<p>The comoving radius of the sphere of matter that we see as the CMBR can be calculated from measured cosmological parameters, and is around 46 billion light years.</p>
<p>The comoving size encompassed by our past light cone back to the beginning of time is not known, but it has to be much larger—at least hundreds of billions of light years across, and potentially vastly larger.</p>
<p>If you calculate the comoving radius of the observable universe back to the end of inflation in ΛCDM cosmology (or back to the non-inflationary big bang—the difference is negligible), then you get a value only slightly larger than the CMBR radius (less than a billion light years more). That's actually a problem, known as the <a href="https://en.wikipedia.org/wiki/Horizon_problem" rel="nofollow noreferrer">horizon problem</a>.</p>
<img src="https://upload.wikimedia.org/wikipedia/commons/c/ce/Horizon_problem.svg" width="300">
<p>As shown in this picture from the Wikipedia article, if the maximum distance that light (or anything else) can have traveled between the big bang and CMBR emission is small, then the causal pasts of distant points in the CMBR sphere have nothing in common, so there's no reason why the CMBR should be as uniform as it's observed to be.</p>
<p>To solve this problem, you need the past light cones of the distant points to overlap substantially. If you imagine that antipodal points should have 99% overlap of their pasts to explain the homogeneity of the CMBR (I don't know whether that's at all reasonable!), then the comoving radius of their causal pasts needs to be around <span class="math-container">$\sqrt{100/2}$</span> times the radius of the CMBR sphere, i.e., ~300 billion light years in present-day comoving distance.</p>
<p>You may have noticed an error here: the causal pasts are only nice circles/spheres if the universe is homogeneous and isotropic, but if it is, then there's nothing to explain. This also affects diagrams like the one in the Wikipedia article: the small circles shouldn't be circular. But the argument is essentially unchanged, just fuzzier. The main takeaway is that the total region encompassed by our past light cone must be a lot larger than the CMBR sphere, not just slightly larger.</p>
<p>Inflation solves this problem by having very large past light cones. If it lasts for 60 e-folds, which is commonly quoted as the minimum possible to explain the observed homogeneity, then the comoving light travel distance during inflation is around <span class="math-container">$e^{60}\approx 10^{26}$</span> times larger than in the radiation-dominated big bang model. This huge blowup applies only to a tiny fraction of a second of expansion in the radiation-dominated model, not 380,000 years worth, but it is still enough to greatly increase the total area covered by the past light cone. This is only a lower bound on the number of e-folds; inflation can easily generate enormously larger homogeneous regions. So the size of our past light cone really is unknown.</p>

CC BY-SA 4.0
2021-10-19T21:48:51.007Z
2021-10-19T21:48:51.007Z
benrg
<p>If I understand you correctly, you want to know the distance from the point from which we observe the CMB, to the edge of the observable Universe.</p>

<p>During inflation, the observable Universe expanded from ridiculously small to some ten meters in radius, so that part can be safely ignored compared to the distances now$^1$.</p>

<p>The distance$^2$ to the "CMB shell"$^3$ is 45.4 billion lightyears ("Glyr"), and the distance to the edge of the observable Universe is 46.3 Glyr. Hence, the shell of the observable Universe that lies beyond the CMB shell has a thickness of only 0.9 Glyr.</p>

<p>Here's a sketch of how I interpret your question (not to scale):</p>

<p><a href="https://i.sstatic.net/QJNB0.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/QJNB0.png" alt="cmb"></a></p>

<p>If you want, I can add details on how to calculate these numbers.</p>

<hr>

<p>$^1$<sub>The <em>relative</em> expansion during inflation was huge, however: The Universe expanded roughly by the same factor that it has expanded afterwards, namely a factor of $\sim10^{26}$.</sub></p>

<p>$^2$<sub>Here, "distance" corresponds to the <em>comoving</em> distance, which is what you would measure if you froze the Universe right now, and started laying out meter sticks.</sub></p>

<p>$^3$<sub>This shell is not infinitely thin, but has a thickness of some 60 <em>million</em> lightyears, so let's ignore that.</sub></p>

