# General relativity potential

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).

• 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 $r = 3M$ – Jerry Schirmer Jun 21 '17 at 18:24

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.