<p>In less mathy words, the Big Crunch scenario occurs if the ratio of the total density of the Universe to its expansion rate is sufficiently large$^\dagger$.</p>

<p>As I understand your question, you're basically asking, "<em>Why would a sufficiently dense universe not simply contract into a clump? Why does it have to pull space itself with it?</em>"</p>

<p>And you basically answer that question yourself: <strong>Yes, space is indeed "tied" to matter</strong>. This is the essence of the Friedmann equation, and of general relativity in general. To my knowledge, there's no "proof" of this other than, well, it's one of the foundations of GR, which has so far proved an immensely successful theory. In moderately overdense regions (galaxy clusters) space expands slower than in underdense regions (voids). In very overdense regions (galaxies, stars, cats, etc.), it doesn't expand at all. And in extremely overdense regions (black holes), space contracts. 
In the case of a black hole, space only contracts locally, but in principle the whole of the Universe could do the same. Only expansion prevents this, and seems to be able to prevent it forever.</p>

<p>We believe that the Universe is homogeneous and isotropic; if this is indeed the case, matter wouldn't be able to contract to a point within the Universe and make one giant black hole in an otherwise expanding Universe, since every bit of matter is attracted the same in all directions. You could perhaps envision clumps of matter contract on very large, yet sub-Universal, scales, to create many supersupermassive black holes, but as it turns out, the expansion rate was simply too large in the early Universe for this to happen, and now it's too late.</p>

<hr>

<p>$^\dagger$<sub>This is analogous to a rock thrown upwards falling back again if the ratio of the gravitational attraction between Earth and rock to the speed with which it is thrown is suffiently large.</sub></p>

CC BY-SA 3.0
2016-12-01T16:35:36.013Z
You're free to use mathematics, I've been exposed to the Friedmann equation and other terms. Anyway, you seem to think I'm proposing some new idea. Rather, I am trying to understand the logic of the Big Crunch hypothesis, which is quite well known. None of the descriptions I have found match what you propose — and while I agree yours makes sense when $\rho_c = \rho$, I'm talking about a universe where $\rho_c < \rho$. That's the important idea of the Big Crunch hypothesis, that the universe would eventually contract. I'm trying to understand how that contraction would work.
Sir Cumference
CC BY-SA 3.0
2016-12-01T16:35:51.453Z
Also, you do not clarify whether or not gravitationally attracting bodies change in comoving distances. I assume that when you say "in overdense regions, space contracts", you're implying that two attracting objects do not decrease in comoving distance?
Sir Cumference
CC BY-SA 3.0
2016-12-02T10:22:51.653Z
@SirCumference: No no, I didn't think you proposed a new theory, I read your question as "why doesn't this happen". And the reason I didn't use math was that it is basically given in Phiteros' answer. Anyway, a $\rho=\rho_c$ universe won't contract, only a $\rho>\rho_c$. And this contraction will take place in comoving coordinates. Moderate _local_ overdensities retard or halt contraction in com. coords (clusters and galaxies), _and_ make stuff collapse in physical coordinates until something prevents this (e.g. radiation pressure in a star). If the overdensity is too large (a black hole), …
pela
CC BY-SA 3.0
2016-12-02T10:23:01.370Z
…then nothing can prevent the collapse from continuing in phys. coords leading to a collapse in com. coords., i.e. a singularity. But this scenario is different from the Big Crunch in the sense that for a BH, only the BH attracts, so matter flows in phys. coords toward to center. For the Universe, there's no central mass, so there will be no movement in phys. coords (except of course for small "peculiar" velocites).
pela
CC BY-SA 3.0
2016-12-03T02:32:09.053Z
Indeed if there were to be a black hole made from all matter with a void space around it, it would mean that the universe has developed a center. From that perspective, one has to consider the information paradox of that kind of scenario. If all waves were again to fold through each other, through the principles of frequency modulation and interaction, it would create so much complexity that the compression of that quantity of matter would cause a time distortion, high temperature, and information knot so vast that it would create more universes than at it's original state of expansion.
bandybabboon
CC BY-SA 3.0
2016-12-01T14:37:05.100Z
2016-12-01T15:14:37.413Z
2016-12-01T15:14:37.413Z
pela
<p>The amount of matter in the universe is directly related to the curvature of space itself.  We can look at the Friedmann Equations to see how this works:
$$
H(t)^2 = \frac{R'(t)^2}{R(t)^2} = \frac{8\pi G}{3}(\rho_{m} + \rho_{r}) + \frac{1}{3}\Lambda - \frac{c^2}{R^2\mathscr{R}^2}
$$
This is the equation for determining the scale factor for finding distances in the universe at any given time.  Here, $H$ is the Hubble Constant, $R$ is the scale factor, $G$ is the gravitational constant, $\rho_m$ is the density of matter (both dark matter and baryonic matter), $\rho_r$ is the density of radiation (photons), $\Lambda$ is the cosmological constant due to dark energy, and $\mathscr{R}$ is the radius of curvature of space.</p>

