I've seen a lot of places where the ISCO of a black hole is given as $R_{ISCO} = 6M$, but it never goes into any particular detail (one source said "as you all know from your relativity courses", haha).
What does the symbol $M$ mean?
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$\begingroup$ Is your question about the what the effective mass is (as the "strength" of the source in a Black hole metric, equal to the energy contained within the horizon) or about the notion of a distance being seemingly expressed as a mass/energy (which kind of thing is very common when dealing with natural units)? $\endgroup$– Selene RoutleyCommented Aug 3, 2016 at 23:56
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$\begingroup$ You can ask the more pertinent question of how do you define radius and circular. $\endgroup$– AHusainCommented Aug 3, 2016 at 23:59
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3 Answers
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In General Relativity so called geometrized units are very common. Especially the ones where the speed of light and the gravitational constant are set to unity $(c=G=1$$)_{\mathrm{GU}}$. In such a unit system one can measure basically everything in powers of one base dimension; often length.
In this setting masses have the same units/dimensions as lengths. So I would say $M$ refers to the gravitational mass of the black hole. One solar mass $M_\odot=1.9884\times10^{30}\mathrm{kg}$ is equal to $1476.57\mathrm{m}$ in gravitational units. The conversion factor is $7.42592\times 10^{^-28}\mathrm{m}/\mathrm{kg}=(G/c^2)_{\mathrm{SI}}$.
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This is 99% of the (better) answer that M.J. provided, but as I started it before I saw that answer, I might as well finish, just in case it adds anything new.
In Newtonian gravity, test particles can stably orbit at arbitrary distances from a central object. In general relativity, however, there exists an innermost stable circular orbit (often called the ISCO), inside of which, any infinitesimal perturbations to a circular orbit will lead to inspiral into the black hole.The location of the ISCO depends on the spin of the black hole, in the case of a Schwarzschild black hole (spin zero) is:
R$_{isco}$ = 3$_{r}$ = 6GM/c$^2$, M is mass, G and c = 1, as in the above answer.
and decreases with increasing spin.
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$\begingroup$ Or increases with increasing spin if you orbit against the spin. $\endgroup$ Commented Aug 4, 2016 at 21:00
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MJ describes well how the mass can be expressed as length in geometrized units. Count_to_10 then explains well what is the meaning of ISCO, inside of which a circular orbit can not be maintained. This just adds a couple points to their correct answers.
For comparison with the $R_{isco} = 6M$ equation, that is actually 3 times the Schwarzchild radius, i.e., the horizon radius, for a spherically symmetric non-rotating BH. That is, $R_{isco} = 3R_s$. The intuitive view is that the tidal forces not too far from the horizon are strong enough to destabilize the circular orbits, i.e., have any minor change cause them to spiral and then plunge in. For Kerr maximally rotating BHs the number is different, it is $R = M$, the very large amount of angular momentum, the maximal allowed rotation for a rotating BH, helps provide stability closer to the BH. For those extreme BH's the horizon is at M.
BTW, this does not mean that BH horizons won't start combining before this innermost circular radius, as was 'seen' in the numerical simulations for the BH merger in 2015.
http://www.phys.lsu.edu/mog/mog9/node10.html describes much more
