cosmology - Supermassive black holes with the density of the Universe

This question was inspired by the answer to the question "If the universe were compressed into a super massive black hole, how big would it be"

Assume that we have a matter with a uniform density $\rho$. Some mass of this matter may forms a black hole with the Schwarzschild radius:

$\large{R_S=c*\sqrt{\frac{3}{8\pi G\rho}}}$

This equation is easy to get from

$\large{R_S=\frac{2GM}{c^2}}$ and $\large{\rho=\frac{3M}{4\pi R_S^3}}$

For universe density ($9.3*10^{-27} kg/m^3$) Schwarzschild radius of the black hole is 13.9 billion light-years. While radius of observable Universe is 46 billion light-years.

We could be located inside such black hole, but we don't observe its singularity and event horizon.

So why there are no supermassive black holes with the density of the Universe?

Is it means, that the whole Universe is infinite and has uniform density?

P.S. Relative link - Is the Big Bang a black hole?

Answer

You are asking a wrong question. Here is the problem with your reasoning.

You are assuming a Schwarzschild metric and a homogenous distribution of mass. But the Schwarzschild geometry describes a vacuum spacetime. So you can't use it for a spacetime filled with matter. For a cosmological spacetime filled with matter, like our universe, the suitable metric to use would be something else, like the FRW for example.

You could only use the Schwarzschild spacetime if you assumed a sphere of some uniform density $\rho$ and vacuum outside the radius of the sphere.

Let me illustrate how things would work out then. As you can see, a particular density corresponds to a particular $R_s$, lets call it $R_s(\rho)$. So if you had a sphere of matter with a radius $R_1$ grater than $R_s(\rho)$, then you couldn't apply the formula $R_s(\rho)=c\sqrt{\frac{3}{8\pi G \rho}}$. You would have to use the Schwarzschild metric only in the vacuum region outside of the sphere. So you would have then $R_s=\frac{8\pi G\rho R_1^3}{3c^2}$. In order to see how the $R_s$ compares with $R_s(\rho)$, you can replace the density with $\rho=\frac{3c^2}{8\pi G R_s(\rho)}$. So you would get that the Schwarzschild radius for a sphere of uniform density $\rho$ and radius $R_1>R_s(\rho)$ is $R_s=\left(\frac{R_1}{R_s(\rho)}\right)^2R_1$, which is grater than the radius of the sphere. So the sphere is inside its Schwarzschild horizon. If on the other hand, the radius $R_1$ is smaller than $R_s(\rho)$, then the corresponding horizon would have to be inside the sphere. But inside the sphere the Schwarzschild metric doesn't apply. So it isn't necessary that there should be a horizon inside the matter distribution.

If you apply these to the universe and assume for example that the radius of the visible universe is the radius $R_1$ of the sphere, then you would have a horizon radius (using your numbers) that would be almost 10 times the radius of the observable universe. So, the entire universe would have to be in a black hole of radius of 460 billion light-years. So the assumption that we should see black holes with horizons of radii of 13.9 billion light-years is not correct.

If one assumes the above point of view, one could say that the universe is a white hole that is exploding.

I hope that all these are helpful and not confusing.