Derivation of geodesic equation from hamiltonian (lagrangian) equations

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SUMMARY

The discussion focuses on deriving the geodesic equations from Hamiltonian and Lagrangian formulations in the context of differential geometry. The Hamiltonian is defined as H = (1/2)g^{\mu\nu}p_{\mu}p_{\nu}, while the Lagrangian is L = g_{\mu\nu}\dot{x}^\mu\dot{x}^\nu. Participants provided insights on applying the Euler-Lagrange equation and suggested corrections to the user's attempts, emphasizing the importance of correctly handling indices and isolating terms to identify the Christoffel symbols. The final form of the geodesic equation is expressed as \ddot{x}^\mu + \Gamma^\mu{}_{\nu \alpha} \dot{x}^\nu \dot{x}^\alpha = 0.

PREREQUISITES
  • Understanding of Hamiltonian mechanics and Lagrangian mechanics
  • Familiarity with differential geometry concepts, specifically metrics and Christoffel symbols
  • Proficiency in tensor calculus and index notation
  • Knowledge of the Euler-Lagrange equation and its application
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  • Study the derivation of the geodesic equation from the Hamiltonian perspective
  • Learn about the properties and applications of Christoffel symbols in general relativity
  • Explore advanced topics in tensor calculus, focusing on index manipulation
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Students and researchers in theoretical physics, particularly those focusing on general relativity, differential geometry, and classical mechanics. This discussion is beneficial for anyone looking to deepen their understanding of geodesic equations and their derivations.

luinthoron
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Homework Statement


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.

Homework Equations



I'm stuck. Apparantely I got lost in the indices. Could anyone please help?

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
 
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luinthoron said:
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

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 p_\nu[\itex] you can always multiply \dot{x}^\alpha = g^{\mu \alpha} p_\mu[\itex] by g_{\nu \alpha}[\itex]
 
Last edited:

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