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?
 A: 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. 
A: 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.
