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I'm trying to find the Ricci tensor in question 3. Here $u=r/R .$

http://imgur.com/gallery/qSAknvz

I found the Christoffel symbols but I can't find the Ricci tensors. On the link, there is also my results for $R_{11}$ and $R_{33}$ but I seem to be far away from the expected result given in the question... I just applied the definition of Ricci tensor, using the Christoffel symbols and $g$ but I'm stuck.

I also can't find the final result for $R ;$ I believe my metric tensor is wrong but I don't why. $$g_{ij} = \operatorname{diag}\left(\mathrm{d}u^2 part, \, \mathrm{d} \theta^2 part , \, \mathrm{d} \phi^2 part\right)$$ And $$g= \det\left(g_{ij}\right)= \text{product of these 3 parts} ?$$

Edit: I did this already for Schwarzschild solution and it worked well. Maybe it's a calculus error but I checked multiple times both of my trials... I believe this is my metric tensor which is wrong

Edit 2: So I tried with Riemann tensor (as @ggcg suggested) rather than direct calculus with Ricci tensor (so no direct use of g) and found the same result for R11. My answer might be the good one, but I don't see then how to go from my answer to the one given in the exercise.

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  • $\begingroup$ You need the Riemann tensor then contract two indices. $\endgroup$ – ggcg Jan 16 at 23:42
  • $\begingroup$ @ggcg So I tried with Riemann tensor rather than direct calculus with Ricci tensor (so no direct use of g) and found the same result for R11. My answer might be the good one, but I don't see then how to go from my answer to the one given in the exercise. $\endgroup$ – Korlek Jan 17 at 5:50
  • $\begingroup$ I don't understand how you didn't use g, that is part of the Christoffel symbol, right? $\endgroup$ – ggcg Jan 17 at 11:36
  • $\begingroup$ Or are you using some reduced formula that makes explicit use of g? $\endgroup$ – ggcg Jan 17 at 11:36
  • $\begingroup$ If $\Gamma_a{}^b{}_c=g^{bm}\Gamma_{amc}$ and $g^{am}g_{mb}=\delta^a_b$, then the expressions given for $\Gamma_a{}^b{}_c$ are not entirely correct. Further I am not sure why one would need $\Gamma_a{}^b{}_c$ anyway and not $\Gamma^a{}_{bc}$. Using $\Gamma^a{}_{bc}$ the non-vanishing components of $R_{abcd}$ are $R_{1212}=\frac{k R^2 u^2}{1-k u^2}$, $R_{1313}=\frac{k R^2 u^2\sin^2\theta}{1-k u^2}$, $R_{2323}=k R^2 u^4\sin^2\theta$ and components related to them via interchange- and skew-symmetry. Using those components one gets $R_{ab}=\frac{2k}{R^2}g_{ab}$. $\endgroup$ – N0va Jan 22 at 1:17

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