# Derivation of geodesic equation from hamiltonian (lagrangian) equations

1. Jul 31, 2011

### luinthoron

1. The problem statement, all variables and given/known data
Hello, I would like to derive geodesics equations from hamiltonian
$H=\frac{1}{2}g^{\mu\nu}p_{\mu}p_{\nu}$
using hamiltonian equations.

A similar case are lagrangian equations. With the definition

$L=g_{\mu\nu}\dot{x}^\mu\dot{x}^\nu$

I tried to solve the Euler-Lagrange equation

$\frac{d}{d\lambda}(\frac{\partial{}L}{\partial\dot{x}^\alpha})-\frac{\partial{}L}{\partial{}x^\alpha}=0$.

2. Relevant equations

3. The attempt at a solution
So far I have

$\dot{x}^\alpha=\frac{\partial{}H}{\partial{}p_{\alpha}}=g^{\alpha\beta}p_\beta$
$\dot{p}_\alpha=-\frac{\partial{}H}{\partial{}x^\alpha}=-\frac{1}{2}(g^{\beta\gamma})_{,\alpha}p_\beta{}p_{\gamma}$

where dot means derivation with respect to a parameter $\lambda$.

I know I should substitute one into another. So to get geodesic equation I write

$\ddot{x}^\alpha=\frac{d}{d\lambda}(g^{\alpha\beta}p_\beta)=(g^{\alpha\beta})_{,\mu}p^{\mu}p_{\beta}-\frac{1}{2}g^{\alpha\beta}(g^{\nu\gamma})_{,\beta}p_\nu{}p_\gamma$

but what do I do now?

For lagrangian problem:

$\frac{d}{d\lambda}(g_{\alpha\nu}\dot{x}^\nu+g_{\mu\alpha}\dot{x}^\alpha)-g_{\mu\nu,\alpha}\dot{x}^\mu\dot{x}^\nu=0$

$g_{\alpha\nu,\beta}\dot{x}^\beta\dot{x}^\nu+g_{\alpha\nu}\ddot{x}^\nu+g_{\mu\alpha,\kappa}\dot{x}^{\kappa}\dot{x}^{\alpha}+g_{\mu\alpha}\ddot{x}^\alpha-g_{\mu\nu,\alpha}\dot{x}^\mu\dot{x}^\nu=0$

2. Aug 4, 2011

### latentcorpse

Your latexing has messed up for your lagrangian one but i think you're almost there.

I find that your 2nd line should be

$g_{\alpha \nu, \mu} \dot{x}^\mu \dot{x}^\nu + g_{\alpha \nu} \ddot{x}^\nu + g_{\mu \alpha , \nu} \dot{x}^\mu \dot{x}^\nu + g_{\mu \alpha} \ddot{x}^\mu - g_{\mu \nu , \alpha} \dot{x}^\mu \dot{x}^\nu=0$

Now the $\ddot{x}$ terms can be combined by relabelling dummy indices. then divide the whole thing by 2 and multiply by an inverse metric so as to "isolate" the $\ddot{x}$ term from the metric it's currently paired with. You should now be able to identify the Christoffel symbol term and rewrite your answer as the familiar

$\ddot{x}^\mu + \Gamma^\mu{}_{\nu \alpha} \dot{x}^\nu \dot{x}^\alpha =0$

I haven't been through your hamiltonian one explicitly but i think you're on the right lines. remember if you need an expression for [itex]p_\nu[\itex] you can always multiply [itex]\dot{x}^\alpha = g^{\mu \alpha} p_\mu[\itex] by [itex]g_{\nu \alpha}[\itex]

Last edited: Aug 4, 2011