<p>Yes, you are right. It refers to the Hubble volume rather than the Hubble radius, specifically it is 251 times the Hubble volume, but based on the volume, we can calculate the radius of the whole universe being 88.2 billion light years..</p>

CC BY-SA 4.0
2023-09-07T19:15:00.840Z
The article seems to claim that the curvature radius is at least 42 Gpc ≈ 137 Gly, not 88 Gly. I don't understand how they got a ratio of 251 from that. 137/88 is close to π/2, but I still don't get it.
benrg
CC BY-SA 4.0
2023-09-07T15:54:53.343Z
2023-09-07T15:54:53.343Z
Arjun
<p>The article <a href="https://academic.oup.com/mnrasl/article/413/1/L91/1747653#25823893" rel="noreferrer"><strong>Applications of Bayesian model averaging to the curvature and size of the Universe</strong></a> cited in the question is from 2011. It is now 2023 and we have much better data, in particular from the Planck satellite measurements: &quot;<a href="https://arxiv.org/abs/1807.06209" rel="noreferrer"><strong>Planck 2018 results. VI. Cosmological parameters</strong></a>&quot;.
According to these data <span class="math-container">$\Omega_k=0.0007 \pm 0.0019$</span></p>
<p>As the central value is <span class="math-container">$\Omega_k&gt;0$</span>, the most favoured result is that, (if the topology of the universe is simple), the universe is infinite.
But since the tolerance in the value of Omega allows it to be negative, we can calculate the value of the minimum size of the universe if it were spherical-finite.</p>
<p>The radius of curvature of a spherical universe <span class="math-container">$R$</span> as a function of the curvature density ratio is:
<span class="math-container">$$R=\dfrac c{H_0} \cdot \dfrac 1{\sqrt{- \Omega_k}}$$</span>
Using the measurements of the Planck Collaboration-2018
<span class="math-container">$$\Omega_k=0.0007 \pm 0.0019$$</span>
If the universe is spherical, its <strong>minimum size</strong> corresponds to:
<span class="math-container">$$\Omega_k=-0.0012$$</span>
If the universe is spherical with <span class="math-container">$H_0=67.66 \ (km/s)/Mpc$</span>, its minimum radius of curvature is:
<span class="math-container">$$R=\dfrac c{H_0} \cdot \dfrac 1{\sqrt{- \Omega_k}}=\dfrac{299792.458}{67.66} \cdot \dfrac 1{\sqrt{0.0012}}=127908 \ Mpc= 417 \ Gly$$</span>
The expression for the 3-volume as a function of the radius of curvature is:
<span class="math-container">$$V=2\pi^2 R^3$$</span>
Therefore, the minimum size of the whole spherical universe is:
<span class="math-container">$$\boxed{V_w=1433486966 \ Gly^3}$$</span>
Let us now calculate the size of the observable universe. The differential volume using spherical coordinates:
<span class="math-container">$$dV=\sqrt{\vert g \vert} \ dr \ d\theta \ d\phi$$</span>
The Friedmann-Lemaitre-Robertson-Walker Metric Tensor:
<span class="math-container">$$g = \begin{bmatrix}
-c^2 &amp; 0 &amp; 0 &amp; 0\\
0 &amp; \dfrac{a^2}{1-Kr^2} &amp; 0 &amp; 0\\
0 &amp; 0 &amp; a^2 r^2 &amp; 0\\
0 &amp; 0 &amp; 0 &amp; a^2 r^2\sin^2\theta \end{bmatrix}$$</span>
The square root of the absolute value of the determinant of the spatial part of the metric tensor:
<span class="math-container">$$\sqrt{\vert g \vert}=a^3 \dfrac{r^2 \sin \theta}{\sqrt{1-K r^2}}$$</span>
<span class="math-container">$$dV=a^3 \ \dfrac{r^2 \sin \theta}{\sqrt{1-K r^2}}dr d\theta d\phi$$</span>
