Is there a formula for volume in a open universe?

Question

Recently while reading a bit about cosmology, I have stumbled upon two different formulas that describe the full volume of a universe with its spatial section $dt=0$.

For instance, for a flat universe ($k=0$), the full volume of its spatial section $dt=0$ can be given by, $$V=\frac{4π}{3}a^3$$ Where $a$ is the cosmic scale factor.

Furthermore for a closed universe, that is k=+1, the full volume of its spatial section $dt=0$ can be given by, $$V=2π^2a^3$$ My question is thus, is there a similarly structured formula for an open universe ($k=-1$)? That is a formula whose general form is $V=c1\times a^3$ where $c1$ is just a constant.

Answer 1

The two equations are telling you different things so you cannot compare them.

A closed universe has a maximum radius (see Why does a flat universe imply an infinite universe? for how to calculate this). Since there is a maximum radius that means the volume is finite and we can calculate it. The calculation is harder that you might think because of the curvature - it is not simply $\tfrac43\pi r^3$ - but if we grind through it we get your equation:

$$ V_{\textrm{closed}}=2π^2a^3 $$

However a flat universe and an open universe are both infinite (assuming they have a simply connected topology) so their volume is infinite. The equation:

$$ V=\frac{4π}{3}a^3 $$

is just telling you how the volume of a unit sphere changes as a flat universe expands. For a flat universe it is conventional to choose our length scale so that $a = 1$ right now, in which case the equation above is just the volume of a sphere of radius one. Then as the universe expands and $a$ increases above one the equation tells you how the volume of the unit sphere increases.

Answer 2

First, note that this is not the volume of the universe, but the volume of an element of radius a - the volume of the shape defined by $0<r<a$. In the first case, this is just the volume of a ball of radius a. The scale factor has no direct physical significance.

So we are actually calculating the volume of a "ball" of radius $a$. Let's set $a=1$ (we can multiply the result by $a^3$ if we wish). The FRW metric has $\sqrt{-g}=\frac{r^2 \ sin{\theta}}{\sqrt{1-kr^2}}$.

The volume is just:

$\int \sqrt{-g} \ dr \ d\theta \ d \phi = 2\pi \int_0^\pi \sin{\theta} \ d\theta \ \int_0^1\frac{r^2 \ dr}{\sqrt{1-kr^2}} = 4\pi \ \int_0^1\frac{r^2\ dr}{\sqrt{1-kr^2}}$.