Download    

Multipole expansion
Introductory example: the potential of two point charges

q
q
 (r )   1    2 
| r  r1 | | r  r2 |
The exact expression is
What is the potential far away from the point charges? Namely, what is the potential when
r  r1 , r2 ? Each of the terms is
1
  
| r  ri |
1
r 2  2rri cos  i  ri 2
with i=1 or 2. Let us denote
 2rri cos i  ri 2
ri 2
2
i 
  ri  rˆ  2
r
r2
r
so that
1
r 2  2rri cos  i  ri 2

1 1
1 
3

 1  i   i2  ...
r 1  i r 
2 8



1 rˆ  ri 3(rˆ  ri ) 2  ri 2
  2 
 ...
r
r
2r 3
Using this expansion, the potential of the two point charges is





q
q
q q
rˆ
1
(r )   1    2   1 2  2  q1r1  q2 r2   3 3q1 (rˆ  r1 ) 2  3q2 (rˆ  r2 ) 2  q1r12  q2 r22
| r  r1 | | r  r2 |
r
r
2r


To see the meaning of this expansion we note that q  q1  q2 is the total charge of the system
generating the potential;



p  q1r1  q2 r2 is the total dipole moment of the system of charges. So,
1
up to order r
2

 q rˆ  p
the potential is  ( r )   2 . The next term is the quarupole moment which
r
r
is a tensor. To see this, we note that

rˆ  ri 
and then
(note that
 rˆ r
 1, 2 , 3
 i ,
,



3(rˆ  ri ) 2  ri 2  3 rˆ rˆm ri , ri ,m  ri 2   rˆ 3ri , ri ,m    ,m ri 2 rˆm
 rˆ
2

 ,m
 ,m
 1). The quadrupole tensor of our example is thus




Qm  q1 3(r1 )  (r1 ) m  r12 m  q2 3(r2 )  (r2 ) m  r22 m
The expansion of the potential to order r
3

is finally

 q rˆ  p rˆ  Q  rˆ
(r )   2 
 ...
r r
2r 3
The general formalism
The potential created by a general charge density is


  (r ' )
(r )   dr '  
| r  r '|



where r  r , , , r '  r ' , ' , ' and the integration is dr '  (r ' dr ' )(sin  ' d ' )d '  r ' dr ' d'
2
2
We look for an expression for the potential far away from the region where the charge density is,
in other words, the potential where r  r ' . This means that now we have to expand
1
r 2  2rr ' sin  sin  ' cos(   ' )  cos cos '  r ' 2
It turns out that this expansion can be written using the formula
2
 r '

1
1
Y*m ( ' , ' )Ym ( , )
   4  1

| r  r '|
2  1 m
 0 r
which is valid for
side.)
r  r ' . ( For r '  r one interchanges the roles of r and r' on the right-hand-
Using this formula, the potential becomes




  (r ' )
 
1
1
(r )   dr '    4  
Ym ( , ) 1  dr '  (r ' )r 'Ym ( ' , ' )
| r  r '|
r
 0 m    2  1
It is therefore convenient to define the following moments of the charge density:


qm   dr ' r 'Y*m ( ' ,  ' )  (r ' )
Then the potential becomes


Y ( , )

  (r ' )
1
(r )   dr '    4  
qm m  1
| r  r '|
r
 0 m    2  1
This is the multipole expansion.
Exercise: A spherical shell of radius a is carrying a uniform surface charge density, given by
q
, except on a cone around the north pole, of angle  . Find the potential inside the spherical
4a 2
shell.
Solution: Since we need the potential inside the spherical shell, we use the formula above with r
interchanged with r', that is
 r

1
1
Y*m ( ' , ' )Ym ( , )
   4  1

| r  r '|
2  1 m
 0 r '
and then




  (r ' )
  1
1
(r )   dr '    4  
Ym ( , )r   dr '  (r ' ) 1 Ym ( ' , ' )
| r  r '|
r'
 0 m    2  1
3
Let us consider the integral here for our case, in which

 (r ' ) 
q
 (r 'a) for  '  
4a 2
The charge density is independent of
Ym ( ' , ' ) 
 ' . Since
(2  1)(  m)! m
P (cos  ' )e im '
4 (  m)!
this means that we are left only with the terms for which m=0,
Y 0 ( ' ) 
(2  1)
P (cos ' )
4
The expressions for the potential inside the spherical shell is now



1
r q
(r )  4 
Y 0 ( ) 1
2 d ' sin  'Y 0 ( ' )
a 4 
 0 2  1


 0

r q
P (cos ) 1  d ' sin  ' P (cos ' )
a 2
4
The first moments of the multipole expansion
The multipole expansion of the potential is given by


Y ( , )

  (r ' )
1
(r )   dr '    4  
qm m  1
| r  r '|
r
 0 m    2  1
with


qm   dr ' r 'Y*m ( ' ,  ' )  (r ' )
and
(2  1)(  m)! m
P (cos  )e im
4 (  m)!
Ym ( , ) 
(Note that Y m ( , )  (1) Ym ( , ) and therefore
m
*
The first moment comes from the term
q,  m  (1) m q*m .)
  0 which implies m=0,



q00   dr 'Y00* ( ' , ' )  (r ' )   dr '