<p>Next, we define a quantity we call the "critical density" ($\rho_c$).  This is the density of matter needed to make the universe go from an "open", hyperbolic geometry to a "closed", spherical geometry.  $\rho_c=\frac{3H^2}{8\pi G}$. We create a value $\Omega=\frac{\rho}{\rho_c}=\frac{8\pi G \rho}{3H^2}$ to take into account this critical density, making our equation:</p>

<p>$$
H^2 = \Omega_m + \Omega_r + \Omega_\Lambda - \frac{c^2}{R^2\mathscr{R}^2}
$$</p>

<p>From here, we input our present-day values for everything.  We define $R$ to be 1 and $H$ to be $H_0$ at the present day.  We also choose to neglect the density of radiation and dark energy.  Simplifying the equation a little gives us:</p>

<p>$$
\frac{c^2}{\mathscr{R}^2} = H_0^2(\Omega_{m,0} - 1)
$$</p>

<p>From here, we can see that the curvature of the universe is dependent on $\Omega_{m,0}$, which is directly related to the density of matter in the universe.  Specifically, for $\Omega_{m,0}\gt1$, $\mathscr{R}$ will be positive, meaning this will be a spherical, closed universe.  If $\Omega_{m,0}\lt1$, $\mathscr{R}$ will be negative, meaning the universe will expand forever, having hyperbolic curvature.  If $\Omega_{m,0}=1$, then $\mathscr{R}=\infty$, which is a flat universe.</p>

<p>So, we can see that having a dense enough universe, gravity will curve space and make it spherical.  You can also use the Friedmann Equation to calculate the deceleration parameter of space - how quickly the expansion is slowing or speeding up:</p>

<p>$$
q_0=\frac{\Omega_{m,0}}{2}
$$</p>

<p>Here, we can see that in a positively curved universe, $q_0&gt;0$, meaning that the universe's expansion will be decelerating.  Eventually, the expansion rate will be negative, and then it will begin collapsing in on itself.</p>

<p>One thing to keep in mind when thinking about this is that gravity can, in a way, be considered a 'fictitious force'. Gravity is the force objects feel as they attempt to travel in straight lines through curved spacetime.  Matter is what is responsible for the curving of spacetime, so by increasing the density of matter in the universe, you increase the curvature, and thus make objects want to move closer together, which will increase the density, which increases the curvature.  Thus you have a feedback cycle, where you will ultimately collect all matter at a single point, and $\mathscr{R}=0$, so spacetime will have a 0 curvature radius, meaning spacetime has collapsed as well.</p>

