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I would like to derive the geodesic equations on a chart on a Riemannian manifold from the Euler-Lagrange equations. What I have so far, is the following computation (basically calculating the things I need in the Euler-Lagrange equation): $$g_{il}\ddot{\gamma}^i + \frac{\partial g_{il}}{\partial x^j}\dot{\gamma}^i\dot{\gamma}^j = \frac{1}{2}\frac{\partial g_{ij}}{\partial x^l}\dot{\gamma}^i \dot{\gamma}^j.$$ Now the book I am following suggest multiplying by $g^{kl}$. Doing this, I end up with: $$\ddot{\gamma}^k + g^{kl}\frac{\partial g_{il}}{\partial x^j}\dot{\gamma}^i\dot{\gamma}^j = \frac{1}{2}g^{kl}\frac{\partial g_{ij}}{\partial x^l}\dot{\gamma}^i \dot{\gamma}^j.$$ Now This should be brought into the form $$\ddot{\gamma}^k + \Gamma^k_{ij}\dot{\gamma}^i\dot{\gamma}^j = 0$$ where $$\Gamma^k_{ij} := \frac{1}{2}g^{kl}\left( \frac{\partial g_{jl}}{\partial x^i} + \frac{\partial g_{il}}{\partial x^j} - \frac{\partial g_{ij}}{\partial x^l}\right).$$ My result looks almost the same, however, I am not able to bring it to the required form, I mean, I have $$\ddot{\gamma}^k + \frac{1}{2}g^{kl}\left(2\frac{\partial g_{il}}{\partial x^j} - \frac{\partial g_{ij}}{\partial x^l}\right)\dot{\gamma}^i \dot{\gamma}^j = 0.$$ But then I do not know how to proceed.

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Let me use a condesed notation, $$\ddot{\gamma}^k + \frac{1}{2}g^{kl}\left(2\partial_j g_{il} - \partial_l g_{ij}\right)\dot{\gamma}^i \dot{\gamma}^j = 0.$$All you have to note is that the expression in parenthesis is symmetric in the indices $i$, $j$ beacause is multiplied by $\dot{\gamma}^i \dot{\gamma}^j$ so you split $2\partial_j g_{il}$ into two term and exchange indices to one.

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  • $\begingroup$ Well...guess I was a bit tired. Thank you! $\endgroup$ – TheGeekGreek Oct 1 '18 at 22:31

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