Seismology
Part III:
Body Waves and Ray Theory
in Layered Medium
Rays in layered medium are simple, and also very useful in a lot of
applications.
Rays within layers are straight lines (wavespeed is constant).
Rays at boundaries refract according to Snell’s law (or, in other words,
they keep the same ray parameter).
Travel time is the length of the straight line path divided by the
wavespeed.
There are three types of path to consider:
1. Direct/transmitted/refracted
2. Critically refracted (head)
3. Reflected
Suppose we have an interface separating two media with wavespeeds
c1 and c2.
Consider a wave in c1 approaching an interface at an angle i w.r.t. the
normal to the interface. The transmitted wave leaves with the angle it,
and the reflected wave with the angle ir. Then
sin(i) sin(ir ) sin(it )
p


c1
c1
c2
And so
ir  i

c2 sin(i) 
it  sin 

 c1 
1

it cannot be greater than 90o for a transmitted wave. We define the
critical angle as:
c2 sin(ic ) 
1  

 c1 
 c1 
ic  sin  
c2 
1

Note that ic can exist only if
c2 > c1. When this happens,

a (head) wave is transmitted
along the interface, traveling
at a speed of c2.
For a single layer over a half space, simple geometry gives:
X
Tdirect 
c1

2h
Treflect 
c1 cos(i)
Thead
 a head wave
Note that for

X  2h tan(ic )
2h


c2
c1 cos(ic )
1
p
c2
c1
sin(ic ) 
 c1 p
c2

2
cos(ic )  1 sin 2(ic )  1 c1 p 2
Substituting in for the expression for head wave time:
Thead
 1
1 c1 p 
sin( ic ) 
2h
 Xp  2h

 Xp 
 

2
2
c1 cos( ic ) c2 cos( ic )
1 c1 p c1 c2 
2 2
2 2


1

c
2h
1  c1 p
1 p
 Xp 
 Xp  2h1

  Xp  2h
2 2
c1
1  c1 p  c1 
where
1 
1
2

p
2
c1
Note that T is proportional to X (minus a constant). Where did we
see that before?
In multiple layers, we can write this as:
Thead
n
n
X
1
tan(ii )
  2 hi

 Xp 2 hii
cn
ci cos(ii )
cn
i 1
i 1
It is generally useful to plot seismograms
in an X vs T plot to identify coherent
arrivals. We interpret these arrivals by
comparing with theory.
For a layer over a half space, the direct
and head waves are straight lines on an
X-T plot, and the reflected wave is a
hyperbola that asymptotically approaches
the head wave.
Note that for the reflected wave
X 2  h 2
2
2
Trefl 
c12
So a plot of T2 vs X2 gives a
straight line, and T vs X is a
hyperbola. There is no exact
extension to multiple layers,
but approximate formulas are
often used.
The refracted arrival will overtake the direct arrival when
2
1 c1 p 2
X X
X
  2h1   2h
c1 c2
c2
c1
c2 1  c1 p 2
c2  c1
X  2h
 2h
c2  c1
c2  c1
2
X  2h
2
2
c2  c1
c2  c1
This crossover distance is useful to
know when planning a refraction
 experiment (i.e., how far away must
sensors be in order to detect a
refracted first motion?).
Some headaches of refraction seismology:
1. Low velocity layers: these cannot be detected and will
give false depths to interfaces. Nothing can be done about
this.
2. Blind zones: Thin layers; too thin for the refracted
arrival to ever arrive first. Can sometimes see in the
background, but usually ambiguous.
3. Dipping Layers: Look just like flat layers but will give
wrong wavespeed and thickness. Remedy is to “reverse” the
profile. Here is how you do it:
Downdip travel time can be shown with simple trig to be:
2hd cos(ic )cos( ) x sin(ic   )
td 

c1
c1
Updip travel time is
2hu cos(ic )cos( ) x sin(ic  )
tu 

c1
c1
where hd and hu are the downdip and
updip depths to the layers, and  is
the dip. Note that both of these X-T
equations are straight lines with
different slopes and intercepts. This
asymmetry is diagnostic of a dipping
interface, and this is why we always
“reverse” the refraction profile.
Note that we can determine the dip, depth, and wavespeeds of the
medium by:
  0.5sin1(c1 / cd ) sin 1(c1 / cu )
ic  0.5sin1(c1 / cd ) sin 1(c1 / cu )
 depth can be determined from the intercepts:
and the


