What is the probability that the Universe is absolutely flat?

[PeterDonis]

Is it necessary or only convenient to reference a 3-space?

If we're talking about the universe, then yes, since spatial slices of the universe are 3-spaces.

I don't know what you're trying to get at with the arcs.

 

[Buzz Bloom]

That's equivalent to what I said: that we currently have no evidence of curvature. If we had evidence of curvature, that would mean the confidence interval for spatial curvature based on the evidence did not include zero curvature.

Hi Peter:

I am finding that your usage for the term "evidence" seems unclear to me. I presume that your usage includes the concept that "evidence" means scientific evidence.

From https://en.wikipedia.org/wiki/Scientific_evidence :
Scientific evidence is evidence which serves to either support or counter a scientific theory or hypothesis. Such evidence is expected to be empirical evidence and interpretation in accordance with scientific method. Standards for scientific evidence vary according to the field of inquiry, but the strength of scientific evidence is generally based on the results of statistical analysis and the strength of scientific controls.

The description of the Ω_k distribution seems to me to satisfy this above definition as support for Ω_k having a non-zero value, and therefore is evidence of curvature, but it is not evidence of flatness. Another source of evidence for curvature is the success of the theory of inflation to explain a variety of otherwise unexplained phenomena. One of these phenomena is the flatness problem.

From https://en.wikipedia.org/wiki/Flatness_problem :
The most commonly accepted solution among cosmologists is cosmic inflation, the idea that the universe went through a brief period of extremely rapid expansion in the first fraction of a second after the Big Bang; along with the monopole problem and the horizon problem, the flatness problem is one of the three primary motivations for inflationary theory.

What seems to me to be missing is evidence of flatness, as opposed to evidence of almost flatness.

I understand that different scientists (and others) have different preferences for what they would like to accept as likely to be true. My own bias is a preference for finiteness. My rationale is admittedly closer to philosophical rather than scientific, and I understand it would be inappropriate to post this rationale in the PFs.

It seems to me that the question of what scientific evidence exists to support a particular shape property of the universe (finite hyperspherical, infinite flat, or infinite hyperbolic) is off topic from my intent for this thread. I think that after a while I will start a thread for this question. I am currently in the process of starting another thread with which I have had some difficulties regarding formatting.

Regards,
Buzz

 

[PeterDonis]

I am finding that your usage for the term "evidence" seems unclear to me.

Then feel free to substitute some other term. I described what I was referring to; you can understand that description without ever having to use the word "evidence" if it bothers you. The important thing is what I described, not the word "evidence".

The description of the Ωk distribution seems to me to satisfy this above definition as support for Ωk having a non-zero value

No, it doesn't. $\Omega_k = 0$ is within the 95% confidence interval. No such confidence interval can support a claim about $\Omega_k$ having a nonzero value. To make that claim, you would have to have a 95% confidence interval (or whatever is the best level of confidence you can get from the data) that did not include $\Omega_k = 0$. See further comments below.

I understand that different scientists (and others) have different preferences for what they would like to accept as likely to be true. My own bias is a preference for finiteness.

To put this in more appropriate terminology: to you, the null hypothesis is that the universe is spatially finite. This is also the null hypothesis that emerges from an inflationary model as a solution to the flatness problem, as @timdeeg argued earlier in the thread. So you view any evidence that does not rule out the null hypothesis as supporting the null hypothesis.

However, there are issues with this reasoning.

First, the universe can be spatially finite even if $\Omega_k = 0$ if it has a non-trivial topology. So just saying that you have a preference for a spatially finite universe, by itself, does not give you any grounds for a null hypothesis that has a nonzero value for $\Omega_k$. You need something else (such as a preference for inflationary models, see next paragraph).

Second, the actual null hypothesis that comes from inflationary models is not $\Omega_k \neq 0$. It's $\Omega_k > 0$. The inflationary model, at least the version that solves the flatness problem without fine-tuning, says that the universe is spatially a 3-sphere, which due to inflation plus expansion since the end of inflation is now so large (much, much larger than our observable universe) that we cannot detect its very small positive spatial curvature.

But the actual data, while it does not rule out $\Omega_k > 0$, also does not rule out $\Omega_k = 0$ or even $\Omega_k < 0$. All three are consistent with the data. So you cannot claim that the data "supports" anyone of those three, since that implies that the data favors one of them over the other two. And it doesn't. Which alternative you favor depends only on which null hypothesis you adopt; the data as it currently stands tells you nothing more.

