Let M be differentiable manifold that represents spacetime, s.t. $\Gamma^{\lambda}_{\mu\nu}$ is the Christoffel symbol/connection coefficient. The general formula for the christoffel symbol is defined as $$\Gamma^{\lambda}_{\mu\nu}=\frac{1}{2}g^{\lambda\sigma}(\partial_{\nu}g_{\sigma\mu}+\partial_{\mu}g_{\sigma\nu}-\partial_{\sigma}g_{\mu\nu})$$ where $\partial_\nu=\frac{\partial}{\partial x^\nu}.$ Now consider the Schwarzschild metric $$ds^2=-\left(1-\frac{2M}{r}\right)dt^2+\left(1-\frac{2M}{r}\right)^{-1}dr^2+r^2d \theta^2+r^2\sin(\theta)d \phi^2.$$ In order to calculate the christoffel symbols for this metric how would we proceed? Would we use the formula $$\Gamma^{\lambda}_{\mu\nu}=\frac{1}{2}g^{\lambda\sigma}(\partial_{\nu}g_{\sigma\mu}+\partial_{\mu}g_{\sigma\nu}-\partial_{\sigma}g_{\mu\nu})$$ or could we use $$\Gamma^{\lambda}_{\mu\nu}=\frac{\partial x^\lambda}{\partial x^\alpha}\frac{\partial^2x^\alpha}{\partial x^{\mu} \partial x^{\nu}}.$$ IF=f we use equation $$\Gamma^{\lambda}_{\mu\nu}=\frac{\partial x^\lambda}{\partial x^\alpha}\frac{\partial^2x^\alpha}{\partial x^{\mu} \partial x^{\nu}}$$ and want to compute say $\Gamma^{\theta}_{r\theta}$ (along with all the other christoffel symbols, this one is just an example, the expression would follow as $$\Gamma^{\theta}_{r\theta}=\frac{\partial \theta}{\partial t}\frac{\partial^2 t}{\partial r \partial \theta}+\frac{\partial \theta}{\partial r}\frac{\partial^2 r}{\partial r \partial \theta}+\frac{\partial \theta}{\partial \theta}\frac{\partial^2 \theta}{\partial r \partial \theta}+\frac{\partial \theta}{\partial \phi}\frac{\partial^2 \phi}{\partial r \partial \theta}.$$ To find this christoffel symbol, would we just find the values of $$\frac{\partial \theta}{\partial t}\frac{\partial^2 t}{\partial r \partial \theta}+\frac{\partial \theta}{\partial r}\frac{\partial^2 r}{\partial r \partial \theta}+\frac{\partial \theta}{\partial \theta}\frac{\partial^2 \theta}{\partial r \partial \theta}+\frac{\partial \theta}{\partial \phi}\frac{\partial^2 \phi}{\partial r \partial \theta}$$ using the equations for each symbol like $r=\sqrt{x^2+y^2}$. Can we even use this method for a specific metric or does it have to be the metric formula? If we can use this equation for any metric what steps would be required for finding $$\Gamma^{\theta}_{r\theta}?$$ Likewise for $$\Gamma^{\phi}_{r\phi}?$$

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    $\begingroup$ The equation that involves the metric is the right one. I have no idea what the other one is. $\endgroup$ Commented Nov 30, 2021 at 18:03

1 Answer 1


The first expression you wrote is the correct one: $$\Gamma^{\lambda}_{\mu\nu}=\frac{1}{2}g^{\lambda\sigma}(\partial_{\nu}g_{\sigma\mu}+\partial_{\mu}g_{\sigma\nu}-\partial_{\sigma}g_{\mu\nu})\tag{1}$$

The other one is meaningless as written (it is identically equal to zero), but I think I know where it came from. Let's say you are working in a coordinate chart with coordinates $x$, in which the Christoffel symbols are given by $\big[\Gamma_{(x)}\big]^\lambda_{\mu\nu}$. I use the subscript $(x)$ to remind us that these are the Christoffel symbols in the $x$ coordinates. If I now change to coordinates $y$, then the Christoffel symbols in the new chart are given by $$\big[\Gamma_{(y)}\big]^\lambda_{\mu\nu} = \frac{\partial x^\alpha}{\partial y^\mu} \frac{\partial x^\beta}{\partial y^\nu} \frac{\partial y^\lambda}{\partial x^\rho} \big[\Gamma_{(x)}\big]^\rho_{\alpha\beta} + \frac{\partial y^\lambda}{\partial x^\rho} \frac{\partial^2 x^\rho}{\partial y^\mu \partial y^\nu}\tag{2}$$

In particular, if you are working in flat spacetime and the $x$ coordinates are Cartesian, then $\big[\Gamma_{(x)}\big]^\lambda_{\mu\nu}=0$ and so you're left with $$\big[\Gamma_{(y)}\big]^\lambda_{\mu\nu} = \frac{\partial y^\lambda}{\partial x^\rho} \frac{\partial^2 x^\rho}{\partial y^\mu \partial y^\nu}\tag{3}$$

This is a very special case, however; it assumes that you're working on a flat manifold so Cartesian coordinates exist, and that the $x$'s in question are those Cartesian coordinates. If for example you were working in the Euclidean plane and chose $(y^1,y^2)=(r,\theta)$ to be the standard polar coordinates, then expression $(3)$ would work.

On a spacetime with curvature, all of this falls apart. You can compute the Christoffel symbols in any particular coordinate chart using $(1)$, and can relate the Christoffel symbols in two different charts using $(2)$, but since there generally are no coordinate systems in which the $\Gamma$'s all vanish everywhere, $(3)$ has to go out the window.


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