1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Euler-Lagrange Equations for geodesics

  1. Sep 10, 2017 #1
    1. The problem statement, all variables and given/known data
    The Lagrange Function corresponding to a geodesic is $$\mathcal{L}(x^\mu,\dot{x}^\nu)=\frac{1}{2}g_{\alpha \beta}(x^\mu)\dot{x}^\alpha \dot{x}^\beta$$

    Calculate the Euler-Lagrange equations

    2. Relevant equations
    The Euler Lagrange equations are $$\frac{\mathrm{d}}{\mathrm{d}s} \left( \frac{\partial \mathcal{L}}{\partial \dot{x}^\mu} \right) =\frac{\partial \mathcal{L}}{\partial x^\mu}$$

    The solution should be
    $$\frac{\mathrm{d}^2 x^\mu}{\mathrm{d}s^2}+\Gamma^\mu_{\;\;\alpha \beta} \frac{\mathrm{d}x^\alpha}{\mathrm{d}s} \frac{\mathrm{d}x^\beta}{\mathrm{d}s}=0$$

    3. The attempt at a solution

    Calculate LHS:

    $$\frac{\partial \mathcal{L}}{\partial \dot{x}^\mu}=\frac{1}{2}\dot{x}^\beta (g_{\mu\beta}+g_{\beta\mu})=\dot{x}^\beta g_{\mu \beta}$$

    $$\frac{\mathrm{d}}{\mathrm{d}s} \left( \frac{\partial \mathcal{L}}{\partial \dot{x}^\mu} \right)=\frac{\mathrm{d}}{\mathrm{d}s} \left(\dot{x}^\beta g_{\mu \beta}\right)=\ddot{x}^\beta g_{\mu\beta}$$

    Calculate RHS:

    $$\frac{\partial \mathcal{L}}{\partial x^\mu}=\frac{1}{2} \partial_m g_{\alpha \beta}(x^\mu)\dot{x}^\alpha\dot{x}^\beta$$

    Equating and multiplying with ##g^{kl}##:

    $$ \ddot{x}^\beta=\frac{1}{2} g^{\mu \beta} \partial_m g_{\alpha \beta}(x^\mu)\dot{x}^\alpha\dot{x}^\beta $$

    This kind of looks like the definition of the Christoffel Symbol that I need, however two derivations of the metric are missing. Where did I go wrong?

  2. jcsd
  3. Sep 10, 2017 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member
    2017 Award

    You did not go wrong. I suggest you check what the missing terms are given that they are contracted with the ##\dot x^\mu##.

    Edit: Note that what you compute with the EL is the affinely parametrised curve that minimises the distance. This is the same as the geodesic only if the connection is the Levi-Civita connection.
  4. Sep 10, 2017 #3
    Ok, I just realized my indecess are off.

    $$\ddot{x}^\beta \underbrace{g^{kl}g_{\mu\beta}}_{\delta^k_{\;\mu} \delta^l_{\; \beta}}=\frac{1}{2} g^{kl} \partial_\mu g_{\alpha \beta}(x^\mu)\dot{x}^\alpha \dot{x}^\beta$$
    $$ \ddot{x}^l = \frac{1}{2} g^{kl} \partial_\mu g_{\alpha \beta}(x^\mu)\dot{x}^\alpha \dot{x}^\beta$$

    However, that is a problem, as I would need the ##k## index to appear in the ## \partial_\mu g_{\alpha \beta} ## term, wouldn't I?

    Then, I'm still missing some terms along the lines of ## \partial_\alpha g_{k \mu} - \partial_k g_{\mu \alpha} ##, and the signs are all wrong.

    I really don't see how to get there!
  5. Sep 10, 2017 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member
    2017 Award

    Ignore what I said in #2. I had some more time to read your post more carefully.

    I suggest you drop the argument of the metric because it seems that its index is confusing you (it is not an actual index in the equation, it is just telling you that the metric depends on the position, which you should already know). In addition, this is not correct:
    What you are looking for is to contract one of the indices of the inverse metric with the free index of the metric, i.e.,
    g^{k\beta} g_{\mu\beta} = \delta^k_\beta.

    Then look at the Christoffel symbols of the Levi-Civita connection
    \Gamma_{\mu\nu}^\lambda = \frac{1}{2} g^{\lambda\rho} (\partial_\mu g_{\rho\nu} + \partial_\nu g_{\mu\rho} - \partial_{\rho}g_{\mu\nu}).
    What happens when you insert this into the geodesic equation
    (\nabla_{\dot\gamma} \dot\gamma)^\mu = \ddot{x}^\mu + \Gamma_{\alpha\beta}^\mu \dot x^\alpha \dot x^\beta = 0?

    Checking it closer, you are actually missing terms on the LHS. Consider this step more carefully:
    In addition, in this step on the RHS
    suddenly an ##m## and a ##\mu## appeared that are unmatched on the LHS, they should be the same (and again, you would probably do better to drop writing out the index on the argument of the metric, it is just confusing you). Also, your index ##\beta## appears three times in the RHS and only once on the LHS. Probably you should use a different index in the inverse metric that you are multiplying by not to confuse the indices.
  6. Sep 10, 2017 #5
    Thanks! I think I get it now!
  7. Sep 11, 2017 #6
    Ok, sorry, but I still don't get it.

    When you refer to missing terms, I assume you mean the product rule in the following equation?

    $$\frac{\mathrm{d}}{\mathrm{d}s} \left( \frac{\partial \mathcal{L}}{\partial \dot{x}^\mu} \right)=\frac{\mathrm{d}}{\mathrm{d}s} \left(\dot{x}^\beta g_{\mu \beta}\right)=\ddot{x}^\beta g_{\mu\beta}+\dot{x}^\beta \frac{\mathrm{d}}{\mathrm{d}s} g_{\mu \beta}$$

    I really don't know what do do with ##\frac{\mathrm{d}}{\mathrm{d}s} g_{\mu \beta}##. Doesn't that bring me even further from the required form?

    And I also can't seem to work out how inserting the Christoffel symbol in terms of the metric into the equation helps me.
  8. Sep 12, 2017 #7


    User Avatar
    Science Advisor

    No, it will bring you closer. Use the chain rule:
    $$\frac{d}{ds} X ~=~ X,_\nu \, \dot x^\nu $$
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted