Has this random youtube crackpot trying to disprove GR inadvertedly stumbled on a genuine prediction of it?

Question

I'm an Astrophysics major. I was watching strange fringe physics crackpots on Youtube to make fun of them, because I'm an acollierastro, planarwalk kinda gal. I came across this guy who thinks Black Holes are fake ('they just created those photos using a secret process, they're not actually real'), Stephen Hawking was a fraud ('I've written books. You can't do that just by moving your eyes to pick out letters, you just can't') and QED is wrong ('nobody is talking about renormalisation'). Anyway, he's obviously insane but he made an argument about black holes that I found legitimately interesting and when I checked the maths I was disturbed to find he seems to be right.

He argued that if you rearrange the Schwartzchild Radius equation to find the critical density of an area of space at which point it becomes the interior of a black hole, you get an inverse squared function of radius, which implies that a very, very large area would only need a small density to be a black hole. He claimed this implies that if you consider the volume of say, the observable universe, its density is trivially larger than this, so it would be a black hole.

He was using this as Reductio Ad Absurdum to say that GR's prediction is wrong and black holes don't exist, and I don't buy this. But I checked the maths with a quick back of the envelope calculation, and he's right. \begin{align} \tag1\frac{2GM}{c^2}&\geqslant R\\ \tag2M&\geqslant\frac{Rc^2}{2G}\\ \tag3\rho V&\geqslant\frac{Rc^2}{2G}\\ \tag4\rho\frac{4\pi R^3}3&\geqslant\frac{Rc^2}{2G}\\ \tag5\rho&\geqslant\frac{3c^2}{8\pi GR^2} \end{align}

This is interesting because sure, the density of the observable universe is vanishingly small because it's huge and mostly empty, so this isn't an issue, but you can show that the density of the entire universe that space's curvature implies is less than the Schwartzchild density.

In other words, as the volume of a space goes to infinity the density required to form an event horizon goes to zero which means that any sufficiently large drawn finite boundary should be able to form a black hole interior and render its contents unable to escape.

This is obviously a difficult issue because the universe doesn't have a finite boundary but you can draw a finite boundary within it that has a Schwartzchild density higher than the average density of the universe, implying that at some point out in the universe any traveller should hit an event horizon that they can't move beyond.

What do you guys think of this idea? Do you think it holds water?

To be clear, I don't think GR is wrong, and I think black holes exist. I think by trying to prove GR wrong this lunatic has accidentally stumbled onto an interesting implication of,its equations that I've never seen anyone discuss before; namely, that if the universe has uniform density on large scales then any sufficiently large volume should satisfy the Schwartzchild radius of its contents and form an enormous black hole around it.

Answer 1

Let’s start by making one thing clear: the Schwarzschild and FLRW spacetimes have different symmetries. Schwarzschild is a spherical vacuum, and FLRW is an isotropic and homogenous non-vacuum solution.

With Birkhoff’s theorem we know that the Schwarzschild solution is the unique spherically symmetric vacuum spacetime. So we can use it in the vacuum outside any spherically symmetric source. So, indeed, a spherically symmetric “dust” surrounded by vacuum can collapse at arbitrarily low densities, given arbitrarily large mass.

This cannot be applied to the universe as a whole because the universe is not surrounded by vacuum. It is represented by the FLRW spacetime, not the Schwarzschild spacetime.