 

[Buzz Bloom]

Hi Peter:

I think I am beginning to understand your reasoning, but there are a few details that need some additional clarification.

First, the universe can be spatially finite even if Ω_k=0 if it has a non-trivial topology.

You are omitting that (as far as we know) no one has described a model for a finite flat isotropic universe. You might argue that there is no evidence that the universe is isotropic beyond the observable universe, but I think the context for the discussion should assume it is isotropic.

Second, the actual null hypothesis that comes from inflationary models is not Ω_k≠0. It's Ω_k>0.

This is just a nit. Ω_k>0 implies a hyperbolic universe. Ω_k<0 implies a hyper-spherical surface universe. The relevant equation is (given r is the radius of curvature):

Ω_k = -A/|r|²​

where

A = (8/3) c² / h₀²,​

r>0 for the finite hypersphere surface universe,​

r is imaginary for the hyperbolic universe, and​

r = ∞ for a flat universe.​

But the actual data, while it does not rule out Ω_k>0, also does not rule out Ω_k=0 or even Ω_k<0. All three are consistent with the data.

I think that you are assuming that it is possible for the calculations leading to the Ω_k distribution to be made with an a priori assumption that the possible values for Ω_k are discrete rather than continuous, and maybe also finite in number. With that prior it would be possible to have distinct values for the probability of each assumed value for Ω_k, and that includes a distinct probability value for Ω_k=0. I agree with this assumption. However, the actual calculation presented is continuous, and that is limited to be evidence that the probability of Ω_k is defined only for any arbitrary specified range (or combination of ranges) of values for Ω_k.

Regards,
Buzz

 

[timmdeeg]

During expansion after the inflation has ended any tiny deviation of $\mid\Omega^{-1}-1\mid$ from zero will increase due to dilution of matter density. It should eventually reach a maximum and decrease in the very far future when the universe is approaching exponential expansion again, if my reasoning is correct.
Has someone an idea how much said maximum value of $\mid\Omega^{-1}-1\mid$ would exceed its value after inflation?

 

[Buzz Bloom]

Hi @timmdeeg:

Please clarify "Ω−1". Generally I understand that Ω is defined as the sum of a combination of model parameters: Ω_r, Ω_m, Ω_k, and Ω_Λ. When all four are summed, the result is Ω=1. If you exclude Ω_k then Ω−1 = -Ω_k. Is this your intended meaning?

Regards,
Buzz

 

[PeterDonis]

You are omitting that (as far as we know) no one has described a model for a finite flat isotropic universe.

A flat 3-torus universe with appropriate parameters is finite, flat, and isotropic. There can also be flat 3-torus geometries that are not isotropic, as discussed earlier, but that does not mean it's impossible for a flat 3-torus to be isotropic.

This is just a nit.

No, it isn't. A correct understanding of what the inflation model, as a solution to the flatness problem, actually implies is essential for this discussion.

I think that you are assuming that it is possible for the calculations leading to the Ωk distribution to be made with an a priori assumption that the possible values for Ωk are discrete rather than continuous, and maybe also finite in number.

It's not the distribution of $\Omega_k$ values that we can infer from evidence that is important here; it's the models. We don't have a continuous distribution of models; we have a discrete set of them. We are trying to assess whether the evidence we have favors one of those models over the other. The models are:

(1) An inflation model. This model says that $\Omega_k > 0$. It makes no concrete prediction for what value $\Omega_k$ should have other than that. It also says that the universe is spatially finite, but makes no concrete prediction for how large it is.

(2) A model that solves the flatness problem by having some special symmetry or other constraint that forces $\Omega_k = 0$. This model has two subtypes:

(a) The simple $\Omega_k = 0$ model with trivial topology, which is spatially infinite;

(b) A flat 3-torus model, which is spatially finite. This model makes no prediction for how large the universe is.

Saying that there is evidence for $\Omega_k$ being nonzero, i.e., for the universe being curved, is saying that the evidence favors model #1 over the others. (Note that there is no model in the set above that has $\Omega_k < 0$.) But it doesn't. It's equally consistent with all of them.

the actual calculation presented is continuous, and that is limited to be evidence that the probability of Ωk is defined only for any arbitrary specified range (or combination of ranges) of values for Ωk.

This is a limitation on the evidence we can gather, but it is not a limitation on models. It's perfectly possible for model #2 above to be true, and thus for $\Omega_k$ to actually be exactly zero, even though we could never collect the infinite amount of evidence that would be required to pin down the value of $\Omega_k$ with that infinite level of precision.

