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Problem b: Insert the expansions (17.25) and (17.26) in the Laplace equation, and use

(17.28) and (17.29) for the Laplacian of the spherical harmonics to show that the

expansion coe cients Vlm (r) of the potential satisfy the following di erential equa-

tion:

1 @ r2 @Vlm (r) l (l + 1) V (r) = 4 G (r) : (17.31)

r2 lm lm

r2 @r @r ;

What we have gained by making the expansion in spherical harmonics is that (17.31)

is an ordinary di erential equation in the variable r whereas the original equation (17.1) is

a partial di erential equation in the variables r, and '. The di erential equation (17.31)

can be solved using the Green's function technique that is described in section 14.3. Let

17.4. THE GRAVITY FIELD OF THE EARTH 249

us rst consider a mass (r r0 ) located at a radius r0 only. The response to this mass is

;

the Green's function Gl that satis es the following di erential equation:

!

l (l + 1) G (r r0 ) = ;r r0 :

1 @ r2 @Gl (r r0 ) (17.32)

r2 l

r2 @r @r ; ;

Note that this equation depends on the angular order l but not on the angular degree m.

For this reason the Green's function Gl (r r0 ) depends on l but not on m.

Problem c: The Green's function can be found 0by rst solving the di erential equation

(17.32) for r = r0 . Show that when r = r the general solution of the di erential

equations (17.32) can be written as Gl = Arl + Br;(l+1) , where the constants A and

6 6

B do not depend on r.

Problem d: In the regions r < r0 and r > r0 the constants A and B will in general have

di erent values. Show that the requirement that the potential is everywhere nite

implies that B = 0 for r < r0 and that A = 0 for r > r0 . The solution can therefore

(l

be written as:

0 ) = Ar ;(l+1) for r < r00

Gl (r r (17.33)

Br for r > r

The integration constants follow in the same way as in the analysis of section 14.3. One

constraint on the integration constants follows from the requirement than the Green's

function is continuous in the point r = r0 . The other constraint follows by multiplying

(17.32) by r2 and integrating the resulting equation over r from r0 " to r0 + ". ;

Problem e: Show that by taking the limit " 0 this leads to the requirement

!

" #0

@Gl (r r0 ) r=r +" = r02 :

r2 @r (17.34)

0 ;"

r=r

Problem f: Use this condition with the continuity of Gl to nd the coe cients A and B

and show that the Green's function is given by:

8

> rl

1 for r < r0

>

0) = < (2l + 1) r0(l;1)

;

Gl (r r > (17.35)

1 r0(l+2) for r > r0

>

: (2l + 1) r(l+1)

;

Problem g: Use this result to derive that the solution of equation (17.31) is given by:

4 G 1 R r (r0 )r0(l+2) dr0

Vlm (r) = l+1 0 lm

(2l + 1) r

;

(17.36)

4 G rl R 1 1

lm (r0 ) r0(l;2) dr0 :

r

(2l + 1)

;

Hint: split the integration over r0 up the interval 0 < r0 < r and the interval r0 > r.

CHAPTER 17. POTENTIAL THEORY

250

Problem h: Let us now consider the potential outside the Earth. Let us denote the

radius of the Earth with the symbol a. Use the above expression to show that the

potential outside the Earth is given by

X X 4 G 1 Za

1l 0 0(l+2) dr0 Ylm ( ') :

V (r ') = lm (r )r (17.37)

l=0 m=;l (2l + 1) r

l+1 0

;

Problem i: Eliminate and show that the potential is nally given by:

lm

X X 4 G 1 Z a 0 0 0 0l

1l

(r ' )r Ylm ( 0 '0 )dV 0 Ylm ( ') :

V (r ') =

l=0 m=;l (2l + 1) r

l+1 0

;

(17.38)

Note that the integration in this expression is over the volume of the Earth rather

than over the distance r0 to the Earth's center.

Let us re ect on the relation between this result and the derivation of upward continu-

ation in a Cartesian geometry of the section 17.2. Equation (17.38) can be compared with

equation (17.14). In (17.14) the potential is written as an integration over wave-number

and the potential is expanded in basis functions exp ikx, whereas in (17.38) the potential

is written as a summation over the degree l and order m and the potential is expanded

in basis functions Ylm . In both expressions the potential is written as a sum over basis

functions with increasingly shorter wavelength as the summation index l or the integration

variable k increases. This means that the decay of the potential with height is in both

geometries faster for a potential with rapid horizontal variations than for a potential with

smooth horizontal variations. In both cases the potential decreases when the height z (or

the radius r) increases. In a at geometry the potential decreases as exp (; z) whereas

in a spherical geometry the potential decreases as r;(l+1) . This di erence in the reduction

jkj

of the potential with distance is due to the di erence in the geometry in the two problems.

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