Why is the Universe (nearly) flat?

[wabbit]

My understanding is that the degree of flatness observed for the spatial universe is quite extraordinary, to the extent of begging for an explanation.
If this is correct, then my question is, what are the (most) plausible mechanisms being considered as an explanation for either an exact, or approximate, spatial flatness?
Thanks

 

[zephyr5050]

The reason such a closely flat universe is so surprising is that once the universe starts out slightly not flat, it will evolve to become extremely not flat extremely quickly (to put it in Layman's terms). Thus we should find that the $\Omega_0 \ll 1$, $\Omega_0 = 1$ (exactly), or $\Omega_0 \gg 1$. These three cases represent the three possible geometries of the universe: open, flat, or closed respectively. We find something odd though. Instead we see that $\Omega_0 \approx 1$. But if it was ever approximately one, and not exactly one it should have long ago deviated from one.

The explanation for how and why this is the case is inflation. Early on the universe's $\Omega$ would have started out very nearly equal to one. The universe expanded exponentially in a very short time essentially locking in the state of this seed universe on a large scale. While $\Omega$ may have tried to diverge, the exponential expansion happened faster and smoothed out any curvatures, leaving us with a very nearly flat universe.

 

[Paulibus]

The explanation for how and why this is the case is inflation.

strikes me as something not answerable by Physics, which doesn't answer "why" questions --it tells only "how" stuff came to be; here via the agency of a scalar inflaton field. But I'd like to know whether such a field is a logically inevitable ingredient of any extensive universe's initial conditions, or "just" something inferred a posteriori from our universe's observed near-flatness?

 

[wabbit]

The reason such a closely flat universe is so surprising is that once the universe starts out slightly not flat, it will evolve to become extremely not flat extremely quickly

Wait, isn't this backward? A k=1 FLRW space is (well, can be) an expanding 3-sphere, and it gets flatter as it expands. This actually explains some of the current flatness, though I thought it wasn't a sufficient explanation for so much flatness.

The explanation for how and why this is the case is inflation. Early on the universe's Ω would have started out very nearly equal to one.

This actually seems to be the same explanation as the expanding sphere though, and the radius of curvature of the sphere at t is given by a(t), not by its (slow or fast, more or less complicated) history, so I am not sure I understand how inflation explains flatness better than regular expansion.

 

[Garth]

Wait, isn't this backward? A k=1 FLRW space is (well, can be) an expanding 3-sphere, and it gets flatter as it expands. This actually explains some of the current flatness, though I thought it wasn't a sufficient explanation for so much flatness.

This actually seems to be the same explanation as the expanding sphere though, and the radius of curvature of the sphere at t is given by a(t), not by its (slow or fast, more or less complicated) history, so I am not sure I understand how inflation explains flatness better than regular expansion.

It depends on the total average density parameter \Omega = \frac{\rho_{total}}{\rho_{cricital}}.

If \Omega> 1 the universe is closed.

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

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

If the universe expands linearly then \Omega remains at its initial value.

Garth

 

[wabbit]

Thanks, I've often seen this argument before but never quite understood it. As the universe expands its density goes down, but the point seems to be (assuming k=1 here), the critical density goes down even faster so that $\Omega$ increases... Still a little hard to reconcile this explanation, which seems to say "the universe gets less flat as it expands" with the decreasing curvature due to expansion - this might be because "less flat" here doesn't mean "a larger radius of curvature" but is a relative statement (e.g involving not the density but its ratio to the critical density). Which is to say, the surprising thing is not the particular radius of curvature we see, but rather the closeness of $\Omega$ to 1.

Assuming this is roughly correct, isn't it still surprising that in "absolute" terms, the current radius of curvature is larger than say the distance to the farthest observable galaxies ? Or is that just a very mundane consequence of 14bn years of increasing radius ?

 

[Chalnoth]

The minimum radius of curvature from current measurements is many times the size of the observable universe.

As for why the curvature is so small, one possible explanation is inflation.

