# Schwarzschild orbits

$L = - m \sqrt{\frac{dx^\mu}{d\tau} \frac{dx_\mu}{d\tau}}$

The interval for the Schwarzschild black hole looks like

$ds^2 = - \left(1- \frac{2M}{r}\right) dt^2 + \left(1-\frac{2M}{r}\right)^{-1} dr^2 + r^2 d\Omega^2$

It is associated with the Lagrangian

$L = - \left(1- \frac{2M}{r}\right) \dot{t}^2 + \left(1-\frac{2M}{r}\right)^{-1} \dot{r}^2 + r^2 (\dot{\theta}^2 + \sin{\theta}^2 \dot{\phi}^2)$

It does not depend explicitly on $$t$$ and $$\phi$$, so $$\frac{dL}{d\dot{t}}$$ and $$\frac{dL}{d\dot{\phi}}$$ will be the constants of motion (from the Euler-Lagrange equations):

$-\frac12 \frac{dL}{d\dot{t}} = \left(1- \frac{2M}{r}\right) \dot{t} = \epsilon \\ \frac12 \frac{dL}{d\dot{\phi}} = r^2 \sin{\theta}^2 \dot{\phi} = \lambda$

Those are related to the energy and the angular momentum of the particle.

And we also know that for a particle with mass $$m$$

$-m^2 = p_\mu p^\mu = m^2 \frac{dx_\mu}{d\tau} \frac{dx^\mu}{d\tau} = m^2 g_{\mu \nu} \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau}$

Euler-Lagrange equation

$\frac{dL}{dx^\mu} = \frac{d}{d\tau} \frac{dL}{d\dot{x}^\mu}$

For an orbit in equatorial plane with a constant angular velocity this becomes

$-m^2 = -m^2 \left(1- \frac{2M}{r}\right) \dot{t}^2 + m^2\left(1-\frac{2M}{r}\right)^{-1} \dot{r}^2 + m^2 r^2 \dot{\phi}^2$

$-m^2 \left(1+\frac{\lambda^2}{r^2}\right) \left(1- \frac{2M}{r}\right) = -m^2 \epsilon^2 + m^2 \dot{r}^2$

$\dot{r}^2 - \frac{2M}{r} + \frac{\lambda^2}{r^2} - \frac{2 M\lambda^2}{r^3} = \epsilon^2 - 1$

$\dot{r}^2 + V(r) = \epsilon^2 - 1$

$V(r) = - \frac{2M}{r} + \frac{\lambda^2}{r^2} - \frac{2 M\lambda^2}{r^3}$