 
1
1
1
P0 (cos  ' )  (r ' ) 
dr '  (r ' ) 
q

4
4
4
where q is the total charge of the system.
Therefore, to order 1/r, the multipole expansion is just
The second moment comes from the term
Y10 ( , ) 

(r )  q / r .
  1 which implies m=1,0,-1. We have
3
3
P1 (cos ) 
cos
4
4
and therefore
q10 


3
d
r
'
r
'
cos

'

(
r
') 
4 


3
d
r
'
z
'

(
r
') 
4 
3
pz
4
Likewise
Y11 ( , )  
3
sin e i
8
and therefore
5
q11  
q11 




3
3
3
dr ' r ' sin  ' e i '  (r ' )  
dr ' ( x'iy ' )  (r ' )  
( p x  ip y )


8
8
4




3
3
i '
d
r
'
r
'
sin

'
e

(
r
'
)

d
r
'
(
x
'

iy
'
)

(
r
') 
8 
8 
3
( px  ip y )
4
so that the term gives
 
4 1  3
z
3 x  iy
3 x  iy
 pr
pz 
( p x  ip y ) 
( p x  ip y )   3

3 r 2  4
r 8 r
8 r
r

where the dipole moment of the charge density is defined by

 
p   dr ' r '  (r ' )
The next contribution, of the terms with
q22 
  2 is a little more complicated


15
1 15
dr ' ( x'iy ' ) 2  (r ' ) 
(Q11  2iQ12  Q22 )

32
12 2
q21  

15 
1 15
dr ' ( x'iy ' ) z '  (r ' )  
(Q13  iQ23 )

8
3 8
q20 


5
1 5
2
2
d
r
'
(
3
z
'

r
'
)

(
r
'
)

Q33
16 
2 4
where Q is the quadrupole moment tensor,


Qij   dr ' (3x'i x' j r '2  ij )  (r ' )
Collecting terms, the multipole expansion is
6
 
xi x j

q pr 1
(r )   3   Qij 5  ...
r
2 ij
r
r
Once we have this expansion, we can find the expansion of the electric field. For example, the
electric field of the dipole term is

moment and r as  , we find

 p  r

Edipole(r )   3 . Taking the angle between the dipole
r



 p  r

 ˆ 1  p cos 
p
3rˆ( p  rˆ)  p
ˆ
E dipole(r )   3  (rˆ  
)
 3 3rˆ cos   (rˆ cos    sin  ) 
r
r 
r
r2
r
r3


Exercise: Given the following charge distribution:

 (r ) 
Cr, r  a
0, r  a
where C is a constant (dimensions: charge over length to the fourth), find the first terms of the
multipole expansion for r>a.
Solution: All the required integrations are now carried only over a spherical volume of radius a.
We find
 
1
a4
q00 
dr '  (r ' ) C 4
4 
4
for the first moment.
q11  


3
d
r
'
(
x
'

iy
'
)

(
r
' ) 0
8 
q10 


3
d
r
'
z
'

(
r
' ) 0
4 
so there is no dipole moment to such a charge distribution. The next terms vanish as well,
q22 



15
15 
dr ' ( x'iy ' ) 2  (r ' ) 0 q21  
dr ' ( x'iy ' ) z '  (r ' ) 0


32
8


5
2
2
q20 
d
r
'
(
3
z
'

r
'
)

(
r
' ) 0
16 
There is no quadrupole moment to this charge distribution.
7
Exercise: Derive the multipole expansion of the charge distribution (dimensionless units)

 (r ) 
1 2 r
r e sin 2 
64
Solution: Firstly, we need to revise a bit the formalism above. Because of the charge density is
independent of  , all the terms with m  0 are zero. SO, we need only the terms with m=0 and
therefore we use
Y 0 
2  1
P (cos  ) .
4
The expansion for any r is then

(r )  4 

r

  
 1

1 Y 0
1 
d
r
'
r
'

(
r
'
)
Y
(

'
,

'
)

4

r
Y
d
r
'

(
r
' )Y 0 ( ' ,  ' )

0
0 
2  1 r  1 0
r '  1
 2  1
r
The Legendre polynomials have the identity
1
 d (1  
1
2
4
8
) P (  )    0 
 2
3
105
(   cos  )
Our potential becomes
r

1
4 Y 0 4
8
( r )  
(  0 
  2 )  dr ' r ' 4 e r '
 1
32  2  1 r
3
105
0

1
4 
4
8
 
r Y 0 (   0 
  2 )  dr ' e  r ' r '3
32  2  1
3
105
r
8
r



1 r

1 1
1
4 r '
3 r '
6 r '
2
r '
  dr ' r ' e   dr ' r ' e  



P
(cos

)
dr
'
r
'
e

r
dr
'
r
'
e
2
3 





24  r 0
r
r
 420
r 0

The terms in the brackets can The terms in the brackets can be now easily calculated. In
particular, as r tends to zero, the first brackets tend to 6, while the second tend to 1.
r



1 r

1 1
1
4 r '
3 r '
6 r '
2
  dr ' r ' e   dr ' r ' e  
 3  dr ' r ' e  r  dr ' r ' e  r ' 

P
(cos

)
2
 420
r

24  r 0
r
r

 0

The terms in the brackets can be now easily calculated. In particular, as r tends to zero, the
first brackets tend to 6, while the second tend to 1.
9
10