Why is the Universe (nearly) flat?

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The discussion centers on the surprising flatness of the universe, specifically why the density parameter Ω is approximately equal to one. It highlights that if the universe were not perfectly flat, it would quickly evolve to a significantly non-flat state, making the observed flatness puzzling. The primary explanation offered is cosmic inflation, which suggests that rapid expansion in the early universe smoothed out any initial curvature. Participants also debate the implications of inflation versus regular expansion and the role of a scalar field in maintaining energy density during inflation. The conversation concludes with considerations of alternative models and the complexities involved in understanding the universe's curvature and density dynamics.
  • #31
Hi PeterDonis:

You very lucid explantion is extremely helpful. Many thanks.
 
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  • #32
Chalnoth said:
Actually, the requirements are about the same for each of them: there must be approximately as much expansion during inflation as has occurred since. This paper has some good detail on the matter:
http://lesgourg.web.cern.ch/lesgourg/Inflation_EPFL.pdf.
Good detail indeed, thanks again for that reference. The "coincidence" I was mentionning actually appears far more natural than I thought.
 
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  • #33
I am struggling to understand the argument for inflation's impact on the relative curvature density.
If inflation is approximately de Sitter as indicated, then it amounts to switching on a high CC during inflation, so the FLRW equation remains, in Planck units,
$$\frac{3}{8\pi}\left(\frac{\dot a}{a}\right)^2=\rho_{\lambda}+\frac{\rho_{k0}}{a^2}+\frac{\rho_{m0}}{a^3}+\frac{\rho_{r0}}{a^4}$$
with ##\rho_{k0}=-\frac{3k}{8\pi},\quad \rho_{\lambda}=\frac{\lambda}{8\pi},\quad\lambda\simeq 10^{-122}## outside inflation but ##\lambda\simeq 1## during inflation.
If this is true then the ratios ##\frac{\rho_k}{\rho_m}\propto a## and ##\frac{\rho_k}{\rho_r} \propto a^2## are affected by inflation exactly as they are by ordinary expansion, so obvioulsy I am missing something... but what?
 
  • #34
wabbit said:
obvioulsy I am missing something... but what?

You're missing the fact that, at the end of inflation, all the energy density that was in the inflaton field (except for the tiny ##10^{-122}## that remains today) got converted into ordinary matter and radiation, so ##\rho_m## and ##\rho_r## got a huge boost. There's no way to convert any other kind of energy density into "curvature energy density", so the latter did not change at the end of inflation.
 
  • #35
Aaaaaaahh yes. Obviously... Thanks I was banging my head on a wall here :smile:

Pretty smart too that scalar field, stealing all the energy to inflate away quietly and then giving it back as if nothing happened:wink:
 
  • #36
Follow-up question: given the similarity of dark energy and inflation, the idea that they might arise from a common mechanism with the CC a residual leftover from the inflaton seems quite natural. I've seen this mentionned under the name "quintessential inflation" - is this ruled out somehow by observations or unlikely? It doesn't seem to be anywhere close to being a standard assumption.
 
  • #37
wabbit said:
Apparently this goes by the name "quintessential inflation"

This isn't quite right. "Quintessence" is the idea that you can have a scalar field (or possibly some other kind of field, although a scalar field is the only one I've seen any treatments of) which causes accelerating expansion, but doesn't have exactly the same equation of state as a cosmological constant. The general expression for the equation of state is ##p = w \rho##, where ##w## is some constant. A cosmological constant has ##w = -1##; but any ##w < - 1/3## will cause accelerating expansion. (Ordinary matter has ##w = 0##, and radiation has ##w = 1/3##.)

wabbit said:
is this ruled out somehow by observations or unlikely?

Observations tell us that the effective ##w## of whatever is causing accelerating expansion in our universe is ##-1##, as best we can tell. So a cosmological constant is the simplest explanation. But it could still turn out that ##w## is not exactly ##-1## when we have more accurate observations.
 
  • #38
I see - and scalar field inflation can't have w=-1?

It's not that I find the idea of non constant CC attractive, but that of having one same cause producing two effects already known to be similar. If we must have an inflaton, at least make it pay its dues:)
 
  • #39
wabbit said:
I see - and scalar field inflation can't have w=-1?

It's not that I find the idea of non constant CC attractive, but that of having one same cause producing two effects already known to be similar. If we must have an inflaton, at least make it pay its dues:)
Well, no. If scalar field inflation had exactly ##w=-1##, then it would never end. Inflation can only be described by some kind of dynamic field.

That said, there are a number of physicists who have tried to come up with models that unify the current accelerated expansion with inflation. Most of these fall under the name of "quintessence". Typically they have a scalar field which, after inflation, retains a fair amount of energy and whose energy "tracks" the energy density of the rest of the universe. When the energy density gets low enough, the scalar field freezes out, and it starts to act like a cosmological constant.