The impact of curvature scales as $1/a^2$. But during inflation, the energy density of our universe was approximately constant. So during this era, the impact of the curvature is reduced. As long as inflation lasted long enough, even a large amount of curvature at the beginning of inflation would make it so that our universe was so flat that we could never hope to measure its spatial curvature.

I'd also like to point out that the era where curvature grew relative to the critical density is over: as the cosmological constant becomes a larger fraction of the energy density of our universe, the energy density (and the critical density) approaches a constant value.

 

[wabbit]

during inflation, the energy density of our universe was approximately constant.

Thanks, this seems to be (at least part of) what I was missing. Being unfamiliar with the proposed mechanism for inflation, this strikes me as odd : how does that density not get diluted during inflation? It cannot come from the cosmological constant, which was negligible then, and matter or radiation should dilute, should they not?

Also, wouldn't the curvature be low now even without inflation, just from the $1/a^2$ scaling? Perhaps detectable though...

 

[Chalnoth]

Thanks, this seems to be (at least part of) what I was missing. Being unfamiliar with the proposed mechanism for inflation, this strikes me as odd : how does that density not get diluted during inflation? It cannot come from the cosmological constant, which was negligible then, and matter or radiation should dilute, should they not?

This arises fairly easily with a scalar field that has some sort of potential energy. If the field is not right at the minimum of its potential, it will tend to "roll" down the potential energy slope. But the expansion acts as a friction to this: the faster the expansion, the more resistance the field has to moving to a lower-energy configuration. Basically, the expansion is so fast during inflation (due to the high energy density) that the scale factor doubles many times before the field has a chance to drop to a lower-energy state.

This effect continues until the scalar field finally reaches its minimum energy, at which point it oscillates around the minimum causing the field to decay into other particles. This causes inflation to end, and the universe to become extremely hot.

There are a great many models of inflation, and usually the main difference is a different shape for the potential energy of the field, though there are more exotic ideas as well. Note that there are alternatives to inflation as well, though I don't know how they solve the flatness problem.

 

[wabbit]

Ah yes that makes sense, thanks - in a way this amounts to postulating a non-diluting energy density (in the form of a scalar field), so you get out what you put in :)

Regarding the current curvature, I was trying to estimate this from $\Omega_k=-0.005\pm0.006$ but from playing with the LCDM/FLRW equations I somehow got to $K=R_H\sqrt{-\Omega_k}$ for the radius of curvature K, which is obviously wrong... would you have a pointer to a source that gives the correct form?

 

[Chalnoth]

The easiest way to get at the radius of curvature is to consider the alternative formulation of the FRW metric, where $k = {-1, 0, 1}$. With this definition, $a(t)$ becomes the radius of curvature.

To translate back to the other convention where $a(t=now) = 1$, we can note that the curvature term in the first FRW equation is $k / a^2$. This means that the radius of curvature $R_c = a/\sqrt{|k|}$, which for today is simply $R_c = 1 / \sqrt{|k|}$.

 

[wabbit]

Thanks. As stated though this is a free parameter; I was trying to relate it to the LCDM parameters, which is where I got the wrong relation $K=R_H\sqrt{-\Omega_k}\simeq 1bn ly$ for $\Omega_k=-0.005$ (based on formulas in http://casa.colorado.edu/~ajsh/phys5770_08/frw.pdf ). I'll redo that later, must be some change of coordinates I missed somewhere.

 

[Chalnoth]

Thanks. As stated though this is a free parameter; I was trying to relate it to the LCDM parameters, which is where I got the wrong relation $K=R_H\sqrt{-\Omega_k}\simeq 1bn ly$ for $\Omega_k=-0.005$ (based on formulas in http://casa.colorado.edu/~ajsh/phys5770_08/frw.pdf ). I'll redo that later, must be some change of coordinates I missed somewhere.

You have to divide, not multiply, the square root of the curvature.

 

[Buzz Bloom]

There are a lot of assumptions in the current standard model. I have not yet been able to find or develop a complete list.

One assumption (speculation as opposed to theory) is the universe is flat. The most significant evidence supporting this assumption is the astronomical determination (using the "standard candle") that in the past the universe was expanding slower than it expands now. However, this is only one way that this acceleration might occur. An alternative that I have not seen discussed anywhere is the following. (I am not proposing that my alternative is "as good as" the current model, only that I am unaware that anyone has any evidence it is not as good, although this alternative does introduce a new parameter whose value would need to be determined.)

