Subfield: Optics

Seismic waves in a stratified medium 

Suppose that a short earthquake happens at the surface of some planet. The seismic waves can be assumed to originate from a line source situated at $z=x=0$, where $x$ is the horizontal coordinate and $z$ is the depth below the surface (Fig. 3). The seismic wave source can be assumed to be much longer than any other length considered in this question.

As a result of the earthquake, a uniform flux of the so-called longitudinal $P$ waves is emitted along all the directions in the $x-z$ plane that have positive component along the $z$-axis. Since the wave theory in a solid is generally complicated, in this problem we neglect all the other waves emitted by the earthquake. The crust of the planet is stratified so that the P-wave speed $v$ depends on depth $z$ according to $v=v_{0}\left(1+z / z_{0}\right)$, where $v_{0}$ is the speed at the surface and $z_{0}$ is a known positive constant.

<img_4472>

Figure 3. Coordinate system used in part B.

The effects of the surface curvature of the planets can be neglected. You might need to use the formula

$$
(1+x)^{\varepsilon} \approx 1+\varepsilon x \text {, when }|x| \ll 1 \text {. }
$$
Context question:
B.1 Consider a single ray emitted by the earthquake that makes an initial angle $0<\theta_{0}<\pi / 2$ with the $z$-axis and travels in the $x-z$ plane. What is the horizontal coordinate $x_{1}\left(\theta_{0}\right) \neq 0$ at which this ray can be detected at the surface of the planet? It is known that the ray path is an arc of a circle. Write your answer in the form $x_{1}\left(\theta_{0}\right)=A \cot \left(b \theta_{0}\right)$, where $A$ and $b$ are constants to be found.
Context answer:
\boxed{$x_{1}(\theta_{0})=2 z_{0} \cot \theta_{0}$}


Extra Supplementary Reading Materials:

If you were unable to find $A$ and $b$, in the following questions you can use the result $x_{1}\left(\theta_{0}\right)=A \cot \left(b \theta_{0}\right)$ as given. Suppose that total energy per unit length of the source released as $\mathrm{P}$ waves into the crust during the earthquake is $E$. Assume that waves are completely absorbed when they reach the surface of the planet from below.
Context question:
B.2 Find how the energy density per unit area $\varepsilon(x)$ absorbed by the surface depends on the distance along the surface $x$.
Context answer:
\boxed{$\varepsilon(x)=\frac{2 E z_{0}}{\pi(4 z_{0}^{2}+x^{2})}$}


Extra Supplementary Reading Materials:

From now on, assume that the waves are instead fuylly reflected when reaching the surface. Imagine a device positioned at $z=x=0$ that has the same geometry as the previously considered earthquake source. The device is capable of emitting $P$ waves in a freely chosen angular distribution. We make the device emit a signal with a narrow range of emission angles. In particular, the initial angle the signal makes with the vertical belongs to the interval $\left[\theta_{0}-\frac{1}{2} \delta \theta_{0}, \theta_{0}+\frac{1}{2} \delta \theta_{0}\right]$, where $0<\theta_{0}<\pi / 2$, $\delta \theta_{0} \ll 1$ and $\delta \theta_{0} \ll \theta_{0}$.

Question: B.3 At what distance $x_{\max }$ along the surface from the source is the furthest point that the signal does not reach? Write your answer in terms of $\theta_{0}, \delta \theta_{0}$ and other constants given above.

Solution:
B.3
Define $x_{-}=x_{1}\left(\theta_{0}-\frac{\delta \theta_{0}}{2}\right)$ and $x_{+}=x_{1}\left(\theta_{0}+\frac{\delta \theta_{0}}{2}\right)$. To the leading order in $\delta \theta_{0}, x_{-} \approx x_{+} \approx x_{1}\left(\theta_{0}\right)$. With each reflection of the signal, the horizontal distance between the points where the edges of

the signal reflect increases by $\left|x_{+}-x_{-}\right|=x_{-}-x_{+}$. When moving along the positive $x$-axis, these zones get wider until they overlap. If this happens after $N$ reflections, then

$$
N \approx \frac{x_{1}\left(\theta_{0}\right)}{x_{-}-x_{+}}
$$

where the approximate sign tends to equality as $\delta \theta_{0} \rightarrow 0$.

The position where the zones start to overlap is at $x_{\max }=N x_{1}\left(\theta_{0}\right)$. Therefore,

$$
x_{\max }=\frac{x_{1}\left(\theta_{0}\right)^{2}}{x_{1}\left(\theta_{0}-\frac{\delta \theta_{0}}{2}\right)-x_{1}\left(\theta_{0}+\frac{\delta \theta_{0}}{2}\right)}
$$

Since $\delta \theta_{0} \ll \theta_{0}$, we can approximate

$$
x_{1}\left(\theta_{0}-\frac{\delta \theta_{0}}{2}\right)-x_{1}\left(\theta_{0}+\frac{\delta \theta_{0}}{2}\right) \approx-\frac{\mathrm{d} x_{1}\left(\theta_{0}\right)}{\mathrm{d} \theta_{0}} \delta \theta_{0}=\frac{A b}{\sin ^{2}\left(b \theta_{0}\right)} \delta \theta_{0}
$$

Combining the last two equations and substituting the $x_{1}\left(\theta_{0}\right)$ expression gives

$$
x_{\max }=\frac{A b \cos ^{2}\left(b \theta_{0}\right)}{\delta \theta_{0}}=\frac{2 z_{0} \cos ^{2} \theta_{0}}{\delta \theta_{0}}
$$


$x_{\max }=\frac{A b \cos ^{2}\left(b \theta_{0}\right)}{\delta \theta_{0}}=\frac{2 z_{0} \cos ^{2} \theta_{0}}{\delta \theta_{0}}$.

Final answer: $x_{\max }=\frac{2 z_{0} \cos ^{2} \theta_{0}}{\delta \theta_{0}}$