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One form of the Robertson-Walker metric is $$ds^2 = c^2dt^2 - a(t)^2[d\chi^2+ S_k(\chi)^2(d\theta^2 + \sin^2\theta ~d\phi^2)]\tag{1}$$ $$\\$$

Considering curvature, where k = 0 , +1, -1 for flat, positive and negative curvatures respectively, then: $$S_k(\chi) = \begin{cases}R\sin(\chi/R)~~~~~~(k=+1)\\\chi~~~~~~~~~~~~~~~~~~~~~~~(k=0)\\R\sinh(\chi/R)~~~~(k=-1)\end{cases}$$ $$\\$$

But another form of this metric is $$ds^2=c^2dt^2 - a(t)^2\left[\frac{dr^2}{1-kr^2} + r^2(d\theta^2+\sin^2~d\phi^2)\right]\tag{2}$$ $$\\$$

How does one go from $(1)$ to $(2)$? (or vice versa) I get that this is due to a switch of choice of coordinates, from radial to co-moving radial, I think? But how can it be shown that they represent the same metric?

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    $\begingroup$ You should use another symbol for the radial coordinate in one of these metrics: $r$ and $\tilde{r}$ for example, since they aren't the same. In the first metric, it is usually $\chi$ instead of $r$. $\endgroup$
    – Cham
    Sep 30, 2019 at 23:55
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    $\begingroup$ Also, consider adding a number to each of your equations, using the LaTeX command \tag{number}. It would help the discussion and answers to come. $\endgroup$
    – Cham
    Sep 30, 2019 at 23:56
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    $\begingroup$ Have you tried the coordinate transformation $r=S_k(\chi)$? $\endgroup$
    – TimRias
    Oct 1, 2019 at 12:23
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    $\begingroup$ Also, $(2)$ needs to have $k$ replace by $k/R^2$ in order to make sense dimensionally. $\endgroup$
    – TimRias
    Oct 1, 2019 at 12:34
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    $\begingroup$ @mmeent, $r$ in (2) doesn't have any units (the scale factor $a(t)$ do have units). So this metric doesn't need $k / R^2$. This version is perfectly standard. I would use $c = 1$, though. Also, $\chi$ shouldn't have any units. The constant $R$ shouldn't be there. $\endgroup$
    – Cham
    Oct 1, 2019 at 13:19

1 Answer 1

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From what I understood in my cosmology course, you simply normalize the metric $(1)$ by taking \begin{equation} \chi(r)=\int\frac{1}{\sqrt{1-kr^2}}\ \mathrm{d}r\tag{int} \end{equation} This integral has 3 different solutions based on the value of $k$ which is your curvature \begin{equation} \chi(r)=\left\{\begin{aligned} &\frac{1}{\sqrt{k}}\sin^{-1}\left(r\sqrt{k}\right)&k>0\\ &r&k=0\\ &\frac{1}{\sqrt{|k|}}\sinh^{-1}\left(r\sqrt{|k|}\right)&k<0 \end{aligned}\right.\tag{sol} \end{equation} Through inversion of $(\mathrm{sol})$ you get your $S_k(\chi)$ function, which gives your $r$ \begin{equation} r(\chi)=S_k(\chi)=\left\{\begin{aligned} &\frac{1}{\sqrt{k}}\sin\left(\chi\sqrt{k}\right)&k>0\\ &\chi&k=0\\ &\frac{1}{\sqrt{|k|}}\sinh\left(\chi\sqrt{|k|}\right)&k<0 \end{aligned}\right.\tag{2} \end{equation} Now, writing our differential we get \begin{equation} \left\{\begin{aligned} \mathrm{d}\chi^2&=\frac{1}{1-kr^2}\mathrm{d}r^2\\\hfill\\ r^2&\left(\mathrm{d}\theta^2+\sin^2(\theta)\mathrm{d}\phi^2\right)=S_k^2(\chi)\left(\mathrm{d}\theta^2+\sin^2(\theta)\mathrm{d}\phi^2\right) \end{aligned}\right. \end{equation} And there you are, it should be what you are searching for if I got the question correctly

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