User:Ageary/Theory of seismic waves

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Chapter 2 Theory of seismic waves

2.1 The basic elastic constants

2.1a Show that when the only nonzero applied stress is , Hooke’s law requires that the normal strains , and that Poisson’s ratio , defined as , satisfies the equation


(2.1a)

Background

Stress is force/unit area and is denoted by , etc., where a force in the -direction acts upon a surface perpendicular to the -axis. The stresses and are, respectively, a normal stress and a shearing stress.

Stresses produce strains (changes in size and/or shape). If the stresses cause a point to have displacements along the coordinate axes, the basic strains are given by derivatives of these displacements as follows:


(2.1b)


(2.1c)

The vector displacement is


(2.1d)

where are unit vectors in the -directions (see Sheriff and Geldart, 1995, problem 15.3). The dilatation is the change in volume per unit volume, i.e.,


(2.1e)

A pressure produces a decrease in volume, the proportionality constant being the bulk modulus :


(2.1f)

Sometimes the compressibility, , is used instead of .

In addition to creating strains, stresses cause rotation of the medium, the vector rotation being equal to


(2.1g)

where , ,

For small strains and an isotropic medium (where properties are the same regard-less of the direction of measurement), Hooke’s law relates the stresses to the strains as follows:


(2.1h)


(2.1i)

where and are Lamé’s constants ( is usually called the modulus of rigidity or the shear modulus).

Solution

Subtracting equation (2.1h) for from the same equation for gives .

Dividing equation (2.1h) for by gives

so

2.1b Show that Young’s modulus , defined as , is given by the equation


(2.1j)

Solution

Adding the three equations (2.1h) for , , and , and recalling that , we get

Dividing both sides by gives

Using equation (2.1a) we get

2.1c A pressure is equivalent to stresses . Derive the following result for the bulk modulus :


(2.1k)

Solution

Since , we add equation (2.1h) for each of the three values , obtaining , so from equation (2.1f),

2.2 Interrelationships among elastic constants

The entries in Table 2.2a express, for isotropic media, the quantities at the heads of the columns in terms of the pairs of elastic constants or velocities at the left ends of the rows. The first three entries in the ninth row are equations (2.1j), (2.1a), and (2.1k) and the next two entries in the same row are formulas for the P- and S- wave velocities and (see problem 2.5). Starting from these five relations, derive the other relations in the table.

Background

For isotropic media, any two of the elastic constants can be considered as independent and the others can be expressed in terms of these two. The P- and S-wave velocities, and , given by the equations


(2.2a)

[see Sheriff and Geldart, 1995, equations (2.28) and (2.29)] can also be expressed in terms of any two elastic constants (plus the density ).

Solution

Denoting the equations by row and column (as for matrix elements) and using a comma instead of a period, we use equations (9,1) to (9,3) to derive the equations that do not involve and , then we use equations (9,6) and (9,7) (see equation (2.2a)) to derive the rest. From equation (9,1),


(3,5)

From equation (9,2),


(5,5)

Solving equation (9,2) for , we have ,


(6,4)

From equation (9,3),


(7,5)

Equating from equations (9,2) and (9,3) [or from equations (6,4) and (7,5)] gives


(8,4)

thus

(6,3)

and

(4,5)

Solving equation (4,5) for gives


(8,2)

Substituting equation (4,5) into equation (8,4), we get,


(4,4)

We use equation (7,5) to eliminate from equation (6,3):

that is,

so


(5,3)

Solving for , we get


(7,2)
Figure 2.2a)  Relations between elastic constants and velocities for isotropic media.

We use equation (8,4) to express in equation (9,1) in terms of . Thus


(8,1)

Solving equation (8,1) for gives


(2,5)

We now eliminate from equation (9,1) using equation (7,5)


(7,1)

Solving equation (7,1) for and gives


(3,3)


(2,4)

Next we use equation (6,4) to eliminate from equation (9,1):


(6,1)

Using equations (9,1) and (5,5) we get


(5,1)

Solving equation (5,1) for and gives


(3,2)


(1,4)

Using equations (4,4) and (4,5) to replace in equation (9,1) by gives


(4,1)

Solving equation (4,1) for and , we get


(1,3)


(2,2)

The last equation (of this group) can be obtained by substituting equation (1,3) into equation (4,5):


