Jackson notes 2018 1 Spherical multipole moments Suppose we have a charge distribution ρ (x) whereallofthechargeiscon- tained within a spherical region of radius R, as shown in the diagram. Then there is no charge in the region r>Randsowemaywritethepotentialinthat region as a solution of Laplace’s equation in spherical coordinates. With the constraint that Φ 0 as r , we have → →∞ +l 1 ∞ 4π qlm Φ (r, θ, φ)= l+1 Ylm (θ, φ) r>R (1) 4πε0 2l +1r l=0 m= l − 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.29): 1 ρ (x) 3 Φ (r, θ, φ)= d x (2) 4πε0 x x | − | andthenexpandthefactor1/ x x in spherical harmonics using J eqn 3.70 | − | l l 1 ∞ 4π r< 3 Φ (r, θ, φ)= ρ (x) l+1 Ylm (θ, φ) Ylm∗ θ, φ d x 4πε0 2l +1r l=0 m= l > − l l 1 ∞ 4π r< 3 = Ylm (θ, φ) ρ (x) l+1 Ylm∗ θ, φ d x 4πε0 2l +1 r l=0 m= l > − 1 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 ≤ 1 4π Ylm (θ, φ) l 3 Φ (r>R,θ, φ)= l+1 ρ (x)(r) Ylm∗ θ, φ d x (3) 4πε0 2l +1 r lm and then comparing equations (1) and (3) we find l 3 qlm = ρ (x)(r) Ylm∗ θ, φ d x (4) These are the spherical multipole moments of the source. With l =1we have the dipole, l =2the quadrupole etc. 2 Cartesian multipole moments 1 This time we expand x x in expression (2) in a Taylor series about x =0: | − 3| 1 1 1 3 ∂ ∂ 1 3 Φ (x)=k ρ (x) + x + xi xj + d x r · ∇ R 2 ∂x ∂x R ··· x3=0 i,j=1 i j x3=0 Now let’s evaluate the derivatives: 1 (x x) x x = x − = x · ∇ R · R3 · r3 x3=0 x3=0 and ∂ ∂ 1 ∂ xj xj δij (xi xi) xj xj = −3 = − 3 +3 − 5 − ∂xi ∂xj R ∂xi R R R x3=0 x3=0 x3=0 δij xixj = − +3 (5) r3 r5 Putting these expressions back into the Taylor series, we get: 1 x 1 x x δ Φ (x)=k ρ (x ) + x + x x 3 i j ij + d3x r r3 2 i j r5 r3 · i,j − ··· We may integrate term by term because the Taylor series is uniformly conver- gent. 1 3 x 3 1 xixj δij 3 Φ (x)=k ρ (x) d x + xρ (x) d x + 3 ρ (x) xx d x + r r3 2 r5 r3 i j · i,j − ··· 2 The first two terms are easily integrated to give: q p x k 2 xi xj 3 Φ (x)=k + k · + 3xixj r δij ρ (x) d x + (6) r r3 r5 − 2 ··· i,j where q is the total charge in the distribution and 3 p ρ (x) x d x (7) ≡ is the Cartesian dipole moment. The last term in (6) needs some work, since we would like to express the result in terms of the quadrupole tensor: 2 3 Qij ρ (x) 3x x (r) δij d x (8) ≡ i j − 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 2 2 2 2 2 2 2 r (x) + r (y) + r (z) =(r) r =(r) x +(r) y +(r) z and so we may interchange prime and unprime and rewrite the integral as: 2 2 3 3 3 ρ (x) xx 3xixj r δij d x = xixj ρ (x) 3x x r δij d x i j − i j − i,j i,j = xixjQij i,j Thus we have for the potential 3 1 q p x 1 xixj Φ (x)= + · + Q + (9) 4πε r r3 2 r5 ij 0 i,j=1 ··· The quadrupole tensor Qij (8) is symmetric, real-valued and traceless. It has three real eigenvalues. The total charge is independent of our choice of origin. In general, the lowest non-zero multipole is independent of origin, but higher order multipoles do depend on the origin. (See Jackson Problem 4.4.) 