1
$\begingroup$

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)?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – SRS Feb 21 '18 at 16:14
  • $\begingroup$ @SRS that's not a sphere, that's flat 3D space in spherical coordinates. $\endgroup$ – Javier Feb 21 '18 at 16:21
  • $\begingroup$ 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 $d\textbf{l}^2=dr^2+r^2(d\theta^2+\sin^2\theta d\phi^2)$? $\endgroup$ – SRS Feb 21 '18 at 16:22
  • $\begingroup$ "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 $d\textbf{l}^2=dr^2+r^2(d\theta^2+\sin^2\theta d\phi^2)$? $\endgroup$ – SRS Feb 21 '18 at 16:23
  • $\begingroup$ @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. $\endgroup$ – Javier Feb 21 '18 at 16:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.