Chapter 3 Partitioning at an interface
3.1 General form of Snell’s law
3.1a A Pwave of amplitude $A_{0}$ is incident at the angle $\theta _{1}$ on a plane interface separating two solid media. This generates reflected and refracted Pwaves and converted reflected and refracted Swaves. Amplitudes, angles of incidence and refraction, and directions of displacements of these waves are shown in Figure 3.1a.
Use Huygens’s principle to show that $\theta _{1}=\theta _{1}^{\prime }$ and that
${\begin{aligned}{\frac {\sin \theta _{1}}{\alpha _{1}}}={\frac {\sin \delta _{1}}{\beta _{1}}}={\frac {\sin \theta _{2}}{\alpha _{2}}}={\frac {\sin \delta _{2}}{\beta _{2}}}=p,\end{aligned}}$


(3.1a)

where $p$ is the raypath parameter. Equation (3.1a) is Snell’s law.
Background
When a wave is incident at the interface between two solid media, four boundary conditions must be satisfied (continuity of normal and tangental displacements and stresses). The velocities are determined by the densities and elastic constants while the angles of incidence, reflection, and refraction are fixed by the velocities [see equation (3.1a)]. So the only remaining parameters that one can adjust in order to satisfy the boundary conditions are the amplitudes of the four waves generated by the incident wave, the reflected and refracted P and Swaves, $A_{1}$, $A_{2}$, $B_{1}$, $B_{2}$ as shown in Figure 3.1a.
Figure 3.1a. Raypaths at solidsolid interface.
Huygens’s principle states that each point on a wavefront acts as a new point source radiating energy in all directions. Subsequent wavefronts can be located by swinging arcs with centers at points on the wavefront and radii equal to the distance traveled in a fixed time interval, the new wavefront being the envelope of the arcs. If the first wavefront and the reflector are planar, only two arcs are necessary, the new wavefront being tangent to the arcs.
Solution
In Figure 3.1b, $AB$ is a wavefront of a planar Pwave approaching a planar interface. When the wavefront reaches the interface, point $A'$ becomes a new source radiating energy upward and downward according to Huygens’s principle. When $B'$ reaches the interface at $R$, the distance $B'R$ being $\alpha _{1}\Delta t$, the wave reflected at $A'$ has traveled upward the same distance $\alpha _{1}\Delta t$. By drawing an arc with center $A'$ and radius $\alpha _{1}\Delta t$ and then drawing a line from $R$ tangent to the arc, we get the reflected wavefront $SR$. The angle of incidence is $\theta _{1}$ and the angle of reflection is $\theta _{1}^{\prime }$. In $\Delta A'B'R$ and $\Delta A'SR,$ the angles at $B'$ and $S$ are $90^{\circ }$ (because rays are perpendicular to wavefronts). Since the triangles have a common side $A'R$, they are equal and $\angle \theta _{1}=\angle \theta _{1}^{\prime }$, that is, the angle of incidence equals the angle of reflection (Law of reflection).
Figure 3.1b. Snell’s law derivation.
In the case of the refracted wave, $\theta _{2}$ is the angle of refraction and $\left(\alpha _{2}\Delta t\right)/A'R=\sin \theta _{2}$ But $\left(\alpha _{1}\Delta t\right)/A'R=\sin \theta _{1}$, so we have
${\begin{aligned}\Delta t/A'R=\sin \theta _{1}/\alpha _{1}=\sin \theta _{2}/\alpha _{2}.\end{aligned}}$
In Figure 3.1b, if we replace $\theta _{1}^{\prime }$ in $\Delta A'SR$ with $\delta _{1}$ (compare with Figure 3.1a) and $\theta _{2}$ in $\Delta A'TR$ with $\delta _{2}$, we arrive at
${\begin{aligned}\Delta t/A'R=\sin \delta _{1}/\beta _{1}=\sin \delta _{2}/\beta _{2}.\end{aligned}}$
Equating the four ratios of sines to velocities, we get equation (3.1a).
3.1b Using the waveform $e^{\mathrm {j} \omega [(lx+nz)/Vt]}$ (see problem 2.5b), where $(l,n)$ are direction cosines of the ray, show that (omitting the factor $e^{\mathrm {j} \omega t}$) the incident, reflected, and refracted waves can be written
${\begin{aligned}\psi _{0}=A_{0}e^{\mathrm {j} \omega \zeta _{0}},\quad \psi _{1}&=A_{1}e^{\mathrm {j} \omega \zeta _{1}},\quad \psi _{2}=A_{2}e^{\mathrm {j} \omega \zeta _{2}},\end{aligned}}$


(3.1b)

${\begin{aligned}\psi _{1}^{'}&=B_{1}e^{\mathrm {j} \omega \zeta _{1}^{'}},\quad \psi _{2}^{'}=B_{2}e^{\mathrm {j} \omega \zeta _{2}^{'}},\end{aligned}}$


(3.1c)

where
${\begin{aligned}\zeta _{0}&=p\left(xz\cot \theta _{1}\right),\quad \zeta _{1}=p\left(x+z\cot \theta _{1}\right),\quad \zeta _{2}=p\left(xz\cot \theta _{2}\right),\end{aligned}}$


(3.1d)

${\begin{aligned}\zeta _{1}^{'}&=p\left(x+z\cot \delta _{1}\right),\quad \zeta _{2}^{'}=p\left(xz\cot \delta _{2}\right).\end{aligned}}$


(3.1e)

Solution
We write $\zeta =\left(lx\pm nz\right)/V$, where the plus sign is used for waves traveling upward (that is, in the positive $z$direction) and the minus for downward traveling waves. The velocity ${\textit {V}}$ is $\alpha$ for Pwaves, $\beta$ for Swaves. We note that for Pwaves, $l=\sin \theta _{i}$, $n=\cos \theta _{i}$, $i=1$, 2. For Swaves we replace $\theta$ with $\delta$ so that $l=\sin \delta _{i}$, $n=\cos \delta _{i}$, $i=1$, 2.
Thus, for Pwaves, $\zeta _{i}=\left(x\sin \theta _{i}\pm z\cos \theta _{i}\right)/\alpha _{i}=p\left(x\pm z\cot \theta _{i}\right)$ while for Swaves we have $\zeta _{i}^{'}=p\left(x\pm z\cot \delta _{i}\right)$. Inserting the amplitudes $A_{i}$, $B_{i}$, we get equations (3.1b,c,d,e).
3.2 Reflection/refraction at a solid/solid interface; Displacement of a free surface
3.2a Derive the Zoeppritz equations for a Pwave incident on a solid/solid interface
Background
The normal and tangential displacements plus the normal and tangential stresses must be continuous when a Pwave 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)

