<p>Based on your comments, I think your confusion comes from having seen the <a href="https://en.wikipedia.org/wiki/Cosmic_microwave_background#/media/File:Ilc_9yr_moll4096.png" rel="nofollow noreferrer">classic rugby ball-shaped image of the CMB</a>.
The CMB <em>we</em> observe is not from the whole cosmos, but only from a thin <em>and completely spherical</em> shell centered on us. This light is redshifted by an amount <span class="math-container">$z=1100$</span>.</p>
<p>The radius of this shell can be calculated integrating the <a href="https://en.wikipedia.org/wiki/Friedmann_equations" rel="nofollow noreferrer">Friedmann equation</a> from us (at <span class="math-container">$z=0$</span>) to <span class="math-container">$z=1100$</span>:
<span class="math-container">$$
R_\mathrm{CMB} = \frac{c}{H_0}\int_0^{1100} 
         \frac{dz}{\sqrt{
         \Omega_r(1+z)^4 + 
         \Omega_m(1+z)^3 + 
         \Omega_k(1+z)^2 + 
         \Omega_\Lambda
         }},
$$</span>
where <span class="math-container">$c$</span> and <span class="math-container">$H_0 \sim 70\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$</span> are the speed of light and the Hubble parameter, respectively, and
<span class="math-container">$\{\Omega_r,\Omega_m,\Omega_k,\Omega_\Lambda\}\sim\{\lt 10^{-4},0.3,0,0.7\}$</span>
are the radiation, matter, curvature, and dark energy density parameters, respectively.</p>
<p>With these approximate numbers, I get <span class="math-container">$R_\mathrm{CMB} = 44.63\,\mathrm{Glyr}$</span>, while using a <a href="http://arxiv.org/abs/1502.01589" rel="nofollow noreferrer">Planck 2015</a> cosmology for the various parameters yields <span class="math-container">$R_\mathrm{CMB} = 45.36\,\mathrm{Glyr}$</span>.</p>
<p>Just as the shell of Earth can be projected onto a rugby-shaped figure using a <a href="https://en.wikipedia.org/wiki/Mollweide_projection" rel="nofollow noreferrer">Mollweide projection</a>, so can the shell of the CMB. Here's a figure from <a href="http://www.universeadventure.org/big_bang/cmb-origins.htm" rel="nofollow noreferrer">Universe Adventure</a> that can help visualize:</p>
<p><a href="https://i.sstatic.net/tSbxw.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/tSbxw.jpg" alt="projection" /></a></p>
<h1 id="the-volume-of-the-observable-universe-t6al">The volume of the observable Universe</h1>
<p>Although we (still) can't see beyond the CMB, the observable Universe is all the way out to redshift <span class="math-container">$z=\infty$</span>. The difference is not big, though; integrating the equation above from <span class="math-container">$0$</span> to <span class="math-container">$\infty$</span> yields <span class="math-container">$R_\mathrm{obs.Uni.} = 46.27\,\mathrm{Glyr}$</span>. Thus, the volume of the <em>observable Universe</em> is
<span class="math-container">$$
V = \frac{4\pi}{3}R_\mathrm{obs.Uni.}^3 = 415,\!065\,\mathrm{Glyr}^3.
$$</span></p>
<p>The volume of the <em>whole</em> Universe is probably much larger, and may in fact easily be infinite.</p>

