# How does the critical density decide the fate of the Universe?

Trying to teach myself cosmology. Given the Friedmann equation $$\left[\frac{1}{R}\frac{dR}{dt}\right]^{2}=\frac{8\pi G}{3}\rho-\frac{kc^{2}}{R^{2}}$$ and the critical density $$\rho_{c}\left(t\right)=\frac{3H^{2}\left(t\right)}{8\pi G}$$

I can see why, if $\rho>\rho_{c}$ then $k=+1.$ But I can't see why this leads to a collapsing universe? Conversely, if $\rho<\rho_{c}$ then $k=-1.$ But why does this lead to an expanding universe? I'm possibly confused because I've seen graphs of other FRW models with no correlation between the sign of $k$ and whether the universe is contracting or expanding. Do I need to learn about the deceleration parameter (I've heard of it and that's all) in order to understand this? Thank you

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## 2 Answers

This is something I, too, found confusing when first learning about cosmology. The correspondence that you mention between the sign of $k$ and the expansion fate of the universe only holds if there is no dark energy. It turns out that in our universe, which does indeed possess dark energy, $k$ might be zero but the universe may continue to experience accelerated expansion for all time (all available cosmological data is consistent with this scenario).

To solve for $a(t)$, you need not only the Friedmann equation but also a statement of energy conservation and an equation of state for each component of $\rho$. The energy conservation equation is $$\dot{\rho} = -3 H/c^2 (\rho c^2 + p)$$ where $p$ is the pressure, related to $\rho$ through an equation of state $$p = w \rho c^2$$

where $w = 0$ for nonrelativistic matter, $w=1$ for radiation (anything traveling at or very near $c$), $w = -1$ for a cosmological constant, and $-1 < w < -1/3$ for any other type of dark energy.

You can now solve the Friedmann equation for different initial densities and values of $k$. Generally you'll need to do this numerically. However, you can get some great insight by examining simple solutions that you can solve by hand. Barbara Ryden's textbook on cosmology is great for going through many of these partial solutions.

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thanks. Ryden uses parametric equations to get her big bang to big crunch graph of a matter only universe on p87. Just out of interest, do you know why I can't set the first term on the rhs of the Friedmann equation to a constant $C$, $k=+1$, $c=1$ to obtain $$\left[\frac{1}{R}\frac{dR}{dt}\right]^{2}=C-\frac{1}{R^{2}}$$ $$\frac{dR}{dt}=\left(R^{2}-1\right)^{1/2}$$ but when I try to solve this, I don't get the nice Big Bang to Big Crunch graph that Ryden does? –  Peter4075 Apr 6 '12 at 8:13
@Peter4075 the first term on the rhs of the Friedmann equation is not constant as the universe expands or contracts. The matter density $\rho$ scales as $a^{-3}$. –  kleingordon Apr 6 '12 at 20:36
of course! Funny how some things are obvious when they're pointed out. I was getting the strangest results trying to feed my equation into the WolframAlpha differential equation calculator. Thanks. –  Peter4075 Apr 7 '12 at 10:14

My answer do not correspond to any official position about this issue, but it's mine ;)
The above equation was written when everyone believed that matter had an infinite age.
Latter, it was found that it is not the case. The issue should be re-thinked. I've some difficulty to accept such equation because the background premises, not written, had changed.

Now, thinking without limitations, if the universe is infinite, if the gravitational interaction evolve at the speed of the light, i.e. is not instantaneous, then the universe has no conditions to collapse. Every points in the universe are equivalent, then there is no central point where to collapse. The universe has no need to be infinite, it suffices that it is larger then the light cone in the perspective of the observer.

If the universe is homogeneous, then the gravitational field is zero everywhere, irrespective of its density. Why should it collapse? What matters is not the absolute density but the gravitational potential, a derivative, a rate of change.
It will not collapse as well it is not expanding, imo. Atoms are shrinking giving us the illusion of space expansion, because we measure with atom properties.

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To the one that is downvoting all my answers (4 in a row in the last minutes), get yourself some time to say something about physics. About this answer, I knew from the first moment that it will be downvoted, that is not a surprise. Say, for instance, that the measurements we have do not support what I say here. If you find that I'm wrong please correct me, for the benefict of the community. Or, may be the case that it's hard to find an argument against my controversial answers. –  Helder Velez Apr 5 '12 at 20:41
There is no contradiction in the collapse of a homogeneous universe. It is a solution of Einstein's field equations for a certain subset of initial conditions. One does not need to worry about collapsing to a "central point" since the collapse process would simply be the time reversed version of the expansion we see today, which also does not rely on a central point of expansion. –  kleingordon Apr 6 '12 at 2:05
'certain subset of initial conditions'? at the time Einstein wrote GR he, and everyone else, believed that the universe was infinitelly aged, then, how is is it possible to discuss 'initial conditions'? Back then there is not any issue related to the light-cone, but now it has to be analysed under this constraint, imposed by Einsten, isn't it? If we measure with atoms Os: OOOOO , then with o: ooooo, and then with a point .: ...., () then space is 'expanding' whitout any expansion. We measure with atoms, and they shrink. () See it in a monospaced font. –  Helder Velez Apr 6 '12 at 3:20
...and a 'collapse of a homogeneous universe' can not happen without the effect of gravitation, that depends not on the density but in change of density from point to point. The universe is not a star, matter inside,... and there is no outside. If the universe is isodense (=homogeneous) it will not move because it will not exist gravitational field, except locally. Concluding : homegeneous->isodense->no gravitational field->no motion. It is not a question of density. –  Helder Velez Apr 6 '12 at 3:33
I suggest that this be the last comment before we move this a separate chat, but I'd just like to point out that your reasoning relies on a Newtonian understanding of gravity. However, in the general relativistic context, which is necessary for understanding cosmology, a homogeneous universe can indeed expand and contract –  kleingordon Apr 6 '12 at 23:47