# Volume of hypothetical closed universe

Consider a physical universe with a beginning in time and space, with a finite amount of mass without the complications of dark energy that we have in the presently open curvature universe, so that the omegas (vacuum and mass) values create a closed universe, a universe with positive curvature. In this way, there will be a maximum (spatial) size that this universe could attain.I am not sure whether the concepts of (spatial) volume or radius would be applicable to this size, but if so, is there a way to calculate what it would be depending on the other variables? Which variables would be necessary -- the omegas, I presume, and what else? Is there a rough rule-of-thumb formula one could use?

• – Yashas Mar 5 '17 at 6:00
• This is pretty mainstream... ...isn't it? – JMLCarter Mar 5 '17 at 6:04
• I don't know, it is unclear. – peterh Mar 5 '17 at 21:29

$$\rm r=\frac{c}{H \sqrt{\Omega_t-1}}$$

where $$\rm H$$ is the Hubble parameter and $$\rm \Omega_t$$ the sum of all energy contributions:

$$\rm \Omega_t=\Omega_r+\Omega_m+\Omega_{\lambda}$$

so the circumference where a straight line closes in on itself is

$$\rm U=2 \pi r$$

The volume of the 3dimensional curved space is the surface area of the corresponding 4D hypercube:

$$\rm A=2 \pi^2 r^3$$

For the general equations for an arbitrary number of spatial dimensions see here.