(h1,h2 ) 
(to1,to2 )cucd cos(ic )cos( )
2cos( ) cu2  cd 2
ALTERNATE REALITIES
A useful way to summarize travel time data is as a -p plot, where p
will be the slope (=1/c) and  will be the intercept. X-T, X-p, and -p
plots all have the same information in the, its just that this
information can be more readily understood in some frames.
Example of shooting across a fault.
Spherical Earth
The ideas above developed for a flat earth are easily extended to a
spherical earth.
Imagine a spherical interface. At the interface, Snell holds:
sin(i1 ) sin(a1 )

c1
c2
Now extend the ray to a deeper
interface. From the diagram, d
is the common distance along a
right triangle formed by the ray
extension, and
d  r1 sin(a1 )  r2 sin(i2 )
This will be true as the width becomes infinitesimally small. So, in
general
r sin( i )
 const .  p
c
Also, near the surface, simple geometry shows:
ro sin( o )
T
 p
co

Remember that "" is in radians!

Travel Time and Distance in a Sphere
Consider a spherical ray segment:
ds  dr  rd
2
and

rd
sin(i) 
ds

r sin( i) r 2 d
p

c
cds
so
and
2
2
4
2
r
d
ds2  dr2  rd  2 2
c p
2

Solve for d:
2 2 2
dr
c p
2
d  4
r  r 2c2 p 2
d 

drcp
1
p

r
r 2  c2 p 2
r
dr
2
r
/
c

p
 
2
Integrate the above from ro to rmin (maximum depth of penetration)
and multiply
by 2 to get total distance:

rmin
  2p 
ro

dr
r
2
r
/
c

p
 
2
We follow similar steps to get travel time by integrating solving for
ds instead of d and then integrating ds/c:
rmin
ds
T 
 2
path c
ro r
r / c
2
r / c
2
 p2
dr
Note that this can be written like a -p equation like in flat earth by
combining the above with the expression for :

rmin
T  p  2 
ro

r / c  p 2
2
r
dr
What happens at an interface:
Energy partitioning at a boundary.
We would like to know what happens when a propagating
displacement encounters a boundary, and specifically how does the
displacement on one side influence that on the other side. To do
so, we need to consider what happens with displacements and
tractions on the boundary.
First, what kinds of boundaries do we have to deal with?
1. Solid-Solid or Welded interface. All points move together.
2. Solid-Liquid (and Liquid-Liquid). Vertical motion continuous
3. Free Surface. No constraint on displacement
Solid-Solid or Welded interface. All points move together. In this
case Displacement and Traction are continuous across the boundary.
Displacement is easy to visualize, here is why traction is:
Imagine a volume V surrounding the interface; V=Adz. From the
homogeneous equation of motion (no sources in V), we have:
 dV 
ij , j
 ui   0
From the divergence theorem:
 dV    dS n 
ij, j

ij j
So
 dV u    dS  n   0
i
ij
j
where n is normal to the surfaces parallel to the interface. Now, as
dz goes to zero, dV goes to zero faster than dS, so it must be that
 dS n   0
ij j
which means that the Tractions on either side must be equal (i.e.,
continuous). 
2. Solid-Liquid (and liquid-liquid). Normal components of
displacement and traction are continuous. Shear displacement is
unconstrained. Shear traction is zero.
3. Free Surface. No displacement constraint. Traction is zero.
(That’s why it’s free!)
Suppose we have two layers, 1 and 2, separated by an interface.
Let's consider the case of a P wave incident in layer 1 upon the
interface.
The potentials in layer 1 will be the sum of incident and reflected P
and the potential due to a reflected SV wave:

layer1
 incident reflected

 layer
1   reflected
 potentials in layer 2 will be the
The
refracted P and SV waves:

layer2
 refracted

layer2
 refracted

We can write the solution to the wave equations as:
incident  A1 exp(i( px1   x3  t))
reflected  A2 exp(i( px1   x3  t))
refracted  A3 exp(i( px1   x3  t))
1
1



2
reflected  B2 exp(i( px1   x3  t))
1
refracted  B3 exp(i( px1   x3  t))
2
The signs
 in the arguments that correspond to the direction of
propagation, and the appropriate choices of the x3 factors. Also, the
is the same in every case because of Snell's law. Now
x1 factor
1 p 2 v 2
cos(i)
v  


v
v
k3
So, in general, we consider the displacements at the interface
substituting the above expressions into the following:

 3 2 
u   

xˆ 1
x1 x2 x3 
 1 3 
 

xˆ 2
x2 x3 x1 
 2 1 
 

xˆ 3
x3 x1 x2 

The continuity of u across the interface means uxi+ = uxi- (as
appropriate). For the traction condition, we convert traction to stress
using

