Math of Multipole Expansion

In this note I explain how to expand

1 1 = √ (1) |R − r| R2 + r2 − 2Rr cos θ

into a power series in (r/R) for r < R, and then apply this expansion to the Coulomb potential.

Let’s start with the few leading terms in this expansion for r R. For the sake of compactness, let’s denote

r α = 1, x = cos θ, β = 2αx − α2 1. R

In these notations,

1 1 1 1 √ = = × √ . (2) R2 + r2 − 2Rr cos θ pR2(1 + α2 − 2αx) R 1 − β √ Next, let’s expand the 1/ 1 − β into powers of β:

∞ 1 X (2n − 1)!! S = √ = 1 + × βn = 1 + 1 β + 3 β2 + 5 β3 + ··· . (3) 1 − β 2n n! 2 8 16 n=1

Now, remember that β = 2αx − α2, plug that into the above expansion, then truncate it to powers of α no larger than 3:

1 2 3 2 2 5 2 3 S = 1 + 2 (2αx − α ) + 8 (2αx − α ) + 16 (2αx − α ) + ··· 1 2 = 1 + αx − 2 α 3 2 2 3 3 + 2 α x − 2 α x + ··· + 5 α3x3 + ··· 2 (4) + ··· 3x2 − 1 5x3 − 3x = 1 + α × x + α2 × + α3 × + ··· 2 2 2 3 = 1 + α × P1(x) + α × P2(X) + α × P3(x) + ··· where P1(x), P2(x), and P3(x) are the Legendre polynomials of respective degrees 1, 2, 3.

1 Plugging this result back into eq. (2), we obtain

1 1 r r2 r3 √ = + ×P1(cos θ) + ×P2(cos θ) + ×P3(cos θ) + ··· . (5) R2 + r2 − 2Rr cos θ R R2 R3 R4

The expansion (5) clearly suggests similar terms for the higher powers of r/R, and indeed there is a theorem:

∞ 1 X r` For r < R, √ = × P (cos θ). (6) 2 2 R`+1 ` R + r − 2Rr cos θ `=0

Proof: Let’s start with the Rodrigues formula for the Legendre polynomials,

1 d` P (x) = (x2 − 1)`, (7) ` 2` `! dx` and the residue method for taking contour integrals in the complex plane,

I n dz f(z) f(z) 1 d f(z) n+1 = Residue n+1 = n (8) 2πi (z − x) (z − x) @z=x n! dz @z=x Γ

provided the contour Γ circles x and that the function f(x) is analytic and has no singularities inside the contour Γ. Applying this method in reverse — i.e., turning nth into a contour integral — to the Rodrigues formula, we obtain

1 I dz (z2 − 1)` P (x) = (9) ` 2` 2πi (z − x)`+1 Γ

where the contour Γ circles x. Now let’s plug this formula into the series on the RHS of

2 eq. (6): ∞ X r` series = × P (x) R`+1 ` `=0 ∞ X r` 1 I dz (z2 − 1)` = × R`+1 2` 2πi (z − x)`+1 `=0 Γ hh putting the sum inside the integral ii ∞ I dz X r` (z2 − 1)` = × 2πi R`+1 2`(z − x)`+1 Γ `=0 (10) ∞ ` I dz 1 X r(z2 − 1) = × 2πi R(z − x) 2R(z − t) Γ `=0 I dz 1 1 = × 2 2πi R(z − x) 1 − r(z −1) Γ 2R(z−x) I dz −2 = . 2πi rz2 − 2Rz + 2Rx − r Γ

Note: before the summation, each term on the third line has poles at z = x and at z = ∞, but after the summation, both poles have moved to the roots of the quadratic equation

rz2 − 2Rz + 2Rx − r = 0, (11) thus √ R ± R2 − 2rRx + r2 2R z = ; for r R, z ≈ → ∞, while z ≈ x. (12) 1,2 r 1 r 2

