Deriving FRW Metric: Ricci Vector Algebra Explained

[binbagsss]

I'm looking at: http://arxiv.org/pdf/gr-qc/9712019.pdf,

deriving the FRW metric, and I don't fully understand how the Ricci Vectors eq 8.5 can be attained from 7.16, by setting $\partial_{0} \beta$ and $\alpha=0$

I see that any christoffel symbol with a $0$ vanish and so so do any Riemann tensors with a $0$, and so only Ricci vectors with $1,2,3$ indices will be non-zero

However, I thought the metric used to compute the Ricci vectors in eq 7.16 - 7.13- would need to reduce to 8.4.
So I see $\beta(t,r) -> \beta(t)$, but I thought also the $dt^{2}$ coefficient would also have to vanish?

Thanks in advance.

 

[PeterDonis]

I thought the metric used to compute the Ricci vectors in eq 7.16 - 7.13- would need to reduce to 8.4.

It does. But that portion of the metric is only the spatial portion, with the $t$ dependence already factored out. See below.

I see $\beta(t,r) -> \beta(t)$ ,

I think you mean $\beta(r)$, correct? In equation 8.4, $\beta$ is a function of $r$ only; the $t$ dependence was already factored out in equation 8.1. See below.

I thought also the $dt^{2}$ coefficient would also have to vanish?

Setting $\alpha = 0$ means the coefficient of $dt^2$, which is $- e^{2 \alpha}$, becomes $-1$. That's what is shown in equation 8.1, which also factors out the $t$ dependence of the spatial part of the metric into the scale factor $a(t)$. Equation 8.4 is then just an equation for what's left in the spatial part in 8.1, i.e., the function $\gamma_{ij}$.