In general relativity it is defined the potential $$ V(r)=-\frac{GM}{r}+\frac{l^{2}}{2r^{2}}-\frac{GMl^{2}}{r^{3}} $$

for massive particles with angular momenta $l$ in a gravitational field created by a mass $M$.

I want to understand why the value of $l^{2}=GM\frac{r^{2}}{r-3GM}$, as well as the value of the energy $e$, go to infinity when the massive particle is in a circular geodesic of radius $r=3GM$ (so there is no possibility of circular geodesic there).

I can't understand this because $V(r)$ has a maximum or a minimum, provided $l>\sqrt{12}GM$, at

$$ r_{{min_{max}}}=\frac{l^{2}}{2GM}\left[1\pm\sqrt{1-12 \left(\frac{GM}{l}\right)^{2}}\right], $$

not at $r=3GM$.

In relation to this, I would appreciate it if someone explains to me where are the stable points for circular orbits, given this massive particle. (I know that for a photon the potential has a maximum at $r=3GM$ and it corresponds to an unstable circular orbit, because $V(r)$ has a maximum there).

  • $\begingroup$ I've never done this, but I bet you would learn something by calculating the 3-velocity of an orbit, relative to a reference frame given by schwarzschild time = constant. My guess is that you'd see the 3-velocity limit to $c$ as the radius of a circular orbit asymptotes to $r = 3M$ $\endgroup$ Jun 21, 2017 at 18:24

1 Answer 1


In the case that $GM \ll l$, we have $$r_{\rm max} = \frac{l^2}{2GM} \left\{1- \left[1 - \frac{1}{2} \times 12 \left( \frac{GM}{l} \right)^2 \right] \right\} + O\left(\left(\frac{GM}{l}\right)^4\right) = 3GM + O\left(\left(\frac{GM}{l}\right)^4\right)$$ So, as $l$ increases, the radius of unstable orbit $r_{\rm max}$ will be closer and closer to the $3GM$. So, we conclude that $r=3GM$ is the innermost unstable orbit.


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