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Problems in Exploration Seismology and their Solutions
Problems-in-Exploration-Seismology-and-their-Solutions.jpg
Series Geophysical References Series
Title Problems in Exploration Seismology and their Solutions
Author Lloyd P. Geldart and Robert E. Sheriff
DOI http://dx.doi.org/10.1190/1.9781560801733
ISBN ISBN 9781560801153
Store SEG Online Store


Chapter 3 Partitioning at an interface

3.1 General form of Snell’s law

3.1a A P-wave of amplitude is incident at the angle on a plane interface separating two solid media. This generates reflected and refracted P-waves and converted reflected and refracted S-waves. Amplitudes, angles of incidence and refraction, and directions of displacements of these waves are shown in Figure 3.1a.

Use Huygens’s principle to show that and that


(3.1a)

where is the raypath parameter. Equation (3.1a) is Snell’s law.

Background

When a wave is incident at the interface between two solid media, four boundary conditions must be satisfied (continuity of normal and tangental displacements and stresses). The velocities are determined by the densities and elastic constants while the angles of incidence, reflection, and refraction are fixed by the velocities [see equation (3.1a)]. So the only remaining parameters that one can adjust in order to satisfy the boundary conditions are the amplitudes of the four waves generated by the incident wave, the reflected and refracted P- and S-waves, , , , as shown in Figure 3.1a.

Figure 3.1a.  Raypaths at solid-solid interface.

Huygens’s principle states that each point on a wavefront acts as a new point source radiating energy in all directions. Subsequent wavefronts can be located by swinging arcs with centers at points on the wavefront and radii equal to the distance traveled in a fixed time interval, the new wavefront being the envelope of the arcs. If the first wavefront and the reflector are planar, only two arcs are necessary, the new wavefront being tangent to the arcs.

Solution

In Figure 3.1b, is a wavefront of a planar P-wave approaching a planar interface. When the wavefront reaches the interface, point becomes a new source radiating energy upward and downward according to Huygens’s principle. When reaches the interface at , the distance being , the wave reflected at has traveled upward the same distance . By drawing an arc with center and radius and then drawing a line from tangent to the arc, we get the reflected wavefront . The angle of incidence is and the angle of reflection is . In and the angles at and are (because rays are perpendicular to wavefronts). Since the triangles have a common side , they are equal and , that is, the angle of incidence equals the angle of reflection (Law of reflection).

Figure 3.1b.  Snell’s law derivation.

In the case of the refracted wave, is the angle of refraction and But , so we have

In Figure 3.1b, if we replace in with (compare with Figure 3.1a) and in with , we arrive at

Equating the four ratios of sines to velocities, we get equation (3.1a).

3.1b Using the waveform (see problem 2.5b), where are direction cosines of the ray, show that (omitting the factor ) the incident, reflected, and refracted waves can be written


(3.1b)


(3.1c)

where


(3.1d)


(3.1e)

Solution

We write , where the plus sign is used for waves traveling upward (that is, in the positive -direction) and the minus for downward traveling waves. The velocity is for P-waves, for S-waves. We note that for P-waves, , , , 2. For S-waves we replace with so that , , , 2.

Thus, for P-waves, while for S-waves we have . Inserting the amplitudes , , we get equations (3.1b,c,d,e).

3.2 Reflection/refraction at a solid/solid interface; Displacement of a free surface

3.2a Derive the Zoeppritz equations for a P-wave incident on a solid/solid interface

Background

The normal and tangential displacements plus the normal and tangential stresses must be continuous when a P-wave is incident at the angle on an interface between two solid media (see problem 2.10).

Solution

We use the functions in equations (3.1b,c,d,e) to represent the displacements of the waves, the positive direction of displacement for the waves being shown in Figure 3.1a. (We omit the factor because the boundary conditions do not depend upon the time , hence this factor cancels out).