CC BY-SA 4.0
2018-08-23T23:37:22.423Z
Very good, you understood correctly, and nice diagram / answer as well!  Interesting to know!
Jonathan
CC BY-SA 4.0
2019-09-28T19:09:55.080Z
The inflationary epoch can't be safely ignored. If it's included, it contributes almost all of the total size, making everything else irrelevant. One way of looking at it is that in any cosmology that solves the horizon problem through past causal contact, our past light cone covers at least the entire homogeneous region. The only ways to get a smaller observable universe are to cut off the integral early (effectively making the cutoff your definition of "observable") or use a cosmology with a horizon problem, like the one that's radiation-dominated back to $a=0$.
benrg
CC BY-SA 4.0
2021-10-18T15:13:27.260Z
Where does the 60 million light years come from? Is that just the co-moving width in time multiplied by $z+1$?
ProfRob
CC BY-SA 4.0
2021-10-19T08:40:48.373Z
@ProfRob Hmm, good question. I think I did it the other way round: Assuming pure hydrogen and solving the Saha equation, I get that the mean free path-to-Hubble distance ratio is roughly $[0.1, 0.3, 1, 3]$ at $z \sim [1200, 1150, 1100, 1050]$, so I have probably assumed a width in redshift of $dz \sim 100$. The comoving distance difference at these redshifts for $dz \sim 100$ — and hence the physical width today — is a little less that 20 Mpc, or just over 60 Mlyr. But thinking about it today, that's probably a bit too wide, so maybe half this width would be better.
pela
CC BY-SA 4.0
2021-10-19T08:49:29.740Z
@benrg I see I missed this comment previously. But I'm not sure I understand what this implies for the distances today. How would you then draw this figure?
pela
CC BY-SA 4.0
2021-10-19T21:49:06.847Z
@pela I wrote it up as another answer.
benrg
CC BY-SA 4.0
2018-08-23T15:34:41.110Z
2018-08-24T10:28:14.230Z
2018-08-24T10:28:14.230Z
pela
<p>What I want to know is how thick the observable universe is from the point of the cosmic microwave background and beyond.</p>

<p>It appears the thickness of the cosmic microwave background itself (the part we can see) is above 100,000 light years, per the following article:
<a href="http://scienceblogs.com/startswithabang/2013/06/19/5-facts-you-probably-dont-know-about-the-cosmic-microwave-background/" rel="nofollow noreferrer">http://scienceblogs.com/startswithabang/2013/06/19/5-facts-you-probably-dont-know-about-the-cosmic-microwave-background/</a></p>

<p>However, I want to know the thickness of that, plus what lies beyond that we cannot see, another way of looking at it would be the distance between the surface of last scattering (cosmic microwave background end) to the beginning (e.g. Big Bang).</p>

<p>According to the following article, it appears that this time from the beginning to the surface of last scattering is about 300,000 years:
<a href="https://ned.ipac.caltech.edu/level5/Glossary/Essay_lss.html" rel="nofollow noreferrer">https://ned.ipac.caltech.edu/level5/Glossary/Essay_lss.html</a></p>

<p>That would imply that the thickness should be about 300,000 light years, but that doesn't take inflation into account. </p>

<p>What is the thickness (in the observable universe) between the beginning (e.g. Big Bang) to the surface of last scattering (Cosmic Microwave Background), including inflation?</p>

CC BY-SA 4.0
2018-08-23T14:35:17.390Z
Beside inflation, shouldn't be further expansion be taken into account, too?
Alchimista
<distances><big-bang-theory><observable-universe><cosmic-microwave-background>
How thick is the cosmic microwave background, including the part we cannot see within the observable universe?
Jonathan
# How thick is the cosmic microwave background, including the part we cannot see within the observable universe?

(Asked by: username_0 on 2018-08-23)

What I want to know is how thick the observable universe is from the point of the cosmic microwave background and beyond.

It appears the thickness of the cosmic microwave background itself (the part we can see) is above 100,000 light years, per the following article:
<http://scienceblogs.com/startswithabang/2013/06/19/5-facts-you-probably-dont-know-about-the-cosmic-microwave-background/>

However, I want to know the thickness of that, plus what lies beyond that we cannot see, another way of looking at it would be the distance between the surface of last scattering (cosmic microwave background end) to the beginning (e.g. Big Bang).

According to the following article, it appears that this time from the beginning to the surface of last scattering is about 300,000 years:
<https://ned.ipac.caltech.edu/level5/Glossary/Essay_lss.html>

That would imply that the thickness should be about 300,000 light years, but that doesn't take inflation into account.

What is the thickness (in the observable universe) between the beginning (e.g. Big Bang) to the surface of last scattering (Cosmic Microwave Background), including inflation?

**Tags:** distances, big-bang-theory, observable-universe, cosmic-microwave-background
> **Comments:**
> **username_1**: Beside inflation, shouldn't be further expansion be taken into account, too?  

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## 2 answers

### Answer by username_3 on 2018-08-23. Score: 7 | Accepted answer

If I understand you correctly, you want to know the distance from the point from which we observe the CMB, to the edge of the observable Universe.

During inflation, the observable Universe expanded from ridiculously small to some ten meters in radius, so that part can be safely ignored compared to the distances now$^1$.

