Mathematical coincidence of the Schwarzschild radius of the Universe? I read here that a black hole with a mass of the observable universe, $M=8.8\times10^{52}kg$, would have a Schwarzschild radius of $r_s=13.7$ billion lightyears.
I immediately noticed that at the speed of light, to travel a distance of $r_s$, it would take nearly exactly the age of the universe. Is that a coincidence, or is there a connection?
Normally, I'd brush this off as a coincidence and go on with my day. However, the top answer to this post details that the previous values are related to the Hubble constant $H_0$; the reciprocal of which is known as Hubble time, only varying in value from the age of the universe (due to the non-linear expansion of the universe), by a dimensionless factor of about $0.96$
Thus, considering that $r_s$ is related to $H_0$, which is related to the age of the universe, is that value of $13.7$ billion (light)years a coincidence, or is there a direct mathematical relationship?
As an undergraduate in math, I'm not extremely familiar with these concepts, and may have glossed over something obvious to the more atrophysically atuned. Hence, this is mere curiosity.
 A: First, the mass contained within the observable universe is found by mutiplying its volume by a density derived from the assumption that the universe is flat with $\Omega_m \simeq 0.3$. It is larger than you say.
Using the critical density, which depends on the Hubble parameter,
$$\rho = \Omega_m \times \left(\frac{3H_0^2}{8\pi G}\right)\ . $$
The (proper) distance to the cosmological horizon in a flat universe is given by
$$D = \frac{c}{H_0} \int^{1100}_{0} (\Omega_m (1+z)^3 + \Omega_r(1+z)^4 + \Omega_{\Lambda})^{-1/2}\ dz \ , $$
where the upper limit to the integral is marked by the redshift corresponding to the epoch of recombination, which is as far as we can abctually observe and the values of $\Omega$ are their present-day values.
For the current best-fit cosmological parameters, these equations yield $\rho \simeq 2.6\times 10^{-27}$ kg/m$^3$ and $D \simeq 13.4$ Gpc. Combining these gives a mass for the observable universe of $7.7\times 10^{53}$ kg.
Thus I am not sure where the number in your question comes from, but it is seems to be incorrect - maybe it is an approximate value for the amount of baryonic mass in the observable universe, which is a factor of 5 smaller than the total mass? Or maybe it used $c/H_0$ or 13.7 billion light years as the radius of the universe (which would be just plain wrong)?
The Schwarzschild radius for this total mass would be 37 Gpc and considerably larger than the observable universe. Note however that although you can define a Schwarzschild radius, it doesn't have the meaning attributed to it in the Schwarzschld metric, since that isn't what governs spacetime in the universe as a whole.
The coincidence between the age of the universe and the inverse of the current Hubble parameter is a cosmic coincidence that is not true for all cosmic times. For a flat universe with no dark energy, the age of the universe is $2/3H$. Our universe apparently does contain dark energy and this means that the age is larger because after the initial period when $\Omega_m$ was dominant, the expansion begins to accelerate, meaning that the age becomes larger than $2/3H$. In fact at later times, $H$ will assume a constant value, and the age of the universe will be much greater than $H^{-1}$. We are at an epoch where the age just happens to be $\sim H^{-1}$ (although there may be some good anthropic reasons for that).
A: It's a very good question, connected to the 'flatness problem' of cosmology.
It can be solved by presuming that, as the universe expands, all objects - people, atoms, galaxies etc... expand too.
This leads to an alternative interpretation of redshift. If the size of atoms and Plancks constant were lower in the past, then from $E=hf$, the energy of photons arriving from a distant star would be lower, hence the redshift.
But crucially it explains the coincidence that you have noted, as follows...
With the symmetric expansion above all physical constants and lengths must change in proportion to the number of length dimensions in them e.g. $h=h_0e^{2Ht}$ where H is a constant related to Hubble's constant.  This type of expansion keeps everything in proportion.
Each particle has total energy $mc^2-GmM/R$  where $m$ is the particles mass and $M$ and $R$ represent the mass and radius of the universe (small constants are omitted for simplicity).
During the expansion this becomes $(mc^2-GmM/R)e^{2Ht}$ and energy can only be conserved if $mc^2-GmM/R = 0$,
i.e. if $G=\frac{Rc^2}M$
There is a fundamental difference with traditional cosmology, with this approach gravity is caused by the expansion and has the value necessary to conserve energy as the expansion occurs.
The alternative approach naturally predicts that the matter density will be measured as 0.25 or 1/3 depending on how it's measured. This seems to be the case.
There is a link to the theory here Here, https://vixra.org/abs/2006.0209
So perhaps the alternative theory answers your question.
