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

#### Buzz Bloom

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Summary
What is the probability/certainty-level that the universe is absolutely flat?
My questions are based on material from the following source.

pg 26 Eqs 27 & 28
In the base CDM model, the Planck data constrain the Hubble constant H0 and matter density Ωm to high precision:​
H0=67.3±1.0 km s−1 Mpc−1
m=0.315±0.013 PlanckTT+lowP. (27)​
With the addition of the BAO measurements, these constraints are strengthened significantly to​
H0=67.6±0.6 km s−1Mpc−1​
m=0.310±0.008 PlanckTT+lowP+BAO. (28)​

P 27 eq 30
The R11 Cepheid data have been reanalysed by Efstathiou (2014, hereafter E14) using the revised geometric maser distance to NGC 4258 of Humphreys et al. (2013). Using NGC 4258 as a distance anchor, E14 finds​
H0=70.6±3.3 km s−1 Mpc−1, NGC 4258, (30)​
which is within 1σ of the Planck TT estimate given in Eq. (27). In this paper we use Eq. (30) as a “conservative” H0 prior.​

P 39 eq 50
The constraint can be sharpened further by adding external data that break the main geometric degeneracy. Combining the Planck data with BAO, we find​
m=0.000±0.005 (95%,PlanckTT+lowP+lensing+BAO). (50)​

This constraint is unchanged at the quoted precision if we add the JLA supernovae data and the H_0 prior of Eq. (30). . . . Our Universe appears to be spatially flat to a 1σ accuracy of 0.25%.​

Q`1: I am not sure I understand what the "95%" means in equation 30. Which of the following does it mean?
a. The mean value 0.000 of the distribution in Equation 30 is accurate with a confidence of 95%.​
b. The error range of ±0.005 of the distribution in Equation 30 is accurate with a confidence of 95%.​
c. Both the mean value the error range of the distribution in Equation 30 is accurate with a confidence of 95%.​
d. It means something else.​
My guess is that the answer is (c).

Q2: Does the final statement referring to "a 1σ accuracy of 0.25%" mean that the following probability statements are all true:
a. PROB{Ωm>+0.005} = 0.25%, and​
b. PROB{Ωm<-0.005} = 0.25%, and​
c. PROB{Ωm>-0.005} AND PROB{Ωm<+0.005} = 99.5%?​
My guess is that the answer "YES".

Assuming both the above guesses are correct, I have a few additional questions motivated by the following goals:
I have been trying to think of an approach to come up with a way of justifying a statement of something similar to the following form:
The probability that the universe is perfectly flat is x (where x is a value close to 1.)​
It is clear that a distribution statement like Equation 50 fails to support such a statement.
Q3: Does anyone know of any scientifically reliable source that includes a statement similar to something similar to the above indented statement?

The best idea I have been able to come up with for what may be a possible scientifically reliable related statement is:
S: The probability that our universe is definitely not perfectly flat is less than x (where x is a small fraction).​

Assuming that no PF participant has a handy "YES" answer to Q3, I offer a fourth question.
Q4: Does the thought experiment I describe below seem reasonable as a theoretical approach that (given sufficiently precision instruments) could lead to a reliable statement of the form S?

I assume that there is some limit to the accuracy with which an astronomer can measure that position in the sky of a distant object. In particular I am thinking of the identifiable patterns in the CBR temperatures. If we arbitrarily define the North Pole, PN, of the sky as the statistically determined point towards which the various CBR telescopes are moving, giving them (by Doppler effects) the center of the area with the largest temperature. Then presumably it can be determined what the angular distance is between the North Pole and a specific spot on some identifiable temperature pattern, say P. It should be possible to choose a point P such that the position of the telescope, O, PN and P form an equilateral triangle. The distance D between O and P is the same as the distance between O and PN. It is approximately the radius of the observable universe (OU). (I understand that the source of the CBR is approximately 400,000 years younger than the age of the universe, which make the distance to the CBR smaller than the radius of the OU, but this is an insignificant reduction with respect to the thought experiment.) The choice of P should be such that the distance (as light travels) between P and PN is D. If the universe is flat, then this is straight forward since the angular distance will be exactly 60 degrees. If the universe is not flat, then the angle with be a very small amount less or more than 60 degrees. This amount depends on the absolute value of the radius of the curvature.
gives the following value for the radius of the OU.
r = Radius of observable universe = 93 Gly = 9.3 x 1010 ly​