This tells us how to choose the integration contour Γ: It should circle around x and have enough room to accommodate the shifting of the pole from x to z2, but it should not include the other pole at z1 which have moved in from the inﬁnity. Consequently, evaluating the integral on the bottom line of eq. (10) by the residue method, we have

I dz −2 −2 = Residue . (13) 2πi rz2 − 2Rz + 2Rx − r rz2 − 2Rz + 2Rx − r @z=z2 Γ

3 Speciﬁcally,

−2 −2 1 2 = × , (14) rz − 2Rz + 2Rx − r r (z − z1)(z − z2)

so the residue of this function at z = z2 is simply

−2 1 −2 r 1 Residue = × = × √ = +√ . (15) r z2 − z1 r −2 R2 − 2rRx + r2 R2 − 2rRx + r2

Thus,

∞ X r` 1 the series = × P (x) = √ , (16) R`+1 ` 2 2 `=0 R − 2rRx + r which proves the theorem (6), quod erat demonstrandum.

Aside: A few words about the convergence of the expansion (6). For x = cos θ ranging between −1 and +1, all the Legendre polynomials take values between −1 and +1. Consequently, the series (6) converges for any r < R. If we analytically continue it to the complex r, it would converge for |r| < R; in other words, √ it has radius of convergence = R. Indeed, as a function of complex r, the 1/ ··· on the LHS of (6) has singularities at

r1,2 = R cos θ ± iR sin θ, |r1,2| = R, and that’s what sets the radius of convergence to |r| < R.

For r > R we may longer expand the inverse distance into powers of r/R. Instead, we may expand it into powers of the inverse ratio R/r:

∞ 1 X R` For r > R, √ = × P`(cos θ), (17) 2 2 r`+1 R + r − 2Rr cos θ `=0

which works exactly like eq. (6) once we exchange r ↔ R.

Physically, the expansion (6) is useful for potentials far outside complicated charged bodies, while the inverse expansion (17) is useful for potentials deep inside a cavity.

4 Multipole Expansion of the Electric Potential

Now consider the Coulomb potential of some continuous charge distribution ρ(~r),

1 ZZZ ρ(r) d3Vol V (R~) = . (18) 4π0 |R − r|

Suppose all the charges are limited to some compact volume, while we want to know the potential far away from that volume, so in the integral (18) we always have r R. Conse- quently, we may expand the denominator in the Coulomb potential according to the Theo- rem (6), thus

ZZZ ∞ ` 1 3 X r V (R) = d Vol ρ(r) × `+1 × P`(cos θ) 4π0 R `=0 (19) ∞ ZZZ X 1 3 ` = `+1 × d Vol ρ(r) × r P`(cos θ) 4π0 R `=0 where θ is the angle between the radius-vectors R and r. In terms of the unit vectors Rb and br in the directions of R and r,

cos θ = Rb · br. (20)

Consequently, we may decompose the potential V (R) of the charges ρ(r) into a series of multipole potentials,

∞ X M`(Rb ) V (R) = `+1 (21) 4π0 R `=0

where ZZZ ` 3 M`(Rb ) = r P`(br · Rb ) × ρ(r) d Vol. (22)

is the 2`–pole moment of the charge distribution ρ(r). Or rather, it’s the component of the 2`–pole moment in the direction Rb .

5 Let’s take a closer look at these components:

• ` = 0 • The monopole moment M0 is simply the net charge of the distribution,

ZZZ 3 net M0 = ρ(r) d Vol = Q , (23)

and it obviously does not depend on the direction Rb , hence isotropic monopole potential, Qnet Vmonopole = . (24) 4π0 R

• ` = 1 • The dipole moment is a vector

ZZZ p = r ρ(r) d3Vol (25)

and the M1(Rb ) in the series (21) is simply

M1(Rb ) = Rb · p, (26)

hence the dipole’s potential p · Rb Vdipole = 2 . (27) 4π0 R

1 To see how this works, note that r × P1(Rb · br) = r(Rb · br) = Rb · r, hence ZZZ ZZZ 3 3 M1(Rb ) = (Rb · r)ρ(r) d Vol = Rb · r ρ(r) d Vol = Rb · p. (28)

• ` = 2 • The quadrupole moment is a 2-index symmetric tensor

ZZZ 3 1 2 3 Qi,j = 2 rirj − 2 δi,jr ρ(r) d Vol (29)

where the indices i, j run over x, y, z, the ri are the components of the vector r, and

δi,j is the Kronecker’s delta (1 for i = j and 0 for i 6= j).

