B What is the probability that the Universe is absolutely flat?

timmdeeg

Gold Member
I have read several times also here in PF that cosmologists tend to believe that the universe is infinite (sorry I have no quotes at hand) but wonder which reasoning supports that.
After some search one example.

@PeterDonis said in post #26:
"Then all your questions boil down to one: what, if any, evidence do we have that bears on the question of whether the universe is infinite or not? The answer to that is that we have plenty of evidence that constrains the parameter space of the standard hot big bang model of cosmology to a pretty narrow range, a range which makes it very unlikely, based on that model, that the universe is spatially finite, and very likely that it is spatially infinite."

So it seems my reasoning in this Thread is not consistent somewhere. What have I overlooked?

PeterDonis

Mentor
I recall that a static model with only matter and Ωm=1 is possible, although it is unstable in GR.
Yes, this model is called the Einstein static universe:

In such a universe a photon in principle could move in any direction, and if it failed to interact with anything, it would return to it's starting point and continue moving in the same direction.
Yes.

I imagine an astronomer seeing the same distant galaxy in two opposite directions
Yes.

Although the distances may differ for different galxies, the sum of the two distances is the same for different galaxies.
Yes.

Any speed measured by Doppler shift will be the same in both direction, but the motions will be in opposite directions.
More precisely, the motions will appear to be in opposite directions, because the observer is looking at them from opposite sides.

do you think a torus universe configuration will be able to have this effect?
It will have similar effects, but whether those effects will be isotropic (i.e., the same in all directions, as they are in the Einstein static universe) will depend on the specific 3-torus configuration. It is possible to have a 3-torus where the "widths" in different directions are different, in which case the space will not be isotropic, and, for example, the sum of distances in your item #1 will be different in different directions.

I visualized that the phenomena would not work for a cube where each point on a side maps onto a corresponding point on the opposite side, and all eight corner point map onto an opposite corner.
The 3-torus does not have "corners"; thinking of it as a cube with opposite sides and corners matched is an artifact of trying to visualize it embedded in 3-D Euclidean space.

Then I visualized a sphere in which every point on the surface maps onto an opposite point on the surface
I don't know if this has the same topology as a torus.

Are you aware of any mathematical model that satisfies these two numbered constraints for a flat finite universe?
The Wikipedia discussion of the flat torus looks like a decent starting point:

PeterDonis

Mentor
So it seems my reasoning in this Thread is not consistent somewhere. What have I overlooked?
The quote from me that you gave was about whether the universe is spatially infinite. This thread, including your reasoning, is about whether the universe is spatially flat. They're not the same thing.

timmdeeg

Gold Member
The quote from me that you gave was about whether the universe is spatially infinite. This thread, including your reasoning, is about whether the universe is spatially flat. They're not the same thing.
Could you please explain which evidence makes it very likely that the universe is spatially infinite.

PeterDonis

Mentor
Could you please explain which evidence makes it very likely that the universe is spatially infinite.
I have looked back through the references I'm aware of, and the only argument I can find for the universe being spatially infinite is the one I was thinking of when I made the post you linked to: that we have no current evidence for any spatial curvature, and the simplest way for there to be no evidence of spatial curvature now is for the universe to be exactly spatially flat, and if the universe is exactly spatially flat, since we also see no evidence of multiple images of objects, the simplest hypothesis is that the universe is spatially infinite. That was my understanding of the standard Lambda-CDM model.

However, that doesn't address the argument you made in post #19, since it ignores the implications of solving the flatness problem with an inflation model.

Unfortunately, I'm not familiar enough with recent literature on the Lambda-CDM model to know if this has been discussed or addressed.

kimbyd

Gold Member
2018 Award
Hi kimbyd and timmdeeg:

This confuses me. Are you assuming that such a universe in not isotropic and homogeneous at large scales? If it is flat, that is Euclidean, and is isotropic and homogeneous, how is it conceptually possible for it to be finite?
Perhaps you meant to say, "Almost a flat space."
Typically, yes, a finite flat space won't be isotropic on very large scales. But we can't tell if our universe is isotropic on scales significantly larger than the horizon, so this is also unanswerable.

A 2D torus can be flat as well. For example, if you ever played the old video game Asteroids, its game screen is a 2D flat torus--if your ship went off any edge of the screen, it would reappear on the opposite edge.

