An infinite, isotropic, homogeneous, and static arrangement collapses under gravity?

In summary: But in short, he's saying that if the stars were evenly distributed, they would stay in the arrangement. But, as I explained above, one slight movement would cause the matrix of stars to collapse. So, we can say that one star explodes, and then the rest would undergo the collapse.
  • #1
Rear Naked
48
0
Can anyone explain this to me?


If we have an infinite amount of balls arranged in a kind of cubic matrix, in an infinite and static space...how the heck would that collapse on itself due to gravity?


Thanks folks
 
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  • #2


It wouldn't - but one slight change in the position of any of the balls would immediate trigger a collapse into lumps. This is because the movement of one ball or create in an imbalance of the gravitational forces acting on the next few balls, causing them to move in the opposite directions. These then alter the movement of the balls near them, and the process continues until the whole setup falls apart.

I'm guessing you are referring to Einstein's static universe. The problem was the same with your infinite arrangement of balls. If one galaxy has a slight peculiar motion relative to the others, it would trigger a collapse as in the previous scenario.
 
  • #3


Mark M said:
It wouldn't - but one slight change in the position of any of the balls would immediate trigger a collapse into lumps. This is because the movement of one ball or create in an imbalance of the gravitational forces acting on the next few balls, causing them to move in the opposite directions. These then alter the movement of the balls near them, and the process continues until the whole setup falls apart.

I'm guessing you are referring to Einstein's static universe. The problem was the same with your infinite arrangement of balls. If one galaxy has a slight peculiar motion relative to the others, it would trigger a collapse as in the previous scenario.


This is all true. It would collapse VERY slowly, though.
 
  • #4


Mark M said:
It wouldn't - but one slight change in the position of any of the balls would immediate trigger a collapse into lumps. This is because the movement of one ball or create in an imbalance of the gravitational forces acting on the next few balls, causing them to move in the opposite directions. These then alter the movement of the balls near them, and the process continues until the whole setup falls apart.

I'm guessing you are referring to Einstein's static universe. The problem was the same with your infinite arrangement of balls. If one galaxy has a slight peculiar motion relative to the others, it would trigger a collapse as in the previous scenario.

This is what my intuition tells me.


This is the quote from Hawking that confused me:

Newton realized that, according to his theory of gravity, the stars should attract each other, so it seemed they could not remain essentially motionless. Would they not all fall together at some point? In a letter in 1691 to Richard Bentley, another leading thinker of his day, Newton argued that his would indeed happen if there were only a finite number of stars distributed over a finite region of space. But he reasoned that if, on the other hand, there were an infinite number of stars, distributed more or less uniformly over infinite space, this would not happen, because there would not be any central point for them to fall to.

This argument is an instance of the pitfalls that you can encounter in talking about infinity. In an infinite universe, every point can be regarded as the center, because every point has an infinite number of stars on each side of it. The correct approach, it was realized only much later, is to consider the finite situation, in which the stars all fall in on each other, and then to ask how things change if one adds more stars roughly uniformly distributed outside this region. According to Newton’s law, the extra stars would make no difference at all to the original ones on average, so the stars would fall in just as fast. We can add as many stars as we like, but they will still always collapse in on themselves. We now know it is impossible to have an infinite static model of the universe in which gravity is always attractive.

So, the qualifiers are that the stars must be "roughly uniformly" distributed?

It seems strange to not further clarify...especially when he says they will "always collapse."
 
  • #5


Rear Naked said:
This is what my intuition tells me.


This is the quote from Hawking that confused me:



So, the qualifiers are that the stars must be "roughly uniformly" distributed?

It seems strange to not further clarify...especially when he says they will "always collapse."

If the stars were evenly distributed, they would stay in the arrangement. But, as I explained above, one slight movement would cause the matrix of stars to collapse. So, we can say that one star explodes, and then the rest would undergo the collapse.

Hawking does have a tendency to do that, especially with the concept of 'imaginary time', that he brings up, but has never explained in any of his books.
 
