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Let us assume that we have have an infinite Newtonian space-time and the universe is uniformly filled with matter of constant density (no fluctuations whatsoever), all of it at rest. By symmetry, the stuff in this universe should not collapse or change position in any way if gravity is the only force acting on it. Now, consider Gauss theorem. It says that within a spherical system (it says more that that but this will suffice), the gravitational force felt by any point will be the same as if all matter between the center and the point were concentrated at the center. The matter outside the sphere does not contribute any force). Thus, in such a system, the matter (stuff, I do not say gas so we can consider it continuous) will collapse towards the center of the sphere. We can apply this argument to any arbitrary point in our previous infinite homogeneous universe, and conclude that matter will collapse towards that point (plus the point is arbitrary). So, why is Gauss's theorem is not valid in this case?

I was signaled that this question is a duplicate and has been answered, however: The best I could get from the redirected question is this quote: "However, the mass can't be negative and the energy density is positive. This would force a violation of the translational symmetry in a uniform Newtonian Universe". It still doesn't give a satisfactory answer. For instance: how is that symmetry broken if we assume that there is no noise nor small density fluctuations in the system? How can you choose then the absolute "origin" that will break the symmetry? Still doesn't make sense to me.

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    Mathematically, the problem is that infinite sums aren't associative, so you can't arbitrarily regroup terms. Physically, what's going on is that this equilibrium is unstable. –  Apr 11 '13 at 03:21
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    @Ben, I already mentioned homogeneity (I know the system is unstable). What I am not sure to understand and you could have a good point is that sums are not associative (and why are they not? I am not sure to understand that). But I am not sure on what part you apply that argument. –  Apr 11 '13 at 03:40
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    See also: "Paradox with Gauss' law when space is uniformly charged everywhere", which is the same thing with electric charge. – Retarded Potential Apr 11 '13 at 18:17
  • The best I could bet from the redirected question is this quote: "However, the mass can't be negative and the energy density is positive. This would force a violation of the translational symmetry in a uniform Newtonian Universe". It still doesn't give a satisfactory answer. For instance: how is that symmetry broken if we assume that there is no noise nor small density fluctuations in the system? How can you choose then the absolute "origin" that will break the symmetry? Still doesn't make sense to me. –  Apr 13 '13 at 18:13
  • @julian I can see why those answers don't prove entirely satisfactory. I made another attempt on that question (avoiding all this translational invariance language): http://physics.stackexchange.com/a/61359/10851 Maybe that will help. –  Apr 17 '13 at 00:53
  • @Chris thanks for the attempt, I'll answer my doubts below your answer. –  Apr 17 '13 at 01:38
  • On p.296-297 in the 1997 ed. of his pop-sci book titled "The Inflationary Universe", Guth gives a simple diagrammatic and algebraic explanation as to why "an infinite distribution of matter under the influence of Newtonian gravity would unquestionably collapse". Its reasoning is simple enough that any paraphrasing of it might approach plagiary, but the book (published by Basic Books) is available thru the libraries of some countries, either directly or via inter-library loan. – Edouard Mar 25 '22 at 03:32

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I think the problem is that your conclusion that 'stuff' outside your Gaussian surface contributes no force is peculiar to the spherical geometry. If you break the pure spherical symmetry of the system, this argument breaks down.

Consider any system that does not have spherical symmetry and you will easily see that objects outside your Gaussian surface CAN exert a force on a point on the edge of your surface. (for example, 2 separate spheres. Surely, any point between them feels a force from both spheres, not just whichever one you arbitrarily put inside your Gaussian surface).

For the case of a truly infinite universe, the symmetry is not really spherical. It may seem like it might be, but a completely infinite homogeneous universe has a symmetry that a finite spherical distribution does not. Namely, that universe in translationally invariant. This extra symmetry is at the heart of your argument that nothing moves and distinguishes it from the spherical case.

