I was working out the christoffel symbols, once where the metric that I am using has (+---) signature and another time where it has (-+++) signature because two books had different signatures and I had to check for any inconsistencies. I had the christoffel symbols the same for both but the curvatures $(R_{\mu\nu})$ has opposite signs. Why is that?


If you stick to one convention out of many other conventions, you should have same results regardless of signature of metric. Here I follow Carroll's conventions: http://amzn.com/0805387323 or http://arxiv.org/abs/gr-qc/9712019.

For Christoffel symbol, we have \begin{equation} \Gamma^\lambda_{\mu\nu}=\frac{1}{2}g^{\lambda\rho}(\partial_\mu g_{\nu\rho}+\partial_\nu g_{\rho\mu}-\partial_\rho g_{\mu\nu})\,. \end{equation} So if you change signature of the metric $g_{\mu\nu}\to -g_{\mu\nu}$, you have same Christoffel symbol. As a same argument, the definition of Ricci curvature is $$ R_{\mu\nu}=R^\lambda_{\phantom{a}\mu\lambda\nu} $$ where Riemann curvature only depends on Christoffel symbol and its derivative. Therefore we have same Ricci curvature, even if we change the signature of metric.

  • $\begingroup$ Just an added question: The Ricci scalar will change sign right? $\endgroup$ – Prahar Jun 10 '15 at 5:44
  • $\begingroup$ @Prahar Yes, Ricci scalar $R=g^{\mu\nu}R_{\mu\nu}$ will change sign. $\endgroup$ – Frame Jun 10 '15 at 5:46
  • $\begingroup$ I sometimes see the Ricci curvature (which is an addition of derivatives if Christoffel symbols and christoffels symbols multiplied by each other) sometimes as $(\partial \Gamma - \partial \Gamma + \Gamma\Gamma -\Gamma\Gamma)$ and other times as $(-\partial \Gamma + \partial \Gamma - \Gamma\Gamma +\Gamma\Gamma)$, where the latter case was used in one of those books with the (+---) convention and the former was used in (-+++) one. May this be the reason for me having two answers of opposite signs in my Ricci curvatures? @Minkyoto $\endgroup$ – Beyond-formulas Jun 10 '15 at 20:48
  • $\begingroup$ @Beyond-formulas Yes, I think so. If you define Riemann curvature depending on signature of metric as you stated, you get same Ricci scalar in both cases as you easily check. $\endgroup$ – Frame Jun 10 '15 at 21:55
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    $\begingroup$ @Beyond-formulas Yes, you're right. $\endgroup$ – Frame Jun 10 '15 at 22:10

The signature is one convention (both in special relativity and in general relativity).

But in general relativity there are many different conventions besides just the signature. The front inside cover of Misner Thorne and Wheeler lists conventions for signature, for the Riemann Tensor, for the Einstein Tensor, and for the use of Greek and Latin indices and lists 34 texts and what conventions they use. And then spells out on the facing side where the signs go.

So it's just another convention and the Ricci tensor inherits the convention from the Riemann tensor but unfortunately also depends on a convention for the Einstein tensor. So even if you know the convention for one of those tensors you still don't know the sign of the Ricci tensor.

So you will have to also pay attention to the convention for the Einstein tensor as well as the convention for the Riemann tensor.

And the Einstein equation itself can look different depending on the convention for the Einstein tensor.

Finally, that book was published in the 1970s so maybe newer books have decided to have even more conventions to break. I can't say I've read absolutely every new book.


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