Reflection/refraction at a solid/solid interface and displacement of a free surface

Problems in Exploration Seismology and their Solutions

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

Problem 3.2a

Derive the Zoeppritz equations for a P-wave incident on a solid/solid interface.

Background

The normal and tangential displacements plus the normal and tangential stresses must be continuous when a P-wave is incident at the angle $\theta _{1}$ on an interface between two solid media (see problem 2.10).

Solution

We use the functions in equations (3.1b,c,d,e) to represent the displacements of the waves, the positive direction of displacement for the waves being shown in Figure 3.1a. (We omit the factor $e^{-\mathrm {j} \omega t}$ because the boundary conditions do not depend upon the time $t$, hence this factor cancels out).

We first derive the equations expressing the continuity of normal and tangential displacements, $w$ and $u$. These equations are obtained by resolving the various wave displacements into $z$- and $x$-components. Thus,

${\begin{aligned}w_{1}&=-A_{0}\cos \theta _{1}e^{\mathrm {j} \omega \zeta _{0}}+A_{1}\cos \theta _{1}e^{\mathrm {j} \omega \zeta _{1}}-B_{1}\sin \delta _{1}e^{\mathrm {j} \omega \zeta _{1}^{\prime }},\end{aligned}}$

(3.2a)

${\begin{aligned}w_{2}&=\qquad \qquad -A_{2}\cos \theta _{2}e^{\mathrm {j} \omega \zeta _{2}}-B_{2}\sin \delta _{2}e^{\mathrm {j} \omega \zeta _{2}^{\prime }},\end{aligned}}$

(3.2b)

${\begin{aligned}u_{1}&=A_{0}\sin \theta _{1}e^{\mathrm {j} \omega \zeta _{0}}+A_{1}\sin \theta _{1}e^{\mathrm {j} \omega \zeta _{1}}+B_{1}\cos \delta _{1}e^{\mathrm {j} \omega \zeta _{1}^{\prime }},\end{aligned}}$

(3.2c)

${\begin{aligned}u_{2}&=\qquad \qquad A_{2}\sin \theta _{2}e^{\mathrm {j} \omega \zeta _{2}}-B_{2}\cos \delta _{2}e^{\mathrm {j} \omega \zeta _{2}^{\prime }}.\end{aligned}}$

(3.2d)

At the interface, $z=0$ and $w_{1}=w_{2}$, $u_{1}=u_{2}$. The exponentials all reduce to $e^{\mathrm {j} \omega x}$, hence cancel out, and we get for the normal and tangential displacements, respectively,

${\begin{aligned}\left(-A_{0}+A_{1}\right)\cos \theta _{1}-B_{1}\sin \delta _{1}=-A_{2}\cos \theta _{2}-B_{2}\sin \delta _{2},\end{aligned}}$

(3.2e)

${\begin{aligned}\left(A_{0}+A_{1}\right)\sin \theta _{1}+B_{1}\cos \delta _{1}=A_{2}\sin \theta _{2}-B_{2}\cos \delta _{2}.\end{aligned}}$

(3.2f)

To apply the boundary conditions for the normal and tangential stresses, we differentiate equations (3.2a,b,c,d) with respect to $x$ and $z$. Equations (3.1d,e) show that the differentiation with respect to $x$ and $z$ multiplies each function by $\mathrm {j} \omega p$ and either $\pm \mathrm {j} \omega p\cot \theta _{i}$ or $\pm \mathrm {j} \omega p\cot \delta _{i}$. The common factor $\mathrm {j} \omega p$ will cancel in the end, so we simplify the derivation by taking

${\begin{aligned}\partial /\partial x=1,\quad \partial /\partial z=\pm \cot \theta _{i}\quad \mathrm {or} \quad \pm \cot \delta _{i}.\end{aligned}}$

(3.2g)

From equations (2.1b,c,e,h,i) we get for the normal and tangential stresses:

${\begin{aligned}\sigma _{zz}&=\lambda \Delta +2\mu \varepsilon _{zz}=\lambda \left(u_{x}+w_{z}\right)+2\mu w_{z}=\lambda u_{x}+\left(\lambda +2\mu \right)w_{z},\\\sigma _{xz}&=\mu \left(u_{z}+w_{x}\right),\end{aligned}}$

where $u_{x}$, $u_{z}$, and $w_{x}$, $w_{z}$ are partial derivatives with respect to $x$ and $z$. This allows us to find the normal and tangential stresses in each medium and equate them at $z=0$. The result for the normal stresses is

${\begin{aligned}\lambda _{1}\left[\left(A_{0}+A_{1}\right)\sin \theta _{1}+B_{1}\cos \delta _{1}\right]+\left(\lambda _{1}+2\mu _{1}\right)\left(A_{0}+A_{1}\right)\cos \theta _{1}\cot \theta _{1}-B_{1}\cos \delta _{1}\\=\lambda _{2}\left(A_{2}\sin \theta _{2}-B_{2}\cos \delta _{2}\right)+\left(\lambda _{2}+2\mu _{2}\right)\left(A_{2}\cos \theta _{2}\cot \theta _{2}+B_{2}\cos \delta _{2}\right).\end{aligned}}$

Writing $\lambda =\left(\lambda +2\mu \right)-2\mu =\rho \alpha ^{2}-2\rho \beta ^{2}$ (see equations (9,6) and (9,7) in Table 2.2a) and recalling that $\sin 2x=2\sin x\cos x$, $\cos 2x=\cos ^{2}x-\sin ^{2}x$, the equation can be changed to the form

${\begin{aligned}\left(A_{0}+A_{1}\right)Z_{1}\cos 2\delta _{1}-B_{1}W_{1}\sin 2\delta _{1}=A_{2}Z_{2}\cos 2\delta _{2}+B_{2}W_{2}\sin 2\delta _{2},\end{aligned}}$

(3.2h)

where $Z_{i}=\rho _{i}\alpha _{i}$, $W_{i}=\rho _{i}\beta _{i}$; $Z_{i}$ and $W_{i}$ are called impedances.

In the same way we get for the tangential stresses the equation

${\begin{aligned}&\mu _{1}\left[2\left(-A_{0}+A_{1}\right)\cos \theta _{1}+B_{1}\left(\cos \delta _{1}\cot \delta _{1}-\sin \delta _{1}\right)\right]\\&\qquad \qquad \qquad =\mu _{2}[-2A_{2}\cos \theta _{2}+B_{2}(\cos \delta _{2}\cot \delta _{2}-\sin \delta _{2})]\end{aligned}}$

This can be simplified using equation (3.1a) to give

${\begin{aligned}&\left(-A_{0}+A_{1}\right)\left(\beta _{1}/\alpha _{1}\right)W_{1}\sin 2\theta _{1}+B_{1}W_{1}\cos 2\delta _{1}&\\&\qquad =-A_{2}\left(\beta _{2}/\alpha _{2}\right)W_{2}\sin 2\theta _{2}+B_{2}W_{2}\cos 2\delta _{2}.\end{aligned}}$

(3.2i)

Equations (3.1e,f,h,i) are known as the Zoeppritz equations. For ease of reference, we have collected them below:

${\begin{aligned}&\left(-A_{0}+A_{1}\right)\cos \theta _{1}-B_{1}\sin \delta _{1}=-A_{2}\cos \theta _{2}-B_{2}\sin \delta _{2},\end{aligned}}$

(3.2e)

${\begin{aligned}&\left(A_{0}+A_{1}\right)\sin \theta _{1}+B_{1}\cos \delta _{1}=A_{2}\sin \theta _{2}-B_{2}\cos \delta _{2},\end{aligned}}$

(3.2f)

