<p>I am interested in this question also, and I think that the interesting fact is that for any mass there is a finite Schwarzschild radius from which nothing can escape not even light from the gravity of the matter within the radius. This is true whether or not the radius aligns with the size of the observable universe. That is what makes it seem like we are living within a black hole.</p>
<p>The difference I think is that the Schwarzschild solution is derived from a spherical mass distribution within an empty space while the universe has an equal density throughout, so at the edge of any Schwarzschild radius there would still be an equal gravity in all directions. But if the universe does have an edge then nothing would be able to escape from it.</p>

CC BY-SA 4.0
2021-02-07T17:22:04.470Z
I once calculated the mass of 1 atmosphere pressure out to Pluto. This was before Pluto's demotion. It turns out that that'd constitute a black hole.
Wayfaring Stranger
CC BY-SA 4.0
2023-10-17T22:38:28.377Z
There is also the consideration that gravitational waves propogate at the speed of light. Not sure what that changes but I feel like it changes things
nog642
CC BY-SA 4.0
2024-02-16T23:10:48.883Z
If the Schwarzschild solution is only for mass within an empty space, doesn't that mean that it cannot be used for any hole with an accretion disk? It also can't apply to star core collapse.
cowlinator
CC BY-SA 4.0
2021-02-04T20:05:34.263Z
2023-12-30T00:01:18.183Z
2023-12-30T00:01:18.183Z
Thomas Tiger
<p>There's an error in your source, but even if there weren't, it wouldn't mean that the Universe is a black hole (see below): The Schwarzschild radius of the observable Universe is <em>not</em> equal to 13.7 Glyr. Wikipedia cites some random, non-refereed paper that uses the Hubble radius as the radius of the observable Universe, which is too small by a factor of <span class="math-container">$\gtrsim3$</span>.</p>

<h3>The Schwarzschild radius of the Universe</h3>

<p>Although the age of the Universe is indeed ~13.8 Gyr, its radius is much larger than 13.8 Glyr, because it expands. In fact, the radius is <span class="math-container">$R \simeq 46.3\,\mathrm{Glyr}$</span>.</p>

<p>The mean density of the Universe is very close to the <em>critical</em><span class="math-container">$^\dagger$</span> density <span class="math-container">$\rho_\mathrm{c} \simeq 8.6\times10^{-30}\,\mathrm{g}\,\mathrm{cm}^{-3}$</span>.
Hence, the total mass (including "normal", baryonic matter, dark matter, and dark energy) is
<span class="math-container">$$
M = \rho_\mathrm{c}V = \rho_\mathrm{c}\frac{4\pi}{3}R^3 \simeq 3.0\times10^{57}\,\mathrm{g},
$$</span>
and the corresponding Schwarzschild radius is
<span class="math-container">$$
R_\mathrm{S} \equiv \frac{2GM}{c^2} \simeq 475\,\mathrm{Glyr}.
$$</span></p>

<h3>The Universe is not a black hole</h3>

<p>Even worse! you might say. If our Universe is much smaller than its Schwarzschild radius, does that mean we live in a black hole?</p>

<p>No, it doesn't. A black hole is a region in space where some mass is squeezed inside its Schwarzschild radius, but the Universe is not "a region in space". There's no "outside the Universe". If anything, you might call it a <em>white hole</em>, the time-reversal of a black hole, in which case you could say that the singularity is not something that everything will fall into in the future, but rather something that everything came from in the past. You may call that singularity Big Bang.</p>

<h3>The error in the source</h3>

<p>You will see questions like this many places on the internet. As I said above, they all assume the Hubble radius, <span class="math-container">$R_\mathrm{H} \equiv c/H_0$</span> for the radius. But this radius is well within our observable Universe, and doesn't really bear any physical significance. In this case, the age in Gyr works out to be exactly equal to the radius in Glyr, by definition.</p>

<p>So, what does that tell us?</p>

<p>Nothing, really. Except that our Universe is flat, i.e. has <span class="math-container">$\rho\simeq\rho_\mathrm{c}$</span>, which we already knew and used in the calculation.</p>

