Barkley and Doppler

Recently we’ve learned a bit about physical modeling of PDEs in my pattern formation class. Here is a video depicting Barkley turbulence, which is a type of reaction-diffusion system.

On an unrelated note, here are a few of my results from implementing a simulation of the Doppler effect. In our first example, a source is traveling at about Mach \( \frac{1}{10} \) horizontally from right to left across a distance of \(200\) meters. When it is directly in front of us, the distance between us and the source is \(10\) meters.

In the second example, the source travels twice as fast, twice as far, and we are twice as close. The shift is correspondingly more dramatic.

One of the more counterintuitive aspects of the Doppler effect is the situation in which the source is coming directly at us. We never really experience this in real life (unless we are actually getting run over by the source). In this situation, the source does not in fact slide continuously from a higher frequency to a lower frequency but rather it does a discrete jump. (This is because the Doppler effect depends on the projections of the velocity vectors of the source and the receiver onto the line connecting the two, so unless the source and receiver are traveling directly toward each other, these projections vary continuously. However, when the source and receiver travel directly toward each other, the projections are constant until they meet, at which point they discretely jump to a new value.) In the particular example I have implemented, the source is traveling at Mach \( \frac{1}{4} \). (Warning: this example might be somewhat uncomfortable to listen to!)

Notice that we heard what sounded like a major sixth! This is actually a pretty straightforward consequence of the mathematics behind the Doppler shift. The emitted frequency, \(f\), is distorted by a frequency factor of \( \frac{c}{c + v_s} \), where \(c\) is the speed of sound in the relevant medium and \(v_s\) is the velocity of the source approaching you, with the convention that it is negative as it approaches you and positive as it recedes. Hence, the frequency ratio between the approaching sound and the receding sound is given by \( \frac{c + |v_s|}{c – |v_s|} \). In particular, at Mach \( \frac{1}{4} \), this ratio is \(\frac{\frac{5}{4}}{\frac{3}{4}} = \frac{5}{3} \), which is a just major sixth. Other musical intervals are, of course, also easily obtainable — Mach \( \frac{1}{3} \) yields an octave, for instance. In general, the reader can verify with some basic algebra that in order to get the frequency ratio \(P:Q\), the source must travel at Mach \( \frac{P – Q}{P + Q} \).

You might have noticed the expression \( c – v_s \) in the denominator above and wondered about the situation in which we are traveling Mach \(1\) or faster, but that’s another story for a future post.