Reflection/refraction at a liquid/solid interface

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Problem 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.


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



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


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


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


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


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





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,


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


where the second equation is obvious from symmetry.


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:

Problem 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).


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 , , .

Problem 3.3c

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


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.
Soft 0.431 0.539 0.244
0.403 0.512 0.408
Hard 0.716 0.227 0.304

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