11

So i was reading about Schwarzschild radius on Wiki and I found a interesting thing written there link.

  • It says that the S. radius of the universe is as big as the size of the universe?

  • How is this possible?

  • Since most the universe is empty space shouldn't the S. radius of our universe be significantly smaller then 13.7 light years?

Iamat8
  • 417
  • 1
  • 4
  • 10
Vajura
  • 221
  • 1
    By definition, the observable universe is given by how far photons could have travelled. Isn't there some similarity with the definition of the S radius? This article seems to suggest they are the same, but I am not sure I fully trust it ... – fffred Apr 07 '15 at 10:04
  • Related: http://physics.stackexchange.com/q/2268/2451 , http://physics.stackexchange.com/q/23118/2451 and links therein. – Qmechanic Apr 07 '15 at 10:04

2 Answers2

13

Firstly we should note that the universe as a whole is not described by the Schwarzschild metric, so the Schwarzschild radius of the universe is a meaningless concept. However if you take the mass of the observable universe you could ask what the Schwarzschild radius of a black hole of this mass is.

For a mass $M$ the Schwarzschild radius is:

$$ r_s = \frac{2GM}{c^2} \tag{1} $$

If the radius of the observable universe is $R$, and the density is $\rho$, then the mass is:

$$ M = \tfrac{4}{3}\pi R^3 \rho $$

and we can substitute in equation (1) to get:

$$ r_s = \frac{8G}{3c^2} \pi R^3 \rho \tag{2} $$

Now we believe that the density of the universe is the critical density, and from the FLRW metric with some hair pulling we can obtain a value for the critical density:

$$ \rho_c = \frac{3H^2}{8\pi G} $$

And we can substitute for $\rho$ in equation (2) to get:

$$ r_s = \frac{H^2}{c^2} R^3 \tag{3} $$

Now, Hubble's law tells that the velocity of a distant object is related to its distance $r$ by:

$$ v \approx Hr $$

and since the edge of the universe, $r_e$, is where the recession velocity is $c$ we get:

$$ r_e \approx \frac{c}{H} $$

and substituting this in equation (3) gives;

$$ r_s = \frac{1}{r_e^2} R^3 \tag{4} $$

If $r_e = R$ then we'd be left with $r_s = R$ and we'd have shown that the Schwarzschild radius of the mass of the observable universe is equal to it's radius. Sadly it doesn't quite work. The dimension $R$ is the current size of the observable universe, which is around 46.6 billion light years, while the size used in Hubble's law, $r_e$, is the current apparent size 13.7 billion light years.

If I take equation (3) and put in $R$ = 46.6 billion light years and $H$ = 68 km/sec/megaParsec I get $r_s$ to be around 500 billion light years or a lot larger than the size of the observable universe.

John Rennie
  • 355,118
  • 3
    The recession velocity at the edge of the Universe (the particle horizon) is not $c$, but more like $3.3c$. Your $r_e$ is simply the same as your $R$, so there's no need to introduce $r_e$. That is, if in Eq. 3 you substitute $c=HR/3.3$, you get that $r_s = 3.3^2R\simeq 11 R \sim 500,\mathrm{Gly}$. – pela Dec 02 '16 at 10:06
  • Why does the Wikipedia value for the radius of the universe differ from your calculation? –  Jan 15 '23 at 00:35
0

Yes, it is. It could be so because the density of a black hole decreases with its squared radius (as does the critical density of a flat universe, by the way). The bigger the S. radius, the lower the density, and there are known supermassive black holes in galactic nuclei that should have an average density lower than the density of the air...

Juan Casado
  • 1
  • Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. – Community Jan 20 '24 at 13:07