Orbits in the Schwarzschild Metric

Start with the dot product of the 4-momentum with itself and use the conserved quantities above to eliminate $\dot{t}$ and $\dot{\varphi}$.

$$-m^2c^2 = p_\mu p_\mu = m^2{dx_\mu\over d\tau}{dx_\mu\over d\tau}$$

$$= -m^2 \left[\left(1 - \frac{r_s}{r} \right) c^2 \dot{t}^2 - \frac{\dot{r}^2}{1-\frac{r_s}{r}} - r^2 \dot{\varphi}^2\right]$$

$$\dot{t} = {E\over mc^2\left(1 - \frac{r_s}{r} \right)}$$

$$\dot{\varphi} = {L_z\over mr^2}$$

$$-m^2c^2 = -m^2 \left[\left(1 - \frac{r_s}{r} \right) {E^2\over m^2c^2\lef... ...r} \right)^2} - \frac{\dot{r}^2}{1-\frac{r_s}{r}} - {L_z^2\over m^2r^2} \right]$$

$$c^2 = {E^2\over m^2c^2\left(1-\frac{r_s}{r}\right)} - \frac{\dot{r}^2}{1-\frac{r_s}{r}} - {L_z^2\over m^2r^2}$$

$$1 = {E^2\over m^2c^4\left(1-\frac{r_s}{r}\right)} -\frac{\dot{r}^2}{c^2\left(1-\frac{r_s}{r}\right)} -{L_z^2\over m^2c^2r^2}$$

$$1-\frac{r_s}{r} = {E^2\over m^2c^4} -\frac{\dot{r}^2}{c^2} -{L_z^2\over m^2c^2r^2}\left(1-\frac{r_s}{r}\right)$$

$${E^2\over m^2c^4}-1 = \frac{\dot{r}^2}{c^2} -\frac{r_s}{r} +{L_z^2\over m^2c^2}\left({1\over r^2}-\frac{r_s}{r^3}\right)$$

$${E^2\over m^2c^4}-1= \frac{\dot{r}^2}{c^2} +V_{eff}(r)$$

$$V_{eff}(r)= -\frac{r_s}{r} +{L_z^2\over m^2c^2}\left({1\over r^2}-\frac{r_s}{r^3}\right)$$

This equation is in many ways similar to the non-relativistic Newtonian gravity problem. If the effective potential has a minimum, the particle can oscillate around that minimum with energy moving back and forth between kinetic in the $\dot{r}^2$ term and the effective potential. We recognize the normal $-{1\over r}$ gravitational potential and the normal $+{1\over r^2}$ angular momentum pseudopotential. There is an additional term proportional to ${r_s\over r^3}$ which makes gravity in GR stronger than Newtonian gravity. We will see that there will be a minimum radius for stable orbits.

The effective potential will increase sharply as we go out from $r=0$, then potentially reach a maximum, drop to a minimum, then increase as $r\rightarrow\infty$. It is around the minimum that there can be a stable bound orbit. As in Newtonian gravity, the particle may have sufficient energy to escape to infinity. In the Schwarzschild solution, it may also have enough energy to go over the angular momentum barrier and fall down to the Schwarzschild radius. As seen from infinity, it takes an infinite amount of time to reach $r_s$, but from the frame of the particle, it can quickly go to $r=0$.

We can find the positions of the maximum and minimum effective potential and the conditions under which there will be a minimum.

$${\partial V\over\partial r}=\frac{r_s}{r^2} -{L_z^2\over m^2c^2}\left({2\over r^3}-\frac{3r_s}{r^4}\right)= 0$$

$$r^2 -{2L_z^2\over m^2c^2r_s}r+{3L_z^2\over m^2c^2}= 0$$

$$r= {L_z^2\over m^2c^2r_s} \pm \sqrt{\left({L_z^2\over m^2c^2r_s}\right)^2-{3L_z^2\over m^2c^2}}$$

This will have two real roots (a maximum and a minimum) if the quantity in the square root is positive and will have no real roots if the quantity is negative. We can find the boundary at which the maximum equals the minimum, where the value inside the square root is zero. That will be the minimum stable circular orbit for a given $L_z$.

$$\left({L_z^2\over m^2c^2r_s}\right)^2 = {3L_z^2\over m^2c^2} L_z$$

$$r = {L_z^2\over m^2c^2r_s} r$$

$$r {L_z^2\over m^2c^2r_s} = {3L_z^2\over m^2c^2} r L_z$$

$$r = 3 r_s$$ radius below which there are no solutions

So the minimum radius stable circular orbit is for $r = 3 r_s$, independent of $L_z$. If particles get too close, they are sucked into the black hole.