${\begin{aligned}&\left(A_{0}+A_{1}\right)Z_{1}\cos 2\delta _{1}-B_{1}W_{1}\sin 2\delta _{1}=A_{2}Z_{2}\cos 2\delta _{2}+B_{2}W_{2}\sin 2\delta _{2},\end{aligned}}$

(3.2h)

${\begin{aligned}&\left(-A_{0}+A_{1}\right)\left(\beta _{1}/\alpha _{1}\right)W_{1}\sin 2\theta _{1}+B_{1}W_{1}\cos 2\delta _{1}\\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad =-A_{2}\left(\beta _{2}/\alpha _{2}\right)W_{2}\sin 2\theta _{2}+B_{2}W_{2}\cos 2\delta _{2}.\end{aligned}}$

(3.2i)

Problem 3.2b

3.2b Derive the equations below for the tangential and normal displacements, $u$ and $w$, of a free surface for an incident P-wave of amplitude $A_{0}$:

${\begin{aligned}u/A_{0}&=\left[2/\left(m+n\right)\right]\left(m\sin \theta +\cos \delta \right)e^{\mathrm {j} \omega \left(px-t\right)},\\w/A_{0}&=\left[-2/\left(m+n\right)\right]\left(n\cos \theta +\sin \delta \right)e^{\mathrm {j} \omega \left(px-t\right)},\end{aligned}}$

(3.2e)

where $m=(\beta /\alpha )\tan 2\delta$, $n=\left(\alpha /\beta \right)\cos 2\delta /\sin 2\theta$, $\theta$ and $\delta$ being the angles of incidence of the P- and S-waves, respectively.

Solution

To determine the displacements at a free surface, we start by disregarding equations (3.2e,f) because there are no constraints on displacements at a free surface. After setting $A_{2}=0=B_{2}$, we are left with

${\begin{aligned}A_{1}Z\cos 2\delta -BW\sin 2\delta &=-A_{0}Z\cos 2\delta ,\end{aligned}}$

(3.2j)

${\begin{aligned}A_{1}\left(\beta /\alpha \right)W\sin 2\theta +BW\cos 2\delta &=A_{0}\left(\beta /\alpha \right)W\sin 2\theta ,\end{aligned}}$

(3.2k)

where $Z=\rho \alpha$, $W=\rho \beta$, $\beta /\alpha =W/Z$, and we have dropped unnecessary subscripts. These equations can be written

${\begin{aligned}A_{1}-mB=-A_{0},\qquad A_{1}+nB=A_{0},\end{aligned}}$

(3.2l)

where

${\begin{aligned}m&=\left(W\sin 2\delta \right)/\left(Z\cos 2\delta \right)=\left(\beta /\alpha \right)\tan 2\delta =\left(\sin \delta \tan 2\delta \right)/\sin \theta ,\end{aligned}}$

(3.2m)

${\begin{aligned}n&=\left(\cos 2\delta \right)/[\left(\beta /\alpha \right)\sin 2\theta ]=\left(\cos 2\delta \right)/\left(2\cos \theta \sin \theta \right).\end{aligned}}$

(3.2n)

The solution of equations (3.2$\ell$) is

${\begin{aligned}A_{1}/A_{0}=\left(m-n\right)/\left(m+n\right),\quad B_{1}/A_{0}=2/\left(m+n\right).\end{aligned}}$

In equations (3.2a,c) we set $z=0$, the factor $e^{\mathrm {j} \omega px}$ drops out, and we get

${\begin{aligned}u/A_{0}&=\left(1+A_{1}/A_{0}\right)\sin \theta +\left(B/A_{0}\right)\cos \delta ,\end{aligned}}$

(3.2o)

${\begin{aligned}w/A_{0}&=\left(-1+A_{1}/A_{0}\right)\cos \theta -\left(B/A_{0}\right)\sin \delta .\end{aligned}}$

(3.2p)

We now reinsert the values of $A_{1}/A_{0}$ and $B_{1}/A_{0}$ in terms of m and $n$, and equations (3.2o,p) become