<span class="math-container">$$\displaystyle V(r)=a^3 \ \int_0^r \dfrac{r^2 dr}{\sqrt{1-K r^2}} \int_0^{\pi} \sin \theta d\theta \int_0^{2\pi} d\phi$$</span>
<span class="math-container">$$\displaystyle V(r)=a^3 \ 4 \pi \int_0^r \dfrac{r^2 dr}{\sqrt{1-K r^2}}$$</span>
<span class="math-container">$$V(r)=a^3 \ 2\pi \left ( \dfrac{\arcsin (r \sqrt K)}{K^{3/2}} - \dfrac{r\sqrt{1-Kr^2}}K \right )$$</span>
The relationship between the coordinate <span class="math-container">$r$</span> and the distance <span class="math-container">$h$</span> is:
<span class="math-container">$$\displaystyle h=\int_0^r \dfrac{dr}{\sqrt{1-Kr^2}}$$</span>
<span class="math-container">$$h=\dfrac{\arcsin (r \sqrt K)}{\sqrt K}$$</span>
<span class="math-container">$$r=\dfrac 1 {\sqrt K} \sin (h\sqrt K)$$</span>
If we substitute in the volume expression:
<span class="math-container">$$V(h)=a^3 \ \dfrac{2\pi}{K^{3/2}} \left ( h\sqrt K-\dfrac{\sin (2 h \sqrt K)}2 \right )$$</span>
Now <span class="math-container">$a=1$</span> and the relationship between the curvature <span class="math-container">$K$</span> and the radius of curvature <span class="math-container">$R$</span> is:
<span class="math-container">$$K=\dfrac 1{R^2}$$</span>
Finally:
<span class="math-container">$$\boxed{V(h)=2\pi R^3 \left ( \dfrac h R-\dfrac{\sin \dfrac{2 h}R}2 \right )}$$</span>
If we perform the Taylor-Maclaurin approximation of this last expression:
<span class="math-container">$$V_T(h) \simeq \dfrac 4 3 \pi h^3 \left [ 1-\dfrac 1 5 \left ( \dfrac h R \right )^2 \ \right ]$$</span>
In which it is clear that, when the distance <span class="math-container">$h$</span> is small with respect to the radius of curvature <span class="math-container">$R$</span>, we recover the expression for the volume of the sphere:
<span class="math-container">$$V=\dfrac 4 3 \pi h^3$$</span>
To calculate the size of the observable universe <span class="math-container">$V_o$</span> we must put as the value of <span class="math-container">$h$</span> the particle horizon <span class="math-container">$h=46.2 \ Gly$</span> in the boxed expression and we obtain:
<span class="math-container">$$\boxed{V_o=412048 \ Gly^3}$$</span>
Note that, if we had used the conventional volume of the sphere, the error we make is small, only 0.25%
<span class="math-container">$$V_0\simeq \dfrac 4 3 \pi \ 46.2^3=413061 \ Gly^3$$</span>
The quotient between the size of the minimum possible spherical universe and the observable universe according to the best measurements currently available is.
<span class="math-container">$$\dfrac{V_w}{V_o}=\dfrac{1433486966}{412048}=3479$$</span>
<strong>If the universe is spherical, its volume is at least 3479 times larger than the observable universe</strong>.</p>
<p>If we prefer to give the result as a function of Hubble spheres, we know that the actual Hubble radius is:
<span class="math-container">$$R_H=14.5 \ Gly$$</span>
Using the boxed expression we calculate the volume of the Hubble sphere:
<span class="math-container">$$V_H=12767 \ Gly^3$$</span>
<span class="math-container">$$\dfrac{V_w}{V_H}=\dfrac{1433486966}{12767}=112281$$</span>
<strong>If the universe is spherical, its volume is at least 112281 times larger than the Hubble Volume</strong>.</p>
<p>Best regards.</p>

