- #1

King Solomon

- 48

- 1

- TL;DR Summary
- The density of the universe was 10^93 kg/cm^3 at 10^-43 (secs) after the Big Bang...doesn't that imply the universe is finite, since 10^-43 occurs before the inflationary epoch?

http://astronomy.swin.edu.au/cosmos/B/Big+Bang

D (density) = m/V

At t = 10^-47, D = 10^93 kg/cm ^3, r = 10^-57 meters, V = 4pi(r^3)/3 which is about (4/3)pi(10^-171)(meters^3)

t = 10^-47 precedes the inflationary epoch at t = 10^-35, this is important since this implies all matter at the big bang remains casually connected at this time.

D = 10^93 kg/cm^3 = 10^99 kg/m^3 = 10^102 g/m^3

DV = (10^102 g/m^3) * (4pi/3)(10^-171 m^3)

m = DV

m = (4/3pi) * 10^-69 g... yes, that was the mass of the universe according to this paper...less than a proton.

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Anyway...let's assume that the density was indeed 10^(102) gm^3 and that due to parameters of which I am not aware, the mass of the universe at time = 10^-47 was m = 10^58grams (made up number for this example), and let's call this mass X.

The mass of the observable universe today is roughly 10^56 grams, let's call this mass Y.

This would imply (in this fictional example) that Y/X is 1/100, or that our observable universe contains 1% of the mass entire universe, and under the assumption of homogeneity, that our observable universe is about 1/100th the size of the entire universe...and also... that there exists an exterior shape to our universe that compromises of all matter (some massive concave polyhedron) in the event that we live in an open or flat universe.

The moment we declare that the mass of the entire universe is finite, and that we do not live in a closed universe...then we have also declared that there exists an exterior to the entirety of the universe (beyond our observable part).

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If I am wrong, then may I ask...how could the universe have a finite density in a very small volume if the mass of the universe was not finite?

If the mass of the universe was finite at the Big Bang, then it must have remain finite, correct?

If the the mass of the entire universe (beyond the observable universe) is finite, and we live an an open/flat universe, then there is strict limit to the number of atoms/energy/particles/etc under the assumption of e = mc^2, correct?

If this strict limit exists, then if all mass were converted to the maximum number of allowable protons (mass of proton = 1.67 *10^-27kg), then we we would have a finite number of protons, and therefore there would some polyhedron (3D shape) that utilized every proton at one of its vertices, and therefore the universe has an exterior shape, correct (assuming the entirety of the universe is open/flat)?

"Grünbaum, B., Hamiltonian polygons and polyhedra, Geombinatorics, 3 (1994) 83-89."

Given a set S of n ≥ 3 points in a space (not all in line in or in the same plane) it is well known that it is always possible to polyhedronize S, i.e., construct a simple polyhedron, P such that the vertices of P are precisely the given points in S, this was proven in 1994 by Gr¨unbaum.

D (density) = m/V

At t = 10^-47, D = 10^93 kg/cm ^3, r = 10^-57 meters, V = 4pi(r^3)/3 which is about (4/3)pi(10^-171)(meters^3)

t = 10^-47 precedes the inflationary epoch at t = 10^-35, this is important since this implies all matter at the big bang remains casually connected at this time.

D = 10^93 kg/cm^3 = 10^99 kg/m^3 = 10^102 g/m^3

DV = (10^102 g/m^3) * (4pi/3)(10^-171 m^3)

m = DV

m = (4/3pi) * 10^-69 g... yes, that was the mass of the universe according to this paper...less than a proton.

------------------------------------------------------------------------------------------------------------------------------------

Anyway...let's assume that the density was indeed 10^(102) gm^3 and that due to parameters of which I am not aware, the mass of the universe at time = 10^-47 was m = 10^58grams (made up number for this example), and let's call this mass X.

The mass of the observable universe today is roughly 10^56 grams, let's call this mass Y.

This would imply (in this fictional example) that Y/X is 1/100, or that our observable universe contains 1% of the mass entire universe, and under the assumption of homogeneity, that our observable universe is about 1/100th the size of the entire universe...and also... that there exists an exterior shape to our universe that compromises of all matter (some massive concave polyhedron) in the event that we live in an open or flat universe.

The moment we declare that the mass of the entire universe is finite, and that we do not live in a closed universe...then we have also declared that there exists an exterior to the entirety of the universe (beyond our observable part).

-----------------------------------------------------

If I am wrong, then may I ask...how could the universe have a finite density in a very small volume if the mass of the universe was not finite?

If the mass of the universe was finite at the Big Bang, then it must have remain finite, correct?

If the the mass of the entire universe (beyond the observable universe) is finite, and we live an an open/flat universe, then there is strict limit to the number of atoms/energy/particles/etc under the assumption of e = mc^2, correct?

If this strict limit exists, then if all mass were converted to the maximum number of allowable protons (mass of proton = 1.67 *10^-27kg), then we we would have a finite number of protons, and therefore there would some polyhedron (3D shape) that utilized every proton at one of its vertices, and therefore the universe has an exterior shape, correct (assuming the entirety of the universe is open/flat)?

"Grünbaum, B., Hamiltonian polygons and polyhedra, Geombinatorics, 3 (1994) 83-89."

Given a set S of n ≥ 3 points in a space (not all in line in or in the same plane) it is well known that it is always possible to polyhedronize S, i.e., construct a simple polyhedron, P such that the vertices of P are precisely the given points in S, this was proven in 1994 by Gr¨unbaum.