3.2b Derive the equations below for the tangential and normal displacements, $u$ and $w$, of a free surface for an incident Pwave of amplitude $A_{0}$:
${\begin{aligned}u/A_{0}&=\left[2/\left(m+n\right)\right]\left(m\sin \theta +\cos \delta \right)e^{\mathrm {j} \omega \left(pxt\right)},\\w/A_{0}&=\left[2/\left(m+n\right)\right]\left(n\cos \theta +\sin \delta \right)e^{\mathrm {j} \omega \left(pxt\right)},\end{aligned}}$


(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 Swaves, 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
${\begin{aligned}A_{1}/A_{0}=\left(mn\right)/\left(m+n\right),\quad B_{1}/A_{0}=2/\left(m+n\right).\end{aligned}}$
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
${\begin{aligned}u/A_{0}&=\left\{\left[1+\left(mn\right)/\left(m+n\right)\left]\sin \theta +\right[2/\left(m+n\right)\right]\cos \delta \right\}e^{\mathrm {j} \omega \left(pxt\right)}\\&=\left\{\left[2/\left(m+n\right)\right]\left(m\sin \theta +\cos \delta \right)\right\}e^{\mathrm {j} \omega \left(pxt\right)},\end{aligned}}$


(3.2q)

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


(3.2r)

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 $u/A_{0}=0$, $w/A_{0}=2$.
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 $e^{\mathrm {j} \omega \left(pxt\right)}$).
${\begin{aligned}u/A_{0}&=\left[2/\left(1.225+0.707\right)\right]\left(1.225\sin 45^{\circ }+\cos 30^{\circ }\right)=1.793,\\w/A_{0}&=\left[2/\left(1.225+0.707\right)\right]\left(0.707\cos 45^{\circ }+\sin 30^{\circ }\right)=1.035.\end{aligned}}$
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 $w/A_{0}=2\cos \theta$.
3.3 Reflection/refraction at a liquid/solid interface
3.3a Derive Zoeppritz’s equations and Knott’s equations for a Pwave 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$\ell$)]; however, we shall derive them from first principles. We use script letters, ${\mathcal {A}}_{i}$ and ${\mathcal {B}}_{i}$ for the amplitudes of the potential functions, reserving italic letters $A_{i}$ and $B_{i}$ 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
${\begin{aligned}\phi _{1}&={\mathcal {A}}_{0}e^{\mathrm {j} \omega \zeta _{0}}+{\mathcal {A}}_{1}e^{\mathrm {j} \omega \zeta _{1}},\quad \chi _{1}={\mathcal {B}}_{1}e^{\mathrm {j} \omega \zeta _{1}^{\prime }},\end{aligned}}$


(3.3a)

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


(3.3b)

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


(3.3c)

Continuity of tangential displacement requires the continuity of $\left(\phi _{x}+\chi _{z}\right)$ [see equation (2.9d)]; this gives the equation
${\begin{aligned}{\mathcal {A}}_{0}+{\mathcal {A}}_{1}+B_{1}\cot \delta _{1}={\mathcal {A}}_{2}{\mathcal {B}}_{2}\cot \delta _{2}.\end{aligned}}$


(3.3d)

The normal stress is given by $\lambda \nabla ^{2}\phi +2\mu \left(\phi _{zz}\chi _{xz}\right)$ [see equation (2.11b)], so
${\begin{aligned}&\lambda _{1}\left(1+\cot ^{2}\delta _{1}\right)\left({\mathcal {A}}_{0}+{\mathcal {A}}_{1}\right)+2\mu _{1}\cot ^{2}\theta _{1}\left({\mathcal {A}}_{0}+{\mathcal {A}}_{1}\right)2\mu _{1}{\mathcal {B}}_{1}\cot \delta _{1}\\&\quad =\lambda _{2}\left(1+\cot ^{2}\theta _{2}\right){\mathcal {A}}_{2}+2\mu _{2}{\mathcal {A}}_{2}\cot ^{2}\theta _{2}+2\mu _{2}{\mathcal {B}}_{2}\cot \delta _{2}.\end{aligned}}$
But $\left(1+\cot ^{2}\theta \right)=1/\sin ^{2}\theta$, $\lambda =\left(\lambda +2\mu \right)2\mu =\rho \alpha ^{2}2\rho \beta ^{2}$ (see equations (9,6) and (9,7) of Table 2.2a); using these relations plus equation (3.1a), we get
${\begin{aligned}&\mu _{1}\left[\left(\cot ^{2}\delta _{1}1\right)\left({\mathcal {A}}_{0}+{\mathcal {A}}_{1}\right)2\mu _{1}{\mathcal {B}}_{1}\cot \delta _{1}\right]\\&\qquad \qquad \qquad \qquad \qquad =\mu _{2}(\cot ^{2}\delta _{2}1){\mathcal {A}}_{2}+2\mu _{2}{\mathcal {B}}_{2}\cot \delta _{2}.\end{aligned}}$


(3.3e)

The tangential stress is $\sigma _{xz}=\mu \left(2\phi _{xz}+\chi _{zz}\chi _{xx}\right)$ from equation (2.11b), so we get
${\begin{aligned}&\mu _{1}[2\left({\mathcal {A}}_{0}+{\mathcal {A}}_{1}\right)\cot \theta _{1}+{\mathcal {B}}_{1}(\cot ^{2}\delta _{1}1)]\\&\quad =\mu _{2}[2{\mathcal {A}}_{2}\cot \theta _{2}+{\mathcal {B}}_{2}(\cot ^{2}\delta _{2}1)].\end{aligned}}$


(3.3f)

