# Infinite lattice scenario

1. Aug 11, 2015

### Helios

Questions I have arrive from other forum by request.
I asked: Why can't an infinite motionless lattice be stable or even possible?
We set the stage with an infinite lattice and an exactly zero cosmo-constant ( which I gather results in infinite Minkowski space ) universe, what happens? Can't a lattice be stable if it is balanced with just the right negative pressure from a vacuum or some other agent? I thought this was stasis in the Gen.Rel. context.
Peter Donis Says:
"(1) If there is no cosmological constant, the stress-energy of the objects in the lattice will produce attractive gravity, meaning the lattice can't remain motionless for more than an instant."
Here I am confused. I must be misunderstanding the cosmological constant. If there is no cosmological constant then it equals zero. If the stress-energy of the objects in the lattice produce attractive gravity, then it does not equal zero. This seems like a contradiction.
Peter Donis Says:
"(2) If there is a cosmological constant, the lattice can be motionless for more than an instant if the repulsive gravity due to the cosmological constant exactly cancels the attractive gravity of the objects in the lattice; but this can only be the case for a finite lattice."

Here I am again confused. If the repulsive gravity due to the cosmological constant exactly cancels the attractive gravity of the objects in the lattice, then why doesn't the cosmological constant equal zero? What afflicts an infinite lattice that does not afflict a finite lattice when a finite lattice can be so large as to be indistinguishable from an infinite lattice?
I gather here that the lattice objects are not mere dust, but carry information requiring nonzero stress-energy. If the vacuum has nonzero stress-energy, does it also carry information? Is there vacuum information and can it stabilize our lattice?

2. Aug 11, 2015

### Staff: Mentor

The cosmological constant is not the same thing as the stress-energy of the objects in the lattice. The cosmological constant (aka "dark energy") produces repulsive gravity. The stress-energy of the objects in the lattice (like any other ordinary matter or energy) produces attractive gravity.

If the cosmological constant is zero, there is no repulsive gravity. See above.

In the previous thread, I referred to the the Einstein static universe, which is the solution of the Einstein Field Equation in which the repulsive gravity due to a positive cosmological constant exactly balances the attractive gravity due to ordinary matter and energy (which is what the objects in the lattice are). The Einstein static universe has a finite spatial volume (spatially it is a 3-sphere, like a closed FRW universe), and its volume is fixed by the density of ordinary matter and energy in it (since that fixes the cosmological constant, because the effects of the two must exactly balance). In the lattice example, this means that once you've fixed the lattice density (how many objects per unit volume, and how much energy each object contains), you've fixed the entire solution, including the volume of the universe.

Furthermore, the volume of the universe in this solution goes down as the lattice density goes up. So a finite lattice that is so large as to be indistinguishable from an infinite lattice must also have a negligibly small lattice density, i.e., the lattice will be indistinguishable from empty space. In other words, the "limiting case" of this solution which is spatially infinite in size is not an infinite lattice; it's infinite empty space.

Yes. (They have to carry information in order to be relevant to the topic of the previous thread, which was the Bekenstein bound on information storage in an object. And they have to have nonzero stress-energy because anything carrying information has to.)

The answer to the second question is easier: no. Whether or not the energy of the vacuum (i.e., the cosmological constant) can carry information does not change the properties of the solution I described above.

As for the first question, I have not seen a good treatment of entropy in curved spacetime that includes a cosmological constant. However, the fact that it is constant at least suggests that it can't carry information; for it to carry information, something would have to vary, and if it's constant nothing can vary. This is just a heuristic argument, though, and there might well be subtleties when quantum effects are taken into account. Unfortunately, we don't have an accepted theory of quantum gravity (yet), so we can't really answer the question at this point.

3. Aug 11, 2015

### Staff: Mentor

I missed this on first reading. If the cosmological constant is zero, there is no way to stabilize the lattice. Any effect that could stabilize the lattice will appear as a nonzero (positive) cosmological constant in the Einstein Field Equation. It doesn't matter whether you call it "energy of the vacuum" or "dark energy" or something else; as far as the EFE is concerned, it's a cosmological constant.

This solution is the Einstein static universe, which, as I described, stabilizes the lattice with a cosmological constant. Again, it doesn't matter whether you call it "negative pressure from a vacuum or some other agent"; it's a CC as far as the EFE is concerned. The properties of this solution are as I described them in my previous post.

