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

$$
(1+x)^{\varepsilon} \approx 1+\varepsilon x \text {, when }|x| \ll 1 \text {. }
$$

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.

Solution:
B.1
Seismic waves in this problem can be treated by using ray theory. Namely, their propagation is described by the Snell's law of refraction

$$
n(0) \sin \theta_{0}=n(z) \sin \theta,
$$

where the refractive index is

$$
n(z)=\frac{c}{v(z)}=\frac{c}{v_{0}\left(1+\frac{z}{z_{0}}\right)}
$$



<img_4564>

Figure 2

and $c$ denotes the seismic wave speed in a material with refractive index $n=1$. From the two equations above we have

$$
v_{0}\left(1+\frac{z}{z_{0}}\right) \sin \theta_{0}=v_{0} \sin \theta
$$

Method 1. Since this describes an arc of a circle, we have that at $\theta=\frac{\pi}{2}, z=R-R \sin \theta_{0}$ (fig. 2), giving

$$
\left(1+\frac{R-R \sin \theta_{0}}{z_{0}}\right) \sin \theta_{0}=1
$$

from where the circle radius $R=\frac{z_{0}}{\sin \theta_{0}}$. From simple geometry we get

$$
x_{1}\left(\theta_{0}\right)=2 R \cos \theta_{0}
$$

i.e. $A=2 z_{0}$ and $b=1$.

Method 2. Implicitly differentiating $v_{0}\left(1+\frac{z}{z_{0}}\right) \sin \theta_{0}=v_{0} \sin \theta$ gives

$$
\frac{\mathrm{d} z}{z_{0}} \sin \theta_{0}=\cos \theta \mathrm{d} \theta
$$

An infinitesimal ray path length $\mathrm{d} l$ is related to the change in the vertical coordinate via

$$
\mathrm{d} z=\mathrm{d} l \cos \theta
$$

giving

$$
\mathrm{d} l=\frac{z_{0}}{\sin \theta_{0}} \mathrm{~d} \theta
$$

This is an equation of an arc of a circle of radius $R=\frac{z_{0}}{\sin \theta_{0}}$

Alternatively, instead of considering an infinitesimal ray path length $d l$, one can obtain the answer by writing

$$
\cot \theta=\frac{\mathrm{d} z}{\mathrm{~d} x}=\frac{\mathrm{d} z}{\mathrm{~d} \theta} \frac{\mathrm{d} \theta}{\mathrm{d} x}
$$



The first derivative can be eliminated via Snell's law, leading to

$$
\cot \theta=\frac{z_{0} \cos \theta}{\sin \theta_{0}} \frac{\mathrm{d} \theta}{\mathrm{d} x}
$$

which can be integrated to get

$$
x_{1}=-\frac{z_{0}}{\sin \theta_{0}} \int_{\text {start }}^{\text {end }} \mathrm{d} \cos \theta=\frac{2 z_{0} \cos \theta_{0}}{\sin \theta_{0}}
$$

where we used Snell's law again to get that the ray has $\cos \theta=-\cos \theta_{0}$ at the point where it reaches the surface.

$x_{1}\left(\theta_{0}\right)=2 z_{0} \cot \theta_{0}$.

Final answer: $x_{1}(\theta_{0})=2 z_{0} \cot \theta_{0}$