# Difference between revisions of "Reflection/refraction at a liquid/solid interface"

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 3.3a

Derive Zoeppritz’s equations and Knott’s equations for a P-wave incident on a liquid/solid interface when the incident wave is (i) in the liquid and (ii) in the solid.

### Background

Knott’s equations differ from the Zoeppritz equations in that they use potential functions instead of displacements. Knott’s equations can be derived directly from the Zoeppritz equations and vice versa [see equation (3.3${\displaystyle \ell }$)]; however, we shall derive them from first principles. We use script letters, ${\displaystyle {\mathcal {A}}_{i}}$ and ${\displaystyle {\mathcal {B}}_{i}}$ for the amplitudes of the potential functions, reserving italic letters ${\displaystyle A_{i}}$ and ${\displaystyle B_{i}}$ for displacements.

To get Knott’s equations for a solid/solid interface, we start with the potential functions in equation (2.9c), the displacements being given by equations (2.9d,e), and apply the boundary conditions of problem 2.11, we write

 {\displaystyle {\begin{aligned}\phi _{1}&={\mathcal {A}}_{0}e^{\mathrm {j} \omega \zeta _{0}}+{\mathcal {A}}_{1}e^{\mathrm {j} \omega \zeta _{1}},\quad \chi _{1}={\mathcal {B}}_{1}e^{\mathrm {j} \omega \zeta _{1}^{\prime }},\end{aligned}}} (3.3a)

 {\displaystyle {\begin{aligned}\phi _{2}&={\mathcal {A}}_{2}e^{\mathrm {j} \omega \zeta _{2}},\qquad \qquad \qquad \chi _{2}={\mathcal {B}}_{2}e^{\mathrm {j} \omega \zeta _{2}^{\prime }},\end{aligned}}} (3.3b)

where ${\displaystyle \zeta _{0}}$, ${\displaystyle \zeta _{1}}$, ${\displaystyle \zeta _{2}}$, ${\displaystyle \zeta _{1}^{\prime }}$ and ${\displaystyle \zeta _{2}^{\prime }}$ are given by equations (3.1d,e). Using equation (2.9e), the continuity of normal displacement requires that ${\displaystyle \left(\phi _{z}-\chi _{x}\right)}$ be continuous at ${\displaystyle z=0}$. Using equations (3.2g), we obtain

 {\displaystyle {\begin{aligned}\left(-{\mathcal {A}}_{0}+{\mathcal {A}}_{1}\right)\cot \theta _{1}-{\mathcal {B}}_{1}=-{\mathcal {A}}_{2}\cot \theta _{2}-{\mathcal {B}}_{2}.\end{aligned}}} (3.3c)

Continuity of tangential displacement requires the continuity of ${\displaystyle \left(\phi _{x}+\chi _{z}\right)}$ [see equation (2.9d)]; this gives the equation

 {\displaystyle {\begin{aligned}{\mathcal {A}}_{0}+{\mathcal {A}}_{1}+B_{1}\cot \delta _{1}={\mathcal {A}}_{2}-{\mathcal {B}}_{2}\cot \delta _{2}.\end{aligned}}} (3.3d)

The normal stress is given by ${\displaystyle \lambda \nabla ^{2}\phi +2\mu \left(\phi _{zz}-\chi _{xz}\right)}$ [see equation (2.11b)], so

{\displaystyle {\begin{aligned}&\lambda _{1}\left(1+\cot ^{2}\delta _{1}\right)\left({\mathcal {A}}_{0}+{\mathcal {A}}_{1}\right)+2\mu _{1}\cot ^{2}\theta _{1}\left({\mathcal {A}}_{0}+{\mathcal {A}}_{1}\right)-2\mu _{1}{\mathcal {B}}_{1}\cot \delta _{1}\\&\quad =\lambda _{2}\left(1+\cot ^{2}\theta _{2}\right){\mathcal {A}}_{2}+2\mu _{2}{\mathcal {A}}_{2}\cot ^{2}\theta _{2}+2\mu _{2}{\mathcal {B}}_{2}\cot \delta _{2}.\end{aligned}}}

