# Difference between revisions of "Reflection/transmission coefficients at small angles and magnitude"

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 3 47 - 77 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 3.9

Show that, when angles in the Zoeppritz equations (3.2e,f,h,i) are small (so that squares and products are negligible), equations (3.6a) and (3.6b) for reflection and transmission at normal incidence are still valid, and that the reflected and transmitted S-waves are given by

{\begin{aligned}\mathbf {{\frac {B_{1}}{A_{0}}}={\frac {2W_{2}q+4Z_{1}r}{\left(W_{1}+W_{2}\right)\left(Z_{1}+Z_{2}\right)}},\;{\frac {B_{2}}{A_{0}}}={\frac {2W_{1}q-4Z_{1}r}{\left(W_{1}+W_{2}\right)\left(Z_{1}+Z_{2}\right)}},} \end{aligned}} where $\mathbf {q=Z_{1}\theta _{2}-Z_{2}\theta _{1}}$ , $\mathbf {r=W_{1}\delta _{1}}$ - $\mathbf {W_{2}\delta _{1}}$ ### Solution

When the angle of incidence $\theta$ is small, $\sin \theta \approx \theta$ and $\cos \theta \approx 1$ and the same is true for $\delta$ . In this case Snell’s law and the Zoeppritz equations (3.2e,f,h,i) become

{\begin{aligned}\theta _{1}/\alpha _{1}\approx \delta _{1}/\beta _{1}\approx \theta _{2}/\alpha _{2}\approx \delta _{2}/\beta _{2},\;\beta _{i}/\alpha _{i}=\delta _{i}/\theta _{i};\\A_{1}-\;\delta _{1}B_{1}+\;A_{2}+\;\delta _{2}B_{2}=\;A_{0},\\\theta _{1}A_{1}+\;B_{1}-\;\theta _{2}A_{2}+\;B_{2}=\;-\theta _{1}A_{0},\\Z_{1}A_{1}-2\delta _{1}W_{1}B_{1}-\;Z_{2}A_{2}-2\delta _{2}W_{2}B_{2}=\;-Z_{1}A_{0},\\2\delta _{1}W_{1}A_{1}+\;W_{1}B_{1}+2\delta _{2}W_{2}A_{2}-\;W_{2}B_{2}=2\delta _{1}W_{1}A_{0}.\end{aligned}} In matrix notation, the Zoeppritz equations are now

 {\begin{aligned}\left\|{\begin{array}{llll}1&-\delta _{1}&1&\delta _{2}\\\theta _{1}&1&-\theta _{2}&1\\Z_{1}&-2\delta _{1}W_{1}&-Z_{2}&-2\delta _{2}W_{2}\\2\delta _{1}W_{1}&W_{1}&2\delta _{2}W_{2}&-W_{2}\end{array}}\right\|\ \left\|{\begin{array}{l}A_{1}\\B_{1}\\A_{2}\\B_{2}\end{array}}\right\|=A_{0}\left\|{\begin{array}{l}1\\-\theta _{1}\\-Z_{1}\\2\delta _{1}W_{1}\end{array}}\right\|\end{aligned}} (3.9a)

To get the amplitude ratios $A_{i}/A_{0}$ and $B_{i}/A_{0}$ , we solve this equation either by inverting the left-hand matrix [see Sheriff and Geldart, 1995, equation (15.20)] or by using Cramer’s rule (see Wylie, 1966, 453). Using the latter method, and neglecting squares and products of the angles, we first calculate the value of det($A$ ), the determinant of the $4\times 4$ matrix in equation (3.9a). We shall expand by elements in the first row [see Sheriff and Geldart, 1995, equation (15.2)]; when we do this we see that the 2nd and 4th $\left(3\times 3\right)\$ determinants in the expansion are multiplied by $+\delta _{1}$ and $-\delta _{2}$ , respectively, and since we are neglecting products and squares of angles, angles inside these two determinants have been replaced with zeros. The expansion about the first row becomes

