# Why is the Universe (nearly) flat?

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1. Mar 21, 2015

### 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

2. Mar 21, 2015

### 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.

3. Mar 22, 2015

### Paulibus

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?

4. Mar 22, 2015

### wabbit

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.

Last edited: Mar 22, 2015
5. Mar 22, 2015

### Garth

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

6. Mar 22, 2015

### 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 ?

7. Mar 23, 2015

### 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.

8. Mar 23, 2015

### wabbit

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...

Last edited: Mar 23, 2015
9. Mar 23, 2015

### Chalnoth

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.

10. Mar 23, 2015

### 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?

Last edited: Mar 23, 2015
11. Mar 23, 2015

### 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 $R_c = 1 / \sqrt{|k|}$.

12. Mar 23, 2015

### 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 [Broken]). I'll redo that later, must be some change of coordinates I missed somewhere.

Last edited by a moderator: May 7, 2017
13. Mar 23, 2015

### Chalnoth

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

Last edited by a moderator: May 7, 2017
14. Mar 23, 2015

### 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.

Last edited: Mar 23, 2015
15. Mar 23, 2015

### Staff: Mentor

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.

16. Mar 23, 2015

### Staff: Mentor

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.

17. Mar 24, 2015

### wabbit

That' it, thanks! Stupid mistake on my part somewhere along the way, the increasing relation made no sense but I couldnt 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)

Last edited: Mar 24, 2015
18. Mar 24, 2015

### Buzz Bloom

Hi PeterDonis:

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?

19. Mar 24, 2015

### Buzz Bloom

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.

Last edited: Mar 24, 2015
20. Mar 24, 2015

### Staff: Mentor

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.

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.