<p>That is, setting <span class="math-container">$R = R_\mathrm{H} \equiv c/H_0$</span>,
<span class="math-container">$$
\begin{array}{rcl}
R_\mathrm{S} &amp; \equiv &amp; \frac{2GM}{c^2}\\
             &amp; =      &amp; \frac{2G}{c^2} \rho V \\
             &amp; =      &amp; \frac{2G}{c^2} \rho \frac{4\pi}{3}R^3 \\
             &amp; =      &amp; \frac{2G}{c^2} \rho \frac{4\pi}{3} \left(\!\frac{c}{H_0}\!\right)^3 \\
             &amp; =      &amp; \frac{8\pi G}{3 H_0^2} \rho \frac{c}{H_0} \\
             &amp; =      &amp; \frac{8\pi G}{3 H_0^2} \rho  R,
\end{array}
$$</span>
so if <span class="math-container">$R=R_\mathrm{S}$</span>, we have
<span class="math-container">$$
\rho = \frac{3H_0^2}{8\pi G},
$$</span>
which is exactly the expression for the critical density you get from the Friedmann equation.</p>

<hr>

<p><span class="math-container">$^\dagger$</span><sub>The density that determines the global geometry of the Universe.</sub></p>

CC BY-SA 4.0
2021-08-03T19:06:46.943Z
_the Universe is not "a region in space". There's no "outside the Universe"._ This seems like a weak argument. Can't you simply consider a region in space with radius $R - \epsilon$ for some arbitrarily small value of $\epsilon$? This region would evade your argument, but still contain nearly all the matter in the observable universe, and thus have a schwarzschild radius much larger than its radius. Why is this region not a black hole?
JSQuareD
CC BY-SA 4.0
2021-08-04T12:46:18.820Z
@JSQuareD I'm not exactly sure I understand; if I use a smaller (or larger) radius than the observable Universe, the mass is also smaller (or larger), and the density will work out the same. But at any rate, the Schwarzschild solution is valid for a mass surrounded by vacuum. In the case of the Universe, there's no vacuum outside outside any volume you pick, there's just more Universe.
pela
CC BY-SA 4.0
2024-02-16T23:04:09.873Z
You state that the radius of the universe is 46.3Glyr. Take a region of space (unrelated to the size of the universe other than being smaller than it) with a radius of 45.9Glry, and call that "region A". Region A is indeed "a region in space". There IS an "outside region A". Region A contains mass and is surrounded by vacuum. I think we can safely estimate that the Schwarzschild radius of Region A is greater than the radius of Region A. Why is Region A not a black hole?
cowlinator
CC BY-SA 4.0
2024-02-17T09:06:06.240Z
@cowlinator But Region A is not surrounded by vacuum; it's surrounded by exactly the same as what's inside. And then the Schwarzschild solution no longer is valid.
pela
CC BY-SA 4.0
2024-05-24T18:40:52.300Z
@pela The vacuum explanation makes a lot of sense. All the stuff outside of region A is constantly tugging on it and counteracting its internal gravity. I would also add that dark energy (which the Schwarzschild radius doesn't take into account as it doesn't affect gravitationally bound systems, which all known black holes were at the time of formation) too has a massive effect on it, and too counteracts its internal gravity.
Max
CC BY-SA 4.0
2024-05-24T19:35:02.763Z
@Max I'm not a general relativist, but it is my understanding that, if you calculate the Schwarzschild solution in a space with a cosmological constant, you'd still get a black hole, but with a slightly larger radius. You can think of its gravity as space flowing into the BH, and a cosmological constant makes space expand, so I suppose this slightly counteracts the inflow.
pela
CC BY-SA 4.0
2024-05-26T13:09:33.227Z
@pela Do you mean that the observable universe would still be expected to form a black hole? What about when gravity from matter outside the observable universe ─ assuming the cosmological principle ─ is taken into account?
Max
CC BY-SA 4.0
2024-05-27T05:52:38.263Z
@Max No, sorry, with my last comment I was just referring to a regular BH in an otherwise empty Universe, except for Λ. This is not a homogenous and isotropic universe. For a hom/iso universe, the calculations in the answer applies, and there is no BH.
pela
CC BY-SA 4.0
2019-06-28T11:42:07.180Z
2019-06-28T12:58:22.027Z
2019-06-28T12:58:22.027Z
pela
<p>The estimated <a href="https://en.wikipedia.org/wiki/Age_of_the_universe" rel="noreferrer">age of the universe</a> is 14 billion years.</p>

