Jackson notes 2017
1
Spherical multipole moments
Suppose we have a charge distribution ρ (x) where all of the charge is contained within a spherical region of radius R, as shown in the diagram. Then
there is no charge in the region r > R and so we may write the potential in that
region as a solution of Laplace’s equation in spherical coordinates. With the
constraint that Φ → 0 as r → ∞, we have
Φ (r, θ, φ) =
1
4πε0
∞
+l
l=0 m=−l
4π qlm
Ylm (θ, φ)
2l + 1 rl+1
r>R
(1)
where the constants 4π/ (2l + 1) have been included for future convenience. We
may also write the potential in terms of the charge density as (Notes 1 eqn.24):
Φ (r, θ, φ) =
1
4πε0
ρ (x ) 3
d x
|x − x |
(2)
and then expand the factor 1/ |x − x | in spherical harmonics using J eqn 3.70
Φ (r, θ, φ) =
=
1
4πε0
1
4πε0
∞
ρ (x )
l
l=0 m=−l
∞
l
l=0 m=−l
l
4π r<
∗
θ ,φ
Ylm (θ, φ) Ylm
l+1
2l + 1 r>
d3 x
l
r<
Y ∗ θ ,φ
l+1 lm
r>
d3 x
4π
Ylm (θ, φ)
2l + 1
1
ρ (x )
The integral is over the entire volume where ρ is not zero, which is inside the
sphere of radius R. Then since r < R and r > R, we have r< = r and r> = r,
so
Φ (r > R, θ, φ) =
1
4πε0
lm
4π Ylm (θ, φ)
2l + 1 rl+1
l
∗
ρ (x ) (r ) Ylm
θ ,φ
d3 x
(3)
and then comparing equations (1) and (3) we find
qlm =
l
∗
θ ,φ
ρ (x ) (r ) Ylm
d3 x
(4)
These are the spherical multipole moments of the source. With l = 1 we have
the dipole, l = 2 the quadrupole etc.
2
Cartesian multipole moments
This time we expand
Φ (x) = k

1
|x−x3 |
in expression (2) in a Taylor series about x = 0:
1
1
ρ (x )  + x · ∇
r
R
+
x3 =0
1
∂ ∂ 1
xx
2 i,j=1 i j ∂xi ∂xj R
Now let’s evaluate the derivatives:
x ·∇
1
R
x3 =0

3
=x ·
(x − x )
R3
x3 =0
=x ·
x3 =0
+ · · ·  d3 x
x
r3
and
∂ ∂ 1
∂xi ∂xj R
=
∂ xj − xj
∂xi
R3
=
−δ ij
xi xj
+3 5
3
r
r
x3 =0
=
x3 =0
(xi − xi ) xj − xj
−δ ij
+3
3
R
R5
Putting these expressions back into the Taylor series, we get:

1
x
1
δ ij
xi xj
Φ (x) = k ρ (x )  + x · 3 +
xx 3 5 − 3
r
r
2 i,j i j
r
r
x3 =0

+ · · ·  d3 x
We may integrate term by term because the Taylor series is uniformly convergent.



1
x
1
δ
x
x
i j
ij
ρ (x ) d3 x + 3 · x ρ (x ) d3 x +
Φ (x) = k
3 5 − 3 ρ (x ) xi xj d3 x + · · ·

r
r
2 i,j
r
r
2
The first two terms are easily integrated to give:
p·x
q
k
Φ (x) = k + k 3 + 5
r
r
r
i,j
3xi xj − r2 δ ij ρ (x )
xi xj 3
d x +···
2
(5)
where q is the total charge in the distribution and
p≡
ρ (x ) x d3 x
(6)
is the Cartesian dipole moment. The last term in (5) needs some work, since
we would like to express the result in terms of the quadrupole tensor:
Qij ≡
2
ρ (x ) 3xi xj − (r ) δ ij d3 x
(7)
but our integral has unprimed variables where we need primes. However, the
first term is symmetric in primed and unprimed variables, while the second is
non-zero only if i = j. The second term is:
2
2
2
2
2
2
2
r2 (x ) + r2 (y ) + r2 (z ) = (r ) r2 = (r ) x2 + (r ) y 2 + (r ) z 2
and so we may interchange prime and unprime and rewrite the integral as:
ρ (x )
i,j
xi xj 3xi xj − r2 δ ij d3 x
=
xi xj
i,j
=
ρ (x ) 3xi xj − r
3
2
δ ij d3 x
xi xj Qij
i,j
Thus we have for the potential


