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(4.12) (which implies that the eld of point charge decays as 1=r2 ). Measurement of the

potential di erences within a hollow spherical shell as described in problem d can be

carried out with very great sensitivity. Experiments based on this principle (usually in

a more elaborated form) have been used to ascertain the decay of the electric eld of a

point charge with distance. Writing the eld strength as 1=r2+" is has now be shown that

" = (2:7 3:1) 10;16 , see section I.2 of Jackson 31] for a discussion. The small value of

" is a remarkable experimental con rmation of equation (4.12) for the electric eld.

CHAPTER 6. THE THEOREM OF GAUSS

54

6.3 A representation theorem for acoustic waves

Acoustic waves are waves that propagate through a gas or uid. You can hear the voice

of others because acoustic waves propagate from their vocal tract to your ear. Acoustic

waves are frequently used to describe the propagation of waves through the earth. Since

the earth is a solid body, this is strictly speaking not correct, but under certain conditions

(small scattering angles) the errors can be acceptable. The pressure eld p(r) of acoustic

waves satisfy in the frequency domain the following partial di erential equation:

!2 p = f :

1 + (6.7)

r rp

In this expression (r) is the mass density of the medium while (r) is the compressibility

(a factor that describes how strongly the medium resists changes in its volume). The right

hand side f(r) describes the source of the acoustic wave. This term accounts for example

for the action of your voice.

We will now consider two pressure elds p1 (r) and p2 (r) that both satisfy (6.7) with

sources f1 (r) and f2 (r) in the right hand side of the equation.

Problem a: Multiply equation (6.7) for p1 with p2, multiply equation (6.7) for p2 with

p1 and subtract the resulting expressions. Integrate the result over a volume V to

show that:

Z Z

1 1

p2 p1 dV = 2 f1 ; p1 f2 g dV : (6.8)

1 2

r rp ; r rp fp

V V

Ultimately we want to relate the wave eld at the surface S that encloses the volume V

to the wave eld within the volume. Obviously, Gauss's law is the tool for doing this.

The problem we face is that Gauss's law holds for the volume integral of the divergence,

1 1 ) with

whereas in expression (6.8) we have the product of a divergence (such as r rp

another function (such as p2 ).

Problem b: This means we have to \make" a divergence. Show that

1 1p 1 (rp

p2 2) :

1 =r (6.9)

2 rp1 1

r rp ; rp

What we are doing here is similar to the standard derivation of integration by parts.

R R

The easiest way to show that ab f(@g=@x)dx = f(x)g(x)]b ab (@f=@x)gdx, is to integrate

a ;

the identity f(@g=@x) = @(fg)=dx (@f=@x)g from x = a to x = b. This last equation

;

has exactly the same structure as expression (6.9).

Problem c: Use expressions (6.8), (6.9) and Gauss's law to derive that

I1 Z

(p2 1 p1 2 ) dS = 2 f1 p1 f2 dV : (6.10)

rp ; rp fp ; g

S V

6.4. FLOWING PROBABILITY 55

This expression forms the basis for the proof that reciprocity holds for acoustic waves.

(Reciprocity means that the wave eld propagating from point A to point B is identical to

the wave eld that propagates in the reverse direction from point B to point A.) To see

the power of expression (6.10), consider the special case that the source f2 of p2 is of unit

strength and that this source is localized in a very small volume around a point r0 within

the volume. This means that f2Rin the right hand side of (6.10) is only nonzero at r0 . The

corresponding volume integral V p1 f2 dV is in that case given by p1 (r0 ). The wave eld

p2 (r) generated by this point source is called the Green's function, this special solution is

denoted by G(r r0 ). (The concept Green's function is introduced in great detail in chapter

14. ) The argument r0 is added to indicate that this is the wave eld at location r due to a

unit source at location r0 . We will now consider a solution p1 that has no sources within

the volume V (i.e. f1 = 0). Let us simplify the notation further by dropping the subscript

"1" in p1 .

Problem d: Show by making all these changes that equation (6.10) can be written as:

I1

(p(r)rG(r r ) G(r r )rp(r)) dS :

p(r ) = (6.11)

0 0 0

;

S

This result is called the "representation theorem" because it gives the wave eld inside

the volume when the wave eld (an its gradient) are speci ed on the surface that bounds

this volume. Expression (6.11) can be used to formally derive Huygens' principle which

states that every point on a wavefront acts as a source for other waves and that interference

of these waves determine the propagation of the wavefront. Equation (6.11) also forms

the basis for imaging techniques for seismic data, see for example ref. 53]. In seismic

exploration one records the wave eld at the earth's surface. This can be used by taking

the earth's surface as the surface S over which the integration is carried out. If the Green's

function G(r r0 ) is known, one can use expression (6.11) to compute the wave eld in the

interior in the earth. Once the wave eld in the interior of the earth is known, one can

deduce some of the properties of the material in the earth. In this way, equation (6.11)

(or its elastic generalization) forms the basis of seismic imaging techniques.

Problem e: This almost sounds too good to be true! Can you nd the catch?

6.4 Flowing probability

In classical mechanics, the motion of a particle with mass m is governed by Newton's law:

mr = F. When the force F is associated with a potential V (r) the motion of the particle

satis es:

d2 r =

m dt2 (r) : (6.12)

;rV

However, this law does not hold for particles that are very small. Microscopic particles

such as electrons are not described accurately by (6.12). It is one of the outstanding

features of quantum mechanics that microscopic particles are treated as waves rather than

particles. The wave function (r t) that describes a particle that moves under the in uence

of a potential V (r) satis es the Schrodinger equation 39]:

h @ (r t) = h2 2 (r t) + V (r) (r t) : (6.13)

i @t 2m

; ; r

CHAPTER 6. THE THEOREM OF GAUSS

56

In this expression, h is Planck's constant h divided by 2 .

Problem a: Check that Planck's constant has the dimension of angular momentum.

Planck's constant has the numerical value h = 6:626 10;34 kg m2 =s. Suppose we are

willing to accept that the motion of an electron is described by the Schrodinger equation,

then the following question arises: What is the position of the electron as a function

of time? According to the Copenhagen interpretation of quantum mechanics this is a

meaningless question because the electron behaves like a wave and does not have a de nite

location. Instead, the wavefunction (r t) dictates how likely it is that the particle is at

location r at time t. Speci cally, the quantity (r t)j2 is the probability density of nding

j

the particle at location r at time t. This implies that the probability PV that the particle

R

is located within the volume V is given by PV = V 2 dV . (Take care not to confuse

j j

the volume with the potential, because they are both indicated with the same symbol V .)

This implies that the wavefunction is related to a probability. Instead of the motion of the

electron, Schrodinger's equation dictates how the probability density of the particle moves

through space as time progresses. One expects that a \probability current" is associated

with this movement. In this section we will determine this current using the theorem of

Gauss.

Problem b: In the following we need the time-derivative of (r t), where the asterisk

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