Suppose that you are given an arbitrary metric $g_{\mu\nu}$ such that you want to calculate all of the Christoffel symbol $\Gamma^{\lambda}_{\mu\nu}$. The equation for a Christoffel symbols are $$\Gamma^{\lambda}_{\mu\nu}=\frac{1}{2}g^{\lambda\sigma}\left(\partial_{\nu}g_{\sigma\mu}+\partial_{\mu}g_{\sigma\nu}-\partial_{\sigma}g_{\mu\nu}\right).$$ It is pretty simple to calculate all 40 of the unique Christoffel symbols, just plug in values of the metric. For instance, given an arbitary metric $$ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$$ where $ds^2$ is arbitary. For arbitrary metrics, the Christoffel symbols $$\Gamma^{\lambda}_{\mu\nu}=\frac{1}{2}g^{\lambda\sigma}\left(\partial_{\nu}g_{\sigma\mu}+\partial_{\mu}g_{\sigma\nu}-\partial_{\sigma}g_{\mu\nu}\right)$$ dummy index $\sigma$ can be $$x^0,x^1,x^2,x^3.$$ How do you determine what value the dummy takes? For instance $\Gamma^{3}_{22}$; this would obviously be $$\Gamma^{3}_{22}=\frac{1}{2}g^{3\sigma}\left(\partial_2g_{\sigma 2}+\partial_2g_{\sigma 2}-\partial_{\sigma}g_{22}\right).$$ How would I choose what coordinate $\sigma$ takes? Would it take all $x^0,x^1,x^2,x^3$ and sum those metric terms up such that it becomes $$\Gamma^{3}_{22}=\frac{1}{2}g^{3 \sigma}\partial_2 g_{\sigma2}+\frac{1}{2}g^{3 \sigma}\partial_2 g_{\sigma2}-\frac{1}{2}g^{3 \sigma}\partial_{\sigma}g_{22}=(\frac{1}{2}g^{3 0}\partial_2 g_{02}+\frac{1}{2}g^{3 1}\partial_2 g_{12}+\frac{1}{2}g^{3 2}\partial_2 g_{22}+\frac{1}{2}g^{3 3}\partial_2 g_{32})+(\frac{1}{2}g^{3 0}\partial_2 g_{02}+\frac{1}{2}g^{3 1}\partial_2 g_{12}+\frac{1}{2}g^{3 2}\partial_2 g_{22}+\frac{1}{2}g^{3 3}\partial_2 g_{32})-(\frac{1}{2}g^{3 0}\partial_{0}g_{22}+\frac{1}{2}g^{3 1}\partial_{1}g_{11}+\frac{1}{2}g^{3 2}\partial_{2}g_{22}+\frac{1}{2}g^{3 3}\partial_{3}g_{33})?$$ If that is the case is there any way to simplify such that the simplification holds for any combination of $x^0,x^1,x^2,x^3?$ Furthermore what is the simplification for diagonal metrics and the naming of the dummy index?

  • 2
    $\begingroup$ It's always a long calculation for the general case. That's why the are computer codes for this stuff. If there is a lot of symmetry it is better to use differential forms rather than tensors with indices --- many of the terms will vanish. $\endgroup$
    – mike stone
    Nov 30, 2021 at 22:00

2 Answers 2


You sum over all $\sigma$, because it's a repeated index. It's called contraction. For example,$$\begin{align}\Gamma_{22}^3&=\frac12g^{30}(\partial_2g_{02}+\partial_2g_{02}-\partial_0g_{22})\\&+\frac12g^{31}(\partial_2g_{12}+\partial_2g_{12}-\partial_1g_{22})\\&+\frac12g^{32}(\partial_2g_{22}+\partial_2g_{22}-\partial_2g_{22})\\&+\frac12g^{33}(\partial_2g_{32}+\partial_2g_{32}-\partial_3g_{22})\end{align}.$$Luckily, when you do these calculations you often find lots of terms vanish, usually due to symmetries. This happens to an especially large number of terms with a symmetry-respecting choice of coordinate system. For diagonal metrics, we unfortunately need to deactivate contraction in our notation to understand the simplification:$$\Gamma_{\mu\nu}^\lambda=\frac12g^{\lambda\lambda}(\color{red}{\partial_\mu g_{\lambda\nu}}+\color{blue}{\partial_\nu g_{\lambda\mu}}-\color{limegreen}{\partial_\lambda g_{\mu\nu}})$$does not sum over $\lambda$. The red term vanishes if $\lambda\ne\nu$; the blue term vanishes if $\lambda\ne\mu$; the green term vanishes if $\mu\ne\nu$. Therefore, cases where $\mu,\,\nu,\,\lambda$ are all different give $\Gamma_{\mu\nu}^\lambda=0$. Terms may also vanish due to a derivative's effect.

  • $\begingroup$ Isnt there a simplification for diagonal metrics? $\endgroup$
    – aygx
    Nov 30, 2021 at 22:01
  • 1
    $\begingroup$ @aygx There certainly is. See if you can work it out. (I'll try to edit it in anyway.) $\endgroup$
    – J.G.
    Nov 30, 2021 at 22:02

I suggest the following method of calculation, which does not require any particular theorem, fits to both diagonal and non-diagonal metrics and can be fully automated. Since it relies on all the possible locations of all the obviously non-zero partial derivatives, it provides immediately all non-zero Christoffel symbols instead of searching for the zero symbols.

