Seismology
Part V:
Surface Waves: Rayleigh
John William Strutt (Lord Rayleigh)
1842 -1919
Why are surface waves important?
The Earth is a finite body and is bounded by a free surface; that is, the
part we live on and build buildings on.
Plus, they are often the 300 pound gorilla on a seismogram!
An Earthquake recorded at
Binghamton, NY
The way to think about how to explain surface waves is to ask if there
are any possible solutions to the wave equation that would trap energy
at the surface. The trick is to see if you can make evanescent waves
somehow, just like in critical refractions.
Thus, we begin our discussion by recalling the free surface boundary
condition:
Free Surface. No displacement constraint. Traction is zero. (That’s
why it’s free!)
Again, if the interface is horizontal, then n = (0,0,1) and
T  T1 ,T2 ,T3   13 , 23 , 33  (0,0,0)
Recall that
u2 u3 
u1 u3 
13    

  23    
x3 x2 
x3 x1 
u1 u2 u3 
u3
 33   

 2
x3
x1 x2 x3 


If we concern ourselves only with motion in the (x1, x3) plane, then
u2 = 0 and d/dx2 = 0, so

u1 u3 
23  0
13    
 0
x3 x1 

u1 u3 
u3
 33   

2

0

x1 x3  x3
(i.e., 23 does not provide any additional constraints).
The potentials in the layer will be the sum of incident and reflected P
and the potential due to a reflected SV wave:
layer  incident reflected
layer  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))
 reflected  B2 exp( i ( px1   x3  t))
1
1
1

The signs in the arguments that correspond to the direction of

propagation, and the appropriate choices of the x3 factors.

The x1 factor is the same in every case because of Snell's law.
Recall that
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 

In the x1, x3 plane this becomes:
 2 
u   
xˆ 1
x1 x3 
 2 
1 3 
 
xˆ 3
xˆ 2  
x3 x1 
x3 x1 
Let's see what happens when we apply the no traction boundary
condition:
 u 

u
u
 33   1  3  2 3  0
x3
x1 x3 
u1
  2  

ip  R   i 1 2
 


x1 x1 x1 x3  x1

  2 p 2  R   2 p 1 2


u3
  2  

i1   R   ip2
 


x3 x3 x3 x1  x3
2
  21   R   p 1 2
 33   2 p 2  R   2 p 2

1

  2   1  R  p 1 2  0
 or

Next


2
2

p 2   1   2    R   2p 1 2  0
2
u1 u3 
13    
 0
x3 x1 
u1


2

i

p



i





p











R
 1 2
1

R
 1 2
x3 x3
u

2
2
3

i




i

p



p




p
2






1

R
2
1

R
x1 x1
so

2
2p1   R  p   1 2  0




Evaluating the above at x3 = 0:


p 2   1   2  A1  A2   2p 1 B2  0
2


2p1 A1  A2  p   1 B  0

2
We
solve for A2/A1 =RPP and B2/A1=RSS (for potentials!)
Eliminating 
B2:

p   
A  A   4 p   A  A 

  2 p    4 p   A 
p      2 p    4 p   A
p 2   1   2  p 2   1
2
2
1
2
R
PP
2
1
1
2
2

  2 p
 1 1
1
2
1

2
1


 p   
 1 1
2

2
2
Hence

2
1

 4 p 
 1 1
p 2   1   2  p 2   1  4 p 2 1 1
2
2
1
2

2
  1

2

 1 1
2
Eliminating A2:


2p p      2 A  A   p      2 p   B  0
4 p p     2A 
p      2p    4p   B  0
2p1 p 2   1   2  A1  A2   4 p 21  1 B2  0
2
1
1
2
2
2
2
1
1
1
2

2
1

2
2
1
1
2
2

2
1
2
1
1
Hence

RPS 



4 p1 p 2   1   2 
2


p 2   1   2  p 2   1  4 p 21  1
2

1

Note that there is an angle of incidence where almost all the reflected
energy is S.
We can repeat the above exercise with an incident SV wave to obtain:


 p   
RSS
RSP 



  2 p    4 p  
4 p p   
  2 p    4 p  
p   1   2  p   1  4 p 2 1 1
2
2

p 2   1
2
2

2
1
1
2

2
2
2
2
1
 1 1

1

1
2
1  1
Because a > b, there will exist an angle of incidence where the P wave
“refracts” along the free surface. Note however that in this case, p >
1/ and n is imaginary = i(p2 – 1/2)1/2 and the total scalar potential
is:
  exp( i( px1  t)) A1 exp(
ˆ1 x3 ) A2 exp(
ˆ1 x3 )