We first derive the equations expressing the continuity of normal and tangential displacements, and . These equations are obtained by resolving the various wave displacements into - and -components. Thus,


(3.2a)


(3.2b)


(3.2c)


(3.2d)

At the interface, and , . The exponentials all reduce to , hence cancel out, and we get for the normal and tangential displacements, respectively,


(3.2e)


(3.2f)

To apply the boundary conditions for the normal and tangential stresses, we differentiate equations (3.2a,b,c,d) with respect to and . Equations (3.1d,e) show that the differentiation with respect to and multiplies each function by and either or . The common factor will cancel in the end, so we simplify the derivation by taking


(3.2g)

From equations (2.1b,c,e,h,i) we get for the normal and tangential stresses:

where , , and , are partial derivatives with respect to and . This allows us to find the normal and tangential stresses in each medium and equate them at . The result for the normal stresses is

Writing (see equations (9,6) and (9,7) in Table 2.2a) and recalling that , , the equation can be changed to the form


(3.2h)

where , ; and are called impedances.

In the same way we get for the tangential stresses the equation

This can be simplified using equation (3.1a) to give


(3.2i)

Equations (3.1e,f,h,i) are known as the Zoeppritz equations. For ease of reference, we have collected them below:

(3.2e)


(3.2f)


(3.2h)


(3.2i)

3.2b Derive the equations below for the tangential and normal displacements, and , of a free surface for an incident P-wave of amplitude :


(3.2e)

where , , and being the angles of incidence of the P- and S-waves, respectively.

Solution

To determine the displacements at a free surface, we start by disregarding equations (3.2e,f) because there are no constraints on displacements at a free surface. After setting , we are left with


(3.2j)


(3.2k)

where , , , and we have dropped unnecessary subscripts. These equations can be written


(3.2l)

where


(3.2m)


(3.2n)

The solution of equations (3.2) is

In equations (3.2a,c) we set , the factor drops out, and we get


(3.2o)


(3.2p)

We now reinsert the values of and in terms of m and , and equations (3.2o,p) become


(3.2q)


(3.2r)

3.2c Show that the displacements of a free surface at normal incidence are

Solution

For normal incidence at the surface, , . Equations (3.2j,k) give , . Substituting in equations (3.2o,p), we get , .

3.2d Show that the displacements of a free surface of a solid, where , km/s, , , are

Solution

For , km/s, , , that is, , so . From the definitions of m and n, we get

Equations (3.2q,r) now give (omitting the factor ).

3.2e Show that the displacements at the surface of the ocean are , .

Solution

In a fluid and equation (3.2j) gives , so equations (3.20,p) show that and .

3.3 Reflection/refraction at a liquid/solid interface

3.3a Derive Zoeppritz’s equations and Knott’s equations for a P-wave incident on a liquid/solid interface when the incident wave is (i) in the liquid and (ii) in the solid.

Background

Knott’s equations differ from the Zoeppritz equations in that they use potential functions instead of displacements. Knott’s equations can be derived directly from the Zoeppritz equations and vice versa [see equation (3.3)]; however, we shall derive them from first principles. We use script letters, and for the amplitudes of the potential functions, reserving italic letters and for displacements.

To get Knott’s equations for a solid/solid interface, we start with the potential functions in equation (2.9c), the displacements being given by equations (2.9d,e), and apply the boundary conditions of problem 2.11, we write


(3.3a)


(3.3b)

where , , , and are given by equations (3.1d,e). Using equation (2.9e), the continuity of normal displacement requires that be continuous at . Using equations (3.2g), we obtain


(3.3c)

Continuity of tangential displacement requires the continuity of [see equation (2.9d)]; this gives the equation


(3.3d)

The normal stress is given by [see equation (2.11b)], so

But , (see equations (9,6) and (9,7) of Table 2.2a); using these relations plus equation (3.1a), we get


(3.3e)

The tangential stress is from equation (2.11b), so we get


(3.3f)

We can write Knott’s equations in a more compact form by substituting , , ; the four equations now become

(3.3g)


(3.3h)


(3.3i)


(3.3j)

To show the correspondence between Knott’s and Zoeppritz’s equations, we calculate the energy density in terms of the displacements and the potential functions used in Knott’s equations. In terms of displacements, the instantaneous kinetic energy density E for a harmonic P-wave is equal to

The total energy density is the maximum kinetic energy density (see problem 3.7), that is,


(3.3k)

The energy density of a P-wave in terms of the potential function [see equation (2.9a)], noting that since there is no S-wave) is

from equations (2.9d,e). Taking the time factor as and reinserting the factor which was deleted to get equation (3.2g), we get , , . Thus, we get for the total energy

Comparing with the expression for in equation (3.3k), we see that


(3.3l)

where the second equation is obvious from symmetry.