3


1
xi xj
q p·x 1
+ 3 +
Φ (x) =
Q
+
·
·
·
ij

4πε0  r
r
2 i,j=1 r5
(8)
The quadrupole tensor Qij (7) is symmetric, real-valued and traceless. It has
three real eigenvalues.
The total charge is independent of our choice of origiin. In general, the
lowest non-zero multipole is independent of origin, but higher order multipoles
do depend on the origin.
3
3
Relations between the two sets of multipoles
Comparing the expressions (1) and (8) allows us to relate the two sets of multipoles. Both expressions are series in increasing powers of 1/r. We’ll need
expressions for the spherical harmonics (Jackson page 109, Lea page 388).
The l = 0 (1/r) term is the monopole term. From equation (4) with
l = m = 0,
q00 =
1
∗
θ , φ d3 x = √
ρ (x ) Y00
4π
q
ρ (x ) d3 x = √
4π
(9)
and the first term in the spherical multipole expansion (1) is
Φ0 =
1
q00
1 q 1 1
1 q
√
√ =
Y00 =
4π
4πε0
r
ε0 4π r 4π
4πε0 r
which is also the first term in expression (8).
With l = 1 ( 1/r2 term) there are three contributions, with m = ±1 and
zero. First note that
ql,−m
l
∗
θ , φ d3 x
ρ (x ) (r ) Yl−m
=
l
= (−1)m
ρ (x ) (r ) Ylm θ , φ d3 x
∗
= (−1)m qlm
(10)
So there are actually only two values to calculate.
With l = 1, m = 0
q10
q10
=
∗
θ , φ d3 x
ρ (x ) r Y10
=
ρ (x ) r
=
3
4π
3
cos θ d3 x
4π
ρ (x ) z d3 x =
3
pz
4π
(11)
where we used the z−component of (6).
With l = 1, m = 1
q11
q11
=
∗
θ , φ d3 x
ρ (x ) r Y11
= −
ρ (x ) r
3
3
sin θ e−iφ d3 x
8π
= −
ρ (x ) r
3
sin θ cos φ − i sin φ
8π
= −
3
8π
=
d3 x
ρ (x ) (x − iy ) d3 x
3
(ipy − px )
8π
(12)
4
Equivalently,
px = −
2π
∗
(q11 + q11
),
3
py =
1
i
2π
∗
(q11 − q11
),
3
4π
q10
3
pz =
(13)
The corresponding l = 1 term in the potential is
Φ1
=
=
=
=
1
4πε0
+1
4π q1m
Y1m (θ, φ)
3 r2
m=−1
1 4π
(q11 Y11 (θ, φ) + q10 Y10 + q1,−1 Y1,−1 )
4πε0 r2 3
1 4π
∗
∗
(q11 Y11 (θ, φ) + q10 Y10 + (−1) q11
(−1) Y11
)
4πε0 r2 3
1
{2 Re [q11 Y11 (θ, φ)] + q10 Y10 }
3ε0 r2
Inserting our expressions for q1m in terms of the components of p,we have
Φ1
=
1
3ε0 r2
=
1
p · r̂
(px sin θ cos φ + py sin θ sin φ + pz cos θ) =
4πε0 r2
4πε0 r2
3
(ipy − px )
8π
2 Re
3
(− sin θ) eiφ +
8π
3
pz
4π
3
cos θ
4π
which is the second term in (8). (See also Notes 1 eqn 28.)
With l = 2 (1/r3 term) we have m = 2, 1, and 0. The multipole moment q20
is
q20
q20
2
=
∗
θ ,φ
ρ (x ) (r ) Y20
=
ρ (x ) (r )
=
1
2
5
4π
=
1
2
5
Q33
4π
2
1
2
d3 x
5
3 cos2 θ − 1 d3 x
4π
2
2
ρ (x ) 3 (z ) − (r )
d3 x
(14)
where we used the (3,3) component of (7). Similarly, we may show that
q21 =
1
3
15
(iQ23 − Q13 )
8π
(15)
15
(Q11 − Q22 − 2iQ12 )
2π
(16)
and
q22 =
1
12
Equivalently:
Q11 − Q22 = Re 12
5
2π
q22
15
and so on. We have 3 independent values of q2m , and q21 and q22 each have a
real and imaginary part, giving us 5 real numbers, but there are 6 independent
