2
$\begingroup$

With the FLRW equations we can get solutions for a matter dominated closed universe in which the finale is an ultimate collapse, but this is only in terms of $a$ (the scale factor) and $t$ (time) and some $q_0$ and $\theta$:

$a = {\frac {q_0}{2 q_0 -1 } } (cosh (\theta) -1) $, (1)

$t = {\frac {q_0}{2 q_0 -1 } } (sinh(\theta) - \theta)$ (2)

where $dt=a d\theta$

Unfortunately, the FLRW does not show that the model should collapse into a black hole. So it only tells you when things might end depending on $q_0$. So I will appreciate if someone can give me an opinion of whether the following solution is appropriate for a universe that has collapsed into a black hole. Please feel free to rip it apart if necessary. And secondly, how can the final black hole equation be factored into the FLRW solution to get a complete model? Here it goes.

First the energy of the matter is simply:

$E_{m}= M c^2$ (3)

The gravitational potential energy $U$ of the universe is (see http://arxiv.org/abs/1004.1035):

$U = - \frac {3 G M^2} {5 R}$ (4)

For a particle away from the event horizon the Newtonian gravity is simply:

$g r^2 = G M $ (5)

where $r$ is the radial coordinate of the observer. If I assume that $U = E$ once the universe has completely collapsed then I get,

$Mc^2 = - \frac {3 G M^2} {5 R}$ (6)

Substituting (5), I get:

$c^2 = - \frac {3 r^2 g}{5R}$ (7)

And finally,

$R = - \frac{3}{5} \frac {r^2 g}{c^2}$ (8)

Now it turns out that the formula for the Schwarzschild radius in Newtonian gravitational fields is remarkably similar http://en.wikipedia.org/wiki/Schwarzschild_radius):

$R_s = 2 \frac {r^2 g} {c^2}$ (9)

The obvious issue here is that I got the constant $-3/5$ instead $2$. So the real question now is, are these computations correct and there is a reason for the difference? (for example, a supermassive BH Universe is not the same as your conventional BH). Or is it this a completely wrong approach that yields a completely wrong answer? If wrong, what is the right answer? Note that I tried to avoid solutions with $M$ and $G$ because that results in absurd answers such as those arrived in If the observable universe were compressed into a super massive black hole, how big would it be? (i.e., the radius of the BH would turn out larger than the current radius of the Universe, etc, etc).

$\endgroup$
  • 1
    $\begingroup$ grwiki.physics.ncsu.edu/wiki/Oppenheimer-Snyder_Collapse $\endgroup$ – Jerry Schirmer Feb 11 '14 at 1:18
  • $\begingroup$ @jerry-schirmer I must say that is an excellent reference with a great answer since it combines the Friedman equation with the Schwarzchild. I am however still a bit curious if the approach that I was taking was correct and, if not, why not. $\endgroup$ – Luis Feb 11 '14 at 6:25
  • $\begingroup$ @jerry-schirmer Also I am not quite sure how to use the Oppenheimer-Snyder model for calculating total time. The way I read it is that R(0) would be the current radius of the Universe and M would be the total mass. So the final t = ... equation gives imaginary numbers because R(0)/2M is an infinitesimal number and subtracting -1 is just $\sqrt-1$. So given the M and R(0) which is easily estimated and the only two parameters I really need other than t'', I am not sure how to use them. Some hint would really help. $\endgroup$ – Luis Feb 11 '14 at 16:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.