We can write Knott’s equations in a more compact form by substituting $a_{i}=\cot \theta _{i}$, $b_{i}=\cot \delta _{i}$, $c_{i}=b_{i}^{2}1$; the four equations now become
${\begin{aligned}a_{1}{\mathcal {A}}_{0}+a_{1}{\mathcal {A}}_{1}{\mathcal {B}}_{1}&=a_{2}{\mathcal {A}}_{2}{\mathcal {B}}_{2},\end{aligned}}$


(3.3g)

${\begin{aligned}{\mathcal {A}}_{0}+{\mathcal {A}}_{1}+b_{1}{\mathcal {B}}_{1}&={\mathcal {A}}_{2}b_{2}{\mathcal {B}}_{2},\end{aligned}}$


(3.3h)

${\begin{aligned}\mu _{1}c_{1}{\mathcal {A}}_{0}+\mu _{1}c_{1}{\mathcal {A}}_{1}2\mu _{1}b_{1}{\mathcal {B}}_{1}&=\mu _{2}c_{2}{\mathcal {A}}_{2}+2\mu _{2}b_{2}{\mathcal {B}}_{2},\end{aligned}}$


(3.3i)

${\begin{aligned}2\mu _{1}a_{1}{\mathcal {A}}_{0}+2\mu _{1}a_{1}{\mathcal {A}}_{1}+\mu _{1}c_{1}{\mathcal {B}}_{1}&=2\mu _{2}a_{2}{\mathcal {A}}_{2}+\mu _{2}c_{2}{\mathcal {B}}_{2}.\end{aligned}}$


(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 Pwave $u=A\cos \omega t$ is equal to
${\begin{aligned}E={\frac {1}{2}}\rho \left({\frac {\partial u}{\partial t}}\right)^{2}={\frac {1}{2}}\rho \omega ^{2}A^{2}\sin ^{2}\omega t.\end{aligned}}$
The total energy density is the maximum kinetic energy density (see problem 3.7), that is,
${\begin{aligned}E={\frac {1}{2}}\rho \omega ^{2}A^{2}.\end{aligned}}$


(3.3k)

The energy density of a Pwave in terms of the potential function $\phi$ [see equation (2.9a)], noting that $\chi =0$ since there is no Swave) is
${\begin{aligned}E={\frac {1}{2}}\rho \left[\left({\frac {\partial u}{\partial t}}\right)^{2}+\left({\frac {\partial w}{\partial t}}\right)^{2}\right]={\frac {1}{2}}\rho \left(\phi _{xt}^{2}+\phi _{zt}^{2}\right),\end{aligned}}$
from equations (2.9d,e). Taking the time factor as $e^{\mathrm {j} \omega t}$ and reinserting the factor $\mathrm {j} \omega p$ which was deleted to get equation (3.2g), we get $\partial /\partial t=\mathrm {j} \omega$, $\partial /\partial x=\mathrm {j} \omega p$, $\partial /\partial z=\pm \mathrm {j} \omega \left(\cos \theta \right)/\alpha$. Thus, we get for the total energy $E$
${\begin{aligned}E={\frac {1}{2}}\rho {\mathcal {A}}^{2}\left\{(\omega ^{2}p)^{2}+\left[{\frac {\omega ^{2}\left(\cos \theta \right)}{\alpha }}\right]^{2}\right\}={\frac {1}{2}}\rho (\omega ^{2}{\mathcal {A}}/\alpha )^{2}.\end{aligned}}$
Comparing with the expression for $E$ in equation (3.3k), we see that
${\begin{aligned}{\mathcal {A}}_{i}=(\alpha _{i}/\omega )^{2}A_{i},\quad {\mathcal {B}}_{i}=(\beta _{i}/\omega )^{2}B_{i},\end{aligned}}$