values Qij in a general symmetric rank 2 tensor. However, in this case the six
values are not all independent as we have the additional constraint that Qij
is traceless, leaving us with 5 independent values. For the extension of this
discussion to higher multipoles, see Jackson Problem 4.3.
Example
A ring of charge of radius a carries linear charge density that varies with angle
φ measured around the ring: λ = λ0 cos φ. Let’s find the multipole moments
and the potential for r > a.
The ring is most easily described in spherical coordinates, so let’s first find
the spherical multipoles. With polar axis along the axis of the ring, the charge
density is (Lea Example 6.7)
ρ (x) =
λ0 cos φ δ (µ) δ (r − a)
r
and thus
qlm
=
∗
ρ (x) rl Ylm
(θ, φ) d3 x
=
λ0 cos φ δ (µ) δ (r − a) l ∗
r Ylm (θ, φ) r2 dµdφdr
r
= λ0
2l + 1 (l − m)!
4π (l + m)!
∞
0
2π
+1
−1
0
cos φ δ (µ) δ (r − a) rl+1 Plm (µ) e−imφ dµdφdr
Using the sifting property, we have immediately
qlm
= λ0
= λ0
2l + 1 (l − m)! l+1 m
a Pl (0)
4π (l + m)!
2l + 1 (l − m)! l+1 m
a Pl (0)
4π (l + m)!
2π
cos φe−imφ dφ
0
2π
0
eiφ + e−iφ
2
e−imφ dφ
The integral over φ is zero unless m = ±1, so
qlm = λ0
2l + 1 (l − m)! l+1 m
a Pl (0) π (δ m1 + δ m,−1 )
4π (l + m)!
Then
ql,1 = λ0 π
2l + 1 l+1
a
4π
(l − 1)! 1
P (0)
(l + 1)! l
We must have l > 0, since with l = 0 there is no m = 1. We also need l + 1 to
be even so that Pl1 (0) is not zero. Thus we have all multipoles of odd order,
and m = 1. We can further simplify using (10):
∗
ql,−1 = −ql1
6
So, since ql1 is real in this case,
2 ∗
Yl1∗ = ql,1 Yl1∗
ql,−1 Yl,−1 = (−1) ql,1
Thus
= ql1 (Yl1 + Yl1∗ ) = 2ql1 Re (Yl1 )
ql1 Yl1 + ql,−1 Yl,−1
= 2λ0 π
2l + 1 l+1
a
4π
(l − 1)! 1
P (0)
(l + 1)! l
2l + 1 (l − 1)! 1
P (µ) cos φ
4π (l + 1)! l
= λ0 al+1
2l + 1 (l − 1)! 1
P (0) Pl1 (µ) cos φ
2 (l + 1)! l
Thus the potential for r > a is (eqn 1)
1
4πε0
Φ =
λ0
2ε0
=
∞
4π
al+1 2l + 1 (l − 1)! 1
λ0 l+1
P (0) Pl1 (µ) cos φ
2l + 1 r
2 (l + 1)! l
l=1, odd
∞
l+1
l=1, odd
a
(l − 1)! 1
P (0) Pl1 (µ) cos φ
rl+1 (l + 1)! l
a4 2 3 3
λ0 cos φ a2 1
sin
θ
+
1 − 5 cos2 θ sin θ + · · ·
2ε0
r2 2
r4 4! 2 2
λ0 sin θ cos φ a2
3 a2
1 − 5 cos2 θ + · · ·
1
+
4ε0
r2
8 r2
=
=
The dominant term is a dipole, because the ring has zero net charge. Comparing
with (8), we may read off the dipole moment:
p = πλ0 a2 x̂
We may also compute p using (6).
∞
p =
0
2π
0
= λ0 a2 πx̂
+1
−1
λ0 cos φ δ (µ) δ (r − a)
r µẑ +
r
1 − µ2 (cos φx̂ + sin φŷ) r2 drdµdφ
The two results agee.
The next term is the octupole (l = 3) term. The potential has the same
azimuthal dependence (cos φ) as the charge density.
Multipole moments are very important in computing the radiation from a
time-dependent charge distribution. (See Ch 9.)
7
4
A surprising result
Let us evaluate the integral:
E (x) d3 x
sphere