  • #6


Mark M said:
If the stars were evenly distributed, they would stay in the arrangement.
No, the whole point that Hawking is talking about is that they would fall together, and quite quickly by cosmological standards. When Hawking says they are roughly uniform, he is not ruling out exactly uniform-- exactly uniform would make his words even more correct, not less so.

Hawking argues it from the standpoint of Newtonian gravity, where the stars would still fall together if they were static and uniform. But it can also be said using Einstein's general relativity, where again they fall together. Indeed, this is exactly why Einstein originally introduced a "cosmological constant," to allow a static distribution of mass to not fall into itself. The cosmological constant is essentially a pervasive antigravity that would offset the tendency to contract. The problem with it is that it would be unstable-- so only if you had an antigravity term would you have to worry about what happens to perturbations in certain regions. That's why Einstein regarded it as a blunder-- his equations told him the universe was dynamic, but he tried to add to them to make the universe static, and he did not recognize that his final result would never be stable.

Ironically, we have since discovered we may need the cosmological constant after all, and of similar magnitude, but for completely different reasons (to get the acceleration of the expansion)!
 
  • #7


How can I make sense of that then?

Newton's logic seems flawless to me.
 
  • #8


Rear Naked said:
How can I make sense of that then?

Newton's logic seems flawless to me.
I believe what Hawking is referring to is that even in the symmetry of complete homogeneity, you can draw an imaginary sphere around any set of mass, and consider that all the mass outside that sphere will produce no net gravitational force within the sphere. So there will only be a force that contracts the sphere, coming from the sphere. Now, the Newtonian model can't really be completely correct, because it is known to be a wrong model for gravity, but it can get the right answer if you treat the problem this way. But instead of a force that is everywhere toward some arbitrarily chosen point, which obviously makes no sense, you simply get a uniform contraction everywhere. Saying the contraction is uniform means you are not picking out any arbitrary centers, it's like the "expanding balloon" analogy in reverse.

It really should be done with GR, but you can get pretty far with Newtonian gravity, as long as you do it the way Hawking intimated. What we know for sure is that Einstein understood GR pretty well, and he needed to invoke a cosmological constant (a form of antigravity) to get the universe to be static (though unstable, as I mentioned).
 
  • #9


Ok, that makes sense.

Thank you for the reply.
 

Related to An infinite, isotropic, homogeneous, and static arrangement collapses under gravity?

1. What is an infinite, isotropic, homogeneous, and static arrangement?

An infinite, isotropic, homogeneous, and static arrangement refers to a theoretical model in astrophysics where the universe is assumed to be infinite in size, with matter distributed uniformly and evenly in all directions. It also assumes that the universe is not expanding or contracting, and is in a state of equilibrium.

2. What does it mean for an arrangement to collapse under gravity?

When an arrangement is said to collapse under gravity, it means that the force of gravity has overcome the outward expansion or pressure of the system, causing it to shrink and become more dense. This often occurs when the mass of the system is too great, causing it to collapse under its own gravitational pull.

3. How does gravity cause an infinite, isotropic, homogeneous, and static arrangement to collapse?

Gravity is a fundamental force in the universe that attracts all objects with mass towards each other. In an infinite, isotropic, homogeneous, and static arrangement, the collective gravitational force of all the matter in the universe would eventually cause it to collapse in on itself, as there is no opposing force to counteract it.

4. What would happen if an infinite, isotropic, homogeneous, and static arrangement were to collapse?

If such an arrangement were to collapse under gravity, it would result in a singularity - a point of infinite density and zero volume. This is similar to what is believed to have occurred during the Big Bang, where all matter in the universe was condensed into a single point before expanding rapidly.

5. Is the collapse of an infinite, isotropic, homogeneous, and static arrangement a realistic scenario?

While the concept of an infinite, isotropic, homogeneous, and static arrangement collapsing under gravity is a theoretical one, it is not considered a realistic scenario. The universe is constantly expanding due to the force of dark energy, and the presence of dark matter also affects the dynamics of gravitational collapse. Additionally, the distribution of matter in the universe is not perfectly uniform, making this scenario unlikely.

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