CC BY-SA 3.0
2016-04-20T19:04:52.890Z
I guess I should read $\mathrm{Gly^3}$ as "cubic giga-light-year," not "billion cubic light-year."
Mike G
CC BY-SA 3.0
2016-04-20T19:11:51.337Z
But still it is unclear why they mostly illustrate the CMG rugby shaped and not just spherical. As I said, for the earth I can imagine why, that is to show all continents in one view.  But there isn't such a reason for the CMB?
Marijn 
CC BY-SA 3.0
2016-04-20T21:41:49.897Z
It may help to think in terms of the [celestial sphere](https://en.wikipedia.org/wiki/Celestial_sphere). Plotting the stars and galaxies on a globe, as if they were a uniform distance from us, helps us make sense of their apparent positions in the sky. We can also transform this globe into a flat map, but most map projections of a whole sphere are not circular, and all have some distortion.
Mike G
CC BY-SA 3.0
2016-04-21T06:52:36.600Z
@MikeG: Yes, $\mathrm{Gly}^3 = (10^9\mathrm{ly})^3 = 10^{27}\mathrm{ly}$. Also, as it seems you spotted, I made an error of a factor $10^3$.
pela
CC BY-SA 3.0
2016-04-21T06:59:23.467Z
@Marijn: The reason is exactly the same as for the Earth: We want to show the entire sky in one view. That's not possible with a spherical projection, unless you invent some crazy projection algorithm that will not only heavily distort the peripheral regions (which to a certain degree will be the case for all projections), but also display some regions multiple times.
pela
CC BY-SA 3.0
2017-10-10T20:28:35.867Z
But that formula is for Euclidean space. Is the universe sufficiently flat for it to be a reasonable approximation?
Acccumulation
CC BY-SA 3.0
2017-10-11T07:08:38.947Z
@Acccumulation: Yes, the observable Universe is extremely (if not completely) flat, with  $|\Omega_k| < 0.005$ ([Planck Collaboration et al. 2016](https://arxiv.org/abs/1502.01589)).
pela
CC BY-SA 3.0
2017-04-24T11:23:25.633Z
NB Only true for a universe with no curvature.
ProfRob
CC BY-SA 4.0
2016-04-20T18:51:33.943Z
2021-07-26T08:23:50.397Z
2021-07-26T08:23:50.397Z
pela
<p>Cosmologists estimate the <a href="https://en.wikipedia.org/wiki/Age_of_the_universe" rel="noreferrer">age of the universe</a> at 13.8 billion years.
The most distant objects we could theoretically observe emitted light that long ago; light from more distant objects hasn't reached us yet.
Due to expansion of the universe, that horizon is now about 46 billion light-years away.
If we assume uniform expansion in all directions, we can approximate the <a href="https://en.wikipedia.org/wiki/Observable_universe" rel="noreferrer">observable universe</a> as a sphere of that radius, consistent with the diameter you'd heard.
I figure its volume as roughly $4 \times 10^{32}$ cubic light-years.
The shape and size of the <em>un</em>observable universe are anyone's guess.</p>

CC BY-SA 3.0
2016-04-20T17:19:01.643Z
2016-04-20T18:58:00.460Z
2016-04-20T18:58:00.460Z
Mike G
<p>I'm assuming you're talking about the visible universe, not the observable universe. If so, the comoving distance we could see is about 45.7 Billion light years. Even though the universe is only 13.8 Billion years old, due to expansion it took longer than that to reach us. It'd be like walking up an escalator thats going down vs walking up stairs, I'll get there but it'll just take much longer. A COMOVING distance would relate better to the person on the stairs, it factors out expansion, thus making calculations much easier. Using the comoving distance, figuring out it's volume is simple math, unless of course you plan on figuring it's volume  without mass, just the space. Then of course you'd have to find each masses volume, subtract and blah, blah, and impossible, since we don't know every object in the visible universe. Using V=4/3πr^3, V = approximately 3.99795E+32ly^3. Or, 399 nonillion Ly^3.</p>

<p>Now imagine the escalator you're on starts running just as fast as you, eventually faster; You'll never reach the top, in fact you'll move away from the top despite running towards it. This is what is happening to our visible universe, it's slowly receding faster than light due to expansion. So the light will continue to travel towards us but will never reach us once it's emitted beyond about 4500 megaparsecs away, it's event horizon, sometimes called particle horizon. 
Point being, the visible universe volume is constantly shrinking. </p>

CC BY-SA 3.0
2017-04-24T07:18:42.553Z
I think your distinguishing between the observable and the visible Universe is not what is normally used. As I understand your answer, you use the term "visible Universe" to refer to the region where light _emitted today_ will be able to reach us. But this is rarely of any interest in a scientific context. Also, note that the 4500 Mpc you're referring to is the comoving distance to the Hubble sphere, outside which stuff recedes faster than the speed of light. But actually light emitted from outside this region right now, up to ~5000 Mpc, will still be able to reach us.
pela
CC BY-SA 3.0
2017-04-24T09:07:18.077Z
As I read your answer, you use the term "visible Universe" for the region bound by the _event horizon_, which is the horizon outside which light emitted today will never reach us. It is true that it is always shrinking in _comoving_ coordinates, but in physical coordinates it actually doesn't. It asymptotically increases toward a finite size (~17 Gly).
pela
CC BY-SA 3.0
2017-04-24T11:22:43.807Z
How does this differ from Pela's answer?
ProfRob
CC BY-SA 3.0
2017-04-24T07:00:54.830Z
2017-04-24T07:00:54.830Z
Mike
<p>The Universe is infinite and always has been. Yes, it is expanding, but what does that mean when it exists everywhere? Yes, it can appear to have exploded from a point, but that "point" is a universe that is infinite in size and has no edge. There is no way to look at it or describe it from "outside".</p>

