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Stellar dynamics
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=== Loss cone === Thirdly a star can be swallowed if coming within a few [[Schwarzschild radius|Schwarzschild radii]] of the black hole. This radius of Loss is given by <math display="block"> s \le s_\text{Loss} = \frac{6 G M_\bullet}{c^2} </math> The loss cone can be visualised by considering infalling particles aiming to the black hole within a small solid angle (a cone in velocity). These particle with small <math> \theta \ll 1</math> have small angular momentum per unit mass <math display="block"> J \equiv r v \sin\theta \le J_\text{loss} = \frac{4G M_\bullet}{c}.</math> Their small angular momentum (due to ) does not make a high enough barrier near <math> s_\text{Loss} </math> to force the particle to turn around. The effective potential <math display="block"> \Phi_\text{eff}(r) \equiv E- {\dot{r}^2 \over 2} = {J^2 \over 2r^2} + \Phi(r) , </math> is always positive infinity in Newtonian gravity. However, in GR, it nosedives to minus infinity near <math> \frac{6 G M_\bullet}{c^2} </math> if <math> J \le \frac{4G M_\bullet}{c}. </math> Sparing a rigorous GR treatment, one can verify this <math> s_\text{loss}, J_\text{loss} </math> by computing the last stable circular orbit, where the effective potential is at an inflection point <math> \Phi''_\text{eff}(s_\text{loss})=\Phi'_\text{eff}(s_\text{loss})=0 </math> using an approximate classical potential of a Schwarzschild black hole <math display="block"> \Phi(r) = - {(4G M_\bullet/c)^2 \over 2r^2} \left[1+{3 (6 G M_\bullet/c^2)^2 \over 8 r^2 }\right] - \frac{G M_\bullet}{r} \left[1 - {(6G M_\bullet/c^2)^2 \over r^2}\right]. </math> <!-- <math> \Phi(r) = - {5(G M_\bullet/c)^2 \over 4 r^2} - {G M_\bullet \over r- 2G M_\bullet/c^2} </math> A rotating Kerr black hole can be approximated with an effective potential <math> \Phi_\text{eff}(r_1,r_2) = E-{(\dot{r}_1 + \dot{r}_2)^2 \over 2 r_1 r_2} = {(2J_z a + 2J_z^2) -5(G M_\bullet/c)^2 \over 4 r_1r_2/a^2}[((r_1-r_2)^2-4a^2)^{-1}-((r_1+r_2)^2-4a^2)^{-1}] - {G M_\bullet}{ \sqrt{(r_1+r_2)^2-4a^2 + (2GM/c^2)^2} - \sqrt{(r_1-r_2)^2-4a^2 + (2GM/c^2)^2} \over (r_1 r_2)} </math> -->
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