CC BY-SA 4.0
2023-09-08T08:27:57.890Z
2023-09-12T07:03:34.427Z
2023-09-12T07:03:34.427Z
Albert
<p>If you search that question on Google, the answers will be &quot;250 times the size of the observable universe&quot; or &quot;more than 15.000.000 the size of the observable universe&quot;. Both of them come from a different interpretation of the same paper (<a href="https://academic.oup.com/mnrasl/article/413/1/L91/1747653#25823893" rel="noreferrer">https://academic.oup.com/mnrasl/article/413/1/L91/1747653#25823893</a>), which estimate the universe size using the Bayesian model. Even if most articles say that the universe is 15.000.000 bigger than the observable, I don't think It's true, since I read the article and from what I understood the minimum size of it is 250 larger than a Hubble sphere (i.e. the observable universe), not 250 time the <em>radius</em> of the Hubble sphere. Am I right or wrong? An answer will be very appreciated, thanks in advance.</p>

CC BY-SA 4.0
2023-09-07T14:37:13.880Z
If you discuss the interpretation of a specific paper, you definitely want to include a link in your question. That is even more true when it deals with so speculative topics like the "size of unobservable universe" which can by definition not be accessible to measurements. Sounds somewhat like discussing the diet preferences of pink elephants, but I'm happy to learn how we can learn about something we cannot observe.
planetmaker
CC BY-SA 4.0
2023-09-07T14:41:28.720Z
Can please link to that paper in the question?
Arjun
CC BY-SA 4.0
2023-09-07T16:01:12.090Z
Yes, you are right. It refers to the Hubble volume rather than the Hubble radius, specifically it is 251 times the Hubble volume, but based on the volume, we can calculate the radius of the whole universe being 88.2 billion light years.. Shall I make this as an answer (Does it answer the question, because I think it is too short)?
Arjun
CC BY-SA 4.0
2023-09-07T18:10:59.853Z
You are right, I edited the question with a link to the paper.
jack_O'Dim
CC BY-SA 4.0
2023-09-07T19:33:38.373Z
According to Ethan Siegal, "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!" I have a link in https://astronomy.stackexchange.com/a/32458/16685 & https://astronomy.stackexchange.com/a/31795/16685
PM 2Ring
CC BY-SA 4.0
2023-09-08T00:30:31.050Z
Google says "must be **at least** 23 trillion light years in diameter" - a lower boundary. Otherwise the answer to the question is *yes* (1,2,...3. Three licks!). *Based on the Hubble volume, we can calculate the radius of the whole universe?* - you can take numbers and put them into equations; doesn't make them mean anything.
Mazura
CC BY-SA 4.0
2023-09-08T12:22:48.043Z
In reality, of course, the universe must be literally infinite. Space can only be endless.
White Prime
CC BY-SA 4.0
2023-09-09T14:23:48.123Z
@WhitePrime no, they need not. Where is the end of the surface of a sphere? It's endless, but finite. Generally any hypersphere is only bound in a space with one dimension higher than the hypersphere's surface.
planetmaker
CC BY-SA 4.0
2023-09-09T17:49:28.543Z
@planetmaker Sigh. A sphere is still a shape; a finite form. Space is not a shape, and it can have no boundary, no edge and no end.
White Prime
<universe><size><estimate>
Size of the Unobservable Universe
jack_O'Dim
# Size of the Unobservable Universe

(Asked by: username_0 on 2023-09-07)

If you search that question on Google, the answers will be "250 times the size of the observable universe" or "more than 15.000.000 the size of the observable universe". Both of them come from a different interpretation of the same paper (<https://academic.oup.com/mnrasl/article/413/1/L91/1747653#25823893>), which estimate the universe size using the Bayesian model. Even if most articles say that the universe is 15.000.000 bigger than the observable, I don't think It's true, since I read the article and from what I understood the minimum size of it is 250 larger than a Hubble sphere (i.e. the observable universe), not 250 time the *radius* of the Hubble sphere. Am I right or wrong? An answer will be very appreciated, thanks in advance.