(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 Pwave is in the liquid, the boundary conditions require the continuity of $w$ and $\sigma _{zz}$ and that $\sigma _{xz}$ in the solid vanish at $z=0$. omitting the factor $e^{\mathrm {j} \omega t}$, we have from equations (3.1b,c,d,e):
${\begin{aligned}\phi _{1}&={\mathcal {A}}_{0}e^{\mathrm {j} \omega \zeta _{0}}+{\mathcal {A}}_{1}e^{\mathrm {j} \omega \zeta _{1}},&\quad \chi _{1}&=0,\\\phi _{2}&=\qquad \qquad {\mathcal {A}}_{2}e^{\mathrm {j} \omega \zeta _{2}},&\quad \chi _{2}&=B_{2}e^{\mathrm {j} \omega \zeta _{2}^{\prime }}.\end{aligned}}$
Continuity of $w=\phi _{z}\chi _{x}$ yields the first equation [note equation (3.2g)]:
${\begin{aligned}\left({\mathcal {A}}_{0}+{\mathcal {A}}_{1}\right)\cot \theta _{1}={\mathcal {A}}_{2}\cot \theta _{2}{\mathcal {B}}_{2}.\end{aligned}}$
Continuity of normal stress requires that $\lambda \nabla ^{2}\phi +2\mu \left(\phi _{z}\chi _{xz}\right)$ be continuous; since $\mu =0$ in the liquid, this results in
${\begin{aligned}\lambda _{1}\left({\mathcal {A}}_{0}+{\mathcal {A}}_{1}\right)\left(1+\cot ^{2}\theta _{1}\right)=\lambda _{2}{\mathcal {A}}_{2}\left(1+\cot ^{2}\theta _{2}\right)+2\mu _{2}\left({\mathcal {A}}_{2}\cot ^{2}\theta _{2}+{\mathcal {B}}_{2}\cot \delta _{2}\right).\end{aligned}}$
Using equations (9,6) and (9,7) in Table 2.2a, also equation (3.1a), we have
${\begin{aligned}\left(\rho _{1}\alpha _{1}^{2}/\sin ^{2}\theta _{1}\right)\left({\mathcal {A}}_{0}+{\mathcal {A}}_{1}\right)&=\left(\lambda _{2}+2\mu _{2}\right)A_{2}\left(1+\cot ^{2}\theta _{2}\right)2\mu _{2}{\mathcal {A}}_{2}+2\mu _{2}{\mathcal {B}}_{2}\cot \delta _{2}\\&=\left(\rho _{2}\alpha _{2}^{2}/\sin ^{2}\theta _{2}2\mu _{2}\right){\mathcal {A}}_{2}+2\mu _{2}{\mathcal {B}}_{2}\cot \delta _{2}\\&=\left(\rho _{2}\beta _{2}^{2}/\sin ^{2}\delta _{2}2\mu _{2}\right){\mathcal {A}}_{2}+2\mu _{2}{\mathcal {B}}_{2}\cot \delta _{2}\\&=\mu _{2}\left(1/\sin ^{2}\delta _{2}2\right){\mathcal {A}}_{2}+2\mu _{2}{\mathcal {B}}_{2}\cot \delta _{2}\\&=\mu _{2}\left(\cot ^{2}\delta _{2}1\right){\mathcal {A}}_{2}+2\mu _{2}{\mathcal {B}}_{2}\cot \delta _{2}.\end{aligned}}$
Continuity of tangential stress $\sigma _{xzz}$ requires that $\left(2\phi _{xz}+\chi _{zz}\chi _{xx}\right)=0$ in the solid when $z=0$, so
${\begin{aligned}2{\mathcal {A}}_{2}\cot \theta _{2}+{\mathcal {B}}_{2}\left(\cot ^{2}\delta _{2}1\right)=0.\end{aligned}}$
Using the coefficients in equations (3.3g,h,i,j), the results become
${\begin{aligned}a_{1}{\mathcal {A}}_{1}+a_{2}{\mathcal {A}}_{2}+{\mathcal {B}}_{2}&=a_{1}{\mathcal {A}}_{0},\\\left(\rho _{1}p^{2}\right){\mathcal {A}}_{1}\mu _{2}c_{2}{\mathcal {A}}_{2}2\mu _{2}b_{2}{\mathcal {B}}_{2}&=\left(\rho _{1}/p^{2}\right){\mathcal {A}}_{0},\\2a_{2}{\mathcal {A}}_{2}c_{2}{\mathcal {B}}_{2}&=0.\end{aligned}}$
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$\ell$) to change the coefficients in Knott’s equations to Zoeppritz’s coeffiicients. Using this latter method, we have ${\mathcal {A}}_{i}=(\omega /\alpha _{i})^{2}A_{i}$, ${\mathcal {B}}_{2}=(\omega /\beta _{2})^{2}B_{2}$. Substituting these, we get the Zoeppritz equations for a liquidsolid interface:
${\begin{aligned}A_{1}\cos \theta _{1}+A_{2}\cos \theta _{2}+B_{2}\sin \delta _{2}&=A_{0}\cos \theta _{1},\\A_{1}Z_{1}A_{2}Z_{2}\cos 2\delta _{2}B_{2}W_{2}\sin 2\delta _{2}&=A_{0}Z_{1},\\A_{2}\left(\beta _{2}/\alpha _{2}\right)\sin 2\theta _{2}B_{2}\cos 2\delta _{2}&=0.\end{aligned}}$
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 $B_{2}=0$ in the liquid, $w$ and $\sigma _{zz}$ are continuous, and $\sigma _{zx}=0$ in the solid at $z=0$.
Equation (3.2e) gives for the normal displacement
${\begin{aligned}\left(A_{0}+A_{1}\right)\cos \theta _{1}B_{1}\sin \delta _{1}=A_{2}\cos \theta _{2}.\end{aligned}}$
The continuity of normal stress is expressed in equation (3.2h), which now becomes