where the integral is over a sphere of radius R that contains all the sources of
E, as shown in the diagram below. We express the electric field in terms of the
potential, and convert to a surface integral:
E (x) d3 x =
sphere
sphere
Φn̂R2 dΩ
−∇Φ dV = − Φn̂ dA = − −
S
S
Insert the integral expression (2) for the potential, and expand 1/|x − x | in
Legendre Polynomials:
sphere
E (x) d3 x = −k
S
= −k
all space
ρ (x )
dV n̂R2 dΩ
|x − x |
ρ (x )
S
all space
8
∞
l=0
l
r<
P (cos γ) dV n̂R2 dΩ
l+1 l
r>
where k = 1/4πε0 and γ is the angle between x and x . Now we interchange
the order of integration, and put the polar axis along the vector x , so that the
polar angle is γ. Then we can write n̂ in Cartesian components as:
n̂ = x̂ sin γ cos φ + ŷ sin γ sin φ + ẑ cos γ
to obtain:
2π
E d3 x = −k
all
space
∞
+1
ρ (x )
0
−1
l=0
l
r<
P µγ (x̂ sin γ cos φ + ŷ sin γ sin φ + ẑ cos γ) R2 dµγ dφdV
l+1 l
r>
The x−and y−components vanish upon integration over φ, leaving:
+1
E d3 x = −k
2π
all space
ρ (x )
−1
∞
l=0
l
r<
P (cos γ) ẑ cos γR2 dµγ dV
l+1 l
r>
Since cos γ = P1 (cos γ) , orthogonality of the Pl requires that only the l = 1
term survive the integration over γ. Then (Jackson eqn 3.21, Lea 8.33) with
l = 1 gives
2
r<
E d3 x = −k
2π ρ (x ) 2 ẑR2 dV
3
r>
all space
Now x is on the surface of the sphere, where |x| = r = R, and |x | = r < r
(because all of the charge is inside the sphere). Also, we chose our polar axis
along x , so ẑ = r̂ . Thus:
sphere
4π
k
3
1
= −
3ε0
E d3 x = −
ρ (x )
all space
all space
r
r̂ R2 dV
R2
ρ (x ) x dV = −
p
3ε0
(17)
where p is the dipole moment with respect to the center of the sphere (eqn 6). .
This is a completely general result: we did not need to say anything about
the details of the charge distribution.
Now let’s look at the usual expression for the potential due to an ideal
“point” dipole at the origin (Notes 1 eqn 28) :
Φ (x) = k
p·x
r3
The electric field is:
E = −∇Φ = −k
x
p
− 3 5p · x
r3
r
(18)
(Here we used the expression for ∇ a · b from J’s front cover, together with
the fact that p is a constant vector, x/r3 = −∇ (1/r) , and the curl of a gradient
is zero. See also Jackson 4.13.) With this electric field, we have:
E d3 x =
sphere
sphere
−k
9
p
x
− 3 5 (p · x) d3 x
r3
r
Put the polar axis along p, so that:
sphere
E d3 x = −
p
r̂
− 3 3 p cos θ r2 drdΩ
r3
r
k
sphere
Now
r̂ = p̂ cos θ + x̂ sin θ cos φ + ŷ sin θ sin φ
Again the x− and y−components vanish when we perform the φ−integration.
We do the θ integration next, to get
R
sphere
E d3 x = −2πk
0
R
= −2πk
0
+1
−1
p
1 − 3µ2 dµdr
r
p
µ − µ3
r
+1
dr
−1
=0
How can this be? We should have obtained −k 4π
3 p (equation 17). The problem
is that when we took the derivatives to get the field (eqn. 18), those operations
are not valid at the origin. There is a missing delta-function! You can also
