
[
3 0 r
r
∞
1
1

3 cos 2 −1 3 [−e−r ' r '55 r ' 420 r '360 r '2120 r '120 ]0r 2 [−e−r ' ]r
150
r
=
[
=

]
1
[−e−r  r46 r −1 6 r −1 ]
3 0
1
3 cos 2 −1 [−e −r 5 r 2060 r −1120 r −2 120 r −3 120 r −3 ]
150
We can now calculate the error:
error = exact −approx.
error =−
1 −r
1
e  r46 r −1 −
3 cos 2 −1 e−r 5 r 2060 r−1120 r −2120 r −3
3 0
15 0
]
Along the positive z-axis this reduces to:
error =0=−
1 −r
e [ r410 r −116 r−216 r−3 ]
0
If we plot this, we get the figure below. The blue line is the error in the potential. The red line is a
scaled plot of the charge distribution along the z-axis. Of note is the fact that the error is reasonably
small even slightly within the charge distribution, but grows exponentially large once deep within the
distribution.
10
0
-10
-20
-30
-40
-50
0
0.5
1
1.5
2
2.5
3
3.5
4