Solution

Since we derived Zoeppritz’s equations in problem 3.2a, we derive Knott’s equations here and then get Zoeppritz’s equations from them using equation (3.31).

i) To derive Knott’s equations when the incident P-wave is in the liquid, the boundary conditions require the continuity of and and that in the solid vanish at . omitting the factor , we have from equations (3.1b,c,d,e):

Continuity of yields the first equation [note equation (3.2g)]:

Continuity of normal stress requires that be continuous; since in the liquid, this results in

Using equations (9,6) and (9,7) in Table 2.2a, also equation (3.1a), we have

Continuity of tangential stress requires that in the solid when , so

Using the coefficients in equations (3.3g,h,i,j), the results become

These are Knott’s equations. We could derive Zoeppritz’s equations as we did equations (3.2e,f,h,i), or we can use equation (3.3) to change the coefficients in Knott’s equations to Zoeppritz’s coeffiicients. Using this latter method, we have , . Substituting these, we get the Zoeppritz equations for a liquid-solid interface:

ii) When the incident wave is in the solid, we shall first derive Zoeppritz’s equations, then change them to Knott’s equations. We have in the liquid, and are continuous, and in the solid at .

Equation (3.2e) gives for the normal displacement

The continuity of normal stress is expressed in equation (3.2h), which now becomes

Finally, at and equation (3.2i) becomes

Using equation (3.3), we get the equivalent Knott’s equations:

3.3b Calculate the amplitudes of the reflected and refracted P-and S-waves when an incident P-wave strikes the interface from a water layer , , g/cm) at when the seafloor is (i) “soft” , m/s, g/cm, and (ii) “hard” , , g/cm).

Solution

i) Where the seafloor is “soft” and the P-wave is incident in the water, we have:

Thus Zoeppritz’s equations become

The solution is: , , .

ii) When the seafloor is “hard”:

The equations are

The solution is , , .

3.3c Repeat part (b) for an angle of incidence of .

Solution

i) For the “soft” bottom and ,

The equations are

The solution is , , .

ii) For the “hard” bottom, , so total reflection occurs.

The results are summarized in Table 3.3a. The table shows that and depend mainly on the hardness of the bottom and only moderately on . However, depends more on the angle of incidence than on the hardness.

Table 3.3a Reflected/transmitted amplitudes for soft/hard bottoms.
Bottom
Soft 0.431 0.539 0.244
0.403 0.512 0.408
Hard 0.716 0.227 0.304

3.4 Zoeppritz’s equations for incident SV- and SH-waves

3.4a Derive Zoeppritz’s equations for an SV-wave incident on a solid/solid interface.

Solution

Figure 3.1a defines the positive directions of displacements except that the incident P-wave is replaced by an incident SV-wave whose positive direction is down and to the left (the same as that of ). Using the same symbols as in equations (3.1b,c), we define the following functions:

where and are the same as in equation (3.1d,e). We get the following expressions for the displacements and using equations (3.2a,b,c,d), where the terms in are replaced with terms in :


The boundary conditions require the continuity of , , and at . Continuity of and gives

For the normal stress, we have


Thus, continuity of requires that

Continuity of the tangential stress, , gives

We can simplify the equations for and by noting that

where is the raypath parameter [see equation (3.1a)]. Also,

We can now write the four equations in the form


3.4b Derive the Zoeppritz equations for an incident SH-wave.

Solution

For an SH-wave traveling in the -plane, the wave motion involves only displacement parallel to the -axis where . We take the incident, reflected, and refracted waves in the form [see equations (3.1b,c,d,e)]


The boundary conditions require that the tangential displacement and tangential stress be continuous at . The first condition gives


(3.4a)

The tangential stress is (note that ), where

Recalling that we can take , [see equation (3.2g)], we get

So


(3.4b)

Solving equations (3.4a,b), we find

The absence of P-waves is important in SH-wave studies.

3.5 Reinforcement depth in marine recording

3.5a For a source at a depth , show that the maximum amplitude of a downgoing incident wave and its reflection at the surface of the sea occurs at the depth , where is the angle of incidence, by expressing the pressure in the form used in equations (3.1b,d) and applying appropriate boundary conditions.

