If we are given a line element;
$$ds^2=dr^2+r^2d\theta^2+r^2sin^2\theta d\varphi^2$$
We can easily then see that the metric and the inverse metric are;
$$g=\begin{pmatrix}1&0&0\\0&r^2&0\\0&0&r^2sin^2\theta\\\end{pmatrix}$$
$$g^{-1}=\begin{pmatrix}1&0&0\\0&r^{-2}&0\\0&0&r^{-2}(\sin\theta)^{-2}\\\end{pmatrix} $$
I have used the formula;
$$\Gamma^m_{ij}=\frac12 g^{ml}(\partial_jg_{il}+\partial_ig_{lj}-\partial_lg_{ji} )$$
Where upper indices indictate the inverse matrix to calculate my Christoffel symbols. My results are as follows:
$$\Gamma^1=\begin{pmatrix}0&0&0\\0&-r&0\\0&0&-rsin^2\theta\\\end{pmatrix}$$$$\Gamma^2=\begin{pmatrix}0&\frac1r&0\\\frac1r&0&0\\0&0&-sin\theta cos\theta\\\end{pmatrix}$$$$\Gamma^3=\begin{pmatrix}0&0&\frac1r\\0&0&cot\theta\\\frac1r&cot\theta&0\\\end{pmatrix}$$
However the results quoted by Wolfram Mathworld () are different, but only in that some entries have been swapped or moved about - all the values in both are the same.
I notice that the article uses the labelling convention $\{r,\varphi,\theta\}$, whereas I have used $\{r,\theta,\varphi\}$ (which is annoying, since the article uses my preferred notation which I have made a conscious effort to move away from due to none of my professors using it), and was wondering whether or not I made a fundamental mistake, or if the different labeling systems could account for the discrepancy.
Thanks in advance!
$\endgroup$ 21 Answer
$\begingroup$The Christoffel symbols you dervied are indeed the correct ones for a spherical coordinate system $(r, \theta, \varphi)$.
If you do the same procedure for a system $(r, \varphi, \theta)$ (in the metric tensor, the entries $(22)$ and $(33)$ are now swapped) you will get the Christoffel symbols as stated on Wolfram Mathworld.
It is simply due to the order of $\theta$ and $\varphi$.
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