# Spherical polar coordinates and the FRW spacetime

The FRW metric of Cosmology is given by $$ds^2=-c^2dt^2+a^2(t)\Big[\frac{dr^2}{1-kr^2}+r^2(d\theta^2+\sin^2\theta d\phi^2)\Big].\tag{1}$$ Now, apart from the overall scale factor $a(t)$, the spatial part of $ds^2$ in Eq. (1), looks very similar to the distance between two infinitesimally closed points with coordinates $(r,\theta,\phi)$ and $(r+dr,\theta+d\theta,\phi+d\phi)$ given by $$d\textbf{l}^2=dr^2+r^2(d\theta^2+\sin^2\theta d\phi^2).\tag{2}$$ In fact, apart from $a(t)$, the spatial part of $ds^2$ becomes exactly same as $d\textbf{l}^2$ of Eq.(2) in the limit $k\to0$.

What does it mean? Does it mean that in an arbitrary curved space the distance between two infinitesimally close points is no longer given by Eq.(2)?

## 1 Answer

That's exactly what it means. The spatial section of FRLW spacetime (that is, at constant $t$) is not necessarily just Euclidean 3D space; in fact, there are three possibilities:

• $k=1$: Space is a 3-sphere. The coordinate $r$ must satisfy $r \le 1$; by defining $r=\sin \psi$, we can make contact with the 3-sphere metric in higher dimensional spherical coordinates: $ds^2 = d\psi^2 + \sin^2\psi\, (d\theta^2 + \sin^2 \theta d\varphi^2)$.
• $k=-1$: Space is what's called hyperbolic 3-space. Topologically it's the same as $\mathbb{R}^3$ (unlike the sphere), but with a different metric. With $r=\sinh \psi$ we get $ds^2 = d\psi^2 + \sinh^2\psi\, (d\theta^2 + \sin^2 \theta d\varphi^2)$. Unfortunately, it can't be visualized as a subset of Euclidean space like the sphere can (it can be visualized as a subset of Minkowski spacetime).
• $k=0$: This is flat Euclidean 3-space.

It's worth noting that even if it might not look like it from the metrics, all three of these spaces are homogeneous and isotropic: every point and every direction looks the same as each other.

• Why can't $k=0$ be the surface of a sphere? On the surface of a sphere, the distance between two points can be written as $d\textbf{l}^2=dr^2+r^2(d\theta^2+\sin^2\theta d\phi^2).$ The surface of a sphere is a curved space, and it's clear that you can write an infinitesimal distance in this fashion. @Javier – SRS Feb 21 '18 at 16:14
• @SRS that's not a sphere, that's flat 3D space in spherical coordinates. – Javier Feb 21 '18 at 16:21
• One can write the metric in any coordinate system in any space. If I want to write the spatial part of the metric of FRW spacetime, why should it be different from $d\textbf{l}^2=dr^2+r^2(d\theta^2+\sin^2\theta d\phi^2)$? – SRS Feb 21 '18 at 16:22
• "that's not a sphere, that's flat 3D space in spherical coordinates." Can I not write the distance between two points on the surface of a sphere as $d\textbf{l}^2=dr^2+r^2(d\theta^2+\sin^2\theta d\phi^2)$? – SRS Feb 21 '18 at 16:23
• @SRS no, the surface of a sphere has constant $r$, so the $dr$ term shouldn't be there. It wouldn't make sense; a sphere is two dimensional, you don't use three coordinates. – Javier Feb 21 '18 at 16:25