# Zoeppritz’s equations for incident SV- and SH-waves

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 3.4a

Derive Zoeppritz’s equations for an SV-wave incident on a solid/solid interface.

### Solution

Figure 3.1a defines the positive directions of displacements except that the incident P-wave is replaced by an incident SV-wave whose positive direction is down and to the left (the same as that of $B_{2})$ ). Using the same symbols as in equations (3.1b,c), we define the following functions:

{\begin{aligned}\chi _{0}=B_{0}e^{\mathrm {j} \omega \zeta _{0}^{\prime }},\qquad \chi _{1}&=B_{1}e^{\mathrm {j} \omega \zeta _{1}^{\prime }},\qquad \chi _{2}=B_{2}e^{\mathrm {j} \omega \zeta _{2}^{\prime }};\\\phi _{1}&=A_{1}e^{\mathrm {j} \omega \zeta _{1}},\qquad \phi _{2}=A_{2}e^{\mathrm {j} \omega \zeta _{2}}.\end{aligned}} where $\zeta _{i}$ and $\zeta _{i}^{'}$ are the same as in equation (3.1d,e). We get the following expressions for the displacements $u_{i}$ and $w_{i}$ using equations (3.2a,b,c,d), where the terms in $A_{0}$ are replaced with terms in $B_{0}$ :

{\begin{aligned}w_{1}&=-\chi _{0}\sin \delta _{1}-\chi _{1}\sin \delta _{1}+\phi _{1}\cos \theta _{1},\\w_{2}&=\qquad \qquad -\chi _{2}\sin \delta _{2}-\phi _{2}\cos \theta _{2};\\u_{1}&=-\chi _{0}\cos \delta _{1}+\chi _{1}\cos \delta _{1}+\phi _{1}\sin \theta _{1},\\u_{2}&=\qquad \qquad -\chi _{2}\cos \delta _{2}+\phi _{2}\sin \theta _{2}.\end{aligned}} The boundary conditions require the continuity of $w$ , $u$ , $\sigma _{zz}$ and $\sigma _{zz}$ at $z=0$ . Continuity of $w$ and $u$ gives

{\begin{aligned}-B_{0}\sin \delta _{1}-B_{1}\sin \delta _{1}+A_{1}\cos \theta _{1}&=-B_{2}\sin \delta _{2}-A_{2}\cos \theta _{2},\\-B_{0}\cos \delta _{1}+B_{1}\cos \delta _{1}+A_{1}\sin \theta _{1}&=-B_{2}\cos \delta _{2}+A_{2}\sin \theta _{2}.\end{aligned}} For the normal stress, we have

{\begin{aligned}\sigma _{zz}&=\lambda \Delta +2\mu \varepsilon _{zz}=\lambda \left({\frac {\partial u}{\partial x}}+{\frac {\partial w}{\partial z}}\right)+2\mu \left({\frac {\partial w}{\partial z}}\right)=\lambda \left(u_{x}+w_{z}\right)+2\mu w_{z}\\&=\left(\lambda +2\mu \right)\left(u_{x}+w_{z}\right)-2\mu u_{x}=\rho \alpha ^{2}\left(u_{x}+w_{z}\right)-2\rho \beta ^{2}u_{x}.\end{aligned}} Thus, continuity of $\sigma _{zz}$ requires that

{\begin{aligned}&\rho _{1}\alpha _{1}^{2}[(-B_{0}\cos \delta _{1}+B_{1}\cos \delta _{1}+A_{1}\sin \theta _{1})+(B_{0}\cot \delta _{1}\sin \delta _{1}\\&\qquad \qquad \quad -B_{1}\cot \delta _{1}\sin \delta _{1}+A_{1}\cot \theta _{1}\cos \theta _{1})]\\&\qquad \quad -2\rho _{1}\beta _{1}^{2}\left(-B_{0}\cos \delta _{1}+B_{1}\cos \delta _{1}+A_{1}\sin \theta _{1}\right)\\&\quad =\rho _{2}\alpha _{2}^{2}\left[\left(-B_{2}\cos \delta _{2}+A_{2}\sin \theta _{2}\right)+\left(B_{2}\cot \delta _{2}\sin \delta _{2}+A_{2}\cot \theta _{2}\cos \theta _{2}\right)\right]\\&\qquad \quad -2\rho _{2}\beta _{2}^{2}\left(-B_{2}\cos \delta _{2}+A_{2}\sin \theta _{2}\right).\end{aligned}} Continuity of the tangential stress, $\sigma _{xz}=\mu \varepsilon _{xz}=\mu \left(u_{z}+w_{x}\right)$ , gives

{\begin{aligned}&\mu _{1}[\left(B_{0}\cot \delta _{1}\cos \delta _{1}+B_{1}\cot \delta _{1}\cos \delta _{1}+A_{1}\cot \theta _{1}\sin \theta _{1}\right)\\&\qquad \qquad \qquad +\left(-B_{0}\sin \delta _{1}-B_{1}\sin \delta _{1}+A_{1}\cos \theta _{1}\right)]\\&\quad =\mu _{2}[\left(B_{2}\cot \delta _{2}\cos \delta _{2}-A_{2}\cot \theta _{2}\sin \theta _{2}\right)+(-B_{2}\sin \delta _{2}-A_{2}\cos \theta _{2})].\end{aligned}} We can simplify the equations for $\sigma _{zz}$ and $\sigma _{xz}$ by noting that

