Conserved quantities in Lagrangian Dynamics

I am given the Lagrangian for a particle of mass $$m$$ orbiting a black hole of radius $$r_h$$:

$$L=\frac{1}{2} m\left\{-F(r) c^{2} \dot{t}^{2}+F(r)^{-1} \dot{r}^{2}+r^{2} \dot{\phi}^{2}\right\}, \quad F(r)=1-\frac{r_{h}}{r}$$

I have derived the momenta to be:

\begin{aligned} p_{t} &=-m c^{2}\left(1-r_{h} / r\right) \dot{t} \\ p_{r} &=m\left(1-r_{h} / r\right)^{-1} \dot{r} \\ p_{\phi} &=m r^{2} \dot{\phi} \end{aligned}

and the Hamiltonian to be

$$H=\frac{1}{2 m}\left\{-\frac{p_{t}^{2}}{1-r_{h} / r}+\left(1-r_{h} / r\right) p_{r}^{2}+\frac{p_{\phi}^{2}}{r^{2}}\right\}$$

How can one determine which quantities are conserved?

(Answer: The conserved quantities are $$p_t$$, $$p_\phi$$ and $$H$$ — I am unsure how this can be seen from the above).

• What does $\color{blue}{\dot{t}}$ mean? – caverac May 21 at 12:39
• Can't you just use Noether's theorem? en.wikipedia.org/wiki/Noether%27s_theorem – E. Bellec May 21 at 12:42
• $p_t$ and $p_\phi$ conservation comes from translation invariance of $L$ regarding $t$ and $\phi$. $H$ comes from time translation invariance of $L$ if your $t$ is not the time variable . – E. Bellec May 21 at 12:45
• @E.Bellec Great, thanks! I've not been introduced to Noether's theorem before. I've wrote an answer to this question for completeness. – Jack G May 21 at 13:18

$$\frac{\mathrm{d}}{\mathrm{d}t}\left( \frac{\partial L}{\partial \dot{t}} \right) = \frac{\mathrm{d} p_t}{\mathrm{d}t} = 0$$
Hence $$p_t$$ is a conserved quantity (and similar for $$p_\phi$$). H has no explicit time dependence so is also conserved. $$\square$$