But ${\displaystyle \left(1+\cot ^{2}\theta \right)=1/\sin ^{2}\theta }$, ${\displaystyle \lambda =\left(\lambda +2\mu \right)-2\mu =\rho \alpha ^{2}-2\rho \beta ^{2}}$ (see equations (9,6) and (9,7) of Table 2.2a); using these relations plus equation (3.1a), we get

 {\displaystyle {\begin{aligned}&\mu _{1}\left[\left(\cot ^{2}\delta _{1}-1\right)\left({\mathcal {A}}_{0}+{\mathcal {A}}_{1}\right)-2\mu _{1}{\mathcal {B}}_{1}\cot \delta _{1}\right]\\&\qquad \qquad \qquad \qquad \qquad =\mu _{2}(\cot ^{2}\delta _{2}-1){\mathcal {A}}_{2}+2\mu _{2}{\mathcal {B}}_{2}\cot \delta _{2}.\end{aligned}}} (3.3e)

The tangential stress is ${\displaystyle \sigma _{xz}=\mu \left(2\phi _{xz}+\chi _{zz}-\chi _{xx}\right)}$ from equation (2.11b), so we get

 {\displaystyle {\begin{aligned}&\mu _{1}[2\left(-{\mathcal {A}}_{0}+{\mathcal {A}}_{1}\right)\cot \theta _{1}+{\mathcal {B}}_{1}(\cot ^{2}\delta _{1}-1)]\\&\quad =\mu _{2}[-2{\mathcal {A}}_{2}\cot \theta _{2}+{\mathcal {B}}_{2}(\cot ^{2}\delta _{2}-1)].\end{aligned}}} (3.3f)

We can write Knott’s equations in a more compact form by substituting ${\displaystyle a_{i}=\cot \theta _{i}}$, ${\displaystyle b_{i}=\cot \delta _{i}}$, ${\displaystyle c_{i}=b_{i}^{2}-1}$; the four equations now become

 {\displaystyle {\begin{aligned}-a_{1}{\mathcal {A}}_{0}+a_{1}{\mathcal {A}}_{1}-{\mathcal {B}}_{1}&=-a_{2}{\mathcal {A}}_{2}-{\mathcal {B}}_{2},\end{aligned}}} (3.3g)

 {\displaystyle {\begin{aligned}{\mathcal {A}}_{0}+{\mathcal {A}}_{1}+b_{1}{\mathcal {B}}_{1}&={\mathcal {A}}_{2}-b_{2}{\mathcal {B}}_{2},\end{aligned}}} (3.3h)

 {\displaystyle {\begin{aligned}\mu _{1}c_{1}{\mathcal {A}}_{0}+\mu _{1}c_{1}{\mathcal {A}}_{1}-2\mu _{1}b_{1}{\mathcal {B}}_{1}&=\mu _{2}c_{2}{\mathcal {A}}_{2}+2\mu _{2}b_{2}{\mathcal {B}}_{2},\end{aligned}}} (3.3i)

 {\displaystyle {\begin{aligned}-2\mu _{1}a_{1}{\mathcal {A}}_{0}+2\mu _{1}a_{1}{\mathcal {A}}_{1}+\mu _{1}c_{1}{\mathcal {B}}_{1}&=-2\mu _{2}a_{2}{\mathcal {A}}_{2}+\mu _{2}c_{2}{\mathcal {B}}_{2}.\end{aligned}}} (3.3j)

To show the correspondence between Knott’s and Zoeppritz’s equations, we calculate the energy density in terms of the displacements and the potential functions used in Knott’s equations. In terms of displacements, the instantaneous kinetic energy density E for a harmonic P-wave ${\displaystyle u=A\cos \omega t}$ is equal to