The distance$^2$ to the "CMB shell"$^3$ is 45.4 billion lightyears ("Glyr"), and the distance to the edge of the observable Universe is 46.3 Glyr. Hence, the shell of the observable Universe that lies beyond the CMB shell has a thickness of only 0.9 Glyr.

Here's a sketch of how I interpret your question (not to scale):

[![cmb](https://i.sstatic.net/QJNB0.png)](https://i.sstatic.net/QJNB0.png)

If you want, I can add details on how to calculate these numbers.

---

$^1$The *relative* expansion during inflation was huge, however: The Universe expanded roughly by the same factor that it has expanded afterwards, namely a factor of $\sim10^{26}$.

$^2$Here, "distance" corresponds to the *comoving* distance, which is what you would measure if you froze the Universe right now, and started laying out meter sticks.

$^3$This shell is not infinitely thin, but has a thickness of some 60 *million* lightyears, so let's ignore that.

> **Comments:**
> **username_0**: Very good, you understood correctly, and nice diagram / answer as well!  Interesting to know!  
> **username_2**: The inflationary epoch can't be safely ignored. If it's included, it contributes almost all of the total size, making everything else irrelevant. One way of looking at it is that in any cosmology that solves the horizon problem through past causal contact, our past light cone covers at least the entire homogeneous region. The only ways to get a smaller observable universe are to cut off the integral early (effectively making the cutoff your definition of "observable") or use a cosmology with a horizon problem, like the one that's radiation-dominated back to $a=0$.  
> **username_4**: Where does the 60 million light years come from? Is that just the co-moving width in time multiplied by $z+1$?  
> **username_3**: @ProfRob Hmm, good question. I think I did it the other way round: Assuming pure hydrogen and solving the Saha equation, I get that the mean free path-to-Hubble distance ratio is roughly $[0.1, 0.3, 1, 3]$ at $z \sim [1200, 1150, 1100, 1050]$, so I have probably assumed a width in redshift of $dz \sim 100$. The comoving distance difference at these redshifts for $dz \sim 100$ — and hence the physical width today — is a little less that 20 Mpc, or just over 60 Mlyr. But thinking about it today, that's probably a bit too wide, so maybe half this width would be better.  
> **username_3**: @benrg I see I missed this comment previously. But I'm not sure I understand what this implies for the distances today. How would you then draw this figure?  
> **username_2**: @pela I wrote it up as another answer.  

---

### Answer by username_2 on 2021-10-19. Score: 2

The comoving radius of the sphere of matter that we see as the CMBR can be calculated from measured cosmological parameters, and is around 46 billion light years.

The comoving size encompassed by our past light cone back to the beginning of time is not known, but it has to be much larger—at least hundreds of billions of light years across, and potentially vastly larger.

If you calculate the comoving radius of the observable universe back to the end of inflation in ΛCDM cosmology (or back to the non-inflationary big bang—the difference is negligible), then you get a value only slightly larger than the CMBR radius (less than a billion light years more). That's actually a problem, known as the [horizon problem](https://en.wikipedia.org/wiki/Horizon_problem).

![](https://upload.wikimedia.org/wikipedia/commons/c/ce/Horizon_problem.svg)

As shown in this picture from the Wikipedia article, if the maximum distance that light (or anything else) can have traveled between the big bang and CMBR emission is small, then the causal pasts of distant points in the CMBR sphere have nothing in common, so there's no reason why the CMBR should be as uniform as it's observed to be.

To solve this problem, you need the past light cones of the distant points to overlap substantially. If you imagine that antipodal points should have 99% overlap of their pasts to explain the homogeneity of the CMBR (I don't know whether that's at all reasonable!), then the comoving radius of their causal pasts needs to be around $\sqrt{100/2}$ times the radius of the CMBR sphere, i.e., ~300 billion light years in present-day comoving distance.

You may have noticed an error here: the causal pasts are only nice circles/spheres if the universe is homogeneous and isotropic, but if it is, then there's nothing to explain. This also affects diagrams like the one in the Wikipedia article: the small circles shouldn't be circular. But the argument is essentially unchanged, just fuzzier. The main takeaway is that the total region encompassed by our past light cone must be a lot larger than the CMBR sphere, not just slightly larger.

Inflation solves this problem by having very large past light cones. If it lasts for 60 e-folds, which is commonly quoted as the minimum possible to explain the observed homogeneity, then the comoving light travel distance during inflation is around $e^{60}\approx 10^{26}$ times larger than in the radiation-dominated big bang model. This huge blowup applies only to a tiny fraction of a second of expansion in the radiation-dominated model, not 380,000 years worth, but it is still enough to greatly increase the total area covered by the past light cone. This is only a lower bound on the number of e-folds; inflation can easily generate enormously larger homogeneous regions. So the size of our past light cone really is unknown.


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