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I understand only a little of general relativity, but that's why I'm here! :)

Consider the hypothetical situation of some extra-terrestrial intelligence pushing all the mass in the universe, every last subatomic particle, into a giant black hole. What is the surface area of such an object? For the sake of argument, let's keep this very idealized. Dark energy, whatever it is, can be captured and thrown into the SMBH; assume there's no Hawking radiation causing it to slowly evaporate, etc.

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There is some problem with "the whole universe" -- probably it is infinite, so would have infinite mass. The right approximation is probably the observable universe. –  mbq Dec 14 '10 at 15:06
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About 73% of the energy of the Universe is stored in dark energy - the cosmological constant, most likely - which has a negative pressure numerically equal to the energy density. Because it is a cosmological "constant", this portion of the mass of the Universe cannot be really compressed. So it is a problematic idea to include the dark energy into the "mass that you want to compress".

Only dark matter and visible matter - whose pressure is nearly zero - can really be "compressed". So that would be about $1\times 10^{54}$ for the visible Universe. $2GM/c^2$ for this mass $M$ produces about 150 billion light years which is about 3 times larger than the radius of the observable Universe.

We obtained the result that even if we only count the "particulate" component of the mass of the Universe, we find out that the Universe is actually smaller than the black hole. It is no contradiction because the Universe is not a static system. It would be impossible for the Universe to be kept this small. However, the galaxies are receding from each other which prevents the formation of a nearby event horizon. Instead, the closest horizon in the reality is the "cosmic horizon" - how far we can actually see because of the finite speed of light and finite age of the Universe. It depends on the observer but effectively makes everything beyond this horizon unphysical.

Our Universe, dominated by the dark energy, is already rather close to an empty de Sitter space which is, from many viewpoints, analogous to a black hole except that the interior of the visible part of the de Sitter space is analogous to the exterior of a normal black hole, and the analogy of the interior of a black hole is everything that is behind the cosmic horizon - where we don't see. It is misleading to create the analogy with the static black holes directly because our Universe is not static in the normal cosmological coordinates.

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Calculation

  1. First I went here, getting the number 3.14×10^54 kg.

  2. Under "the size of a black hole" I understand the Schwarzschild radius. Which is $r=\frac{2 G M}{c^2}$.

  3. Then I just did the math:
    http://www.wolframalpha.com/input/?i=2G%2A(3.14E54+kg)%2Fc%5E2
    Obtaining ~5 diameters of observable Universe...

The Problem

I felt really uneasy about the result. Indeed: I took the mass of observable Universe and obtained 5 times radius of the observable Universe. But how on Earth the observable Universe is smaller than the black hole you can make from it?!

To elaborate, let's also calculate how the "density" of the black hole depends on its mass:
$\rho(M) = \frac{3M}{4\pi r^3}=\frac{3Mc^6}{4\pi\cdot 8 G^3 M^3} = \frac{3 c^6}{32\pi G^3}\cdot\frac{1}{M^2}$
We got that with the growth of mass of the black hole its "density" decreases. At really high masses it gets so disperse that it is even gets smaller than the density of the Universe as we see it.

Interpretation and the answer

Now, how do I understand this. The black hole is "a region of space from which nothing, not even light, can escape". So, in order to make sense of the question we got to imagine some (presumably, flat) empty space. And then we are putting all the matter we gathered into it. It turns out, that even if we do not "compress" this stuff (we can even make it $\simeq5^3=125$ times more disperse), the gravitation attraction of the system would be strong enough for "nothing, not even light, could escape" from the system.

So, as far as I understand the situation now, answer to the question is: aliens don't need to do anything with the Universe in order to make a black hole (in a sense) of it.

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Well, I feel really uneasy about my result... –  Kostya Dec 14 '10 at 15:20
    
Why do you feel uneasy? –  Vagelford Dec 14 '10 at 15:32
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It should be noted that if one would live under (but close to) the horizon of this huge black hole, they actually wouldn't notice it (because curvature is inversely proportional to the mass of the BH). I think it was actually proposed in some papers that our universe is actually the inside of some BH (and this means that observable universe would end in a Big crunch in a finite time) but I don't think this is consistent with observation. –  Marek Dec 14 '10 at 17:41
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I should point out that a black hole solution is a vacuum solution of GR. On the other hand, the universe is a uniform density solution. That changes things in the sense that you should address the two problems differently. –  Vagelford Dec 14 '10 at 18:04
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There are the swiss cheese models, where there are patches of the universe that are replaced with Schw solutions with the same mass ( you could have a look at this thesis astro.utoronto.ca/theses/thesis05.mcclure.pdf ). But I don't think that these solutions are unstable. –  Vagelford Dec 14 '10 at 19:03
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Why would you need to compress the universe?

The radius of the observable universe equals the Schwarzschild radius of the total mass of the observable universe. In terms of density the universe is dense enough to be a black hole. But it isn't a black hole, it behaves like a white hole. If you want to get a black hole, you'd rather change the direction of time so as to transform the Hubble expansion into a Hubble contraction.

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