4. Aug 11, 2015

### PAllen

Relevant to this discussion is the following fascinating paper Markus Hanke posted in a recent thread:

http://arxiv.org/abs/1208.1411

This only covers the case of no cosmological constant, but just like an FLRW solution, the lattice expands.

5. Aug 12, 2015

### pervect

Staff Emeritus
I'm not sure I know what specifically is meant by an "infinite motionless lattice" without some references or discussion to make sure everyone's using words to mean the same thing. But what I can say is that the current Lambda-CDM cosmology has, among other features, an event horizon, unlike Minkowskii space, which does not have an event horizon.

An event horizon is defined as the region that light emitted at time t can reach given infinte time, see for example http://arxiv.org/abs/astro-ph/0310808. In Minkowskii space, the flat space-time of special relativity, there is no event horizon. Given infinite time, a light signal emitted at time t will reach all points in space. In the current consensus model of cosmology, the Lambda-CDM model, there is an event horizon, a diagram of which can be found in the referenced paper.

At the moment I'm not sure about the "why" part, I'm trying to clarify the question, and provide a proof that it is, in fact, our universe can't be described by an "infinite motionless lattice". As a sub-point, I'm restricting the question to our universe, both for ease of addressing the problem and because I rather suspect that Minkowskii space would be a counter-example, so that it would be too broad to say that infinite motionless lattices can't exist, but it's not too broad to say that our universe (according to the current consensus cosmological model) doesn't qualify.

6. Aug 12, 2015

### Staff: Mentor

No, it wouldn't. Minkowski spacetime contains zero stress-energy; the "lattice" being asked about is composed of objects with nonzero stress-energy.

7. Aug 13, 2015

### votingmachine

Hmmm. In my head, it seems that infinite makes that not true. There is no center of mass. The pull in any direction for any piece is identical.

It is not a possible REAL thing, but he did say infinite ...

8. Aug 13, 2015

### Staff: Mentor

Only if there is a valid solution of the applicable equations to which "infinite" applies and that is also motionless. There isn't.

If what he said isn't a valid solution to the applicable equations, then it isn't a valid solution. Just "saying" it's infinite does not make it valid.

9. Aug 13, 2015

### votingmachine

Of course. But if someone says something correct, it is important not to say it is incorrect, but to say that it is a meaningless question.

A truly infinite matrix of identical elements, stationary with respect to each other would not collapse. Consider that the atom at the center of a uniform sphere is at the center of mass, and is uniformly attracted in every direction. In an infinite matrix, EVERY point is at the center of an infinitely large sphere.

That is definitely a bit off-topic. I agree that a "negligible" attraction for ANY finite distribution of elements stationary with respect to each other would eventually collapse to the center of mass. When I read the comment, the "infinite" part was still in my mind, and that is different from finite. I'm not sure why that difference is not exactly what I said. If there is no valid math solution, then there is no valid math solution. That would still create a small need for a cautionary note, if only to say that there is no valid math solution to suggest an infinite matrix would collapse. I think it wouldn't.

But there is a strict prohibition in this forum against thinking thoughts the venture into philosophy, so regard my comment as more that the original formulation of the matrix as infinite is one which is not one that can exist, and therefore, any real matrix has an eventual gravitational attraction problem. I can see that he gave a correct answer to an incorrect question. I thought that was all implied in the "Hmmm."

10. Aug 13, 2015

### Staff: Mentor

Um, what?

You are incorrect. If you try to find a solution of the Einstein Field Equation that has this property, you will find that you can't. If you disagree, please show me such a solution, explicitly.

I understand that this reasoning seems intuitively plausible to you. Unfortunately, in physics, intuitive plausibility is not enough. You need to actually work out the details, using the physical theory that applies. If you do, you will see that the above reasoning is incorrect, however intuitively plausible it seems.

No, he gave an incorrect answer to a question that has a correct answer--the one I gave just above.

(And, by the way, the reason for what I'm saying in all the above is not that a spatially infinite universe is impossible in GR. It isn't; according to our best current model, our own universe is spatially infinite. But it is also expanding, i.e., it is not "motionless". It's the "infinite and motionless" part that is impossible, not the "infinite" part by itself.)

Last edited: Aug 13, 2015
11. Aug 13, 2015

### votingmachine

I could not begin to work out the Einstein Field Equations. So if they contradict this, then I accept that. I was thinking in terms of the Newton equations for gravitation ... and I thought that equation would apply here, with low attraction and zero velocity (stationary relative to each other).