**Tags:** universe, size, estimate

> **Comments:**
> **username_1**: If you discuss the interpretation of a specific paper, you definitely want to include a link in your question. That is even more true when it deals with so speculative topics like the "size of unobservable universe" which can by definition not be accessible to measurements. Sounds somewhat like discussing the diet preferences of pink elephants, but I'm happy to learn how we can learn about something we cannot observe.  
> **username_2**: Can please link to that paper in the question?  
> **username_2**: Yes, you are right. It refers to the Hubble volume rather than the Hubble radius, specifically it is 251 times the Hubble volume, but based on the volume, we can calculate the radius of the whole universe being 88.2 billion light years.. Shall I make this as an answer (Does it answer the question, because I think it is too short)?  
> **username_0**: You are right, I edited the question with a link to the paper.  
> **username_3**: According to Ethan Siegal, "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!" I have a link in https://astronomy.stackexchange.com/a/32458/16685 & https://astronomy.stackexchange.com/a/31795/16685  
> **username_4**: Google says "must be **at least** 23 trillion light years in diameter" - a lower boundary. Otherwise the answer to the question is *yes* (1,2,...3. Three licks!). *Based on the Hubble volume, we can calculate the radius of the whole universe?* - you can take numbers and put them into equations; doesn't make them mean anything.  
> **username_5**: In reality, of course, the universe must be literally infinite. Space can only be endless.  
> **username_1**: @WhitePrime no, they need not. Where is the end of the surface of a sphere? It's endless, but finite. Generally any hypersphere is only bound in a space with one dimension higher than the hypersphere's surface.  
> **username_5**: @username_1 Sigh. A sphere is still a shape; a finite form. Space is not a shape, and it can have no boundary, no edge and no end.  

---

## 2 answers

### Answer by username_7 on 2023-09-08. Score: 11 | Accepted answer

The article [**Applications of Bayesian model averaging to the curvature and size of the Universe**](https://academic.oup.com/mnrasl/article/413/1/L91/1747653#25823893) cited in the question is from 2011. It is now 2023 and we have much better data, in particular from the Planck satellite measurements: "[**Planck 2018 results. VI. Cosmological parameters**](https://arxiv.org/abs/1807.06209)".
According to these data $\Omega\_k=0.0007 \pm 0.0019$

As the central value is $\Omega\_k>0$, the most favoured result is that, (if the topology of the universe is simple), the universe is infinite.
But since the tolerance in the value of Omega allows it to be negative, we can calculate the value of the minimum size of the universe if it were spherical-finite.

