So, I am doing coursework for maths, and I'm trying to find out how I could calculate the Christoffel symbols for the Schwarzschild metric; however, I have realized that calculating them is too difficult for the level of maths I currently have, as I haven't even finished high school. So I was wondering if anyone knows where I could find them. So far this is the data I have:

The Schwarzschild metric $$\begin{pmatrix} c^2-\frac{2GM}{r} & 0 & 0 & 0 \\ 0 & \frac{-1}{1-\frac{2GM}{rc^2}} & 0 & 0 \\ 0 & 0 & -r^2 & 0 \\ 0 & 0 & 0 & -r^2\sin^2\theta \end{pmatrix}$$ and some of the Christoffel symbols $$\begin{align} \Gamma^t_{tt} &= 0 \\ \Gamma^t_{tr} &= \frac{GM}{r^2c^2}\left(\frac{-1}{1-\frac{2GM}{rc^2}}\right) \\ \Gamma^t_{rt} &= \frac{GM}{r^2c^2}\left(\frac{-1}{1-\frac{2GM}{rc^2}}\right) \\ \Gamma^t_{rr} &= 0 \\ \Gamma^r_{tt} &= \frac{GM}{r^2}\left(1-\frac{2GM}{rc^2}\right) \\ \Gamma^r_{tr} &= 0 \\ \Gamma^r_{rt} &= 0 \\ \Gamma^r_{rr} &= \frac{-GM}{r^2c^2}\left(\frac{-1}{1-\frac{2GM}{rc^2}}\right) \end{align}$$ I want to use some Christoffel symbols I couldn't find. Also, if anything in the formulas is wrong, could you tell me?

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    $\begingroup$ I'm a bit worried about giving suggestion to HS students, who don't have the required proficiency in the mathematical tools. This is quite an advanced topics in Physics and should require some mathematical tools like tensors and tensor fields. It's quite a long jump from HS to the evaluation of Christoffel symbols of Schwarzschild metric...I could tell you that it's just algebra (on mathematical objects that it's likely you can't completely understand), but it's better if I ask you what you exactly need and why $\endgroup$
    – basics
    Commented Oct 23, 2022 at 15:54
  • $\begingroup$ Look into "Lots of Calculations in General Relativity" on physicssusan.mono.net/9035/General%20Relativity, Chapter 4 and/or 11. It should help you. If not then you should specify your need, as @basics has already said. $\endgroup$
    – JanG
    Commented Oct 23, 2022 at 16:33
  • $\begingroup$ @JanGogolin Good reference for very boring algebra. elena, Physics goes well beyond albegra, and in-depth knowledge of tensor calculus (needed for properly doing GR) goes well beyond algebra, as well $\endgroup$
    – basics
    Commented Oct 23, 2022 at 16:37
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    $\begingroup$ Here is a good reference: Catalogue of Spacetimes, pg. 23. Keep in mind that Christoffel symbols are symmetric in lower indices. Those symbols not explicitly listed (up to a symmetry) are assumed zero. But I agree with basics, it may be better to ask about an actual problem you are interested in rather than about specific calculation details if you do not have necessary mathematical background. $\endgroup$
    – A.V.S.
    Commented Oct 23, 2022 at 17:07
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    $\begingroup$ calculating them is too difficult for the level of maths I currently have as I haven't even finished high school All you have to be able to do is differentiate. Have you learned how to do that? The rest is just algebra, and you surely know algebra. $\endgroup$
    – Ghoster
    Commented Oct 23, 2022 at 18:39

1 Answer 1



First of all, let's focus on the notation used in the question. The Schwarzschild metric is, in $(−+++)$ sign convention and units of $c=G=1$: $$ d{s}^2= -\left(1-\dfrac{2M}{r} \right) d{t}^2 + \left(1-\dfrac{2M}{r} \right)^{-1} d{r}^2 +r^2(d{\theta}^2+\sin^2{\theta}d{\phi}^2) $$ The reasons why it is convenient to put both the speed of light and Newton gravitational constant equal to $1$ are various and you can find an exhaustive list of the possible choices of the natural units in the following site.

