Observable universe equals its Schwarzschild radius (event horizon)?

Question

The estimated age of the universe is 14 billion years.

The estimated Schwarzschild radius (event horizon) of the observable universe is 14 billion light-years.

What are the ramifications?

Answer 1

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.}

Answer 2

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.