Please check my calculation of the radius of a finite universe

[Buzz Bloom]

TL;DR

I have used a formula for calculating the radius of a finite universe based on the value of Ω[SUB]k[/SUB]. When I started this project I made several mistakes which I had trouble correcting. However, I think I have the correct answer now which I will present below.

My source for Ω_k and H₀ is

https://www.cosmos.esa.int/documents/387566/387653/Planck_2018_results_L06.pdf ,​

page 40, equation 47b. Ω_k = 0.0007.​

Page 15 equation 13 has H₀ = 67.27 km/(s Mpc).​

The formula I used is

R = (c/H₀) (1/Ω_k)^1/2 .​

( I did have a reference for this, but I misplaced it, and I can't find it again.)
I assume c = 1 ly/y. Using the values

Mpc = 3.08568 x 10¹⁹ km, and​

1 yr = 31557600 sec,​

I get​

c/H₀ = 4.587 x 10¹⁷ s = 14.535 billion years,​

and

(1/Ω_k)^1/2 = 37.8.​

The result is then

R = 549 x 10⁹ ly.​

 

[PeterDonis]

I have used a formula for calculating the radius of a finite universe based on the value of Omega_k. The result I get is not reasonable.

What result do you get? Please show your work.

 

[PeterDonis]

My source of Omega_k is

https://www.cosmos.esa.int/documents/387566/387653/Planck_2018_results_L06.pdf​

page 40,​

You do realize that that page is saying that the values given in equation 46 are only an "apparent" detection of curvature, right? Further analysis corrects those values to the ones given in equation 47, which are consistent with a spatially flat universe.

 

[Buzz Bloom]

Further analysis corrects those values to the ones given in equation 47, which are consistent with a spatially flat universe.

Hi @PeterDonis:

I scanned through the rest of the Planck_2018 (pages 40-60) and I was unable to find any later "corrected" (or any other) value for Ω_k. Can you help me find the correction you mentioned?

My purpose in this exercise was to get a sense of the relative volumes of a plausible finite universe and the observable universe. It was not to intend that the given Ω_k value was accurate. It was just a plausible value for the comparison.

ADDED
The volume ratio is about 8700.

MORE ADDED
I confess I do not understand with any confidence the last sentence on page 40 just before section 7.4. Here is the sentence.

The joint results suggests our Universe is spatially flat to a 1 σ accuracy of 0.2 %.​

What I don't get is the particular value that has an accuracy of 0.2%. My guess is this value is the confidcence level for the standard deviation (σ) of the distribution of values for Ω_k=0.

Regards,
Buzz

 

[PeterDonis]

I scanned through the rest of the Planck_2018 (pages 40-60) and I was unable to find any later "corrected" (or any other) value for Ωk. Can you help me find the correction you mentioned?

I already gave you the equation number in post #3. It's on the same page you referenced, right-hand column.

The volume ratio is about 8700.

Based on what formula? If you're not going to tell us what formula you're using and where you got it from, you can't expect any useful feedback.

I confess I do not understand with any confidence the last sentence on page 40 just before section 7.4.

The values for $\Omega_K$ in all the equations on that page are given as a point estimate plus or minus one standard deviation, i.e., $1 \sigma$. The sentence you refer to is talking about the value given in equation 47b just above it. The value of $\sigma$ in that equation is $0.0019$, which rounds to $0.002$, or 0.2 percent.

 

[Bandersnatch]

Based on what formula?

I recognise the number ballpark. He's taking the ratio of the surface area of a hypersphere with the calculated minimal radius of curvature for the given accuracy of $\Omega_k$, and the volume of the observable universe.

 

[Buzz Bloom]

Based on what formula?

Hi PeterDonis:

The volume of the three dimensional boundary of a 3D hyper-sphere is
V_fu = 2π²r_fu³ ,
where the "fu" designates the object as the "finite universe".

The volume of the observable universe is
V_ou = (4/3)πr_ou³ ,
where the "ou" designates the object as the observable universe.

The ratio is

V_fu/V_ou​

= (2π² * (3/4π)) * r_fu³ / r_ou³​

= (3π/2) * (r_fu / r_ou)³ .​

r_fu = 549 Gly.​

r_ou = 45.7 Gly​

​

V_fu/V_ou = 3π/2 = 4.712​

r_fu/r_ou = 549/45.7 = 12.013​

V_fu/V_ou = 4.712 * 12.013³ = 8169​

Regards,
Buzz

 

[PeterDonis]

The volume of the three dimensional boundary of a 3D hyper-sphere

A 3D hypersphere doesn't have a 3-dimensional boundary. I assume you mean a 4D hypersphere, but then the question is, what 4D hypersphere? A closed universe is not a 4D hypersphere in spacetime. It's a sequence of 3D hyperspheres, one for each instant of time. In the presence of a positive cosmological constant, the sequence can be infinite.

