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.
Pierre de Fermat
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.
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:
2

 u1
t
2
   u1 u 2 u 3



 
x1   x1 x 2 x 3

u1 

 2 
x1 

   u1 u 2    u1 u 3



 
 

x 2   x 2 x1  x 3  x 3 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:
  2 2
2
2


    
 

with the solution:
 ( x ,t )  Ae
i ( k x  t )
and the wavespeed is:



|k |

  2

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:
   ( x )  
2
 ( x ,t )  A( x )e
2
i ( W ( x ) /  o  t )
where
so
 W ( x ) / o  k  x

2
1  
i ( W ( x ) /  o  t )
2
i ( W ( x ) /  o  t )
A(
x
)e


A(
x
)e



2
2 
 ( x ) t
Substitution into the wave equation above gives, for d/dx1:
i ( W ( x ) /  o  t ) 
  ( x ) i ( W ( x ) /  o  t )
e
)

e
 (x)


2
x1
x1  x1
x1

2



  ( x ) i ( W ( x ) /  o  t )
i ( W ( x ) /  o  t ) i  W ( x )

e
  ( x )e


x1  x1
 o x1 
2 
 2
2


  ( x )   ( x ) W ( x )
i ( W ( x ) /  o   t )




e


2
2
o

 x1  
 x1

2
2  ( x ) W ( x )
  W ( x )  i ( W ( x ) /  o   t )
 i
 (x)
e
2
x1
 o x1 
 o x1

And similarly for the other two directions.

If we add all three equations together, we get for the real part:
2
 (x)
2
1
x
2

 (x)
x
2
2
2

 (x)
2
 x3
2
2
2 

2





  ( x ) W ( x )
W ( x )
W ( x )



 ( x)
  
  
   2
2
co

 x1   x 2   x 3  
  ( x )
2
or

2
2
2 

2







(
x
)
W
(
x
)
W
(
x
)
W
(
x
)


2

 (x)
 ( x)
  
  
   2
2
o

 x1   x 2   x 3  
  ( x )
2
and for the imaginary part:
2  ( x ) W ( x )  ( x ) W ( x )  ( x ) W ( x ) 




 o  x1
x1
x 2
x 2
x 3
x 3 
 2W ( x )  2W ( x )  2W ( x ) 

 ( x )


  0
2
2
2
o
x 2
x 3 
 x1

or

 ( x ) W ( x )  ( x ) W ( x )  ( x ) W ( x ) 
2
2


  ( x ) W ( x )  0
x1
x 2
x 2
x 3
x 3 
 x1
Rearrange the real part:
2
2
2 

2
2





W ( x )
W ( x )
W ( x )
o
o
2

 2
 (x)
  
  
  2

 x1   x 2   x 3  
  ( x )   ( x)
or
o
o
2
W ( x )  W ( x ) 
 (x)
2
2

2
  ( x)
2
 (x)
Let's define a reference wavelength as: o 

Then
o
2
W ( x )  W ( x ) 
o   ( x )
2


 ( x ) 2
2
2
 ( x)
2  o

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
2 ( x )  W ( x )   ( x ) W ( x )  0
so
2
 W ( x )  2

2
 W (x)


 W ( x)

( x )  W ( x )
(x)
 (x )
(x )

If the term on the right is small, then
2
2
2
2
W ( x )  W ( x )  W ( x ) 
o

  
  
  2
 (x)
  x1   x 2   x 3 
so
 W ( x) 

o
( x )
and
  o  ( x )   ( x ) 1
 (x )
 ( x )


 2

 W ( x)
(x)
 o ( x )
 ( x ) 1 ( x )
( x )
2
 W (x)
Hence, requiring the right hand side to be small is equivalent to
satisfying:
o   ( x )
2
2
2
 ( x)
o   ( x )
2

2  ( x )
 1
If we estimate the gradient in wavespeed over one 
 ( x ) 
o     ( x ) o
2

or
( x )

 ( x )

1     ( x )
2
( 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
 is true (and note that it will be more precise for
extent that this
short wavelength/high f waves), then
2
2
2
2
W ( x )  W ( x )  W ( x ) 
o

  
  
