Musings

Consider an incompressible fluid ($\rho=\text{const}$), with an interface, $h(x,y,t)$, to the air. At the interface, the pressure must equal atmospheric pressure.

Define $$\tilde{p} = \frac{p-p_0}{\rho}$$ The Navier-Stokes and continuity equations (again, neglecting viscosity) become

If the channel has a depth, $D$, $\vec{v}$ obeys the boundary condition

At the surface, it obeys

In the linearized approximation, we can simplify this latter condition to

We’ll set $v_y=0$, and assume everything is $y$-independent, taking an ansatz of the form $$v_z(x,z,t) =f(z) \sin(k x-\omega t)$$ and similarly for $v_x$ and $\tilde{p}$. Linearizing means neglecting the convective term, $(\vec{v}\cdot\vec{\nabla})\vec{v}$, in (2). Plugging into the linearized equations, we get a solution (satisfying the boundary condition (3)),

We next need to satisfy the interface condition $$\tilde{p}(x,z=h(x,t),t) = 0$$ To the linear order to which we are working, when we set $z=h$ in (6), we are free to approximate $$\begin{gathered} \cosh\left(k(h+D)\right) \to \cosh(k D)\\ \sinh\left(k(h+D)\right) \to \sinh(k D)\\ \end{gathered}$$ so that the height of the interface (our surface wave) is

Satisfying (5), gives the dispersion relation

or, equivalently, a $k$-dependent phase velocity $$v_{\text{phase}} = \frac{\omega}{k} = \sqrt{\frac{g}{k}\tanh(k D)}$$ In deep water ($D\gg\lambda$), we can neglect the $\tanh(k D)$, and have the approximation $$v_{\text{phase}} = \sqrt{\frac{g}{k}}$$ which might be familiar to you (it is the sort of thing I might mention in a Freshman-Physics discussion of waves).

What about those conservation laws? There are local conservation laws, but for our purposes, it’s illuminating to consider the semi-local ones, integrated over the $z$-direction.

The conservation of mass equation takes the form $$0 =\frac{\partial h}{\partial t} + \vec{\nabla}\cdot \vec{\mathcal{M}}$$ where $$\vec{\mathcal{M}}(x,y,t) = {\int^{h(x,y,t)}_{-D} (v_x\hat{x} + v_y\hat{y}) d z}$$ and the conservation of energy equation takes the form $$0 =\frac{\partial \mathcal{E}}{\partial t} + \vec{\nabla}\cdot \vec{\mathcal{I}}$$ where $$\begin{split} \frac{\mathcal{E}}{\rho} &= {\int^{h(x,y,t)}_{-D} \left(\tfrac{1}{2} v^2 + g z\right) d z} \\ &= -\tfrac{1}{2} g D^2 + \tfrac{1}{2} g h^2 + \tfrac{1}{2} {\int^{h(x,y,t)}_{-D} v^2 d z} \end{split}$$ and $$\frac{\vec{\mathcal{I}}}{\rho} = {\int^{h(x,y,t)}_{-D} \left( \tfrac{1}{2} v^2 + +\tilde{p}+g z\right)(v_x\hat{x} + v_y\hat{y}) d z}$$

Note that, (up to an irrrelevant overall constant in $\mathcal{E}$), both $\mathcal{E}$ and $\vec{\mathcal{I}}$ are quadratic in fluctuations, as expected.

So, no, there’s no funny business in this system.