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

13 u1

x3

u3

x1

23 u2

x3

u3

x2

33 u1

x1

u2

x2

u3

x3

2

u3

x3

If we concern ourselves only with motion in the (x1, x3) plane, then u2 = 0 and d/dx2 = 0, so

33 u1

x1

u3

x3

2

u3

x3

0

13 u1

x3

u3

x1

0

23 0

(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 1x3 t))

reflected A2 exp(i( px1 1x3 t))

reflected B2 exp( i( px1 1x3 t))

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

v k3

cos(i)v

1 p2v2

v

So, in general, we consider the displacements at the interface substituting the above expressions into the following:

u x1

3

x2

2

x3

x 1

x2

1

x3

3

x1

x 2

x3

2

x1

1

x2

x 3

In the x1, x3 plane this becomes:

u x1

2

x3

x 1

1

x3

3

x1

x 2

x3

2

x1

x 3

Let's see what happens when we apply the no traction boundary condition:

33 u1

x1

u3

x3

2

u3

x3

0

u1

x1

x1

x1

2

x3

x1

ip R i 12

2 p2 R 2 p 12

u3

x3

x3

x3

2

x1

x3

i1 R ip2

21

2 R p 12

33 2 p2 R 2 p 12

2 21

2 R p 12 0

p21

2 2 R 2p 12 0

or

13 u1

x3

u3

x1

0

Next

u1

x3

x3

ip R i 12 2 p1

R 1

2

u3

x1

x1

i1 R ip2 2 p1

R p22

2p1 R p2 1

2 0so

Evaluating the above at x3 = 0:

p21

2 2 A1 A2 2p 1B2 0

2p1A1 A2 p2 1

B 0

We solve for A2/A1 =RPP and B2/A1=RSS (for potentials!)

Eliminating B2:

p21

2 2 p2 1

A1 A2 4p2 11

A1 A2

p21

2 2 p2 1

4p2 11 A1

p21

2 2 p2 1

4p2 11 A2

RPP p21

2 2 p2 1

4p2 11

p21

2 2 p2 1

4p2 11

Hence

Eliminating A2:

2p1p21

2 2 A1 A2 4p21 1

B2 0

2p1p21

2 2 A1 A2 p21

2 2 p2 1

B 0

4 p1p2 1

2 2 A1

p2 1

2 2 p2 1

4p211 B 0

RPS 4 p1

p21

2 2 p21

2 2 p 2 1

4p21 1

Hence

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:

RSS p21

2 2 p2 1

4p2 11

p21

2 2 p2 1

4p2 11

RSP 4p 1

p2 1

p21

2 2 p2 1

4p21 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 1x3 ) A2 exp( ˆ 1

x3 )

To prevent this from blowing up with large x3, we must have A1 = 0.

If we require

13 u1

x3

u3

x1

2p1

R p2 1

2 0

If A1 = 0, then in the absence of an SV wave we have:

2p1R 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 1x3 )

B exp(i( px1 t))exp( ˆ 1x3 )

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:

33 u1

x1

u3

x3

2

u3

x3

0

u1

x1

x1

x1

2

x3

x1

ip ˆ 12

2 p2 i 2 p ˆ 12

u3

x3

x3

x3

2

x1

x3

ˆ 1 ip2

2 ˆ 1

2 i 2 p ˆ 12

33 2 p2 i 2 p ˆ 12

2 2 ˆ 1

2 i 2 p ˆ 12 0

Evaluating at x3 = 0

p2A ip ˆ 1B 2 ˆ 1

2A ip ˆ 1B 0

A p2 2 ˆ 1

2 B 2 ip ˆ 1 ip ˆ 1

A p2 2 ˆ 1

2 Bi2p ˆ 1

Remember that and , so

i ˆ

2 ˆ 2

A p2 2 1

2 B2p 1

Now apply the other traction condition

13 u1

x3

u3

x1

0

u1

x3

x3

ip ˆ 12 2 ip ˆ 1

ˆ 1

2

u3

x1

x1

ˆ 1 ip2 2 ip ˆ 1

p22

Thus, evaluating at x3 = 0

ip ˆ 1A ˆ 1

B ip ˆ 1A p2B 2ip ˆ 1

A p2 ˆ 1

B 0

2p1A p2 1

B 0

or

The question now is: is there a nontrivial choice for A and B that satisfies these two equations? We write them in matrix form:

p2 2 1

2 2p 1

2p1p2 1

AB

00

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

p2 2 1

2 p2 1

2 4p2 11

0

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

p2 2 1

2

p2 1

2 4

p2 11

0

p2 2 2 2 21

2 p2 1

2 4 2 p2 11

0

2 p2 1

2 2p2 2 p2 1

2 4 2 p2 11

0

2 11

2

p2

2 2

1

1

2

p2

4 2 11

p2 0

factor out p4:

v

p

1 p2v2

pv 1/ pv 2 1 i 1 1/ pv 2 i 1 c2 / v2

v

p

2

1 c2 / v2 c2 / v2 1

where c = 1/p is the apparent horizontal velocity. So

c2 2 2 2c2

2

4 2 1 c2 /2 1 c2 / 2 0

we need to solve this for c. Factor a 2 from the above:

c2

2 2

2

c2

2

4 1 c2 /2 1 c2 / 2 0

2c2

2

2

4 1 c2 /2 1 c2 / 2 0

2c2

2

2

4 1 c2 /2 1 c2 / 2

Multiply the above by

To get

2c2

2

4

16 1 c2 /2 1 c2 / 2 0

c4

4 4c2

2 4

2

c8

8 4c6

6 4c4

4 4c6

6 16c4

4 16c2

2

The first term on the left becomes:

4c4

4 16c2

2 16

c8

8 8c6

6 24c4

4 32c2

2 16

16 1 c2 /2 1 c2 / 2 16 1 c2 /2 c2 / 2 c4

2 2

The second term on the left is:

c8

8 8c6

6 24c4

4 16c2

2 16c2

2 16c4

2 2 0

Combining

c2

2

c6

6 8 c4

4 24 c2

2 1616 2

2 16 c2

2

0

c6

6 8c4

4 c2 24 2

162

16 1

2

2

0

A solution to the above can be found for any and .

c2

2 4,223

,223

which has roots

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

As an example, assume a Poisson solid () in which case 2 = 32:

c6

6 8 c4

4 c2 24 2

163 2

16 2

3

0

c6

6 8c4

4 c2

2

563

323

0

What are the particle motions?

We need to solve for displacement (u):

u1 x1

2

x3

u3 x3

2

x1

u1 ip ˆ 12

u3 ˆ 1 ip2

at x3=0. From the 33 = 0 condition above, we found that

A p2 2 1

2 B2p 1

B Ap2 2 1

2

2p 1

Ap2 2 2 2 21

2

2 2 p 1

or

Ap22 2 2 2 c2 /2 1

2 2 p 1

A 2 c2 3 2

2 2c 1

A

c2

2 2

2c 1

A exp(i( px1 t))exp( ˆ 1x3 )

Recall that

B exp(i( px1 t))exp( ˆ 1x3 )

The shear term for u1 is then

ˆ 1A

c2

2 2

2c 1

exp(i( px1 t))exp( ˆ 1x3 )

ˆ 12

A exp(i( px1 t))ip ˆ 1

ip 1

c2

2 2

2c

exp( ˆ 1x3 )

A exp(i( px1 t))ip1p

c2

2 2

2c

exp( ˆ 1x3 )

A exp(i( px1 t))ip12

c2

2 2

exp( ˆ 1

x3 )

u1 ip ˆ 12

Combining:

A exp(i( px1 t))ip exp( ˆ 1x3 )

12

c2

2 2

exp( ˆ 1

x3 )

ip2 ipA

c2

2 2

2c 1

exp( i( px1 t))exp( ˆ 1x3 )

The shear term for u3 is:

A exp(i( px1 t))

c2

2 2

2c2 ˆ 1

exp( ˆ 1x3 )

u3 ˆ 1 ip2

Combining:

A exp(i( px1 t)) ˆ 1exp( ˆ 1

x3 )

c2

2 2

2c2 ˆ 1

exp( ˆ 1x3 )

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 be the Rayleigh wave number, then and and

ˆ 10.85

ˆ 10.39

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 0.62Ak exp(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))

A exp(i( px1 x3 t))

B exp(i( px1 x3 t))

In the half space, we have

33 u1

x1

u3

x3

2

u3

x3

u1

x1

u3

x3

0

At the surface of the liquid, the boundary condition is

u x1

x 1

x3

x 3

u1

x1

x1

x1

x1

ip R 2 p2 R

u3

x3

x3

x3

x3

iw I R 2w2 I R

33 2 p2 I R 2w2 I R 0

p2 I R w2 I R

Which will be true for all p only if I = -R. Evaluating at x3 = -H gives

C1 exp( iwH ) C2 exp( iwH )

C1 C2 exp( 2iwH )