To prevent this from blowing up with large x3, we must have A1 = 0.
If we require
u1 u3 

2
13    
  2p1   R   p   1 2  0
x3 x1 


 
If A1 = 0, then in the absence of an SV wave we have:
2p1 R  0
Which means that A2 = 0! So, these evanescent (exponentially
decaying) P waves cannot propagate by themselves.

To trap energy at the surface, we need to generate interference
between P and SV waves, and this is what Lord Rayleigh figured out
in 1887.
The way to do this is to presume that an appropriate solution might
exist and then see if it really does.
So, let's presume that we can trap energy at the surface by allowing
both evanescent P and SV waves to exist simultaneously. In this case,
the potentials are:
  A exp( i ( px1  t))exp(
ˆ x3 )
1
  B exp( i ( px1  t))exp(
ˆ x3 )
1
Note that in this case p > 1/ > 1/which means the that the

horizontal apparent velocity is less than both the shear and
compressional
wave velocities.

Now let's apply the traction free boundary conditions:
u1 u3 
u3
 33   
0
 2
x3
x1 x3 



  2 p 2  i 2 p
ˆ 1 2
u3
  2  


ˆ1  ip2
 


x3 x3 x3 x1  x3
  2
ˆ1   i 2 p
ˆ 1 2
2
 33   2 p 2  i 2 p
ˆ 2

1


u1
  2  

ip 
ˆ 1 2
 


x1 x1 x1 x3  x1


  2   2
ˆ1   i 2 p
ˆ 1 2  0
2
Evaluating at x3 = 0


 p A  ip
ˆ 1 B    2  
ˆ1 A  ip
ˆ 1 B  0
2
2

 B  2 ipˆ
A p     2 
ˆ  Bi2p
ˆ
A  p     2 
ˆ1

2
2
2
2
1
1
1
2
2




ˆ
, so
ˆ and
Remember that   i

A p 2    2 1




2
 B2p
1
 ip
ˆ 1 
Now apply the other traction condition
u1 u3 
13    
 0
x3 x1 
u1


2

i

p







ip




ˆ
ˆ
ˆ

 1 2
1
 1 2
x3 x3
u3 
2
2






i

p


ip



p
2 
ˆ
ˆ


1
2
1
x1 x1





Thus, evaluating at x3 = 0



ip
ˆ1 A 
ˆ 1 B  ip
ˆ1 A p 2B 2ip
ˆ1 A p 2  
ˆ 1 B  0

or



2p1 A  p   1 B  0
2

The question now is: is there a nontrivial choice for A and B that
satisfies these two equations? We write them in matrix form:
p 2     2  2
1

2p1


2p 1 A 0
   
 
2
p   1 
B 0
The above system of equations will have nontrivial solutions for A
and B only if the determinate of the 2x2 matrix is 0. Thus, we
requirethat

p 2    2 1
2


p 2   1  4 p 2 11  0
2
This term is the same as the denominator of the all the reflection
coefficients (RPP, RPS, RSS, RSP), and is called the Rayleigh
denominator.
Note that if we were dealing with real reflected waves

the amplitude would be infinite! But remember that we started off
assuming that we would be dealing with evanescent waves trapped at
the surface.
Let's divide this equation by  so we can talk about wavespeeds
instead of elastic moduli. Also remember that
  2  2  2  2 2  2  2 2
p 2     2 1
2
 2
p    4 p    0



 2 2
2
2
p   2 2 2 p 2    4  2 p 2   0
2
2
1


 p
2

 2p  p
1
2
 1
2
2
2
1
2
  1

 4 
2
2
1
1
1
1
p 2 1 1  0
 factor out p4:

   2 
  2  4  2 
 1 1
2 1 12  2 2 1  12 
0
2

  p 

p 
p

 

v
p

1 p 2 v 2
pv
 1/ pv  1  i 1 1/ pv  i 1 c 2 / v 2
2
2
v 2
2
2
2
2
   1 c / v   c / v  1
 p 

where c = 1/p is the apparent horizontal velocity. So
2 

c
c2  2 22  2  4  2 1 c 2 / 2 1 c2 /  2  0


we need to solve this for c.