Suppose there is no dark energy, and that dark matter has a half-life. (This half-life is the parameter needing a value.) Suppose the particles of dark matter disintegrate into two energy particles with an exponential probability distribution for the occurrence of disintegration. As the universe expands, the energy density of each the energy particles would get smaller in inverse proportion to universe's scale, commonly represented as a(t). Thus while the mass density of matter (including dark matter) is inversely proportional to a(t)^3, the density of the dark matter's disintegration particles would be inversely proportional to a(t)^4. Thus the total mass density would decline faster than that of the combination of ordinary matter and dark matter (without any disintegration). The result would be that the rate of expansion now would be greater than it was is the past, when Omega would be greater than it is now. Perhaps a suitable value for the hypothetical half life of dark matter would result in a model that matches the value of the deceleration parameter q found by recent astronomical investigations.

 

[PeterDonis]

Suppose the particles of dark matter disintegrate into two energy particles

If this were happening, we would see the radiation produced (which is what I assume you mean by "energy particles"--particles with zero rest mass), just as we see the CMBR from the early universe. But the radiation from such an ongoing process, unlike the CMBR, would not look like black body radiation at a single temperature. It would have fairly constant intensity over a wide range of wavelengths, since it would have been emitted from dark matter decays throughout the universe's history, and so would have been redshifted by widely varying amounts, depending on when it was emitted. (The CMBR was emitted over a very short time period when the universe was a few hundred thousand years old, so it all has the same redshift and therefore looks like a black body at a single temperature, just under 3 K.)

We don't see anything like this kind of radiation, so no decay process such as you suggest could have taken place with enough intensity to significantly affect the overall energy density of the universe.

 

[PeterDonis]

The result would be that the rate of expansion now would be greater than it was is the past, when Omega would be greater than it is now.

No, that's not what the result would be. The result would be that the expansion would be decelerating more slowly now than it was in the past. In a universe with no dark energy, the density of ordinary matter and radiation determines how fast the expansion decelerates; it can never cause the expansion to accelerate.

 

[wabbit]

You have to divide, not multiply, the square root of the curvature.

That' it, thanks! Stupid mistake on my part somewhere along the way, the increasing relation made no sense but I couldn't see where it was wrong.

So I get a radius of about 200 bn ly (or rather, $\geq$ 130 bn ly taking into account the confidence interval)

 

[Buzz Bloom]

If this were happening, we would see the radiation produced (which is what I assume you mean by "energy particles"--particles with zero rest mass), just as we see the CMBR from the early universe.

Hi PeterDonis:

I responded to your corresponding comment in the other thread:
Dark energy and the Cosmological Constant

Suppose that the hypothetical energy particles into which the hypothetical dark matter particle decayed were also dark -- that is these energy particles also would have no interaction with photons. Could that speculative possibility not explain why we don't see the radiation you mentioned?

 

[Buzz Bloom]

The result would be that the rate of expansion now would be greater than it was is the past

The result would be that the expansion would be decelerating more slowly now than it was in the past.

Hi PeterDonis:

Sorry, I did not write what I intended.

Consider a very large sphere at the present time, t0, with radius R(t0). At some time in the past, t1, this sphere would have a radius R(t1) = R(t0) * a(t1)/a(t0), where a(t) is the scale of the universe. Consider the ratio alpha = a(t0)/a(t1) for two different assumed conditions: (a) Lambda = 0, (b) Lambda /= 0 corresponding to the current model in which the expansion of the universe is accelerating. The ratio alpha for (a) is less than the ratio alpha for (b).

Now consider a third condition (c): Lambda = 0, and the total amount of dark matter (DM) and its decay particles in S reduces with time, due to the continuing decaying of DM particles into energy particles. Assume that S contains the same the identical t1 total amount of mass/energy, the total matter and energy density in S at time t0 will be less for case (c) than for case (a). Therefore the gravitonal influence of matter to slow the expansion will be less over time for case (c) than for case (a). Therefore the value of alpha will also be less for (a) then for (c).

