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From http://www.spacetimetravel.org/ssm/ssm.html :

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?

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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

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up vote 5 down vote accepted

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 $R/R_S > 9/8$.

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Can you explain this further? I don't understand why it would collapse at this particular size. –  Charles Dec 10 '11 at 0:15
    
Yes, of course. I've just expanded the original answer. –  Omar Dec 10 '11 at 0:33
    
In case it's not clear from this answer, the 9/8 value is unrelated to the TOV limit. –  Peter Shor Dec 10 '11 at 17:16
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This answer basically explains it. The relevant reference is Chandrasekhar, ApJ 140, 417 (1964). –  Warrick May 4 '12 at 10:21

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