 

[PeterDonis]

The volume of the observable universe is

Yes, that part's fine, but the 3-volume to compare it with is not a "3D boundary of a 4D hypersphere", it's just a 3D hypersphere with a radius of curvature equal to the one implied by the $\Omega_K$ value you are using. The formula you are using for $V_\text{fu}$ looks like it could apply to that, but the words you are using don't.

 

[Buzz Bloom]

H i @PeterDonis:

I confess I have had difficulty in the nomenclature for multidimensional objects. Various sources use different names where n has a different meaning. I am OK using your usage, which matches Wikipedia.: an n-sphere is an n-dimensional volume in an n-dimensional Euclidean space where all points of this volume are less than or equal to a specified distance from a point specified as the center. I am assuming that a 3D finite universe can be considered to be the collection of all points at a specified distance from a center point of a 4-dimensional hyper-sphere.. This 3D volume has no boundary, just as the 2D boundary (like the surface of the Earth) of a 3-sphere has no boundary.

it's just a 3D hypersphere with a radius of curvature equal to the one implied by the value you are using.

If I am understanding your nomenclature correctly, a 3D hyper-sphere is a 3-dimesional space with curvature where all points can be considered to be equidistant from a center point in a 4-dimensional Euclidean space. Is this correct? The volume of such a space is 2π²r³ where r is the radius of curvature, which is the common radial distance.

However, the quote above seems to be describing a 3D space with a 2D boundary. Surely a finite universe also has no boundaries. It is a curved space in the same sense as the Earth is a 2D surface with curvature.

Regards,
Buzz

 

[PeterDonis]

I am OK using your usage, which matches Wikipedia.: an n-sphere is an n-dimensional volume in an n-dimensional Euclidean space where all points of this volume are less than or equal to a specified distance from a point specified as the center.

No, that is not my usage. The correct term for what you are quoting here from Wikipedia is an n-ball, not an n-sphere. (More precisely, a closed n-ball, since it says "less than or equal to"; if it were just "less than" it would be an open ball.)

A common definition of an n-sphere is that it is the n-dimensional boundary of an n+1-ball. However, that requires the n-sphere to be embedded in a higher dimensional space. That is not the case for our universe. So to have a definition that makes sense for our universe, we need to avoid anything that requires an embedding.

I am assuming that a 3D finite universe can be considered to be the collection of all points at a specified distance from a center point of a 4-dimensional hyper-sphere.

This definition requires an embedding, which we need to avoid. See above. For a "no embedding" definition, see the end of this post.

If I am understanding your nomenclature correctly

You're not. See above.

the quote above seems to be describing a 3D space with a 2D boundary.

No, it isn't. It's describing a 3D space with a finite volume but no boundary.

Surely a finite universe also has no boundaries.

Correct. See above.

To come up with a "no embedding" definition, first consider the case of a 2-sphere. This is a 2-dimensional surface with a finite area but no boundary. Its area can be expressed as $4 \pi r^2$, where $r$ is usually called the "radius of curvature" of the 2-sphere. But we have to be careful, because "radius" naturally suggests an embedding in 3-dimensional Euclidean space, which is what we want to avoid. However, there is an alternative choice: $r$ is the Gaussian curvature of the 2-sphere, i.e., its intrinsic curvature, considered as a manifold in its own right. We can measure this intrinsic curvature by measuring geodesic deviation on the 2-sphere. (In GR terms, the Riemann tensor for a 2-dimensional manifold has only one independent component, the Gaussian curvature of the surface, which in the general case can be a function of the coordinates: but if the one component is positive and constant, i.e., independent of the coordinates, then we have a 2-sphere, and the one constant independent component is $r$.)

The "no embedding" definition of the 3-sphere is similar: the only difference is that now the 3-volume of the 3-sphere is $2 \pi^2 r^3$, where $r$ is the constant Gaussian (intrinsic) curvature of the 3-sphere. Here the connection with the Riemann tensor is a bit more complicated, because the Riemann tensor for a 3-dimensional manifold in the general case has 6 independent components; but for a 3-sphere those 6 components are all equal, and the Gaussian curvature $r$ is the Ricci scalar derived from that Riemann tensor.

 

[Buzz Bloom]

the 3-volume of the 3-sphere is 2π²r², where is the constant Gaussian (intrinsic) curvature of the 3-sphere.

I think I am getting that a professional does not want to discuss the geometry in the terms of a 4D space. I understand that it is not necessary to introduce the 4D space, but I find it makes my visualization easier. What I see as important is that both views give the very same volume formula.

Regards,
Buzz

 

[PeterDonis]

I understand that it is not necessary to introduce the 4D space, but I find it makes my visualization easier.

Really? You can visualize a 4-D Euclidean space? I can't.

Also, even if you think you can visualize a 4-D Euclidean space, I doubt your powers of visualization will extend to a 4-D spacetime.

That's why I think it's highly preferable to get yourself used to the "intrinsic curvature" viewpoint for the simpler cases, so you get used to not having to lean on the crutch of an embedding, since you won't have it for the more complicated cases anyway.