Multipole expansions
We have frequently referred to our RULE 1: far enough away from any
charge distribution with net charge Q, the potential is approximately that due
to a point charge Q located in the distribution. We also found from our very
¤rst example that the next correction is a dipole. The dipole potential falls o¤
faster (/ 1=r2 ) than the point charge (or monopole) potential (/ 1=r): Now
wezd like to make these ideas more precise.
We have a charge distribution with charge density ½ (~r 0 ) : We put the origin
somewhere inside the distribution, and we put the polar axis in a spherical
coordinate system through a point P at which we want to ¤nd the potential.
The entire charge distribution is located inside a sphere of radius r0 and P is
outside that sphere.
Then the potential at P is
V (~r ) =
where
j~r ¡ ~r0 j =
1
4¼"0
q
Z
½ (~r0 )
d¿ 0
j~r ¡ ~r 0j
2
r2 + (r 0) ¡ 2rr0 cos µ 0
Since r > r0 and r 0 < r0 ; we factor out the r to get
s
µ 0 ¶2
p
r
r0
0
j~r ¡ ~r j = r 1 +
¡ 2 cos µ 0 = r 1 + "
r
r
where
¯ 0µ 0
¶¯
¯r r
¯
0 ¯
¯
j"j = ¯
¡ 2 cos µ ¯ < 1
r r
1
(1)
In fact, as P gets farther from the charge distribution, " becomes much less
than 1. So we expand the square root
¡ 1¢¡ 3¢
¡2 ¡2 2
1
1
¡1=2
p
= (1 + ")
= 1¡ "+
" +¢¢¢
2
2
1+ "
1
3
5 3
= 1 ¡ " + "2 ¡
" +¢¢¢
2
8
16
So
1
j~r ¡ ~r 0j
=
=
=
=
"
#
µ
¶
µ ¶2 µ 0
¶2
µ 0 ¶3 µ 0
¶3
1
1 r0 r0
3 r0
r
5
r
r
0
0
0
1¡
¡ 2 cos µ +
¡ 2 cos µ
¡
¡ 2 cos µ
+ ¢ ¢¢
r
2r
r
8 r
r
16 r
r
¶ 3
2
³ ´2
³ 0 ´2 µ ³ 0 ´2
0
r0 cos µ 0
1 r0
3
r
r
r0
2 0
¡2 r
+8 r
¡ 2 r cos µ + 4 cos µ
16 1+
r
r
7
4
5
³
´
³
´
3
3
0
0
r
0
5
r
r
¡ 16
¡
2
cos
µ
+
¢
¢
¢
r
r
2
3
³
´2 ¡
¢
r0 cos µ0
1
r0
2 µ0 ¡ 1 +
1
+
+
3
cos
r
2
16
7
¶r
³ 0 ´3 ³ 0
´3
4 ³ 0 ´2 µ ³ 0 ´2
5
0
0
0
3 r
r
r
5
r
r
r
¡
2
cos
µ
¡
¡
2
cos
µ
+
¢
¢
¢
8
r
r
r
16
r
r
"
#
µ
¶
2
1
r0
r0
1 + P 1 (¹0 ) +
P 2 (¹0 ) + ¢ ¢ ¢
r
r
r
Izll let you check the next few. In fact
1
1
1X
=
j~r ¡ ~r 0j
r l=0
µ
r0
r
¶l
Pl (¹0 )
Now we put this result into our integral (1) for the potential:
V (~r ) =
=
1 µ ¶l
1 X r0
Pl (¹0) d¿ 0
r
r
l=0
Z
1
X
1 1
1
l
½ (~r0 ) (r 0) Pl (¹0 ) d¿ 0
4¼"0 r
rl
1
4¼"0
Z
½ (~r0 )
(2)
l=0
This is the multipole expansion of the potential at P due to the charge distribution. The ¤rst few terms are:
Z
1 1
Q
l=0:
½ (~r0 ) d¿ 0 =
4¼"0 r
4¼ "0 r
This is our RULE 1. The monople moment (the total charge Q) is indendent of
our choice of origin. The potential does depend on the origin (because r does)
but only weakly if r À r0 :
Z
1 1
l=1:
½ (~r 0 ) r0 cos µ 0 d¿ 0
4¼"0 r2
2
This is the dipole potential.
l=2:
1 1
4¼"0 r 3
Z
2
½ (~r 0) (r0 ) P 2 (¹0 ) d¿ 0
This is the quadrupole potential.
Each succeeding term decreases faster with r and so becomes less important
as P gets further from the origin.
These expressions depend on the particular coordinate system that we have
chosen. Most importantly, P is on the polar axis. So letzs see if we can write
the results in a coordinate independent way. Note that
cos µ0 =
r^ ¢ ~r 0
r0
so the dipole potential is
V dip ole (~r) =
=
=
Z
1 1
^ ¢ ~r0
0
0r
½
(~
r
)
r
d¿ 0
4¼"0 r2
r0
Z
1 1
^r ¢ ½ (~r 0) ~r0 d¿ 0
4¼" 0 r2
1
r^ ¢ ~p
4¼" 0 r2
where
p~ =
Z
½ (~r0 ) ~r0 d¿ 0
(3)
(4)
is the dipole moment of the charge distribution. Notice that this integral may
depend on the choice of origin, because ~r0 does. However, if the total charge Q
is zero, then ~p is independent of origin. To see this, let ~p 1 be the dipole moment
with respect to origin 1, ~p 2 with respect to origin 2, and let ~r12 be the position
of origin 2 with respect to origin 1. Then
3
~p 1
=
Z
Z
½ (~r0 ) ~r01 d¿ 0 =
½ (~r0 ) (~r12 + ~r02 ) d¿ 0
Z
Z
0
0
= ~r12 ½ (~r ) d¿ + ½ (~r0 ) ~r20 d¿ 0
=
Q~r12 + ~p 2
(5)
So if Q = 0; then ~p 1 = ~p2 :
This is an example of a more general result:
The ¤rst non-zero multipole moment is independent of origin.
Result (5) also explains the results we obtained in our very ¤rst example for
the dipole moment of the two point charges. A charge that is not at the origin
but at position ~rQ contributes a dipole moment Q~rQ with respect to that origin.
An ideal or "pure" dipole is located at a single point. That is, it is the
dipole moment of two equal and opposite point charges separated by a distance
d in the limit that d ! 0: In order that p~ not be zero, we have to let q ! 1:
~p = lim lim q d~
q!1 d!0
where, as we take the limit, we hold the product qd = p constant. The vector
d~ points from the negative charge to the positive charge in the pair. (Gri¢ths
uses the term "pure", but I donzt like it. I think "ideal" is more appropriate.)
Equation (2) gives the potential at a point on the polar axis as a series in
powers of 1=r: It does not give us the potential at other points. However, the
dipole potential (3) is valid everywhere. It may be written
V dip ole (r; µ) =
1
1
r^ ¢ ~p =
p cos µ
4¼"0 r2
4¼"0 r2
where µ is the angle between ~p and ~r ; that is, it is the polar angle in a coordinate
system with polar axis along p~: Then we have
V d ip ole (r; µ) =
p
P 1 (cos µ)
4¼"0 r 2
(6)
The dipole itself, remember, has z¡component (from 2 with l = 1)
Z
p z = ½ (~r 0) r0 P1 (¹0 ) d¿ 0
and it is not coincidental that P 1 shows up again in (6). In fact, if our charge
distribution has azimuthal symmetry about an axis that we choose as our polar
axis, then we may write:
Z
1
1 X Pl (cos µ)
l
V (r; µ) =
½ (~r0 ) (r0 ) P l (¹0 ) d¿ 0
4¼"0 r
rl
l=0
4
(7)
We still do not have a completely general result, and that will have to wait until
Physics 704.
Example:
2
A hemisphere of radius a contains charge density ½ = ½0 za + ½1 ar2 : Find the
monople, dipole and quadrupole moments of this charge distribution, and hence
¤nd the potential at distance r À a from the hemisphere.
The monopole is the total charge.
Z
Q =
½ (r; µ) 2¼r2 sin µdµdr
¶
Z a
Z 1µ
r¹
r2
2
= 2¼
r dr
½0
+ ½1 2 d¹
a
a
0
0
µ
¶¯ 1
Z a
2
2
¯
r¹
r
2
= 2¼
r dr ½0
+ ½1 2 ¹ ¯¯
2a
a
0
0
µ
¶
4
5
r
r
= 2¼ ½0
+ ½1 2
8a
5a
³½
´
½
= 2¼a3 0 + 1
8
5
The dipole moment is
Z
Z
~p =
½~rd¿ = ½ (r; µ) (z z^ + x x^ + y^
y ) r2 sin µdµdÁdr
Only the z¡component is non-zero, because
x = r sin cos Á
and
Z
2¼
cos ÁdÁ = 0
0
The y¡component vanishes similarly.
