- #1

sergiokapone

- 302

- 17

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

\begin{equation}

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

\end{equation}

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?