If the right value is taken for the half-life of DM, then the value of alpha for case (c) can be made equal to the value of alpha for case (b). I think this means that given the best fit value for the half-life of DM, the astronomical data from which the deceleration parameter q was calculated could be made to fit (Least Mean Squared) the shape of a(t) for case (c) to get the same value for q.

If this calculation were performed, it is possible that the LMS error would be greater for the for using a case(c) model than the fit using the case (b) model, and that this greater LMS error might be statistically significant. That would be evidence that case (b) is a better model than case (c). On the other hand, maybe it would turn out that the case (c) model would have a better fit, or I think more likely, the fits would be statistically equivalent.

 

[PeterDonis]

Suppose that the hypothetical energy particles into which the hypothetical dark matter particle decayed were also dark -- that is these energy particles also would have no interaction with photons. Could that speculative possibility not explain why we don't see the radiation you mentioned?

Speculatively, yes, although that now adds two types of new particles--the dark matter particles and the non-interacting energy particles. (Btw, the key thing is not that they wouldn't interact with photons, but that they wouldn't interact with ordinary matter the way photons do; we couldn't see them with telescopes or detect them with X-ray detectors or radio antennas, etc., etc.) Also, there are no indications even in theoretical extensions of the standard model of particle physics of any such non-interacting energy particles, whereas there are candidates for particles that could make up dark matter.

Finally, none of this changes what I said in post #16, that you can't get accelerating expansion this way. See below.

If the universe is "decelerating more slowly" in the past as it expands, then it must be expanding faster then than it is now.

Since the terminology here can get confusing, let me use math instead, which is unambiguous. The key mathematical quantity here is $\ddot{a}$, the second derivative of the scale factor with respect to "comoving" time (i.e., with respect to the proper time of "comoving" observers). Any model without dark energy, i.e., that only contains ordinary matter and radiation, or dark matter and your hypothetical "energy particles", can only have $\ddot{a} < 0$. But we actually observe that $\ddot{a} > 0$ starting a few billion years ago. Dark energy is the only way of getting $\ddot{a} > 0$.

The specific behavior of $\ddot{a}$ can depend on what particular kind of ordinary matter or radiation or dark matter or hypothetical "enegy particles" you have. So having a bunch of matter (ordinary or dark) transmute into a bunch of radiation (ordinary or your hypothetical "energy particles") can change the detailed behavior of $\ddot{a}$. But it can't make $\ddot{a} > 0$. Only dark energy can do that.

 

[wabbit]

By itself I can't say I find the inflation explanation truly convincing. At least in part it seems ad hoc that the inflation mechanism would produce just the right amount of flatness, unless things were flat at the outset. Is there a proposed mechanism that would automatically calibrate things? e.g. an inflation that would require curvature to happen, and would die out of itself once that curvature fell below some natural scale?

Edit: perhaps (speculatively) there might be hope for something of that kind in quantum cosmology.

 

[Buzz Bloom]

But it can't make a¨>0\ddot{a} > 0. Only dark energy can do that.

Hi PeterDonis:

Thank you very much for your explanation. I have just done the math to educate my self that the sign of the deceleration parameter is the opposite of the sign of a-double-dot.

q = - (a * a-double-dot) / (a-dot)^2.

I would very much like to see a plot of distance (based on the "standard candle" observations) vs. redshift, and the fit of the data by a best fit curve whose shape depends on the value of q. Do you know where I can find a diagram with such a plot? I was able to find it by searching the internet.
I also understand that the Fourier transform analysis of the microwave background radiation angular irregularities support the flat universe conclusion, but I admit I am as yet not able to understand this evidence. Can you recommend a source for a good clear explanation for a non-physics professional?

 

[PeterDonis]

the sign of the deceleration parameter is the opposite of the sign of a-double-dot.

Correct.

Do you know where I can find a diagram with such a plot?

Try these articles for a start:

http://en.wikipedia.org/wiki/Lambda-CDM_model

http://www.astro.ucla.edu/~wright/cosmology_faq.html#CC

Can you recommend a source for a good clear explanation for a non-physics professional?

