Seismology

Part III:

Body Waves and Ray Theoryin 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

p sin(i)

c1

sin( ir )

c1

sin(it )

c2

ir i

it sin 1 c2 sin(i)

c1

And so

it cannot be greater than 90o for a transmitted wave. We define the critical angle as:

1c2 sin(ic )

c1

ic sin 1 c1

c2

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:

Tdirect X

c1

Treflect 2h

c1 cos(i)

Thead X 2h tan(ic )

c2

2h

c1 cos(ic )

p 1

c2

sin(ic )c1

c2

c1p

Note that for a head wave

cos(ic ) 1 sin 2(ic ) 1 c12 p2

Thead Xp 2h1

c1 cos(ic )

sin(ic )

c2 cos(ic )

Xp

2h

1 c12 p2

1

c1

c1p

c2

11

221

1

221

221

21

21

1

2 hXpc

pchXp

c

pc

pc

hXp

22

1

1

1p

c

Substituting in for the expression for head wave time:

where

Note that T is proportional to X (minus a constant). Where did we see that before?

Thead X

cn

2 hi

1

ci cos(ii )

tan(ii )

cni1

n

Xp 2 hiii1

n

In multiple layers, we can write this as:

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.

Trefl2

X 2 2 h2

c12

Note that for the reflected wave

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

X

c1

X

c2

2h1 X

c2

2h1 c1

2 p2

c1

12

21

22

12

2212 2

12

cc

cch

cc

pcchX

X 2hc2 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:

td 2hd cos(ic )cos( )

c1

x sin(ic )

c1

tu 2hu cos(ic )cos( )

c1

x sin(ic )

c1

Updip travel time is

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 ) and the depth can be determined from the intercepts:

(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:

2

1

1

1 )sin()sin(

c

a

c

i

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

)sin()sin( 2211 irard

This will be true as the width becomes infinitesimally small. So, in general

pconstc

r i .)sin(

Also, near the surface, simple geometry shows:

ro sin( o )

co

p T

Remember that "" is in radians!

Travel Time and Distance in a Sphere

Consider a spherical ray segment:

ds2 dr 2 rd 2

sin(i)rdds

and

p r sin(i)

c

r 2dcds

so

ds2 dr 2 rd 2 r 4d2

c2 p2

and

d2 dr 2c2 p2

r 4 r 2c2 p2

Solve for d:

d drcp

r

1

r 2 c2 p2

p

r

dr

r / c 2 p2

Integrate the above from ro to rmin (maximum depth of penetration) and multiply by 2 to get total distance:

2pdr

r r / c 2 p2ro

rmin

We follow similar steps to get travel time by integrating solving for ds instead of d and then integrating ds/c:

T ds

cpath

2r / c 2

r r / c 2 p2ro

rmin

dr

T p 2r / c 2 p2

rro

rmin

dr

Note that this can be written like a -p equation like in flat earth by combining the above with the expression for :

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:

0, ijij udV

dV ij, j dS ijnj

From the divergence theorem:

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 ijnj 0

0 jiji ndSudV

So

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

reflectedlayer 1

The potentials in layer 2 will be the refracted P and SV waves:

layer 2 refracted

layer 2 refracted

We can write the solution to the wave equations as:

incident A1 exp(i( px11x3 t))

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

refracted A3 exp(i( px1 2x3 t))

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

refracted B3 exp(i( px1 2x3 t))

The signs in the arguments that correspond to the direction of propagation, and the appropriate choices of the x3 factors. Also, the x1 factor is the same in every case because of Snell's law. Now

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

The continuity of u across the interface means uxi+ = uxi

- (as appropriate). For the traction condition, we convert traction to stress using

Ti ijnj

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 x1

ˆ x 1x3

ˆ x 3

At a liquid-liquid interface, only the vertical component is continuous, so

x3

i1A1 A2 exp(i( px11

x3 t))

x3

i 2A3 exp(i( px1 2

x3 t))

1A1 A2 exp(i1

x3 ) 2A3 exp(i 2

x3 )

Thus

or, for convenience we choose x3 = 0, then

1A1 A2 2

A3

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 + 233 u 2

because = 0 in a liquid.

121 2

22

The scalar potential

121

1

12

21

t2

21

12 1

1

12 (A1A2 )

2

22 A3

Thus

1(A1A2 )2A3

So

A1A2 2

1

A3

A1 A2 2

1

A3

2A1 2

1

2

1

A3

2112

11

A3

A3

A1

211

2112

Sum the above

Eliminating A3 from the above:

1

2

A1A2 1

2

A1 A2

A2

1

2

1

2

A2

12 21

22

A1

1

2

1

2

A1

21 12

22

A2

A1

21 12

12 21

or

Now, note that these relationships are for the amplitudes of the scalar potential function, not of the displacement. To recover displacement, recall that

u x3

ˆ x 3

1221

1221

11

21

Ai

Ai

u

uR

incident

reflect

2112

21

11

32 2

Ai

Ai

u

uT

incident

refract

Note that at normal incidence, p = 0, so 1= 1/1, 2= 1/2 and so in this case

2211

2211

1221

1221

//

//

incident

reflect

u

uR

1122

11

2112

21 2

//

/2

incident

refract

u

uT

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 p22

2

2

1/22 p2 1/2

2 sin2(i1)/12

when i = ic, n2 = 0.

When i1 > ic, n2 is imaginary:

2 i p2 1/22

This means that the transmitted wave will decay exponentially with distance away from the interface.

The reflection coefficient becomes:

ureflect

uincident

1i2 21

1i2 21

21 1i2

211i2

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:

shift 2tan 1 12

21

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 1x3 t / ))

which means that the phase of the wave (constant argument) is a function of frequency.

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

22

21

21

22

12212

12

2

2112

2112

2112

21 22

i

i

i

i

i

u

uT

incident

refract

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.

ˆ t t / (t / )

The term