The Thinker 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.

Wow, thanks for putting in so much effort. I guess you are right that the situation with surface wave is simpler, the term linear in the amplitude is removed simply by shifting the zero of the potential (although just as in the density wave case, it corresponds to shifting the energy density by a term linear in the density).

I guess the exercise also shows that the alternating term really has no physical significance: By putting the zero of my potential energy at the bottom of the Mariana trench, I can make it as large as I wish, and this energy can only be extracted by putting a hole in the bottom of the sea, not by doing anything at the surface.

Re: Surface Waves

Jacques,

I really enjoyed your two posts about the Navier Stokes equation.
Could you perhaps please write a 3rd post outlining briefly the proof that a smooth solution actually exists?