{\displaystyle {\begin{aligned}E={\frac {1}{2}}\rho \left({\frac {\partial u}{\partial t}}\right)^{2}={\frac {1}{2}}\rho \omega ^{2}A^{2}\sin ^{2}\omega t.\end{aligned}}}

The total energy density is the maximum kinetic energy density (see problem 3.7), that is,

 {\displaystyle {\begin{aligned}E={\frac {1}{2}}\rho \omega ^{2}A^{2}.\end{aligned}}} (3.3k)

The energy density of a P-wave in terms of the potential function ${\displaystyle \phi }$ [see equation (2.9a)], noting that ${\displaystyle \chi =0}$ since there is no S-wave) is

{\displaystyle {\begin{aligned}E={\frac {1}{2}}\rho \left[\left({\frac {\partial u}{\partial t}}\right)^{2}+\left({\frac {\partial w}{\partial t}}\right)^{2}\right]={\frac {1}{2}}\rho \left(\phi _{xt}^{2}+\phi _{zt}^{2}\right),\end{aligned}}}

from equations (2.9d,e). Taking the time factor as ${\displaystyle e^{\mathrm {j} \omega t}}$ and reinserting the factor ${\displaystyle \mathrm {j} \omega p}$ which was deleted to get equation (3.2g), we get ${\displaystyle \partial /\partial t=\mathrm {j} \omega }$, ${\displaystyle \partial /\partial x=\mathrm {j} \omega p}$, ${\displaystyle \partial /\partial z=\pm \mathrm {j} \omega \left(\cos \theta \right)/\alpha }$. Thus, we get for the total energy ${\displaystyle E}$

{\displaystyle {\begin{aligned}E={\frac {1}{2}}\rho {\mathcal {A}}^{2}\left\{(\omega ^{2}p)^{2}+\left[{\frac {\omega ^{2}\left(\cos \theta \right)}{\alpha }}\right]^{2}\right\}={\frac {1}{2}}\rho (\omega ^{2}{\mathcal {A}}/\alpha )^{2}.\end{aligned}}}

Comparing with the expression for ${\displaystyle E}$ in equation (3.3k), we see that

 {\displaystyle {\begin{aligned}{\mathcal {A}}_{i}=(\alpha _{i}/\omega )^{2}A_{i},\quad {\mathcal {B}}_{i}=(\beta _{i}/\omega )^{2}B_{i},\end{aligned}}} (3.3l)

where the second equation is obvious from symmetry.

### Solution

Since we derived Zoeppritz’s equations in problem 3.2a, we derive Knott’s equations here and then get Zoeppritz’s equations from them using equation (3.31).

i) To derive Knott’s equations when the incident P-wave is in the liquid, the boundary conditions require the continuity of ${\displaystyle w}$ and ${\displaystyle \sigma _{zz}}$ and that ${\displaystyle \sigma _{xz}}$ in the solid vanish at ${\displaystyle z=0}$. omitting the factor ${\displaystyle e^{\mathrm {j} \omega t}}$, we have from equations (3.1b,c,d,e):

{\displaystyle {\begin{aligned}\phi _{1}&={\mathcal {A}}_{0}e^{\mathrm {j} \omega \zeta _{0}}+{\mathcal {A}}_{1}e^{\mathrm {j} \omega \zeta _{1}},&\quad \chi _{1}&=0,\\\phi _{2}&=\qquad \qquad {\mathcal {A}}_{2}e^{\mathrm {j} \omega \zeta _{2}},&\quad \chi _{2}&=B_{2}e^{\mathrm {j} \omega \zeta _{2}^{\prime }}.\end{aligned}}}

Continuity of ${\displaystyle w=\phi _{z}-\chi _{x}}$ yields the first equation [note equation (3.2g)]:

{\displaystyle {\begin{aligned}\left(-{\mathcal {A}}_{0}+{\mathcal {A}}_{1}\right)\cot \theta _{1}=-{\mathcal {A}}_{2}\cot \theta _{2}-{\mathcal {B}}_{2}.\end{aligned}}}