Given equation 50, we can derive from probabilty statements of Q2 a,b, and c, the following regarding the radius of curvature R:
ρcrit = 3h02/8πG​
ρk = c2/πG|R2|​
I am unable to find the source where I got this formula. The URL I had for this source has become lost, and I cannot find it again. Does any reader know a source for this formula can be found?​
Ωk = ρkcrit
|R2| = (8/3) c2 / (Ωkh02)​
PROB{|R|< 1.254 x 1019 ly} = 0.5%​
The area A of a flat equilateral traingle with sides of length r is
A = (1/4) √π r2
α = sum of angles - π​
A = α x R2
α= (1/4) √π (r/R)2 = (2.44 10-17) radians​
1 radian = 57.296 degrees = 206,265 arcsecs = 2.06265x105 arcsecs​
α= 5.03 x 10-12 arcsecs​

If it is currently technically impossible to distinguish two angles which are different by 5 trillionths of an arcsec, is it then reasonably to conclude that the following S-like sentence is true?
The probability that our universe (using current technology) is distinguishable from a perfectly flat universe is less than 0.5%.​

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#### Arman777

Gold Member
P 27 eq 30

The R11 Cepheid data have been reanalyzed by Efstathiou (2014, hereafter E14) using the revised geometric maser distance to NGC 4258 of Humphreys et al. (2013). Using NGC 4258 as a distance anchor, E14 findsH0=70.6±3.3 km s−1 Mpc−1, NGC 4258, (30)which is within 1σ of the Planck TT estimate given in Eq. (27). In this paper we use Eq. (30) as a “conservative” H0 prior

In recent study the Hubble constant found to be as $H_0=74.03\pm 1.42kms^{-1}Mpc^{-1}$. Also, the study was done for the same galaxy, NGC4258. In the paper it is claimed that $H_0=70.6\pm 3.3 km s^{-1} Mpc^{-1}$ is a conservative value which you also quoted it.

This constraint is unchanged at the quoted precision if we add the JLA supernovae data and the H_0 prior of Eq. (30). . . .
I am not sure what would be the new curvature parameter if we take $H_0=74.03\pm 1.42kms^{-1}Mpc^{-1}$. I think that's important. The conventionalist approach in the article is bad actually.

Summary: What is the probability/certainty-level that the universe is absolutely flat?

S: The probability that our universe is definitely not perfectly flat is less than x (where x is a small fraction).
I cannot much say about your thought experiment. But I believe that for OU we can claim some statement like S, which is also given by the Planck data. Of course, we cannot claim such a thing for the Universe as you probably know.

#### Orodruin

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There are two main type of statistics used in physics: Frequentist and Bayesian. The one most commonly used in cosmology is Bayesian statistics, which Planck also states that they are using at the beginning of page 4.

In Bayesian statistics (unlike frequentist statistics) you do assign a probability to different models or model parameters. However, this comes at the price of having to assume a prior probability, an assumption on your degree of belief (or probability) that the model is true and your final answer is going to depend on that prior. Ideally, an experiment is accurate enough for your conclusions to be the same regardless of the prior, but in those cases you do not need statistics anyway.

Unless you put some restrictions on your prior, you can get any answer you want. However, with the prior that Planck uses, which are continuous (and discussed throughout the paper), the posterior probability distribution will also be continuous, which means that the probability of getting exactly flat is zero.

#### kimbyd

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2018 Award
For Q1, the interpretation of the 95% is that 0.000 is the "best fit" parameter value, and there's a 95% probability that the true value lies within 0.005 of that best-fit value.

There are some caveats to that claim, as Orodruin started explaining. In practice you can't usually take these probabilities too seriously, which is why one of my university professors was fond of saying, "I don't get out of bed for less than four sigma," which was his way of saying that he won't start thinking something interesting might be happening unless an experiment disagrees with standard theory by more than 99.99% probability.

For Q2, I think the concept is accurate but you're not using the right variables or numbers.

For Q3, it's impossible to assign a probability to the universe being perfectly flat. You can only say that the curvature lies between some error bounds with some probability, and observations will never be able to reduce those error bounds to zero.

#### Buzz Bloom

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Hi @Arman777, @Orodruin, and @kimbyd:

Thank you all for your posts. I have just completed my post #1, and also edited some errors. I plan to also reply to your posts later this evening if circumstances permit. After that I will be away from PFs for the weekend.

Regards,
Buzz

#### Buzz Bloom

Gold Member
I am not sure what would be the new curvature parameter if we take H0=74.03±1.42kms−1Mpc−1H_0=74.03\pm 1.42kms^{-1}Mpc^{-1}.
Of course, we cannot claim such a thing for the Universe as you probably know.
Hi Arman:

Ωk varies inversely with H0. The differences you quote are sufficiently small to avoid any significant changes to my thought experiment.