<p>The diagrams you often see of the history of the Big Bang show a poor concept of expansion and are just misleading. They imply an outside view is possible.</p>

<p>See the excellent answer to a similar question here:  <a href="https://physics.stackexchange.com/questions/136860/did-the-big-bang-happen-at-a-point">https://physics.stackexchange.com/questions/136860/did-the-big-bang-happen-at-a-point</a></p>

CC BY-SA 3.0
2017-10-13T07:42:23.147Z
This answer has a number of highly contentious statements (eg the universe "always has been" and it "exploded"). At the very least you need to back them up with references.
Chappo Hasn't Forgotten
CC BY-SA 3.0
2017-10-14T01:54:11.030Z
I you can show that time existed before the Big Bang, then I will retract "always has been".
C. Towne Springer
CC BY-SA 3.0
2017-10-14T22:40:15.587Z
No, that's not how science works. It's up to you to explain how your hypothesis resolves all possible scenarios (eg a cyclic universe).
Chappo Hasn't Forgotten
CC BY-SA 3.0
2017-10-16T16:10:28.107Z
Are you two perhaps actually agreeing? It seems @Chappo objects to the "always has been" statement because this sounds like the Universe has existed forever. But maybe C.T. Springer just means "has been infinite since time started at BB" (which nevertheless is a somewhat bold statement, since this is not really known, although it's true if the rest of the Universe is like the obsevable Universe).
pela
CC BY-SA 3.0
2017-10-17T04:20:44.447Z
"No, that's not how science works." Every measurement, experiment, and observation indicates an infinite universe and the existence of time. The current theory of course has to fit these conditions. A singularity at a Big Bang is an inference. Your objection to "always has been"  places the burden on you to show why you claim something outside the evidence is true.
C. Towne Springer
CC BY-SA 3.0
2017-10-12T05:29:47.140Z
2017-10-14T01:52:53.523Z
2017-10-14T01:52:53.523Z
C. Towne Springer
<p>Of course it is not exactly known how big the universe is, but I thougt that the universe is about 100 billion lightyears in diameter. But if that is true can you also make an estimation of the volume of the universe?</p>

CC BY-SA 3.0
2016-04-20T08:54:29.733Z
Just use your good old $V = \frac{4\pi}{3}R^3$ (and remember that we're talking about the _observable_ Universe; the rest may be infinite).
pela
CC BY-SA 3.0
2016-04-20T10:10:26.293Z
What you give is the volume of a round sphere but in this picture https://en.wikipedia.org/wiki/File:Ilc_9yr_moll4096.png it is always not shown as as perfect globe. Why is that?
Marijn 
CC BY-SA 3.0
2016-04-20T10:46:38.813Z
You can thank [Karl Mollweide](https://en.wikipedia.org/wiki/Mollweide_projection) for that :)
pela
CC BY-SA 3.0
2016-04-20T12:22:14.267Z
Fun exercise:  Once you have calculated the volume, you can then calculate the average approximate density, too.

(Estimated mass of an average star (in sols)) * (Estimated number of stars in an average galaxy) * (Total estimated number of galaxies in observable universe) / Volume of observable universe.
EastOfJupiter
CC BY-SA 3.0
2016-04-20T13:51:42.653Z
I think the reason to use a mollweide ellips is to get the whole earth on a map with similar proportions but I can't see why this reason should be valid for the cosmos?
Marijn 
CC BY-SA 3.0
2016-04-21T10:13:03.333Z
@TexasTubbs: That will give you the density of stars only, which is only $\sim0.5$% of the total mass of the Universe.
pela
<universe>
What is the volume of the universe?
Marijn 
# What is the volume of the universe?

(Asked by: username_0 on 2016-04-20)

Of course it is not exactly known how big the universe is, but I thougt that the universe is about 100 billion lightyears in diameter. But if that is true can you also make an estimation of the volume of the universe?

**Tags:** universe
> **Comments:**
> **username_1**: Just use your good old $V = \frac{4\pi}{3}R^3$ (and remember that we're talking about the _observable_ Universe; the rest may be infinite).  
> **username_0**: What you give is the volume of a round sphere but in this picture https://en.wikipedia.org/wiki/File:Ilc_9yr_moll4096.png it is always not shown as as perfect globe. Why is that?  
> **username_1**: You can thank [Karl Mollweide](https://en.wikipedia.org/wiki/Mollweide_projection) for that :)  
> **username_2**: Fun exercise:  Once you have calculated the volume, you can then calculate the average approximate density, too.  (Estimated mass of an average star (in sols)) * (Estimated number of stars in an average galaxy) * (Total estimated number of galaxies in observable universe) / Volume of observable universe.  
> **username_0**: I think the reason to use a mollweide ellips is to get the whole earth on a map with similar proportions but I can't see why this reason should be valid for the cosmos?  
> **username_1**: @TexasTubbs: That will give you the density of stars only, which is only $\sim0.5$% of the total mass of the Universe.  

