Conserved Energy and Angular Momentum in the Schwarzschild Metric

Let us work in the equatorial plane $(\theta={\pi\over 2})$. In the Schwarzschild solution, with $\theta={\pi\over 2}$, the Lagrangian is

$$L = -mc\sqrt{\left(1 - \frac{r_s}{r} \right) c^2 \dot{t}^2 - \frac{\dot{r}^2}{1-\frac{r_s}{r}} - r^2 \dot{\varphi}^2}$$

again with the dot denoting differentiation with respect to proper time $\tau$. The square root does not really complicate things in this case since when we differentiate it we get a square root in the denominator but this just introduces a multiplicative constant into our equations. Its the derivative of what's inside the square root that ends up in the equations of motion.

Since the Lagrangian does not depend on $t$ or on $\varphi$, the momenta corresponding to those variables are conserved. For $t$, this is the energy. For $\phi$ its the $z$ component of angular momentum.

$${\partial L\over\partial \dot{t}} = {-mc \ove... ...ight) c^2 =-mc^2\left(1 - \frac{r_s}{r} \right)\dot{t}=const.\equiv -E$$

This goes to $-\gamma mc^2$ as $r_s\rightarrow 0$ which agrees with the Energy in Special Relativity.

Similarly for $\varphi$,

$${\partial L\over\partial \dot{\varphi}}={-mc \over \sqrt{-v_\mu v_\mu}} r^2\dot{\varphi}=-mr^2\dot{\varphi} =const.=-L_z$$

Again this agrees with the angular momentum conserved in Special Relativity.

With these two constants of the motion, we can write an equation in one variable $r$ and its proper time derivative which will describe the orbits of particles in the Schwarzschild metric.