(1,5)

Equations (10,1) to (10,3) express , , and in terms of the P- and S-wave velocities, and (see equation 2.2a). To introduce and , we write equations (10,1) to (10,3) in terms of and . Thus,


(10,1)


(10,2)


(10,3)

Equation (10,4) is the second equation in equation (2.2a). To get equation (10,5), we write


(10,5)

To verify column 6, we start with equation (9,6) and express and in terms of the required pair of constants. Thus, using equations (1,4) and (1,5) we get


(1,6)

Following the same procedure, using equations (2,4) and (2,5), we get


(2,6)

In the same way, we get the following results:


(3,6)


(4,6)


(5,6)


(6,6)


(7,6)


(8,6)

Column 7 is merely the square root of column 4 after dividing by . Column 8 is obtained by dividing column 7 by column 6.

2.3 Magnitude of disturbance from a seismic source

2.3a Firing an air gun in water creates a pressure transient a small distance away from the air gun with peak pressure of 5 atmospheres ( Pa). If the compressibility of water is /Pa, what is the peak energy density?

Background

Air guns (see problem 7.7) suddenly inject a bubble of high‐pressure air into the water to generate a seismic wave.

Stresses acting upon a medium cause energy to be stored as strain energy, because the stresses are present while the medium is being displaced. Strain energy density (energy/unit volume) is equal to [see Sheriff and Geldart, 1995, equation (2.22)]


(2.3a)

Solution

From problem 2.1c, we see that Pa. Also, for water, so . From equation (7,5) of Table 2.2a we find that when . Also, (see equation (2.1f)), so


Using equation (2.3a) we find that


[The dimensions of are the same as those of stress, since strains are dimensionless. Thus, stress units are .]

2.3b If the same wave is generated in rock with Pa, what is the peak energy density? Assume a symmetrical -wave with for .

Solution

We have , , so equation (2.3a) becomes


Equation (9,3) in Table 2.2a gives so that


2.4 Magnitudes of elastic constants

To illustrate the relationships and magnitudes of the elastic constants, complete Table 2.4a.

[Note that these values apply to specific specimens; the elastic constants for rocks range considerably, especially as porosity and pressure change.]

Solution

We use the row-column notation to designate equations from Table 2.2a.

Water: Since water is a fluid we know that . Equation (4,1) shows that also. From equation (4,5) we get Pa.

Table 2.4a. Magnitudes of elastic constants and velocities.
Constant Water Stiff mud Shale Sandstone Limestone Granite
16 54 50
2.1
0.50 0.43 0.38 0.34 0.25 0.20
1.0 1.5 1.8 1.9 2.5 2.7
1.5 1.6 3.2

Stiff mud: Because , stiff mud is equivalent to a solid, hence . From and we get using equation (9,6), while equation (9,2) expresses in terms of and , thus enabling us to find both and .

We have:


Solving the two equations, we get Pa, Pa. Using equation (6,1),


Equation (6,3) gives


Finally, to get we note that , , and are in , and so we must express in appropriate units of , i.e., . We now have


Shale. As with stiff mud, we have been given , , and , so that again and give us and gives us so that we can solve these equations for and , then find and using equations (9,3) and (9,1). Thus,


Solving the two equations gives Pa, Pa.

From equation (9,3), Pa, and equation (9,1) gives


Finally,


Sandstone: We are given the elastic constants and (plus ), so we get , , , , using equations (1,3) to (1,7) in Table 2.2a. Thus


[We could also have obtained and by using equations (9,6) and (9,7).]

Limestone: We solve in the same way as with sandstone since we are given the same constants:


Granite: Again the solution is the same as for sandstone.


Table 2.4b summarizes the results.

Table 2.4b Magnitudes of elastic constants.
Constant Water Stiffmud Shale Sandstone Limestone Granite
0 1.4 9.9 16 54 50
2.1 3.2 13 17 36 28
0 0.47 3.6 6.0 22 21
2.1 2.9 11 13 22 14
0.50 0.43 0.38 0.34 0.25 0.20
1.0 1.5 1.8 1.9 2.5 2.7
1.5 1.6 3.2 3.6 5.1 4.5
0 0.56 1.4 1.8 2.9 2.8


2.5 General solutions of the wave equation

2.5a Verify that and are solutions of the wave equation (2.5b).

