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Aug 19, 2016 at 15:08 history edited user3264392 CC BY-SA 3.0
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Aug 19, 2016 at 15:02 history edited user3264392 CC BY-SA 3.0
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Aug 19, 2016 at 14:51 history edited user3264392 CC BY-SA 3.0
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Aug 19, 2016 at 14:49 comment added user3264392 I apologise, I competely didn't realise that I had the universal expansion factor $R(t)$ in the same notation as the Ricci scalar $R$. I was trying to start from $k$ as an easy way of seeing positive and negative curvature, going to $K$ which is slightly more involved, finally to the Ricci scalar R which I believe to be the closest thing to a numerical value that determines the nature of the local spacetime. I was basing my answer off chapters in "Relativity, Gravitation and Cosmology" by Robert J. A. Lambourne.
Aug 16, 2016 at 16:47 comment added user4552 This doesn't answer the question, and in any case is nonsense. The R in your equation is not the Ricci scalar, it's a unitless cosmological scale factor. The easiest way to see that your R can't be the Ricci scalar is that this metric has flat spacetime as a special case. In flat spacetime, the Ricci scalar vanishes, but this metric does not produce a flat-spacetime metric when we set R=0.
Aug 16, 2016 at 12:23 history answered user3264392 CC BY-SA 3.0