1000 Solved Problems in Modern Physics

(Grace) #1

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 )−

(^32)


=


(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
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