<p>Confusingly, positive <span class="math-container">$Ω_k$</span> implies negative curvature and vice versa, so you really want <span class="math-container">$Ω_{k,0}=-0.005$</span>.</p>
<p>Plugging <span class="math-container">$ρ=(1-Ω_{k,0})ρ_\mathrm{crit,0}$</span> (and <span class="math-container">$Λ=0$</span> because I'm including that energy in <span class="math-container">$ρ$</span>) into the first Friedmann equation, you get <span class="math-container">$k=-Ω_{k,0} H_0^2/c^2$</span>. The corresponding radius of curvature is <span class="math-container">$1/\sqrt{|k|} = c/H_0\sqrt{|Ω_{k,0}|}$</span>. For <span class="math-container">$Ω_{k,0}=-0.005$</span> that's around 200 billion light years.</p>
<p>The diameter would be <span class="math-container">$2π$</span> times that if the spatial geometry is spherical, or <span class="math-container">$π$</span> times that if it's elliptical (a sphere with opposite points identified). But the universe is expanding, so that isn't the distance that a light beam would travel when circumnavigating the universe. If the current data implying accelerating expansion is correct, then a light beam emitted now will never return even if the spatial curvature is positive.</p>
<p>To get a circumference of 14 trillion light years you'd need <span class="math-container">$|Ω_{k,0}| &lt; 10^{-4}$</span>. I think that's just an error in the blog post.</p>
<p>Even if we knew the circumference to be over 14 trillion light years, it wouldn't mean that the universe is at least that size (as the blog post also claims), since we don't know that it's homogeneous out to that distance.</p>

CC BY-SA 4.0
2021-10-20T19:50:28.030Z
2021-10-20T19:50:28.030Z
benrg
<p>According to this Wikipedia article: <a href="https://en.wikipedia.org/wiki/Shape_of_the_universe" rel="noreferrer">https://en.wikipedia.org/wiki/Shape_of_the_universe</a></p>
<blockquote>
<p>Results of the Planck mission released in 2015 show the cosmological curvature parameter, ΩK, to be 0.000±0.005, consistent with a flat universe.</p>
</blockquote>
<p>If we assume the curvature of the universe is in fact at the upper bound of this measurement (ΩK = 0.005) then what would the diameter of the universe be?</p>
<p>More precisely, if we assume the universe is a closed 3-sphere with ΩK=0.005, then how far would a light beam travel before returning to its source?</p>
<p>For reference, this article implies that the universe should be at least 14 trillion light years in diameter based on curvature estimates from the earlier WMAP mission:
<a href="https://scienceblogs.com/startswithabang/2012/07/18/how-big-is-the-entire-universe" rel="noreferrer">https://scienceblogs.com/startswithabang/2012/07/18/how-big-is-the-entire-universe</a></p>
<blockquote>
<p>If the Universe does curve back and close on itself, its radius of curvature is at least 150 times as large as the part that's observable to us! Meaning that -- even without speculative physics like cosmic inflation -- we know that the entire Universe extends for at least 14 trillion light years in diameter, including the part that's unobservable to us today.</p>
</blockquote>

<cosmology>
Minimum size of closed universe that satisfies ΩK < 0.005
JeffB
# Minimum size of closed universe that satisfies ΩK < 0.005

(Asked by: username_0 on 2021-10-20)

According to this Wikipedia article: <https://en.wikipedia.org/wiki/Shape_of_the_universe>

> Results of the Planck mission released in 2015 show the cosmological curvature parameter, ΩK, to be 0.000±0.005, consistent with a flat universe.

If we assume the curvature of the universe is in fact at the upper bound of this measurement (ΩK = 0.005) then what would the diameter of the universe be?

More precisely, if we assume the universe is a closed 3-sphere with ΩK=0.005, then how far would a light beam travel before returning to its source?

For reference, this article implies that the universe should be at least 14 trillion light years in diameter based on curvature estimates from the earlier WMAP mission:
<https://scienceblogs.com/startswithabang/2012/07/18/how-big-is-the-entire-universe>

> If the Universe does curve back and close on itself, its radius of curvature is at least 150 times as large as the part that's observable to us! Meaning that -- even without speculative physics like cosmic inflation -- we know that the entire Universe extends for at least 14 trillion light years in diameter, including the part that's unobservable to us today.

**Tags:** cosmology


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## 1 answers

### Answer by username_1 on 2021-10-20. Score: 5 | Accepted answer

Confusingly, positive $Ω\_k$ implies negative curvature and vice versa, so you really want $Ω\_{k,0}=-0.005$.

Plugging $ρ=(1-Ω\_{k,0})ρ\_\mathrm{crit,0}$ (and $Λ=0$ because I'm including that energy in $ρ$) into the first Friedmann equation, you get $k=-Ω\_{k,0} H\_0^2/c^2$. The corresponding radius of curvature is $1/\sqrt{|k|} = c/H\_0\sqrt{|Ω\_{k,0}|}$. For $Ω\_{k,0}=-0.005$ that's around 200 billion light years.

The diameter would be $2π$ times that if the spatial geometry is spherical, or $π$ times that if it's elliptical (a sphere with opposite points identified). But the universe is expanding, so that isn't the distance that a light beam would travel when circumnavigating the universe. If the current data implying accelerating expansion is correct, then a light beam emitted now will never return even if the spatial curvature is positive.

To get a circumference of 14 trillion light years you'd need $|Ω\_{k,0}| < 10^{-4}$. I think that's just an error in the blog post.

Even if we knew the circumference to be over 14 trillion light years, it wouldn't mean that the universe is at least that size (as the blog post also claims), since we don't know that it's homogeneous out to that distance.


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