Fundamentals of potential theory - Physical Geodesy - page 9

It can be shown that Laplace's operator in these coordinates is

∂
∂q 1

h 2 h 3
h 1

+ h 3 h 1

h 2

+ h 1 h 2

h 3

.

1
h 1 h 2 h 3

∂V
∂q 1

∂
∂q 2

∂V
∂q 2

∂
∂q 3

∂V
∂q 3

∆ V =

(1-32)
For spherical coordinates we have q 1 = r , q 2 = ϑ , q 3 = λ .Comparisonof
(1-30) and (1-31) shows that

h 1 =1 ,h 2 = r, h 3 = r sin ϑ. (1-33)

Substituting these relations into (1-32) yields

r 2 ∂V
∂r

+ sin ϑ ∂V

∂ϑ

+

∂ 2 V
∂λ 2

1
r 2

∂
∂r

1
r 2 sin ϑ

∂
∂ϑ

1
r 2 sin 2 ϑ

∆ V = . (1-34)

Performing the differentiations we find

∂ 2 V
∂r 2

∂ 2 V
∂ϑ 2

∂ 2 V
∂λ 2

∂V
∂r

+ cot ϑ
r 2

∂V
∂ϑ +

+ 2
r

1
r 2

1
r 2 sin 2 ϑ

∆ V ≡ + =0 , (1-35)

which is Laplace's equation in spherical coordinates . An alternative expres-
sion is obtained when multiplying both sides by r 2 :

r 2 ∂ 2 V
∂r 2

+ ∂ 2 V
∂ϑ 2

∂ 2 V
∂λ 2

+2 r ∂V
∂r

+cot ϑ ∂V

1
sin 2 ϑ

∂ϑ +

=0 . (1-36)

This form will be somewhat more convenient for our subsequent develop-
ment.

1.5 Spherical harmonics

We attempt to solve Laplace's equation (1-35) or (1-36) by separating the
variables r, ϑ, λ using the trial substitution

V ( r, ϑ, λ )= f ( r ) Y ( ϑ, λ ) , (1-37)

where f is a function of r only and Y is a function of ϑ and λ only. Performing
this substitution in (1-36) and dividing by fY ,weget

∂ 2 Y
∂ϑ 2 ∂λ 2 ,

∂ 2 Y 1 1

Y

+cot ϑ ∂Y

1
sin 2 ϑ f ( r 2 f +2 rf )=

−

∂ϑ +

(1-38)

where the primes denote differentiation with respect to the argument ( r ,in
this case). Since the left-hand side depends only on r and the right-hand side