---

## 4 answers

### Answer by username_1 on 2016-04-20. Score: 7

Based on your comments, I think your confusion comes from having seen the [classic rugby ball-shaped image of the CMB](https://en.wikipedia.org/wiki/Cosmic_microwave_background#/media/File:Ilc_9yr_moll4096.png).
The CMB *we* observe is not from the whole cosmos, but only from a thin *and completely spherical* shell centered on us. This light is redshifted by an amount $z=1100$.

The radius of this shell can be calculated integrating the [Friedmann equation](https://en.wikipedia.org/wiki/Friedmann_equations) from us (at $z=0$) to $z=1100$:
$$
R\_\mathrm{CMB} = \frac{c}{H\_0}\int\_0^{1100}
\frac{dz}{\sqrt{
\Omega\_r(1+z)^4 +
\Omega\_m(1+z)^3 +
\Omega\_k(1+z)^2 +
\Omega\_\Lambda
}},
$$
where $c$ and $H\_0 \sim 70\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$ are the speed of light and the Hubble parameter, respectively, and
$\{\Omega\_r,\Omega\_m,\Omega\_k,\Omega\_\Lambda\}\sim\{\lt 10^{-4},0.3,0,0.7\}$
are the radiation, matter, curvature, and dark energy density parameters, respectively.

With these approximate numbers, I get $R\_\mathrm{CMB} = 44.63\,\mathrm{Glyr}$, while using a [Planck 2015](http://arxiv.org/abs/1502.01589) cosmology for the various parameters yields $R\_\mathrm{CMB} = 45.36\,\mathrm{Glyr}$.

Just as the shell of Earth can be projected onto a rugby-shaped figure using a [Mollweide projection](https://en.wikipedia.org/wiki/Mollweide_projection), so can the shell of the CMB. Here's a figure from [Universe Adventure](http://www.universeadventure.org/big_bang/cmb-origins.htm) that can help visualize:

[![projection](https://i.sstatic.net/tSbxw.jpg)](https://i.sstatic.net/tSbxw.jpg)

# The volume of the observable Universe

Although we (still) can't see beyond the CMB, the observable Universe is all the way out to redshift $z=\infty$. The difference is not big, though; integrating the equation above from $0$ to $\infty$ yields $R\_\mathrm{obs.Uni.} = 46.27\,\mathrm{Glyr}$. Thus, the volume of the *observable Universe* is
$$
V = \frac{4\pi}{3}R\_\mathrm{obs.Uni.}^3 = 415,\!065\,\mathrm{Glyr}^3.
$$

The volume of the *whole* Universe is probably much larger, and may in fact easily be infinite.

> **Comments:**
> **username_3**: I guess I should read $\mathrm{Gly^3}$ as "cubic giga-light-year," not "billion cubic light-year."  
> **username_0**: But still it is unclear why they mostly illustrate the CMG rugby shaped and not just spherical. As I said, for the earth I can imagine why, that is to show all continents in one view.  But there isn't such a reason for the CMB?  
> **username_3**: It may help to think in terms of the [celestial sphere](https://en.wikipedia.org/wiki/Celestial_sphere). Plotting the stars and galaxies on a globe, as if they were a uniform distance from us, helps us make sense of their apparent positions in the sky. We can also transform this globe into a flat map, but most map projections of a whole sphere are not circular, and all have some distortion.  
> **username_1**: @MikeG: Yes, $\mathrm{Gly}^3 = (10^9\mathrm{ly})^3 = 10^{27}\mathrm{ly}$. Also, as it seems you spotted, I made an error of a factor $10^3$.  
> **username_1**: @Marijn: The reason is exactly the same as for the Earth: We want to show the entire sky in one view. That's not possible with a spherical projection, unless you invent some crazy projection algorithm that will not only heavily distort the peripheral regions (which to a certain degree will be the case for all projections), but also display some regions multiple times.  
> **username_5**: NB Only true for a universe with no curvature.  
> **username_4**: But that formula is for Euclidean space. Is the universe sufficiently flat for it to be a reasonable approximation?  
> **username_1**: @Acccumulation: Yes, the observable Universe is extremely (if not completely) flat, with  $|\Omega_k| < 0.005$ ([Planck Collaboration et al. 2016](https://arxiv.org/abs/1502.01589)).  

