Proof of a generalized reciprocal method relation

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
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 ${\displaystyle \left(\xi _{i}-\xi _{j}\right)\approx 0}$ for all values of ${\displaystyle i}$ and ${\displaystyle j}$.

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 ${\displaystyle z_{Ai}}$ and ${\displaystyle z_{Bi}}$, ${\displaystyle \alpha _{i}}$ and ${\displaystyle \beta _{i}}$ are angles of incidence, those for the deepest interface being critical angles, ${\displaystyle \xi _{i}}$ is the dip of the interface at the top of the ${\displaystyle i^{\rm {th}}}$ layer. To get the traveltime from ${\displaystyle A}$ to ${\displaystyle B}$, ${\displaystyle t_{AB}}$, we consider a plane wavefront ${\displaystyle PQ}$ that passes through ${\displaystyle A}$ at time ${\displaystyle t=0}$ 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 ${\displaystyle C}$ at time ${\displaystyle t_{AC}}$ and ${\displaystyle R}$ at time ${\displaystyle t_{AR}}$ where

{\displaystyle {\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 ${\displaystyle T}$ to ${\displaystyle B}$ in time

{\displaystyle {\begin{aligned}t_{BT}=\sum \limits _{i=1}^{3}z_{Bi}\cos \beta _{i}/V_{i}.\end{aligned}}}

Since ${\displaystyle \alpha _{3}}$ is the critical angle,

{\displaystyle {\begin{aligned}t_{AB}=\sum \limits _{i=1}^{3}(z_{Ai}+z_{Bi})/V_{i}+RV/V_{4}.\end{aligned}}}

Generalizing, we get for ${\displaystyle n}$ layers

{\displaystyle {\begin{aligned}t_{AB}=\sum \limits _{i=1}^{n-1}(z_{Ai}+z_{Bi})/V_{i}+RV/V_{n}.\end{aligned}}}

But ${\displaystyle RV=YJ=EJ\cos \left(\xi _{3}-\xi _{2}\right)=AB\cos \xi _{1}\cos \left(\xi _{2}-\xi _{1}\right)\cos \left(\xi _{3}-\xi _{2}\right)}$; for ${\displaystyle n}$ layers, we get

{\displaystyle {\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

 {\displaystyle {\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, ${\displaystyle \left(\xi _{i}-\xi _{j}\right)\approx 0}$. 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

{\displaystyle {\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, ${\displaystyle \xi _{j}-\xi _{j-1}\approx 0}$. 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 ${\displaystyle \cos \left(\xi _{n-1}-\xi _{m}\right)}$ and expand:

{\displaystyle {\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,

{\displaystyle {\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 ${\displaystyle \cos[\left(\xi _{n-2}-\xi _{n-3}\right)+(\xi _{n-3}-\xi _{m})]}$. We now expand the factor and drop the sine term. Continuing in this way we eventually arrive at the result

{\displaystyle {\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 ${\displaystyle \xi _{m}=0}$ and the result is equation (11.7a).

Figure 11.8a.  Illustrating delay time.