Different Ricci tensors for the same metric?

Question

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:

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

Answer 1

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,

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