{\begin{aligned}\mathrm {det} \left(A\right)=\left|{\begin{array}{ccc}{1}&{-\theta _{2}}&{1}\\{-2\delta _{1}W_{1}}&{-Z_{2}}&{-2\delta _{2}W_{2}}\\{W_{1}}&{2\delta _{2}W_{2}}&{-W_{2}}\end{array}}\right|+\delta _{1}\left|{\begin{array}{ccc}{0}&{0}&{1}\\{Z_{1}}&{-Z_{2}}&{0}\\{0}&{0}&{-W_{2}}\end{array}}\right|\\+\left|{\begin{array}{ccc}{\theta _{1}}&{1}&{1}\\{Z_{1}}&{-2\delta _{1}W_{1}}&{-2\delta _{2}W_{2}}\\{2\delta _{1}W_{1}}&{W_{1}}&{-W_{2}}\end{array}}\right|-\delta _{2}\left|{\begin{array}{ccc}{0}&{1}&{0}\\{Z_{1}}&{0}&{-Z_{2}}\\{0}&{W_{1}}&{0}\end{array}}\right|\\=\left(Z_{2}W_{2}+Z_{2}W_{1}\right)+\left(Z_{1}W_{1}+Z_{1}W_{2}\right)=\left(Z_{1}+Z_{2}\right)\left(W_{1}+W_{2}\right).\end{aligned}} Next we calculate the values of $\mathrm {det} \left(A_{i}\right)$ and $\mathrm {det} \left(B_{i}\right)\$ , $i=1$ , 2, where $\mathrm {det} \left(A_{1}\right)$ is $\mathrm {det} \left(A\right)$ with column 1 replaced with the elements of the right-hand matrix in equation (3.9a), etc. (see Cramer’s rule in Sheriff and Geldart, 1995, problem 15.2j). Expanding $\mathrm {det} \left(A_{1}\right)$ about the first row and setting the angles in the 2nd and 4th $3\times 3$ determinant equal to zero as before, the expansion becomes

{\begin{aligned}\mathrm {det} \left(A\right)=\left|{\begin{array}{ccc}{1}&{-\theta _{2}}&{1}\\{-2\delta _{1}W_{1}}&{-Z_{2}}&{-2\delta _{2}W_{2}}\\{W_{1}}&{2\delta _{2}W_{2}}&{-W_{2}}\end{array}}\right|+\delta _{1}\left|{\begin{array}{ccc}{0}&{0}&{1}\\{Z_{1}}&{-Z_{2}}&{0}\\{0}&{0}&{-W_{2}}\end{array}}\right|\\+\left|{\begin{array}{ccc}{\theta _{1}}&{1}&{1}\\{Z_{1}}&{-2\delta _{1}W_{1}}&{-2\delta _{2}W_{2}}\\{2\delta _{1}W_{1}}&{W_{1}}&{-W_{2}}\end{array}}\right|-\delta _{2}\left|{\begin{array}{ccc}{0}&{1}&{0}\\{Z_{1}}&{0}&{-Z_{2}}\\{0}&{W_{1}}&{0}\end{array}}\right|\\=\left(Z_{2}W_{2}+Z_{2}W_{1}\right)+\left(Z_{1}W_{1}+Z_{1}W_{2}\right)=\left(Z_{1}+Z_{2}\right)\left(W_{1}+W_{2}\right).\end{aligned}} The second and fourth determinants are zero, so

{\begin{aligned}\mathrm {det} \left(A_{1}\right)=\left(Z_{2}W_{2}+Z_{2}W_{1}\right)-\left(Z_{1}W_{2}+Z_{1}W_{1}\right)=\left(W_{1}+W_{2}\right)\left(Z_{2}-Z_{1}\right).\end{aligned}} Dividing by $\mathrm {det} \left(A\right)$ , we get

{\begin{aligned}R=A_{1}/A_{0}=\left(Z_{2}-Z_{1}\right)/\left(Z_{2}+Z_{1}\right),\end{aligned}} which is the same as equation (3.6a). Similarly, we find that

{\begin{aligned}T=A_{2}/A_{0}=2Z_{1}/\left(Z_{2}+Z_{1}\right),\\B_{1}/A_{0}=\left(2W_{2}q+4Z_{1}r\right)/\left(Z_{1}+Z_{2}\right)\left(W_{1}+W_{2}\right),\\B_{2}/A_{0}=\left(2W_{1}q-4Z_{1}r\right)/\left(Z_{1}+Z_{2}\right)\left(W_{1}+W_{2}\right),\end{aligned}} where $q=\left(Z_{1}\theta _{2}-Z_{2}\theta _{1}\right)$ , $r=\left(W_{1}\delta _{1}-W_{2}\delta _{2}\right)$ . Note that

{\begin{aligned}A_{1}+A_{2}=A_{0},\end{aligned}} {\begin{aligned}B_{1}+B_{2}=2qA_{0}/\left(Z_{1}+Z_{2}\right)={\frac {2A_{0}\left(Z_{1}\theta _{2}-Z_{2}\theta _{1}\right)}{\left(Z_{1}+Z_{2}\right)}}\\=0\qquad {\text{for normal incidence}}.\end{aligned}} Also, when $\rho _{1}=\rho _{2},$ {\begin{aligned}\theta _{2}=\left(\alpha _{2}/\alpha _{1}\right)\theta _{1}=\left(Z_{2}/Z_{1}\right)\theta _{1},\end{aligned}} so q $=0$ and $B_{1}=-B_{2}=4Z_{1}r/[\left(W_{1}+W_{2}\right)(Z_{1}+Z_{2})].$ 