I'm not familiar enough with this myself to recommend a source, but others on this forum might be able to.

 

[Chalnoth]

By itself I can't say I find the inflation explanation truly convincing. At least in part it seems ad hoc that the inflation mechanism would produce just the right amount of flatness, unless things were flat at the outset. Is there a proposed mechanism that would automatically calibrate things? e.g. an inflation that would require curvature to happen, and would die out of itself once that curvature fell below some natural scale?

Edit: perhaps (speculatively) there might be hope for something of that kind in quantum cosmology.

Inflation makes the universe more flat the longer it lasts. So it just needs to last long enough. The normal expectation among inflation theorists is that inflation will make the observable universe so flat that we could never hope to detect any spatial curvature.

 

[Buzz Bloom]

Do you know where I can find a diagram with such a plot?

Try these articles for a start:

http://www.astro.ucla.edu/~wright/cosmology_faq.html#CC

Thanks again PeterDonis:

I found three interesting figures in the astro.ucla article: (a) cz vs. Luminosity Distance, (b) DeltaDM {-0.5,+0.5} vs Redshift {0.0,2.0}, and (c) DeltaDM {-2,+2} vs Redshift {0,7}. However, I am at a lost to see how a value for q is derived from any of these charts, or even that q is positive. Can you help me?

 

[wabbit]

Inflation makes the universe more flat the longer it lasts. So it just needs to last long enough. The normal expectation among inflation theorists is that inflation will make the observable universe so flat that we could never hope to detect any spatial curvature.

My admittedly vague understanding is that inflation serves (at least) two purposes: flattening the curvature, and smoothing out inhomogeneities; would it be correct to assume that the second one is the primary determinant of its parameters, so that flatness comes out as a by-product, or in other words homogeneity is in a way the reason for flatness, at least within "typical" inflation models?

 

[Chalnoth]

My admittedly vague understanding is that inflation serves (at least) two purposes: flattening the curvature, and smoothing out inhomogeneities; would it be correct to assume that the second one is the primary determinant of its parameters, so that flatness comes out as a by-product?

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

The relevant parts are sections 1.2.1 and 1.2.2, starting on page 6.

Inflation can last much longer and have no problem, but if it doesn't last long enough to provide about a factor of $10^{30}$ of expansion (less than a second in most models of inflation), then inflation can't solve either the flatness or horizon problems.

 

[wabbit]

Thanks, will look at this. Weird coincidence too...

 

[Chalnoth]

The primary other thing that inflation does is it produces density perturbations with a nearly scale-invariant power spectrum. It's generally not perfectly scale-invariant because inflation has to end, and if there was no tilt in the power spectrum inflation couldn't end, because the inflaton field would remain at the same energy. And this is exactly what we see. There are other possible models that can produce a nearly scale-invariant spectrum, however.

One big thing that inflation can produce that most other models cannot is gravity waves. Sadly, it is entirely possible for the gravity waves produced by inflation to be far too small to ever measure. If that were the case, we could not ever use gravity waves to distinguish between, e.g., inflation and a LQC bounce.

 

[PeterDonis]

I am at a lost to see how a value for q is derived from any of these charts

Yes, it's hard to read it off of those graphs, and unfortunately Wright's comment on this in the article is very terse: "The data show an accelerating Universe at low to moderate redshifts but a decelerating Universe at higher redshifts, consistent with a model having both a cosmological constant and a significant amount of dark matter."

The first chart (cz vs. luminosity distance) is probably the best one to look at. The redshift z tells you by what factor the universe has expanded since the light being seen was emitted, i.e., it gives you the ratio of the scale factor now (when we see the light) to the scale factor then (when it was emitted). (More precisely, it is 1 + z that gives this ratio.) Using cz instead of z just converts this to units of velocity, which is how cosmologists usually quote the numbers. The luminosity distance can be thought of as the distance the light had to travel to get to us.