With that gravitational math, every element is symmetrically surrounded by the same attraction in every direction. There is no center to an infinite lattice. I get that it is an impossible hypothetical, I thought that was well answered in the preceding thread.

I completely agree that any finite lattice has a center of mass, and the gravitational pull, no matter how negligible will move the elements to that center.

While I don't have the math background to solve the Einstein Field equations, what is the basis for the force imbalance that leads to collapse? If every element feels a symmetrical force, it has no net force, and won't move.

12. Aug 13, 2015

### votingmachine

I think there is still some misunderstanding about what I was saying in the prior thread. There was a comment that a very large finite lattice would approximate an infinite lattice. There were some very good comments made. Then there was a comment that I interpreted to answer the question about a "very large finite lattice" correctly, but in the context of "an infinite lattice".

The correct answer (the one given) for the finite lattice is that it eventually collapses to the center of mass. But a large finite lattice will never be the same as an infinite lattice. And it surprises me that a hypothetical infinite lattice is not stable.

13. Aug 13, 2015

### PAllen

The paper I linked in #4 discusses precisely the class of solution for infinite (or unbounded, in some cases) lattice massive objects without cosmological constant. It shows that each such solution matches the behavior of an FLRW solution with nearly the same overall energy density. Thus, it either expands forever or expands, then contracts, depending on density. There is no possible static solution (again, without cosmological constant).

14. Aug 13, 2015

### votingmachine

That makes sense to me. But I can't reconcile that with the idea that the infinite lattice is gravitationally unstable.

I fully agree that the hypothetical is impossible. I keep getting tripped up by the "so large as to be indistinguishable from an infinite" parts. I guess I keep seeing a distinction. Your relegation of the mass to nothing seems to be a good way to wipe out the distinction. But a mass-less thing should not be susceptible to gravitational collapse (should it?).

15. Aug 13, 2015

### PAllen

Maybe the way to break the Newtonian intuition that an infinite lattice could be stable is to realize that Newtonian gravity is based on fixed background space, while in GR, the geometry of spacetime is dynamic and coupled to matter/energy. Einstein went looking for exactly such a static solution, based on Newtonian expectation, and was (initially) horrified to discover there could not be such a solution of the field equations without introducing a cosmological constant.

Another counter-intuitive consequence in GR is that once you give up that an infinite lattice could be stable, you would think that collapse is the likely outcome (not towards a center, which doesn't exist, but of the the whole 'universe' collectively, that is a big crunch cosmology). However, this intuition is again wrong. Any maximally extended solution must start out expanding, with later collapse depending on average density of the lattice.

16. Aug 13, 2015

### votingmachine

That does help. I know it is a strange and impossible hypothetical, but it seemed putting infinity as the dimensions of the uniform lattice, it made it essentially the Newtonian space. I was not very interested in the information side of the discussion, and the practical objections seemed well covered.

This is one of those times when the insight into the difference between Newton and Einstein seems particularly stimulating. MANY thanks to people behind the comments made.

17. Aug 13, 2015

### Staff: Mentor

It doesn't. "Low attraction" and low speeds are necessary conditions for Newtonian gravity to be a good approximation, but they are not sufficient. You also need an isolated system of finite extent surrounded by empty space. If you have an infinite lattice, you obviously don't meet that latter requirement.

A "mass-less thing" is empty space. As soon as you postulate a "lattice" made of something, anything at all, it isn't empty space. You can't have your cake and eat it too: if you want "something" to be present, you can't assume it has no mass; more precisely, you can't assume it has no stress-energy. And if it has stress-energy, it has gravitational effects.

18. Aug 13, 2015

### votingmachine

I see that now. The simplicity of the small masses at 1 light year spacing, led me to seeing them as a pure Euclidean system, with the masses at every integer location. A Newtonian approximation has a built in error, and in this case the error is important.

I see that now. It is again a case for not using an approximation. Things need to be as simple as possible, but not simpler. I took it a step too far.

In another thread there was a link to Bell's Spaceship Paradox, and I had exactly the reaction in that story, as the consensus of the scientists, who also said the thread was fine. And then you go thru the math, and it breaks. I see that this is another case where one has to see the full story ... no shortcuts. Moments where a thread provides an insight are rare, and this makes twice ... so hurrah for these boards.