# Different Ricci tensors for the same metric?

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:

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

• In your homework are there hats on the indices? Jan 13, 2021 at 18:10
• No, it's just like I wrote in the post. Jan 13, 2021 at 18:15
• 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 $\sin^2\theta$). Jan 13, 2021 at 18:38
• I'm going to edit the post and put some information regarding the metric that is given in the sheet. Jan 13, 2021 at 18:45
• @G.Smith Now,what do you think? Jan 13, 2021 at 18:51

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.
$$R_{(x)00} = e^{2U}R_{(\omega)00}$$ $$R_{(x)11} = e^{2V}R_{(\omega)11}$$ $$R_{(x)22} = r^2R_{(\omega)22}$$ $$R_{(x)33} = r^2\sin^2(\theta)R_{(\omega)33}$$
• @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 $e^{2U}$. Jan 13, 2021 at 22:59