Limits on phase speed for a growing instability? When analyzing the Rayleigh-Taylor instability relativistically, a growing instability only occurs for $ \frac {\omega^2} {k^2 c^2} \ll 1 $. Why must the phase speed of the instability be sufficiently slow from a physical standpoint?
 A: Think about what the perturbation wavenumber $k$ and the growth rate parameter $\omega$ mean in the real, physical world.

$k$ is a measure of the size of the growing perturbation at the boundary of the two fluids; $\omega$ is a measure of the speed at which it grows.  Thus, we expect information about the newly created perturbation to travel outward at a velocity of order $\omega/k$.  Since information cannot move faster than light, an approximate limit on the phase velocity, $\omega/k < c$, becomes obvious.
Must $\omega/k$ be much less than $c$ rather than merely less than $c$?  First of all, it's worth noting that the upper bound on the phase velocity actually isn't all that far from $c$.  It looks like you're getting the idea that $\omega^2/k^2 c^2 \ll 1$ from section 4.1 of the Allen and Hughes paper you mentioned in the comments.  The only justification they provide is their equation 27, but they don't provide the actual solutions to the equation; in fact, the positive root is located at $\omega^2/k^2 c^2 = \frac{1}{12}(7+\sqrt{97}) \approx 1.40$, indicating that the maximum possible phase velocity is about 1.18$c$.  Intuitively, this seems reasonable; the perturbation takes more than just one time constant to reach depth $z$, so it makes sense that the phase velocity can be a little bit larger than $c$.
