# R=6k or R=3k confused, FRW metric derivation, Maximally symm

1. Apr 5, 2015

### binbagsss

I'm looking at Tod and Hughston Introduction to GR and writing the metric in the two forms;

[1]$ds^{2}=dt^{2}-R^{2}(t)(\frac{dr^{2}}{1-kr^{2}}+r^{2}(d\theta^{2}+sin^{2}\theta d\phi^{2}))$

[2] $ds^{2}=dt^{2}-R^{2}(t)g_{ij}dx^{i}dx^{j}$

where

$g_{ij}dx^{i}dx^{j}=d\chi^{2}+\chi^{2}(d\theta^{2}+sin^{2}\theta d\phi^{2})$ for $k=0$
$=d\chi^{2}+sin^{2}\chi{2}(d\theta^{2}+sin^{2}\theta d\phi^{2})$ for $k=1$
$=d\chi^{2}+sinh^{2}\chi{2}(d\theta^{2}+sin^{2}\theta d\phi^{2})$ for $k=-1$

Now in solving for the form [2] Tod computes the Ricci scalar of $ds^{2}=d\chi^{2}+f^{2}(\chi)(d\theta^{2}+sin^{2}\theta d\phi^{2})$ and finds $R=-(2\frac{f''}{f}-\frac{1}{f^{2}}+\frac{(f')^{2}}{f^{2}}$ then integrates, uses $R=3k$ and solves for all 3 cases $k=0,\pm 1$.

My question

$R=3k$ doesn't seem right to me, since in 3-d space we can write $R_{ab}=2kg_{ab}$. Of course you could just define a constant $K=2k$, but it uses the constant $k$ in the FRW metric of the form [1] not $k$, comparing to Introduction to GR lecture notes by sean M.Caroll,I thought that this should be $R=6k$

...In Caroll's notes he uses $R_{ab}=2kg_{ab}$ in the derivation and gives the FRW metric the same as in form [1] with small $k$. So it doesn't look as though Tod has used $K=2k$.
Can anyone help explain how Tod uses $R=3k$?

Thanks.

2. Apr 5, 2015

### Staff: Mentor

That's what I get when I compute the Ricci scalar for the spatial part of [1] (i.e., the part inside the parentheses after $R^2(t)$ ).

When I compute the Ricci scalar for this, I get twice the answer you are giving from Tod. So it looks to me like Tod is dividing by 2 somewhere in order to equate his answer for [2] with R = 3k for [1].

3. Apr 6, 2015

### binbagsss

Thanks for your reply. I can't see at all where Tod is dividing by 2 though.
Could the answer possibly lie in the fact that simulatenously rescaling:
$k \to k/| k/$
$r \to \sqrt{| k|} r$
$a \to a/\sqrt{|k|}$
The FRW metric is unchanged?
If it could however, I'm unsure how to justify explicitly.
Thanks.

4. Apr 6, 2015

### Staff: Mentor

I don't see how that would account for the factor of 2. I don't have Tod and Hughston, so I can't say where the discrepancy arises.

5. Apr 20, 2015

### binbagsss

I don't suppose anyone else has any ideas on where this extra factor of 2 is coming from?
Is it possible that $\chi$ could have been re-scaled such that solving the case $R=3k$ is the same as solving for $R=6k$.

Last edited: Apr 20, 2015