{\begin{aligned}&\rho \alpha ^{2}(-\cos \delta +\cot \delta \sin \delta )+2\rho \beta ^{2}\cos \delta =2\rho \beta ^{2}\cos \delta =\left(W/p\right)\sin 2\delta ,\\&\rho \alpha ^{2}\left(\sin \theta +\cot \theta \cos \theta \right)-2\rho \beta ^{2}\sin \theta =\rho \left[\left(\alpha ^{2}/\sin \theta \right)-2\beta ^{2}\sin \theta \right]\\&\quad =\rho \sin \theta \left(1/p^{2}-2\sin ^{2}\delta /p^{2}\right)=\left(\rho \alpha /p\right)\cos 2\delta =\left(Z/p\right)\cos 2\delta ,\end{aligned}} where $p$ is the raypath parameter [see equation (3.1a)]. Also,

{\begin{aligned}\mu \left(\cot \delta \cos \delta -\sin \delta \right)=\rho \beta ^{2}\left({\frac {\cos ^{2}\delta -\sin ^{2}\delta }{\sin \delta }}\right)=\left(W/p\right)\cos 2\delta .\end{aligned}} We can now write the four equations in the form

{\begin{aligned}A_{1}\cos \theta _{1}-B_{1}\sin \delta _{1}+A_{2}\cos \theta _{2}+B_{2}\sin \delta _{2}&=B_{0}\sin \delta _{1},\\A_{1}\sin \theta _{1}+B_{1}\cos \delta _{1}-A_{2}\sin \theta _{2}+B_{2}\cos \delta _{2}&=B_{0}\cos _{l},\\A_{1}Z_{1}\cos 2\delta _{1}-B_{1}W_{1}\sin 2\delta _{1}-A_{2}Z_{2}\cos 2\delta _{2}-B_{2}W_{2}\sin 2\delta _{2}&=-B_{0}W_{1}\sin 2\delta _{1},\\\left(\beta _{1}/\alpha _{1}\right)A_{1}W_{1}\sin 2\theta _{1}+B_{1}W_{1}\cos 2\delta _{1}+\left(\beta _{2}/\alpha _{2}\right)A_{2}W_{2}\sin 2\theta _{2}&-B_{2}W_{2}\cos 2\delta _{2}\\&=-B_{0}W_{1}\cos 2\delta _{1}.\end{aligned}} ## Problem 3.4b

Derive the Zoeppritz equations for an incident SH-wave.

### Solution

For an SH-wave traveling in the $xz$ -plane, the wave motion involves only displacement $v$ parallel to the $y$ -axis where $v=v\left(x,\;z,\;t\right)$ . We take the incident, reflected, and refracted waves in the form [see equations (3.1b,c,d,e)]

{\begin{aligned}v_{1}&=C_{0}e^{\mathrm {j} \omega p\left(x-z\cot \delta _{1}\right)}+C_{1}e^{\mathrm {j} \omega p\left(x+z\cot \delta _{1}\right)},\\v_{2}&=C_{2}e^{\mathrm {j} \omega p\left(x-z\cot \delta _{2}\right)}.\end{aligned}} The boundary conditions require that the tangential displacement and tangential stress be continuous at $z=0$ . The first condition gives

 {\begin{aligned}C_{0}+C_{1}=C_{2},\quad \mathrm {or} \quad C_{1}-C_{2}=-C_{0}.\end{aligned}} (3.4a)

The tangential stress is $\sigma _{yz}$ (note that $\sigma _{xz}=0$ ), where

{\begin{aligned}\sigma _{yz}=\mu \varepsilon _{yz}=\mu \left({\frac {\partial v}{\partial z}}+{\frac {\partial w}{\partial y}}\right)=\mu {\frac {\partial v}{\partial z}}.\end{aligned}} Recalling that we can take $\partial /\partial x=+1$ , $\partial /\partial z=\pm \cot \delta _{i}$ [see equation (3.2g)], we get

{\begin{aligned}\mu _{1}\left(-C_{0}\cot \delta _{1}+C_{1}\cot \delta _{1}\right)=-\mu _{2}C_{2}\cot \delta _{2},\end{aligned}} So

 {\begin{aligned}\mu _{1}C_{1}\cot \delta _{1}+\mu _{2}C_{2}\cot \delta _{2}=\mu _{1}C_{0}\cot \delta _{1}.\end{aligned}} (3.4b)

Solving equations (3.4a,b), we find

{\begin{aligned}{\frac {C_{1}}{C_{0}}}&={\frac {\rho _{1}\beta _{1}^{2}\cot \delta _{1}-\rho _{2}\beta _{2}^{2}\cot \delta _{2}}{\rho _{1}\beta _{1}^{2}\cot \delta _{1}+\rho _{2}\beta _{2}^{2}\cot \delta _{2}}}={\frac {\left(W_{1}\cos \delta _{1}-W_{2}\cos \delta _{2}\right)}{\left(W_{1}\cos \delta _{1}+W_{2}\cos \delta _{2}\right)}},\\{\frac {C_{2}}{C_{0}}}&={\frac {2\mu _{1}\cot \delta _{i}}{\left(\mu _{1}\cot \delta _{i}+\mu _{2}\cot \delta _{i}\right)}}={\frac {2W_{1}\cos \delta _{1}}{\left(W_{1}\cos \delta _{1}+W_{2}\cos \delta _{2}\right)}}.\end{aligned}} The absence of P-waves is important in SH-wave studies.