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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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You can introduce the Christoffel symbols alternatively through the parallel transport and the covariant derivative.
The geodesic is characterized by its normalized velocity $dx^{i}/ds$ being parallel-transported (i.e. it's constant in magnitude and direction under the metric), meaning its covariant derivative vanish.
You can work out the geodesic differential equations from this, and the Christoffel symbols will appear through the covariant derivative.

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If you REALLY want to solve for the christoffel symbols this way, you'll basically always end up with a bunch of second derivative terms, and a bunch of products of first derivative terms. You can always just factor them into a 4x4 matrix $A_{ij}$ so that your four variation equations look like

$A_{ij}\frac{d^{2}x^{i}}{ds^{2}} = \left({\rm products\, of\, first\, derivative\, terms}\right)_j$

Then, just invert the matrix A, and you have your four geodesic equations.

It's going to be less work to just use the general formula for the Christoffel symbols, though, except maybe in some very specialized cases.

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