${\begin{aligned}\left(A_{0}+A_{1}\right)Z_{1}\cos 2\delta _{1}B_{1}W_{1}\sin 2\delta _{1}=A_{2}Z_{2}\cos 2\delta _{2}.\end{aligned}}$
Finally, $\sigma _{xz}=0$ at $z=0$ and equation (3.2i) becomes
${\begin{aligned}\left(A_{0}+A_{1}\right)\left(\beta _{1}/\alpha _{1}\right)\sin 2\theta _{1}+B_{1}\cos 2\delta _{1}=0.\end{aligned}}$
Using equation (3.3$\ell$), we get the equivalent Knott’s equations:
${\begin{aligned}\left({\mathcal {A}}_{0}+{\mathcal {A}}_{1}\right)\cot \delta _{1}{\mathcal {B}}_{1}&={\mathcal {A}}_{2}\cot \theta _{2},\\\left({\mathcal {A}}_{0}+{\mathcal {A}}_{1}\right)\left(\lambda _{1}+2\mu _{1}\right)\cot ^{2}\theta _{1}{\mathcal {B}}_{1}\cot \delta _{1}&=\lambda _{2}{\mathcal {A}}_{2}\left(1+\cot ^{2}\theta _{2}\right),\\2\left({\mathcal {A}}_{0}+{\mathcal {A}}_{1}\right)\cot \theta _{1}+{\mathcal {B}}_{1}(\cot ^{2}\delta _{1}1)&=0.\end{aligned}}$
3.3b Calculate the amplitudes of the reflected and refracted Pand Swaves when an incident Pwave strikes the interface from a water layer $(\alpha =1500m/s$, $\beta =0$, $\rho =1.00$ g/cm$^{3}$) at $20^{\circ }$ when the seafloor is (i) “soft” $(\alpha =2000m/s$, $\beta =1000$ m/s, $\rho =2.00$ g/cm$^{3})$, and (ii) “hard” $(\alpha =4000m/s$, $\beta =2500m/s$, $\rho =2.50$ g/cm$^{3}$).
Solution
i) Where the seafloor is “soft” and the Pwave is incident in the water, we have:
${\begin{aligned}\sin \theta _{2}&=\left(2.00/1.50\right)\sin 20^{\circ },&\theta _{2}&=27.1^{\circ };2\theta _{2}=54.2^{\circ },\\\cos \theta _{2}&=0.890,&\sin 2\theta _{2}&=0.811;\\\sin \delta _{2}&=\left(1.00/1.50\right)\sin 20^{\circ }=0.228,&\delta _{2}&=13.2^{\circ },2\delta _{2}=26.4^{\circ },\\\sin 2\delta _{2}&=0.445,&\cos 2\delta _{2}&=0.896;\\\cos \theta _{1}&=0.940,\qquad \qquad Z_{1}=1.500,&Z_{2}&=4.000,W_{2}=2.000.\end{aligned}}$
Thus Zoeppritz’s equations become
${\begin{aligned}0.940A_{1}+0.890A_{2}+0.228B_{2}&=0.940A_{0},\\1.500A_{1}3.584A_{2}0.890B_{2}&=1.500A_{0},\\0.406A_{2}0.896B_{2}&=0.\end{aligned}}$
The solution is: $A_{1}/A_{0}=0.431$, $A_{2}/A_{0}=0.539$, $B_{2}/A_{0}=0.244$.
ii) When the seafloor is “hard”:
${\begin{aligned}\sin \theta _{2}&=\left(4.00/1.50\right)\sin 20^{\circ },&\theta _{2}&=65.8^{\circ },2\theta _{2}=131.6^{\circ },\\\cos \theta _{2}&=0.410;&\sin 2\theta _{2}&=0.748;\\\sin \delta _{2}&=\left(2.50/1.50\right)\sin 20^{\circ }=0.570,&\delta _{2}&=34.8^{\circ },2\delta _{2}=69.6^{\circ },\\\sin 2\delta _{2}&=0.937,&\cos 2\delta _{2}&=0.349;\\\cos \theta _{1}&=0.940,\qquad \qquad Z_{1}=1.50,&Z_{2}&=10.0,W_{2}=6.25.\end{aligned}}$
The equations are
${\begin{aligned}0.94A_{1}+0.41A_{2}+0.57B_{2}&=0.94A_{0},\\1.50A_{1}3.49A_{2}5.86B_{2}&=1.50A_{0},\\0.468A_{2}0.349B_{2}&=0.\end{aligned}}$
The solution is $A_{1}/A_{0}=0.716$, $A_{2}/A_{0}=0.227$, $B_{2}/A_{0}=0.30$.
3.3c Repeat part (b) for an angle of incidence of $30^{\circ }$.
Solution
i) For the “soft” bottom and $\theta _{1}=30^{\circ }$,
${\begin{aligned}\sin \theta _{2}&=\left(2.00/1.50\right)\sin 30^{\circ },&\theta _{2}&=41.8^{\circ },2\theta _{2}=83.6^{\circ },\\\cos \theta _{2}&=0.745,&\sin 2\theta _{2}&=0.994;\\\sin \delta _{2}&=\left(1.00/1.50\right)\sin 30^{\circ }=0.333,&\delta _{2}&=19.5^{\circ };\\\sin 2\delta _{2}&=0.629,&\cos 2\delta _{2}&=0.777.\end{aligned}}$
The equations are
${\begin{aligned}0.866A_{1}+0.745A_{2}+0.333B_{2}&=0.866A_{0},\\1.50A_{1}3.11A_{2}1.26B_{2}&=1.50A_{0},\\0.621A_{2}0.778B_{2}&=0.\end{aligned}}$
The solution is $A_{1}/A_{0}=0.40$, $A_{2}/A_{0}=0.51$, $B_{2}/A_{0}=0.41$.
ii) For the “hard” bottom, $\sin \theta _{2}=(4.00/1.50)\sin 30^{\circ }=1.33>1$, so total reflection occurs.
The results are summarized in Table 3.3a. The table shows that $A_{1}$ and $A_{2}$ depend mainly on the hardness of the bottom and only moderately on $\theta _{1}$. However, $B_{2}$ depends more on the angle of incidence than on the hardness.
Table 3.3a Reflected/transmitted amplitudes for soft/hard bottoms.
Bottom

