# Diving waves

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 4 79 - 140 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## 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 $x$ and with traveltime $t$ given by equations (4.20a,b), and that the maximum depth of penetration $h_{m}$ is given by equation (4.20c).

 {\begin{aligned}x&=(2V_{0}/a)\sin h(at/2),\end{aligned}} (4.20a)

 {\begin{aligned}t&=(2/a)\ln[\cot(i_{0}/2)],\end{aligned}} (4.20b)

 {\begin{aligned}h_{m}&=(V_{0}/a)[\cos \mathrm {h} (at/2)-1].\end{aligned}} (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.

### Solution

Because the raypath is symmetrical about the midpoint, we can find $x$ and $t$ for the midpoint and double the values to get $x$ and $t$ for the entire path. At the midpoint the ray is traveling horizontally, hence $i=90^{\circ }$ . Substituting in equations (4.17b,c), we get

 {\begin{aligned}x&=(2/pa)\cos i_{0}=(2V_{0}/a)\cot i_{o},\\t&=(2/a)\ln(\cot i_{0}/2),\quad \cot i_{0}/2=e^{at/2}.\end{aligned}} (4.20d)

Using the identity $\cot \,2\phi ={\frac {1}{2}}(\cos \phi -\tan \phi )$ , equation (4.20d) becomes

{\begin{aligned}x=(2V_{0}/a)(e^{at/2}-e^{-at/2})/2=(2V_{0}/a)\sinh at/2.\end{aligned}} The maximum depth $h_{m}$ is the value of $z$ when $i=90^{\circ }$ , so from equation (4.17f) we find that

{\begin{aligned}h_{m}&=(1/pa)(\sin i-\sin i_{0})=(1/pa)(1-\sin i_{0})\\&=(V/a)(1/\sin i_{0}-1).\end{aligned}} Using the identity: $\sin x=2\tan(x/2)/[1+\tan ^{2}(x/2)]$ , we can write

{\begin{aligned}1/\sin i_{o}=\cot(i_{o}/2)[1+\tan ^{2}(i_{o}/2)]/2\\={\frac {1}{2}}[\cot(i_{o}/2)+\tan(i_{o}/2)].\end{aligned}} Equation (4.20d) now gives

{\begin{aligned}h_{m}&=(V/a)[(e^{at/2}+e^{-at/2})/2-1]\\&=(V/a)(\cosh at/2-1).\end{aligned}}  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 $p$ (see problem 3.1a) becomes

 {\begin{aligned}p^{\prime }=r_{n}\sin i_{n}/V_{n}.\end{aligned}} (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, $i_{2}\neq i_{2}^{*}$ . But $OP=r_{3}\sin i_{2}^{*}=r_{2}\sin i_{2}$ , and Snell’s law becomes $\sin i_{2}^{*}/V_{2}=\sin i_{3}/V_{3}$ , or

{\begin{aligned}{\frac {r_{2}\sin i_{2}}{V_{2}}}={\frac {r_{3}\sin i_{3}}{V_{3}}}=p^{\prime }.\end{aligned}} ## Problem 4.20c

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

 {\begin{aligned}t=2\int _{R_{e}-h_{m}}^{R_{e}}{\frac {\mathrm {d} r}{V(r)\{1-[p^{\prime }V(r)/r]^{2}\}^{1/2}}},\end{aligned}} (4.20f)

 {\begin{aligned}\Delta =2\int _{R_{e}-h_{m}}^{R_{e}}{\frac {[p^{\prime }V(r)/r]\mathrm {d} r}{r\{1-[p^{\prime }V(r)/r]^{2}\}^{1/2}}}.\end{aligned}} (4.20g)

### Solution

Using equation (4.20e) and $\Delta ABC$ in Figure 4.20c, we get

 {\begin{aligned}p^{\prime }={\frac {r\sin i}{V}}=\left({\frac {r}{V}}\right)\left({\frac {r\delta \Delta }{V\delta t}}\right)=\left({\frac {r}{v}}\right)^{2}\left({\frac {\delta \Delta }{\delta t}}\right).\end{aligned}} (4.20h)

$\Delta ABC$ also gives

 {\begin{aligned}(V\delta t)^{2}=(\delta r)^{2}+(r\delta \Delta )^{2}.\end{aligned}} (4.20i)

Eliminating $\delta \Delta$ between equations (4.20h,i) gives

{\begin{aligned}&(r\delta \Delta )^{2}=(V\delta t)^{2}-(\delta r)^{2}=(p^{\prime }V^{2}\delta t/r)^{2},\\&(\delta t)^{2}[V^{2}-(p^{\prime }V^{2}/r)^{2}]=(\delta r)^{2},\\\mathrm {so} \qquad &\delta t=(\delta r/V)[1-(p^{\prime }V/r)^{2}]^{-1/2}.\end{aligned}} Integrating from $(R_{e}-h_{m})$ to $R_{e}$ and multiplying by 2 to allow for the return path gives equation (4.20f).

Eliminating $\delta t$ between equations (4.20h,i), we have

{\begin{aligned}&(\delta t)^{2}=(1/V^{2})[(\delta r)^{2}+(r\delta \Delta )^{2}]=(r/V)^{4}(\delta \Delta /p^{\prime })^{2},\\&(\delta r/r)^{2}+(\delta \Delta )^{2}=(r/p^{\prime }V)^{2}(\delta \Delta )^{2}.\end{aligned}} Again we multiply by 2 and integrate, obtaining equation (4.20g).

### Alternative solution

Equation (4.20f) is a relation between $t$ and $r$ , so we use $\Delta ABC$ in Figure 4.20c to get a relation between $\delta t$ and $\delta r$ , then integrate to get $t$ . From $\Delta ABC$ and equation (4.20e), we have

{\begin{aligned}\cos i=(\delta r/V\delta t),\quad \delta t=(\delta r/V)[1-(p^{\prime }/r)^{2}]^{-1/2},\end{aligned}} since $\sin i=p^{\prime }V/r$ . Integrating this expression for $\delta t$ gives equation (4.20f).

Equation (4.20g) expresses $\Delta$ in terms of $r$ , so we use $\Delta ABC$ to get the relation $\tan i=r\delta \Delta /\delta r$ . Using equation (4.20e), this becomes

{\begin{aligned}\delta \Delta =(\delta r/r)(p^{\prime }V/r)[1-(p^{\prime }/r)^{2}]^{1/2}.\end{aligned}} Multiplying by 2 and integrating gives equation (4.20g).