C2 C1 exp(2iwH )

or

or

At the interface, we require that u3 is continuous:

u3

w

x3

iw R

u3

T

x3

x1

iT ip

iw R iT ipSo

w C C1 exp(2iwH ) A pB

Evaluating at x3 = 0 gives:

1 exp(2iwH ) 1 cos(wH ) i sin(wH ) 2

Note that

1 cos2(wH ) sin 2(wH ) 2i cos(wH )sin(wH )

2cos2(wH ) 2i cos(wH )sin(wH )

2cos(wH )exp(iwH )

Cw 1 exp(2iwH ) 2wC1 cos(wH )exp(iwH )So

2wC1 cos(wH )exp(iwH ) A pB

33 w

u1

x1

u3

x3

33

u1

x1

u3

x3

2

u3

x3

Continuity of the 33 gives:

33 w

u1

x1

u3

x3

2 p2w I R 2ww

2 I R

From above:

u1

x1

x1

T

x1

x3

x1

ipT i

For the half space:

2 p2T 2 p

u3

x3

x3

T

x3

x1

x3

iT ip

22T 2 p

33 2 p2T 2 p 2 21

2T 2 p

2 p2w I R 2ww2 I R

Thus

2 p2T 2 p 2 21

2T 2 p

w C1 C2 p2 w2

Evaluating this equation at x3 = 0 gives:

A p2 2 2 B p 2 p

C2 C1 exp(2iwH )

Note that

C1 C2 C1(1 exp(2iwH ))

1 exp(2iwH ) 1 cos(wH ) i sin(wH ) 2

1 cos2(wH ) sin 2(wH ) 2i cos(wH )sin(wH )

2sin2(wH ) 2i cos(wH )sin(wH )

2i sin(wH ) cos(wH ) i sin(wH )

2i sin(wH )exp(iwH )

w C1 C2 p2 w2 2iwC1 p2 w

2 sin(wH )exp(iwH )

So

A p2 2 2 A p2 2

2 On the right side:

B p 2 p B2p

2iwC1 p2 w2 sin(wH )exp(iwH ) A p2 2

2 Hence

B 2p

w p2 w2

w

w2 w

2iwC1 sin(wH )exp(iwH ) A p2 22

So

Finally, we have to satisfy the shear stress condition on the interface:

13 u1

x3

u3

x1

is zero in the liquid, so we must have 13 = 0 at the interface.

u1

x3

x3

ipT i 1 2 p1

T 1

u3

x1

x1

i1T ip 2 p1

T p2

p1T 1

p1T p2 0

So

T 2p1 1

p2 0

A 2p1 B p2 1

0

Evaluating the above at x3 = 0

We now have 3 equations with 3 unknowns (A, B, and C1). Writing these in matrix form:

2p p2 2 0

p2 22 2p 2iw sin( )exp(i )

p 2w cos( )exp(i )

ABC1

0

wH

4 p exp(i )( 2pw cos( ) piw sin( ))

Non trivial solutions exist when the determinate of the above matrix is zero:

2 p2 2 exp(i ) ( p2 2

2 )w cos( ) i w sin( ) 0

2p ( 2pw piw tan( ))

p2 2 ( p2 2

2 )w i w tan( ) 0

or

solving for tan():

2p piw i w p2 2 tan( )

tan()iw 2p2 p2 2

tan( )i w p2 2 tan( )

ˆ w

2

w 4p2 p2 2 ( p2 2

2 )

tan( ) 2w

w ˆ 4p2 p2

2 ( p2 22 )

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

4

p2 11

p2 2 1

2

p2 1

2

The term in the brackets becomes:

4 2 p2 11

p2 2 2 2 21

2 p2 1

2

4 2 p2 11

2 p2 1

2 2p2 2 p2 1

2

p4 4 2 11

p2 2 11

2

p2

2 2

1

1

2

p2

factor out p4:

v

p

1 p2v2

pv 1/ pv 2 1 i 1 1/ pv 2 i 1 c2 / v2

v

p

2

1 c2 / v2 c2 / v2 1

where c = 1/p is the apparent horizontal velocity. So the bracket term becomes

p4 c2 2 2 2c2

2

4 2 1 c2 /2 1 c2 / 2

p 4 2 c2

2 2

2

c2

2

4 1 c2 /2 1 c2 / 2

Factor a 2 from the above to get:

p4 2 4 1c2

2 1c2

2 2c2

2

2

or

w

4

c4

w

ˆ 4 1

c2

2 1c2

2 2c2

2

2

(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.

tan( ) 2w

w ˆ p4 2 4 1

c2

2 1c2

2 2c2

2

2

Thus