# Hamilton EOM for the the Schwarzschild metrics

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1. Jun 15, 2015

### sergiokapone

I have a problem (this is not homework)
Based on covariant Lagrangian $\mathcal {L} = \frac {m}{2} \frac{dx^{\mu}}{ds} \frac {dx _ {\mu}}{ds}$ record the equations of motion in Hamiltonian form for a particle in the Schwarzschild metric (SM).

Based on Legandre transformations

H =\frac {m}{2} p_{\mu}g^{\mu\nu}p_{\nu}

EOM in Hamilton form (as shown in Goldstein)
\begin{align}
\frac{dx^{\mu}}{ds} = g^{\mu\nu} \frac{\partial H}{\partial p^{\nu}}\\
\frac{dp^{\mu}}{ds} = -g^{\mu\nu} \frac{\partial H}{\partial x^{\nu}}
\end{align}

Canonical momentum is $p_{\mu}$, but not $p^{\mu}$. How it is possible to apply this equations for the
Schwarzschild metric?

Again, if I write

\begin{align}
\frac{dx_{\mu}}{ds} = g_{\mu\nu} \frac{\partial H}{\partial p_{\nu}}\\
\frac{dp_{\mu}}{ds} = -g_{\mu\nu} \frac{\partial H}{\partial x_{\nu}} \label{5}
\end{align}

The RHS of equation $\eqref{5}$ in SM for any component allways will give $0$, because $H$ does not depend on $x^{\mu}$. But one know the $p_{r}$ does not conserve, $dp_r/d\tau \neq 0$ for SM.

My question, what is the correct view of EOM in Hamilton form for GR in general, or for the SM in specific?

2. Jun 15, 2015

### George Jones

Staff Emeritus
Careful. In the definition of H in (1), the g^{\mu \nu} have explicit coordinate dependence, and thus H does also.

3. Jun 15, 2015

### sergiokapone

Ok, thanx, I try.

4. Jun 15, 2015

### Mentz114

If you want to see this worked out have a look at http://arxiv.org/abs/1201.5611v1.pdf.