pansion for the potential as a Legendre polynomial series in a neighbourhood

of the origin.

8.2. SPHERICAL HARMONICS 225

| R’r|

r

R

θ

O

Geometry for generating function.

Let start by assuming that |r| < |R|. We know that in this region the point

charge potential 1/|r ’ R| is a solution of Laplace™s equation , and so we can

expand

∞

1 1

Al r l Pl (cos θ).

≡√2 = (8.34)

|r ’ R| r + R2 ’ 2rR cos θ l=0

We also know that the coe¬cients Bl are zero because • is ¬nite when r = 0.

We can ¬nd the coe¬cients Al by setting θ = 0 and Taylor expanding

2

1 1 1 r r

+··· ,

= = 1+ + r < R. (8.35)

|r ’ R| R’r R R R

By comparing the two series, we ¬nd that Al = R’l’1 . Thus

∞ l

1 1 r

√ = Pl (cos θ), r < R. (8.36)

R R

r 2 + R2 ’ 2rR cos θ l=0

This last expression is the generating function formula for Legendre polyno-

mials. It is also a useful formula to have in your long-term memory.

If |r| > |R|, then we must take

∞

1 1

Bl r ’l’1 Pl (cos θ),

≡√2 = (8.37)

|r ’ R| 2 ’ 2rR cos θ

r +R l=0

because we know that • tends to zero when r = ∞. We now set θ = 0 and

compare with

2

1 1 1 R R

+··· ,

= = 1+ + R < r, (8.38)

|r ’ R| r’R r r r

to get

l

∞

1 1 R

√ = Pl (cos θ), R < r. (8.39)

r r

r 2 + R2 ’ 2rR cos θ l=0

226 CHAPTER 8. SPECIAL FUNCTIONS I

Example: A planet is spinning on its axis and so its shape deviates slightly

from a perfect sphere. The position of its surface is given by

R(θ, φ) = R0 + ·P2 (cos θ). (8.40)

Observe that, to ¬rst order in ·, this deformation does not alter the volume

of the body. Assuming that the planet has a uniform density ρ0 , compute

the external gravitational potential of the planet.

θR

R0

Deformed planet.

The gravitational potential obeys Poisson™s equation

2

φ = 4πGρ(x), (8.41)

where G is Newton™s gravitational constant. We decompose the gravitating

mass into a uniform undeformed sphere, which has external potential

43 G

φ0,ext = ’ πR0 ρ0 , r > R0 , (8.42)

3 r

and a thin spherical shell of areal mass-density

σ(θ) = ρ0 ·P2 (cos θ). (8.43)

The thin shell gives rise to a potential

φ1,int (r, θ) = Ar 2 P2 (cos θ), r < R0 , (8.44)

and

1

φ1,ext (r, θ) = B P2 (cos θ), r > R0 . (8.45)

r3

8.2. SPHERICAL HARMONICS 227

At the shell we must have φ1,int = φ1,ext and

‚φ1,ext ‚φ1,int

’ = 4πGσ(θ). (8.46)

‚r ‚r

’5

Thus A = BR0 , and

4 4

B = ’ πG·ρ0 R0 . (8.47)

5

Putting this together, we have

4 31 4 4 P2 (cos θ)

+ O(· 2 ),

φ(r, θ) = ’ ’

πGρ0 R0 πG·ρ0 R0 r > R0 .

3

3 r5 r

(8.48)

8.2.2 Spherical Harmonics

When we do not have axisymmetry, we need the full set of spherical harmon-

ics. These involve solutions of

m2

d 2d

(1 ’ x ) + l(l + 1) ’ ¦ = 0, () (8.49)

1 ’ x2

dx dx

l

which is the associated Legendre equation and has solutions, P|m| (x), for inte-

ger l and m. By substituting y = (1’x2 )m/2 z(x) into ( ), and comparing the

resulting equation for z(x) with the m-th derivative of Legendre™s equation,

we ¬nd that

m

2 m/2 d

l m

P|m| (x) = (’1) (1 ’ x ) Pl (x). (8.50)

dxm

l

Since Pl is a polynomial of degree l we observe that P|m| (x) = 0 if m > l.

For each l, the allowed values of m are ’l, ’(l ’ 1), . . . , (l ’ 1), l, a total of

2l + 1 possibilities.

The spherical harmonics are the normalized product of these associated

Legendre functions with the corresponding eimφ :

Ym (θ, φ) ∝ P|m| (cos θ)eimφ .

l l

(8.51)

The ¬rst few are

√1