Background

When unbalanced stresses act upon a medium, the strains are propagated throughout the medium according to the general wave equation


(2.5a)


being a disturbance such as a compression or rotation. is propagated with velocity (see Sheriff and Geldart, 1995, Section 2.2). The disturbance is the result of unbalanced normal stresses, shearing stresses, or a combination of both. When normal stresses create the wave, the result is a volume change and is the dilitation [see equation (2.1e)], and we get the P-wave equation, becoming the P-wave velocity . Shearing stresses create rotation in the medium and is one of the components of the rotation given by equation (2.lg) ; the result is an S-wave traveling with velocity . Various expressions for and in isotropic media are given in Table 2.2a.

In one dimension the wave equation (2.5a) reduces to


(2.5b)

Solution

We use subscripts to denote partial derivatives and primes to denote derivatives with respect to the argument of the function. Then, writing , we have


Substituting in equation (2.5b), we get the identity so is a solution. We get the same result when . A sum of solutions is also a solution, so is a solution.

2.5b Verify that is a solution of equation (2.5a), where are direction cosines.

Solution

Let . We now must show that is a solution of equation (2.5a). Proceeding as before, we have


In the same way we get


But (see Sheriff and Geldart, 1995, problem 15.9a), so .

Following the same procedure we find that thus verifying that is a solution of equation (2.5a).

2.5c Show that


is a solution of the wave equation in spherical coordinates (see problem 2.6b) when the wave motion is independent of and :


(2.5c)

Solution

The wave equation in spherical coordinates is given in problem 2.6b. When we drop the derivatives with respect to and , the equation reduces to equation (2.5c). Writing , we proceed as in part (a). Starting with the right-hand side, we ignore for the time being and obtain


Substitution in equation (2.5c) shows that is a solution. In the same way we can show that is also a solution, hence the sum is a solution.

2.6 Wave equation in cylindrical and spherical coordinates

2.6a Show that the wave equation (2.5a) can be written in cylindrical coordinates (see Figure 2.6a as


(2.6a)

Solution

Figure 2.6a.  Cylindrical coordinates.

We shall solve by direct substitution. We have , , , and , . The following solution is lengthy, so we use subscripts to denote partial derivatives and write


We shall require the derivatives:


To replace and with derivatives with respect to and , we write:


Then,


Thus


so


2.6b Transform the wave equation into spherical coordinates (see Figure 2.6b), showing that it becomes


(2.6b)

Solution

Spherical coordinates and rectangular coordinates are related as follows (see Figure 2.6b):


We continue to use subscripts to denote derivatives and letters to denote sines and cosines:


The derivatives of , , and now become:



Figure 2.6b  Spherical coordinates.

Summarizing these results, we have


We now calculate the derivatives , etc.:


Adding the three derivatives, we get


Substituting the values of , , , and , we get for the wave equation


This is often written in the more compact form


2.7 Sum of waves of different frequencies; Group velocity

2.7a A pulse composed of two frequencies, , can be represented by factors involving the sum and difference of the two frequencies. If the two components have the same amplitudes, we can write


where ,

Show that the composite wave is given approximately by the expression


where .

Background

When different frequency components in a pulse have different phase velocities (the velocity with which a given frequency travels), the pulse changes shape as it moves along. The group velocity is the velocity with which the envelope of the pulse travels.

The envelope of a pulse comprises two mirror-image curves that are tangent to the waveform at the peaks and troughs, and therefore define the general shape of the pulse.

Solution

Adding the two components and using the identity


we get for the composite wave


2.7b Why do we regard as the amplitude? Show that the envelope of the pulse is the graph of plus its reflection in the -axis.

Solution

The solution in (a) can be written . We regard as the amplitude for two reasons: (1) repeats every time that increases by or each time that increase by . But , , so must repeat more slowly than . (2) Each time that attains its limiting values, , has the value and therefore never exceeds ; thus the curves of and pass through the maxima and minima of and therefore constitute the envelope.

2.7c Show that the envelope moves with the group velocity where


(2.7a)

(see Figure 2.7a).

Solution

If we consider the quantity as a wave superimposed on the primary wavelet , comparison with the basic wave type of problem 2.5a shows that takes the place of , i.e., is the velocity with which the envelope travels. In the limit, is given by