  2
 (x)
  x1   x 2   x 3 
which is a form of the eikonal equation and forms the basis for ray
theory.
Recall that
 W ( x ) / o  k  x
so
W ( x )

x1

k1 o


o
1
and the eikonal equation becomes

2
2
2
 1   1   1 
 k1   k 2   k 3 
1
                  2
     
 (x)
1   2   3 
2
2
2
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

2
2
2
  t    t    t 
1
   
  
  2
 (x)
 x1   x 2   x 3 
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:
2
2
2
 k1   k 2   k 3 
1
         2
     
 (x)
or

2
2
k1  k 2  k 3
2
2
 2  

 2
 
  k
 ( x )  ( x ) 
2
2
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
d sˆ 
k
k
so
 ( k1 ,k 2 ,k 3 )
k1



( x )
2
2
ds 1
( x ) 2  f
 ( ds1 ,ds 2 ,ds 3 )

ds 1
( x )
Also, we can write these components of ds as direction cosines:
ds1 

dx 1

ds

dx 1
 k1

ds

 W
 o dx 1
o

is the index of refraction n:
n
dx 1
ds

W
dx 1
We can figure out what the ray path is by examining how the
direction cosines change as we progress along the ray (s):

d  dx 1  d  W 
n
 


ds  ds  ds  dx 1 
Change order of integration and use chain rule:
d  W    dW    W dx 1  W dx 2  W dx 3 



 

 



ds  dx 1   x1  ds   x1  dx 1 ds
dx 2 ds
dx 3 ds 
2
2
2 
 

dx 1   dx 2   dx 3  




n 
n
  
  
  

 x1  
 ds   ds   ds  
  x1
Then
d  dx 1   n
n
 
ds  ds  dx 1
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 x
2 and x3). The dx/ds derivatives are direction cosines,
and n is the index of refraction:
n
o
( x )
o is the mean wavespeed, and (x) is the variation of wavespeed
with position. Combining the above with the other two components
gives:


 1 
d  1 dx 

   

ds   ( x ) ds 
  (x) 
which is the general raypath equation.
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  dx 1 
dx 1
 c1
n
  0; n
ds  ds 
ds
d  dx 2 
dx 2
 c2
n
  0; n
ds  ds 
ds


d  dx 3  dn
n
 
ds  ds  dx 3
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:
dx 3
ds
dx 1
ds
 cos( i )
 sin( i )

where i is called the angle of incidence. Then

n
dx 1
ds
 n 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  dx 3  d
dn
 n cos( i ) 
n
 
ds  ds  ds
dx 3
d
ds
 n cos( i ) 
 n sin( i )

  n sin( i )
di
dn
 cos( i )
ds
di
ds
2
 cos ( i )
ds
  n sin( i )
di
ds
dn
dx 3
So
dn

dx 3
  n sin( i )
dn
dx 3
1  cos
2
di
ds
2
 cos ( i )
( i ) 
dn
dx 3
2
dn
dx 3
sin ( i )   n sin( i )
di
ds
 cos( i )
dn dx 3
dx 3 ds
di
ds
 
dn sin( i )
dx 3
n
 

d(  o  ( x ))
sin( i )
dx 3
o ( x )
d (1  ( x )) sin( i )
dx 3


1
d  ( x ) sin( i )
 ( x ) dx 3 1  ( x )
2

1  (x)
d  ( x ) sin( i )
dx 3
( x )
or, again

di
ds

d  ( x ) sin( i )
dx 3
( x )
 p
d ( x )
dx 3
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:
dx 1
 sin( i )   p
ds
dx 1  ds sin( i )  ds  p





dx 3
ds
 cos( i ) 
ds 
dx 3
cos( i )

dx 1  ds  p 
2
1  sin ( i ) 
dx 3
1  p
2
2
p
1  p
2
2
dx 3
1  p
2
2
So, to calculate the total distance traveled by the ray, integrate:
X 

zmax
dx 1  2

p
1  p
2
0
2
dx 3
And the total travel time is

T 

dt 

ds

zmax
 2

0
z max
dx 3
 cos( i )
 2

0
dx 3
 1  p
2
2
We can combine the two to give:
zmax
T  pX  2


  p dx 3
2
2
0
where 1. 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.

To reiterate the important findings from these derivations - In a onedimensional medium, the following are true:
sin( i )

 const .  p
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.