If anyone has a derivation of this result, it would be appreciated greatly.
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1$\begingroup$ These lecture notes by Mike Guidry appear to address it on pages 785-790. $\endgroup$ – Mike G Jun 30 '18 at 16:53
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There's a fairly full treatment in these lecture notes but it's not really amenable to a summary in a stackexchange answer. However they end up with the following equation for the radius of a circular orbit of a particle with angular momentum $L$ and mass (ie not a photon).
$$GMr^2 - L^2r + 3GML^2 = 0$$
For suitable values of $L$ this has two solutions for $r$ and, one inside the ISCO radius (which is $6GM$ in these units) and one outside. Further analysis shows that the inner orbit is unstable and the outer one stable. See around equations 7.55 for more details.
Photons have an unstable circular orbit at $3GM$ and the event horizon is at $2GM$.
answered Jun 30 '18 at 19:15
