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).

  • $\begingroup$ What does $\color{blue}{\dot{t}}$ mean? $\endgroup$
    – caverac
    May 21, 2019 at 12:39
  • $\begingroup$ Can't you just use Noether's theorem? en.wikipedia.org/wiki/Noether%27s_theorem $\endgroup$
    – E. Bellec
    May 21, 2019 at 12:42
  • $\begingroup$ $p_t$ 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 . $\endgroup$
    – E. Bellec
    May 21, 2019 at 12:45
  • 1
    $\begingroup$ @E.Bellec Great, thanks! I've not been introduced to Noether's theorem before. I've wrote an answer to this question for completeness. $\endgroup$
    – Jack G
    May 21, 2019 at 13:18

1 Answer 1


Via Noether's theorem (thanks, E. Bellec!) and the Euler-Lagrange equations,

$$\frac{\mathrm{d}}{\mathrm{d}t}\left( \frac{\partial L}{\partial \dot{t}} \right) = \frac{\mathrm{d} p_t}{\mathrm{d}t} = 0$$

Hence $p_t$ is a conserved quantity (and similar for $p_\phi$). H has no explicit time dependence so is also conserved. $\square$


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