Continuity of normal stress requires that ${\displaystyle \lambda \nabla ^{2}\phi +2\mu \left(\phi _{z}-\chi _{xz}\right)}$ be continuous; since ${\displaystyle \mu =0}$ in the liquid, this results in

{\displaystyle {\begin{aligned}\lambda _{1}\left({\mathcal {A}}_{0}+{\mathcal {A}}_{1}\right)\left(1+\cot ^{2}\theta _{1}\right)=\lambda _{2}{\mathcal {A}}_{2}\left(1+\cot ^{2}\theta _{2}\right)+2\mu _{2}\left({\mathcal {A}}_{2}\cot ^{2}\theta _{2}+{\mathcal {B}}_{2}\cot \delta _{2}\right).\end{aligned}}}

Using equations (9,6) and (9,7) in Table 2.2a, also equation (3.1a), we have

{\displaystyle {\begin{aligned}\left(\rho _{1}\alpha _{1}^{2}/\sin ^{2}\theta _{1}\right)\left({\mathcal {A}}_{0}+{\mathcal {A}}_{1}\right)&=\left(\lambda _{2}+2\mu _{2}\right)A_{2}\left(1+\cot ^{2}\theta _{2}\right)-2\mu _{2}{\mathcal {A}}_{2}+2\mu _{2}{\mathcal {B}}_{2}\cot \delta _{2}\\&=\left(\rho _{2}\alpha _{2}^{2}/\sin ^{2}\theta _{2}-2\mu _{2}\right){\mathcal {A}}_{2}+2\mu _{2}{\mathcal {B}}_{2}\cot \delta _{2}\\&=\left(\rho _{2}\beta _{2}^{2}/\sin ^{2}\delta _{2}-2\mu _{2}\right){\mathcal {A}}_{2}+2\mu _{2}{\mathcal {B}}_{2}\cot \delta _{2}\\&=\mu _{2}\left(1/\sin ^{2}\delta _{2}-2\right){\mathcal {A}}_{2}+2\mu _{2}{\mathcal {B}}_{2}\cot \delta _{2}\\&=\mu _{2}\left(\cot ^{2}\delta _{2}-1\right){\mathcal {A}}_{2}+2\mu _{2}{\mathcal {B}}_{2}\cot \delta _{2}.\end{aligned}}}

Continuity of tangential stress ${\displaystyle \sigma _{xzz}}$ requires that ${\displaystyle \left(2\phi _{xz}+\chi _{zz}-\chi _{xx}\right)=0}$ in the solid when ${\displaystyle z=0}$, so

{\displaystyle {\begin{aligned}-2{\mathcal {A}}_{2}\cot \theta _{2}+{\mathcal {B}}_{2}\left(\cot ^{2}\delta _{2}-1\right)=0.\end{aligned}}}

Using the coefficients in equations (3.3g,h,i,j), the results become

{\displaystyle {\begin{aligned}a_{1}{\mathcal {A}}_{1}+a_{2}{\mathcal {A}}_{2}+{\mathcal {B}}_{2}&=a_{1}{\mathcal {A}}_{0},\\\left(\rho _{1}p^{2}\right){\mathcal {A}}_{1}-\mu _{2}c_{2}{\mathcal {A}}_{2}-2\mu _{2}b_{2}{\mathcal {B}}_{2}&=-\left(\rho _{1}/p^{2}\right){\mathcal {A}}_{0},\\2a_{2}{\mathcal {A}}_{2}-c_{2}{\mathcal {B}}_{2}&=0.\end{aligned}}}

These are Knott’s equations. We could derive Zoeppritz’s equations as we did equations (3.2e,f,h,i), or we can use equation (3.3${\displaystyle \ell }$) to change the coefficients in Knott’s equations to Zoeppritz’s coeffiicients. Using this latter method, we have ${\displaystyle {\mathcal {A}}_{i}=(\omega /\alpha _{i})^{2}A_{i}}$, ${\displaystyle {\mathcal {B}}_{2}=(\omega /\beta _{2})^{2}B_{2}}$. Substituting these, we get the Zoeppritz equations for a liquid-solid interface:

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

ii) When the incident wave is in the solid, we shall first derive Zoeppritz’s equations, then change them to Knott’s equations. We have ${\displaystyle B_{2}=0}$ in the liquid, ${\displaystyle w}$ and ${\displaystyle \sigma _{zz}}$ are continuous, and ${\displaystyle \sigma _{zx}=0}$ in the solid at ${\displaystyle z=0}$.