Factor a 2 from the above:
c 2
 c 2 
2
2
2
2
 2  22 2  4 1 c /  1 c /   0

  
 c 2 2
2
2
2
2
2 2   4 1 c /  1 c /   0
  
Multiply the above by
 c 2 2
2
2
2
2
2

4
1
c
/

1
c
/



2 
  
To get


 c 2 4
2
2
2
2
2 2   161 c /  1 c /    0
  
The first term on the left becomes:
2
2
c 4

c
c8
c6
c4
c6
c4
c2
 4  4 2  4   8  4 6  4 4  4 6 16 4  16 2 







 
c4
c2
4 4  16 2 16



c8

8
8
c6

6
 24
c4

4
 32
c2

2
16

The second term on the left is:

4 

c
161 c 2 / 2 1 c 2 /  2  161 c 2 / 2  c 2 /  2  2 2 
  

Combining
c8



8
8
c6

6
 24
c4

4
 16
c2

2
16
c2

2
 16
c4

2
2
0
c 2 c 6
c4
c2
2
c 2 
 8 4  24 2  1616 2  16 2  0
2  6
 



 
2 
c 6



c4
24
16

 6  8 4  c 2  2  2 161 2  0

    


A solution to the above can be found for any  and .

As an example, assume a Poisson solid () in which case 2 = 32:
c 6

c4
16  2 
2 24
16  0
 6  8 4  c  2 
2 

3  3 


c 6
c 4 c 2 56  32 
 6  8 4  2    0
   3  3 


which has roots
c2

2
2
 4,2
,2
2

3
3
Of these, only the last satisfied c < , and in this case

c = 0.9194
generally c is in the range of 0.9 to 0.95
What are the particle motions?
We need to solve for displacement (u):
 2 
u1   

x1 x3 
 2 
u3   

x3 x1 
u1  ip 
ˆ 1 2
u3  
ˆ1  ip2


at x3=0. From the 33 = 0 condition above, we found that


2
2
A p     2 1  B2p 1


or
p     2 1
2
B A
2p 1
2
p   2   1
2
A
2
2
2
2
2 2 p 1
 c 2

 2  2
2
2
2
2
2
2
2
2
  2   c /   1
  c  3 


2
 Ap
A
 A
2
2

2 p 1
2 c 1
2c 1
Recall that
  A exp( i ( px1  t))exp(
ˆ x3 )
  B exp( i ( px1  t))exp(
ˆ x3 )
1
1


The shear term for u1 is then
 c 2

 2  2


exp( i ( px1  t)) exp(
ˆ 1 A
ˆ 1 x3 )

ˆ 1 2  
2c 1



 c 2


 2  2

ˆ 1 


 A exp( i ( px1  t))ip 
exp(
ˆ 1 x3 )
 ip 

2c
1




  c 2

  2  2
1 


 A exp( i ( px1  t))ip
exp(
ˆ 1 x3 )
 p

2c





1  c 2
 A exp( i ( px1  t))ip  2  2exp(
ˆ 1 x3 )
2 

Combining:
u1  ip 
ˆ 1 2




1 c 2
 A exp( i ( px1  t))ip exp(
ˆ1 x3 )  2  2exp(
ˆ 1 x3 )
2 



The shear term for u3 is:
 c 2

 2  2


ip2   ipA
exp( i ( px1  t))exp(
ˆ 1 x3 )
2c 1
 c 2

 2  2


 A exp( i ( px1  t))
exp(
ˆ 1 x3 )
2
2c 
ˆ 1
Combining:
u3  
ˆ1  ip2


 c 2



 2  2




  A exp( i ( px1  t)) 
exp(


x
)
ˆ1 exp(
ˆ1 x3 )
ˆ
 1 3 
2

2c 
ˆ 1




Now, because of the "i" in front of u1, the u1 motion is 90 degrees out
of phase with that of u3 (i = e i/2). Another way to see this is to
consider the real parts of u1 and u2. u1 will have a sine for the first
term, and u2 a cosine.
For a Poisson solid, c = 0.9194 = 0.531. If we also let k=p=/c
ˆ 1  0.85 and 
ˆ1  0.39 and
be the Rayleigh wave number, then 
u1  iAk exp( i(kx1  t))exp(0.85kx3 ) 0.58exp(0.39kx3 )
u3  Ak exp( i(kx1  t))0.85exp(0.85kx3 ) 1.47exp(0.39kx3 )


