While I was walking my dog this morning, I passed over a canal filled with boats, barges, and kayaks all of different masses and moving at different speeds. I noticed that all of these vessels left behind wakes, and the waves of these wakes moved through the water at different rates, though my ability to make good observations was limited. This made me wonder: what determines the velocity of these waves?

I know that sound travels through water at 1,480 m/s or so, but surface waves travel at different rates (tsunamis travel faster than the ripples I make in my bowl of soup when I blow on it). In particular, how is the velocity of wake waves determined?


1 Answer 1


The complete answer to that question is an open problem in fluid mechanics, as exact closed form solutions to the irrotational surface gravity water wave equations are unknown. However, under certain asymptotic approximations, we can estimate the speed of these waves.

Irrotational inviscid surface waves are governed by Laplace's equation, i.e.

$$\nabla^2 \phi = 0$$

where $\phi$ is the velocity potential. This governing equation, together with the boundary conditions

$$\phi_t+\frac{1}{2}(\nabla \phi)^2 +gz = 0$$

$$\eta_t +(\nabla \phi)\cdot (\nabla \eta) = \phi_z$$

where $\eta$ is the free surface displacement, and these equations are evaluated at the free surface, i.e. $z = \eta$, and the bottom boundary condition $\phi_z = 0$ at $z=-h$, with h the depth of water, constitute the complete set of equations. Also, here $g$ is the acceleration due to gravity.

The governing equation is linear, i.e. Laplace's equations, but the BC are nonlinear, and furthermore are evaluated at a point that we must solve for, which makes these equations very difficult to solve.

To make any kind of analytic progress, we make asymptotic approximations. Depending on whether you are describing deep or shallow water waves, different dimensionless parameters come into play. For linear waves however, they both share the common small parameter $\epsilon \equiv ak$, which describes the wave slope.

In this case, the governing equations, to $\mathcal{O}(\epsilon)$ are

$\nabla^2\phi = 0$, with $\phi_t+ g\eta = 0$ and $\eta_t = \phi_z$, both evaluated at $z=0$, while $\phi_z = 0$ at $z=-h$.

For simplicity let's consider waves of two dimensions, where $x$ is the horizontal direction, and $z$ is the vertical coordinate. Assuming the solutions are permanent progressive waves of the form $$\eta = a\ cos(kx-\omega t)$$ with $a$ the amplitude, $k$ the wavenumber and $\omega$ the frequency, we find that the linear governing equations imply

$$\omega^2 = gk \tanh(kh)$$

Now, if we follow waves of constant phase $\theta = kx -\omega t$, we see that these waves travel at a speed $c = \omega/k$. In shallow water, $kh \gg 1$, so that

$$\omega^2 \approx ghk^2\\ (\text{and}) \\c = \sqrt{gh},$$

while in deep water, $kh \ll 1$ so that

$$\omega^2 \approx gk\\ (\text{and}) \\ c= \sqrt{\frac{g}{k}}$$

The first thing we notice is that in deep water, the waves are dispersive, meaning the phase velocity depends on the wavenumber. This is why, for instance, when swells come to shore, it is the longest waves that arive first. In shallow water, to first order, the waves are not dispersive.

To first order, wakes are nothing more than linear superposition of waves, due to a moving point disturbance. This is known as the Kelvin ship wake problem, and I discussed a way to derive this result here.

Now, the above description barely constitutes a summary. For instance, there are many interesting effects that happen when capillary effects are included. For deep water capillary waves

$$\omega^2 = Tk^3$$

where $T$ is the surface tension of water. We see that for these waves, the velocity increases with wavenumber, in contrast with gravity waves. However, this equation is academic, as any description of capillary waves must necessarily include dissipation, which is significantly more difficult to model (and has only been done for nonlinear cases numerically).

Second order effects (eg in shallow water solitons, in deep water Stokes effects and the nonlinear Schrödinger equation) are really interesting but involve heavier lifting.

  • $\begingroup$ I gave you a pointS for a good answer but Is there a simpler description? Thanks $\endgroup$ Feb 10, 2018 at 22:14

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