Solution

Since the interface is liquid/vacuum, only two waves exist, the incident and reflected P-waves. Taking the z-axis positive downward, we take in the form

There is only one boundary condition, namely that at . This gives Using Euler’s formulas (see Sheriff and Geldart, 1995, problem 15.12a), we get

upon inserting the time factor. The amplitude of the combined incident and reflected waves is

It is a maximum when , that is, when

3.6 Complex coefficient of reflection

3.6a Using the expression to represent a plane wave incident on a plane interface, show that a complex coefficient of reflection,

R [defined by equation (3.6a) below] corresponds to a reduction in amplitude by the factor and an advance in phase by .

Background

When a plane P-wave is incident perpendicularly on a plane interface, the tangential displacements and stresses vanish, so equations (3.2f,i) are not constraining and we are left with equations (3.2e,h). Moreover, , so the equations reduce to

, , being impedances (see problem 3.2). The solution of these equations is


(3.6a)


(3.6b)

and being the coefficient of reflection and coefficient of transmission, respectively. Although equations (3.6a,b) hold only for normal incidence, the definitions and are valid for all angles of incidence. A negative value of means that in equation (3.6a) is opposite in sign to . Since , the minus sign is equivalent to adding to the phase of the waveform in part (a), that is, reversing the phase. Note that (except for phase reversal) is independent of the direction of incidence on the interface; however, the magnitude of depends upon this direction, and when necessary, we shall write and to distinguish between the two values. Note the following relations:


(3.6c)

where is the fraction of energy transmitted as defined in equation (3.7a).

Euler’s formulas (see Sheriff and Geldart, 1995, p. 564) express and as


(3.6d)

The hyperbolic sine and cosine are defined by the relations


(3.6e)

Solution

Writing , we have

But where (see Sheriff and Geldart, 1995, section 15.1.5), so

Since , the amplitude is reduced by this factor and the phase is advanced by

3.6b Show that an imaginary angle of refraction (see Figure 3.1b) in equations (3.2e,f,h,i) leads to a complex value of and hence to phase shifts.

Solution

Let , where is real. Then, using equations (3.6d,e), we have

hence some of the coefficients in equations (3.2e,f,h,i) are imaginary and and will in general be complex, so phase shifts will occur.

3.6c Show that when is negative, .

Solution

If is negative, in equation (3.6a), so . Therefore, from equation (3.6b) we see that , and since , .

3.7 Reflection and transmission coefficients

3.7a Calculate the reflection and transmission coefficients, and of equations (3.6a,b), for a sandstone/shale interface (for a wave incident from the sandstone) for the following:

  1. km/s, km/s, g/cm, and g/cm;
  2. km/s, km/s,
      g/cm, and  g/cm.
    
  3. What are the corresponding values in nepers and decibels?

Background

A wave has kinetic energy due to the velocity of the medium and potential energy due to the strains in the medium. For a harmonic wave , the particle velocity is and the kinetic energy/unit volume is , being the density. As the wave progresses, the energy changes back and forth between kinetic and potential. When the potential energy is zero, the kinetic energy is a maximum and therefore equals the total energy. The maximum value comes when , so the total energy/unit volume, where is the energy density (problem 2.3).

The coefficients and in equations (3.6a,b) give ratios of the relative amplitudes of the reflected and transmitted waves. We denote the fractions of the incident energy that are reflected and transmitted by and ; “energy” as used here denotes the amount of energy flowing through a unit area normal to the wave direction per unit time (the intensity). The energy density (i.e., energy/unit volume) is given by equation (3.3k). The energy flowing through a unit area normal to the wave direction per unit time is equal to the energy density times the velocity, that is, for a P-wave. Therefore


(3.7a)

The coefficients and are sometimes referred to as reflection and transmission energy coefficients to distinguish them from and . Note that both and are independent of the direction of travel through the interface.

Solution

i) Using equations (3.6a,b), we have

(where the units are g. km/cm3. s). Then,

(The minus sign denotes a phase reversal; see problem 3.6.) Because the incident wave is in the sandstone, we have

Note that the amplitude of the transmitted wave is larger than that of the incident wave when is negative (see problem 3.6c).


ii) .