<p>The estimated <a href="https://en.wikipedia.org/wiki/Schwarzschild_radius#Parameters" rel="noreferrer">Schwarzschild radius</a> (event horizon) of the observable universe is 14 billion light-years.</p>

<p>What are the ramifications?</p>

CC BY-SA 4.0
2019-07-01T14:07:46.630Z
Related: [How can the Schwarzschild radius of the universe be 13.7 billion light years?](https://physics.stackexchange.com/questions/174665/how-can-the-schwarzschild-radius-of-the-universe-be-13-7-billion-light-years)
John Rennie
<black-hole><universe><cosmology>
Observable universe equals its Schwarzschild radius (event horizon)?
Dirk Helgemo
# Observable universe equals its Schwarzschild radius (event horizon)?

(Asked by: username_0 on 2019-06-27)

The estimated [age of the universe](https://en.wikipedia.org/wiki/Age_of_the_universe) is 14 billion years.

The estimated [Schwarzschild radius](https://en.wikipedia.org/wiki/Schwarzschild_radius#Parameters) (event horizon) of the observable universe is 14 billion light-years.

What are the ramifications?

**Tags:** black-hole, universe, cosmology

> **Comments:**
> **username_1**: Related: [How can the Schwarzschild radius of the universe be 13.7 billion light years?](https://physics.stackexchange.com/questions/174665/how-can-the-schwarzschild-radius-of-the-universe-be-13-7-billion-light-years)  

---

## 2 answers

### Answer by username_6 on 2019-06-28. Score: 14 | Accepted answer

There's an error in your source, but even if there weren't, it wouldn't mean that the Universe is a black hole (see below): The Schwarzschild radius of the observable Universe is *not* equal to 13.7 Glyr. Wikipedia cites some random, non-refereed paper that uses the Hubble radius as the radius of the observable Universe, which is too small by a factor of $\gtrsim3$.

### The Schwarzschild radius of the Universe

Although the age of the Universe is indeed ~13.8 Gyr, its radius is much larger than 13.8 Glyr, because it expands. In fact, the radius is $R \simeq 46.3\,\mathrm{Glyr}$.

The mean density of the Universe is very close to the *critical*$^\dagger$ density $\rho\_\mathrm{c} \simeq 8.6\times10^{-30}\,\mathrm{g}\,\mathrm{cm}^{-3}$.
Hence, the total mass (including "normal", baryonic matter, dark matter, and dark energy) is
$$
M = \rho\_\mathrm{c}V = \rho\_\mathrm{c}\frac{4\pi}{3}R^3 \simeq 3.0\times10^{57}\,\mathrm{g},
$$
and the corresponding Schwarzschild radius is
$$
R\_\mathrm{S} \equiv \frac{2GM}{c^2} \simeq 475\,\mathrm{Glyr}.
$$

### The Universe is not a black hole

Even worse! you might say. If our Universe is much smaller than its Schwarzschild radius, does that mean we live in a black hole?

No, it doesn't. A black hole is a region in space where some mass is squeezed inside its Schwarzschild radius, but the Universe is not "a region in space". There's no "outside the Universe". If anything, you might call it a *white hole*, the time-reversal of a black hole, in which case you could say that the singularity is not something that everything will fall into in the future, but rather something that everything came from in the past. You may call that singularity Big Bang.

### The error in the source

You will see questions like this many places on the internet. As I said above, they all assume the Hubble radius, $R\_\mathrm{H} \equiv c/H\_0$ for the radius. But this radius is well within our observable Universe, and doesn't really bear any physical significance. In this case, the age in Gyr works out to be exactly equal to the radius in Glyr, by definition.

So, what does that tell us?

Nothing, really. Except that our Universe is flat, i.e. has $\rho\simeq\rho\_\mathrm{c}$, which we already knew and used in the calculation.

