Suppose that you want to take a covariant derivative $$\nabla$$ of some arbitrary (2,0)-tensor $$T^{\mu\nu}$$such that $$\nabla_{\alpha}T^{\mu\nu}=\partial_{\alpha}T^{\mu\nu}+\Gamma^{\mu}_{\alpha\beta}T^{\beta\nu}+\Gamma^{\nu}_{\alpha\beta}T^{\mu\beta}.$$ In order to calculate the Christoffel symbol we must use metric, such that $$\Gamma^{\mu}_{\alpha\beta}=\frac{1}{2}g^{\mu\sigma}(\partial_{\beta}g_{\sigma\alpha}+\partial_{\alpha}g_{\sigma\beta}-\partial_{\sigma}g_{\alpha\beta}).$$ My question is; is there any other method to calculate $\Gamma^{\mu}_{\alpha\beta}$ (without resorting to computers) or is the metric the only way?
General Relativity – Christoffel Symbols Calculation
differential-geometrygeneral-relativitytensor-calculus
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@Prahar is right, the variation of the Christoffel symbol is a tensor, even if the Christoffel itself is not. We have
$\delta \Gamma^\rho_{\mu\nu}=\frac{1}{2}\delta\bigg(g^{\rho\alpha}(2\partial_{(\mu}g_{\nu)\alpha}-\partial_\alpha g_{\mu\nu})\bigg)=\frac{1}{2}\delta g^{\rho\alpha}(2\partial_{(\mu}g_{\nu)\alpha}-\partial_\alpha g_{\mu\nu})+ \frac{1}{2}g^{\rho\alpha}(2\partial_{(\mu}\delta g_{\nu)\alpha}-\partial_\alpha \delta g_{\mu\nu})$
where $A_{(\mu\nu)}=\frac{1}{2}(A_{\mu\nu}+A_{\nu\mu})$. Using $\delta g^{\rho\alpha}=-g^{\rho\gamma}g^{\alpha\delta}\delta g_{\gamma\delta}$ we have:
$\delta \Gamma^\rho_{\mu\nu}=\frac{1}{2}g^{\rho\alpha}(2\partial_{(\mu}\delta g_{\nu)\alpha}-\partial_\alpha \delta g_{\mu\nu}-2\Gamma_{\mu\nu}^\beta\delta g_{\alpha\beta})$
The Christoffel then combines nicely with the standard derivative to give a covariant tensor (the other Christoffel symbols cancel each other)
$\delta \Gamma^\rho_{\mu\nu}=\frac{1}{2}g^{\rho\alpha}(2\nabla_{(\mu}\delta g_{\nu)\alpha}-\nabla_\alpha \delta g_{\mu\nu})$.
So to answer the original question, we finally have:
$\nabla_\mu V_\nu=\nabla_\mu \delta V_\nu-\frac{1}{2}g^{\rho\alpha}(2\nabla_{(\mu}\delta g_{\nu)\alpha}-\nabla_\alpha \delta g_{\mu\nu})A_\rho$
Remember that we did not assume anything on $V_\mu$. Depending on the problem, it is then possible to integrate by parts to isolate $\delta g_{\mu\nu}$ and obtain the energy momentum tensor.
It took me a couple of days to figure it out, but I can now answer my own question:
The problem with directly substituting the expression for $\delta g_{\mu\nu}$ into that for ${\delta \Gamma^\alpha}_{\beta\mu}$ is that there is no obvious source for the term $\xi^\sigma \partial_\sigma { \Gamma^\alpha}_{\beta\mu}$ that has to be the leading term in the Lie derivative of the connection. We can however proceed as follows $$ \delta {\Gamma^{\alpha}}_{\beta\mu}= \frac 12 g^{\alpha\lambda}(\nabla_\beta(\nabla_\lambda \xi_\mu+\nabla_\mu\xi_\lambda)+ \nabla_\mu(\nabla_\beta \xi_\lambda+\nabla_\lambda\xi_\beta) -\nabla_\lambda(\nabla_\beta \xi_\mu+\nabla_\mu\xi_\beta)) $$ $$ =\frac 12 (\nabla_\beta\nabla_\mu+\nabla_\mu\nabla_\beta)\xi^\alpha +\frac 12 g^{\lambda\alpha}([\nabla_\beta,\nabla_\lambda]\xi_\mu + [\nabla_\mu,\nabla_\lambda]\xi_\beta) $$ $$ =\frac 12 (\nabla_\beta\nabla_\mu+\nabla_\mu\nabla_\beta)\xi^\alpha +\frac 12 g^{\lambda\alpha}(-{R^\sigma}_{\mu\beta\lambda}- {R^\sigma}_{\beta\mu\lambda})\xi_\sigma $$ $$ =\frac 12 (\nabla_\beta\nabla_\mu+\nabla_\mu\nabla_\beta)\xi^\alpha +\frac 12 \xi^\sigma({R^\alpha}_{\beta\sigma \mu}+ {R^\alpha}_{\mu\sigma \beta} ) $$