5
Then
pz
=
2¼
Z
a
0
=
=
=
=
Z
Z
1
0
a
Z
1
µ
µ
r¹
r2
½0
+ ½1 2
a
a
¶
(r¹) r 2 d¹dr
r 4 ¹2
r5 ¹
+ ½1 2
a
a
0
0
Z aµ
¶
4
5
r
r
2¼
½0
+ ½1 2 dr
3a
2a
0
³½
½ ´
2¼a 4 0 + 1
15
12
2¼a4 ³ ½0
½ ´
+ 1
3
5
4
2¼
½0
¶
d¹dr
Finally the quadrupole is
Z
qz z =
½ (~r ) r2 P 2 (¹) d¿
¶
Z aZ 1µ
¢
r¹
r2
1¡
= 2¼
½0
+ ½1 2 r 2 3¹2 ¡ 1 r 2 drd¹
a
a
2
0
0
Z aZ 1µ
¶
¢
¢
r5 ¡ 3
r6 ¡ 2
= ¼
½0
3¹ ¡ ¹ + ½1 2 3¹ ¡ 1 drd¹
a
a
0
0
µ
µ
¶
¶
Z a
r5 3
1
r6
= ¼
½0
¡
+ ½1 2 (1 ¡ 1) dr
a 4
2
a
0
µ ¶
½
1
½
= ¼a5 0
= ¼a5 0
6 4
24
Thus the potential is
½
¾
1
Q
p z cos µ
qz z
V (r; µ) =
+
+ 3 P2 (cos µ) + ¢ ¢ ¢
4¼"0 r
r2
r
½
¾
³
´
3
¢
¼a
½0
½1
2a ³ ½0
½1 ´
½0 a2 1 ¡
2
=
2
+
+
+
cos µ +
3 cos µ ¡ 1 + ¢ ¢ ¢
4¼"0 r
8
5
3r 5
4
24 r2 2
½
¾
¢
¼a3
½0
½1
2a ³ ½0
½1 ´
½ 0 a2 ¡
2
=
+2
+
+
cos µ +
3 cos µ ¡ 1 + ¢ ¢ ¢
4¼"0 r 4
5
3r 5
4
48 r2
Are there more terms? Yes there are.
¶
Z µ
Z
Z 1h
i
r¹
r2
2¼ a l+3
r
½0
+ ½1 2 rl P l (¹) d¿ =
r dr
½0 ¹P l (¹) + ½1 Pl (¹) d¹
a
a
a 0
a
0
Now if l is even, then P l (¹) is an even function of ¹; but ¹Pl (¹) is odd. But if
l is odd, then ¹P l (¹) is even. For an even function
Z
1
f (¹) d¹ =
0
6
1
2
Z
1
f (¹) d¹
¡1
Thus for l even
Z 1
0
1
P l (¹) d¹ =
2
Z
1
P 0 (¹) P l (¹) d¹ = 0 unless l = 0
¡1
We already found that this is true for l = 2 above. Then for l odd
Z 1
Z
1 1
¹P l (¹) d¹ =
P 1 (¹) P l (¹) d¹ = 0 unless l = 1
2 ¡1
0
Thus there are higher multipoles, but for l > 1; all even multipoles involve only
½0 but all odd multipoles involve only ½1:
Plot
½
µ
¶
¾
¢
¼a3 ½0 a 1
½1
2a 1
½1
1 a2 ¡
2
V (r; µ) =
+2
+
+
cos µ +
3 cos µ ¡ 1 + ¢ ¢ ¢
4¼"0 ar 4
5½0
3r 5
4½0
48 r2
Thus
V (r; µ)
¼ a3 ½0
4¼"0a
=
µ
1
½
+2 1
4
5½0
¶
a 2a2
+ 2
r
3r
µ
1
½
+ 1
5
4½0
¶
cos µ +
¢
1 a3 ¡
3 cos 2 µ ¡ 1 + ¢ ¢ ¢
3
48 r
Now suppose ½1 = ½0 =2: Then
µ
¶
µ
¶
¢
V (r; µ)
1
1 a 2a2 1
1
1 a3 ¡
=
+
+
+
cos
µ
+
3 cos2 µ ¡ 1 + ¢ ¢ ¢
3
¼ a ½0
4
5 r
3r 2 5
8
48 r3
4¼ "0a
=
¢
9 a 13 a2
1 a3 ¡
+
cos
µ
+
3 cos2 µ ¡ 1 + ¢ ¢ ¢
2
3
20 r
60 r
48 r
10
8
6
z/a
4
2
0
-2
2
4 s/a 6
8
10
-4
-6
-8
-10
Green 0.15, Black 0.1 red 0.05
Notice how the equipotential surfaces get more spherical as distance from
the hemisphere (in blue) increases.
7