# Ricci tensor for FRW Metric [closed]

I am attempting to prove that the FLRW metric given by $$ds^2 = -dt^2 + g_{ij}dx^idx^j = -dt^2 + a^2(t)\left(d\vec{x}^2+K\frac{(\vec{x}\cdot d\vec{x})^2}{1-K\vec{x}^2}\right)$$ has $$R_{ij} = \left[\frac{a^{\prime\prime}}{a} + 2 \left(\frac{a^\prime}{a}\right)^2+\frac{K}{a^2}\right]g_{ij}.$$ The other cases, namely $$R_{00}$$ and $$R_{0i}$$ are easy enough to derive, however I am struggling to see how the above result is obtained. I am yet to find a source online that explicitly derives the result for the Ricci tensor in the above case. I am self-taught, and rely entirely upon online resources, so if someone could show me how this was derived, I would very much appreciate it. I think my confusion lies in calculating the Christoffel symbols in this case. It seems like it would be quite messy in Cartesian coordinates. There must surely be an easier way?

• It'd be useful to see where your calculation with the Christoffel symbols has got to (and why it hasn't worked)? Apr 23, 2021 at 18:02
• I hasn't got far at all, other than a general statement of $\Gamma_{jk}^i$ in terms of the metric function (this is all that many texts seem to give). I can calculate them fine when one or more index is 0, but when all three are spatial I seem to have a mental block. Where do I go from here? I await your answer to this problem, given your interest in it. Apr 23, 2021 at 19:43
• Sorry, missed your laste sentence. Why don't you try spherical coordinates? As far as I remember, it works out quite nicely in spherical coordinates. Apr 23, 2021 at 19:50
• Would that give the same answer? As far as I was aware, the $i$, $j$ etc. notation referred to the $x$, $y$ and $z$ components in Cartesian coordinates Apr 23, 2021 at 19:57
• @wrb98 latin indices just refer to the first three spatial components whatever they may be, Greek refer to all four values of the index which includes the temporal pieces. Apr 23, 2021 at 20:51