understand this by looking at the field line diagram for a pair of equal and
opposite point charges. As the separation goes to zero, the field lines between
the charges (whose direction is opposite p) get packed into zero space— the field
becomes infinite!
For consistency, the dipole field must be:
Edip ole = −k
p
x
4π
kpδ (x)
− 3 5p · x −
r3
r
3
(19)
To see why this result makes sense, let’s look again at the derivative of the
potential:
Φ=k
p·x
= −kp · ∇
r3
1
r
Thus, with z−axis along p,
1
r
= kp
1
r
= −4πδ (x) ,
−∇Φ = k∇ p · ∇
∂
∇
∂z
1
r
We already have the relation
∇2
but
1
=∇· ∇
r
∂
∇
∂z
(20)
1
r
is only one of the three second derivatives in the standard delta-function result
(20), which helps us to understand the factor of 1/3 in equation (19). We can
understand the direction by looking at the field line diagram.
10
To understand the magnitude another way, look at the elecric field half way
between two point charges ±q separated by a displacement d, where p = q d.
E
kq
ˆ = − kq d
3
(d/2)
(d/2)
kp
4πk
p
= −
3 =− 3 4
d
(d/2)
3π 2
= −2
2d
3
3
→ pδ (x) and
Now in the limit d → 0 the dipole moment density p/ 4π(d/2)
3
so
4πk
pδ (x)
E→−
3
which is what we obtained in (19).
We can write our result (19) as a purely mathematical statement:
∂
∇
∂z
1
r
=∇
∂ 1
∂z r
x
4π
ẑ
δ (x) ẑ
− 3 5 ẑ · x −
3
r
r
3
=−
(21)
You’ll need this result in Problem 6.20.
5
Energy
If our charge distribution is now placed in an external field Eext = −∇Φext , the
energy of the system is (Notes 2 eqn 3. Do you understand why there is no
factor of 1/2 here?)
ρ (x) Φext (x) d3 x
U=
V
To exhibit the result in terms of the multipoles of the charge distribution, we
expand the external potential in a Taylor series about the origin;
U
=
V
ρ (x) Φext (0) + x · ∇Φext
= qΦext (0) − p · Eext (0) −
0
1
∂ ∂
+ xi xj
Φext
2
∂xi ∂xj
1 ∂Eext,j
2 ∂xi
0
V
0
+···
d3 x
ρ (x) xi xj d3 x + · · ·
where p is the dipole moment with respect to the same origin O. We want to
express the third term using the Qij , but we are missing a term. We can add
it in because it is zero! Since the sources of the external field Eext are not in
the volume V that contains the charge density ρ, then
∇ · Eext = 0 in V
Thus
∂Eext,j 2
∂Eext,j 2
r δ ij =
r = ∇ · Eext r2 = 0
∂xi
∂xj
11
and so
1 ∂Eext.j
2 ∂xi
ρ (x) xi xj d3 x =
0
V
=
Thus
1 ∂Eext.j
2 ∂xi
1 ∂Eext.j
6 ∂xi
0
V
ρ (x) xi xj −
r2
δ ij
3
d3 x
Qij
0
1
∂Eext.j
U = qΦext (0) − p · Eext (0) − Qij
6
∂xi
0
+···
(22)
The second term shows that a dipole is in stable equilbrium in an external field
when it is aligned parallel to Eext . (This is the minimum energy state.)
Similar techniques may be used to express the force and torque on a charge
distribution in terms of its multipoles, as in J Problem 4.5.
12