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Friday, February 17, 2012

Schwarzschild radius

estimate the Schwarzschild-radius $r_S$ of the Universe!

The Schwarzschild radius corresponding to a mass M is given by

R_s = \frac{2GM}{c^2}

Let's assume the mass M of the universe as the mass enclosed by a homogeneous sphere with the critical density and the Hubble radius R_H = \frac{c}{H_0}.

M = \frac{4 \pi}{3} \rho_{cr} R_H^3

The critical density is given by

\rho_{cr} = \frac{3H_0^2}{8 \pi G} with H_0 denoting the Hubble constant.

the light rays should just propagate towards us, the Hubble radius is 3 Gpc/h away from us, that's about 1e10 light years. please keep in mind that the question was just aiming at the fact that the Universe is a system that needs to be described using general relativity. nowhere this would imply that the metric of the Universe would be a Schwarzschild metric - in fact this metric is only applicable to a gravitational field generated outside a spherically symmetric non-rotating static matter distribution (with no charge). a sensible starting point for the metric describing the Universe is the Robertson-Walker metric, which is necessarily time-dependent.

According to general relativity the universe could never expand beyond it schwarzschild radius. Time dilatation caused by the universe's gravity would stop any object from moving over its R_s. Any objects closing in on R_s would be infinitely red shifted too. Red shift is no prove or disprove of the big bang as any objects will be red shifted as farer away they are from the center of the universe. This is the logic of the big bang theory. As we don't see many blue shifted galaxies we may assume that we are close to this center. If we assume the universe is infinite than there would be no time dilatation (because no center of gravity) and no gravitational red shift neither. But as there is a red shift it would contradict a pure steady state universe which is infinite. It would however allow one which is finite and is smaller than R_s. In any case the schwarzschild radius of the universe is very important. Its shows that the universe is either infinite in time and space or limited at least in space. Of course I assume that our understanding of gravity is correct...

The considerations by Stephan Q, June 11, 2012, 1:48 AM are not correct - or better phrased: not relevant to cosmology. The first sentence is simply wrong.

The Schwarzschild radius corresponding to a mass M is given by

ReplyDeleteR_s = \frac{2GM}{c^2}

Let's assume the mass M of the universe as the mass enclosed by a homogeneous sphere with the critical density and the Hubble radius R_H = \frac{c}{H_0}.

M = \frac{4 \pi}{3} \rho_{cr} R_H^3

The critical density is given by

\rho_{cr} = \frac{3H_0^2}{8 \pi G} with H_0 denoting the Hubble constant.

Thus

R_s = \frac{2GM}{c^2} = \frac{8\pi G}{3c^2} \frac{3 H_0^2}{8\pi G} R_H^3 = \frac{H_0^2}{c^2} R_H^3 = R_H

Thus with the assumptions made at the beginning, the Schwarzschild radius of the universe is given by its Hubble radius.

Given that analogy to black holes: What happens to light rays which are emitted by a source close to the Hubble radius?

ReplyDeletethe light rays should just propagate towards us, the Hubble radius is 3 Gpc/h away from us, that's about 1e10 light years. please keep in mind that the question was just aiming at the fact that the Universe is a system that needs to be described using general relativity. nowhere this would imply that the metric of the Universe would be a Schwarzschild metric - in fact this metric is only applicable to a gravitational field generated outside a spherically symmetric non-rotating static matter distribution (with no charge). a sensible starting point for the metric describing the Universe is the Robertson-Walker metric, which is necessarily time-dependent.

ReplyDeleteAccording to general relativity the universe could never expand beyond it schwarzschild radius. Time dilatation caused by the universe's gravity would stop any object from moving over its R_s. Any objects closing in on R_s would be infinitely red shifted too.

ReplyDeleteRed shift is no prove or disprove of the big bang as any objects will be red shifted as farer away they are from the center of the universe. This is the logic of the big bang theory.

As we don't see many blue shifted galaxies we may assume that we are close to this center.

If we assume the universe is infinite than there would be no time dilatation (because no center of gravity) and no gravitational red shift neither. But as there is a red shift it would contradict a pure steady state universe which is infinite. It would however allow one which is finite and is smaller than R_s.

In any case the schwarzschild radius of the universe is very important. Its shows that the universe is either infinite in time and space or limited at least in space. Of course I assume that our understanding of gravity is correct...

The considerations by Stephan Q, June 11, 2012, 1:48 AM are not correct - or better phrased: not relevant to cosmology. The first sentence is simply wrong.

ReplyDeleteUli Bastian.