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PHY481 - Lecture 13: Solutions to Laplace’s equationGriffiths: Chapter 3

Spherical polar co-ordinatesThe Laplacian in spherical polar co-ordinates is

∇2V =1r2

∂

∂r(r2

∂V

∂r) +

1r2sinθ

∂

∂θ(sinθ

∂V

∂θ) +

1r2sin2θ

∂2V

∂φ2= 0 (1)

We only consider the case where there is no dependence on φ, so V (r, θ, φ) → V (r, θ) = R(r)Θ(θ). Note thatin Quantum mechanics solutions to the time independent Schrodinger equation with a spherical potential lead tosolutions of the form R(r)Y m

l (θ, φ). The solutions we find here are for the special case m = 0, so that the LegendrePolynomials we find below are related to the spherical Harmonics through Y 0

l (θ, φ) = alPl(cosθ), where al is a constantthat is needed due to a different choice of normalization for the spherical harmonics. Assuming no dependence on φand making the subsitution V (r, θ) = R(r)Θ(θ) into Laplace’s equation, then dividing through by RΘ gives,

∇2V =1R

∂

∂r(r2

∂R

∂r) = − 1

Θsinθ∂

∂θ(sinθ

∂Θ∂θ

) = C = l(l + 1) (2)

The reason why we choose the separation constant C = l(l+ 1) will become clear later. With this choice, we have thetwo equations,

∂

∂r(r2

∂R

∂r)− l(l + 1)R = 0 (3)

and

1sinθ

∂

∂θ(sinθ

∂Θ∂θ

) + l(l + 1)Θ = 0 (4)

The R equation is solved by R(r) = A(l)rl + B(l)/rl+1. The Θ equation is more interesting and requires a seriessolution. First we make the substitution, u = cosθ and P (u) = Θ(θ), which imply that

∂

∂θ= −sinθ ∂

∂u(5)

so that,

d

du[(1− u2)

∂P (u)∂u

] + l(l + 1)P (u) = 0 (6)

This equation is call Legendre’s equation and is solved using a series solution, P (u) =∑

n Cnun.

Substituting this series into Legendre’s equation we find that,

∂

∂u

∞∑n=1

(1− u2)Cnnun−1 +

∞∑n=0

l(l + 1)Cnun = 0 (7)

and,

∞∑n=2

Cnn(n− 1)un−2 −∞∑

n=1

Cnn(n+ 1)un +∞∑

n=0

l(l + 1)Cnun = 0 (8)

The coefficient of un in this equation must be zero, implying that,

Cn+2(n+ 2)(n+ 1) + Cn[l(l + 1)− n(n+ 1)] = 0; hence Cn+2 = Cnn(n+ 1)− l(l + 1)

(n+ 1)(n+ 2)(9)

For a given value of l, we get two series of solutions, one starting with a value of C0 and producing C2, C4... andthe other starting with C1 and producing C3, C5... The key physical observation is that if l is an integer, the seriesterminates at n = l, leading to a finite polynomial solution. In contrast if l is not an integer, the series does notterminate and the coefficients remain finite at infinity, a solution that does not have physicial meaning. The constants

2

C0 and C1 are fixed by requiring that the polynomials be normalized on the interval [−1, 1], which corresponds to[−π, π] in the original variable θ. The normalization and orthogonality conditions are,∫ 1

−1

Pl(u)Pm(u)du =2

2l + 1δml (10)

The first few Legendre polynomials are,

P0(u) = 1; P1(u) = u; P2(u) = (3u2 − 1)/2; P3(u) = (5u3 − 3u)/2 (11)

Even l correspond to even functins, while odd l correspond to odd functions. Instead of using Eq. (8) to find thecoefficients and Eq. (9) to normalize them, it is more convenient to use a direct recursion formula called Bonnet’sformula,

(l + 1)Pl+1(u) = (2l + 1)uPl(u)− lPl−1(u) with P0(u) = 1;P1(u) = u (12)

Setting up Problem 3.22We need to find the potential inside and outside a sphere of radius R. There is a charge density σ0 on the upper

half of the spherical surface and a charge density on the lower half of the sphere surface. The general solution is,

V (r, θ) =∑l=0

(Alrl +

Bl

rl+1)Pl(cosθ) (13)

For r > R convergence at infinity requires that Al = 0, while for r < R convergence requires that Bl = 0, so we have,

V (r > R, θ) =∑l=0

Bl

rl+1Pl(cosθ); V (r, θ) =

∑l=0

AlrlPl(cosθ) (14)

Now we need to impose the boundary conditions. First impose the condition that the potential is continuous at r = R(or equivalently that the parallel electric field is continuous), so that∑

l=0

Bl

Rl+1Pl(cosθ) =

∑l=0

AlRlPl(cosθ) so that

Bl

Rl+1= AlR

l. (15)

Next we need to impose the relation between the perpendicular electric field and the charge density,

Er(r = R+, θ)− Er(r = R−, θ) =σ

ε0(16)

where σ = σ0 for θ < π/2 and σ = −σ0 for π/2 < θ < π. Due to the symmetry of σ we only need to keep odd termsin the sum over l.

— Aside — To prove orthogonality, consider Legendre’s equation for two polynomials Pm and Pn as follows,

Pn[(1− u2)P ′′m − 2uP ′m +m(m+ 1)Pm] = 0; Pm[(1− u2)P ′′n − 2uP ′n + n(n+ 1)Pn] = 0 (17)

where the primes indicate a derivative with respect to u. Subtracting the first from the second of these two equationsand using d/du(PnP

′m − PmP

′n) = PnP

′′m − PmP

′′n gives,

(1− u)2d

du(PnP

′m − PmP

′n)− 2u(PnP

′m − PmP

′n) + (m(m+ 1)− n(n+ 1))PnPm = 0 (18)

which is equal to,

d

du[(1− u2)(PnP

′m − PmP

′n)] + (m(m+ 1)− n(n+ 1))PnPm = 0 (19)

Now we integrate this equation over the interval [−1, 1]. This integral produces zero for the first term of the equation,so the second must also be zero. Therefore if m 6= n the integral of PnPm over this interval must be zero, provingothogonality. — End of Aside —

Since Pl(cosθ) are an orthogonal (and complete) set, we can use them to expand any piecewise continuous function.In spherical co-ordinates, we use them as the basis for a Fourier-type analysis.