# How can the Schwarzschild radius of the universe be 13.7 billion light years?

• 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?

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.

• 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 2 '16 at 10:06