Then,


iii) From problem 2.17, we have nepers ln(amplitude ratio) and 1 neper dB. Therefore,

for (i),

For (ii),

Negative values of nepers and merely mean that the values are less than unity.

3.7b Calculate the energy coefficients and for cases (i) and (ii) in part (a).

Solution

We use equation (3.7a) to get,

for (i),


For (ii),

Note that (within the accuracy of the calculations).


3.8 Amplitude/energy of reflections and multiples

3.8a Assume horizontal layering (as shown in Figure 3.8a), a source just below interface , and a geophone at the surface. Calculate (ignoring absorption and divergence) the relative amplitudes and energy densities of the primary reflections from and and the multiples , , and (where the letters denote the interfaces involved). Compare traveltimes, amplitudes, and energy densities of these five events for normal incidence.

Background

Multiples are events that have been reflected more than once. They are generally weak because the energy decreases at each reflection, but where the reflection coefficients are large, multiples may be strong enough to cause problems. Multiples are of two kinds as shown in Figure 3.8b: long-path multiples which arrive long enough after the primary reflection that they appear as separate events, and short-path multiples which arrive so soon after the primary wave that they add to it and change its shape. The most important short-path multiples are two in number: (i) ghosts (Figure 3.8b) where part of the energy leaving the source travels upward and is reflected downward either at the base of the LVL (see problem 4.16) or at the surface, (ii) peg-leg multiples resulting from the addition to a primary reflection of energy reflected from both the top and bottom of a thin bed, either on the way to or on the way back from the principal reflecting horizon. Short-path near-surface multiples are also called ghosts and long-path interformational multiples are also called peg-leg multiples. A notable example of the latter occurs in marine work when wave energy bounces back and forth within the water layer.

The energy density of a wave (see problem 3.7) decreases continuously as the wave progresses because of two factors: absorption and spreading or divergence. The energy density is proportional to the square of the amplitude, so both effects are usually expressed in terms of the decrease in amplitude with distance.

Figure 3.8a.  A layered model.
Figure 3.8b.  Types of multiples.


Absorption causes the amplitude to decrease exponentially, the relation being where the amplitude decreases from to over a distance ; the absorption coefficient is often expressed in terms of per wavelength,

For a point source in an infinite constant-velocity medium, divergence causes the energy density to decrease inversely as the square of the distance from the source, the amplitude decreasing inversely as the first power of the distance from the source.

Nepers and decibels are defined in problem 2.17.

Solution

We first calculate the impedances for each layer, the coefficients of reflection and downgoing and upgoing transmission , (see problem 3.6), and the reflected and transmitted energy coefficients, and , for each interface. The results are shown in Table 3.8a.

Table 3.8a. Reflection and transmission coefficients.
Interface
S 1.000 0.000 0.000 1.000 0.000
Layer 1 0.870
0.733 0.267 1.733 0.537 0.463
Layer 2 5.640
0.207 0.793 1.207 0.043 0.957
Layer 3 8.576
0.034 0.966 1.034 0.001 0.999
Layer 4 9.180
* Signs are for incidence from above.


Assuming unit amplitude and unit energy density for the downgoing wave incident on interface and neglecting absorption and divergence, we arrive at the following values:

Reflection :

Amplitude of reflection Energy density Arrival time

Reflection

Amplitude

Energy density

Arrival time

Multiple BSA

Amplitude

Energy density

Arrival time

Multiple BAB

Amplitude

Energy density

Arrival time

Multiple BSB

Amplitude

Energy density

Arrival time

The results are summarized in Table 3.8b.

BSA arrives 33 ms after (one period for a 33-Hz wave) with reversed polarity and about 75% of the amplitude and 50% of the energy of , so BSA will significantly alter the waveshape of B. BSA involves an extra bounce at the surface and is a type of ghost whose effect is mainly that of changing waveshape rather than showing up as a distinct event.

and BAB arrive simultaneously with opposite polarities, being slightly stronger than BAB; the multiple will obscure and significantly alter the waveshape of the primary reflection.


Table 3.8b. Amplitude/energy density of primary/multiple reflections.
Event Amplitude Energy