At the surface (x3 = 0):
u1  0.42iAk exp( i(kx1  t))
u3  0.62Akexp( i(kx1  t))
The motion described by the
above is retrograde elliptical.
At a depth of about /5, the
motion goes to zero, and
reverses to a prograde motion
at deeper depths.
Notes:
1. There is no tangential motion (u2 = 0)
2. The rate of decay depends on k, which means that longer
wavelength waves penetrate deeper into the earth.
3. Wavespeed (c) does not depend intrinsically on frequency, so
these waves are not dispersive in a half space. However, if  and/or
 increase with depth, the waves will disperse with the longer
wavelengths coming in first.
4. Two Dimensional spreading means these guys are large amplitude
for long distances. Scorpions use them to locate prey.
5. Wave curvature is required to generate Rayleigh waves, which
means that deep sources generally do not produce them.
Example of Dispersion of Rayleigh waves
Rayleigh waves are not naturally dispersive, but become so if there are
vertical variations in wavespeed. A simple but useful example of what
happens can be had by considering what happens in a fluid layer over a
half space.
Let the boundary between the fluid and the solid be at x3 = 0, and the
top of the fluid at x3 = -H. There are no S waves in the fluid, and the P
wave potentials are:
w  C1 exp( i( px1  wx3  t))C2 exp( i( px1  wx3  t))
In the half space, we have
  Aexp( i( px1   x3  t))
  B exp( i ( px1   x3  t))

At the surface of the liquid, the boundary condition is
u1 u3 
u1 u3 
u3
 33   
  
 2
 0
x3
x1 x3 
x1 x3 



   
u   xˆ 1   xˆ 3
x1  x3 
u1
   

ip R    2 p 2 R 
 

 x1 x1 x1  x1
u3
   
2
2


i









 

w
I
R 
w  I   R 
x3 x3 x3  x3
 33   2 p 2I R    2w2I R   0
 p 2 I R   w I R 
2
Which will be true for all p only if I = -R. Evaluating at x3 = H gives
C1 exp( iwH )  C2 exp(iwH )
or
or

C1  C2 exp(2iwH )
C2  C1 exp( 2iwH )
At the
interface, we require that u3 is continuous:


w
u 
 iw R 
x3
T 

u3 

 iT  ip
x3 x1

3
So
iw  R   i T  ip
Evaluating at x3 = 0 gives:
wC  C1 exp( 2iwH )   A  pB

Note that

1 exp( 2iwH )  1 cos(wH ) i sin( wH )
2
 1 cos2 (wH ) sin 2(wH ) 2i cos(wH )sin( wH )



 2cos2 (wH ) 2i cos(wH )sin( wH )
 2cos(wH )exp( iwH )
So
C w1 exp( 2iwH )  2wC1 cos(wH )exp( iwH )
2wC1 cos(wH )exp( iwH )   A pB
 Continuity of the  gives:
33



 33

 u u 
u1 u3 
u3 

1
3
 w 

   33   
 2
x3 
x1 x3 
 x1 x3 
From above:


33
u1 u3 
2
2 2
2
 w 



p









w I
R
w w  I   R 
x1 x3 
For the half space:
u1
 T   


ip T  i 



x1 x1  x1 x3  x1
  2 p 2T  2 p 


u3
 T   


iT  ip


x3 x3 x3 x1  x3
   T   2 p 
2
2

33   p T  p    2   T   2 p 


2
2
2
2
2
1
Thus
 2 p 2 wI R    2 ww I R 
2


  2 p 2T  2 p    2   21 T   2 p 
2
Evaluating this equation at x3 = 0 gives:


