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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?

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Nobody ever said that there is a singularity or an event horizon. –  David Z Dec 26 '10 at 8:53
    
@David, The question about singularity of such black hole is rhetoric of course. There is no any evidence of that. The question is why? –  voix Dec 26 '10 at 9:21
    
@voix: (since I can't post an answer) Your second equation is correct, but the third one isn't, so your first equation is not correct either. The third one assumes that the universe is a sphere, and that its radius is equal to its Schwarzschild radius; neither of these statements is correct. –  Bruce Connor Dec 26 '10 at 19:56
    
@Bruce , third equation refers only to the part of the Universe and that part may be spherical. –  voix Dec 26 '10 at 20:12
    
@voix: true, my mistake. I'm thinking of a couple other possibilities. Why was this closed @David ? –  Bruce Connor Dec 26 '10 at 20:32

1 Answer 1

up vote 2 down vote accepted

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.

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radius in my first equation is a critical radius. Spherical uniform matter with radius more than the critical one should turns into the black hole. But if mass of the uniform matter is enough, it may turns into 3, 5, 10 black holes. You don’t take this into account in your answer. The whole clusters of stars are forming from the uniform matter, why the black holes in my case isn't? –  voix Dec 27 '10 at 17:34
    
First of all, you do understand that you can’t use the Schwarzschild metric to identify a horizon inside the sphere of uniform density, right? If now you would like to treat the problem of the collapse of a sphere of uniform density and zero pressure, then the geometry is described by a Schwarzschild geometry outside and an FRW inside the sphere, with the appropriate matching conditions on the surface. Due to the symmetry of this system, the whole thing will collapse simultaneously and you will not have any fragmentation of the type that you are proposing. –  Vagelford Dec 28 '10 at 11:50
    
So, in the example that I am using the consistent description is that you have a Schwarzschild geometry outside the radius of the observable universe, with a horizon at a radius 10 times that and the interior geometry is FRW without horizons. –  Vagelford Dec 28 '10 at 11:51
    
Now, regarding the fragmentation issue. In order to have it there should be density perturbations and the scale of fragmentation would depend on the wavelength of the perturbations. There is a stability criterion, called the Jeans criterion that tells us what is the minimum scale of stable perturbations. That minimum scale (wavelength) depends on the density and on the speed of sound in the medium. But since you have zero pressure, you also have zero speed of sound and thus the minimum wavelength for stable perturbations is zero. That means that all perturbations are unstable and collapse. –  Vagelford Dec 28 '10 at 11:54
    
So the fragmentation would depend on the spectrum of the perturbations and nothing else. –  Vagelford Dec 28 '10 at 11:55

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