 

[Buzz Bloom]

No, it isn't. A correct understanding of what the inflation model, as a solution to the flatness problem, actually implies is essential for this discussion.

Hi Peter:

There is more to discuss, but I thought we might be able to resolve a simple misunderstanding first.

I understand that the inflation model assumes a finite universe with a positive radius of curvature. Before inflation the radius was very small and the absolute value of the curvature was very large. After inflation, the radius was very large and the absolute value of the curvature was very small, in fact sufficiently small that it is difficult (maybe impossible) to distinguish the properties of this curved universe from that of a flat universe.

Do you disagree with this?

The following is the equation relating curvature with the radius of curvature (for a finite universe).

Ω_k = -(8/3) c² / (h₀² r²)​

Do you disagree with this?

Regards,
Buzz

 

[PeterDonis]

I understand that the inflation model assumes a finite universe with a positive radius of curvature. Before inflation the radius was very small and the absolute value of the curvature was very large. After inflation, the radius was very large and the absolute value of the curvature was very small, in fact sufficiently small that it is difficult (maybe impossible) to distinguish the properties of this curved universe from that of a flat universe.

Yes, this is basically my understanding as well.

The following is the equation relating curvature with the radius of curvature (for a finite universe).

This looks right, yes.

 

[timmdeeg]

Hi Buzz Bloom:

Hi @timmdeeg:

Please clarify "Ω−1". Generally I understand that Ω is defined as the sum of a combination of model parameters: Ω_r, Ω_m, Ω_k, and Ω_Λ. When all four are summed, the result is Ω=1. If you exclude Ω_k then Ω−1 = -Ω_k. Is this your intended meaning?

Regards,
Buzz

I had a typo, its correctly $\mid\Omega^{-1}-1\mid$, I've fixed that in my previous posts.
For some background please read https://en.wikipedia.org/wiki/Flatness_problem.

 

[Buzz Bloom]

Hi

This looks right, yes.

OK. Then the following is what you appeared to disagree with in your post #42.

This is just a nit. Ω_k>0 implies a hyperbolic universe. Ω_k<0 implies a hyper-spherical surface universe.​

What I called a nit was

Second, the actual null hypothesis that comes from inflationary models is not Ω_k≠0. It's Ω_k>0.

You are now agreeing that

the actual null hypothesis that comes from inflationary models is not Ω_k≠0. It's Ω_k<0.​

Regards,
Buzz

 

[Buzz Bloom]

I had a typo, its correctly $\mid\Omega^{-1}-1\mid$, I've fixed that in my previous posts.
For some background please read https://en.wikipedia.org/wiki/Flatness_problem.

Hi timmdeeg:

Thanks for clarifying. The model parameter that corresponds to your Ω is Ω_m.

Regards,
Buzz

 

[PeterDonis]

You are now agreeing that
the actual null hypothesis that comes from inflationary models is not Ωk≠0. It's Ωk<0.

Actually, looking through more literature, it's not clear to me that there is a consistent sign convention for $\Omega_k$. So I would rather state it as: the null hypothesis that comes from inflationary models is that the spatial curvature is positive. That implies a total value of $\Omega$ (i.e., including all forms of energy plus curvature) that is greater than 1.

 

[Buzz Bloom]

Actually, looking through more literature, it's not clear to me that there is a consistent sign convention for Ω_k.

Hi Peter:

I think that is unfortunate. Usually when I try to explore cosmology I mostly work with the Friedmann equation.

In this context, the assumption that the sum of the four Ω parameters equals 1 implies that Ω_k is negative for a closed finite universe. This is very clear in particular for the case when

Ω_R = Ω_Λ = 0.​

Then for a closed finite universe

Ω_M > 1​

which implies

Ω_k = 1 - Ω_M < 0​

so

Ω_M + Ω_k = 1.​

Regards,
Buzz

 

[PeterDonis]

In this context, the assumption that the sum of the four Ω parameters equals 1 implies that Ωk is negative for a closed finite universe.

Yes. I'm fine with using that convention for this discussion.

 

[Buzz Bloom]

A flat 3-torus universe with appropriate parameters is finite, flat, and isotropic. There can also be flat 3-torus geometries that are not isotropic, as discussed earlier, but that does not mean it's impossible for a flat 3-torus to be isotropic.

