Size of universe after inflation?

Question

Wikipedia states the period of inflation was from $10^{-36}$sec to around $10^{-33}$sec or $10^{-32}$sec after Big Bang, but it doesn't say what the size of the universe was when inflation ended. Just saw a Brian Greene show on the Multiverse and I thought I heard him say size was galactic scales when inflation ended. However I've also read size was about a basketball.

Are there multiple theories with different resulting sizes? Does 'size' even mean anything in this context?

Unless I'm missing something there doesn't seem to be a consensus. However the question still seems sensible to me- at the time of the big bang everything was really tiny and it started expanding. Why can't you know the volume before inflation and after? — Art Hays, Jul 27 '12 at 17:49
The universe being "the size of a basketball" may be a misreading of "the Universe [a local one in a multiverse] having the shape of the skin of a basketball", which was a description used by Nikodem Poplawski in his "cosmology with torsion" model (described in numerous preprints that can be found by his name on Arxiv), that is usually considered to be a version of inflation (asymptotically exponential spatial expansion). — Edouard, Feb 18 '24 at 18:35

score 30 · Answer 1 · edited Aug 11 '14 at 15:53

30

With the proper definition of the "size" of the universe, this question does make sense. The standard model of cosmology would say that the universe is infinite which therefore does not have a "size". However, if we take into account that the big bang occurred $13.7 \pm 0.17$ billion years ago we can define a meaningful size for the observable universe. You might, for example, define the size of the observable universe as the distance a photon could have traveled since the big bang.

Consider, for example, a cosmic microwave background (CMB) photon that was emitted as visible light about 379,000 years after the big bang and is just now hitting our microwave detectors (the redshift is z=1089): that photon has been traveling for 13.7 billion years so it has traveled a distance of 13.7 billion light years. So you might imagine that the current radius of the observable universe is 13.7 billion light years. However, during this time the universe has been expanding, so the current position of the matter that emitted that photon will now be 46.5 billion light years away. (By now, the little $10^{-5}$ bumps on the CMB will have condensed into galaxies and stars at that distance.) This gives a diameter of the current observable universe of 93 billion light years. Note that as time passes, the size of the observable universe will increase. In fact it will increase by significantly more than two (to convert radius to diameter) light years per year because of the continued (accelerating) expansion of the universe. Also note that we will not be able to use photons (light) to explore the universe earlier than 379,000 years after the big bang since the universe was opaque to photons at that time. However, in the future we could conceivably use neutrinos or gravitational wave telescopes to explore the earlier universe.

So given a size of the current observable universe, we can ask how big was that volume at any particular time in the past. According to this paper at the end of inflation the universe's scale factor was about $10^{-30}$ smaller than it is today, so that would give a diameter for the currently observable universe at the end of inflation of 0.88 millimeters which is approximately the size of a grain of sand (See calculation at WolframAlpha).

It is believed that inflation needed to expand the universe by at least a factor of 60 e-foldings (which is a factor of $e^{60}$). So using WolframAlpha again we find that the diameter of the universe before inflation would have been $7.7 \times 10^{-30}$ meters which is only about 480,000 Planck lengths.

Perhaps Brian Greene was talking about the size of the observable universe at the time when the CMB photons started traveling towards us. That happened 379,000 years after the big bang at a redshift of 1098 which means the universe was about 84.6 million light years in diameter which, per WolframAlpha, is about half the diameter of the local super cluster of galaxies or about 840 times the diameter of our galaxy.

edited Aug 11 '14 at 15:53

Community

1

answered Jul 26 '12 at 19:00

FrankH

10,676

2

Nice answer. Also, according to the APOD scales of the universe (http://apod.nasa.gov/apod/ap120312.html), a human ovum is more like 0.12 millimeters. Not that it matters... I just like the applet. :) – AdamRedwine Jul 26 '12 at 19:20
1

@AdamRedwine Thanks. Wikipedia agrees with APOD so I changed the comparison to a grain of sand. – FrankH Jul 26 '12 at 22:32
This answer is using a bad notion of the radius of the observable universe. You should measure the radius along a past light cone, without extrapolation to "now", extrapolation is not the right way to describe the physics. The universe was 380,000 light-years across at the time of photon decoupling, measured along the back light-cone, and this is the physical size, and it's the size of a galaxy. The "now" extrapolations are unphysical and arbitrary. – Ron Maimon Jul 27 '12 at 04:50
1

