Conserved Energy and Angular Momentum in the Schwarzschild Metric

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

\bgroup\color{black}$ \displaystyle 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} $\egroup
again with the dot denoting differentiation with respect to proper time \bgroup\color{black}$ \tau$\egroup. 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 \bgroup\color{black}$ t$\egroup or on \bgroup\color{black}$ \varphi$\egroup, the momenta corresponding to those variables are conserved. For \bgroup\color{black}$ t$\egroup, this is the energy. For \bgroup\color{black}$ \phi$\egroup its the \bgroup\color{black}$ z$\egroup component of angular momentum.

\bgroup\color{black}$\displaystyle {\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 $\egroup

This goes to \bgroup\color{black}$ -\gamma mc^2$\egroup as \bgroup\color{black}$ r_s\rightarrow 0$\egroup which agrees with the Energy in Special Relativity.

Similarly for \bgroup\color{black}$ \varphi$\egroup,

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

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 \bgroup\color{black}$ r$\egroup and its proper time derivative which will describe the orbits of particles in the Schwarzschild metric.

Jim Branson 2012-10-21