---

### Answer by username_3 on 2016-04-20. Score: 5

Cosmologists estimate the [age of the universe](https://en.wikipedia.org/wiki/Age_of_the_universe) at 13.8 billion years.
The most distant objects we could theoretically observe emitted light that long ago; light from more distant objects hasn't reached us yet.
Due to expansion of the universe, that horizon is now about 46 billion light-years away.
If we assume uniform expansion in all directions, we can approximate the [observable universe](https://en.wikipedia.org/wiki/Observable_universe) as a sphere of that radius, consistent with the diameter you'd heard.
I figure its volume as roughly $4 \times 10^{32}$ cubic light-years.
The shape and size of the *un*observable universe are anyone's guess.


---

### Answer by username_6 on 2017-04-24. Score: 2

I'm assuming you're talking about the visible universe, not the observable universe. If so, the comoving distance we could see is about 45.7 Billion light years. Even though the universe is only 13.8 Billion years old, due to expansion it took longer than that to reach us. It'd be like walking up an escalator thats going down vs walking up stairs, I'll get there but it'll just take much longer. A COMOVING distance would relate better to the person on the stairs, it factors out expansion, thus making calculations much easier. Using the comoving distance, figuring out it's volume is simple math, unless of course you plan on figuring it's volume without mass, just the space. Then of course you'd have to find each masses volume, subtract and blah, blah, and impossible, since we don't know every object in the visible universe. Using V=4/3πr^3, V = approximately 3.99795E+32ly^3. Or, 399 nonillion Ly^3.

Now imagine the escalator you're on starts running just as fast as you, eventually faster; You'll never reach the top, in fact you'll move away from the top despite running towards it. This is what is happening to our visible universe, it's slowly receding faster than light due to expansion. So the light will continue to travel towards us but will never reach us once it's emitted beyond about 4500 megaparsecs away, it's event horizon, sometimes called particle horizon.
Point being, the visible universe volume is constantly shrinking.

> **Comments:**
> **username_1**: I think your distinguishing between the observable and the visible Universe is not what is normally used. As I understand your answer, you use the term "visible Universe" to refer to the region where light _emitted today_ will be able to reach us. But this is rarely of any interest in a scientific context. Also, note that the 4500 Mpc you're referring to is the comoving distance to the Hubble sphere, outside which stuff recedes faster than the speed of light. But actually light emitted from outside this region right now, up to ~5000 Mpc, will still be able to reach us.  
> **username_1**: As I read your answer, you use the term "visible Universe" for the region bound by the _event horizon_, which is the horizon outside which light emitted today will never reach us. It is true that it is always shrinking in _comoving_ coordinates, but in physical coordinates it actually doesn't. It asymptotically increases toward a finite size (~17 Gly).  
> **username_5**: How does this differ from Pela's answer?  

---

### Answer by username_7 on 2017-10-12. Score: 1

The Universe is infinite and always has been. Yes, it is expanding, but what does that mean when it exists everywhere? Yes, it can appear to have exploded from a point, but that "point" is a universe that is infinite in size and has no edge. There is no way to look at it or describe it from "outside".

The diagrams you often see of the history of the Big Bang show a poor concept of expansion and are just misleading. They imply an outside view is possible.

See the excellent answer to a similar question here: <https://physics.stackexchange.com/questions/136860/did-the-big-bang-happen-at-a-point>

> **Comments:**
> **username_8**: This answer has a number of highly contentious statements (eg the universe "always has been" and it "exploded"). At the very least you need to back them up with references.  
> **username_7**: I you can show that time existed before the Big Bang, then I will retract "always has been".  
> **username_8**: No, that's not how science works. It's up to you to explain how your hypothesis resolves all possible scenarios (eg a cyclic universe).  
> **username_1**: Are you two perhaps actually agreeing? It seems @Chappo objects to the "always has been" statement because this sounds like the Universe has existed forever. But maybe C.T. Springer just means "has been infinite since time started at BB" (which nevertheless is a somewhat bold statement, since this is not really known, although it's true if the rest of the Universe is like the obsevable Universe).  
> **username_7**: "No, that's not how science works." Every measurement, experiment, and observation indicates an infinite universe and the existence of time. The current theory of course has to fit these conditions. A singularity at a Big Bang is an inference. Your objection to "always has been"  places the burden on you to show why you claim something outside the evidence is true.  

---