6 To see the relation between this tensor and the M2(Rb ) in the series (21), let’s expand the second Legendre polynomial P2(cos θ) = P2(Rb · br):

3 2 1 P2(Rb · br) = 2 (Rb · br) − 2 , (30) !2 !   3 2 3 X 3 X X (Rb · r) = Rbiri = Rbiri × Rbjrj 2 b 2 b 2 b  b  i i j 3 X = RbiRbjrirj , (31) 2 b b i,j

1 1 2 = 2 Rb · Rb hh since Rb is a unit vector ii 1 X 1 X = RbiRbi = δi,j × RbiRbj , (32) 2 2 i i,j

X 3 1 hence P2(Rb · br) = RbiRbj 2 rbirbj − 2 δij , (33) i,j

2 X 3 1 2 and r × P2(Rb · br) = RbiRbj 2 rirj − 2 r δij . (34) i,j

Plugging the last line here into eq. (22) for ` = 2, we obtain

ZZZ X 3 1 2 3 M2(Rb ) = RbiRbj 2 rirj − 2 r δij × ρ(r) d Vol i,j ZZZ X 3 1 2 3 = RbiRbj × 2 rirj − 2 r δij × ρ(r) d Vol (35) i,j X = RbiRbj × Qi,j , i,j hence the quadrupole potential

P i,j Qi,jRbiRbj Vquadrupole(R) = 3 . (36) 4π0 R

7 • ` ≥ 3 • The higher multipoles are `–index symmetric tensors. For example, the octupole moment is the 3-index tensor

ZZZ 5 1 1 1 3 Oi,j,k = 2 rirjrk − 2 δi,jrk − 2 δi,krj − 2 δj,kri × ρ(r) d Vol (37)

whose potential is P i,j,k Oi,j,kRbiRbjRbk Voctupole(R) = 4 . (38) 4π0 R Likewise, for higher ` the potential has form

P (`) i,j,...,n Mi,j,...,nRbiRbj ··· Rbn V2`−pole(R) = `+1 (39) 4π0 R

where ZZZ homogeneous polynomial M(`) = × ρ(r) d3Vol (40) i,j,...,n of degree ` in x, y, z

where the speciﬁc form of the degree–` polynomial follows from the P`(cos θ).

Axial Symmetry

For the axially symmetric charge distributions ρ(r, θ, φ) = ρ(r, θ only), we may re-express the angular dependence of the multipole expansion using the following Lemma: Let (θ, φ) be the spherical angles of the direction Rb while (θ0, φ0) are spherical angles of the direction br, then

2π Z 0 dφ 0 P`(Rb · r) = P`(cos θ) × P`(cos θ ). (41) 2π b 0

Consequently, for an axially symmetric charge distribution

ZZZ ` 0 2 0 0 0 M`(Rb ) = r P`(Rb · br) × ρ(r, θ ) × r sin θ dr dθ dφ (`) = P`(cos θ) × Mz···z (42)

8 ZZZ (`) ` 0 0 2 0 0 0 where Mz···z = r P`(cos θ ) × ρ(r, θ ) × r sin θ dr dθ dφ (43)

` is the z, . . . , z component of the 2 –pole vector or tensor, for example pz, Qz,z, or Oz,z,z. For the axially symmetric charge distribution it’s the only independent component, and it’s also the only component we need for expanding the potential:

(`) Mz···z P`(cos θ) V (R, θ) = × `+1 . (44) 4π0 R

You should see examples of such expansion in your homework.