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I'm studing the FLRW metric using Daniel Baumann's book, Cosmology (2022), in this book the author derived the FRLW metric using the following equations:

$$ dl^2 = \textbf{dx}^2 \pm du^2 $$ and $$ \textbf{x}^2 \pm u^2 = \pm R_0$$

Where $u$ is a auxiliary fourth dimensional variable, $+R_0$ means for the spherical geometric, the radius of a three-sphere $S^3$ embedded in four-dimensional Euclidean space $E^4$, and $R_{0}^{2}$ means for the hyperbolic geometry a constant determining the curvature of the hyperboloid.

My question sits at the final form of the FLRW metric derived:

$$ ds^2 = -c^2dt^2 + a^2(t) \left[ \frac{dr^2}{1- \frac{kr^2}{R_{0}^{2}}} + r^2d\Omega \right] $$

Where $d\Omega$ is the usual, $k$ has values ${-1,0,+1}$, as normal, but he says now that $R_0$ is the physical curvature scale today, what's that supposed to means?

I'm aware that there is different forms of writing the FRLW metric, one of does includes the $R(t)$ scale factor instead of the $a(t)$, i know too that the $k$ factor comes from the Gaussian curvature factor $K$, is the $R_0$ coming from the gaussian factor? Is the interpretation that when $k = 1$, $R_0$ means the radius of the 3-sphere, correct? If so, what exactly that means? can we measure this radius?

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Using the notation you've defined...

If $k=+1$, then at a fixed time, our Universe is described by a three-dimensional sphere, $S^3$. The radius of that sphere at an arbitrary time $t$ is $a(t)R_0$. When $t$ is equal to the current age of the Universe, then $a(t)=1$, so the radius of the three sphere is $R_0$. If we could determine that $k=+1$, then we would expect to be able to measure the length scale $R_0$. It would be a fundamental scale describing the geometry of our Universe.

Current observations are completely consistent with a flat Universe, $k=0$. Because of this, $R_0$ is not a physical quantity. Note that when $k=0$, $R_0$ does not appear in your expression for the metric $ds^2$ at all. This makes sense; in a flat space with zero curvature, there is no length scale describing the curvature. Therefore, in practice, unless future observations discover some small amount of curvature that has not been observable to date, we don't expect that there is an $R_0$ to measure.

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  • $\begingroup$ Thank you for the answer, in this case the scale factor $R(t)$ seen in other literatures that carries dimension of the length would be $R(t) = a(t)R_0$ ? $\endgroup$ Commented Jan 21, 2023 at 19:42
  • $\begingroup$ @GustavoHenriqueMagro Of course it depends on context, so I can't say as a blanket statement that $R(t)=a(t)R_0$. You need to follow what definitions and conventions each author uses. However, that is common choice of notation. $\endgroup$
    – Andrew
    Commented Jan 21, 2023 at 20:02

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