Assuming that the Christoffel symbols are given by $\quad \Gamma_{ij}^{k}=\frac 1 2 g^{km}(g_{mj,i}+g_{mi,j}-g_{ij,m})\quad $ :

  1. list all the non-zero partial derivatives, beginning with the diagonal metrics ; let n be their number,

  2. build a table with n+1 rows and 3 columns and write :

    • in row 1 : column 1 : title "NZPD" for "non-zero partial derivatives" ; column 2 : literal expression of the first term inside parentheses $g_{jm,i}$; a column with the second term $g_{im, j}$ is useless since the permutation of the values of the i,j indices yields equal symbols, so column 3 : literal expression $g_{ij, m}$

    • from row 2 up to n+1 in column 1 : successively all partial derivatives given by step 1.

  3. in each of the empty cells give to each of the indices of the expression at the head of the column the values of the indices located at the same place at the head of the row,

  4. from the value of m get all the possible values for k from the (known) non-zero $g^{km}$,

  5. from the current values of k, i and j get the "names" of all the non-zero Christoffel symbols.

The non diagonal metrics lead to obvious redundancies because of previous occurences of the same upper index k, but note pragmatically that they can easier be understood afterwards than predicted through a somewhat sophisticated specific rule …

As an example, consider a horizontal rotating disc with constant velocity $\omega$ as a reference frame.

The square of the space-time distance with respect to polar coordinates system $\left[\begin{array} {c}dx^1\\dx^2\\dx^0\end {array}\right]=\left[\begin{array} {c}dr\\d\vartheta\\cdt\end {array}\right]$ reads $ds^2 = dr^2 + r^2 d\vartheta^2 + 2 r^2 \omega d\vartheta dt - \left ( 1 - \frac{r^2 \omega^2}{c^2} \right)c^2 dt^2$ which yields the non-diagonal metrics $$\left[ g_{km}\right]=\left[\begin{array}{cc}1&0&0\\0&r^2&\frac{r^2\omega}{c}\\0&\frac{r^2\omega}{c}&-\left(1-\frac{r^2\omega^2}{c^2}\right)\end{array}\right]$$

The non-zero $g^{km}$ are $\quad g^{11}=1\quad g^{22}=\frac{1}{r^2}\quad g^{00}=-\frac{1}{\left(1-\frac{r^2\omega^2}{c^2}\right)}\quad g^{20}=\frac{c}{r^2\omega}=g^{02}$

The list of the non-zero partial derivatives is obvious : $$\quad g_{22,1}= 2r\quad g_{00,1}= 2r\frac{\omega^2}{c^2}\quad g_{20,1}= 2r \frac{\omega}{c}$$ and the table becomes : \begin{array}{c|c} NZPD & g_{mj,i} & g_{ij,m} \\\hline g_{22,1} & \begin{cases}i=1, \ j=2\\ m=2\Rightarrow \begin{cases}\Gamma^2_{{12}}={\Gamma^2_{21}}\neq 0 & \text{for k=2} \\{\Gamma^0_{12}}={\Gamma^0_{21}}\neq 0 & \text{for k=0} \end{cases} \end{cases}& \begin{cases} i=2, \ j=2\\ m=1\Rightarrow k=1 \Rightarrow {\Gamma^1_{22}}\neq 0 \end{cases} \\ \hline g_{00,1} & \begin{cases}i=1, \ j=0\\m=0\Rightarrow \begin{cases}{\Gamma^0_{10}}={\Gamma^0_{01}}\neq 0 & \text{for k=0} \\{\Gamma^2_{10}={\Gamma^2_{01}}}\neq 0 & \text{for k=2} \end{cases} \end{cases} & \begin{cases}i=0, \ j=0\\ m=1\Rightarrow k=1 \Rightarrow{\Gamma^1_{00}}\neq 0 \end{cases} \\ \hline g_{20,1} & \begin{cases}i=1, \ j=0\\ m=2\Rightarrow \begin{cases}{redundant} \\{\Gamma^2_{10}}={\Gamma^2_{01}}\neq 0 & \text{for k=2} \\{\Gamma^0_{10}}={\Gamma^0_{01}}\neq 0 & \text{for k=0} \end{cases} \end{cases} & \begin{cases}i=2, \ j=0\\ m=1\Rightarrow k=1 \Rightarrow{\Gamma^1_{20}}={\Gamma^1_{02}}\neq 0 \end{cases} \end{array}

12 non redundant Christoffel symbols have been found, among which 7 are independent because of lower indices symmetries.

$$\begin{array}{l} \Gamma^1_{22} = -r&\Gamma^1_{20}=\Gamma^1_{02}=-\frac{r\omega}{c} &\Gamma^1_{00}=-\frac{ r %oméga^2} {c^2} \\\Gamma^2_{12}=\Gamma^2_{21} = \frac{1}{r}&\Gamma^2_{10}=\Gamma^2_{01}= \frac{\omega}{r\,c} \\\Gamma^0_{12}=\Gamma^0_{21} = \frac{r \omega}{g_{00}\,c} &\Gamma^0_{10}=\Gamma^0_{01} = \frac{r \omega^2}{g_{00}\,c^2} \end{array}$$

  • $\begingroup$ That's actually very helpful! Thank you! $\endgroup$
    – aygx
    Nov 7, 2022 at 2:35

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