These models are interesting ideas, but they're a little bit complicated, generally more complicated than just positing a small cosmological constant and a separate inflaton. And none of them, to my knowledge, is well-motivated by any proposed model of physics beyond the standard model. That is, they're all ad-hoc models explicitly to explain the dark energy problem, and generally aren't connected with any other branch of physics.
 
  • #40
Thanks PeterDonis and Chalnoth. It doesn't sound so attractive if you need to tweak the model and add more parameters to get the right prediction...

I also found this review paper about this, a quick look at which seems to suggest the tweaks are rather heavy:
Unification of inflation and dark energy à la quintessential inflation, Md. Wali Hossain, R. Myrzakulov, M. Sami, Emmanuel N. Saridakis (Submitted on 22 Oct 2014)
 
  • #41
Hi @Garth:

Garth said:
It depends on the total average density parameter Ω=ρtota l ρcricital .

If Ω>1 the universe is closed.

In a decelerating universe, without inflation or dominant dark energy, whatever value it starts out with, <>1, Ω will be driven further away from 1, it will become more open or closed. .

To get a current value of unity Ω must originally equal 1 exactly, or be driven down onto the value 1 by cosmic acceleration - this is one reason for inflation.

I am confused by the use of Ω (with no subscript) in the quote. I understood that by definition Ω = 1, being the sum of all the contributing relative density ratio, for example as in:

equation2-png.81011.png


Please re-explain your quote with respect to this equation. Why can't this equation be used to show how a finite closed empty universe would expand:
k > 0, Ωm = Ωr = 0, ΩDE = 1- Ωk)?
 
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  • #42
Buzz Bloom said:
Hi @Garth:
I am confused by the use of Ω (with no subscript) in the quote. I understood that by definition Ω = 1, being the sum of all the contributing relative density ratio, for example as in:

equation2-png.81011.png


Please re-explain your quote with respect to this equation.
I should have made it clear; \Omega without a subscript referred to the total mass content: \Omega = \Omega_M + \Omega_r + \Omega_\Lambda + \Omega_k, where \Omega_M = \Omega_b + \Omega_{DM}.

Garth
 
  • #43
Hi @Garth:

I am still confused about your point regarding the effect of the expanding universe on the radius of curvature. What do you mean by the following:
"To get a current value of unity Ω must originally equal 1 exactly, or be driven down onto the value 1 by cosmic acceleration - this is one reason for inflation."​
You must have had some different idea in mind about what Ω represents in this quote since this Ω can not be defined to be exactly 1.
Also, "If Ω>1 the universe is closed" seems to be referring to Ωk rarther than the sum of all the density ratios.

Thanks for your very prompt response,
Buzz
 
  • #44
The best current estimated value I can find for Ωk is from the following:
Astronomy & Astrophysics, manuascript: Planck 2015 results. XIII. Cosmological parameters, February 9, 2015
(http://xxx.lanl.gov/pdf/1502.01589v2.pdf ).​
On pg 31, table 5 gives Ωk = 0.0008 +0.0040 / -0.0039. The 0.0008 corresponds to a current curvature radius Rc = 487 Gly. The confidence level for the +0.0040 / -0.0039 is specified as 95%, and assuming Gausian errors, the probabilitiy that the universe is close and finite is then approx. 0.66.

https://en.wikipedia.org/wiki/Inflation_(cosmology) gives the following:
The inflationary epoch lasted from 10−36 seconds after the Big Bang to sometime between 10−33 and 10−32 seconds.​
Perhaps someone can calculate values of a(t) for t = 10-36 seconds for both (a) the inflation assumption and (b) no inflation assumption, and then from those values calculate the two corresponding values for Rc. With those two values of Rc, can someone explain why the inflation value is better than the no inflation value?
 
  • #45
Buzz Bloom said:
Hi @Garth:

I am still confused about your point regarding the effect of the expanding universe on the radius of curvature. What do you mean by the following:
"To get a current value of unity Ω must originally equal 1 exactly, or be driven down onto the value 1 by cosmic acceleration - this is one reason for inflation."​
You must have had some different idea in mind about what Ω represents in this quote since this Ω can not be defined to be exactly 1.
Also, "If Ω>1 the universe is closed" seems to be referring to Ωk rarther than the sum of all the density ratios.

Thanks for your very prompt response,
Buzz
Hi Buzz,

\Omega could be anything - in fact it is observed to be nearly unity, but why?

An extra complication is that in a decelerating universe the value of \Omega, if initially slightly larger or smaller than unity, would be driven further away from unity, so at this epoch its value should be vastly different to its observed value.

Conversely an accelerating universe would drive the value of \Omega closer to unity.

There are three possibilities, either \Omega was exactly equal to unity at the earliest, Planck, time or that the universe is not decelerating or that there was a short period of inflation that drove an initial value of \Omega onto unity so closely that subsequent decelerating expansion has not prised it away from that value today.

