Charge and Current Distributions
2.1 Multipole Moments
2.1.1 The Cartesian Multipole Expansion
In general the term 1/|r−r| in the expression for the electric potential may be written as
1
|r−r| = 1
√r2+r2−2r·r = 1 r>
1 +
r<
r>
2
−2r>·r<
r>2
(2–5)
where we have definedr> as the greater ofr andr and r< as the lesser ofr and r. This slightly cumbersome notation is necessary to ensure that the binomial ex- pansion of the radical converges. We proceed to expand the radical by the binomial expansion (33) to obtain
1
|r−r| = 1 r>
1−1
2 r2<
r2> −2r>·r<
r2>
+3
2 1 2
1 2!
r<2
r>2 −2r>·r<
r2>
2
+. . .
(2–6)
= 1 r>
+r>·r<
r>3 −1 2
r<2 r>3 + 3
8r>
4(r>·r<)2
r4> +. . . (2–7) where the neglected terms are of order (r</r>)3 or higher. We evaluate the two terms inr2< in such a manner as to separate ther< andr> terms.
3 8r>
4(r>·r<)2 r4> −1
2 r<2 r>3 =
!3(r>·r<)2−r<2r>2"
2r>5 = 3(r·r)2−r2r2
2r5> (2–8)
= 1
2r>5 3,3
i=1j=1
3(xixi)(xjxj)−xixixjxj (2–9)
= 1 2r>5
3,3
i=1j=1
xixj(3xixj−δijr2) (2–10) The primed and unprimed coordinates could, of course, have been exchanged. To be explicit we assume that the source coordinates r are consistently smaller than r (only then does the multipole expansion make sense), and we can write
1
|r−r| =1
r +r·r r3 + 1
2r5 3,3
i=1j=1
xixj
3xixj−δijr2
+. . . (2–11)
leading to
V(r) = 1 4πε0
ρ(r)
|r−r|d3r (2–12)
= 1 4πε0
ρ(r)d3r r +r·
ρ(r)rd3r r3 +
#xixj
ρ(r)!
3xixj−δijr2"
d3r 2r5 +. . .
(2–13)
= 1
4πε0 q
r+r·p r3 +
#xixjQij 2r5 +. . .
(2–14) Hereq is the total charge of the source,
p=
ρ(r)rd3r (2–15)
is the dipole moment of the distribution and Qij ≡
ρ(r)(3xixj−δijr2)d3r (2–16) is theij component of the Cartesian (electric) quadrupole moment tensor. In prin- ciple this expansion could be continued to higher order, but it is rarely fruitful to do so. Thetrace (or sum of diagonal elements) of the quadrupole moment tensor is easily seen to vanish:
Q11+Q22+Q33=
ρ!
(3x2−r2) + (3y2−r2) + (3z2−r2)"
d3r
=
ρ(3x2+ 3y2+ 3z2−3r2)d3r = 0 (2–17) To confuse matters, when a quadrupole has azimuthal symmetry, thezz com- ponent,Q33=−2Q11=−2Q22, is often referred to asthe quadrupole moment.
The multipole moment of a charge distribution generally depends on the choice of origin. Nonetheless, the lowest order nonzero moment is unique, independent of the choice of origin.
Example 2.2: Find the dipole moment of a line charge of lengthaand charge density ρ(r) =λzδ(x)δ(y) forz ∈(−a/2, a/2) and 0 elsewhere.
Solution: Thex and y components of the dipole moment clearly vanish, and the z component is easily obtained from the definition (2–15).
px=py= 0 (Ex 2.2.1)
pz= a/2
−a/2λz2dz= λz3 3
a/2
−a/2
= λa3
12 (Ex 2.2.2)
We note that as the net charge is zero, the dipole moment should be unique.
Shifting the origin to−12aalong thez axis, for instance, gives us a new charge den- sity,ρ=λ(z−12a)δ(x)δ(y) and the dipole moment with the shifted origin becomes
pz= a
0
λ(z−12a)zdz= λa3
12 (Ex 2.2.3)
Example 2.3: Find the quadrupole moment of the charge distribution shown in Figure 2.2.
Figure 2.2: Two positive charges occupy diagonally opposite corners, and two equal, negative charges occupy the remaining corners.
Solution: Thexx(1,1) component of the quadrupole tensor is again easily evaluated from its definition (2–16) replacing the integral over the source by a sum over the source particles:
Qxx= (+q)
3 a
2 2
− a2
2
+ (−q)
3 −a
2 2
− a2
2
+(+q)
3 −a
2 2
− a2
2
+ (−q)
3 a
2 2
− a2
2
= 0 (Ex 2.3.1) In the same manner,Qyy andQzz are found to be 0. Qxy=Qyx is
Qxy= 3q a
2 ·a 2
−3q a
2 · −a 2
+ 3q
−a 2 ·−a
2
−3q −a
2 ·a 2
(Ex 2.3.2)
= 3qa2 (Ex 2.3.3)
andQxz=Qzx vanishes.
