Is Newtonian gravity consistent with an infinite universe?

Question

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.

Answer 1

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.

Answer 2

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...

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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"...

Answer 3

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.