Hi Peter:

Do you know of any particular specified parameters that mathematically defines a theoretical 3-torus topologically shaped universe model which is finite, flat, isotropic, and homogeneous? For the purpose of defining such a model the homogeneous requirement can just be assumed.

I found some suggestions on the internet that some people believe it is possible, but I could not find any actual example. Nor could I find any argument supporting this concept that seemed rational to me.

In the absence of any even partially defined plausible model, I feel it necessary (for the present) to disbelieve it is possible. But, perhaps it is just a difficult problem, like the Riemann hypothesis, and someday I might be persuaded to change my mind.

Regards,
Buzz

 

[George Jones]

I understand that the inflation model assumes a finite universe with a positive radius of curvature.

No.

So I would rather state it as: the null hypothesis that comes from inflationary models is that the spatial curvature is positive.

Again, no.

See, for example Baumann's lectures on inflation,

https://arxiv.org/abs/0907.5424
Equations (43), (44), (45), as well as the discussion preceding these equation, all use absolute values. Other examples that don't restrict inflation to universe that have positive spatial curvature: "Cosmology" by Weinberg; "The Primordial Density Perturbation: Cosmology, Inflation, and the Origin of Structure" by Lyth and Liddle.

That implies a total value of $\Omega$ (i.e., including all forms of energy plus curvature) that is greater than 1.

As @Buzz Bloom notes, the sum of all the $\Omega$ s, including curvature is always unity. This means that $\Omega_k > 0$ for open (non-flat) universes, and $\Omega_k < 0$ for closed universes. Always.

 

[PeterDonis]

Do you know of any particular specified parameters that mathematically defines a theoretical 3-torus topologically shaped universe model which is finite, flat, isotropic, and homogeneous?

Sure, the specified parameters that the space is a 3-torus which is finite, flat, isotropic, and homogeneous. Unless you have a proof that these parameters are not all consistent with each other, then such a model must exist.

[Edit: Erroneous portion deleted.]

 

[PeterDonis]

If you then assume "homogeneous", then the space must also be isotropic

Oops, sorry, I got this backwards. Isotropy (strictly speaking, isotropy about every point) implies homogeneity, but not the other way around. I have edited the previous post to remove this portion.

 

[PeterDonis]

Equations (43), (44), (45), as well as the discussion preceding these equation, all use absolute values.

Yes, but those are general constraints from known data, not parameters specific to any inflation model.

The actual models discussed in this reference don't seem to make any assumption about spatial geometry, though, so they would indeed be consistent with either a small positive or a small negative spatial curvature today, as far as I can tell.

the sum of all the $\Omega$ s, including curvature is always unity.

Ah, yes, I misstated it. The sum of all the $\Omega$ s except curvature is greater than 1 if the spatial curvature is positive.

 

[George Jones]

Generally I understand that Ω is defined as the sum of a combination of model parameters: Ω_r, Ω_m, Ω_k, and Ω_Λ. When all four are summed, the result is Ω=1.

I don't think that I have ever seen this convention for an unsubscripted $\Omega$.

If you exclude Ω_k then Ω−1 = -Ω_k. Is this your intended meaning?

This is the convention with which I am familiar.

Yes, but those are general constraints from known data, not parameters specific to any inflation model.

And thus to be consistent with known data, models of inflation must work in universes that have negative spatial curvature. The Lyth and Liddle reference that I gave above explicitly notes that inflation smooths inhomogeneities in both $\Omega > 1$ and $\Omega < 1$ universes.

 

[Buzz Bloom]

Sure, the specified parameters that the space is a 3-torus which is finite, flat, isotropic, and homogeneous. Unless you have a proof that these parameters are not all consistent with each other, then such a model must exist.

Hi Peter:

I think you are using the term "parameters" in a difference sense than I use it. The terms flat, finite and isotropic are resulting behaviors/properties of a model. I use "parameters" to mean the variables (and possibly inital values) which are inputs to a model and from which the resulting behavior can be determined. For example, the models based on the Friedmann equation has five parameters: 4 Ωs and H₀. From values for these parameters ons can calculate radius of curvature as a function of time, the scale factor "a" as a function of time, age as the length of time from a=0 to a=1, etc. I also believe isotropy can also be deduced, but I am not up to explaining how in general, but I can for special cases, such as rhe one i describe in my post #25.

Regards,
Buzz

 

[PeterDonis]

And thus to be consistent with known data, models of inflation must work in universes that have negative spatial curvature.

