So the FRW metric is $$ds^2=-c^2dt^2+a(t)^2\left(\frac{dr^2}{1-kr^2}+r^2(d\theta^2+\sin^2\theta d\phi^2)\right). $$

I have seen books mention that $k=1,0,-1$ - depending one of the values listed above ding on curvature type - and $a(t_0)=1$. Are both of these conventions - $k$ being normalised to one of those three values and $a(t_0)$ being normalised to 1 - allowed at the same time (it seems to me that scaling to change the scale factor would also change the curvature)?

edit: a better explanation of the questions after Qmechanic's comment.


The answer is no; you can't have both equal to 1 in the equation you gave, but you can if you introduce $R_0$, as follows.

Friedmann-Lemaitre-Robertson-Walker metric is $$ { ds^2 = -c^2 dt^2 + a^2(t) \left( \frac{dr^2}{1 - K_0 r^2} + r^2\left(d\theta^2 + \sin^2(\theta) d\phi^2\right) \right) } $$ where $K_0$ is a constant. $K_0$ is equal to the Gaussian curvature of space at the time when $a=1$. (The metric is called FLRW since as I understand it, the history here is such that Lemaitre has as much right to be named in this list of names as the others.)

If $K_0 \ne 0$ then we can introduce $$ k = K_0 /\, |K_0| \\ \bar{r} = \sqrt{|K_0|} \, r \\ R = a / \sqrt{|K_0|} $$ and obtain $$ { ds^2 = -c^2 dt^2 + R^2(t) \left( \frac{d\bar{r}^2}{1 - k \bar{r}^2} + \bar{r}^2 \left(d\theta^2 + \sin^2(\theta) d\phi^2 \right) \right) } $$ where now $k$ has one of the values $-1,0,1$. Finally there is nothing to stop us writing $R(t) = R_0 a(t)$ where $a(t_0)=1$. Then we have $$ { ds^2 = -c^2 dt^2 + R_0^2 a^2(t) \left( \frac{d\bar{r}^2}{1 - k \bar{r}^2} + \bar{r}^2 \left(d\theta^2 + \sin^2(\theta) d\phi^2 \right) \right) . } $$ You can now drop the bar if you only want to use this form.

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