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Mathematical coincidence of the Schwarzschild radius of the Universe?

Physics Asked on December 29, 2021

I read here that a black hole with a mass of the observable universe, $M=8.8times10^{52}kg$, 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.

One Answer

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.

Answered by John Hunter on December 29, 2021

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