Gone with the wind

Let us assume that the distance between the boat and the observer is $2x_0$ . The sound waves propagate on a parabolic orbit, $y= \frac{1}{2} a (x_0^2-x^2)$ . The slope of this function at $x=-x_0$ is the tangent of the angle of emission (relative to the horizontal). By taking the derivative we get $y’ = -ax$ and $\tan \alpha=ax_0$ .

First, let us look at this situation qualitatively. Obviously, the larger is $a$ , the longer is the arc of the parabola. Without the wind shear, the propagation time would be longer as well. With the wind shear, however, the time lost at the beginning and at the end of the travel may be compensated by the time gained at the top of the path. This is because at the beginning and at the end the horizontal component of the velocity is less than $c$ , but at the top of the path the propagation is helped by the wind. [We should also point out that the “real” path of the sound waves are not parabolic, but the parabola is the simplest function that exhibits the necessary properties, i.e. it points slightly upwards at the beginning and it curves back to the observer. For sound waves emitted at a few degrees upwards relative to the horizontal, this is going to be a really good approximation.]

In order to apply Fermat’s principle, we calculate the time of travel as

$T=\int dt = \int_{-x_0}^{x_0} \frac{1}{v_x} dx = \frac{1}{c}\int_{-x_0}^{x_0} \left( 1-\frac{\delta}{c}\right) dx =\frac{2x_0}{c} -\frac{1}{c^2}\int_{-x_0}^{x_0} \delta(x) dx$ .

Here we expressed the horizontal component of the velocity as $v_x=c+\delta$ , and we used a first order expansion, approximating $\frac {1}{v_x}= \frac {1}{c+\delta} \approx \frac{1}{c} \left( 1-\frac{\delta}{c}\right)$ . In order to apply Fermat’s principle, we have to investigate the dependence of the propagation time on the parameter $a$ ( or $\tan \alpha = ax_0$ .) Since the first term $\frac{2x_0}{c}$ does not depend on $a$ , we have to work on

$\Delta T = -\frac{1}{c^2}\int_{-x_0}^{x_0} \delta(x) dx$ ,

and look for its extremum in $a$ or make it independent of $a$ .

In order to find the function $\delta(x)$ we need two identities. First, the horizontal and vertical components of the velocity are related to the direction of propagation by

$\frac{v_y}{v_x}=\frac{dy}{dx}= y’=-ax$ .

Second, we need to implement the way the wind speed helps the propagation of the sound. The vector sum of the sound velocity $\vec c$ and the wind velocity $\vec {v_s}$ is the true velocity, $\vec v$ : $\vec{v}= \vec{c}+\vec{v_s}$ or $\vec{c}= \vec{v}-\vec{v_s}$ . Since the wind velocity is purely horizontal, Pythagoras’ theorem yields

$c^2= (v_x-v_s)^2+v_y^2 = (1+y’^2)v_x^2-2v_x v_s + v_s^2$ .

We express $v_x=c+\delta$ and we get

$(1+y’^2)(c+\delta)^2-2(c+\delta)v_s +v_s^2-c^2 = 0$ ,

or

$c^2+2 c \delta +\delta^2+ y’^2c^2+2c y’ \delta +y’2 \delta^2 - 2c v_s -2 \delta v_s +v_s^2 - c^2$ = 0 .

This looks terrible, but there are ways to simplify it. The two $c^2$ terms cancel each other. We know that $\delta<<c$ and $v_s<<c$ . At this point we drop all terms containing $\delta^2$ , and $v_s^2$ . We get a much simplified expression:

$\delta = v_s - \frac{1}{2} c y’^2 c$ .

Note that for $x=\pm x_0$ we get a negative $\delta = -\frac{1}{2} c y’^2$ (as expected) and for $x=0$ we get a positive $\delta = v_s$ (as expected). This shows that dropping the second order terms is OK, and we still have the correct physics represented in the equations. We plug in $v_s = v_0 (y/h_0)^2$ , $y= \frac{1}{2} a (x_0^2-x^2)$ and $y’=-ax$ . The integrand is an even function, so we can do the integration from $0$ to $x_0$ and double the result. With all of these, the correction to the propagation time is

$\Delta T = \frac {2}{c^2} \int_0^{x_0} \left[\frac {1}{2}c a^2 x^2 -v_0 \left( \frac{(1/2) a(x_0^2-x^2)}{h_0}\right)\right] dx$ .

The first part yields $\frac{1}{3}\frac{a^2x_0^3}{c}$ , the second part is $\frac {4}{15} \frac{v_0 a^2x_0^5}{c^2 h_0^2}$ . We get

$\Delta T = \frac {a^2}{c} \left( \frac {1}{3} x_0^3- \frac {4}{15} \frac{v_0 x_0^5}{c h_0^2}\right)$

This quantity is independent of $a$ when the term in the parenthesis is zero. (Note: It is somewhat of an accident, related to the quadratic form of the wind shear factor, that we got a result that has such a simple quadratic dependence on the parameter $a$ . If the wind shear function is not quadratic, we get a more complex function in $a$ , and Fermat’s principle would require to make the first derivative with respect to $a$ equal to zero.) Solving the equation we get

$x_0= h_0 \sqrt{\frac{15}{12} \frac{c}{v_0}}$ .

With the numbers given in the problem we get the range, $2 x_0 = 922m \approx 1$ km .

NOTES

1. For the readers who doubt that dropping the terms containing $\delta^2$ and $v_s^2$ is a good idea, I solved the problem without dropping those terms. In that case we have a quadratic equation for $\delta$ . In the two figures here I compare the solution of the quadratic equation to the approximate expression used above. For angles smaller than about $20 \deg$ there is not much difference between the two. Accordingly, for small angles our solution is good.

2. For larger angles several of the other assumptions we made are also violated (the parabolic path, the particular form of the wind shear), so it does not make any sense to push for more accuracy at large angles. But it is remarkable that in the range of angles from 0 to about $15 \deg$ there is a good focusing effect.

3. The focusing effect only works in the vertical plane. In the horizontal direction the sound waves spread out approximately uniformly. That is why the sound reaching the observer is much weaker than the sound on the boat.