First, look at the green "empty model" curve on the chart. This is a universe with no matter or energy of any kind at all; so its deceleration parameter $q$ is zero. (In fact this model is just flat Minkowski spacetime in unusual coordinates, but we can ignore that interesting sidebar here.) This model serves as the "reference" on the chart to which you can compare the other curves to see what their value of $q$ is.

Now look at the blue curve that the key says is the "de Sitter model". This is a universe with only dark energy in it, nothing else. Note how the slope of this curve is always flatter than that of the "empty model" curve. This indicates $q < 0$ (i.e., $\ddot{a} > 0$), which is accelerating expansion.

At the other extreme, look at the red curve that the key says is the "closed matter only model". Note that the slope of this curve is always steeper than that of the "empty model" curve. This indicates $q > 0$, or decelerating expansion.

Now look at the curves which are good fits to the actual data. (The best fits in this chart are the "closed dark energy model" and the "evolving SNe".) Note how these curves are flatter than the "empty model" curve at smaller distances, but steeper at larger distances. This indicates that $q < 0$ at smaller distances (i.e., accelerating expansion more recently, since smaller distances mean light emitted closer to now), but $q > 0$ at larger distances (i.e., decelerating expansion at earlier times).

In other words, to read off the value of $q$ for a given model at a given luminosity distance, we compare the slope of that model's curve with the slope of the "empty model" curve at that distance. Flatter slope means $q < 0$; steeper slope means $q > 0$.

 

[Buzz Bloom]

Hi PeterDonis:

You very lucid explantion is extremely helpful. Many thanks.

 

[wabbit]

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.

 

[wabbit]

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?

 

[PeterDonis]

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.

 

[wabbit]

Aaaaaaahh yes. Obviously... Thanks I was banging my head on a wall here

Pretty smart too that scalar field, stealing all the energy to inflate away quietly and then giving it back as if nothing happened

 

[wabbit]

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.

 

[PeterDonis]

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 $w = 1/3$.)

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.

 

[wabbit]

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:)

 

[Chalnoth]

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.

 

[wabbit]

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)

 

[Buzz Bloom]

Hi @Garth:

It depends on the total average density parameter Ω=ρ_tota l ρ_cricital .

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:

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:
(Ω_k > 0, Ω_m = Ω_r = 0, Ω_DE = 1- Ω_k)?

 

[Garth]

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:

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

 

[Buzz Bloom]

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 Ω_k rarther than the sum of all the density ratios.

Thanks for your very prompt response,
Buzz

 

[Buzz Bloom]

The best current estimated value I can find for Ω_k 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 Ω_k = 0.0008 +0.0040 / -0.0039. The 0.0008 corresponds to a current curvature radius R_c = 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 R_c. With those two values of R_c, can someone explain why the inflation value is better than the no inflation value?

 

[Garth]

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 Ω_k 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

 

[Buzz Bloom]

Hi @Garth:

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* = Ω_k + Ω_m + Ω_r + Ω_Λ = 1.​

I am guessing that the Ω in the quote from your post #45 is Ω = Ω* - Ω_k. Is this correct? If so, then I interpret your question as

"Why is Ω_k 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 = t_i = 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 Ω_k at t=t_i would have been much much larger than the it's value with inflation, Therefore the current value of Ω_k 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

 

[Garth]

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* = Ω_k + Ω_m + Ω_r + Ω_Λ = 1.​

I am guessing that the Ω in the quote from your post #45 is Ω = Ω* - Ω_k. Is this correct? If so, then I interpret your question as

"Why is Ω_k 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* = Ω_k + Ω_m + Ω_r + Ω_Λ = 1 is an identity.
​

I also get that extrapolation backwards to just before inflation at t = t_i = 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 Ω_k at t=t_i would have been much much larger than the it's value with inflation, Therefore the current value of Ω_k 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

 

[George Jones]

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

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:

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.

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.

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.

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.

 

[Tanelorn]

I was trying to follow the arguments in this thread and I found these links:
https://en.wikipedia.org/wiki/Flatness_(cosmology)
https://en.wikipedia.org/wiki/Flatness_problem

What I am not getting is if the flatness is approx 1 now, how could it have been approx 1 in the past when the universe was denser, before it underwent dark energy expansion?