Of course there seems to be a recent period of acceleration, caused by DE, but this has not been for long enough to negate the need for Inflation in a normally decelerating universe.

If the value of total \Omega is close to unity, as it seems to be then the universe is nearly spatially flat and \Omega_k will be very small.

Garth
 
  • #46
Hi @Garth:

Garth said:
Ω could be anything - in fact it is observed to be nearly unity, but why?

For the purpose of this post, I define Ω* = by definition to 1, and also equal to the sum of all constituent Ωs. That is, with respect to the equation in my post #41,
Ω* = Ωk + Ωm + Ωr + ΩΛ = 1.​

I am guessing that the Ω in the quote from your post #45 is Ω = Ω* - Ωk. Is this correct? If so, then I interpret your question as
"Why is Ωk measured to be so small if it is not exactly equal to 0?"​

I also get that extrapolation backwards to just before inflation at t = ti = 10-36 seconds, the the value of a(t) without inflation would have been much much larger than it's value at t = 10-36 seconds with inflation. Therefore, the value of Ωk at t=ti would have been much much larger than the it's value with inflation, Therefore the current value of Ωk would also be much large now than it's small current value of approx. 0.0008.

My question about this is, what other cosmological difference would no inflation have made with respect to the universe we now have?

Thanks for your discussion,
Buzz
 
  • #47
Buzz Bloom said:
Hi @Garth:
For the purpose of this post, I define Ω* = by definition to 1, and also equal to the sum of all constituent Ωs. That is, with respect to the equation in my post #41,
Ω* = Ωk + Ωm + Ωr + ΩΛ = 1.​

I am guessing that the Ω in the quote from your post #45 is Ω = Ω* - Ωk. Is this correct? If so, then I interpret your question as
"Why is Ωk measured to be so small if it is not exactly equal to 0?"​
Correct

The fundamental Friedman equation
\dot{a}^2 + k = \frac{8\pi G\rho}{3} a^2

means that Ω* = Ωk + Ωm + Ωr + ΩΛ = 1 is an identity.
I also get that extrapolation backwards to just before inflation at t = ti = 10-36 seconds, the the value of a(t) without inflation would have been much much larger than it's value at t = 10-36 seconds with inflation. Therefore, the value of Ωk at t=ti would have been much much larger than the it's value with inflation, Therefore the current value of Ωk would also be much large now than it's small current value of approx. 0.0008.

My question about this is, what other cosmological difference would no inflation have made with respect to the universe we now have?

Thanks for your discussion,
Buzz
That's a good question.

If the BICEP2 polarised light discovery is proven by post-Planck experiments to be caused totally by dust so there is no significant signal from primordial gravity waves then Inflation theories will have been constrained again, and again, and again....

If there are no further discoveries that confirm the Inflation paradigm of theories then it will be important to study alternatives and the leading problem will be to identify those that predict a similar universe to the one we observe, but without Inflation. It might be an interesting ride...

Garth
 
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  • #48
Buzz Bloom said:
My question about this is, what other cosmological difference would no inflation have made with respect to the universe we now have?

Garth said:
If there are no further discoveries that confirm the Inflation paradigm of theories then it will be important to study alternatives and the leading problem will be to identify those that predict a similar universe to the one we observe, but without Inflation. It might be an interesting ride...

Including anisotropies in the CMB and in the galaxy survey data that are consistent with baryon acoustic oscillations. According to recent grad/research level texts, inflation, via quantum fluctuations, gives the most plausible mechanism for the the generation of perturbations:

Weinberg "Cosmology" (2008) p. 208 said:
The most serious of the above three problems is the horizon problem. As we have seen, there are possible solutions of the flatness and monopole problems that do not rely on inflation.

Weinberg "Cosmology" (2008) p. 469 said:
The most exciting aspect of the inflationary cosmological theories described in chapter 4 is that they provide a natural quantum mechanical mechanism for the origin of the cosmological fluctuations observed in the cosmic microwave background and in the large scale structure of matter, and that may in the future be observed in gravitational waves.

Lyth and Liddle "The Primordial Density Perturbation" (2009) p. 307 said:
In the modern view, by far the most important function of inflation is to generate the primordial curvature perturbation ... It may generate other primordial perturbations too, including the isocurvature and tensor perturbations ... However, the historical motivation for inflation was rather different, and arose largely on more philosophical grounds concerning the question of whether the initial conditions required for the unperturbed Big Bang seem likely or not.

Padmanabhan "Gravitation: Foundations and Frontiers" (2010) p. 631 said:
Originally inflationary scenarios were suggested as a pseudo-solution to certain pseudo-problems; these are only of historical interest today and the only reason to take the possibility of an inflationary phase in the early universe seriously is because it provides a mechanism for generation the initial perturbations.
 

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