Seismology

Part II: Body Waves and Ray Theory

Some definitions:

Body Waves: Waves that propagrate through the "body" of a medium (in 3 dimensions)

WRONG!

Wavefront: The solution to the wave equation a contant time.

Rays: Normals to the wavefront, or in direction of maximum change in time. Direction of wave propagation.

.We think of energy traveling along rays, but it is important to remember that the reality is waves, not rays. Rays are an approximation (high f), but they are so useful that we tend to make use of them whenever possible.

Two useful ideas from Classical Optics:

Huygen's Principle: An expanding wavefront is generated as the sum of contributions of individual point sources.

Christian Huygens

Fermat's principle: (1) Energy will follow an extremal time path (usually, but not always, minimum) and (2) 1st order deviations in ray path result in 2nd order deviations in travel time.

Now, if all we had to deal with was a homogeneous, isotropic space, we would be done. But we would not have anything very useful to apply to the Earth.

Pierre de Fermat

Let's start relaxing these assumptions by allowing heterogeneity. We can do this in two ways:

1. Allow for small gradients in , , and

2. Allow for large localized gradients (i.e., interfaces)

The first option is very common in seismology, but a good explanation is a bit cumbersome. It is gone over in detail in the next several (and optional) “green slides”.

The second one is a bit easier to follow and we’ll come back to it in a bit.

Allowing for gradients in , , and

Recall that the homogeneity assumption allowed us to take and out from the spatial derivatives, for example in:

2u1

t2

x1

u1

x1

u2

x2

u3

x3

2 u1

x1

x2

u1

x2

u2

x1

x3

u1

x3

u3

x1

If we let and be a function of space, but require that their gradients are much less than du/dx, then we can still be approximately right in ignoring them. du/dx is related to the wavelength of the wave, so basically this means that the wavelength is short compared to the variations in elastic moduli.

We recall that for scalar wave potential equation for a plane wave is:

222 2

(x,t)Aei (kxt )

with the solution:

| k |

2

and the wavespeed is:

If , , and are functions of position, then so is and k. We also should allow the amplitude A to be a function of position as well.

Let o be the mean wavespeed, so that ko is a mean wavenumber and ko = /o and k = W(x)ko. We attempt to find a solution analogous to the plane wave solution:

22 )( x

(x,t)A(x)e i (W (x) / o t )

W (x)/o k x

where

1

2(x)

2

t2A(x)e i (W (x) / o t ) 2 A(x)e i (W (x) / o t )

so

Substitution into the wave equation above gives, for d/dx1:

2x1

2

x1

(x)

x1

e i (W (x) / o t ) (x)e i (W (x) / o t ) )

x1

x1

(x)

x1

e i (W (x) / o t ) (x)e i (W (x) / o t ) io

W (x)

x1

i2o

(x)

x1

W (x)

x1

(x)o

2W (x)

x12

e i (W (x) / ot )

2(x)

x12

2(x)

o2

W (x)

x1

2

e i (W (x) / ot )

And similarly for the other two directions.

If we add all three equations together, we get for the real part:

2(x)

x12

2(x)

x22

2(x)

x32

2(x)

co2

W (x)

x1

2

W (x)

x2

2

W (x)

x3

2

2

2(x)(x)

2(x) 2(x)

o2

W (x)

x1

2

W (x)

x2

2

W (x)

x3

2

2

2(x)(x)

or

and for the imaginary part:

2o

(x)

x1

W (x)

x1

(x)

x2

W (x)

x2

(x)

x3

W (x)

x3

o

(x)2W (x)

x12

2W (x)

x22

2W (x)

x32

0

2(x)

x1

W (x)

x1

(x)

x2

W (x)

x2

(x)

x3

W (x)

x3

(x)2W (x)0

or

Rearrange the real part:

W (x)

x1

2

W (x)

x2

2

W (x)

x3

2

o2

2(x)

o2

2(x)2(x)

W (x)W (x)o

2

2(x)

o2

2(x)2(x)

or

Let's define a reference wavelength as:

o 2o

W (x)W (x)o

2

2(x)

o2

22(x)

(x)

Then

We seek the conditions where the right side of this equation is small and so can be neglected. The original form suggests this should be true at high frequencies. But we can be more precise. The imaginary part of the solution shows that

2(x)W (x)(x)2W (x)0

2W (x) 2(x)W (x)

(x)

2W (x)

W (x)(x)

(x)

so

W (x)

x1

2

W (x)

x2

2

W (x)

x3

2

o

2

2(x)

If the term on the right is small, then

W (x) o

(x)

2W (x)

W (x)(x)

(x)o (x) o (x)

(x)

2(x)

1

1 (x)(x)

(x)

so

and

Hence, requiring the right hand side to be small is equivalent to satisfying:

o2

22(x)

(x)

o2

2(x)

(x)1

If we estimate the gradient in wavespeed over one

(x) (x)

o

2 (x) o

(x)

1

2 (x) (x)

1

(x) 2(x)

In other words, the change in wavespeed gradient over a distance of a wavelength is small compared to the wavespeed. To the extent that this is true (and note that it will be more precise for short wavelength/high f waves), then

W (x)

x1

2

W (x)

x2

2

W (x)

x3

2

o

2

2(x)

or

which is a form of the eikonal equation and forms the basis for ray theory.