${\begin{aligned}u/A_{0}&=\left\{\left[1+\left(m-n\right)/\left(m+n\right)\left]\sin \theta +\right[2/\left(m+n\right)\right]\cos \delta \right\}e^{\mathrm {j} \omega \left(px-t\right)}\\&=\left\{\left[2/\left(m+n\right)\right]\left(m\sin \theta +\cos \delta \right)\right\}e^{\mathrm {j} \omega \left(px-t\right)},\end{aligned}}$

(3.2q)

${\begin{aligned}w/A_{0}&=\left\{\left[-2/\left(m+n\right)\right]\left(n\cos \theta +\sin \delta \right)\right\}e^{\mathrm {j} \omega \left(px-t\right)}.\end{aligned}}$

(3.2r)

Problem 3.2c

3.2c Show that the displacements of a free surface at normal incidence are

${\begin{aligned}u/A_{0}=0,\qquad w/A_{0}=-2.\end{aligned}}$

Solution

For normal incidence at the surface, $z=0$, $\theta =0=\delta$. Equations (3.2j,k) give $A_{1}/A_{0}=-1$, $B/A_{0}=0$. Substituting in equations (3.2o,p), we get $u/A_{0}=0$, $w/A_{0}=-2$.

Problem 3.2d

3.2d Show that the displacements of a free surface of a solid, where $\theta =45^{\circ }$, $\alpha =3.0$ km/s, $\beta /\alpha =1/{\sqrt {2}}$, $\sigma =1/{\sqrt {2}}$, are

${\begin{aligned}u/A_{0}=1.793,\quad w/A_{0}=-1.035.\end{aligned}}$

Solution

For $\theta =45^{\circ }$, $\alpha =3$ km/s, $\left(\beta /\alpha \right)=1/{\sqrt {2}}$, $\sin \delta =\left(\beta /\alpha \right)\sin \theta =(1/{\sqrt {2}})^{2}$, that is, $\sin \delta =1/2$, so $\delta =30^{\circ }$. From the definitions of m and n, we get

${\begin{aligned}m&=\left(\beta /\alpha \right)\tan 2\delta =(1/{\sqrt {2}})\tan 60^{\circ }=1.225,\\n&=\left(\alpha /\beta \right)(\cos 2\delta /\sin 2\theta )={\sqrt {2}}\left(\cos 60^{\circ }/\sin 90^{\circ }\right)=0.707.\end{aligned}}$

Equations (3.2q,r) now give (omitting the factor $e^{\mathrm {j} \omega \left(px-t\right)}$).

${\begin{aligned}u/A_{0}&=\left[2/\left(1.225+0.707\right)\right]\left(1.225\sin 45^{\circ }+\cos 30^{\circ }\right)=1.793,\\w/A_{0}&=\left[-2/\left(1.225+0.707\right)\right]\left(0.707\cos 45^{\circ }+\sin 30^{\circ }\right)=1.035.\end{aligned}}$

Problem 3.2e

3.2e Show that the displacements at the surface of the ocean are $u/A_{0}=0$, $w/A_{0}=-2\cos \theta$.

Solution

In a fluid $B=0$ and equation (3.2j) gives $A_{1}/A_{0}=-1$, so equations (3.20,p) show that $u/A_{0}=0$ and $w/A_{0}=-2\cos \theta$.

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Table of Contents (book)

Also in this chapter

• General form of Snell’s law
• Reflection/refraction at a liquid/solid interface
• Zoeppritz’s equations for incident SV- and SH-waves
• Reinforcement depth in marine recording
• Complex coefficient of reflection
• Reflection and transmission coefficients
• Amplitude/energy of reflections and multiples
• Reflection/transmission coefficients at small angles and magnitude
• Magnitude
• AVO versus AVA and effect of velocity gradient
• Variation of reflectivity with angle (AVA)

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