Example 2.4: Find the quadrupole moment of a uniformly charged ellipsoid of revo- lution, with semi-major axisaalong thez axis and semi-minor axisb bearing total chargeQ.
Solution: We begin by finding thezz component of the quadrupole moment tensor.
The equation of the ellipsoid can be expressed as s2
b2 +z2
a2 = 1 (Ex 2.4.1)
wheres=
x2+y2is the cylindrical radius. Then
Qzz =
volume
(3z2−r2)ρd3r= a z=−a
b√
1−(z/a)2
s=0
ρ(2z2−s2)2πsdsdz (Ex 2.4.2)
= 2πρ a
−a
z2b2
1−z2
a2
−1 4b4
1−z2
a2 2
dz (Ex 2.4.3)
=158πρab2(a2−b2) (Ex 2.4.4)
The total charge in the volume is 43πρab2, so that Qzz = 25Q(a2−b2). Rather than compute thexx andyycomponents, we make use of the fact that the trace of the quadrupole tensor vanishes, requiring Qxx =Qyy =−12Qzz. The off-diagonal elements vanish.
The net charge on the ellipsoid is not zero, hence we conclude that the com- puted quadrupole moment is not unique. The identical calculation has considerable application in gravitational theory with mass replacing charge.
The vector potentialA can be expanded in precisely the same fashion asV. Let us consider the expression for the vector potential of a plane current loop (1–52)
A( r) = µ0 4π
J(r)d3r
|r−r|
= µ0I 4π
d
|r−r| (2–18)
in which we expand the denominator using the binomial theorem (33) as A( r) =µ0I
4πr d+ µ0I
4πr3 d(r·r) +. . . (2–19) The first term vanishes identically, as the sum of vector displacements around a loop is zero. The second term is the contribution from the magnetic dipole moment of the loop. Using identity (17),
(r·r)d =
dS×∇(r·r) =−
r×dS (2–20) we find for the vector potential of a small current loop
A( r) = µ0I 4πr3
dS×r= µ0 4πr3
I
dS
×r (2–21)
The bracketed term,
I
dS ≡m (2–22)
is the magnetic dipole moment of the loop. We have then, to first order, the magnetic vector potential
A( r) = µ0
4πr3m ×r (2–23)
Other useful forms for the magnetic moments are easily obtained by generalizing to not necessarily planar current loops:
m=I
dS= I
2 r×d= 1 2
r×J(r)d3r (2–24) Example 2.5: Find the magnetic dipole moment of a uniformly charged sphere of
radiusarotating with angular velocityω about thez axis.
Solution: The magnetic moment may be found from m = 12
(r×J)d3r. The integrand can be rearranged as follows.
r×J=r×ρv=ρr×(ω×r) =ρ!
r2ω−(r·ω)r"
(Ex 2.5.1)
=ρω
$
r2kˆ−(rcosθ)r
%
(Ex 2.5.2)
=ρω
r2ˆk−r2cos2θkˆ−r2sinθcosθˆs
(Ex 2.5.3) where ˆsis the unit cylindrical radial basis vector. The last term of the preceding expression (the ˆsterm) vanishes when integrated over ϕ, leaving
m= 12ρωˆk a
0
4π
0
r2r2drdΩ
− a
0
π
0
2π
0
r2cos2θr2drsinθdθdϕ
( Ex 2.5.4)
= 12ρ ωkˆ 4πa5
5 −2 3
2π a5 5
=4πρωa5
15 ˆk Ex 2.5.5)
To obtain the magnetic induction field of a magnetic dipole at the origin, we merely take the curl of the vector potential (2–23) yielding
B(r) =∇ × A = µ0 4π
−m
r3 +3(m ·r)r r5
(2–25) The magnetic induction field of a small loop is illustrated in Figure 2.3. The dipole field is obtained as the area of the loop is shrunk to zero. The expression (2–25) can also be written as
B( r) =−µ0∇ m ·r
4πr3
(2–26) Recalling that outside current sources B can be expressed as B = −µ0∇V m, we deduce that the scalar magnetic potential for a small current loop located at the origin is
Vm(r) = m ·r
4πr3 (2–27)
Figure 2.3:The magnetic induction field of a small current loop.
This form of the magnetic scalar potential allows us to easily generate our earlier result (1–64) for the arbitrary current loop. If the loop is considered as a sum of small bordering loops (each of which is adequately described as a dipole) situated atr (Figure 2.4), we find
Figure 2.4: The scalar magnetic potential of the loop may be found as the sum of the potentials of each of the small loops. The countercurrents along the boundary of the inner loops are fictitious and cancel exactly.
Vm=
dVm=
d m(r)·(r−r) 4π|r−r|3 =
Id A·R
4πR3 =−IΩ
4π (2–28)
Although only the current along the outside of the loop is real, we supply cancelling currents along the boundaries of contiguous loops. The fact that we need look at no further terms than those of the magnetic dipole to generate the correct form of the magnetic scalar potential suggests there is little to be gained from considering higher order terms.