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 😉
UPDATE 1.
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.
UPDATE 2.
According to the comments partly from @VladimirKalitvianski a few points may be helpful:
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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.
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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.
Best 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.