3 3 Relations between the two sets of multipoles Comparing the expressions (1) and (9) allows us to relate the two sets of mul- tipoles. 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, 3 1 3 q q00 = ρ (x) Y ∗ θ, φ d x = ρ (x) d x = (10) 00 √4π √4π and the first term in the spherical multipole expansion (1) is 1 q00 1 q 1 1 1 q Φ0 = 4π Y00 = = 4πε0 r ε0 √4π r √4π 4πε0 r which is also the firstterminexpression(9). With l =1( 1/r2 term) there are three contributions, with m = 1 and zero. First note that ± l 3 ql, m = ρ (x)(r) Yl∗ m θ, φ d x − − m l 3 =(1) ρ (x)(r) Ylm θ, φ d x − m =(1) q∗ (11) − lm So there are actually only two values to calculate. With l =1,m=0 3 q10 = ρ (x) rY10∗ θ, φ d x 3 3 = ρ (x ) r cos θ d x 4π 3 3 q = ρ (x ) z d3x = p (12) 10 4π 4π z whereweusedthez component of (7). With l =1,m=1− 3 q11 = ρ (x) rY11∗ θ, φ d x 3 iφ3 3 = ρ (x) r sin θe− d x − 8π 3 3 = ρ (x) r sin θ cos φ i sin φ d x − 8π − 3 3 = ρ (x)(x iy) d x − 8π − 3 q = (ip p ) (13) 11 8π y x − 4 Equivalently, 2π 1 2π 4π p = (q + q ) ,p= (q q ) ,p= q (14) x 3 11 11∗ y i 3 11 11∗ z 3 10 − − The corresponding l =1term in the potential is 1 +1 4π q Φ = 1m Y (θ, φ) 1 4πε 3 r2 1m 0 m= 1 − 1 4π = 2 (q11Y11 (θ, φ)+q10Y10 + q1, 1Y1, 1) 4πε0r 3 − − 1 4π = 2 (q11Y11 (θ, φ)+q10Y10 +( 1) q11∗ ( 1) Y11∗ ) 4πε0r 3 − − 1 = 2 2Re[q11Y11 (θ, φ)] + q10Y10 3ε0r { } Inserting our expressions for q1min terms of the components of p, we have 1 3 3 3 3 Φ = 2Re (ip p ) ( sin θ) eiφ + p cos θ 1 3ε r2 8π y x 8π 4π z 4π 0 − − 1 p rˆ = 2 (px sin θ cos φ + py sin θ sin φ + pz cos θ)= · 2 4πε0r 4πε0r which is the second term in (9). (See also Notes 1 eqn 33.) 3 With l =2(1/r term) we have m =2, 1, and 0. The multipole moment q20 is 2 3 q20 = ρ (x)(r) Y20∗ θ, φ d x 2 1 5 2 3 = ρ (x)(r) 3cos θ 1 d x 2 4π − 1 5 2 2 3 = ρ (x) 3(z) (r) d x 2 4π − 1 5 q = Q (15) 20 2 4π 33 where we used the (3,3) component of (8). Similarly, we may show that 1 15 q = (iQ Q ) (16) 21 3 8π 23 13 − and 1 15 q = (Q Q 2iQ ) (17) 22 12 2π 11 22 12 − − Equivalently: 2π Q Q =Re 12 q 11 22 15 22 − 5 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 Aringofchargeofradiusa 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) λ0 cos φδ(µ) δ (r a) ρ (x)= − r and thus l 3 qlm = ρ (x) r Ylm∗ (θ, φ) d x λ0 cos φδ(µ) δ (r a) l 2 = − r Y ∗ (θ, φ) r dµdφdr r lm 2π +1 2l +1(l m)! ∞ l+1 m imφ = λ0 − cos φδ(µ) δ (r a) r Pl (µ) e− dµdφdr 4π (l + m)! 0 0 1 − − Using the sifting property, we have immediately 2π 2l +1(l m)! l+1 m imφ q = λ − a P (0) cos φe− dφ lm 0 4π (l + m)! l 0 2π iφ iφ 2l +1(l m)! l+1 m e + e− imφ = λ0 − a P (0) e− dφ 4π (l + m)! l 2 0 The integral over φ is zero unless m = 1, so ± 2l +1(l m)! l+1 m qlm = λ0 − a Pl (0) π (δm1 + δm, 1) 4π (l + m)! − Then 2l +1 l+1 (l 1)! 1 ql,1 = λ0π a − Pl (0) 4π (l +1)! We must have l>0, since with l =0there is no m =1.