ñòð. 31 |

According to (9.21) the function (r) does not depend on frequency. Note that equa-

tion (9.22) for the amplitude does not contain any frequency dependence either. This

means that the amplitude also does not depend on frequency: A = A(r). This has im-

portant consequences for the shape of the wave- eld in the ray-geometric approximation.

Suppose that the wave- eld is excited by a source-function s(t) in the time domain that is

represented in the frequency domain by a complex function S(!). (The forward and back-

ward Fourier-transform is de ned by the equations (11.42) and (11.43) of section 11.5.)

In the frequency domain the response is given by expression (9.15) multiplied with the

source function S(!). Using that A and do not depend on frequency the pressure in the

time domain can be written as:

Z1

A(r)ei! (r) e;i!t S(!)d! :

p(r t) = (9.23)

;1

Problem h: Use this expression to show that the pressure in the time domain can be

written as:

p(r t) = A(r)s(t (r)) : (9.24)

;

This is a very important result because it implies that the time-dependence of the wave-

eld is everywhere given by the same source-time function s(t). In a ray-geometric ap-

proximation the shape of the waveforms is everywhere the same. The are no frequency-

dependent e ects in a ray geometric approximation.

Problem i: Explain why this implies that geometric ray theory can not be used to explain

why the sky is blue.

The absence of any frequency-dependent wave propagation e ects is both the strength

and the weakness is ray theory. It is a strength because the wave- elds can be computed

in a simple way once (r) and A(r) are known. The theory also tells us that this is an

adequate description of the wave- eld as long as the frequency is su ciently high that

9.3. GEOMETRIC RAY THEORY 97

LA and L . However, many wave propagation phenomena are in practice

frequency-dependent, it is the weakness of ray theory that it cannot account for these

phenomena.

According to expression (9.24) the function (r) accounts for the time-delay of the

waves to travel to the point r. Therefore, (r) is the travel time of the wave- eld. The

travel time is described by the di erential equation (9.21), this equation is called the

eikonal equation.

Problem j: Show that it follows from the eikonal equation that can be written as:

r

= n=c

^ (9.25)

r

where n is a unit vector. Show also that n is perpendicular to the surface =constant.

^ ^

The vector n de nes the direction of the rays along which the wave energy propagates

^

through the medium. Taking suitable derivatives of expression (9.25) one can derive the

equation of kinematic ray-tracing. This is a second-order di erential equation for the

position of the rays, details are given by Virieux 63] or Aki and Richards 2].

Once (r) is known, one can compute the amplitude A(r) from equation (9.22). We

have not yet applied any scale-analysis to this expression. We will not do this, because

it can be solved exactly. Let us rst simplify this di erential equation by considering the

dependence on the density in more detail.

Problem k: Write A= B, where the constant is not yet determined. Show that the

transport equation results in the following di erential equation for B(r):

) + Br2 = 0 :

) B + 2 (rB

(2 1) (r (9.26)

; r r

Choose the constant in such a way that the gradient of the density disappears from

;B 2 = 0.

the equation and show that the remaining terms can be written as r r

Show nally using (9.25) that this implies the following di erential equation for the

amplitude:

1 A2 n = 0 :

c^ (9.27)

r

;

Equation (9.27) states that the divergence of the vector A2 = c n vanishes, hence the ux

^

of this vector through any closed surface that does not contain the source of the wave- eld

;

vanishes, see section 6.1. This is not surprising, because the vector A2 = c n accounts

^

for the energy ux of acoustic waves. Expression (9.27) implies that the net ux of this

vector through any closed surface is equal to zero. This means that all the energy that

ows in the surface must also ow out through the surface again. The transport equation

in the form (9.27) is therefore a statement of energy conservation. Virieux 63] or Aki and

Richards 2] show how one can compute this amplitude once the location of rays is known.

An interesting complication arises when the energy is focussed in a point or on a surface

in space. Such an area of focussing is called a caustic. A familiar example of a caustic is

the rainbow. One can show that at a caustic, the ray-geometric approximation leads to

an in nte amplitude of the wave- eld 63].

CHAPTER 9. SCALE ANALYSIS

98

Problem l: Show that when the amplitude becomes in nite in a nite region of space

LA must be violated.

the condition

This means that ray theory is not valid in or near a caustic. A clear account of the physics

of caustics can be found in refs. 9] and 34]. The former reference contains many beautiful

images of caustics.

9.4 Is there convection in the Earth's mantle?

The Earth is a body that continuously looses heat to outer space. This heat is a remnant

of the heat that has been converted from the gravitational energy during the Earth's

formation, but more importantly this heat is generated by the decay of unstable isotopes

in the Earth. This heat is transported to the Earth's surface, and the question we aim to

address here is: is the heat transported by conduction or by convection?

Convection

Conduction

Cold Cold

Hot Hot

Figure 9.4: Two alternatives for the heat transport in the Earth. In the left panel the

material does not move and heat is transported by conduction. In the right panel the

material ows and heat is tranported by convection.

If the material in the Earth would not ow, heat could only be transported by con-

duction. This means that it is the average transfer of the molecular motion from warm

regions to cold regions that is responsible for the transport of heat. On the other hand, if

the material in the Earth would ow, heat could be carried by the ow. This process is

called convection.

The starting point of the analysis is the heat equation (8.26) of section 8.4. In the

absence of source terms this equation can for the special case of a constant heat conduction

coe cient be written as:

@T + (vT ) = 2 T : (9.28)

@t r r

2 T accounts

The term (vT ) describes the convective heat transport while the term

r r

for the conductive heat transport.

Problem a: Let the characteristic velocity be denoted by V , the characteristic length

scale by L, and the characteristic temperature perturbation by T. Show that the

ratio of the convective heat transport to the conductive heat transport is of the

following order:

ñòð. 31 |