Physical meaning of relativistic saturation of an instability? In the derivation of the Rayleigh-Taylor instability when the fluid is in the extreme relativistic limit ($\rho_0 c^2 \ll p$) and there is a large effective gravity ($ g \gg kc^2$), where $\rho_0$, $p$, $g$, and $k$ are the rest mass fluid density, pressure, effective acceleration, and wave vector for the instability, respectively, then the instability growth rate is observed to be independent of $g$. What phenomenologically permits this saturation in growth rate to not further increase? Cannot the instability continue increasing if energy is available through $g$?
 A: The Rayleigh-Taylor instability is driven by a difference in potential energy density between two fluids.  If the interface between the fluids is disturbed by a small perturbation of size $z$, then the pressure from above changes by $\rho_1 g z$, while the pressure from below changes by $\rho_2 g z$.  If $\rho_1$ is greater than $\rho_2$, then the perturbation grows.  The growth rate is determined by $\omega^2 = gk \eta$ where $\eta$ is the Atwood number
$$
\eta = \frac{\rho_1 - \rho_2}{\rho_1 + \rho_2}
$$
Physically, the numerator represents the pressure difference that drives the instability, while the denominator represents the inertia of the fluid that has to move as the instability grows.  We can draw a loose analogy with Newton's law $F = ma$, with $\rho_1 - \rho_2$ as the "force," $\rho_1 + \rho_2$ as the "mass," and the Atwood number $\eta$ as the "acceleration."
In the Newtonian regime, that's all we need to worry about.  But in the relativistic regime, the energy of the fluid becomes a significant contributor to its mass-energy.  As a result, the fluid has a higher-than-expected inertia and the denominator of the Atwood number increases.  We can take this into account by correcting the Atwood number to
$$
\eta_{rel} = \frac{\rho_1 - \rho_2}{8p/c^2 + \rho_1 + \rho_2}
$$
In the highly relativistic regime, the $8p/c^2$ term overwhelms the rest of the denominator, so we get a very small Atwood number and the Rayleigh-Taylor instability grows much more slowly than we'd expect.
That still isn't a full explanation, though, because the corrected growth rate $\omega^2 = gk\eta_{rel}$ still depends on $g$.  Indeed, the growth rate does not saturate for incompressible fluids.  When the fluid is compressible and $g$ is very high, however, another effect comes into play.  As the instability grows, the local density in the instability must rise because the fluid is compressed by the $g$-force.  (Just as Earth's atmosphere becomes denser with lower altitude.)  Without this change in density, the pressure would not be able to rise sufficiently to fuel further instability growth.  But the density can only increase if molecules are pulled in from other parts of the fluid.  The rate at which this can happen is limited by the speed of sound in the fluid (since the molecules are attracted towards the instability by pressure differences rather than the $g$-force).  That's what causes the growth rate to saturate.
Aside: In relativity, of course, there is a hard upper bound on the growth rate because molecules cannot move faster than the speed of light.  Note that saturation happens before this limit is reached.
As you correctly pointed out in your comment, the same thing can happen in the non-relativistic case.  (You can see this from the equations given by Livescu for the Rayleigh-Taylor instability in compressible non-relativistic fluids.)  In practice, this is not especially significant because we rarely see situations with very high $g$ and low Atwood number.  Because relativity decreases the effective Atwood number, however, saturation becomes somewhat more important in the relativistic regime (though it's still not especially relevant in most astrophysical situations, for the reasons given by Allen and Hughes).
Sorry if that all seems a bit "hand-wavey," but I don't think I can make it any more specific without the seriously complex math in the Allen and Hughes paper.