CC BY-SA 3.0
2016-12-01T08:44:59.147Z
You showed that the curvature of the Universe is dependent on its mass density. Can you elaborate on how this would cause space to contract, if not due to gravity? In fact, I may be very mistaken, but when $ρ_c<ρ$, shouldn't gravity be strong enough to halt and reverse the expansion?
Sir Cumference
CC BY-SA 3.0
2016-12-01T13:42:32.680Z
@SirCumference I think your right, if the density of the universe is greater than the critical density then gravity will eventually win and reverse expansion.
Dean
CC BY-SA 3.0
2016-12-01T16:52:12.323Z
@SirCumference That's actually built into $\Omega$.  Since $\Omega=\frac{\rho}{\rho_c}$, when $\rho\gt\rho_c$, $\Omega\gt1$.
Phiteros
CC BY-SA 3.0
2016-12-01T16:53:27.450Z
@Phiteros, I know this. I'm asking why you said gravity isn't relevent in the contraction of the universe
Sir Cumference
CC BY-SA 3.0
2016-12-01T16:55:48.033Z
@SirCumference Yeah, I don't know why I worded it like that.  I wrote this up late at night.  Let me revise it.
Phiteros
CC BY-SA 3.0
2016-12-01T17:10:02.500Z
@SirCumference I added some information about the deceleration parameter as well.  Does that help?
Phiteros
CC BY-SA 3.0
2016-12-02T18:01:32.843Z
@SirCumference I expanded on how this relates to gravity, in that gravity is defined as the force objects feel when attempting to travel through curved spacetime.  I hope this helps you understand it.
Phiteros
CC BY-SA 3.0
2016-12-05T23:38:23.113Z
@Phiteros Sorry, I don't follow with the last part. Can you show how $\mathscr{R}=0$?
Sir Cumference
CC BY-SA 3.0
2016-12-06T01:19:52.930Z
@SirCumference As $\Omega_m\to\infty$, $\frac{c}{\mathscr{R}}\to\infty$, so $\mathscr{R}\to0$, because $c$ is a constant and $\frac{const}{0}\to\infty$.
Phiteros
CC BY-SA 3.0
2016-12-06T01:25:24.893Z
@Phiteros Yeah, somehow I didn't realize that.
Sir Cumference
CC BY-SA 3.0
2016-12-08T04:27:59.030Z
Sorry, but you relate the deceleration parameter with $\Omega_m$, but you don't relate it with any other $\Omega$s. Are you sure that doesn't just apply to an ordinary-matter dominated universe?
Sir Cumference
CC BY-SA 3.0
2016-12-08T18:03:50.147Z
@SirCumference Yes, it does rely on the $\Omega_r$ and $\Omega_\Lambda$, but I neglected those since it makes it more complicated to solve for, and here we were only talking about the big crunch scenario.
Phiteros
CC BY-SA 3.0
2016-12-08T18:31:22.437Z
@Phiteros Well wouldn't those be important factors?
Sir Cumference
CC BY-SA 3.0
2016-12-09T04:00:46.730Z
@SirCumference well, $\Omega_\Lambda$ will counteract deceleration, so if we want to talk about the Big Crunch, we should neglect it.  And $\Omega_r$ will be similar to $\Omega_m$, with some different factors out front.  But if all we want to do is talk about how the Big Crunch happens, we can get that picture just with $\Omega_m$
Phiteros
CC BY-SA 3.0
2016-12-13T06:56:52.720Z
@Phiteros Just one more question. You say that because $q_0 > 0$, $\ddot{R} < 0$ and $\dot{R}$ will eventually be negative. But are we sure $\ddot{R}$ is constant? Couldn't $\ddot{R}$ decrease and approach zero, but not reach it? Can we know the sign of $\dddot{R}$ or if $\dddot{R} = 0$ solely by knowing $q_0$?
Sir Cumference
CC BY-SA 3.0
2016-12-13T07:27:51.940Z
@SirCumference I don't know how $q_0$ relates to $\dddot{R}$.  But $q_0 \varpropto -\ddot{R}$, so if $q_0 \gt 0$, $\ddot{R}$ must be less than 0.
Phiteros
CC BY-SA 3.0
2016-12-13T07:43:14.987Z
@Phiteros Even so, can we ascertain that $\dot{R}$ will eventually be negative without knowing $\dddot{R}$?
Sir Cumference
CC BY-SA 3.0
2016-12-14T00:37:52.557Z
As long as $q_0$ is positive, $\dot{R}$ will decrease.  If $q_0$ remains constant, that will mean that $\dot{R}$ will become negative.
Phiteros
CC BY-SA 3.0
2016-12-01T08:03:09.887Z
2016-12-26T07:42:50.870Z
2016-12-26T07:42:50.870Z
Phiteros
<p>According to modern cosmology, space is expanding, causing proper distances (but not comoving distances) to increase between galaxies. In the Big Crunch hypothesis, gravity halts and reverses the expansion of the Universe, causing all matter to collide and eventually form a single black hole. This gives way to other oscillating universe hypotheses, which generally propose that the conditions in a compressed Universe would be the same as those during the Big Bang, leading a cycle of expanding and contracting universes.</p>

