1000 Solved Problems in Modern Physics

1.3 Solutions 71

1.80 Legendre’s equation is

(1−x^2 )

∂^2 Pn(x)

∂x^2

− 2 x

∂Pn(x)

∂x

+n(n+1)Pn(x)=0(1)

Putx=cosθ, Eq. (1) then becomes

sin^2 θ

∂^2 Pn

∂cos^2 θ

−2 cosθ

∂Pn

∂cosθ

+n(n+1)Pn=0(2)

For largen,n(n+1)→n^2 , and cosθ→1 for smallθ,

sin^2 θ

∂^2 Pn

∂cos^2 θ

− 2

∂Pn

∂cosθ

+n^2 Pn=0(3)

Now, Bessel’s equation of zero order is

x^2

d^2 J 0 (x)

dx^2

+x

d

dx

J 0 (x)+x^2 J 0 (x)=0(4)

Lettingx= 2 nsinθ/ 2 =nsinθ,in(4)forsmallθ, and noting that cosθ→

1, after simple manipulation we get

sin^2 θ

d^2 J 0 (nsinθ)

dcos^2 θ

− 2 d

dJ 0 (nsinθ)

dcosθ

+n^2 J 0 (nsinθ)=0(5)

Comparing (5) with (3), we conclude that

Pn(cosθ)→J 0 (nsinθ)

1.81 T(x,s)=(1− 2 sx+s^2 )−^1 /^2 =

∑

pl(x)sl

(1)

(a) Differentiate (1) with respect tos.

∂T

∂s

=(x−s)(1− 2 sx+s^2 )−

3

2

=

∑

(x−s)(1− 2 sx+s^2 )−^1 pl(x)sl

=

∑

lpl(x)sl−^1

Multiply by (1− 2 sx+s^2 )

∑

(x−s)plsl=

∑

lplsl−^1 (1− 2 sx+s^2 )

Equate the coefficients ofsl

xpl−pl− 1 =(l+1)pl+ 1 − 2 xlpl+(l−1)pl− 1

or (l+1)pl+ 1 =(2l+1)xpl−lpl− 1