# Why can't a stable star have radius 1 < r < 9/8 its Schwarzschild radius?

A mass of 1.78 [in geometric units] corresponds to a ratio of radius to Schwarzschild radius of 9/8. Theory predicts that a smaller ratio is not possible for a stable star.

What causes this cutoff? Is this the TOV limit or something else?

-
Far outside my competence, but I'll bet dollars to donuts it is related to the lack of stable orbits inside $3/2 r_0$. Things are just goofy when the curvature of space gets that high. –  dmckee Dec 10 '11 at 0:02
@dmckee: Perhaps, but 9/8 isn't that close to 3/2, so something else would have to be at work as well. –  Charles Dec 10 '11 at 0:16
TOV limit is related to maximum mass that can be supported by the neutron degeneracy pressure. The $9/8$ ratio comes from General Relativistic arguments which state that any star that fills the Schwarzschild volume any more than $88\%$ will collapse under its gravitational field to end up as a Black Hole.
The $88\%$ limit ultimately comes from stellar stability to radial oscillations. The stability under these oscillations can be measured using the adiabatic exponent $\Gamma$ which is $\frac{\Delta P/P}{\Delta \rho/\rho}$. This would be in Newtonian dynamics. However, when dealing with General Relativity, we know that anything compactified to a radius less than its $R_S$ will collapse to a Black Hole. So even if $\Gamma\rightarrow \infty$, a star can still collapse. Interestingly, $\Gamma\rightarrow \infty$ corresponds to $R/R_S > 9/8$.