Kevin Driscoll
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  • I thought about that, but in an infinite universe, event if euclidean, is translation invariant. And you can always define spheres as large as you want. Then what it is left outside, can also be divided into spheres. If I am not wrong, this is pretty common assumption when we make infinite integrals, either in math or in physics –  Apr 11 '13 at 03:37
  • In the case of integrals, there is a rigorous way to talk about 'closing a path by circle at infinity'. Basically it involves taking a limit. My point was that your theoretical universe IS translationally invariant. Any finite spherical distribution is not. So your application of Gauss' Law does not respect the symmetry of the problem. This is why its conclusions don't hold here. – Kevin Driscoll Apr 11 '13 at 03:45
  • Thanks, I'll try to find that, in the meantime is there any reference that you can give me? I would be very happy to accept your answer if that seems to be the mistake in my argument! –  Apr 11 '13 at 03:51
  • To be more direct, it seems that an 'infinite sphere' is somehow different from a finite one, precisely because the symmetries are different. Also, in the case of integrals extending a path/surface to infinity is usually useful when the contribution to the integral from that path/surface goes to 0. In that sense, the exact shape of the path/surface doesn't matter. – Kevin Driscoll Apr 11 '13 at 03:52
  • Sorry, I don't think I can give you a reference. What I wrote is just my intuition about the problem. – Kevin Driscoll Apr 11 '13 at 03:56
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We can say that the gravitational flux is proportional to mass enclosed by the universe. So, this can be "thought" of as if the whole mass of universe is concentrated at the center (i.e.) It can be concluded that the effect on any other massive body (if any, or if you put any) outside the sphere is the same.

But, the mass cannot be dragged towards the center. Because, we've already assumed the whole mass to be concentrated at the center (like a big bang singularity). Either, there's no mass other than the singularity or some mass is outside our Gaussian sphere which can disturb the system we've assumed, and it can't be spherical...

When we're applying this to our present universe, there are a couple of issues to be noted, that sums up as a disadvantage. The universe (as we know), is not spherical. The observable universe may be. But, we don't know whether there's a boundary or atleast the spherical geometry outside the observable universe. But to support Friedmann's assumptions, we can say that the universe is symmetric.

Then, there are these density fluctuations all over the universe (in fact, Planck's results have also proved that), which you've assumed to be absent. Next, the most necessary observation: the expanding universe. We've already concluded that the expansion of universe can't be stopped just by gravity (i.e) gravity can't overtake our assumption - "dark-energy"...

Waffle's Crazy Peanut
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  • Thanks! but I tried to make clear that I was only interested in a purely theoretical issue (which doesn't apply to our universe because it is not Newtonian). Plus no fluctuations. Think as if I had asked this question in Newton's time! –  Apr 11 '13 at 03:33
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All measures we have about the Large Scale Structure (LSS) of the universe say 'isotropic, infinite and flat' . There is not a single hint that it is not flat. GR apply locally.
In pure Newtonian gravity there is a feature that we can now drop: the speed of propagation of gravity is infinite. Indeed if it is infinite then the universe is unstable.
The other feature of Newtonian gravity we can drop now is 'the universe is infinitely old'.
The universe is 'infinite' or at least the radius is large enough in such a way that the propagation of gravity from the most distant objects are the same as if it is truly infinite. In other words there is no need to consider an enclosing sphere with radius greater than the observed max radius inferred by age of the universe.

Under the pure Newtonian gravity picture the universe is unstable but it is stable if corrected by the present knowledge.
The truth that the LSS is influenced by GR is a preconception theoretically motivated. Data is pointing in other direction. We know that since long time ago exists a huge decouple between data and theory (data account only for 4% of matter, theory for 96% of DE, DM).
'A self-similar model of the Universe unveils the nature of dark energy (pdf)' is a document at vixra that presents formally a model without artifacts, i.e. perfect agreement between data and model.

The model is deducted from two observational results (expansion of space and invariance of constants) and has just one parameter, the Hubble parameter.

Helder Velez
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