<p>According to the Friedmann equations of the ΛCDM model, the simplest way to get to know the curvature of the universe is to measure the density of the universe:</p>
<ul>
<li><p>If the density coincides with the critical density, the universe is flat and infinite. The density ratio, defined as the ratio of the measured density to the critical density, is then <span class="math-container">$\Omega=1$</span>.</p>
</li>
<li><p>If the density is less than the critical density, the universe is hyperbolic and infinite. The density ratio is <span class="math-container">$\Omega &lt; 1$</span>.</p>
</li>
<li><p>If the density is greater than the critical density, the universe is spherical and finite. The density ratio is <span class="math-container">$\Omega &gt; 1$</span>.</p>
</li>
</ul>
<p>The best measurements we currently have of the <strong>current density ratio</strong> <span class="math-container">$\Omega_0$</span> are those made by the Planck Mission, the results of which are:</p>
<p><span class="math-container">$$\Omega_0 = 0.9993 \pm 0.0019$$</span></p>
<p>Measurement that does not exclude any of the 3 possibilities. Source: <a href="https://arxiv.org/abs/1807.06209" rel="noreferrer"><em><strong>Planck 2018 results. VI. Cosmological parameters</strong></em></a></p>
<p><em><strong>Imagine</strong></em> that the real density ratio of the universe was <span class="math-container">$\Omega_0 = 1.0012$</span> Then it would be a spherical universe, whose radius of curvature can be easily calculated by means of:</p>
<p><span class="math-container">$$R=\dfrac c{H_0 \cdot \sqrt{\Omega - 1}}$$</span></p>
<p><span class="math-container">$c=299792458$</span> m/s</p>
<p>According to the best value of the Planck Mission:</p>
<p><span class="math-container">$H_0=67.66$</span> (km/s)Mpc</p>
<p>And with this hypothetical value of <span class="math-container">$\Omega_0=1.0012$</span> we would obtain that the universe would now have a radius of curvature of 417 billion light years.</p>
<p>Note that in this case, the maximum value of the density ratio provided by the error bar has been chosen for the calculation:</p>
<p><span class="math-container">$\Omega_0=0.9993+0.0019=1.0012$</span></p>
<p>And therefore we have obtained the minimum possible value (according to Planck's measurements) of the radius of curvature of the Universe:</p>
<p><span class="math-container">$R=417$</span> billion light years</p>
<p>As the radius of the Observable Universe is known to be 46.2 billion light years, we have obtained that, (again, according to the Planck Mission measurements), <strong>the radius of curvature of the Total Universe is at least 9 times larger than the radius of the Observable Universe</strong>.</p>
<p>You may be interested in the related thread: <a href="https://astronomy.stackexchange.com/a/54658/31201"><strong>Size of the Unobservable Universe</strong></a></p>
<p>Best regards.</p>

CC BY-SA 4.0
2023-02-15T13:56:18.283Z
2024-07-01T12:56:08.103Z
2024-07-01T12:56:08.103Z
Albert
<p>Suppose we were in a closed universe likened with a spherical surface such as that of the Earth's surface.</p>
<p>Assume also that the limits, represented by the limits of the observable universe, corresponds to a continent which would not have information from other continents.
Obviously the inhabitants of any continent could not send a light beam out to be observed coming back.</p>
<p>Is this a correct description of the real situation and are there other less dramatic experiments we might make to determine the universe’s possible curvature?</p>

CC BY-SA 4.0
2023-02-15T13:28:22.370Z
I'm not even sure I understand your argument, less your reasoning. Can you elaborate a bit more? You seem to describe some kind of though experiment, but to me big parts of that are missing. The curvature of the universe *has been* and *is being* measured...
planetmaker
CC BY-SA 4.0
2023-02-16T23:19:43.367Z
@planetmaker My issue is simply how we can make determination about part of a universe - open or closed - outside its observable part.
Mikael Jensen
<astrophysics><cosmology><expansion><space-geometry>
Can the curvature of a closed universe be measured?
Mikael Jensen
# Can the curvature of a closed universe be measured?

(Asked by: username_0 on 2023-02-15)

Suppose we were in a closed universe likened with a spherical surface such as that of the Earth's surface.

Assume also that the limits, represented by the limits of the observable universe, corresponds to a continent which would not have information from other continents.
Obviously the inhabitants of any continent could not send a light beam out to be observed coming back.

Is this a correct description of the real situation and are there other less dramatic experiments we might make to determine the universe’s possible curvature?

**Tags:** astrophysics, cosmology, expansion, space-geometry
> **Comments:**
> **username_1**: I'm not even sure I understand your argument, less your reasoning. Can you elaborate a bit more? You seem to describe some kind of though experiment, but to me big parts of that are missing. The curvature of the universe *has been* and *is being* measured...  
> **username_0**: @planetmaker My issue is simply how we can make determination about part of a universe - open or closed - outside its observable part.  

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## 1 answers

### Answer by username_2 on 2023-02-15. Score: 8 | Accepted answer

According to the Friedmann equations of the ΛCDM model, the simplest way to get to know the curvature of the universe is to measure the density of the universe:

* If the density coincides with the critical density, the universe is flat and infinite. The density ratio, defined as the ratio of the measured density to the critical density, is then $\Omega=1$.
* If the density is less than the critical density, the universe is hyperbolic and infinite. The density ratio is $\Omega < 1$.
* If the density is greater than the critical density, the universe is spherical and finite. The density ratio is $\Omega > 1$.

The best measurements we currently have of the **current density ratio** $\Omega\_0$ are those made by the Planck Mission, the results of which are:

$$\Omega\_0 = 0.9993 \pm 0.0019$$

Measurement that does not exclude any of the 3 possibilities. Source: [***Planck 2018 results. VI. Cosmological parameters***](https://arxiv.org/abs/1807.06209)

***Imagine*** that the real density ratio of the universe was $\Omega\_0 = 1.0012$ Then it would be a spherical universe, whose radius of curvature can be easily calculated by means of:

$$R=\dfrac c{H\_0 \cdot \sqrt{\Omega - 1}}$$

$c=299792458$ m/s

According to the best value of the Planck Mission:

$H\_0=67.66$ (km/s)Mpc

And with this hypothetical value of $\Omega\_0=1.0012$ we would obtain that the universe would now have a radius of curvature of 417 billion light years.

Note that in this case, the maximum value of the density ratio provided by the error bar has been chosen for the calculation:

$\Omega\_0=0.9993+0.0019=1.0012$

And therefore we have obtained the minimum possible value (according to Planck's measurements) of the radius of curvature of the Universe:

$R=417$ billion light years

As the radius of the Observable Universe is known to be 46.2 billion light years, we have obtained that, (again, according to the Planck Mission measurements), **the radius of curvature of the Total Universe is at least 9 times larger than the radius of the Observable Universe**.

You may be interested in the related thread: [**Size of the Unobservable Universe**](https://astronomy.stackexchange.com/a/54658/31201)

Best regards.


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