I was looking at the second-order equations of motion in the Schwarzschild metric and compared them to the Christoffel symbols however they seemed slightly different. I know that the Geodesic Equation is $$\frac{d^2x^\mu}{ds^2} + \Gamma^\mu_{\alpha\beta}\frac{dx^\alpha}{ds}\frac{dx^\beta}{ds} = 0$$ and therefore $\frac{d^2x^\mu}{ds^2} = -\Gamma^\mu_{\alpha\beta}\frac{dx^\alpha}{ds}\frac{dx^\beta}{ds}$, however when looking at the second-order equations of motion in the Schwarzschild metric, particularly the $r$ component, I noticed that the Christoffel symbols were not the same as the written coefficients.

The $r$ equation of motion is $$\ddot{r} = -\frac{GM}{r^2}(1-\frac{2GM}{rc^2})\dot{t}^2 + \frac{GM}{rc^2}\left(\frac{1}{1-\frac{2GM}{rc^2}}\right)\dot{r}^2 + r(1 - \frac{2GM}{c^2})\dot{\theta}^2 + r(1 - \frac{2GM}{c^2})\sin^2(\theta)\dot{\phi}^2.$$

However the first term, which corresponds to $-\Gamma^r_{tt}\dot{t}^2$, where according to Wikipedia: $$\Gamma^r_{tt} = -\frac{GM}{r^2c^2}\left(1-\frac{2GM}{rc^2}\right),$$ and from what I can see, the $\dot{t}$ term would be the negative of the value stated by the geodesic equation and I would like to know why it is negated when the Christofffel symbol is not. I can understand why the $c^2$ in the denomiator goes due to $\dot{t}$ actually being $c\dot{t}$ due to $\frac{d}{d\tau}(ct) = c\dot{t}$, but I do not understand where the minus comes from and why this negative (and also a factor of $c$) is not used on $\Gamma^t_{rt}$.

At first I thought it was because $\dot{t}$ was actually $ic\dot{t}$ and the $ic$ coefficient cancels in $\ddot{t}$ as it would also have that constant imaginary factor. Whereas in $\ddot{r}$, the $ic\dot{t}$ is squared, becoming $-c^2\dot{t}^2$, which would explain the selective $c$ multiplication, and the arrival of the negative. However, when inspecting other metrics' equations of motion, I saw that this would not work in say the Reissner–Nordström metric (according to Wikipedia) or Kerr metric as the imaginary terms would not cancel, so I am left still confused.

I am fairly new to General Relativity and I'm sorry if this question seems very poorly asked, because I recognise that I may not have worded this too well, nor well explained my issue, but I would just like an explanation as to why the selective negation and multiplication of $c$ occurs in some places and not others, as well as this explanation extending to other metrics. Perhaps, I may have misread the Christoffel symbols or the equations I claimed that my initial explanation would not work for, in which case this question is pointless. However, I am extremely young and I would like a firmer understanding on the equations given, as well as study some geodesics. Sorry, for this "essay" but any help would be appreciated!

  • 2
    $\begingroup$ Hey, welcome to the site! The answer to your question is to never, ever, ever use Wikipedia as a source for anything besides trivia. It is collectively written by thousands of volunteers who make tons of mistakes. In this case, they screwed up the sign on the Christoffel symbol. $\endgroup$
    – knzhou
    Commented Sep 9, 2023 at 1:04
  • 1
    $\begingroup$ In the long run, you'll save yourself a lot of time and frustration if you go directly to a good book or set of lecture notes. (In this case, Hartle's book or Tong's lecture notes would be a good fit.) The only good use of a Wikipedia page is the "further reading" section at the bottom, which can direct you to such books. $\endgroup$
    – knzhou
    Commented Sep 9, 2023 at 1:06
  • $\begingroup$ Do you know the formula for calculating the Christoffel symbols from the metric? Will help you verify. $\endgroup$
    – R. Romero
    Commented Sep 9, 2023 at 2:50
  • $\begingroup$ Thank you very much! From now on I will calculate the christoffel symbols by myself using $\Gamma^\lambda_{\mu\nu} = \frac{1}{2}g^{\lambda\sigma}(\partial_\mu g_{\sigma\nu} + \partial_\nu g_{\sigma\mu} - \partial_\sigma g_{\mu\nu})$ $\endgroup$ Commented Sep 9, 2023 at 9:48
  • $\begingroup$ I'm not sure if I went too far but, I edited the Schwarzschild geodesics Wikipedia page to correct the sign on the $\Gamma^r_{tt}$ christoffel symbol to help anyone else who might have this confusion. $\endgroup$ Commented Sep 9, 2023 at 11:47

1 Answer 1


As what the comments have advised, I will not turn to Wikipedia as a source for anything but trivia, as it requires the input of several volunteers who can make mistakes.

So the answer to my question is that $\Gamma^r_{tt} = \frac{GM}{r^2c^2}\left(1-\frac{2GM}{rc^2}\right)$ and Wikipedia made a sign mistake. The "selective" multiplication of $c$ comes from $\dot{t}$ being $c\dot{t}$ and $\ddot{t}$ being $c\ddot{t}$, which in Schwarzschild geodesics, the $c$ coefficients cancel as you can divide both sides by $c$, which keeps the coefficient of $\dot r \dot t$ equal to $-2\Gamma^t_{rt}$. However in $\ddot{r}$, $\Gamma^r_{tt}$ is negated, then multiplied by $c^2\dot{t}^2$ which becomes: $-\frac{GM}{r^2}\left(1-\frac{2GM}{rc^2}\right)\dot{t}^2$, which is the correct term, and explained with no imaginary weirdness.

I can calculate the Christoffel symbols using: $\Gamma^\lambda_{\mu\nu} = \frac{1}{2}g^{\lambda\sigma}(\partial_\mu g_{\sigma\nu} + \partial_\nu g_{\sigma\mu} - \partial_\sigma g_{\mu\nu})$

So in short: Don't always turn to Wikipedia as a source for the maths, find the formula or a book/paper then do it yourself.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.