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September 18, 2009

Surface Waves

In the comments to my previous post, Thomas Schaefer suggested that I look at the same issue (of the energy density having a linear term, rather than being quadratic in the amplitude of the wave) for the case of surface waves. I don’t think that’s actually the case.

But surface waves are the simplest example I know where one obtains a (highly) nontrivial dispersion relation from relatively simple physics. So it’s fun to review them, anyway.

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

(1)p(x,y,z=h(x,y,t),t)=p 0p\left(x,y,z=h(x,y,t),t\right) = p_0

Define p˜=pp 0ρ \tilde{p} = \frac{p-p_0}{\rho} The Navier-Stokes and continuity equations (again, neglecting viscosity) become

(2)vt+(v)v+gz^=p˜ v=0\begin{gathered} \frac{\partial\vec{v}}{\partial t} + (\vec{v}\cdot\vec{\nabla})\vec{v} + g \hat{z} = -\vec{\nabla}\tilde{p}\\ \vec{\nabla}\cdot\vec{v} = 0 \end{gathered}

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

(3)v z(z=D)=0v_z(z=-D) = 0

At the surface, it obeys

(4)DhDtht+v x(z=h)hx+v y(z=h)hy=v z(z=h) \frac{D h}{D t} \equiv \frac{\partial h}{\partial t} + v_x(z=h) \frac{\partial h}{\partial x} + v_y(z=h) \frac{\partial h}{\partial y}= v_z(z=h)

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

(5)v z(z=h)=htv_z(z=h) = \frac{\partial h}{\partial t}

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

(6)v z =Asinh(k(z+D))sin(kxωt) v x =Acosh(k(z+D))cos(kxωt) p˜+gz =ωkAcosh(k(z+D))cos(kxωt) \begin{split} v_z &= A \sinh\left(k(z+D)\right)\sin(k x -\omega t)\\ v_x &= A \cosh\left(k(z+D)\right)\cos(k x -\omega t)\\ \tilde{p} + g z &= \frac{\omega}{k}A \cosh\left(k(z+D)\right)\cos(k x -\omega t)\\ \end{split}

We next need to satisfy the interface condition p˜(x,z=h(x,t),t)=0 \tilde{p}(x,z=h(x,t),t) = 0 To the linear order to which we are working, when we set z=hz=h in (6), we are free to approximate cosh(k(h+D))cosh(kD) sinh(k(h+D))sinh(kD) \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

(7)h(x,t)=1gωkAcosh(kD)cos(kxωt)h(x,t) = \frac{1}{g}\frac{\omega}{k} A\cosh(k D)\cos(k x -\omega t)

Satisfying (5), gives the dispersion relation

(8)ω 2(k)=gktanh(kD)\omega^2(k) = g k \tanh(k D)

or, equivalently, a kk-dependent phase velocity v phase=ωk=gktanh(kD) v_{\text{phase}} = \frac{\omega}{k} = \sqrt{\frac{g}{k}\tanh(k D)} In deep water (DλD\gg\lambda), we can neglect the tanh(kD)\tanh(k D), and have the approximation v phase=gk 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 zz-direction.

The conservation of mass equation takes the form 0=ht+ 0 =\frac{\partial h}{\partial t} + \vec{\nabla}\cdot \vec{\mathcal{M}} where (x,y,t)= D h(x,y,t)(v xx^+v yy^)dz \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=t+ 0 =\frac{\partial \mathcal{E}}{\partial t} + \vec{\nabla}\cdot \vec{\mathcal{I}} where ρ = D h(x,y,t)(12v 2+gz)dz =12gD 2+12gh 2+12 D h(x,y,t)v 2dz \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 ρ= D h(x,y,t)(12v 2++p˜+gz)(v xx^+v yy^)dz \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.

Posted by distler at September 18, 2009 12:29 PM

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3 Comments & 0 Trackbacks

Re: Surface Waves

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.

Posted by: thomas on September 18, 2009 3:02 PM | Permalink | Reply to this

Re: Surface Waves

Wow, thanks for putting in so much effort.

One of the reasons for having this blog is so that I never have to work my way through the above derivation again!

I did, particularly, want to reproduce the dispersion relation (8), which always seemed deeply mysterious to me. It’s marvelous to see how elementary the computation is that leads to it.

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.

Exactly! That’s the point I was trying to emphasize in our previous discussion.

Posted by: Jacques Distler on September 18, 2009 3:19 PM | Permalink | PGP Sig | Reply to this

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?

Posted by: wolfgang on September 19, 2009 5:57 AM | Permalink | Reply to this

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