Rayleigh-wave relationships

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Problem 2.14a

Show that, when , the Rayleigh‐wave potentials, equation (2.14c), become

and

and that the displacements and at depth are


(2.14a)


(2.14b)

Background

When a medium is divided by the -plane into two semi-infinite media having different properties, surface waves are propagated parallel to the -plane, the amplitude decreasing with increasing distance from the plane. When one medium is a solid and the other a vacuum, the surface wave is known as a Rayleigh wave. The near-equivalent at the surface of the real earth, a pseudo-Rayleigh wave, is called ground roll.

We take the potential functions of equation (2.9b) in the form


(2.14c)

where the -axis is positive downward, and are real positive constants (so that the amplitudes decrease as increases) and is the Rayleigh-wave velocity. We can take either the real or the imaginary parts of the functions as a solution, the only difference being the phase. When we substitute these functions in the P- and S-wave equations [equation (2.5a) with replaced with and , respectively), we find that and must satisfy the equations


(2.14d)

Because both and must be real, .

When we apply the boundary conditions of problem 2.10b, we find that

and that the velocity is a root of the equation


(2.14e)

where from equation (1,8) in Table 2.2a.

The left-hand side of equation (2.14e) is negative when , positive for , so a root must exist in the range . When , the root is .

The angle in Figure 2.14a is given by the equation


(2.14f)

(the minus sign is necessary because we have taken positive downward).

Solution

When , , and . Using these values, equation (2.14d) gives , ; also, ; the j indicates that and are out of phase. The potential functions are now

and

From equations (2.9d) and (2.9e) we have

Taking the real part of the solution (see Sheriff and Geldart, 1995, Section 15.1.5) we obtain


(2.14g)


(2.14h)

Problem 2.14b

What are the values of , , and (i) when  ; (ii) when ; (iii) when ?

Solution

i) At the surface , and equations (2.14g,h) give

From equation (2.14f),

ii) When , equations (2.14f,g,h) give for , , and

iii) For , we get

Problem 2.14c

Is the motion retrograde for all values of ?

Solution

When and are fixed, the argument of decreases as increases, and so the cotangent increases, that is, the angle in Figure 2.14a also increases. Because the wave is progressing in the positive direction of the -axis, this counter-clockwise rotation is said to be retrograde. For the motion not to be retrograde, must change sign, and thus must also change sign, that is, either or but not both must change signs. For , is negative while is positive. For to change sign, it must pass through zero; in this case


or For , is positive. For to change sign,

or . Since is always positive, can never be zero. Consequently, the motion is retrograde in the interval and prograde when .

Figure 2.14a  Retrograde motion.

Problem 2.14d

What are the values of , the Rayleigh-wave velocity, when and when ? What are the corresponding values of the constants in the expressions for and in equation (2.14g,h)?

Solution

From Figure 2.14b we find that when and when . To get more accurate values, we solve equation (2.14e) using one of the standard methods of solving cubic equations.

Figure 2.14b  as a function of .

i) For . Equation (2.14e) now becomes

where , , . Next we eliminate the -term by substituting . This gives , where To check on the nature of the roots, we calculate ; the value is , that is, it is positive, so we have one real root and two complex ones.

We now calculate the quantities

The three roots of the equation are . The last two roots are complex, so we are left with only the first root, that is, . Substituting the values of and , we get , . Thus, , , so .

To get the values of the constants in part (a) for , we have

ii) For , . This gives

Let , so , where , .

The discriminant , so there are three real unequal roots; in this case a trigonometric solution is convenient. We find the value of

Next we calculate . The roots are now

But , so the only valid root is

Hence, for

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