I agree that we cannot extrapolate from the OU to the rest of the universe except for the standard assumption that for large scales the statistical properties of the universe as a whole are similar everywhere.

Regards,
Buzz

#### Buzz Bloom

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which means that the probability of getting exactly flat is zero.
Hi Orodruin:

I understand that this is so with respect to the possible conclusions from the Planck 2015 Results XIII. That is why I came up with the thought experiment I introduced in the thread, and the idea of the S-like proability statement at the end of post #1.

Regards,
Buzz

#### Orodruin

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Hi Orodruin:

I understand that this is so with respect to the possible conclusions from the Planck 2015 Results XIII. That is why I came up with the thought experiment I introduced in the thread, and the idea of the S-like proability statement at the end of post #1.

Regards,
Buzz
Since the probability of the Universe being flat in this scenario (continuous prior) is zero. The Universe has a 100% probability to not be flat. This follows directly from the prior. No experiment needed.

#### kimbyd

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Since the probability of the Universe being flat in this scenario (continuous prior) is zero. The Universe has a 100% probability to not be flat. This follows directly from the prior. No experiment needed.
Not quite. That's taking Bayesian inference a bit too seriously. Bayesian inference of this kind can only tell us the probability of various outcomes given the data and within a specific theoretical framework. In principle it's possible for there to be a theory which requires spatial curvature to be identically zero due to some symmetry or other that the theory obeys. The probability of whether or not the universe is flat would become an exercise in model selection, which carries different properties and could very easily lead to a non-zero probability for the perfectly-flat model.

Ultimately, the perfectly-flat model could only be considered likely if it also predicted something else about our universe that is very hard for other theories to replicate. Then the high probability of perfect flatness would stem from the other independent evidence supporting the theory.

Finally, it's very possible that the spatial curvature of our universe is so tiny that it could never be measured, even in principle. For example, if $\Omega_k = 0.000001$, then there'd basically be no way to confirm that number. It isn't quite zero, but it might as well be for all practical purposes. The probability most definitely isn't zero for a curvature small enough that it doesn't matter that it isn't perfectly flat.

#### Orodruin

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In principle it's possible for there to be a theory which requires spatial curvature to be identically zero due to some symmetry or other that the theory obeys.
Of course, this is why I specified a continuous prior on the curvature. If you break this assumption by introducing a model that has a discrete prior, the overall prior on measured curvature is no longer continuous and therefore breaks that assumption. I think this is clear from my post.

Finally, it's very possible that the spatial curvature of our universe is so tiny that it could never be measured, even in principle. For example, if Ωk=0.000001, then there'd basically be no way to confirm that number. It isn't quite zero, but it might as well be for all practical purposes. The probability most definitely isn't zero for a curvature small enough that it doesn't matter that it isn't perfectly flat.
This is something else entirely.

#### Buzz Bloom

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Hi kimbyd:

I very much appreciate your responces to my questions. I have some follow-up questions/comments below.

For Q1, the interpretation of the 95% is that 0.000 is the "best fit" parameter value, and there's a 95% probability that the true value lies within 0.005 of that best-fit value.
I believe I understand it now. The 0.005 value is 1.96×σ, where σ is the standard deviation of the probability distribution. (I looked it up in my CRC tables: The area A of the standard Gaussian distribution with mean=0 and σ=1 between x=0 and x=1.96 is A(1.96) = 0.475. Tis implies the ares between x=-1.96 and +1.96 is 95%.)

. . . his way of saying that he won't start thinking something interesting might be happening unless an experiment disagrees with standard theory by more than 99.99% probability.
I have seen such quotes before, and I am sympathetic to the concept.
BTW, CRC gives by interpolation A(3.865)= 0.49995. Assuming Ωk has a Gaussian distribution, to make a statement with 99.99% confidence
the error range of Ωk is ±0.005×3.865 ≈±0.02.​
That suggests the following S-like statement.
The probability is 99.99% that a flat universe cannot be distinguished by current technological methods from any non-flat universe with a value of |Ωk|<0.02.​
In the context of my thought experiment, I have not yet done the math to figure out how many decimal digits of precision are required in distinguising the angles of a equilateral triangle with sides of length equal to the radius our observable universe from a universe with |Ωk|= 0.02.

For Q2, I think the concept is accurate but you're not using the right variables or numbers.
I would much appreciate your being specific about one (or more) variables with the wrong value, and also particularly appreciate a correct value.

