In an ocean, described with $x-y$ Cartesian coordinates where $y=0$ on the sea bed, the sound speed varies linearly with $y$: $$v(y)=c+by$$

A textbook says that, if sound wave is emitted on the sea bed, at an angle $\theta$ upward, then the sound wave will travel a circular path, of radius $$R=\frac{c}{b\sin\theta}$$

However, my calculations do not agree with this statement.

Assume the path of sound is described by $y=f(x)$.

By Snell’s law, $$\frac{\sin \theta_{\text{incident}}}{v(y)}=\text{constant}=k$$

Since $\tan\theta_{\text{incident}}=f’$, $$kv=\frac{f’}{\sqrt{1+f’^2}}\implies f’^2=\frac{(kv)^2}{1-(kv)^2}\implies 1+f’^2=\frac1{1-(kv)^2}$$

Differentiating both sides w.r.t. $x$, $$2f’f’’= \frac{2k^2v}{(1-(kv)^2)^2}\frac{dv}{dx}= \frac{2k^2v}{(1-(kv)^2)^2}\frac{dv}{dy}\frac{dy}{dx}= \frac{2k^2vb}{(1-(kv)^2)^2}f’$$

Thus, $$f’’=\frac{k^2bv}{(1-(kv)^2)^2}$$

Calculating the radius of curvature: $$R=\frac{(1+f’^2)^{3/2}}{f’’}=\frac{\sqrt{1-(kv)^2}}{k^2bv}\ne \text{a constant}$$ Therefore the path cannot be a circle.

  1. Are my calculations correct?
  2. If yes, why is there a discrepancy between the textbook’s statement and my calculations?

I believe that your statement of Snell's Law is incorrect because you only take into account an incident angle. Furthermore, your $\theta$ is the angle that the tangent makes with the x-axis, which is implied when you say $\tan(\theta)=f'$, but the refractive index varies with $y$, this means that for the purposes of this law, the incident angle would be $\phi=\pi/2-\theta$. I think it would be more accurate to write something like this

$$ n(y)\sin(\phi) = n(y+\Delta y)\sin(\phi + \Delta \phi) $$

Of course, this is only true as $\Delta y \rightarrow 0$. I offer an alternative to solve this problem. Snell's law is a consequence of Fermat's Principle

the optical length of the path followed by light between two fixed points, A and B, is an extremum. The optical length is defined as the physical length multiplied by the refractive index of the material.

Wikipedia does a good job of explaining how to apply this principle, so you should be able to easily adapt it for the case of sound waves

$$L(x, y, y')= \frac{1}{v(y)}\sqrt{1+y'^2}$$

Then you can apply the Euler-Lagrange equation

$$ \frac{\partial L}{\partial y} - \frac{d}{dx}\frac{\partial L}{\partial {y'}} = 0 $$

If you go through this calculation, you'll identify the formula for the curvature, whose inverse results in

$$ R = \frac{v(y)}{b\sin(\theta)}$$

Here, I already accounted for the fact that $\phi$ and $\theta$ differ by $\pi/2$. Note that when $y=0$ and $\theta=\theta_0$, this equation results in the equation your textbook gave you. The radius of curvature need not be constant unless it is explicitly stated that it is a uniform circular motion.


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