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Today I was reading Carrol's book on General Relativity and got a bit confused. In the book, we are given the following metric $$ds^2 = - e^{-2 U(t,r)} dt^2 + e^{2 V(t,r)} dr^2 + r^2 d\Omega^2$$ from which the author calculates the following components of the Ricci, using the Cartan's structural equations:

$$ R_{00} = [\partial_t^2 V + (\partial_t V)^2 - \partial_t U \partial_t V] + e^{2(U - V)}[\partial_r ^2U + (\partial_r U)^2 - \partial_rU \partial_rV + \frac{2}{r} \partial_r U] \\ R_{rr} = -[\partial_r^2U + (\partial_r U)^2 -\partial_rU \partial_rV -\frac{2}{r}\partial_rV] \\ R_{tr} = \frac{2}{r}\partial_t V \\ R_{\theta \theta} = e^{-2V}[r(\partial_rV - \partial_rU) - 1] + 1 \\ R_{\phi \phi} = R_{\theta \theta} \sin^2\theta $$.

Assuming as in my homework that U and V dependent only on r, the components of the Ricci tensor would reduce to $$ R_{00} = e^{2(U - V)}[\partial_r ^2U + (\partial_r U)^2 - \partial_rU \partial_rV + \frac{2}{r} \partial_r U] \\ R_{rr} = -[\partial_r^2U + (\partial_r U)^2 -\partial_rU \partial_rV -\frac{2}{r}\partial_rV] \\ R_{tr} = 0 \\ R_{\theta \theta} = e^{-2V}[r(\partial_rV - \partial_rU) - 1] + 1 \\ R_{\phi \phi} = R_{\theta \theta} \sin^2\theta $$

However in my homework exercise, for the same metric the professor gives us the following components of the Ricci tensor: $$ R_{r0} = 0 \\ R_{00} =e^{-(U + V)}[\partial_r ( e^{(-V + U)} \partial_rU)] + \frac{2}{r} e^{-2V} \partial_rU \\ R_{rr} = -e^{-(U+V)}[\partial_r(e^{(-V + U)} \partial_rU)] + \frac{2}{r} e^{-2V} \partial_rV \\ R_{\theta \theta} = R_{\phi \phi} =\frac{e^{-2V}}{r}(\partial_rV - \partial_rU) + \frac{1}{r^2} (1 - e^{-2V}) $$

How can the same space have different Ricci tensors for the same components?

Edit:

Metric information:

$$ds^2 = g_{\mu\nu}dx^{\mu}dx^{\nu} = \eta_{\mu \nu} \omega^\mu \otimes \omega^\nu $$

where:

$$ \begin{align} \omega^0 = e^{U(r)dt} \\ \omega^1 = e^{V(r)dt} \\ \omega^{\theta} = rd\theta \\ \omega^{\phi} = r\sin\theta d\phi \end{align} $$

After some calculations I got:

$$ds^2 = - e^{2U(r)}dt^2 + e^{2V(r)}dr^2 +r^2d\theta^2 +r^2sin(\theta)^2 d\phi^2 $$

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  • $\begingroup$ In your homework are there hats on the indices? $\endgroup$
    – G. Smith
    Jan 13, 2021 at 18:10
  • $\begingroup$ No, it's just like I wrote in the post. $\endgroup$
    – RFeynman
    Jan 13, 2021 at 18:15
  • $\begingroup$ The fact that $R_{\theta\theta}=R_{\phi\phi}$ indicates to me that your homework is using an orthonormal tetrad (or whatever the right terminology is) rather than plain spherical coordinates (where these components are not equal but related by a factor of $\sin^2\theta$). $\endgroup$
    – G. Smith
    Jan 13, 2021 at 18:38
  • $\begingroup$ I'm going to edit the post and put some information regarding the metric that is given in the sheet. $\endgroup$
    – RFeynman
    Jan 13, 2021 at 18:45
  • $\begingroup$ @G.Smith Now,what do you think? $\endgroup$
    – RFeynman
    Jan 13, 2021 at 18:51

1 Answer 1

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If I'm reading things correctly, it looks to me as though Carroll provides the components of the Ricci tensor in the $dx^\mu$ coordinate basis, while your instructor is providing the components of the Ricci tensor in the orthonormal $\omega^\mu$ basis:

$$\mathbf R = R_{(x) \mu\nu} \big(dx^\mu\otimes dx^\nu\big) = R_{(\omega)\mu\nu} \big(\omega^\mu \otimes \omega^\nu\big)$$

where $R_{(x)\mu\nu}$ and $R_{(\omega)\mu\nu}$ are the Ricci components in the $dx^\mu$ and $\omega^\mu$ bases, respectively.

In this case, the Ricci tensor is diagonal in both bases and the respective basis vectors are related to each other by simple scaling factors. As a result,

$$R_{(x)00} = e^{2U}R_{(\omega)00}$$ $$R_{(x)11} = e^{2V}R_{(\omega)11}$$ $$R_{(x)22} = r^2R_{(\omega)22}$$ $$R_{(x)33} = r^2\sin^2(\theta)R_{(\omega)33}$$

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  • $\begingroup$ It might be off-topic but using Einstein equations to find the metric associated with each components of the Ricci tensor isn't harder in the components that the my professor gave? Can I transform each component the components of the Ricci tensor given by my professor to the ones given by Carroll to make the calculations easier? $\endgroup$
    – RFeynman
    Jan 13, 2021 at 21:17
  • $\begingroup$ @RFeynman It doesn't really make a difference once you're this far. The Einstein equations will be almost identical in either basis - they will differ by the same overall factors as the Ricci tensor components. In other words, the $00$ equation in the $dx^\mu$ basis will be identical to the equation in the $\omega^\mu$ basis, except that every term will be multiplied by $e^{2U}$. $\endgroup$
    – J. Murray
    Jan 13, 2021 at 22:59

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