W (x)/o k x

W (x)

x1

k1o

o

1

Recall that

so

and the eikonal equation becomes

k1

2

k2

2

k3

2

1

1

2

1

2

2

1

3

2

1

2(x)

We can think of the components of wavespeed in these directions as the rate that travel time changes along the coordinate axes:

t

xi

1

i

This is known as “apparent” slowness. Thus

t

x1

2

t

x2

2

t

x3

2

1

2(x)

Which is a form that commonly appears in various texts and has proven to be very useful in solving the problem of calculating travel times in three dimensional media.

Now, we can use this result to tell us about the ray path (where the ray goes in space). Note that the following form of the eikonal equation:

k1

2

k2

2

k3

2

1

2(x)

k12 k2

2 k32

2

2(x)

2(x)

2

k2

or

So, the eikonal equation is really about the direction of wave propagation, parallel to k, the components of which are (k1, k2, k3) are proportional to direction cosines (by 2) for angles between the propagation direction and the Cartesian frame.

If we consider a raypath moving a distance ds in the k direction, then

dˆ s k

k(k1,k2 ,k3 )

(x)

2(ds1,ds2 ,ds3 )

k1

2(x)

ds1

2f

ds1

(x)

so

Also, we can write these components of ds as direction cosines:

ds1 dx1

ds

dx1

dsk1

o

W

dx1

o

is the index of refraction n:

ndx1

dsW

dx1

We can figure out what the ray path is by examining how the direction cosines change as we progress along the ray (s):

d

dsn

dx1

ds

d

ds

W

dx1

Change order of integration and use chain rule:

d

ds

W

dx1

x1

dW

ds

x1

W

dx1

dx1

dsW

dx2

dx2

dsW

dx3

dx3

ds

x1

ndx1

ds

2

dx2

ds

2

dx3

ds

2

x1

n

d

dsn

dx1

ds

n

dx1

Then

If you skipped the green slides, the above equation relates a short segment of a ray path (ds) in one of the cardinal directions (x1 – the other two are x2 and x3). The dx/ds derivatives are direction cosines, and n is the index of refraction:

n o

(x)

)(

1

)(

1

xx ds

xd

ds

d

which is the general raypath equation.

o is the mean wavespeed, and (x) is the variation of wavespeed with position. Combining the above with the other two components gives:

It will be useful to consider what happens when the wavespeed changes in one direction (one-dimensional Earth), because most of the variation in elastic moduli is vertical. Let’s suppose that wavespeed changes only in the x3 direction:

d

dsn

dx1

ds

0;n

dx1

dsc1

d

dsn

dx2

ds

0;n

dx2

dsc2

d

dsn

dx3

ds

dn

dx3

The fact that the direction cosines in the 1 and 2 directions are constant means that those angles are constant and hence there is no change in orientation of the ray relative to those axes. The ray this therefore confined to a plane that is perpendicular to the (1,2) plane).

Let’s suppose that that plane is parallel to the 1 axis. Along any point of the ray, we have:

dx3

dscos(i)

dx1

ds sin(i)

where i is called the angle of incidence. Then

ndx1

dsn sin(i)

o

sin(i)const.

Which means that

sin(i)

const. p

and p is called the ray parameter. It is a characteristic of the entire ray. It is also a statement of Snell’s law, which is a consequence of Fermat’s principle.

If we look at the cosine part:

d

dsn

dx3

ds

d

dsn cos(i) dn

dx3

d

dsn cos( i) n sin( i)

di

ds cos( i)

dn

ds n sin( i)

di

ds cos( i)

dn

dx3

dx3

ds

n sin(i)di

ds cos2(i)

dn

dx3

dn

dx3

n sin(i)di

ds cos2(i)

dn

dx3

So

dn

dx3

1 cos2( i) dn

dx3

sin2(i) n sin( i)di

ds

di

ds

dn

dx3

sin(i)

n

d(o (x))

dx3

sin(i)

o (x)

)(1

)sin())(1(

3 x

x

i

dx

d

1

2(x)

d(x)

dx3

sin( i)

1 (x)

d(x)

dx3

sin( i)

(x)

di

ds

d(x)

dx3

sin(i)

(x) p

d(x)

dx3

or, again

So, if speed increases with depth, so does the angle w.r.t. the x3 axis, which means that the ray is curving up. Likewise, if the speed decreases with depth, the ray curves down.

We can calculate the time and distance for a ray (surface to surface). First, let’s derive a relation between dx1 and dx3:

dx1

ds sin(i)p

dx1 dssin(i)dsp

dx3

dscos(i) 1 sin2(i) 1 2 p2

dsdx3

cos(i)

dx3

1 2 p2

dx1 dsp p

1 2 p2dx3

So, to calculate the total distance traveled by the ray, integrate:

X dx1 2p

1 2 p2dx3

0

zmax

T dt ds

2dx3

cos(i)0

zmax

2dx3

1 2 p20

zmax

And the total travel time is

T pX 2 2 p2 dx30

zmax

We can combine the two to give:

where . Note then that

dT

dX p

Which means that we can deduce the ray parameter from surface observations, as p determines the rate of change of T.

1. The raypath is characterized by the ray parameter “p”, which is the ratio of the sin of the incidence angle to the local wavespeed

2. We can deduce the ray parameter from surface observations, because p determines the rate of change of T. In fact, you can think of p as the apparent “slowness” of the wave.

3. If speed increases with depth, so does the angle w.r.t. the x3 axis, which means that the ray is curving up. Likewise, if the speed decreases with depth, the ray curves down.

To reiterate the important findings from these derivations - In a one-dimensional medium, the following are true:

sin(i)

const. p