classical mechanics - Why doesn't the wake angle of a boat obey the relation $tan(theta)=c/v$?

I set up the situation as a boat that leaves a trail of point like waves which expand out with radius depending on the speed $c$. Intuitively, I would naively expect the answer to be $$\tan(\theta)=\frac{c}{v},$$ because the outermost edges do not interfere destructively so make an angle $ct/vt$ where $v$ is the speed of the boat.

It is obviously a great deal more complicated than this, however the answer is also much stranger $\sin(\theta)=1/3$: see Ship Kelvin Wake at wikiwaves.

I have not yet grasped the level of calculus that this seems to require. All I'm asking is to pick apart any fundamental errors in derivation or assumptions in my answer.

Answer

The reason that your naive guess of $$ \tan(\theta)=\frac{c}{v} $$ fails is that deep-water waves are dispersive, or, in other words, the speed $c=c(\lambda)$ of the wave depends on the wavelength: longer waves are faster than shorter waves, a fact which is evident if you drop a stone into a sufficiently deep pond.

Now, the source for Kelvin wakes is the boat, which can mostly be considered as a point disturbance, and which certainly does not have a well-defined wavelength associated with it. The way you deal with this is via a Fourier transform (that's the calculus-looking maths in the page you linked to) which is just fancy language for saying

1. that every point disturbance can be understood as a superposition of regular wavetrains of well-defined wavelength, and


2. that once we've done that decomposition, we can propagate the individual waves on their own, which is usually much simpler, and then re-assemble them to form the disturbance's wake pattern at any later point.

The reason the Kelvin wake angle is independent of the velocity, over reasonable ranges, is that the boat's disturbance contains many different wavelengths, and the wave propagation picks out the wavelengths that travel at the correct speed to match the boat's motion. Those waves will interfere constructively and will form a set pattern, while on the other hand waves that are longer or shorter will be too fast or too slow, drift out of phase with the waves produced earlier, interfere destructively with themselves, and drop out of the race.

The rest is just math, but the essentials are now there: the calculation of the wake pattern is just a matter of taking all the component waves, moving at different speeds, and figuring out the only possible interference pattern that will be invariant with respect to the boat's motion at all times.

And, moreover, this has physical consequences that can be readily checked: going twice as fast is not going to make the wake angle smaller, but it will make the spatial scale of the pattern increase by a factor of $\sqrt{2}$.