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

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?

## 2 Answers

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.

• Please have a look at the Wiki page for the FLRW metric: en.wikipedia.org/wiki/…. The section labeled Hyperspherical coordinates doesn't appear to agree with your second definition. Is the Wiki page wrong (I'm perfectly happy to accept that, I'm just looking for clarity here)? May 9 '20 at 0:47
• W/r/t my original question: how does $\kappa$ capture the positive or negative curvature? It's just an integer. For example, for $\kappa = 1$, your equation has the same metric for that of a 3-sphere where the curvature is equal to the radius. How does this metric deal with, for example, an open universe where $\kappa \ne -\frac{1}{r^2}$ May 9 '20 at 0:54
• It says "In hyperspherical or curvature-normalized coordinates the coordinate r". So it actually uses another coordinate system but writes the coordinate as $r$ which is confusing May 9 '20 at 7:32
• @GluonSoup You can look at theory.uchicago.edu/~liantaow/my-teaching/dark-matter-472/… page 9, section 1.1.2 May 9 '20 at 7:40
• Thanks, but this still doesn't address my original question: how does an integer (-1, 0, +1) give us anything other than a perfectly round curvature (or flat plane for 0)? Said differently, if you select +1 for $\kappa$, the formula reflects a perfect sphere ($\kappa=\frac{1}{r^2}$). How does this metric allow for, say, a hyperbolic plane with an arbitrary curvature? May 9 '20 at 14:29

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.