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Incompressible approximation of fluid flow is usually known to be lame in modeling the propagation of any disturbance in it, predicting a speed equal to infinity for the propagation of the disturbance. However, very recently I have entered the field of water waves and I just wondered when mentioned that they are easily obtaining finite speeds for the propagation of waves in water starting from the very incompressible equations! The speed of wave propagation is called the group speed which is defined by the relation $v_g=\tfrac{d\omega}{d\kappa}$ wherein, $\omega$ is the frequency and $\kappa$ is the wave number. For example consider the gravity waves, their governing equation is $\nabla^2\phi=0$ which compared to the equation of sound waves $\nabla^2\phi=\tfrac{1}{c^2}\tfrac{\partial^2\phi}{\partial t^2}$ should perhaps predict the speed $c\rightarrow\infty$ for the propagation of the waves in water but all the books that I saw use the relation already expressed for the group speed and no infinity appears anywhere in their calculations! How is it understandable? Maybe the question can be reduced to why the propagation speed is obtained from the relation given for $v_g$?

Thanks in advance ;)


In the case it may help let me add that the answer usually used for $\phi$ in the gravity wave equation $\nabla^2\phi=0$ is a superposition of the harmonic modes of the form $$\phi(\vec{x},t)=Re\left[A\,e^{i (\vec{\kappa}\cdot\vec{x}-\omega t)}\right]$$ so that it contains $t$ although there is no time derivative present in the equation.


According to the comments partly from @VladimirKalitvianski a few points may be helpful:

  • We have $∇⋅v⃗ =0$ which together with inviscid assumption in a small wave easily give the general equation $∇^2 ϕ=0$ having no time derivative in it. It's a kinematic equation like in potential flow modeling. It does not contain any steadyness assumption.

  • In the first glance, in an incompressible liquid there may occur transversal waves (like surface waves), since they do not need the liquid to be compressed like it is the case for longitudinal waves (like sound waves), but instead due to liquid motion. But suppose a fluid particle is to move upward, the motion is not due to expansion thus immediately another particle should fill its previous position, and this will immediately affect its own neighborhoods, and this way particles one after another but immediately will sense the presence of the disturbance. Now look from a far distance to see again the wave has been propagated everywhere immediately with the infinity speed, even if the wave was transversal.

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In an incompressible liquid there may wall be transversal waves. They do not need the liquid to be compressed, but are due to liquid motion (surface waves, for example). –  Vladimir Kalitvianski Mar 4 '13 at 13:08
@VladimirKalitvianski, you are right, I had almost forgotten the classification of the waves into longitudinal and transeversal waves. Maybe that;s why Landau has mentioned in his book that every small-amplitude disturbances propagate with the speed of sound except for entropy (caused by shear?) and vorticity (again caused by shear)? However, surface waves like gravity waves shouldn\t still propagte at the infinity speed when there is no time-derivative in a wave equation like that of gravity waves? Can the incompressible theory capture the transient problem? –  owari Mar 4 '13 at 13:18
Well, maybe I expressed myself too sloppily. I meant when there is a fluid discontinuity, a surface of some kind, an interface with other (compressible) medium. Then there is a freedom to move there. –  Vladimir Kalitvianski Mar 4 '13 at 13:44
Hi @owari. Do you want me to merge your new account with your old account for you? –  Qmechanic Mar 4 '13 at 20:27
@owari there's a new procedure for merging accounts; just go to this page and follow the instructions. –  David Z Mar 4 '13 at 23:12
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1 Answer

Having looked again on the formulation of the gravity waves in this link I used @VladimirKalitvianski's comment and changed the upper boundary condition from $$\frac{\partial\eta}{\partial t}=\frac{\partial\phi}{\partial z}\mbox{, at}\;z=\eta(x,t)$$ to $$\frac{\partial\phi}{\partial z}=0\mbox{, at}\;z=\eta=const.$$ Having done such now assuming the solution to equation $$\nabla^2\phi=0 \mbox{ with the BCs: } \tfrac{\partial\phi}{\partial z}=0\quad\mbox{ at}\;z=\eta=const.\mbox{ and }z=-h$$ having a form $$\phi(x,z,t)=A \cos(\kappa_x x + \kappa_z z - \omega t)$$ and substituting it in the equation we will obtain: $$-(\kappa_x^2 + \kappa_z^2)\,\phi=0$$ which itself yields in $$\kappa_x^2 + \kappa_z^2=0\quad\Rightarrow\quad\kappa_x=0 \mbox{ and } \kappa_z=0$$ which by itself means the solution will have the form $$\phi(x,z,t)=A \cos(- \omega t)$$ whose propagation speed is clearly infinity. Therefore, the boundary condition $$\frac{\partial\eta}{\partial t}=\frac{\partial\phi}{\partial z}\mbox{, at}\;z=\eta(x,t)$$ is critical in obtaining a finite wave propagation speed in an incompressible medium, that is, a free surface.

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@VladimirKalitvianski, thanks for your comment it helped me a lot :D –  owari Mar 5 '13 at 20:39
If the medium is stratified then you can have a rigid lid (no free surface) and still have internal waves with finite speed. I was also thinking about another solution, since $k$ is the wave number, $k = 0$ means there are no waves, therefore, $\omega$ is also zero and the wave speed is indeterminate, not infinity. –  Isopycnal Oscillation Oct 5 '13 at 3:10
@IsopycnalOscillation, about the stratified medium maybe it is not as restricted as the incompressible flow? But about $k=0$ yielding into $\omega=0$ then what about the standing (non-traveling) waves? The speed of information in a standing wave is infinity if I am not mistaking, since a little change in some quantity will be traveled instantaneously to everywhere else. (though I'm not sure) –  owari Oct 5 '13 at 11:19
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