Fields, Potentials and Equations of Motion
51
then the first term is
1
r
∂
∂r
r
∂V
∂r
=
1
r
∞
k=1
∞
l=0
M k,l (s) cos (lφ + θ k,l ) k
2 r
k−1
=
∞
k=0
∞
l=0
M k,l (s) cos (lφ + θ k,l ) k
2 r
k−2 ,
where we let the sum start at k = 0 in the last step, since there is no contribution anyway because of the factor k
2 . The second term is
1
r 2
∂
2 V
∂φ 2 = −
1
r 2
∞
k=0
∞
l=0
M k,l (s) cos (lφ + θ k,l ) l
2 r
k
= −
∞
k=0
∞
l=0
M k,l (s) cos (lφ + θ k,l ) l
2 r
k−2 .
The third term is
∂
2 V
∂s 2 =
∞
k=0
∞
l=0
M
k,l (s) cos (lφ + θ k,l ) r
k
=
∞
k=2
∞
l=0
M
k−2,l (s) cos (lφ + θ k−2,l ) r
k−2
=
∞
k=0
∞
l=0
M
k−2,l (s) cos (lφ + θ k−2,l ) r
k−2 ,
where we let the sum start at k = 2 in the second step, and, further, in the last
step, we used the convention that the coefficient M k,l (s) vanish for negative
indices. Recognizing that all the terms have the common summations and
the factor r
k−2 , we obtain the Laplacian for the Fourier-Taylor expansion of
the potential (3.3) as
ΔV =
∞
k,l=0
M k,l (s) cos (lφ+θ k,l )
k
2
− l
2
+ M
k−2,l (s) cos (lφ+θ k−2,l )
r
k−2 .
To satisfy Laplace’s equation, we obtain a set of conditions for k, l ≥ 0,
M k,l (s) cos (lφ + θ k,l )
k
2
− l
2
+ M
k−2,l (s) cos (lφ + θ k−2,l ) = 0,
where the second term vanishes for k = 0, 1 because of the negative indices
for M k,l .
We begin the analysis of Laplace’s equation by studying the case k = 0,
where only the first term matters. Apparently M 0,0 and θ 0,0 can be chosen
freely because k
2
− l
2 = 0 for k = l = 0. For l ≥ 1, we infer M 0,l = 0.