<p>Ignoring the <a href="https://astronomy.stackexchange.com/questions/13615/are-there-any-studies-that-prove-disprove-the-oscillating-universe-theory/18569#18569">problems with entropy</a> in reverting the Universe to Big Bang conditions, how can gravity be the cause of a Big Crunch in the first place? Specifically, gravity (to my knowledge) only curves space; the idea that it can revert the Universe to the conditions of the Big Bang seems to imply that gravity actually can contract space. Is this actually the case?</p>

<p>If not, the gravitating objects should be moving through a comoving coordinate system, so space itself would not actually be contracting. As far as I can tell, we would have all the Universe's matter compressed within a single point in space, instead of space itself contracting. This should be completely unlike the Big Bang, when space was far less dilated than it is now. If this is in fact what the Big Crunch hypothesis describes, than I'm utterly confused as to how an oscillating universe could work in such a situation.</p>

<p>Am I mistaken, or do the Big Crunch and oscillating universe hypotheses imply that gravity actually contracts space (as in, the comoving distances of gravitationally attracting objects would not change)? If not, how could gravity possibly lead to these scenarios?</p>

CC BY-SA 3.0
2016-11-30T22:01:28.440Z
Unless you have an idea for how expansion would stop, gravity could not do anything anyway.
called2voyage
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2016-11-30T22:26:36.283Z
@called2voyage That doesn't fix the confusion. How could the Big Crunch be possible due to gravity?
Sir Cumference
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2016-11-30T22:28:15.117Z
I need more clarification on that as well, just noting that the Big Crunch is not likely to occur anyway.
called2voyage
CC BY-SA 3.0
2016-11-30T22:47:38.143Z
If the curvature is sufficient then the origin joins with the end. 

The idea was (I'm not sure many think it stands up to the data these days), that the original impulse from the Big Bang was subject to gravitational deceleration and eventually the universe will start to contract as gravitational drag applies.
adrianmcmenamin
CC BY-SA 3.0
2016-12-01T01:45:49.287Z
This isn't an answer, but in the history of our universe, gravity has already slowed the expansion of space. Up until 5 billion years ago, our universe was slowing down in it's expansion, due to gravity. It wasn't until then that it became large enough that dark energy could overpower gravity and force the expansion to accelerate. Besides, don't you think "curving" space time involves contracting it locally? You can't curve 3D space in 3D without expanding part of it and contracting another part.
zephyr
CC BY-SA 3.0
2016-12-01T04:15:59.277Z
@zephyr The difference is whether the comoving distances of gravitationally attracting objects decrease or not. To my knowledge, they do, so the idea that "curved space contracts locally" seems unrelated to the spacial expansion and contraction I'm talking about. :/
Sir Cumference
<gravity><cosmology><general-relativity><expansion><fate-of-universe>
How can gravity lead to the Big Crunch scenario?
Sir Cumference
# How can gravity lead to the Big Crunch scenario?

(Asked by: username_0 on 2016-11-30)

According to modern cosmology, space is expanding, causing proper distances (but not comoving distances) to increase between galaxies. In the Big Crunch hypothesis, gravity halts and reverses the expansion of the Universe, causing all matter to collide and eventually form a single black hole. This gives way to other oscillating universe hypotheses, which generally propose that the conditions in a compressed Universe would be the same as those during the Big Bang, leading a cycle of expanding and contracting universes.

Ignoring the [problems with entropy](https://astronomy.stackexchange.com/questions/13615/are-there-any-studies-that-prove-disprove-the-oscillating-universe-theory/18569#18569) in reverting the Universe to Big Bang conditions, how can gravity be the cause of a Big Crunch in the first place? Specifically, gravity (to my knowledge) only curves space; the idea that it can revert the Universe to the conditions of the Big Bang seems to imply that gravity actually can contract space. Is this actually the case?

