<p>Surprisingly small!</p>

<p>(To me, at least.)</p>

<p>The paper referred can be found on the arXiv as <a href="https://arxiv.org/abs/1911.02087" rel="noreferrer">Di Valentino et al. (2019)</a>.</p>

<p>As is customary with Planck results, the exact values of the cosmological parameters depend on how much confidence you put in auxiliary data, such as baryonic acoustic oscillation data (from surveys such as the 6dF Galaxy Survey, SDSS, and BOSS), supernova data, and Big Bang nucleosynthesis models. In this answer, I'm going to assume that the inferred value of the curvature parameter is given by their "<span class="math-container">$\Lambda\mathrm{CDM}$$+$$\Omega_K\!$</span>" model, with their 99% confidence limits of <span class="math-container">$-0.007&gt;\Omega_K&gt;-0.095$</span>, i.e. I'll use
<span class="math-container">$$
\Omega_K = -0.0438^{+0.0368}_{-0.0512}.
$$</span>
The dynamics of the Universe is given by the <a href="https://en.wikipedia.org/wiki/Friedmann_equations" rel="noreferrer">Friedmann equations</a>, which can be rearranged like this:
<span class="math-container">$$
\Omega_K = -\frac{kc^2}{R_0^2\,a(t)^2 H(t)^2},
$$</span>
where <span class="math-container">$k=+1$</span> for a closed universe, <span class="math-container">$a\equiv1$</span> today, <span class="math-container">$H=H_0$</span> is the Hubble constant today, and <span class="math-container">$R_0$</span> is the radius of curvature, and <span class="math-container">$c$</span> is the speed of light. The value of <span class="math-container">$H_0$</span> depends somewhat on the assumed value of <span class="math-container">$\Omega_K$</span>; here I'll use <span class="math-container">$H_0 \sim 70\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$</span>.</p>

<p>I this case, we get that the radius of the Universe is roughly
<span class="math-container">$$
R_0 \sim 67_{-21}^{+100}\,\mathrm{billion\,light\text{-}years}.\qquad (99\%\,\mathrm{C.L.})
$$</span></p>

<p>Accepting the "mainstream" cosmological parameter the observable Universe has a radius of <span class="math-container">$46.3\,\mathrm{Glyr}$</span>. This result will change somewhat in the case of a closed Universe, but I can't seem to identify the author's preferred set of cosmological parameters. If we just use this value nonetheless, it means that we are currently able to observe a volume fraction of
<span class="math-container">$$
f = \frac{V_\mathrm{obs}}{V_\mathrm{tot}} \simeq \left(\frac{46.3}{67_{-21}^{+100}}\right)^3 \sim 33_{-31}^{+67}\%
$$</span>
of the total Universe, i.e. we see somewhere between a few percent and everything, but most likely "a third".</p>