Equation (3.2e) gives for the normal displacement

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

The continuity of normal stress is expressed in equation (3.2h), which now becomes

{\displaystyle {\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}.\end{aligned}}}

Finally, ${\displaystyle \sigma _{xz}=0}$ at ${\displaystyle z=0}$ and equation (3.2i) becomes

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

Using equation (3.3${\displaystyle \ell }$), we get the equivalent Knott’s equations:

{\displaystyle {\begin{aligned}\left(-{\mathcal {A}}_{0}+{\mathcal {A}}_{1}\right)\cot \delta _{1}-{\mathcal {B}}_{1}&=-{\mathcal {A}}_{2}\cot \theta _{2},\\\left({\mathcal {A}}_{0}+{\mathcal {A}}_{1}\right)\left(\lambda _{1}+2\mu _{1}\right)\cot ^{2}\theta _{1}-{\mathcal {B}}_{1}\cot \delta _{1}&=\lambda _{2}{\mathcal {A}}_{2}\left(1+\cot ^{2}\theta _{2}\right),\\2\left(-{\mathcal {A}}_{0}+{\mathcal {A}}_{1}\right)\cot \theta _{1}+{\mathcal {B}}_{1}(\cot ^{2}\delta _{1}-1)&=0.\end{aligned}}}

## Problem 3.3b

Calculate the amplitudes of the reflected and refracted P-and S-waves when an incident P-wave strikes the interface from a water layer ${\displaystyle (\alpha =1500m/s}$, ${\displaystyle \beta =0}$, ${\displaystyle \rho =1.00}$ g/cm${\displaystyle ^{3}}$) at ${\displaystyle 20^{\circ }}$ when the seafloor is (i) “soft” ${\displaystyle (\alpha =2000m/s}$, ${\displaystyle \beta =1000}$ m/s, ${\displaystyle \rho =2.00}$ g/cm${\displaystyle ^{3})}$, and (ii) “hard” ${\displaystyle (\alpha =4000m/s}$, ${\displaystyle \beta =2500m/s}$, ${\displaystyle \rho =2.50}$ g/cm${\displaystyle ^{3}}$).

### Solution

i) Where the seafloor is “soft” and the P-wave is incident in the water, we have:

{\displaystyle {\begin{aligned}\sin \theta _{2}&=\left(2.00/1.50\right)\sin 20^{\circ },&\theta _{2}&=27.1^{\circ };2\theta _{2}=54.2^{\circ },\\\cos \theta _{2}&=0.890,&\sin 2\theta _{2}&=0.811;\\\sin \delta _{2}&=\left(1.00/1.50\right)\sin 20^{\circ }=0.228,&\delta _{2}&=13.2^{\circ },2\delta _{2}=26.4^{\circ },\\\sin 2\delta _{2}&=0.445,&\cos 2\delta _{2}&=0.896;\\\cos \theta _{1}&=0.940,\qquad \qquad Z_{1}=1.500,&Z_{2}&=4.000,W_{2}=2.000.\end{aligned}}}

Thus Zoeppritz’s equations become

{\displaystyle {\begin{aligned}0.940A_{1}+0.890A_{2}+0.228B_{2}&=0.940A_{0},\\1.500A_{1}-3.584A_{2}-0.890B_{2}&=-1.500A_{0},\\0.406A_{2}-0.896B_{2}&=0.\end{aligned}}}