Moreover, the convention of choosing the signature $(-+++)$ instead of the alternative $(+---)$ relies on the fact that it's the most used in modern textbooks on General Relativity and personally I find myself more comfortable with it. I'm sure you will be able to adjust this notation to your specific needs.

In our case, the covariant metric components form a diagonal matrix, namely: $$ g_{\mu\nu}=\begin{pmatrix} -\left(1-\dfrac{2M}{r} \right) & 0 & 0 & 0\\ 0 & \left(1-\dfrac{2M}{r} \right)^{-1}& 0 & 0\\ 0 & 0 & r^2 & 0\\ 0 & 0 & 0 & r^2\sin^2{\theta} \end{pmatrix} $$


In general, Christoffel symbols are related to the metric in the following way: $$ \Gamma^\sigma_{\mu \nu}=\dfrac{1}{2}g^{\sigma\rho}\left(\partial_\mu g_{\nu\rho}+\partial_\nu g_{\rho\mu}-\partial_\rho g_{\mu\nu} \right)\,\text{,} \qquad \qquad \partial_\mu \equiv \dfrac{\partial}{\partial x^\mu} $$

So, as I suggested in the comments below your question, instead of blindly calculating all $64$ symbols, we can use their symmetry properties and cleverly noting that some of them are identically equal to zero. For example, the Schwarzschild metric doesn't depend on $t$ or $\phi$.

Thus, by setting appropriately the index $\sigma =t$ from the equation above, also $\rho =t$ must be valid because the metric $g^{\sigma \rho}$ doesn't have off-diagonal elements. For the same reason, the last term $\partial_\rho g_{\mu\nu}$ is always zero for every value of $\mu$ and $\nu$ since nothing depends explicitly on time $t$. You have to then force $\mu=t$ in order to have diagonal elements of the metric and since this term only depends on $r$, there's only one possible choice for $\nu=r$. Doing this, you can compute $\Gamma^{t}_{rt}=\Gamma^{t}_{tr}$ and understand that all the other symbols for $\sigma=t$ are vanishing.

This might seem tricky and overcomplicated, but it allows us to discard by simple inspection all the symbols which are trivially equal to zero. An equivalent reasoning might be also implemented for $\sigma =\phi$ with the difference that now the element $g_{\phi\phi}$ depends both on $r$ and $\theta$, so we have two different non-zero symbols to consider.

After that, with a little work it's possible to identify the nonzero Riemann-Christoffel symbols: \begin{equation} \begin{matrix} \Gamma^r_{tt}=\dfrac{M(r-2M)}{r^3}\,\text{,} & ~ & \Gamma^r_{rr}=-\dfrac{M}{r(r-2M)}\,\text{,} & ~ & \Gamma^r_{\theta \theta}=-(r-2M)\,\text{,}\\ ~ & ~ & ~ & ~ & ~\\ \Gamma^r_{\phi \phi}=-(r-2M)\sin^2{\theta}\,\text{,} & ~ & \Gamma^t_{rt} = \dfrac{M}{r(r-2M)}\,\text{,} & ~ & \Gamma^\theta_{r\theta}=\dfrac{1}{r}\,\text{,}\\ ~ & ~ & ~ & ~ & ~\\ \Gamma^\theta_{\phi\phi}=-\sin{\theta}\cos{\theta}\,\text{,} & ~ & \Gamma^\phi_{r\phi}=\dfrac{1}{r}\,\text{,} & ~ & \Gamma^\phi_{\theta\phi} = \dfrac{\cos{\theta}}{\sin{\theta}}\,\text{.}\\ ~ & ~ & \end{matrix} \end{equation}

  • $\begingroup$ Thank you so much $\endgroup$
    – elena
    Commented Oct 26, 2022 at 6:38
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    $\begingroup$ No problem. Hope it was useful. $\endgroup$
    – Gianluca
    Commented Oct 26, 2022 at 18:21
  • $\begingroup$ I actually calculated the time Christoffel Symbols. All that dizzying algebra to tell me— yep it's zero. $\endgroup$ Commented Jan 2 at 9:17

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