@RonMaimon - Sorry, but I disagree. We are detecting CMB photons that were emitted by real atoms and while the photons were on the way here for the last 13.7Blyr, we can confidently say that those atoms have formed galaxies and stars that are now about 46Blyrs away from us. Taking the scale factor a(t) back in time to when the universe was 379,000 years old the universe was 1/1098 of it's current size so those same atoms, at that time, were spread out over a volume with a radius of 42Mlyrs. (...cont'd...) – FrankH Jul 27 '12 at 18:47
(...cont'd...) Yes, the atoms of my body, at that time, would have only have had a past light cone of size 379,000 light years but the other atoms did exist which we can tell because they have NOW come into our past light cone. There is a universe outside the observable universe and all we are doing is to take the CURRENTLY observable universe and calculate the history of those currently observable atoms/quarks/strings back in time to say where they were then. – FrankH Jul 27 '12 at 18:50
@FrankH: This is extrapolation to a current time-slice, and it is the wrong way to talk about horizons. If you don't use a past light cone, you might as well use a fully extended solution and say the universe is infinite, and forget about the distance to the horizon. There is no sensible cutoff in the t=const time slice, your heuristic of following atoms inside the horizon to "now" doesn't work, it fails even for pure deSitter space. – Ron Maimon Jul 27 '12 at 23:11
Please forgive me, I am a lowly computer programmer with an interest in physics, please can you explain what you mean by the universe being only 0.88mm at the end of inflation... Are you saying that after approx. 400000 years of inflation (which to my mind is part of the big bang explosion) that this is the size of the universe? Sorry if this is a stupid question. – Mr. Mr. Oct 22 '12 at 13:20
2

@Monkieboy - Where did you get 400000 years of inflation? Inflation ended at about $10^{-32}$ seconds after the big bang - that is when the universe was only 0.88mm. The 400000 years age is when the universe had expanded and cooled enough to become transparent - that is when the CMB photons started traveling towards us 13.7 billion years ago. At that time the universe was about 42 million light years in diameter. – FrankH Oct 22 '12 at 15:37
@FrankH - Thank you I misread the question where it says, '... the period of inflation was from $10^{−36}$ sec to around $10^{-33}$ sec or $10^{−32}$ sec after Big Bang' to be that it was approximated to have started at those moments, not that it started at $10^{−36}$ sec and ended at $10^{−32}$ secs. – Mr. Mr. Oct 22 '12 at 15:48
@FrankH : In your second paragraph, I think you are incorrect, when you say that the observable universe will grow in size due to the accelerated expansion of the universe. A strong-energy-violating fluid, such as dark energy, domination era causes the Hubble Sphere (that is, the observable universe) to shrink in size. So the observable universe is getting smaller, not bigger. As the rate of expansion of some distant galaxy accelerates to an expansion rate which is faster than the limiting speed of light we stop being able to observe said galaxy. – Flint72 May 09 '14 at 15:20
@Flint72 - I agree that the amount of matter in the observable universe will shrink as the accelerated expansion increases - which I think is what you mean. Is it? So at say, 100 billion years post big bang, there will still be CMB photons that will be redshifted down to a very long wavelength. Those photons traveled along our past light cone from a distance of 100 billion light years, right? – FrankH May 09 '14 at 21:35
@Flint72 .... Now in the rest frame of those CMB photons, the matter that emitted those photons will now be at a distance that is much more than 100 billion light years away from us at our current time. They will never be observed because of the accelerated expansion (if DE=0 they would eventually come back into our Hubble volume). So that is what I meant. Is this right? – FrankH May 09 '14 at 21:37

pela · Answer 2 · 2020-05-16T21:04:39.997

Of the order 10 meter.

The size of the Universe can be calculated by integrating the Friedman equation, which is a function of the densities of the components of the Universe (radiation, matter, dark energy, curvature, as well as more exotic components), as well as their equations of state. In general, there is no analytical result, but in certain epochs the history of the Universe, its dynamics are completely dominated by one or two of these components.

The early Universe was dominated by relativistic matter, i.e. radiation and neutrinos (early matter was also relativistic, but didn't contribute significantly to the energy density). In this case, integration yields the following relation between the scale factor $a$ (i.e. ratio between lengths at that time and today) and time $t$: $$ a(t) \simeq \left( 2 \sqrt{\Omega_\mathrm{r,0}} H_0 t \right)^{1/2}, $$ where $\Omega_\mathrm{r,0}$ is today's value of the energy density of radiation relative to the critical density, and $H_0$ is the Hubble constant. For a Planck Collaboration et al. (2016) cosmology, at $t\sim10^{-32}\,\mathrm{s}$, this yields $2\times10^{-26}$.

That is, if inflation ended after $10^{-32}\,\mathrm{s}$, everything was $5\times10^{25}$ times closer to each other, or roughly 60 e-folding$^\dagger$.

The total Universe may or may not be infinite, but what we usually refer to when talking about the Universe, is the observable Universe, which is the part of the Universe from which light has had the time to reach us since Big Bang. The Universe is 13.8 Gyr old, but because it has expanded in the meantime, the observable Universe is more than 13.8 Glyr in radius — in fact $R_0 = 46.3\,\mathrm{Glyr}$.

