I need help computing the effect of curvature on the FRW metric

Question

Apparently there are different forms of the FLRW metric. I'm focusing on Anti-de Sitter space, so I'll just give the hyperbolic version of the function. $$ds^2=-c^2dt^2+a^2(t)\left[dr^2+R_0\space \sinh\left(\frac{r}{R_0}\right)d\Omega^2\right]\tag 1$$ $$d\Omega^2=d\theta^2+sin^2 \theta\space d\phi^2$$ Here, $d\Omega$ is the angular separation of two points in the sky, but I'm not interested in two points in the sky. I'm analyzing SNe Ia data, so I'm just working in line-of-sight measurements, so as I understand this, the $d\theta$ and $d\phi$ terms go to zero (that is, there's no change in the angle), so the whole $d\Omega$ term is zero. This leaves us with: $$ds^2=-c^2dt^2+a^2(t)dr^2$$ Which doesn't seem right. Is the distance in a hyperbolic plane the same as a flat plane or closed surface if you're not dealing with angular separation? Am I interpreting the metric correctly?

EDIT: The other form of the FRW metric seems to suggest the curvature changes the length of a line-of-sight measurement. $$ds^2=-c^2dt^2+a^2(t)\left[\frac{dr^2}{1+k\space r^2}+r^2d\Omega^2\right]\tag 2$$ Where $k$ is either a scalar (1 for closed, 0 for flat, -1 for saddle) or the Gaussian Curvature (still not sure how that's used). Setting the angular separation to zero, you get: $$ds^2=-c^2dt^2+a^2(t)\frac{dr^2}{1+k\space r^2}$$ This seems to fly in the face of the version in (1), so I'm missing some major concept here. Can anyone tell me what that is?

Answer 1

The correct equation should be

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

Here $r$ is the usual radial coordinate.

We may re-define the radial coordinate by taking $$d\chi = \frac{dr}{\sqrt{1-\kappa r^2}}$$ such that

$$r = \begin{cases} sinh(\chi), & \kappa = -1 \\ \chi & \kappa = 0 \\ sin(\chi) & \kappa = 1 \end{cases}$$

This implies $$ds^2 = -c^2dt^2 + a^2(t)[d\chi^2 + S_{\kappa}^2(\chi)d\Omega^2]$$

where

$$S_{\kappa}(\chi) = \begin{cases} sinh(\chi), & \kappa = -1 \\ \chi & \kappa = 0 \\ sin(\chi) & \kappa = 1 \end{cases}$$

So in both cases when $d\Omega = 0$, we have the same result.

Answer 2

Alright, there's a major concept that seems to be missing from all of the papers I read on this subject: in closed and open geometries, the radius is not related to the circumference by $2 \pi$. This means that angles don't work the way you would expect them to. Obviously standard trig operations wouldn't work.

To illustrate this, let's take a walk on this sphere from the north pole to point P.

Let's say $\chi$ is $30^o$ and a is 1. The distance from the pole to P is then $\frac{30}{360}\times 1\times 2\pi=\frac{1}{6}\pi$. The circumference of a circle with this radius is, $\frac{1}{6}\pi \times 2\pi=\frac{1}{3}\pi^2$. However, the actual circumference at latitude P is $Sin(30)\times 1\times 2\pi=\pi$. We see that $\frac{1}{3}\pi^2\gt\pi$.

So in order to get the angles to work like they do in Euclidean space, the actual distance travelled in space is reduced so that it's $2\pi$ of the circumference. So the big missing concept here is that the radial coordinate is not the distance in $$ds^2=-c^2dt^2+a^2(t)\left[\frac{dr^2}{1+k\space r^2}+r^2d\Omega^2\right]$$

but it is the actual comoving distance in this version of the metric:

$$ds^2=-c^2dt^2+a^2(t)\left[dr^2 + S_k^2(r)d\Omega^2\right]$$

This doesn't negate, in any way, the answer by Reign, it's just intended to give a more intuitive explanation of why the two formulas are different. This is the difference between Reduced-Circumference Polar Coordinates and Hyperspherical Coordinates.