# Proof of a generalized reciprocal method relation

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 11 415 - 468 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 11.7

Prove equation (11.7a), assuming that $\left(\xi _{i}-\xi _{j}\right)\approx 0$ for all values of $i$ and $j$ .

### Background

The generalized reciprocal method (GRM) can be used with beds of different dips provided all have the same strike. Figure 11.7a shows a series of such beds. Depths normal to the beds are denoted by $z_{Ai}$ and $z_{Bi}$ , $\alpha _{i}$ and $\beta _{i}$ are angles of incidence, those for the deepest interface being critical angles, $\xi _{i}$ is the dip of the interface at the top of the $i^{\rm {th}}$ layer. To get the traveltime from $A$ to $B$ , $t_{AB}$ , we consider a plane wavefront $PQ$ that passes through $A$ at time $t=0$ in a direction such that it will be totally refracted at one of the interfaces, the third in the case of Figure 11.7a. The wavefront reaches $C$ at time $t_{AC}$ and $R$ at time $t_{AR}$ where

{\begin{aligned}t_{AC}=\left(z_{A1}\cos \alpha _{1}/V_{1}\right),\quad t_{AR}=\sum \limits _{i=1}^{3}z_{Ai}\cos \alpha _{i}/V_{i}.\end{aligned}} The same wavefront will travel upward from $T$ to $B$ in time

{\begin{aligned}t_{BT}=\sum \limits _{i=1}^{3}z_{Bi}\cos \beta _{i}/V_{i}.\end{aligned}} Since $\alpha _{3}$ is the critical angle,

{\begin{aligned}t_{AB}=\sum \limits _{i=1}^{3}(z_{Ai}+z_{Bi})/V_{i}+RV/V_{4}.\end{aligned}} Generalizing, we get for $n$ layers

{\begin{aligned}t_{AB}=\sum \limits _{i=1}^{n-1}(z_{Ai}+z_{Bi})/V_{i}+RV/V_{n}.\end{aligned}} But $RV=YJ=EJ\cos \left(\xi _{3}-\xi _{2}\right)=AB\cos \xi _{1}\cos \left(\xi _{2}-\xi _{1}\right)\cos \left(\xi _{3}-\xi _{2}\right)$ ; for $n$ layers, we get

{\begin{aligned}t_{AB}=\sum \limits _{i=1}^{n-1}(z_{Ai}+z_{Bi})/V_{i}+AB\left(S_{n}/V_{n}\right),\end{aligned}} where

 {\begin{aligned}S_{n}=\cos \xi _{1}\cos \left(\xi _{2}-\xi _{1}\right)\ldots \cos \left(\xi _{n-1}-\xi _{n-2}\right)\approx \cos \xi _{n-1},\end{aligned}} (11.7a)

all differences in dip being small, that is, $\left(\xi _{i}-\xi _{j}\right)\approx 0$ . We shall not carry the derivation of the GRM formulas beyond this point; those who are interested should consult Sheriff and Geldart, 1995, Section 11.3.3, or Palmer (1980).

### Solution

We are asked to prove that

{\begin{aligned}\cos \xi _{1}\cos \left(\xi _{2}-\xi _{1}\right)\cos \left(\xi _{3}-\xi _{2}\right)\ldots \cos \left(\xi _{n-1}-\xi _{n-2}\right)\approx \cos \xi _{n-1},\end{aligned}} where the differences in dip are all small, that is, $\xi _{j}-\xi _{j-1}\approx 0$ . We start with the single cosine on the right-hand side of the equation and try to express it as a product of cosines. We write it as $\cos \left(\xi _{n-1}-\xi _{m}\right)$ and expand:

{\begin{aligned}\cos \left(\xi _{n-1}-\xi _{m}\right)=\cos[\left(\xi _{n-1}-\xi _{n-2}\right)+(\xi _{n-2}-\xi _{m})].\end{aligned}} Since all differences in dip are small, we expand the right-hand side and set the products of the sines equal to zero. Thus,

{\begin{aligned}\cos \left(\xi _{n-1}-\xi _{m}\right)\approx \cos \left(\xi _{n-1}-\xi _{n-2}\right)\cos \left(\xi _{n-2}-\xi _{m}\right).\end{aligned}} Next we treat the right-hand cosine in the same way, writing it as $\cos[\left(\xi _{n-2}-\xi _{n-3}\right)+(\xi _{n-3}-\xi _{m})]$ . We now expand the factor and drop the sine term. Continuing in this way we eventually arrive at the result

{\begin{aligned}\cos \left(\xi _{n-1}-\xi _{m}\right)&\approx \cos \left(\xi _{n-1}-\xi _{n-2}\right)\cos \left(\xi _{n-2}-\xi _{n-3}\right)\;\ldots \\&\qquad \times \cos \left(\xi _{2}-\xi _{1}\right)\cos \left(\xi _{1}-\xi _{m}\right).\end{aligned}} We now take $\xi _{m}=0$ and the result is equation (11.7a).