ñòð. 84 |

17.3 Upward continuation in a at geometry in 3D

The analysis of the previous section is valid for a at geometry in two dimensions. How-

ever,the theory can readily be extended to three dimensions by including another horizon-

tal coordinate y in the derivation.

Problem a: Show that the theory of the previous section up to equation (17.16) can

be generalized by carrying out a Fourier transform over both x and y. Show in

particular that in three dimensions:

1

ZZ

H 3D (x x0 y y0 z)V (x0 y0 z = 0)dx0 dy0

V (x y z) = (17.18)

; ;

;1

with 1

1 Z Z e; kx+ky z e;i(kx x+ky y) dk dk :

p

H 3D (x y z) = (17.19)

2 2

xy

2

(2 ) ;1

The only di erence with the case in two dimensions is that the integral (17.19) leads to a

di erent upward continuation function than the integral (17.16) for the two-dimensional

case. The integral can be solved by switching the k-integral to cylinder coordinates. The

product kx x + ky y can be written as kr cos ' where k and r are the length of the k-vector

and the position vector in the horizontal plane.

Problem b: Use this to show that H 3D can be written as:

1 Z 1 Z 2 ke;kz e;ikrcos' d'dk :

H 3D (x y z) = (17.20)

(2 )2 0 0

Problem c: The Bessel function has the following integral representation:

1 Z2 eix sin d :

J0 (x) = 2 (17.21)

0

Use this result to write the upward continuation lter as:

Z

3D (x y z) = 1 1 e;kz J0 (kr)kdk :

H (17.22)

20

It appears that we have only made the problem more complex, because the integral of

the Bessel function is not trivial. Fortunately, books and tables exist with a bewildering

collection of integrals. For example, in equation (6.621.1) of Gradshteyn and Ryzhik 25]

R

you can nd an expression for the following integral: 01 e; x J ( x)x ;1 dx.

17.4. THE GRAVITY FIELD OF THE EARTH 247

Problem d: What are the values of , , and if we would like to use this integral to

solve the integration in (17.22)?

Take a look at the form of integral in Gradshteyn and Ryzhik 25]. You will probably be

discouraged by what you nd because the result is expressed in hypergeometric functions,

which means that you now have the new problem to nd out what these functions are.

There is however a way out, because in expression (6.611.1) of 25] gives the following

integral:

; np 2 + 2 o

Z1

e; xJ ( x)dx = p2 2

;

: (17.23)

+

0

This is not quite the integral that we want, because it does not contain a term that

corresponds to the factor k in the integral (17.22). However, we can introduce such a

factor by di erentiating expression (17.23) with respect to .

Problem e: Do this to show that the upward continuation operator is given by

H 3D (x y z) = 21 2 2z 2 3=2 : (17.24)

(x + y + z )

You can make the problem simpler by rst inserting the appropriate value of in

(17.23).

Problem f: Compare the upward continuation operator (17.24) for three-dimensions with

the corresponding operator for two dimensions in equation (17.17). Which of these

operators decays more rapidly as a function of the horizontal distance? Can you

explain this di erence physically?

Problem g: In section 17.2 you showed that the integral of the upward continuation

operator over the horizontal distanceRRis equal to one. Show that the same holds

1

in three dimensions, i.e. show that ;1 H 3D (x y z)dxdy = 1. The integration

simpli es by using cylinder coordinates.

Comparing the upward continuation operators in di erent dimensions one nds that

these operators are di erent functions of the horizontal distance in the space domain.

However, a comparison for (17.16) with (17.19) shows that in the wave-number domain

the upward continuation operators in two and three dimensions have the same dependence

on wave-number.. The same is actually true for the Green's function of the wave equation

in 1,2 or 3 dimensions. As you can see in section 15.4 these Green's functions are very

di erent in the space domain, but one can show that in the wave-number domain they are

in each dimension given by the same expression.

17.4 The gravity eld of the Earth

In this section we obtain an expression for the gravitational potential outside the Earth

for an arbitrary distribution of the mass density (r) within the Earth. This could be

done be using the Green's function that is appropriate for the Laplace equation (17.1).

As an alternative we will solve the problem here by expanding both the mass density and

CHAPTER 17. POTENTIAL THEORY

248

the potential in spherical harmonics and by using a Green's function technique for every

component in the spherical harmonics expansion separately.

When using spherical harmonics, the natural coordinate system is a system of spherical

coordinates. For every value of the radius r both the density and the potential can be

expanded in spherical harmonics:

1l

XX

(r ') = ')

lm (r)Ylm ( (17.25)

l=0 m=;l

and

1l

XX

V (r ') = Vlm (r)Ylm ( ') : (17.26)

l=0 m=;l

Problem a: Show that the expansion coe cients for the density are given by:

Z

Ylm ( ') (r ')d

lm (r) = (17.27)

R

where ( ) d denotes an integration over the unit sphere.

Equation (17.1) for the gravitational potential contains the Laplacian. The Laplacian in

spherical coordinates can be decomposed as

2= 1 @ @

r2 @r + r12 2 (17.28)

1

r2 @r

r r

with 2 the Laplacian on the unit sphere:

1r

@ sin @ + 1 @ 2 :

1

2

r1 = (17.29)

sin2 @'2

@ @

sin

The reason why an expansion in spherical harmonics is used for the density and the

2 (see p.

potential is that the spherical harmonics are the eigenfunctions of the operator 1

r

379 of Butkov 14]):

2 Ylm = l (l + 1) Ylm : (17.30)

1

ñòð. 84 |