$\theta _{1}$

$A_{1}/A_{0}$

$A_{2}/A_{0}$

$B_{2}/A_{0}$

Soft

$20^{\circ }$

0.431

0.539

0.244


$30^{\circ }$

0.403

0.512

0.408

Hard

$20^{\circ }$

0.716

0.227

0.304

3.4 Zoeppritz’s equations for incident SV and SHwaves
3.4a Derive Zoeppritz’s equations for an SVwave incident on a solid/solid interface.
Solution
Figure 3.1a defines the positive directions of displacements except that the incident Pwave is replaced by an incident SVwave 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
${\begin{aligned}A_{1}\cos \theta _{1}B_{1}\sin \delta _{1}+A_{2}\cos \theta _{2}+B_{2}\sin \delta _{2}&=B_{0}\sin \delta _{1},\\A_{1}\sin \theta _{1}+B_{1}\cos \delta _{1}A_{2}\sin \theta _{2}+B_{2}\cos \delta _{2}&=B_{0}\cos _{l},\\A_{1}Z_{1}\cos 2\delta _{1}B_{1}W_{1}\sin 2\delta _{1}A_{2}Z_{2}\cos 2\delta _{2}B_{2}W_{2}\sin 2\delta _{2}&=B_{0}W_{1}\sin 2\delta _{1},\\\left(\beta _{1}/\alpha _{1}\right)A_{1}W_{1}\sin 2\theta _{1}+B_{1}W_{1}\cos 2\delta _{1}+\left(\beta _{2}/\alpha _{2}\right)A_{2}W_{2}\sin 2\theta _{2}&B_{2}W_{2}\cos 2\delta _{2}\\&=B_{0}W_{1}\cos 2\delta _{1}.\end{aligned}}$
3.4b Derive the Zoeppritz equations for an incident SHwave.
Solution
For an SHwave 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)]
${\begin{aligned}v_{1}&=C_{0}e^{\mathrm {j} \omega p\left(xz\cot \delta _{1}\right)}+C_{1}e^{\mathrm {j} \omega p\left(x+z\cot \delta _{1}\right)},\\v_{2}&=C_{2}e^{\mathrm {j} \omega p\left(xz\cot \delta _{2}\right)}.\end{aligned}}$
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 Pwaves is important in SHwave studies.
3.5 Reinforcement depth in marine recording
3.5a For a source at a depth $h$, show that the maximum amplitude of a downgoing incident wave and its reflection at the surface of the sea occurs at the depth $\lambda /\left(4\cos \theta \right)$, where $\theta$ is the angle of incidence, by expressing the pressure $P$ in the form used in equations (3.1b,d) and applying appropriate boundary conditions.
Solution
Since the interface is liquid/vacuum, only two waves exist, the incident and reflected Pwaves. Taking the zaxis positive downward, we take ${\mathcal {P}}$ in the form
${\begin{aligned}{\mathcal {P}}=A_{0}e^{\mathrm {j} \omega p\left(x+z\cot \theta \right)}+A_{1}e^{\mathrm {j} \omega p\left(xz\cot \theta \right)}.\end{aligned}}$
There is only one boundary condition, namely that ${\mathcal {P}}=0$ at $z=0$. This gives $A_{1}=A_{0}$ Using Euler’s formulas (see Sheriff and Geldart, 1995, problem 15.12a), we get
${\begin{aligned}{\mathcal {P}}&=A_{0}e^{\mathrm {j} \omega px}\left(e^{\mathrm {j} \omega pz\cot \theta }e^{\mathrm {j} \omega pz\cot \theta }\right)\\&=2\mathrm {j} A_{0}e^{\mathrm {j} \omega \left(pxt\right)}\sin \left(\omega pz\cot \theta \right)\end{aligned}}$
upon inserting the time factor. The amplitude of the combined incident and reflected waves is
${\begin{aligned}2A_{0}\sin[\left(\omega pz\cot \theta \right]=2A_{0}\sin \left[\left(\omega z/\alpha \right)\cos \theta \right].\end{aligned}}$
It is a maximum when $\left(\omega z/\alpha \right)\cos \theta =\pi /2$, that is, when
${\begin{aligned}z=\left(\pi /2\right)\alpha /\left(\omega \cos \theta \right)=\lambda /\left(4\cos \theta \right).\end{aligned}}$
3.6 Complex coefficient of reflection
3.6a Using the expression $\psi =Ae^{\mathrm {j} \omega \left(r/Vt\right)}$ to represent a plane wave incident on a plane interface, show that a complex coefficient of reflection,
${\begin{aligned}R=a+\mathrm {j} b,\quad a^{2}+b^{2}<1,\end{aligned}}$
R [defined by equation (3.6a) below] corresponds to a reduction in amplitude by the factor $(a^{2}+b^{2})^{1/2}$ and an advance in phase by $\tan ^{1}\left(b/a\right)$.
Background
When a plane Pwave is incident perpendicularly on a plane interface, the tangential displacements and stresses vanish, so equations (3.2f,i) are not constraining and we are left with equations (3.2e,h). Moreover, $\delta _{1}=0=\delta _{2}$, so the equations reduce to
${\begin{aligned}A_{1}+A_{2}&=A_{0},\\Z_{1}A_{1}Z_{2}A_{2}&=Z_{1}A_{0},\end{aligned}}$
$Z_{1}$, $Z_{2}$, being impedances (see problem 3.2). The solution of these equations is
${\begin{aligned}R&={\frac {A_{1}}{A_{0}}}={\frac {\rho _{2}\alpha _{2}\rho _{1}\alpha _{1}}{\rho _{2}\alpha _{2}+\rho _{1}\alpha _{1}}}={\frac {Z_{2}Z_{1}}{Z_{2}+Z_{1}}},\end{aligned}}$


(3.6a)

${\begin{aligned}T&={\frac {A_{2}}{A_{0}}}={\frac {2\rho _{1}\alpha _{1}}{\rho _{2}\alpha _{2}+\rho _{1}\alpha _{1}}}={\frac {2Z_{1}}{Z_{2}+Z_{1}}},\end{aligned}}$


(3.6b)

$R$ and $T$ being the coefficient of reflection and coefficient of transmission, respectively. Although equations (3.6a,b) hold only for normal incidence, the definitions $R=A_{1}/A_{0}$ and $T=A_{2}/A_{0}$ are valid for all angles of incidence. A negative value of $R$ means that $A_{1}$ in equation (3.6a) is opposite in sign to $A_{0}$. Since $e^{\mathrm {j} \pi }=1$, the minus sign is equivalent to adding $\pi$ to the phase of the waveform in part (a), that is, reversing the phase. Note that (except for phase reversal) $R$ is independent of the direction of incidence on the interface; however, the magnitude of $T$ depends upon this direction, and when necessary, we shall write $T\downarrow$ and $T\uparrow$ to distinguish between the two values. Note the following relations:
${\begin{aligned}R+T\downarrow =1,\quad T\uparrow +T\downarrow =2,\quad T\uparrow T\downarrow =E_{T},\end{aligned}}$


(3.6c)

where $E_{T}$ is the fraction of energy transmitted as defined in equation (3.7a).
Euler’s formulas (see Sheriff and Geldart, 1995, p. 564) express $\sin x$ and $\cos x$ as
${\begin{aligned}\sin x=\left(e^{\mathrm {j} x}e^{\mathrm {j} x}\right)/2\mathrm {j} ,\quad \cos x=\left(e^{\mathrm {j} x}+e^{\mathrm {j} x}\right)/2.\end{aligned}}$


(3.6d)

The hyperbolic sine and cosine are defined by the relations
${\begin{aligned}\sinh x=\left(e^{x}e^{x}\right)/2,\quad \cosh x=\left(e^{x}+e^{x}\right)/2.\end{aligned}}$


(3.6e)