Hence, the radius of what comprises "our" Universe today, was at time $t$ only $R(t) = a(t) R_0$, so at the end of inflation $$ \begin{array}{rcl} r(10^{-32}\,\mathrm{s}) & = & a(10^{-32}\,\mathrm{s}) \, R_0 \\ & = & 2\times10^{-26} \, \times \, 46.3\,\mathrm{Glyr} \\ & = & 9\,\mathrm{m}. \end{array} $$

If you think that inflation ended already after $10^{-33}\,\mathrm{s}$, you'll get $r=3\,\mathrm{m}$ instead.

$^\dagger$_{Coincidentally (I think) roughly the same number of e-foldings as inflation itself.}

I think you're using the wrong $t$. Your formula needs the time since the singularity of the non-inflationary model, which is much smaller than the actual duration of inflation. In a toy model with radiation-dominated expansion from a singularity followed by $e^{Ht}$ followed by radiation-dominated again, the correct $t$ is actually the time of the start of inflation, since that's equal to $1/(2H)$ and the end time isn't. — benrg, Oct 12 '21 at 06:48
I'm not sure any of the times are trustworthy anyway – the energy scale of inflation isn't known last I heard, nor is the duration, and the "time of the start of inflation" may not even be meaningful if the universe wasn't FLRW... — benrg, Oct 12 '21 at 06:51
But my time is just the "regular" time, extrapolated back to just after inflation. I'm not concerned with what happened during, or before, inflation. I don't understand why that's not appropriate, but I'd be happy to see your calculation on this (in writing, this looks like a sarcastic "Just write your own damn answer", but that's not how I mean it :) ). And yes, I assume a mainstream FLRW universe (partly because this site is mainstream, but mostly because I'm not enough of a cosmologist to do otherwise). — pela, Oct 12 '21 at 12:29

John Rennie · Answer 3 · 2012-07-26T16:37:32.210

In the simplest model of the universe, the FLRW metric, the universe is infinite and has always been infinite right back to the Big Bang. Inflation doesn't change this assumption.

So it makes sense to ask, for example, how big a Planck volume became during inflation, but it doesn't make sense to ask how big the whole universe is. (Depending on what you take as the inflation scale factor a Planck volume ended up about $10^{-27}m^3$ and this is a lot smaller than a basketball.)

Having said this, Don Page has suggested a lower bound for the size of the whole universe at the end of inflation, and his answer is $10^{10^{10^{122}}}$ cubic megaparsecs. However I think you should regard this as extremely speculative.

How would you know it's infinite? The FLRW metric is only a predictive metric for this patch, and any extension past the horizon is speculation about unobservable things. — Ron Maimon, Jul 27 '12 at 04:51

score -1 · Answer 4 · answered Mar 24 '14 at 20:10

I'm just a lowly aerospace engineer. But when I think of inflation to a certain volume, my Euclidean me says that it had to be at least large enough at the end of inflation for the 13.7 billion light year distance from our present position to the opposite end of universe since that is the farthest light we have detected. If this is the case can't we just back calculate to the radius given the accelerated expansion (assuming constant acceleration rate) since the end of inflation? In my thinking if the universe was millimeters in size at the end of expansion, then the photons emitted from the "other side" of the universe would have passed us already. So even with an accelerating expansion from that time to today at velocities much less than the speed of light means that we cannot possibly be seeing the start of the universe after cooling.

Perhaps it is just me, but this answer doesn't really seem to answer the question. — Kyle Kanos, Mar 24 '14 at 20:31

score -9 · Answer 5 · edited Aug 05 '13 at 02:26

Answer to the inflation riddle is simple, it is staring everyone in the face.

You can not explain the early universe in classical terms, the size/volume whatever you want to think of before and after the inflationary period is irrelevant.
The before and after were completely different to each other, why? Because in the beginning there was nothing, then there was something, this something did not come fully formed as we see the universe now with it's multiple dimensions, at that beginning point there was only one dimension, what happen after in the inflationary period was the expanding or addition of the extra dimensions, two then three, and so on, that is why the expansion seems so incomprehensibly large and seems to move at impossible speeds far beyond what the speed of light would allow. The number of dimensions is infinite on a scale of infinities beyond reasonable comprehension, of which the observable universe is only three and was fully formed at the end of the inflationary period, these infinite dimensions never stop growing exponentially.

BUT Time it's self is the first dimension, a framework of sorts, in which all the other dimensions rest, it is the container, not a point, like the exponentially dividing cell when human life begins. Time is nothing more than exponential dimension addition/growth.

Size of universe after inflation?

5 Answers5

Linked

Related