1

For a free particle in curved spacetime with signature $(-,+,+,+)$, consider the Hamiltonian $$H = \frac{1}{2}\left(g^{\mu\nu}p_\mu p_\nu + m^2\right).\tag{0}$$ Hamilton's equations are given by:

$$\dot{x}^\mu = \frac{\partial H}{\partial p_\mu} \qquad \dot{p}^\mu = - \frac{\partial H}{\partial x^\mu}~. \tag{1}$$

I want to show that this is equivalent to the geodesic equation

$$\frac{d^2 x^\mu}{d\tau^2} + \Gamma^\mu_{\alpha \beta}\dot{x}^\alpha \dot{x}^\beta = 0 =\dot{p}^\mu + \Gamma^\mu_{\alpha \beta}p^\alpha p^\beta~. \tag{2}$$

Direct computation from the Hamiltonian gives $\dot{x}^\mu = p^\mu$. The other equation gives me

$$\dot{p}^\mu = g^{\mu\nu}\dot{p}_\nu = g^{\mu\nu}\Gamma^{\rho}_{\nu\beta}p^\beta p_\rho~. \tag{3}$$

This must be wrong, unless $g^{\mu\nu}\Gamma^{\rho}_{\nu\beta}p^\beta p_\rho = -\Gamma^\mu_{\alpha \beta} p^\beta p^\alpha$, but I am unable to prove this.

Connor Behan
  • 10,659
newtothis
  • 727

2 Answers2

1

In the Hamiltonian formalism, the first equation is \begin{align} \dot{x}^\mu = \frac{\partial H}{\partial p_\mu} = g^{\mu\nu} p_\nu \equiv p^\mu \end{align} as you say. The second equation is \begin{align} \dot{p}_\mu = -\frac{\partial H}{\partial x^\mu} = -\frac{1}{2} p_\rho p_\sigma \partial_\mu g^{\rho \sigma}. \end{align} Now we have to see how these are implied by the geodesic equation.

This equation is \begin{align} \frac{d \dot{x}^\mu}{d\tau} &= -\Gamma^\mu_{\alpha \beta} p^\alpha p^\beta \\ \frac{d}{d\tau}(g^{\mu\nu} p_\nu) &= -\Gamma^\mu_{\alpha \beta} p^\alpha p^\beta \\ g^{\mu\nu} \dot{p}_\nu + p_\nu \dot{x}^\alpha \partial_\alpha g^{\mu\nu} &= -\Gamma^\mu_{\alpha \beta} p^\alpha p^\beta \\ \dot{p}^\mu &= -\Gamma^\mu_{\alpha \beta} p^\alpha p^\beta - p_\nu p^\alpha \partial_\alpha g^{\mu\nu}. \end{align} Out task now is to lower the free index $\mu$ on both sides, expand the Christoffel symbol and make use of $g^{\alpha \rho} \partial_\sigma g_{\rho \beta} + g_{\rho \beta} \partial_\sigma g^{\alpha \rho} = \partial_\sigma \delta^\alpha_\beta = 0$. This leads to \begin{align} \dot{p}_\mu &= -g_{\mu\nu} \Gamma^\nu_{\alpha \beta} p^\alpha p^\beta - g_{\mu\nu} p_\beta p^\alpha \partial_\alpha g^{\nu\beta} \\ &= -\frac{1}{2} p^\alpha p^\beta \left ( 2\partial_\alpha g_{\mu\beta} - \partial_\mu g_{\alpha\beta} \right ) - g_{\mu\nu} p_\beta p^\alpha \partial_\alpha g^{\nu\beta} \\ &= -\frac{1}{2} p_\rho p_\sigma g^{\rho\alpha} g^{\rho\beta} \left ( 2\partial_\alpha g_{\mu\beta} - \partial_\mu g_{\alpha\beta} \right ) - g_{\mu\nu} p_\beta p^\alpha \partial_\alpha g^{\nu\beta} \\ &= \frac{1}{2} p_\rho p_\sigma g^{\rho\alpha} \left ( 2g_{\mu\beta} \partial_\alpha g^{\sigma\beta} - g_{\alpha\beta} \partial_\mu g^{\sigma\beta} \right ) - g_{\mu\nu} p_\beta p^\alpha \partial_\alpha g^{\nu\beta} \\ &= -\frac{1}{2} p_\rho p_\sigma g^{\rho\alpha} g_{\alpha\beta} \partial_\mu g^{\sigma\beta} \\ &= -\frac{1}{2} p_\rho p_\sigma \partial_\mu g^{\rho \sigma} \end{align} as desired.

Connor Behan
  • 10,659
0

  1. For what it's worth, OP's Hamiltonian (0) is the $e=1$ gauge of the Hamiltonian $$ \begin{align}H~=~ &\frac{e}{2}(p^2+(mc)^2),\cr p^2~:=~&g^{\mu\nu}(x)~ p_{\mu}p_{\nu}\end{align}\tag{A}$$ for a massive relativistic point particle on a curved manifold, which is known to follow a timelike geodesic.

  2. For more details, see e.g. my related Phys.SE answer here and links therein.

Qmechanic
  • 220,844