For Q3, it's impossible to assign a probability to the universe being perfectly flat. You can only say that the curvature lies between some error bounds with some probability, and observations will never be able to reduce those error bounds to zero.
I understand the natural assumption that the distribution of Ωk is a continuous distribuition, no single val;ue has a meaningful probability value, except to say it is infinitesimal. The entire purpose of this thread is an attempt to find a meaningful, and also possibly correct, statement of the S-like type. The S-like statements seek to provide a specific probability for a range of value for Ωk for which it is correct to say that all universes with values of Ωk in the specified range are not distinguishable from each other with current technology.

Regards,
Buzz

#### kimbyd

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Hi kimbyd:

I very much appreciate your responces to my questions. I have some follow-up questions/comments below.

I believe I understand it now. The 0.005 value is 1.96×σ, where σ is the standard deviation of the probability distribution. (I looked it up in my CRC tables: The area A of the standard Gaussian distribution with mean=0 and σ=1 between x=0 and x=1.96 is A(1.96) = 0.475. Tis implies the ares between x=-1.96 and +1.96 is 95%.)
Small quibble: the 95% percentiles are typically $2\sigma$. They're just rounded to 95%.

I have seen such quotes before, and I am sympathetic to the concept.
BTW, CRC gives by interpolation A(3.865)= 0.49995. Assuming Ωk has a Gaussian distribution, to make a statement with 99.99% confidence the error range of Ωk is ±0.005×3.865 ≈±0.02.
The reason is probably different from what you're thinking. Actual errors are generally not Gaussian, and in particular they tend to have fatter tails, meaning high-sigma observations are more common than you would expect from assuming a Gaussian. Gaussian statistics should not be taken seriously beyond 1-2$\sigma$ in most cases.

The Gaussian case might suggest only a 0.01% chance of something occurring, but in reality it might be closer to 0.1% or so. Plus there are issues like publication bias and systematic errors that make it even more likely for erroneous high-significance results to appear. So an easy rule of thumb is to simply not take a result too seriously unless it has both high significance and can be independently verified.

#### timmdeeg

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For example, if $\Omega_k = 0.000001$, then there'd basically be no way to confirm that number. It isn't quite zero, but it might as well be for all practical purposes.
According to $(\Omega^{-1}-1)\rho{a}^2=-\frac{3c^2}{8\pi{G}}k$ inflation predicts an even much smaller value for $\Omega_k$, but not $\Omega_k=0$.

Doesn't this mean that $\Omega_k=0$ today requires $\Omega_k=0$ before inflation started?
But if true how likely is that at all? Often people talk about an infinite universe as if this possibility is suggested by our present knowledge. As if we had reason to think that $k=0$.

#### kimbyd

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According to $(\Omega^{-1}-1)\rho{a}^2=-\frac{3c^2}{8\pi{G}}k$ inflation predicts an even much smaller value for $\Omega_k$, but not $\Omega_k=0$.

Doesn't this mean that $\Omega_k=0$ today requires $\Omega_k=0$ before inflation started?
But if true how likely is that at all? Often people talk about an infinite universe as if this possibility is suggested by our present knowledge. As if we had reason to think that $k=0$.
I don't know of any model which has a symmetry forcing $\Omega_k=0$. I'm not sure one exists. My statement was more about trying to show that the statement that a continuous probability predicts zero probability for any particular value is not sufficient. My motivation was more about displaying that this sort of thing might potentially happen in a variety of contexts in science, rather than establishing a likely scenario in this particular case.

It is, of course, still theoretically possible for such a model to exist, just because you can't prove a negative. That we haven't yet conceived of such a model may indicate it's highly unlikely, but it's not really possible to put anything approaching a precise probability on that statement. Is the statement, "Nobody has thought of a way to make that work despite ample opportunity to investigate the issue," a good guide to the truth? Possibly. It's hard to say. I think it's probably not the case that a zero-curvature model exists, but I don't believe there's a way to make that statement objective. And it's generally worth considering such things to be at least possible, because we don't want to limit our future investigations by possible personal biases.

#### timmdeeg

Gold Member
I think it's probably not the case that a zero-curvature model exists, but I don't believe there's a way to make that statement objective. And it's generally worth considering such things to be at least possible, because we don't want to limit our future investigations by possible personal biases.
Thanks.

Regardless whether or not such a model exists would you agree that we can't conclude, neither from observation nor from the effect of inflation, that almost flat is an indication that the universe is infinite?

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.

#### kimbyd

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Thanks.