 Ap     2   Bp     2  p 

wC1 C2 p 2  w
2
2

2



Note that
C2  C1 exp( 2iwH )
C1 C2  C1(1 exp( 2iwH ))
1 exp( 2iwH )  1 cos(wH ) i sin( wH )


 1 cos2 (wH ) sin 2(wH ) 2i cos(wH )sin( wH )
2
 2sin 2 (wH ) 2i cos(wH )sin( wH )
 2i sin( wH )cos(wH ) i sin( wH )


 2i sin( wH )exp( iwH )

So




2
2
2
2

 wC1 C2 p  w  2iwC1 p  w sin( wH )exp( iwH )
On the right side:

A p     2 
2
2
 Ap     
2
2
2

Bp     2  p   B2p

Hence



2iwC1 p  w sin( wH )exp( iwH )  A p    
2
2
2
2
2
 B2p 

w
wp  w  2  w
w
2
2

So

2iwC
1 sin( wH )exp( iwH )  A p    

2
2
2

Finally, we have to satisfy the shear stress condition on the interface:
u1 u3 
13    

x3 x1 
 is zero in the liquid, so we must have 13 = 0 at the interface.
u


2
1

i

p


i





p





T
1 
1 T
1 
x3 x3
u3

2
2

i



i

p



p



p




1 T
1 T
x1 x1


So



 p1T   1    p1T  p 2 0

T 2p    p 2 0

1

1

Evaluating the above at x3 = 0

A2p1  B p 2   1

 0
We now have 3 equations with 3 unknowns (A, B, and C1). Writing
these in matrix form:

 2p
A 
p 2  2
0
 2
 
2 2
p     2p 2i w sin(  )exp( i )B  0


p
2w cos( )exp( i )
C1



  wH



Non trivial solutions exist when the determinate of the above matrix
is zero:
4 p exp( i )(2pw cos( ) piw sin(  ))
2p 2  2 exp( i )( p 2  22 )w cos( ) i w sin(  )  0
or
2p (2pw  piw tan( ))
p 2  2 ( p 2  22 )w  i w tan( ) 0
solving for tan():


2p piw  i wp 2  2  tan( )  tan(  )i  w 2 p 2  p 2  2 


ˆ 
 tan( )i  wp 2  2   tan( )  2 w





 w4 p 2  p 2  2 ( p 2  22 )

  2w
tan( ) 
4 p 2   p 2  2 ( p 2    22 )
 w
ˆ


The term in the brackets is just the (negative) Raleigh denominator
again. Recalling the substitutions we made before:

  2  2  2  2 2  2  2 2
The term in the brackets becomes:
2
2
 
p




2




1
2
2
2


4
p



p



 1 1
1







  4  2 p 2 1 1  p 2 2  2 2   21
2






p 2   1
2





  4  2 p 2 1 1   2 p 2  1  2p 2  2 p 2   1
2
2

factor out p4:
4  2 
   2 
  2 
 1 1
1
1
2
2







 p 4 


1

2

1
2
2
2

  p 

p
p



 



v

p

1 p 2 v 2
pv
 1/ pv  1  i 1 1/ pv  i 1 c 2 / v 2
2
2
v 2
2
2
2
2


1
c
/
v

c
/
v
1
 


 p 
where c = 1/p is the apparent horizontal velocity. So the bracket
term becomes
2 



c
4
2
2
2
2
2
2
2
 p c  2 2 2  4  1 c /  1 c /  
  


Factor a 2 from the above to get:
2 
 c 2



c
4 2
2
2
2
2
 p   2  22 2  4 1 c /  1 c /  
  





or

2
2
2 2 

c
c
c 
4 2 
 p  4 1 2 1 2  2 2 









Thus

2
2
2 2 

 w 4 2 
c
c
c 
tan( ) 
p  4 1 2 1 2  2 2  
w

    
ˆ

2

2
2
2 2 

  w 
c
c
c 

4 1 2 1 2  2 2 
4

w c 

    
ˆ 

4
(NB: For readers of the textbook by Lay and Wallace, this is their
equation 4.3.6 except that the density terms are inverted. Note

that there is an extra 1/ term in their equation 4.3.6 which is
obviously wrong since the tangent MUST be dimensionless.