Diving waves

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Problem 4.20a

When the velocity is a linear function of depth only, as in equation (4.17a), show that the wave will return to the surface again at offset and with traveltime given by equations (4.20a,b), and that the maximum depth of penetration is given by equation (4.20c).


(4.20a)


(4.20b)


(4.20c)

Background

When the velocity increases continuously with depth, a wave will eventually return to the surface (see Figures 4.20a,c): such waves are known as diving waves. When the velocity layers are horizontal, as in Figure 4.20a, or spherically symmetrical as in Figure 4.20c, the raypath is symmetrical about the midpoint and the maximum depth occurs at this point.

Figure 4.20a.  Raypaths where velocity increases linearly with depth.

Solution

Because the raypath is symmetrical about the midpoint, we can find and for the midpoint and double the values to get and for the entire path. At the midpoint the ray is traveling horizontally, hence . Substituting in equations (4.17b,c), we get


(4.20d)

Using the identity , equation (4.20d) becomes

The maximum depth is the value of when , so from equation (4.17f) we find that

Using the identity: , we can write

Equation (4.20d) now gives

Figure 4.20b.  Raypath parameter for concentric spherical layering.

Problem 4.20b

Show that when the constant velocity layers are concentric spherical shells, the raypath parameter (see problem 3.1a) becomes


(4.20e)

Solution

Equation (4.20e) is a modification of equation (3.1a) to take into account concentric spherical shells instead of plane parallel interfaces. The angle between a ray and the radii changes as the wave travels downward (see Figure 4.20b) so that the angle of entry into a layer does not equal the incident angle at the base of the layer, that is, . But , and Snell’s law becomes , or

Problem 4.20c

For concentric spherical layering in the Earth, show that diving waves will return to the surface at the time at the angular distance , where is the angle subtended at the center of the Earth by the ray , and is the radius of the Earth; (see Sheriff and Geldart, 1995, 99);

Figure 4.20c.  Spherical layering raypaths.


(4.20f)


(4.20g)

Solution

Using equation (4.20e) and in Figure 4.20c, we get


(4.20h)

also gives


(4.20i)

Eliminating between equations (4.20h,i) gives

Integrating from to and multiplying by 2 to allow for the return path gives equation (4.20f).

Eliminating between equations (4.20h,i), we have

Again we multiply by 2 and integrate, obtaining equation (4.20g).

Alternative solution

Equation (4.20f) is a relation between and , so we use in Figure 4.20c to get a relation between and , then integrate to get . From and equation (4.20e), we have

since . Integrating this expression for gives equation (4.20f).

Equation (4.20g) expresses in terms of , so we use to get the relation . Using equation (4.20e), this becomes

Multiplying by 2 and integrating gives equation (4.20g).

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