Ti   ij nj
and then go from stress to strain using the linear elastic constitutive
equation, and finally to displacement by taking spatial derivatives.
Note that if the interface is horizontal, then n = (0,0,1) and
T  T1,T2 ,T3  13 , 23 , 33
The “answer” is to determine the relation between the incident and
reflected/refracted energy. For the most part this is simply a
question ofalgebra. Let’s look at a simple case of P waves at a
liquid-liquid interface ( = 0 so no S waves) in the x1-x3 plane (i.e.,
no displacements in the x2 direction). In this case the displacement
field is


u
xˆ 1 
xˆ 3
x1
x3

At a liquid-liquid interface, only the vertical component is continuous,
so
 
 i1 A1  A2 exp(i ( px1  1 x3  t))
x3
Thus
 
 i 2 A3 exp(i ( px1   2 x3  t))
x3

 A1  A2 exp(i x3 )   A3 exp(i x3 )

or, for convenience we choose x3 = 0, then
1
1

2
 A1  A2   A3
1

2
2
Continuity of traction. In a liquid/liquid case, only the normal
component is continuous:
T  13 , 23 , 33  0,0, 33 
and from our constitutive equation:

 33  u + 2 33  u  2
because  = 0 in a liquid.
121  222

The scalar potential

2
2




1
2
1
1
1 1  2 2   2 1
1 t
1
Thus
1
2
(A1  A2 )  2 A3
2
1
2
So

1(A1  A2 )   2 A3
2
A1  A2 
A3
1
2

A1  A2 
A3
1
Sum the above
2 2 
21  12 

2A1    A3  
A3
1 1 
 11 


A3
211

A1 21  12
Eliminating A3 from the above:
1
1
A1  A2  A1  A2
2
2
 1 1 
12  21 
1 1 
21  12 
A2   A2
 A1   A1

2 2 
 22 
2 2 
 22 

or
A2 21  12

A1 12  21
Now, note that these relationships are for the amplitudes of the
scalar potential function,
not of the displacement. To recover

displacement, recall that

u
xˆ 3
x3
R
ureflect
uincident
T
urefract
uincident
 i1 A2 12   21


i1 A1
12   21
i2 A3
212


i1 A1  21  12
Note that at normal incidence, p = 0, so 1= 1/1, 2= 1/2 and so in
this case
ureflect 1 /  2   2 / 1 11  2 2
R


uincident 1 /  2  2 / 1 11  2 2
T
urefract
uincident

21 /  2
211

2 / 1  1 /  2 2 2  11
The product  is called the seismic impedance. Note that it is the
impedance contrast that is most important in determining how much
energy is reflected. There is a simple general relationship between the
reflection and transmission coefficients that holds for any interface
and for any angle:
T = 1+R
R can vary between –1 and 1, which means that T goes between 0 and
2. R = 1 at a free surface, which means that we generally will record a
wave at twice its normal amplitude.
These expressions allow us to understand quantitatively what happens
at an interface when the angle of incidence is greater than or equal to
the critical angle.
Recall that
2 
1 p 22
2
2
 1/ 2  p 2  1/ 2  sin 2 (i1 )/ 1
2
2
2
when i = ic, n2 = 0.
 When i1 > ic, n2 is imaginary:
2  i p 2 1/22
This means that the transmitted wave will decay exponentially with
distance away from the interface.
 coefficient becomes:
The reflection
ureflect
 i  21
    i
 1 2
 2 1 1 2
uincident 1i2  21
21  1i2
This is just a complex number divided by its complex conjugate. We
know immediately then than the magnitude of R is 1, and that there
will be a phase shift  in the amount:


1 12
shift  2tan 

21 
So, beyond the critical angle, we have total internal reflection (R =
1) and there will be some distortion because of the phase shift. To
quantify this distortion,
note that we can write the potential of the

postcritical reflection as:
reflected  A2 exp(i( px1   x3  t  / ))
1
which means that the phase of the wave (constant argument) is a
function of frequency.

The term
tˆ  t  /  (t   / )
can be thought of as an apparent time. Since  < 0, lower frequencies
will have earlier arrival times than higher frequencies, which means the

wave is dispersed (spread out).
The transmission coefficient is
T
urefract
uincident
2i1 2  21  i1 2
 2 1  i1 2  21

2
 21  i12  21  i12
 2 212  12 2 2
2
2
Note that in general, an incident P wave will generate (Prefl, Srefl,
Ptrans, Strans) so we get 4 waves generated for every one incident.
Same with SV. SH produces only an SH wave.