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(sin + r sin @r (rv' )

@ @' @' @

r ;

(5.16)

Problem c: Show that in cylinder coordinates (r ' z) the curl is given by:

n 1 @vz @v' o n @vr @vz o 1 n @ o

v = ^ r @' @z + ' @z @r + ^ r @r (rv') @'

r ^ z @vr (5.17)

r ; ; ;

p2 2

with r = x + y .

Problem d: Use this result to re-derive (5.8) for vector elds of the form v = v(r)^ .

'

Hint: use the same method as used in the derivation of (5.14) and treat the

three components of the curl separately.

CHAPTER 5. THE CURL OF A VECTOR FIELD

50

Chapter 6

The theorem of Gauss

In section (4.5) we have determined the gravitational eld in N-dimensions using as only

ingredient that in free space, where the mass density vanishes, the divergence of the

gravitational eld vanishes (r g) = 0. This was su cient to determine the gravitational

eld in expression (4.21). However, that expression is not quite satisfactory because it

contains a constant A that is unknown. In fact, at this point we have no idea how this

constant is related to the mass M that causes the gravitational eld! The reason for this

is simple, in order to derive the gravitational eld in (4.21) we have only used the eld

equation (4.17) for free space (where = 0). However, if we want to nd the relation

between the mass and the resulting gravitational eld we must also use the eld equation

(r g) = G at places where the mass is present. More speci cally, we have to

;4

integrate the eld equation in order to nd the total e ect of the mass. The theorem of

Gauss gives us an expression for the volume integral of the divergence of a vector eld.

6.1 Statement of Gauss' law

In section (4.2) it was shown that the divergence is the ux per unit volume. In fact,

equation (4.4) gives us the outward ux d through an in nitesimal volume dV d =

(r v)dV . We can immediately integrate this expression to nd the total ux through the

surface S which encloses the total volume V :

I Z

v dS = (r v)dV : (6.1)

S V

In deriving this expression (4.4) has been used to express the total ux in the left hand

side of (6.1). This expression is called the theorem of Gauss..

Note that in the derivation of (6.1) we did not use the dimensionality of the space,

this relation holds in any number of dimensions. You may recognize the one-dimensional

version of (6.1). In one dimension the vector v has only one component vx , hence (r v) =

@xvx . A \volume" in one dimension is simply a line, let this line run from x = a to x = b.

The \surface" of a one-dimensional volume consists of the endpoints of this line, so that

the left hand side of (6.1) is the di erence of the function vx at its endpoints. This implies

that the theorem of Gauss is in one-dimension:

Z b @vx

vx (b) vx(a) = @x dx : (6.2)

;

a

51

CHAPTER 6. THE THEOREM OF GAUSS

52

This expression will be familiar to you. We will use the 2-dimensional version of the

theorem of Gauss in section (7.2) to derive the theorem of Stokes.

Problem a: Compute the ux of the vector eld v(x y z) = (x + y + z)^ through a

z

sphere with radius R centered on the origin by explicitly computing the integral

that de nes the ux.

Problem b: Show that the total ux of the magnetic eld of the earth through your skin

is zero.

Problem c: Solve problem a without carrying out any integration explicitly.

6.2 The gravitational eld of a spherically symmetric mass

In this section we will use Gauss's law (6.1) to show that the gravitational eld of a body

with a spherically symmetric mass density depends only on the total mass but not on

the distribution of the mass over that body. For a spherically symmetric body the mass

density depends only on radius: = (r). Because of the spherical symmetry of the mass,

the gravitational eld is spherically symmetric and points in the radial direction

g(r) = g(r)^ :

r (6.3)

Problem a: Use the eld equation (4.17) for the gravitational eld and Gauss's law

(applied to a surface that completely encloses the mass) to show that

I

g dS = 4 GM (6.4)

;

S

where M is the total mass of the body.

Problem b: Use a sphere with radius r as the surface in (6.4) to show that the gravita-

tional eld is in three dimensions given by

g(r) = GM ^ :

r2 r (6.5)

;

This is an intriguing result. What we have shown here is that the gravitational eld

depends only the total mass of the spherically symmetric body, but not on the distribution

of the mass within that body. As an example consider two bodies with the same mass.

One body has all the mass located in a small ball near the origin and the other body

has all the mass distributed on a thin spherical shell with radius R, see gure (6.1).

According to expression (6.5) these bodies generate exactly the same gravitational eld

outside the body. This implies that gravitational measurements taken outside the two

bodies cannot be used to distinguish between them. The non-unique relation between the

gravity eld and the underlying mass-distribution is of importance for the interpretation

of gravitational measurements taken in geophysical surveys.

6.2. THE GRAVITATIONAL FIELD OF A SPHERICALLY SYMMETRIC MASS 53

g(r) g(r)

Same Mass

R

M

Figure 6.1: Two di erent bodies with a di erent mass distribution that generate the same

gravitational eld for distances larger than the radius of the body on the right.

Problem c: Let us assume that the mass is located within a sphere with radius R, and

that the mass density within that sphere is constant. Integrate equation (4.17) over

a sphere with radius r < R to show that the gravitational eld within the sphere is

given by:

g(r) = MGr ^ :

R3 r (6.6)

;

Plot the gravitational eld as a function from r when the distance increases from

zero to a distance larger than the radius R. Verify explicitly that the gravitational

eld is continuous at the radius R of the sphere.

Note that all conclusions hold identically for the electrical eld when we replace the

mass density by the charge density, because expression (4.12) for the divergence of the

electric eld has the same form as equation (4.17) for the gravitational eld. As an example

we will consider a hollow spherical shell with radius R. On the spherical shell electrical

charge is distributed with a constant charge density: = const.

Problem d: Use expression (4.12) for the electric eld and Gauss's law to show that

within the hollow sphere the electric eld vanishes: E(r) = 0 for r < R.

This result implies that when a charge is placed within such a spherical shell the electrical

eld generated by the charge on the shell exerts no net force on this charge the charge

will not move. Since the electrical potential satis es E = , the result derived in

;rV

problem d implies that the potential is constant within the sphere. This property has

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