CC BY-SA 4.0
2019-11-12T17:21:53.730Z
So the circumference of the whole universe is 4.52 times larger than the diameter of the observable universe?
user1781498
CC BY-SA 4.0
2019-11-12T17:30:27.593Z
How big the would the observable universe be compared to the total universe?
user1781498
CC BY-SA 4.0
2019-11-12T21:45:56.377Z
Geometry for a closed universe is a little different than for a flat one, but if I've calculated correctly, the difference is on the ~1% level, so roughly, with a diameter of the obs. Universe of ~2×46 Glyr, and a circumference of the whole Universe of $C\simeq2\pi R_0\simeq434$ Glyr, you're right. The obs. Universe then comprises roughly $(43/67)^3 \sim 25\%$ of the total Universe.
pela
CC BY-SA 4.0
2019-11-14T08:40:29.357Z
@user1781498 Observable universe is not ~14 Glyr - the light we see from distant objects is more like 46 Glyr away now - the light has been traveling for nearly 14 glyr but the distant object has been moving away from us all that time and is now very much further away. See Davis and Lineweaver for a nice discussion. https://arxiv.org/pdf/astro-ph/0310808.pdf
Tom Mozdzen
CC BY-SA 4.0
2019-11-17T19:07:02.180Z
@pela This calculation gives the Radius of the observable universe isnt it ?
seVenVo1d
CC BY-SA 4.0
2019-11-17T21:08:08.563Z
@Reign No, this (i.e. $R_0$) should be the radius of the _whole_ Universe.
pela
CC BY-SA 4.0
2019-11-17T21:25:26.007Z
@pela Something bothers me. Let us assume we live on a 2D sphere. So we are 2D creatures. Radius of the observable universe has a meaning since we can measure it and say its x ly away.  But 'radius of the universe' for a 2D sphere seems not much meaningful. We can measure the radius of the curvatue and put some meaning on it. But we cannot show any direction or assign a distance for radius of the universe since its not in the same space as we live in ? And the same logic goes for the 3d sphere. What am I missing ?
seVenVo1d
CC BY-SA 4.0
2019-11-18T11:15:32.460Z
@Reign I must admit that I'm not a differential geometrist, but remember that a closed 3D universe is not like a "normal" 3D sphere with a boundary. It's true that you can't point in some direction and say "That's where the radius of curvature goes", but that doesn't mean it's not there. Any point in space has a radius of curvature—a scalar—and if space is homogeneous and isotropic, it's the same everywhere. I think if an $n$D space is embedded in an $(n+1)$D space, (eg a normal ball in normal 3D space) then you _can_ assign a direction, but that's not (necessarily) the case for our Universe.
pela
CC BY-SA 4.0
2019-11-18T12:11:49.877Z
@pela Well yes, indeed that the radius of curvature has some meaning but in your post you are claiming that "the radius of the universe is x ly". I dont think thats possible since you cannot define distance for "the radius of the universe". Simply radius of the universe and radius of  curvature of the universe are vastly two different thing.
seVenVo1d
CC BY-SA 4.0
2019-11-21T09:53:52.460Z
@Reign Hmm… no, that's not correct: The radius of curvature $R_0$ is a scalar with the dimension of distance. For a positively curved universe, R0 can be identified as the actual radius of the Universe. There's a good introduction to this in [Ryden (2003)](https://ui.adsabs.harvard.edu/abs/2003itc..book.....R).
pela
CC BY-SA 4.0
2019-11-21T09:54:09.700Z
@Reign It's true that $R_0$ is not to be thought of as the radius of, say, Earth, since Earth is a 3D ball in a 3D space. I suppose a better analogy is to think of $R_0$ as ½ the circumference of Earth, i.e. 20k km, such that if you look 20k km to the left, and 20k km to the right, you'll see the same location. But because not enough time has passed, at the moment you can only see 7800 km, making 33% of Earth's surface observable to you (I erroneously wrote 25% in the comment above, because I used 43 Glyr instead of 46 Glyr).
pela
CC BY-SA 4.0
2019-11-30T05:05:26.313Z
so the observable universe is 33% of the whole universe instead of 25% of the whole universe?
user1781498
CC BY-SA 4.0
2019-11-30T21:39:46.710Z
@user1781498 Yes, if we use the 67 Glyr estimate, but note the error bars, as well as the fact that using other constraints (BAOs, SN-Ia, …) together with the Planck data gives somewhat other figures. Also, I should say that I haven't gone into the statistical methods of this paper — it would be fair to say that the results are not undisputed.
pela
CC BY-SA 4.0
2019-11-12T13:53:22.800Z
2020-03-09T08:46:16.380Z
2020-03-09T08:46:16.380Z
pela
<p>This study found that the universe has a positive curvature <a href="https://www.nature.com/articles/s41550-019-0906-9" rel="noreferrer">https://www.nature.com/articles/s41550-019-0906-9</a>. I didn't really want to buy it and see if it says how big the universe might be. I know it didn't have a low enough p-value to be considered confirmed. Is there a big margin of error for the whole size of the universe? </p>

<universe><space-time>
Based off of the study that found the universe has a positive curvature, how big would the whole universe be?
user1781498
# Based off of the study that found the universe has a positive curvature, how big would the whole universe be?

(Asked by: username_0 on 2019-11-11)

This study found that the universe has a positive curvature <https://www.nature.com/articles/s41550-019-0906-9>. I didn't really want to buy it and see if it says how big the universe might be. I know it didn't have a low enough p-value to be considered confirmed. Is there a big margin of error for the whole size of the universe?

**Tags:** universe, space-time


---

## 1 answers

### Answer by username_1 on 2019-11-12. Score: 9 | Accepted answer

Surprisingly small!

(To me, at least.)

The paper referred can be found on the arXiv as [Di Valentino et al. (2019)](https://arxiv.org/abs/1911.02087).

As is customary with Planck results, the exact values of the cosmological parameters depend on how much confidence you put in auxiliary data, such as baryonic acoustic oscillation data (from surveys such as the 6dF Galaxy Survey, SDSS, and BOSS), supernova data, and Big Bang nucleosynthesis models. In this answer, I'm going to assume that the inferred value of the curvature parameter is given by their "$\Lambda\mathrm{CDM}$$+$$\Omega\_K\!$" model, with their 99% confidence limits of $-0.007>\Omega\_K>-0.095$, i.e. I'll use
$$
\Omega\_K = -0.0438^{+0.0368}\_{-0.0512}.
$$
The dynamics of the Universe is given by the [Friedmann equations](https://en.wikipedia.org/wiki/Friedmann_equations), which can be rearranged like this:
$$
\Omega\_K = -\frac{kc^2}{R\_0^2\,a(t)^2 H(t)^2},
$$
where $k=+1$ for a closed universe, $a\equiv1$ today, $H=H\_0$ is the Hubble constant today, and $R\_0$ is the radius of curvature, and $c$ is the speed of light. The value of $H\_0$ depends somewhat on the assumed value of $\Omega\_K$; here I'll use $H\_0 \sim 70\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$.