If not, the gravitating objects should be moving through a comoving coordinate system, so space itself would not actually be contracting. As far as I can tell, we would have all the Universe's matter compressed within a single point in space, instead of space itself contracting. This should be completely unlike the Big Bang, when space was far less dilated than it is now. If this is in fact what the Big Crunch hypothesis describes, than I'm utterly confused as to how an oscillating universe could work in such a situation.

Am I mistaken, or do the Big Crunch and oscillating universe hypotheses imply that gravity actually contracts space (as in, the comoving distances of gravitationally attracting objects would not change)? If not, how could gravity possibly lead to these scenarios?

**Tags:** gravity, cosmology, general-relativity, expansion, fate-of-universe

> **Comments:**
> **username_1**: Unless you have an idea for how expansion would stop, gravity could not do anything anyway.  
> **username_0**: @username_1 That doesn't fix the confusion. How could the Big Crunch be possible due to gravity?  
> **username_1**: I need more clarification on that as well, just noting that the Big Crunch is not likely to occur anyway.  
> **username_2**: If the curvature is sufficient then the origin joins with the end.   The idea was (I'm not sure many think it stands up to the data these days), that the original impulse from the Big Bang was subject to gravitational deceleration and eventually the universe will start to contract as gravitational drag applies.  
> **username_3**: This isn't an answer, but in the history of our universe, gravity has already slowed the expansion of space. Up until 5 billion years ago, our universe was slowing down in it's expansion, due to gravity. It wasn't until then that it became large enough that dark energy could overpower gravity and force the expansion to accelerate. Besides, don't you think "curving" space time involves contracting it locally? You can't curve 3D space in 3D without expanding part of it and contracting another part.  
> **username_0**: @username_3 The difference is whether the comoving distances of gravitationally attracting objects decrease or not. To my knowledge, they do, so the idea that "curved space contracts locally" seems unrelated to the spacial expansion and contraction I'm talking about. :/  

---

## 2 answers

### Answer by username_6 on 2016-12-01. Score: 6 | Accepted answer

The amount of matter in the universe is directly related to the curvature of space itself. We can look at the Friedmann Equations to see how this works:
$$
H(t)^2 = \frac{R'(t)^2}{R(t)^2} = \frac{8\pi G}{3}(\rho\_{m} + \rho\_{r}) + \frac{1}{3}\Lambda - \frac{c^2}{R^2\mathscr{R}^2}
$$
This is the equation for determining the scale factor for finding distances in the universe at any given time. Here, $H$ is the Hubble Constant, $R$ is the scale factor, $G$ is the gravitational constant, $\rho\_m$ is the density of matter (both dark matter and baryonic matter), $\rho\_r$ is the density of radiation (photons), $\Lambda$ is the cosmological constant due to dark energy, and $\mathscr{R}$ is the radius of curvature of space.

Next, we define a quantity we call the "critical density" ($\rho\_c$). This is the density of matter needed to make the universe go from an "open", hyperbolic geometry to a "closed", spherical geometry. $\rho\_c=\frac{3H^2}{8\pi G}$. We create a value $\Omega=\frac{\rho}{\rho\_c}=\frac{8\pi G \rho}{3H^2}$ to take into account this critical density, making our equation:

$$
H^2 = \Omega\_m + \Omega\_r + \Omega\_\Lambda - \frac{c^2}{R^2\mathscr{R}^2}
$$

From here, we input our present-day values for everything. We define $R$ to be 1 and $H$ to be $H\_0$ at the present day. We also choose to neglect the density of radiation and dark energy. Simplifying the equation a little gives us:

$$
\frac{c^2}{\mathscr{R}^2} = H\_0^2(\Omega\_{m,0} - 1)
$$

From here, we can see that the curvature of the universe is dependent on $\Omega\_{m,0}$, which is directly related to the density of matter in the universe. Specifically, for $\Omega\_{m,0}\gt1$, $\mathscr{R}$ will be positive, meaning this will be a spherical, closed universe. If $\Omega\_{m,0}\lt1$, $\mathscr{R}$ will be negative, meaning the universe will expand forever, having hyperbolic curvature. If $\Omega\_{m,0}=1$, then $\mathscr{R}=\infty$, which is a flat universe.

