# 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.

• So... I was searching this up, and I saw very many conflicting sources on what the mass of the observable universe is. Wikipedia "observable universe" page says that the 10^53 figure is for ordinary matter only, but then I've seen other stuff that say that that figure is for all matter. And then there's the fundamental problem that there are multiple competing definitions for mass-energy in GR because of various technicalities about energy being dependent on reference frame, gravitational potential energy being weird, etc. Jun 30 '20 at 7:36
• I guess being within a few orders of magnitude of the Schwarzchild radius would still be an interesting coincidence, though. Jun 30 '20 at 7:38
• @qwyxivi I forgot how complicated it must be to get a figure for the mass for the of the observable universe. I wonder how the zero-energy hypothesis muddles the matter more. Jul 1 '20 at 1:31
• Jul 1 '20 at 11:29
• This is answered here. astronomy.stackexchange.com/questions/32443/…. Also, the relationship the OP mentions (as Pela explains in the link) is the universe is flat. Sep 1 '20 at 11:48

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