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What is the Mean Expected Volume of a Finite Universe?

  • Thread starter Buzz Bloom
  • Start date

Buzz Bloom

Gold Member
Problem 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
as discussed in the thread
The distribution of the value of Ωk 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
Ωk ∈ (-∞, 0).​

Relevant Equations:
Note: I used the online tool (OLT)
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 = ∫ab x f(x) dx​
It is of course assumed (and I verify it) that
F = ∫ab f(x) dx = 1.​
In particular, for
x = Ωk, 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. Mx is the mean expected value of x.
Mx = ∫0 x f(x)dx​
= mean expected value of Ωk.​
OLT gives Mx = 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.​
A = -k c2 / h02
= -1.86709*1046 km2 .​
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 rs be the radius for the simplest solution. From Eq. 2 we get
rs = √(-A/Mx)​
= ( √(1.86709 x 1046/0.00199471) ) km​
= 3.05944*1024 km​
= 3.23384*1011 ly​

Let Vsi be the simplest solution volume of the universe in km3 units. From Eq 5 we get
Vsi = 5.65271*1074 km3.​
This is the simplest solution for part i.

Let Vsii be the simplest solution volume of the universe in ly3 units.
1 ly = 9460730000000 km = 9.46073*1012
Vsii = Vsi / (9.460733*1036)​
= 6.67548*1035 ly3
This is the simplest solution for part ii.

For part iii we need the value for Rou, the radius of the observable universe from any observation position in comoving coordinates.
From https://en.wikipedia.org/wiki/Observable_universe :
Rou = 9.3x1010 ly .​
Note that value is quite close to rs. The ratio of radii is
rs/Rou = 3.48 .​
The ratio of volumes is
Vsii/Vou = 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​
Vsiv = Vsi/(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.
Last edited:

Buzz Bloom

Gold Member
The first and simplest solution is now completed.

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