Solution
Writing $\psi 0=A_{0}e^{\mathrm {j} \omega \left(r/Vt\right)}=A_{0}e^{\mathrm {j} \left(\kappa r\omega t\right)}$, we have
${\begin{aligned}\psi _{1}=R\psi _{0}=\left(a+\mathrm {j} b\right)A_{0}e^{\left(\kappa r\omega t\right)}.\end{aligned}}$
But $\left(a+\mathrm {j} b\right)=(a^{2}+b^{2})^{1/2}e^{\mathrm {j} \phi }$ where $\tan \phi =b/a$ (see Sheriff and Geldart, 1995, section 15.1.5), so
${\begin{aligned}\psi _{1}=(a^{2}+b^{2})^{1/2}A_{0}e^{\mathrm {j} (kr\omega t+\phi )}.\end{aligned}}$
Since $(a^{2}+b^{2})^{1/2}<1$, the amplitude is reduced by this factor and the phase is advanced by $\phi$
3.6b Show that an imaginary angle of refraction $\theta _{2}$ (see Figure 3.1b) in equations (3.2e,f,h,i) leads to a complex value of $R$ and hence to phase shifts.
Solution
Let $\theta _{2}=\pi /2\mathrm {j} \theta$, where $\theta$ is real. Then, using equations (3.6d,e), we have
${\begin{aligned}\sin \theta _{2}=\cos \left(\mathrm {j} \theta \right)=\cosh \theta ,\qquad \cos \theta _{2}=\sin \left(\mathrm {j} \theta \right)=\mathrm {j} \sinh \theta ,\end{aligned}}$
hence some of the coefficients in equations (3.2e,f,h,i) are imaginary and $R$ and $T$ will in general be complex, so phase shifts will occur.
3.6c Show that when $R$ is negative, $A_{2}>A_{0}$.
Solution
If $R$ is negative, $Z_{1}>Z_{2}$ in equation (3.6a), so $2Z_{1}>\left(Z_{1}+Z_{2}\right)$. Therefore, from equation (3.6b) we see that $T>1$, and since $T=A_{2}/A_{0}$, $A_{2}>A_{0}$.
3.7 Reflection and transmission coefficients
3.7a Calculate the reflection and transmission coefficients, $R$ and $T$ of equations (3.6a,b), for a sandstone/shale interface (for a wave incident from the sandstone) for the following:
 $V_{ss}=2.43$ km/s, $V_{sh}=2.02$ km/s,
$\rho _{ss}=2.08$ g/cm$^{3}$, and $\rho _{sh}=2.23$ g/cm$^{3}$;
 $V_{ss}=3.35$ km/s, $V_{sh}=3.14$ km/s,
$\rho _{ss}=2.21$ g/cm$^{3}$, and $\rho _{sh}=2.52$ g/cm$^{3}$.
 What are the corresponding values in nepers and decibels?
Background
A wave has kinetic energy due to the velocity of the medium and potential energy due to the strains in the medium. For a harmonic wave $\psi =A\sin \omega t$, the particle velocity is $\partial \psi /\partial t=\omega A\cos \omega t$ and the kinetic energy/unit volume is ${\frac {1}{2}}\rho \omega ^{2}A^{2}\cos ^{2}\omega t$, $\rho$ being the density. As the wave progresses, the energy changes back and forth between kinetic and potential. When the potential energy is zero, the kinetic energy is a maximum and therefore equals the total energy. The maximum value comes when $\cos ^{2}\omega t=+1$, so the total energy/unit volume${}=E={\frac {1}{2}}\rho \omega ^{2}A^{2}$, where $E$ is the energy density (problem 2.3).
The coefficients $R$ and $T$ in equations (3.6a,b) give ratios of the relative amplitudes of the reflected and transmitted waves. We denote the fractions of the incident energy that are reflected and transmitted by $E_{R}$ and $E_{T}$; “energy” as used here denotes the amount of energy flowing through a unit area normal to the wave direction per unit time (the intensity). The energy density (i.e., energy/unit volume) is given by equation (3.3k). The energy flowing through a unit area normal to the wave direction per unit time is equal to the energy density times the velocity, that is, $\left({\frac {1}{2}}\rho \omega ^{2}A^{2}\right)\alpha$ for a Pwave. Therefore
${\begin{aligned}E_{R}=R^{2},\qquad E_{T}={\frac {\rho _{2}\alpha _{2}\omega ^{2}A_{2}^{2}}{\rho _{1}\alpha _{1}\omega ^{2}A_{0}^{2}}}=\left({\frac {Z_{2}}{Z_{1}}}\right)T^{2}={\frac {4Z_{1}Z_{2}}{(Z_{1}+Z_{2})^{2}}}.\end{aligned}}$


(3.7a)

The coefficients $E_{R}$ and $E_{T}$ are sometimes referred to as reflection and transmission energy coefficients to distinguish them from $R$ and $T$. Note that both $E_{R}$ and $E_{T}$ are independent of the direction of travel through the interface.
Solution
i) Using equations (3.6a,b), we have
${\begin{aligned}Z_{ss}=2.08\times 2.43=5.05,\qquad Z_{sh}=2.23\times 2.02=4.50\end{aligned}}$
(where the units are g. km/cm^{3}. s). Then,
${\begin{aligned}R=\left(4.505.05\right)/\left(4.50+5.05\right)=0.55/9.55=0.058.\end{aligned}}$
(The minus sign denotes a phase reversal; see problem 3.6.) Because the incident wave is in the sandstone, we have
${\begin{aligned}T=2\times 5.05/9.55=1.06.\end{aligned}}$
Note that the amplitude of the transmitted wave is larger than that of the incident wave when $R$ is negative (see problem 3.6c).
ii) $Z_{ss}=2.21\times 3.35=7.40,\ Z_{sh}=2.52\times 3.14=7.91$.
Then,
${\begin{aligned}R&=\left(7.917.40\right)/\left(7.91+7.40\right)=0.51/15.3=0.033,\\T&=2\times 7.40/15.3=0.967.\end{aligned}}$
iii) From problem 2.17, we have nepers ln(amplitude ratio) and 1 neper ${}=8.686$ dB. Therefore,
for (i),
${\begin{aligned}R&=\ln \left(0.058\right)=2.8\ \mathrm {nepers} =24\ \mathrm {dB} ,\\T&=\ln \left(1.06\right)=0.058\ \mathrm {nepers} \ =0.51\ \mathrm {dB} .\end{aligned}}$
For (ii),
${\begin{aligned}R&=\ln \left(0.033\right)=3.4\ \mathrm {nepers} \ =30\ \mathrm {dB} ,\\T&=\ln \left(0.967\right)=0.034\ \mathrm {nepers} \ =0.29\ \mathrm {dB} .\end{aligned}}$
Negative values of nepers and $dB$ merely mean that the values are less than unity.
3.7b Calculate the energy coefficients $E_{R}$ and $E_{T}$ for cases (i) and (ii) in part (a).
Solution
We use equation (3.7a) to get,
for (i),
${\begin{aligned}E_{R}&=R^{2}=(0.058)^{2}=0.0034,\\E_{T}&=\left(Z_{2}/Z_{1}\right)T^{2}=\left(4.50/5.05\right)\times 1.06^{2}=1.00.\end{aligned}}$
For (ii),
${\begin{aligned}E_{R}&=0.033^{2}=0.001,\\E_{T}&=(\left(7.91/7.40\right)\times 0.967^{2}=1.00.\end{aligned}}$
Note that $E_{R}+E_{T}=1$ (within the accuracy of the calculations).
3.8 Amplitude/energy of reflections and multiples
3.8a Assume horizontal layering (as shown in Figure 3.8a), a source just below interface $A$, and a geophone at the surface. Calculate (ignoring absorption and divergence) the relative amplitudes and energy densities of the primary reflections from $B$ and $C$ and the multiples $BSA$, $BAB$, and $BSB$ (where the letters denote the interfaces involved). Compare traveltimes, amplitudes, and energy densities of these five events for normal incidence.
Background
Multiples are events that have been reflected more than once. They are generally weak because the energy decreases at each reflection, but where the reflection coefficients are large, multiples may be strong enough to cause problems. Multiples are of two kinds as shown in Figure 3.8b: longpath multiples which arrive long enough after the primary reflection that they appear as separate events, and shortpath multiples which arrive so soon after the primary wave that they add to it and change its shape. The most important shortpath multiples are two in number: (i) ghosts (Figure 3.8b) where part of the energy leaving the source travels upward and is reflected downward either at the base of the LVL (see problem 4.16) or at the surface, (ii) pegleg multiples resulting from the addition to a primary reflection of energy reflected from both the top and bottom of a thin bed, either on the way to or on the way back from the principal reflecting horizon. Shortpath nearsurface multiples are also called ghosts and longpath interformational multiples are also called pegleg multiples. A notable example of the latter occurs in marine work when wave energy bounces back and forth within the water layer.
The energy density of a wave (see problem 3.7) decreases continuously as the wave progresses because of two factors: absorption and spreading or divergence. The energy density is proportional to the square of the amplitude, so both effects are usually expressed in terms of the decrease in amplitude with distance.
Figure 3.8a. A layered model.
Figure 3.8b. Types of multiples.
Absorption causes the amplitude to decrease exponentially, the relation being $A=$ $A_{0}e^{\eta x}$ where the amplitude decreases from $A_{0}$ to $A$ over a distance $x$; the absorption coefficient $\eta$ is often expressed in terms of per wavelength, $\lambda .$
For a point source in an infinite constantvelocity medium, divergence causes the energy density to decrease inversely as the square of the distance from the source, the amplitude decreasing inversely as the first power of the distance from the source.
Nepers and decibels are defined in problem 2.17.
Solution
We first calculate the impedances $Z_{i}$ for each layer, the coefficients of reflection and downgoing and upgoing transmission $R,$ $T\downarrow$, $T\uparrow$ (see problem 3.6), and the reflected and transmitted energy coefficients, $E_{R}$ and $E_{T}$, for each interface. The results are shown in Table 3.8a.
Table 3.8a. Reflection and transmission coefficients.
Interface

