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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!

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    $\begingroup$ 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
    Sep 9, 2023 at 1:04
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    $\begingroup$ 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
    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
    Sep 9, 2023 at 2:50
  • $\begingroup$ Thank you very much! From now on I will calculate the christoffel symbols by myself using $\Gamma^\lambda_{\mu\nu} = \frac{1}{2}g^{\lambda\sigma}(\partial_\mu g_{\sigma\nu} + \partial_\nu g_{\sigma\mu} - \partial_\sigma g_{\mu\nu})$ $\endgroup$ 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. $\endgroup$ Sep 9, 2023 at 11:47

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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.

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