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Hamilton EOM for the the Schwarzschild metrics

  1. Jun 15, 2015 #1
    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
    \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?
     
  2. jcsd
  3. Jun 15, 2015 #2

    George Jones

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    Staff Emeritus
    Science Advisor
    Gold Member

    Careful. In the definition of H in (1), the g^{\mu \nu} have explicit coordinate dependence, and thus H does also.
     
  4. Jun 15, 2015 #3
    Ok, thanx, I try.
     
  5. Jun 15, 2015 #4

    Mentz114

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    Gold Member

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