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In the case of a diagonal metric, \begin{align} \mathrm{d}s^2=g_{\mu\nu}\mathrm{d}{x}^\mu\mathrm{d}{x}^\nu, \end{align} it is relatively straightforward to find the Christoffel symbols by comparing the Euler-Lagrange equation \begin{align} \frac{\mathrm{d}}{\mathrm{d}\tau}\left(\frac{\partial L}{\partial \dot{x}^\mu}\right)-\frac{\partial L}{\partial x^\mu}=0, \end{align} where $L=\frac{1}{2}g_{\mu\nu}\dot{x}^\mu\dot{x}^\nu$ and $\dot x^\mu=\mathrm{d}x^\mu/\mathrm{d}\tau$, to the geodesic equation \begin{align} \ddot{x}^\mu+\Gamma_{\rho\sigma}^\mu\dot x^\rho\dot x^\sigma=0. \end{align}

However, this becomes less straightforward for a metric with non-diagonal terms. Additional cross terms in the line element will make not one but two second order derivative terms appear in the Euler-Lagrange equation, making a direct comparison to the geodesic equation less insightful.

Consider for illustrative purposes a 2 dimensional metric \begin{align} \mathrm{d}s^2=f\mathrm{d}t^2+g\mathrm{d}t\mathrm{d}r+h\mathrm{d}r^2, \end{align} with arbitrary functions $f=f(t,r),g=g(t,r),h=h(t,r)$.

In this case the $\mu=t$ and $\mu=r$ components give for the Euler-Lagrange equations respectively \begin{align} 2f\ddot t+g\ddot r+2\left(\frac{\partial f}{\partial t}\dot t+\frac{\partial f}{\partial r}\dot r\right)\dot t+\left(\frac{\partial g}{\partial t}\dot t+\frac{\partial g}{\partial r}\dot r\right)\dot r-\frac{\partial f}{\partial t}\dot t^2-\frac{\partial g}{\partial t}\dot t\dot r-\frac{\partial h}{\partial t}\dot r^2=0\\ 2h\ddot r+g\ddot t+2\left(\frac{\partial h}{\partial t}\dot t+\frac{\partial h}{\partial r}\dot r\right)\dot r+\left(\frac{\partial g}{\partial t}\dot t+\frac{\partial g}{\partial r}\dot r\right)\dot t-\frac{\partial f}{\partial r}\dot t^2-\frac{\partial g}{\partial r}\dot t\dot r-\frac{\partial h}{\partial r}\dot r^2=0. \end{align}

Now the additional $g\ddot r$ in the first and $g\ddot t$ in the second equation forbid a direct comparison to the geodesic equation and subsequently finding the Christoffel symbols.

How do we in general find the Christoffel symbols for a metric with non diagonal terms this way? Is it as simple as substituting one Euler-Lagrange equation in the other to eliminate either of the second order derivative terms?

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  • $\begingroup$ Are your Christoffel symbols different from the Levi-Civita Christoffel symbols? $\endgroup$ – Qmechanic Mar 1 at 18:15
  • $\begingroup$ Why don't you work through how the equation $\ddot x^\mu+\Gamma^\mu_{\rho \sigma} \dot x^\rho \dot x^\sigma$ arsises as the Euler Langrange equation from $\int d\tau g_{mu\nu} \dot x^\mu\dot x^\nu$. That will answer your question for you. (Hint: use $g^{\mu\nu}$) $\endgroup$ – mike stone Mar 1 at 19:00
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If we assume no torsion, and metric compatibility with the connection ($\nabla_{\mu}g_{\alpha\beta}=0$) there is the formula of: $$ \Gamma^{\mu}_{\alpha\beta}=\frac{1}{2}g^{\mu\rho}(\partial_{\alpha}g_{\rho\beta}+\partial_{\beta}g_{\rho\alpha}-\partial_{\rho}g_{\alpha\beta})$$

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  • $\begingroup$ Your geodesic equation is not quite right. It should have a $\Gamma^\mu_{\rho \sigma}$. But more to the point, what do you mean the first geodesic equation only holding for a "diagonal metric"? Thr geodesic eq is the Euler Lagrange eq. Your orgininal $g_{\mu \nu} dx^\mu dx^\nu$ is quite general and not "diagonal". $\endgroup$ – mike stone Mar 1 at 14:38
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    $\begingroup$ I think you meant to commend this on the question? $\endgroup$ – Aylon Pinto Mar 1 at 14:40
  • $\begingroup$ I mean that the geodesic equation is the Euler Lagrange equation for $\int d\tau g_{\mu,\nu}(x) \dot x^\mu \dot x^\nu$. If you don't get it, you are making a mistake. $\endgroup$ – mike stone Mar 1 at 14:43
  • $\begingroup$ Yes I know this, I did not ask the question, I awnsered it... $\endgroup$ – Aylon Pinto Mar 1 at 16:59
  • $\begingroup$ Yes. Sorry! I meant to comment on the question, not your answer! $\endgroup$ – mike stone Mar 1 at 17:14

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