# Is the mass of the universe finite (collection of objects)?

The observable universe has a finite volume.
Density is (locally) the ratio of mass to volume.
For the mass of the universe to be infinite, the local density of the universe must be (somewhere) infinite.
At the location where infinite density is to be found, the curvature of space-time (i.e. gravity) must also be infinite.
Since the space-time curvature follows an inverse-square law, if there is an infinite curvature *anywhere* there must be an infinite curvature *everywhere*.
[Infinity]/R^2 = [Infinity] for all real values of R.
There is a place where space-time curvature is finite.
Therefore space-time curvature is finite *everywhere*.

PeterDonis
Mentor
2019 Award
For the mass of the universe to be infinite, the local density of the universe must be (somewhere) infinite.
Nobody is claiming that the mass of the observable universe is infinite.

space-time curvature follows an inverse-square law
No, it doesn't. Don't confuse Newtonian gravity with GR.

The observable universe has a finite volume.
Density is (locally) the ratio of mass to volume.
For the mass of the universe to be infinite, the local density of the universe must be (somewhere) infinite.
At the location where infinite density is to be found, the curvature of space-time (i.e. gravity) must also be infinite.
Since the space-time curvature follows an inverse-square law, if there is an infinite curvature *anywhere* there must be an infinite curvature *everywhere*.
[Infinity]/R^2 = [Infinity] for all real values of R.
There is a place where space-time curvature is finite.
Therefore space-time curvature is finite *everywhere*.
Do I detect a whiff of non sequitur?
I might assume that you have a cogent support for your first statement, though it seems to assume that the infinite mass of the universe is of the same or greater order than the volume of the universe. If the volume is of a greater order, you will have to explain why the infinity of mass must fill it. If the volume is in fact of the same order and its average density is everywhere the same on the same scale as we observe, then it would have infinite mass of the same order as the volume of space, without anywhere having infinite density (except possibly in some black holes somewhere FAIK).
Density is not mass, nor vice versa.

Hmmm... I hadn't thought of that polyhedral thing, though it has a certain attraction. Mind you, It is not clear to me that universal shape means much in a closed universe, especially if we are speaking of more than three dimensions. It seems to me that in a closed and finite universe, most plausible views would suggest a universe full of dirty space (the dirt including things like hydrogen, stars, and planetary dust particles like Earth and their world lines, time being one of the dimensions, etc etc) much as a balloon would be full of gas, irrespective of its pressure.

Talking about universes is an activity fraught with flypapers for the unwary intellect, I am tempted to reflect...
Speaking of the shape of the universe, does the 1994 Polyhedronization proof apply to non-euclidean space?

Another interesting question: Assume the Big Rip is occurring, can set of points S , whose distances diverge to infinity in finite time, still be claimed to form a polyhedron?

I cannot tell you how often I lament that I was born too early. Then again, I'd also lament if I was born so late that every imaginable question had already been answered. Given a choice, I'd choose the former (our current situation). :)