$Z$

$R^{*}$

$T\downarrow$

$T\uparrow$

$E_{R}$

$E_{T}$

S


1.000

0.000

0.000

1.000

0.000

Layer 1

0.870






${\textit {A}}$


0.733

0.267

1.733

0.537

0.463

Layer 2

5.640






${\textit {B}}$


0.207

0.793

1.207

0.043

0.957

Layer 3

8.576






${\textit {C}}$


0.034

0.966

1.034

0.001

0.999

Layer 4

9.180






* Signs are for incidence from above.

Assuming unit amplitude and unit energy density for the downgoing wave incident on interface $B$ and neglecting absorption and divergence, we arrive at the following values:
Reflection $B$:
Amplitude of reflection $B=R_{B}T_{A}\uparrow =0.207\times 1.733=0.359.$
Energy density $=E_{RB}E_{TA}=0.043\times 0.463=0.020.$
Arrival time $t_{B}=2\times 0.600/2.400+0.010/0.600=0.517s.$
Reflection $C$
Amplitude $=T_{B}\downarrow R_{C}T_{B}\uparrow T_{A}\uparrow =0.793\times 0.034\times 1.207\times 1.733=0.056.$
Energy density $=E_{TB}^{2}E_{RC}E_{TA}=0,0004.$
Arrival time $t_{C}=t_{B}+2\times 0.800/3.20=1.017s.$
Multiple BSA
Amplitude $=R_{B}T_{A}\uparrow \left(R_{S}\right)R_{A}=0.207\times 1.733\times 1.000\times 0.733=0.263.$
Energy density $=0.043\times 0.463\times 1.000\times 0.537=0.011.$
Arrival time $t_{BSA}=2\times 0.600/2.40+3\times 0.010/0.600=0.550s.$
Multiple BAB
Amplitude $=(R_{B})^{2}\left(R_{A}\right)T_{A}\uparrow =0.207^{2}\times \left(0.733\right)\times 1.733=0.0544.$
Energy density $=0.043^{2}\times 0.537\times 0.463=0.0005.$
Arrival time $t_{BAB}=4\times 0.600/2.40+0.010/0.600=1.017s.$
Multiple BSB
Amplitude $=(R_{B})^{2}(T_{A}\uparrow )^{2}T_{A}\downarrow \left(R_{S}\right)=0.0344.$
Energy density $=(E_{RB})^{2}(E_{TA})^{3}\left(E_{RS}\right)=0.0002.$
Arrival time $t_{BSB}=4\times 0.600/2.400+3\times 0.010/0.600=1.050.$
The results are summarized in Table 3.8b.
BSA arrives 33 ms after $B$ (one period for a 33Hz wave) with reversed polarity and about 75% of the amplitude and 50% of the energy of $B$, so BSA will significantly alter the waveshape of B. BSA involves an extra bounce at the surface and is a type of ghost whose effect is mainly that of changing waveshape rather than showing up as a distinct event.
$C$ and BAB arrive simultaneously with opposite polarities, $C$ being slightly stronger than BAB; the multiple will obscure and significantly alter the waveshape of the primary reflection.
Table 3.8b. Amplitude/energy density of primary/multiple reflections.
Event

$t$

Amplitude

$20\log \left(A/A_{B}\right)$

Energy

$B$

$0.517s$

$0.359$

$0dB$

$0.020$

$BSA$

$0.550$

$0.263$

$2.7$

$0.011$

$C$

$1.017$

$0.056$

$16.1$

$0.0004$

$BAB$

$1.017$

$0.054$

$$ 