Regardless whether or not such a model exists would you agree that we can't conclude, neither from observation nor from the effect of inflation, that almost flat is an indication that the universe is infinite?

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.
Correct. It's possible for a flat space to be finite.

That said, I don't think that it's accurate at all to state that "cosmologist tend to believe that the universe is infinite". A more precise statement would be that cosmologists rarely think about whether the universe is finite or infinite because it's not really something that is answerable. Most cosmologists tend to try to stay within the bounds of answerable questions that have an impact on observations we can potentially make.

#### timmdeeg

Gold Member
A more precise statement would be that cosmologists rarely think about whether the universe is finite or infinite because it's not really something that is answerable.
Ok, that makes sense, thanks.

#### Buzz Bloom

Gold Member
Hi kimbyd and timmdeeg:

Correct. It's possible for a flat space to be finite.
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."

would you agree that we can't conclude, neither from observation nor from the effect of inflation, that almost flat is an indication that the universe is infinite?
Absolutely correct. I am currently in the process of posting a new thread
based on the following about the curvature parameter.
[The following is based on
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).​
Regards,
Buzz

#### timmdeeg

Gold Member
The flatness problem in a nutshell:

Flatness Problem - Wikipedia
This tiny value [means $\mid\Omega^{-1}-1\mid$ is the crux of the flatness problem. If the initial density of the universe could take any value, it would seem extremely surprising to find it so 'finely tuned' to the critical value $\rho_c$
The problem was solved by assuming inflation which forces $\mid\Omega^{-1}-1\mid$ down to almost zero.

If I see it correctly to speculate about an infinite universe with euclidean geometry would require not only fine tuning as in the case above but $\mid\Omega^{-1}-1\mid$ to be identically zero initially. This would seem to suggest $\rho=\rho_c$ as an intrinsic property of the big bang.

As @kimbyd said finite or infinite is not "not really something that is answerable." But doesn't infinite seem to be extremely unlikely from the above?

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#### George Jones

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I am not sure what "spatially flat" means for our universe, which is neither homogeneous nor isotropic, i.e., is not a FLRW universe. On very large scales, to good a approximation, our universe is homogeneous and isotropic, so does "spatially flat" mean when averaged over some appropriate scale? Then, what does "exactly spatially flat" mean?

#### PeterDonis

Mentor
Are you assuming that such a universe in not isotropic and homogeneous at large scales?
No.

If it is flat, that is Euclidean, and is isotropic and homogeneous, how is it conceptually possible for it to be finite?
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.

#### Buzz Bloom

Gold Member
Because it could have a topology like a 3-torus, which would give it a finite 3-volume.
H i Peter:
Thank you. I had forgotten about the torus possibility. I have never seen any math for a torus universe, so I am unfamiliar with how to visualize a 2D torus analog with equal curvature and radii in two right angle directions.

Regards,
Buzz

#### timmdeeg

Gold Member
I am not sure what "spatially flat" means for our universe, which is neither homogeneous nor isotropic, i.e., is not a FLRW universe.
" To the best of our capabilities, we measure it to be perfectly flat."

I think if this is true 380000 years after the big bang it should be true at any time, regardless structure formation.

#### PeterDonis

Mentor
I am unfamiliar with how to visualize a 2D torus analog with equal curvature and radii in two right angle directions.
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.

#### Buzz Bloom

Gold Member
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.
Hi Peter:

Thanks for your reply. Aftere some thought over lunch I realized that I described my understanding of the problem badly. I hope you will forgive me and let me try again.

To simplify I will assume that the universe is static. I recall that a static model with only matter and Ωm=1 is possible, although it is unstable in GR. 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.

Consequently I imagine an astronomer seeing the same distant galaxy in two opposite directions, and (assuming that transverse motion is negligible) he will then discover that this phenomenon has two properties.
1. Although the distances may differ for different galxies, the sum of the two distances is the same for different galaxies.
2. Any speed measured by Doppler shift will be the same in both direction, but the motions will be in opposite directions.

If you agree that these phenomena are reasonable to expect, do you think a torus universe configuration will be able to have this effect?

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.

Then I visualized a sphere in which every point on the surface maps onto an opposite point on the surface. If a photon starts at the center and moves radially it will return to the center having moved one diameter. However, it the photon starts at a point halfway between the center and the surface and moves perpendicular to a radius, it will travel √2 times the diameter. This shows that this universe is not isotropic.

Are you aware of any mathematical model that satisfies these two numbered constraints for a flat finite universe?

Regards,
Buzz

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

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