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I calculated the formula for the scalar curvature for a spherically symmetric metric, in particular of the form \begin{equation*} g= h(r) dr^2 + r^2(d\theta^2 + \sin^2(\theta) d\phi^2) \end{equation*} and after a lengthy calculation I got the scalar curvature $R$ to be \begin{equation} R(r)=\frac{2h'(r)}{rh^2(r)} - \frac{2}{r^2 h(r)} + \frac{2}{r^2}. \end{equation} Could someone please let me know if this formula is correct? In the case of $h(r)\equiv 1$ the metric is the flat Euclidean metric in spherical coordinates and the formula does indeed give $0$ as it should.

Many thanks!

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Your formula gives the correct answer, $ R=\frac{6}{a^2} $, for the isotropic space also: $$ h=\frac{1}{1-\frac{r^2}{a^2}} $$ (see "classical theory o fields" by Landau and Lifshtz). Moreover I obtained the same result, but this fact is not very important.

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