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

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

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

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

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

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

This effective potential basically leaves us with one-dimensional problem on bound movement. Free-falling orbit is possible if there is a minimum of the potential.

As we should find the circular orbit, \(\dot{r}\) necessarily is \(0\)