So, we can see that having a dense enough universe, gravity will curve space and make it spherical. You can also use the Friedmann Equation to calculate the deceleration parameter of space - how quickly the expansion is slowing or speeding up:

$$
q\_0=\frac{\Omega\_{m,0}}{2}
$$

Here, we can see that in a positively curved universe, $q\_0>0$, meaning that the universe's expansion will be decelerating. Eventually, the expansion rate will be negative, and then it will begin collapsing in on itself.

One thing to keep in mind when thinking about this is that gravity can, in a way, be considered a 'fictitious force'. Gravity is the force objects feel as they attempt to travel in straight lines through curved spacetime. Matter is what is responsible for the curving of spacetime, so by increasing the density of matter in the universe, you increase the curvature, and thus make objects want to move closer together, which will increase the density, which increases the curvature. Thus you have a feedback cycle, where you will ultimately collect all matter at a single point, and $\mathscr{R}=0$, so spacetime will have a 0 curvature radius, meaning spacetime has collapsed as well.

> **Comments:**
> **username_0**: You showed that the curvature of the Universe is dependent on its mass density. Can you elaborate on how this would cause space to contract, if not due to gravity? In fact, I may be very mistaken, but when $ρ_c<ρ$, shouldn't gravity be strong enough to halt and reverse the expansion?  
> **username_7**: @username_0 I think your right, if the density of the universe is greater than the critical density then gravity will eventually win and reverse expansion.  
> **username_6**: @username_0 That's actually built into $\Omega$.  Since $\Omega=\frac{\rho}{\rho_c}$, when $\rho\gt\rho_c$, $\Omega\gt1$.  
> **username_0**: @username_6, I know this. I'm asking why you said gravity isn't relevent in the contraction of the universe  
> **username_6**: @username_0 Yeah, I don't know why I worded it like that.  I wrote this up late at night.  Let me revise it.  
> **username_6**: @username_0 I added some information about the deceleration parameter as well.  Does that help?  
> **username_6**: @username_0 I expanded on how this relates to gravity, in that gravity is defined as the force objects feel when attempting to travel through curved spacetime.  I hope this helps you understand it.  
> **username_0**: @username_6 Sorry, I don't follow with the last part. Can you show how $\mathscr{R}=0$?  
> **username_6**: @username_0 As $\Omega_m\to\infty$, $\frac{c}{\mathscr{R}}\to\infty$, so $\mathscr{R}\to0$, because $c$ is a constant and $\frac{const}{0}\to\infty$.  
> **username_0**: @username_6 Yeah, somehow I didn't realize that.  
> **username_0**: Sorry, but you relate the deceleration parameter with $\Omega_m$, but you don't relate it with any other $\Omega$s. Are you sure that doesn't just apply to an ordinary-matter dominated universe?  
> **username_6**: @username_0 Yes, it does rely on the $\Omega_r$ and $\Omega_\Lambda$, but I neglected those since it makes it more complicated to solve for, and here we were only talking about the big crunch scenario.  
> **username_0**: @username_6 Well wouldn't those be important factors?  
> **username_6**: @username_0 well, $\Omega_\Lambda$ will counteract deceleration, so if we want to talk about the Big Crunch, we should neglect it.  And $\Omega_r$ will be similar to $\Omega_m$, with some different factors out front.  But if all we want to do is talk about how the Big Crunch happens, we can get that picture just with $\Omega_m$  
> **username_0**: @username_6 Just one more question. You say that because $q_0 > 0$, $\ddot{R} < 0$ and $\dot{R}$ will eventually be negative. But are we sure $\ddot{R}$ is constant? Couldn't $\ddot{R}$ decrease and approach zero, but not reach it? Can we know the sign of $\dddot{R}$ or if $\dddot{R} = 0$ solely by knowing $q_0$?  
> **username_6**: @username_0 I don't know how $q_0$ relates to $\dddot{R}$.  But $q_0 \varpropto -\ddot{R}$, so if $q_0 \gt 0$, $\ddot{R}$ must be less than 0.  
> **username_0**: @username_6 Even so, can we ascertain that $\dot{R}$ will eventually be negative without knowing $\dddot{R}$?  
> **username_6**: As long as $q_0$ is positive, $\dot{R}$ will decrease.  If $q_0$ remains constant, that will mean that $\dot{R}$ will become negative.  