$$ =\xi^\sigma \partial_\sigma {\Gamma^{\alpha}}_{\beta\mu} +(\partial_\beta \xi^\lambda ){\Gamma^\alpha}_{\lambda\mu}+(\partial_\mu\xi^\lambda) {\Gamma^\alpha}_{\beta \lambda}- (\partial_\lambda \xi^\alpha) {\Gamma^{\lambda }}_{\beta\mu}+ \partial^2_{\beta\mu} \xi^\alpha $$ $$ \equiv {({\mathcal L}_\xi \Gamma)^{\alpha}}_{\beta\mu} $$ In passing from the third line to the fourth we have used some symmetries of the Riemann tensor. We have then (somewhat tediously) substituted the usual expression for the curvature and the covariant derivatives in the last line. I end up with almost the same expression as Bardeen and Zumino, but with a sign difference in the inhomogenous part of the Christoffel symbol transformation
The net result is that first transforming the metric under the diffeomorphism and then computing the connection gets the same result as first computing the connection and then transforming it under a diffeomorphism. This had to be true of course, but there are many places where one can get signs wrong...!
Best Answer
There are other methods. I'll show you one of my favorites and provide references to other techniques. This one is particularly useful when the metric is diagonal on the coordinate system you are using. Take, for example, the metric on the unit sphere, $$g_{\mu\nu} = \begin{pmatrix} 1 & 0 \\ 0 & \sin^2\theta \end{pmatrix}.$$
As you know, the geodesic equation is $$\frac{\textrm{d}^2 x^{\mu}}{\textrm{d} \tau^2} + \Gamma^{\mu}{}_{\nu\sigma} \frac{\textrm{d} x^{\nu}}{\textrm{d} \tau}\frac{\textrm{d} x^{\sigma}}{\textrm{d} \tau} = 0.$$
We also know the geodesic equation can be obtained from the Lagrangian $$L = \frac{1}{2} g_{\mu\nu} \frac{\textrm{d} x^{\mu}}{\textrm{d} \tau}\frac{\textrm{d} x^{\nu}}{\textrm{d} \tau},$$ as shown, e.g., on pp. 43–45 of Bob Wald's General Relativity and also discussed on a recent question you and I are familiar with: Different Lagrangian formulas for geodesics.
Let us consider the particular case for the sphere. The Lagrangian becomes $$L = \frac{\dot{\theta}^2}{2} + \frac{\sin^2\theta \dot{\phi}^2}{2},$$ where the dots denote differentiation with respect to the affine paremeter $\tau$, which here we use just to help with the calculation. The Euler–Lagrange equations are $$\ddot{\theta} - \sin\theta \cos\theta \dot{\phi}^2 = 0$$ and $$\ddot{\phi} + 2 \cot\theta \dot{\theta}\dot{\phi}= 0.$$
If we open up the components of the geodesic equation, it reads $$\ddot{\theta} + \Gamma^{\theta}{}_{\theta\theta}\dot{\theta}\dot{\theta} + 2\Gamma^{\theta}{}_{\phi\theta}\dot{\phi}\dot{\theta} + \Gamma^{\theta}{}_{\phi\phi}\dot{\phi}\dot{\phi} = 0$$ and similarly for $\phi$. Hence, comparing the Euler–Lagrange equations and the geodesic equation (which must coincide), we find that $$\Gamma^{\theta}{}_{\phi\phi} = - \sin\theta \cos\theta, \quad \text{and} \quad \Gamma^{\phi}{}_{\phi\theta} = \cot\theta$$ and all other Christoffel symbols vanish. This is precisely what you would get if you went through the painful calculation.
This is my favorite technique, and works quite well for diagonal metrics, which lead to simple Euler–Lagrange equations. It also works for other metrics, but it gets more cumbersome.
There are still other techniques for computing curvatures. Wald's book discusses a tetrad-based technique on Sec. 3.4b, which I don't remember well enough to discuss here. He also gives an example calculation when dealing with the Schwarzschild metric on Chap. 6. Still on Sec. 3.4b (p. 52, to be exact) he also mentions yet another approach based on spinors, and provides references to it.