What you can't do is embed a flat 2D torus in Euclidean 3-space; to do that, the torus has to be curved (and that's what people will typically visualize when asked to visualize a 2D torus). I believe having the two curvatures equal is indeed not possible for a 2D torus embedded in Euclidean 3-space.
This example is also useful for explaining why closed space isn't necessarily finite and open space not necessarily infinite: the 2D torus embedded in 3D isn't flat everywhere, but its average curvature can be thought of as flat, in that for every positively-curved region there is a corresponding negatively-curved region (these would be on the outer and inner surfaces, respectively). Overall the space is finite, but some parts of that space have negative curvature. Similarly, if an infinite space had large-scale irregularities, then it too could have positive curvature in places and still remain infinite.

Curvature and topology are only related if you take the simplest situation of constant curvature and a simple manifold.

Buzz Bloom

Gold Member
Typically, yes, a finite flat space won't be isotropic on very large scales. But we can't tell if our universe is isotropic on scales significantly larger than the horizon, so this is also unanswerable.
Hi kimbyd:

There are some questions I am OK with the idea that it cannot be known because some similar alternative question can be known. On the other hand, if some other question cannot be known, and no one has proposed any similar question that can be known, then (in my opinion) the inability to know is worthless.

An example of the first kind is: Is the universe flat (i.e., Euclidean)? The similar question is: If our universe is not flat, is it plausibly sufficiently close to flat that it is highly likely not distinguishable, using current technology, from one that is flat? I think this alternative question is demonstrably true.

An example of the second kind is: Is the universe flat and finite and isotropic? As far as I know from the discussion so far in this thread, although it has not been proved to be impossible, there is no current model proposed by anyone that suggests that it is possible.

Regards,
Buzz

Buzz Bloom

Gold Member
we have no current evidence for any spatial curvature
Hi Peter:

Maybe it is a matter of interpretation, but I find the conclusion in
regarding the probability distribution of possible values for Ωk implies that the present state of scientific knowledge is that:
1. we cannot know if the universe is flat, but
2. we are highly confident (95%) that although it may be curved it is very close to being flat, and
3. if it is curved, it is (approximately) equally likely to be finite or infinite.

Regards,
Buzz

PeterDonis

Mentor
the present state of scientific knowledge is that:
1. we cannot know if the universe is flat, but
2. we are hight confident (95%) that although it may be curved it is very close to being flat.
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.

metastable

Because it could have a topology like a 3-torus, which would give it a finite 3-volume. Having a Euclidean metric does not require that the space have the same topology as Euclidean 3-space.
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 View attachment sphere-spiral-n-orbits-3-gif.gif

PeterDonis

Mentor
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

Gold Member
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 Ωk distribution seems to me to satisfy this above definition as support for Ωk 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

Mentor
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" any one 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

Gold Member
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 Ωk=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 Ωk≠0. It's Ωk>0.
This is just a nit. Ωk>0 implies a hyperbolic universe. Ωk<0 implies a hyper-spherical surface universe. The relevant equation is (given r is the radius of curvature):
Ωk = -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 Ωk>0, also does not rule out Ωk=0 or even Ωk<0. All three are consistent with the data.
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. With that prior it would be possible to have distinct values for the probability of each assumed value for Ωk, and that includes a distinct probability value for Ωk=0. I agree with this assumption. However, 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.

Regards,
Buzz

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timmdeeg

Gold Member
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?

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Buzz Bloom

Gold Member
Hi @timmdeeg:

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

Regards,
Buzz

PeterDonis

Mentor
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

Gold Member
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).
Ωk = -(8/3) c2 / (h02 r2)​

Do you disagree with this?

Regards,
Buzz

PeterDonis

Mentor
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

Gold Member
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: Ωr, Ωm, Ωk, and ΩΛ. When all four are summed, the result is Ω=1. If you exclude Ωk then Ω−1 = -Ωk. 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.

Buzz Bloom

Gold Member
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. Ωk>0 implies a hyperbolic universe. Ωk<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 Ωk≠0. It's Ωk>0.
You are now agreeing that
the actual null hypothesis that comes from inflationary models is not Ωk≠0. It's Ωk<0.​

Regards,
Buzz

Gold Member

PeterDonis

Mentor
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

Gold Member
Actually, looking through more literature, it's not clear to me that there is a consistent sign convention for Ωk.
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 Ωk is negative for a closed finite universe. This is very clear in particular for the case when
ΩR = ΩΛ = 0.​
Then for a closed finite universe
ΩM > 1​
which implies
Ωk = 1 - ΩM < 0​
so
ΩM + Ωk = 1.​

Regards,
Buzz

PeterDonis

Mentor
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.

"What is the probability that the Universe is absolutely flat?"

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