The solution is: ${\displaystyle A_{1}/A_{0}=0.431}$, ${\displaystyle A_{2}/A_{0}=0.539}$, ${\displaystyle B_{2}/A_{0}=0.244}$.

ii) When the seafloor is “hard”:

{\displaystyle {\begin{aligned}\sin \theta _{2}&=\left(4.00/1.50\right)\sin 20^{\circ },&\theta _{2}&=65.8^{\circ },2\theta _{2}=131.6^{\circ },\\\cos \theta _{2}&=0.410;&\sin 2\theta _{2}&=0.748;\\\sin \delta _{2}&=\left(2.50/1.50\right)\sin 20^{\circ }=0.570,&\delta _{2}&=34.8^{\circ },2\delta _{2}=69.6^{\circ },\\\sin 2\delta _{2}&=0.937,&\cos 2\delta _{2}&=0.349;\\\cos \theta _{1}&=0.940,\qquad \qquad Z_{1}=1.50,&Z_{2}&=10.0,W_{2}=6.25.\end{aligned}}}

The equations are

{\displaystyle {\begin{aligned}0.94A_{1}+0.41A_{2}+0.57B_{2}&=0.94A_{0},\\1.50A_{1}-3.49A_{2}-5.86B_{2}&=-1.50A_{0},\\0.468A_{2}-0.349B_{2}&=0.\end{aligned}}}

The solution is ${\displaystyle A_{1}/A_{0}=0.716}$, ${\displaystyle A_{2}/A_{0}=0.227}$, ${\displaystyle B_{2}/A_{0}=0.30}$.

## Problem 3.3c

3.3c Repeat part (b) for an angle of incidence of ${\displaystyle 30^{\circ }}$.

### Solution

i) For the “soft” bottom and ${\displaystyle \theta _{1}=30^{\circ }}$,

{\displaystyle {\begin{aligned}\sin \theta _{2}&=\left(2.00/1.50\right)\sin 30^{\circ },&\theta _{2}&=41.8^{\circ },2\theta _{2}=83.6^{\circ },\\\cos \theta _{2}&=0.745,&\sin 2\theta _{2}&=0.994;\\\sin \delta _{2}&=\left(1.00/1.50\right)\sin 30^{\circ }=0.333,&\delta _{2}&=19.5^{\circ };\\\sin 2\delta _{2}&=0.629,&\cos 2\delta _{2}&=0.777.\end{aligned}}}

The equations are

{\displaystyle {\begin{aligned}0.866A_{1}+0.745A_{2}+0.333B_{2}&=0.866A_{0},\\1.50A_{1}-3.11A_{2}-1.26B_{2}&=-1.50A_{0},\\0.621A_{2}-0.778B_{2}&=0.\end{aligned}}}

The solution is ${\displaystyle A_{1}/A_{0}=0.40}$, ${\displaystyle A_{2}/A_{0}=0.51}$, ${\displaystyle B_{2}/A_{0}=0.41}$.

ii) For the “hard” bottom, ${\displaystyle \sin \theta _{2}=(4.00/1.50)\sin 30^{\circ }=1.33>1}$, so total reflection occurs.

The results are summarized in Table 3.3a. The table shows that ${\displaystyle A_{1}}$ and ${\displaystyle A_{2}}$ depend mainly on the hardness of the bottom and only moderately on ${\displaystyle \theta _{1}}$. However, ${\displaystyle B_{2}}$ depends more on the angle of incidence than on the hardness.

Table 3.3a Reflected/transmitted amplitudes for soft/hard bottoms.
Bottom ${\displaystyle \theta _{1}}$ ${\displaystyle A_{1}/A_{0}}$ ${\displaystyle A_{2}/A_{0}}$ ${\displaystyle B_{2}/A_{0}}$
Soft ${\displaystyle 20^{\circ }}$ 0.431 0.539 0.244
${\displaystyle 30^{\circ }}$ 0.403 0.512 0.408
Hard ${\displaystyle 20^{\circ }}$ 0.716 0.227 0.304