---

### Answer by username_4 on 2016-12-01. Score: 2

In less mathy words, the Big Crunch scenario occurs if the ratio of the total density of the Universe to its expansion rate is sufficiently large$^\dagger$.

As I understand your question, you're basically asking, "*Why would a sufficiently dense universe not simply contract into a clump? Why does it have to pull space itself with it?*"

And you basically answer that question yourself: **Yes, space is indeed "tied" to matter**. This is the essence of the Friedmann equation, and of general relativity in general. To my knowledge, there's no "proof" of this other than, well, it's one of the foundations of GR, which has so far proved an immensely successful theory. In moderately overdense regions (galaxy clusters) space expands slower than in underdense regions (voids). In very overdense regions (galaxies, stars, cats, etc.), it doesn't expand at all. And in extremely overdense regions (black holes), space contracts.
In the case of a black hole, space only contracts locally, but in principle the whole of the Universe could do the same. Only expansion prevents this, and seems to be able to prevent it forever.

We believe that the Universe is homogeneous and isotropic; if this is indeed the case, matter wouldn't be able to contract to a point within the Universe and make one giant black hole in an otherwise expanding Universe, since every bit of matter is attracted the same in all directions. You could perhaps envision clumps of matter contract on very large, yet sub-Universal, scales, to create many supersupermassive black holes, but as it turns out, the expansion rate was simply too large in the early Universe for this to happen, and now it's too late.

---

$^\dagger$This is analogous to a rock thrown upwards falling back again if the ratio of the gravitational attraction between Earth and rock to the speed with which it is thrown is suffiently large.

> **Comments:**
> **username_0**: You're free to use mathematics, I've been exposed to the Friedmann equation and other terms. Anyway, you seem to think I'm proposing some new idea. Rather, I am trying to understand the logic of the Big Crunch hypothesis, which is quite well known. None of the descriptions I have found match what you propose — and while I agree yours makes sense when $\rho_c = \rho$, I'm talking about a universe where $\rho_c < \rho$. That's the important idea of the Big Crunch hypothesis, that the universe would eventually contract. I'm trying to understand how that contraction would work.  
> **username_0**: Also, you do not clarify whether or not gravitationally attracting bodies change in comoving distances. I assume that when you say "in overdense regions, space contracts", you're implying that two attracting objects do not decrease in comoving distance?  
> **username_4**: @username_0: No no, I didn't think you proposed a new theory, I read your question as "why doesn't this happen". And the reason I didn't use math was that it is basically given in username_6' answer. Anyway, a $\rho=\rho_c$ universe won't contract, only a $\rho>\rho_c$. And this contraction will take place in comoving coordinates. Moderate _local_ overdensities retard or halt contraction in com. coords (clusters and galaxies), _and_ make stuff collapse in physical coordinates until something prevents this (e.g. radiation pressure in a star). If the overdensity is too large (a black hole), …  
> **username_4**: …then nothing can prevent the collapse from continuing in phys. coords leading to a collapse in com. coords., i.e. a singularity. But this scenario is different from the Big Crunch in the sense that for a BH, only the BH attracts, so matter flows in phys. coords toward to center. For the Universe, there's no central mass, so there will be no movement in phys. coords (except of course for small "peculiar" velocites).  
> **username_5**: Indeed if there were to be a black hole made from all matter with a void space around it, it would mean that the universe has developed a center. From that perspective, one has to consider the information paradox of that kind of scenario. If all waves were again to fold through each other, through the principles of frequency modulation and interaction, it would create so much complexity that the compression of that quantity of matter would cause a time distortion, high temperature, and information knot so vast that it would create more universes than at it's original state of expansion.  

---