The radius of curvature of a spherical universe $R$ as a function of the curvature density ratio is:
$$R=\dfrac c{H\_0} \cdot \dfrac 1{\sqrt{- \Omega\_k}}$$
Using the measurements of the Planck Collaboration-2018
$$\Omega\_k=0.0007 \pm 0.0019$$
If the universe is spherical, its **minimum size** corresponds to:
$$\Omega\_k=-0.0012$$
If the universe is spherical with $H\_0=67.66 \ (km/s)/Mpc$, its minimum radius of curvature is:
$$R=\dfrac c{H\_0} \cdot \dfrac 1{\sqrt{- \Omega\_k}}=\dfrac{299792.458}{67.66} \cdot \dfrac 1{\sqrt{0.0012}}=127908 \ Mpc= 417 \ Gly$$
The expression for the 3-volume as a function of the radius of curvature is:
$$V=2\pi^2 R^3$$
Therefore, the minimum size of the whole spherical universe is:
$$\boxed{V\_w=1433486966 \ Gly^3}$$
Let us now calculate the size of the observable universe. The differential volume using spherical coordinates:
$$dV=\sqrt{\vert g \vert} \ dr \ d\theta \ d\phi$$
The Friedmann-Lemaitre-Robertson-Walker Metric Tensor:
$$g = \begin{bmatrix}
-c^2 & 0 & 0 & 0\\
0 & \dfrac{a^2}{1-Kr^2} & 0 & 0\\
0 & 0 & a^2 r^2 & 0\\
0 & 0 & 0 & a^2 r^2\sin^2\theta \end{bmatrix}$$
The square root of the absolute value of the determinant of the spatial part of the metric tensor:
$$\sqrt{\vert g \vert}=a^3 \dfrac{r^2 \sin \theta}{\sqrt{1-K r^2}}$$
$$dV=a^3 \ \dfrac{r^2 \sin \theta}{\sqrt{1-K r^2}}dr d\theta d\phi$$
$$\displaystyle V(r)=a^3 \ \int\_0^r \dfrac{r^2 dr}{\sqrt{1-K r^2}} \int\_0^{\pi} \sin \theta d\theta \int\_0^{2\pi} d\phi$$
$$\displaystyle V(r)=a^3 \ 4 \pi \int\_0^r \dfrac{r^2 dr}{\sqrt{1-K r^2}}$$
$$V(r)=a^3 \ 2\pi \left ( \dfrac{\arcsin (r \sqrt K)}{K^{3/2}} - \dfrac{r\sqrt{1-Kr^2}}K \right )$$
The relationship between the coordinate $r$ and the distance $h$ is:
$$\displaystyle h=\int\_0^r \dfrac{dr}{\sqrt{1-Kr^2}}$$
$$h=\dfrac{\arcsin (r \sqrt K)}{\sqrt K}$$
$$r=\dfrac 1 {\sqrt K} \sin (h\sqrt K)$$
If we substitute in the volume expression:
$$V(h)=a^3 \ \dfrac{2\pi}{K^{3/2}} \left ( h\sqrt K-\dfrac{\sin (2 h \sqrt K)}2 \right )$$
Now $a=1$ and the relationship between the curvature $K$ and the radius of curvature $R$ is:
$$K=\dfrac 1{R^2}$$
Finally:
$$\boxed{V(h)=2\pi R^3 \left ( \dfrac h R-\dfrac{\sin \dfrac{2 h}R}2 \right )}$$
If we perform the Taylor-Maclaurin approximation of this last expression:
$$V\_T(h) \simeq \dfrac 4 3 \pi h^3 \left [ 1-\dfrac 1 5 \left ( \dfrac h R \right )^2 \ \right ]$$
In which it is clear that, when the distance $h$ is small with respect to the radius of curvature $R$, we recover the expression for the volume of the sphere:
$$V=\dfrac 4 3 \pi h^3$$
To calculate the size of the observable universe $V\_o$ we must put as the value of $h$ the particle horizon $h=46.2 \ Gly$ in the boxed expression and we obtain:
$$\boxed{V\_o=412048 \ Gly^3}$$
Note that, if we had used the conventional volume of the sphere, the error we make is small, only 0.25%
$$V\_0\simeq \dfrac 4 3 \pi \ 46.2^3=413061 \ Gly^3$$
The quotient between the size of the minimum possible spherical universe and the observable universe according to the best measurements currently available is.
$$\dfrac{V\_w}{V\_o}=\dfrac{1433486966}{412048}=3479$$
**If the universe is spherical, its volume is at least 3479 times larger than the observable universe**.

If we prefer to give the result as a function of Hubble spheres, we know that the actual Hubble radius is:
$$R\_H=14.5 \ Gly$$
Using the boxed expression we calculate the volume of the Hubble sphere:
$$V\_H=12767 \ Gly^3$$
$$\dfrac{V\_w}{V\_H}=\dfrac{1433486966}{12767}=112281$$
**If the universe is spherical, its volume is at least 112281 times larger than the Hubble Volume**.

Best regards.


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### Answer by username_2 on 2023-09-07. Score: 5

Yes, you are right. It refers to the Hubble volume rather than the Hubble radius, specifically it is 251 times the Hubble volume, but based on the volume, we can calculate the radius of the whole universe being 88.2 billion light years..

> **Comments:**
> **username_6**: The article seems to claim that the curvature radius is at least 42 Gpc ≈ 137 Gly, not 88 Gly. I don't understand how they got a ratio of 251 from that. 137/88 is close to π/2, but I still don't get it.  

---

