What is the Mean Expected Volume of a Finite Universe?

[Buzz Bloom]

Homework Statement

Given the assumptions below (in the main section of this post), calculate the expected value of the volume of the universe. That is, calculate the mean of the volume distribution using each of the following four units:
i. km$^3$
ii. ly$^3$ = one light-year cubed,
iii. ou$^3$ = radius of the observable universe cubed,
iv. pl$^3$ = one Planck length cubed.
The assumptions are in the main section due to formatting problems I have in this section.

Relevant Equations

I have entered to Relevant Equations in the main section because of formatting problems I have in this section.

Problem Statement Assumptions:

a. The universe is finite. That is, it is (approximately) a 3D boundary of a 4D hyper-sphere of radius r.

b. [The following is based on

https://arxiv.org/pdf/1502.01589.pdf​

as discussed in the thread

https://www.physicsforums.com/threads/questions-re-flat-universe.971984/ .]​

The distribution of the value of Ω_k is twice the negative half of a Gaussian distribution with

mean = 0, and​

standard deviation: σ = 0.0025.​

So, the range of possible values is

Ω_k ∈ (-∞, 0).​

Relevant Equations:
Note: I used the online tool (OLT)

https://www.dcode.fr/definite-integral​

to calculate definite integrals.

1. The mean, M, of a probability distribution f(x) is the integral over the range (a,b) of f(x) of the integrand:

x f(x).​

M = ∫_a^b x f(x) dx​

It is of course assumed (and I verify it) that

F = ∫_a^b f(x) dx = 1.​

In particular, for

x = Ω_k, and​

f(x) = (((√2/π)/σ) *e^{((-x2)/(2*σ2))},​

OLT gives F = 1 for the INTEGRAL between 0 and ∞ of

(sqrt(2/%pi)/0.0025)*exp((-x^2)/(2*0.0025^2)) .​

​

2. M_x is the mean expected value of x.

M_x = ∫₀^∞ x f(x)dx​

= mean expected value of Ω_k.​

OLT gives M_x = 0.00199471

3. Let r = the radius of a 4D hyper-sphere whose 3D hyper-surface is the finite universe.
The values for constants are

1 Mpc = 3.0856776x10¹⁹ km,​

c = 2.9979258 x 10⁵ km/s,​

1/h₀ = (1/67.7) s Mpc/km = (1/67.7) x 3.0856776x10¹⁹ s​

= 4.5579x10¹⁷ s​

= 1.44529*10¹⁰ yr.​

The relationship between x and r is:

x = A/r².​

where

A = -k c² / h₀²​

= -1.86709*10⁴⁶ km² .​

Source:​

https://www.easycalculation.com/physics/astrodynamics/cosmology-curvature-density-php​

Note: Sometimes the above URL works and sometimes not. I do not understand why.​

The assumption a means k=+1 and A<0, consistent with x<0.

4. Let V₃ represent the volume of a 3D sphere.

V₃ = (4π/3)r³.​

5. Let V₄ represent the 3D volume of the hyper-surface of a 4D hypersphere.

V₄ = (2π²)r³.​

The four equations above are sufficient for the simplest method of calculating the (approximate) solution to the problem statement. Later I intend to calculate more precise solutions since I am curious about how much they differ from the simplest solution. I will post later some additional relevant equations as needed for these later solutions.

Simplest Solution:
Let r_s be the radius for the simplest solution. From Eq. 2 we get

r_s = √(-A/M_x)​

= ( √(1.86709 x 10⁴⁶/0.00199471) ) km​

= 3.05944*10²⁴ km​

= 3.23384*10¹¹ ly​

Let V_si be the simplest solution volume of the universe in km³ units. From Eq 5 we get

V_si = 5.65271*10⁷⁴ km³.​

This is the simplest solution for part i.

Let V_sii be the simplest solution volume of the universe in ly³ units.

1 ly = 9460730000000 km = 9.46073*10¹²​

V_sii = V_si / (9.46073³*10³⁶)​

= 6.67548*10³⁵ ly³​

This is the simplest solution for part ii.

For part iii we need the value for R_ou, the radius of the observable universe from any observation position in comoving coordinates.
From https://en.wikipedia.org/wiki/Observable_universe :

R_ou = 9.3x10¹⁰ ly .​

Note that value is quite close to r_s. The ratio of radii is

r_s/R_ou = 3.48 .​

The ratio of volumes is

V_sii/V_ou = 198.​

This is quite odd to me, but that is what the math seems to be saying, unless I made an error I can't find.

For part iv we need the value for Planck length.
From https://en.wikipedia.org/wiki/Planck_length

1 pl = 1.616229*10^-35 m​

= 1.616229*10^-38 km​

V_siv = V_si/(1.616229^3 * 10^-114) pl³​

= 5.83045*10¹⁸⁸ pl³​

Note: I made this calculation because I was curious if I could find a number greater than one googol which corresponded to something real (as apposed to a number of combinations of something).

I plan to calculate a more precise solution in a later post based on the expected value of the distribution of the radius of curvature.

 

[Buzz Bloom]

The first and simplest solution is now completed.