I this case, we get that the radius of the Universe is roughly
$$
R\_0 \sim 67\_{-21}^{+100}\,\mathrm{billion\,light\text{-}years}.\qquad (99\%\,\mathrm{C.L.})
$$

Accepting the "mainstream" cosmological parameter the observable Universe has a radius of $46.3\,\mathrm{Glyr}$. This result will change somewhat in the case of a closed Universe, but I can't seem to identify the author's preferred set of cosmological parameters. If we just use this value nonetheless, it means that we are currently able to observe a volume fraction of
$$
f = \frac{V\_\mathrm{obs}}{V\_\mathrm{tot}} \simeq \left(\frac{46.3}{67\_{-21}^{+100}}\right)^3 \sim 33\_{-31}^{+67}\%
$$
of the total Universe, i.e. we see somewhere between a few percent and everything, but most likely "a third".

> **Comments:**
> **username_0**: So the circumference of the whole universe is 4.52 times larger than the diameter of the observable universe?  
> **username_0**: How big the would the observable universe be compared to the total universe?  
> **username_1**: Geometry for a closed universe is a little different than for a flat one, but if I've calculated correctly, the difference is on the ~1% level, so roughly, with a diameter of the obs. Universe of ~2×46 Glyr, and a circumference of the whole Universe of $C\simeq2\pi R_0\simeq434$ Glyr, you're right. The obs. Universe then comprises roughly $(43/67)^3 \sim 25\%$ of the total Universe.  
> **username_2**: @username_0 Observable universe is not ~14 Glyr - the light we see from distant objects is more like 46 Glyr away now - the light has been traveling for nearly 14 glyr but the distant object has been moving away from us all that time and is now very much further away. See Davis and Lineweaver for a nice discussion. https://arxiv.org/pdf/astro-ph/0310808.pdf  
> **username_3**: @username_1 This calculation gives the Radius of the observable universe isnt it ?  
> **username_1**: @Reign No, this (i.e. $R_0$) should be the radius of the _whole_ Universe.  
> **username_3**: @username_1 Something bothers me. Let us assume we live on a 2D sphere. So we are 2D creatures. Radius of the observable universe has a meaning since we can measure it and say its x ly away.  But 'radius of the universe' for a 2D sphere seems not much meaningful. We can measure the radius of the curvatue and put some meaning on it. But we cannot show any direction or assign a distance for radius of the universe since its not in the same space as we live in ? And the same logic goes for the 3d sphere. What am I missing ?  
> **username_1**: @Reign I must admit that I'm not a differential geometrist, but remember that a closed 3D universe is not like a "normal" 3D sphere with a boundary. It's true that you can't point in some direction and say "That's where the radius of curvature goes", but that doesn't mean it's not there. Any point in space has a radius of curvature—a scalar—and if space is homogeneous and isotropic, it's the same everywhere. I think if an $n$D space is embedded in an $(n+1)$D space, (eg a normal ball in normal 3D space) then you _can_ assign a direction, but that's not (necessarily) the case for our Universe.  
> **username_3**: @username_1 Well yes, indeed that the radius of curvature has some meaning but in your post you are claiming that "the radius of the universe is x ly". I dont think thats possible since you cannot define distance for "the radius of the universe". Simply radius of the universe and radius of  curvature of the universe are vastly two different thing.  
> **username_1**: @Reign Hmm… no, that's not correct: The radius of curvature $R_0$ is a scalar with the dimension of distance. For a positively curved universe, R0 can be identified as the actual radius of the Universe. There's a good introduction to this in [Ryden (2003)](https://ui.adsabs.harvard.edu/abs/2003itc..book.....R).  
> **username_1**: @Reign It's true that $R_0$ is not to be thought of as the radius of, say, Earth, since Earth is a 3D ball in a 3D space. I suppose a better analogy is to think of $R_0$ as ½ the circumference of Earth, i.e. 20k km, such that if you look 20k km to the left, and 20k km to the right, you'll see the same location. But because not enough time has passed, at the moment you can only see 7800 km, making 33% of Earth's surface observable to you (I erroneously wrote 25% in the comment above, because I used 43 Glyr instead of 46 Glyr).  
> **username_0**: so the observable universe is 33% of the whole universe instead of 25% of the whole universe?  
> **username_1**: @username_0 Yes, if we use the 67 Glyr estimate, but note the error bars, as well as the fact that using other constraints (BAOs, SN-Ia, …) together with the Planck data gives somewhat other figures. Also, I should say that I haven't gone into the statistical methods of this paper — it would be fair to say that the results are not undisputed.  

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