I don't think a model that didn't work in a universe with negative spatial curvature would be inconsistent with the known data (since the known data does not rule out zero or positive spatial curvature). But it might be considered more unlikely given the known data. However, that appears to be a moot point for inflation models, since the reference you give notes that they do in fact work with either positive or negative spatial curvature.

 

[timmdeeg]

In this context, the assumption that the sum of the four Ω parameters equals 1 implies that Ω_k is negative for a closed finite universe.

This follows directly from $\Omega_k=-\frac{kc^2}{H_0^2}$

 

[timmdeeg]

The model parameter that corresponds to your Ω is Ω_m.

No, $\Omega$ in $(\Omega^{-1}-1)\rho{a}^2=-\frac{3c^2}{8\pi{G}}k$ (post #13) is the ratio of the actual density to the critical density. From this $k=0$ requires $\Omega =1$ which is definitively not a prediction of inflation.

 

[metastable]

Is it necessary or only convenient to reference a 3-space?

I define an arc with n=arbitrarily high # turns. I make a straight line from point t=0 to point t=1. Next I bisect line t=0, t=1 creating a point A. The vector is created first by a line from point A, through the referenced spiral point B to point C, a point in space. Points in space are referenced as Vector: t=0.14563... Distance: 686,739,974.97969...km

If we're talking about the universe, then yes, since spatial slices of the universe are 3-spaces.

I don't know what you're trying to get at with the arcs.

wolframalpha.com:

ParametricPlot3D [ { [//math:cos(2*pi*10^27*t)*sin(pi*t)//] , [//math:sin(2*pi*10^27*t)*sin(pi*t)//] , [//math:cos(pi*t)//] } , { t , [//number:0//] , [//number:1//] } ]

if on this scale 1 = Earth radius, loops = 10^27

unless I've made a mistake in my calculation we have:

~0.000020037... femtometers position specification accuracy & separation distance between loops at Earth radius

 

[Ibix]

...so you are trying to use distance along your spiral to specify angular position and radius as in regular spherical polars? How do you specify a point that does not lie on a line through your spiral?

 

[metastable]

How do you specify a point that does not lie on a line through your spiral?

I make a straight line from point t=0 to point t=1. Next I bisect line t=0, t=1 creating a point A. The vector is created first by a line from point A, through the referenced spiral point B to point C, a point in space.

^With a point A at the bisection of a straight line from point t=0 to t=1

 

[Buzz Bloom]

the ratio of the actual density to the critical density

Hi timmdeeg:

The above quote is the definition of Ω_M.
Please see

https://en.wikipedia.org/wiki/Friedmann_equations​

in the section Density parameter.

Regards,
Buzz

 

[Buzz Bloom]

And thus to be consistent with known data, models of inflation must work in universes that have negative spatial curvature.

However, that appears to be a moot point for inflation models, since the reference you give notes that they do in fact work with either positive or negative spatial curvature.

Hi George and Peter:

I apologize for my error. My bad. My previous careless exposure to discussions of inflation have involved explanatory pictures of finite shapes getting bigger. I failed to realize that the metaphor in the pictures intended the inclusion of both kinds of non-flat universes.

https://en.wikipedia.org/wiki/Inflation_(cosmology)#/media/File:History_of_the_Universe.svg​

http://www.ctc.cam.ac.uk/outreach/origins/inflation_zero.php
Regards,
Buzz

 

[Ibix]

^With a point A at the bisection of a straight line from point t=0 to t=1

And the position of A along that line is your third number.

 

[timmdeeg]

The above quote is the definition of Ω_M.
Please see

https://en.wikipedia.org/wiki/Friedmann_equations​

in the section Density parameter.

I'm not sure how you come to this conclusion. This article states "The density parameter (useful for comparing different cosmological models) is then defined as:" ... and describes thereafter $\Omega$ as the ratio of the actual density to the critical density.

If $\Omega$ in the expression in brackets (post #60) were $\Omega_m$ then how could this expression be identical zero in case of exact flatness?

 

[metastable]

And the position of A along that line is your third number.

Are you saying if I want to specify 3 chosen subsequent particle positions it would take me more than 6 numbers? (3 spiral points and 3 distances from point A)

 

[Ibix]

Are you saying if I want to specify 3 chosen subsequent particle positions it would take me more than 6 numbers? (3 spiral points and 3 distances from point A)

Yes. Because you cannot specify a position that does not lie on a line through your spiral.

 

[metastable]

Because you cannot specify a position that does not lie on a line through your spiral.

In a 3 space can I specify any irrational position?