That is, setting $R = R\_\mathrm{H} \equiv c/H\_0$,
$$
\begin{array}{rcl}
R\_\mathrm{S} & \equiv & \frac{2GM}{c^2}\\
& = & \frac{2G}{c^2} \rho V \\
& = & \frac{2G}{c^2} \rho \frac{4\pi}{3}R^3 \\
& = & \frac{2G}{c^2} \rho \frac{4\pi}{3} \left(\!\frac{c}{H\_0}\!\right)^3 \\
& = & \frac{8\pi G}{3 H\_0^2} \rho \frac{c}{H\_0} \\
& = & \frac{8\pi G}{3 H\_0^2} \rho R,
\end{array}
$$
so if $R=R\_\mathrm{S}$, we have
$$
\rho = \frac{3H\_0^2}{8\pi G},
$$
which is exactly the expression for the critical density you get from the Friedmann equation.

---

$^\dagger$The density that determines the global geometry of the Universe.

> **Comments:**
> **username_7**: _the Universe is not "a region in space". There's no "outside the Universe"._ This seems like a weak argument. Can't you simply consider a region in space with radius $R - \epsilon$ for some arbitrarily small value of $\epsilon$? This region would evade your argument, but still contain nearly all the matter in the observable universe, and thus have a schwarzschild radius much larger than its radius. Why is this region not a black hole?  
> **username_6**: @username_7 I'm not exactly sure I understand; if I use a smaller (or larger) radius than the observable Universe, the mass is also smaller (or larger), and the density will work out the same. But at any rate, the Schwarzschild solution is valid for a mass surrounded by vacuum. In the case of the Universe, there's no vacuum outside outside any volume you pick, there's just more Universe.  
> **username_5**: You state that the radius of the universe is 46.3Glyr. Take a region of space (unrelated to the size of the universe other than being smaller than it) with a radius of 45.9Glry, and call that "region A". Region A is indeed "a region in space". There IS an "outside region A". Region A contains mass and is surrounded by vacuum. I think we can safely estimate that the Schwarzschild radius of Region A is greater than the radius of Region A. Why is Region A not a black hole?  
> **username_6**: @username_5 But Region A is not surrounded by vacuum; it's surrounded by exactly the same as what's inside. And then the Schwarzschild solution no longer is valid.  
> **username_8**: @username_6 The vacuum explanation makes a lot of sense. All the stuff outside of region A is constantly tugging on it and counteracting its internal gravity. I would also add that dark energy (which the Schwarzschild radius doesn't take into account as it doesn't affect gravitationally bound systems, which all known black holes were at the time of formation) too has a massive effect on it, and too counteracts its internal gravity.  
> **username_6**: @username_8 I'm not a general relativist, but it is my understanding that, if you calculate the Schwarzschild solution in a space with a cosmological constant, you'd still get a black hole, but with a slightly larger radius. You can think of its gravity as space flowing into the BH, and a cosmological constant makes space expand, so I suppose this slightly counteracts the inflow.  
> **username_8**: @username_6 Do you mean that the observable universe would still be expected to form a black hole? What about when gravity from matter outside the observable universe ─ assuming the cosmological principle ─ is taken into account?  
> **username_6**: @username_8 No, sorry, with my last comment I was just referring to a regular BH in an otherwise empty Universe, except for Λ. This is not a homogenous and isotropic universe. For a hom/iso universe, the calculations in the answer applies, and there is no BH.  

---

### Answer by username_2 on 2021-02-04. Score: 5

I am interested in this question also, and I think that the interesting fact is that for any mass there is a finite Schwarzschild radius from which nothing can escape not even light from the gravity of the matter within the radius. This is true whether or not the radius aligns with the size of the observable universe. That is what makes it seem like we are living within a black hole.

The difference I think is that the Schwarzschild solution is derived from a spherical mass distribution within an empty space while the universe has an equal density throughout, so at the edge of any Schwarzschild radius there would still be an equal gravity in all directions. But if the universe does have an edge then nothing would be able to escape from it.

> **Comments:**
> **username_3**: I once calculated the mass of 1 atmosphere pressure out to Pluto. This was before Pluto's demotion. It turns out that that'd constitute a black hole.  
> **username_4**: There is also the consideration that gravitational waves propogate at the speed of light. Not sure what that changes but I feel like it changes things  
> **username_5**: If the Schwarzschild solution is only for mass within an empty space, doesn't that mean that it cannot be used for any hole with an accretion disk? It also can't apply to star core collapse.  

---

