Buzz Bloom
Gold Member
 2,035
 330
 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 lightyear 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 hypersphere of radius r.
b. [The following is based on
as discussed in the thread
The distribution of the value of Ω_{k} is twice the negative half of a Gaussian distribution with
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:
3. Let r = the radius of a 4D hypersphere whose 3D hypersurface is the finite universe.
The values for constants are
4. Let V_{3} represent the volume of a 3D sphere.
5. Let V_{4} represent the 3D volume of the hypersurface of a 4D hypersphere.
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
Let V_{si} be the simplest solution volume of the universe in km^{3} units. From Eq 5 we get
Let V_{sii} be the simplest solution volume of the universe in ly^{3} units.
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 :
For part iv we need the value for Planck length.
From https://en.wikipedia.org/wiki/Planck_length
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.
a. The universe is finite. That is, it is (approximately) a 3D boundary of a 4D hypersphere of radius r.
b. [The following is based on
as discussed in the thread
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 isΩ_{k} ∈ (∞, 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 = ∫_{a}^{b} x f(x) dx
It is of course assumed (and I verify it) thatF = ∫_{a}^{b} f(x) dx = 1.
In particular, forx = 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} = 0.001994713. Let r = the radius of a 4D hypersphere whose 3D hypersurface is the finite universe.
The values for constants are
1 Mpc = 3.0856776x10^{19} km,
c = 2.9979258 x 10^{5} km/s,
1/h_{0} = (1/67.7) s Mpc/km = (1/67.7) x 3.0856776x10^{19} s
= 4.5579x10^{17} s
= 1.44529*10^{10} yr.
The relationship between x and r is:x = A/r^{2}.
whereA = k c^{2} / h_{0}^{2}
= 1.86709*10^{46} km^{2} .
Source:
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 V_{3} represent the volume of a 3D sphere.
V_{3} = (4π/3)r^{3}.
5. Let V_{4} represent the 3D volume of the hypersurface of a 4D hypersphere.
V_{4} = (2π^{2})r^{3}.
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 10^{46}/0.00199471) ) km
= 3.05944*10^{24} km
= 3.23384*10^{11} ly
Let V_{si} be the simplest solution volume of the universe in km^{3} units. From Eq 5 we get
V_{si} = 5.65271*10^{74} km^{3}.
This is the simplest solution for part i.Let V_{sii} be the simplest solution volume of the universe in ly^{3} units.
1 ly = 9460730000000 km = 9.46073*10^{12}
V_{sii} = V_{si} / (9.46073^{3}*10^{36})
= 6.67548*10^{35} ly^{3}
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} = 9.3x10^{10} ly .
Note that value is quite close to r_{s}. The ratio of radii isr_{s}/R_{ou} = 3.48 .
The ratio of volumes isV_{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}) pl^{3}
= 5.83045*10^{188} pl^{3}
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: