What is the depth of the gravitational potential of the universe?

[Halc]

I've not seen this question posed before, and I've approached the question from different perspectives yielding different answers.

What is the difference in gravitational potential from here on Earth to zero potential (the hypothetical potential of an empty universe)?

From a mathematical standpoint, it seems to be infinitely negative.  The escape velocity from a 1D infinite length rod drops off at 1/r, and so if gravity is X at the rod, it is 1/2x at twice the radius of the rod. The series ∑ 1/n does not converge.  With a 2D infinite sheet, gravity does not fall off at all with distance, and the series ∑ n certainly does not converge. All these abstract cases have infinite gravitational depth. The universe on the other hand is a 3D uniform distribution of matter at a current density of around 6 protons per cubic meter.

Imagine we choose a random orientation for a plane passing through Earth and all mass on one side of that plane is put on the other side where its reflection appears.  This does not change the potential at Earth one big since all matter is at the same distance it was before. But now there is infinite gravity (weight) at the plane, and it doesn't drop off with altitude. Hence the absolute depth of anywhere being infinitely negative.

There doesn't seem to be an event horizon involved in any of the cases above since there's no altitude where some finite escape velocity exists.

Is my argument sound?  I can think of a couple different ways to counter it:

1) Gravitational potential is only a function of the mass within the visible universe, a mass that is continuously growing as the visible universe encompasses new material, but a depth that perhaps is lessening due to the reduction in density over time. This approach implies that gravity (as opposed to gravitational waves) 'travels' and therefore the mass outside the visible universe does not contribute to Earth's gravitational potential.

2) I see arguments against the big bang citing that it violates the conservation of energy with all this insane energy density coming from nowhere. But if everything is at -infinite potential, every gram of mass actually has infinitely negative PE and only finite positive energy, so the question becomes: "If energy is conserved, where did all the energy go?  Why is it so negative?".  If this argument holds water, and I'm not sure it does because we have zero evidence of a different energy level from whatever precipitated the big bang, then the negative potential can probably be computed from whatever is needed to exactly cancel all the positive energy (mostly dark energy I'd imagine) we can find.

Edited February 27, 2021 by Halc

[swansont]

I think you are over-complicating this with your non-convergent series. It’s a 3D problem. The potential due to mass M at r is -GM/r. It’s zero at infinity. 

 

[Halc]

1 hour ago, swansont said:

I think you are over-complicating this with your non-convergent series. It’s a 3D problem. The potential due to mass M at r is -GM/r. It’s zero at infinity. 

You're not doing the mathematics at all. Sure, I illustrated infinite potential even with the linear and planar case (even Newton knew this), but the uniform distribution of mass in 3D is even worse.

Your -GM/r is for a point mass (or spherical) from which you can distance yourself. Yes, there is finite potential for any one finite mass, but the mass of the universe is not finite and not at some fixed radius.

OK, we can do it another way using Newton's shell theorem.  At a large enough scale, matter is distributed uniformly in the universe so the shell theorem is valid.  Imagine a thin shell of material going from radius r to r+1.  Per the formula, it yields a potential of -GM/r where M is the mass contained in that shell.  Call this potential X.  Consider a different shell that goes from 2r to 2r+1.  That shell has 4 times the mass of the first shell and only twice the radius, so it has -G4M/2r potential which is twice the potential.  The further away the mass is, the more it's collective influence on our potential, which results in the infinity-squared I suggesting in the OP.

Edited February 28, 2021 by Halc

[MigL]

The gravitational potential of the universe would be manifest  by the global curvature.
Global curvature is, to my knowledge, very nearly flat.
( or within error bars of any measurements made, so far )

And I don't think you can make a distinction between observable and total universe.
The global curvature would have been established when no longer observable parts of the univere were in causal contact.
The gravitational field has no reason to propagate; only changes do.

[Markus Hanke]

7 hours ago, Halc said:

Yes, there is finite potential for any one finite mass, but the mass of the universe is not finite and not at some fixed radius.

The notion of ‘gravitational potential’ can be meaningful defined only in spacetimes which are (among other requirements) stationary, i.e. in spacetimes that, in mathematically precise terms, admit a time-like Killing vector field. The universe in its entirety is approximately described by an FLRW spacetime, which does not fulfil this crucial condition. So the concept of ‘gravitational potential of the universe’ is meaningless, which is why you weren’t able to find anything on this topic. 

[swansont]

11 hours ago, Halc said:

You're not doing the mathematics at all. Sure, I illustrated infinite potential even with the linear and planar case (even Newton knew this), but the uniform distribution of mass in 3D is even worse.

 

A uniform mass distribution would not have a change in gravitational potential as your position changed. 

[SergUpstart]

16 hours ago, Halc said:

I see arguments against the big bang citing that it violates the conservation of energy with all this insane energy density coming from nowhere.

The Heisenberg uncertainty relation also violates the law of conservation of energy for time delta t.

[Halc]

12 hours ago, Markus Hanke said:

The notion of ‘gravitational potential’ can be meaningful defined only in spacetimes which are (among other requirements) stationary, i.e. in spacetimes that, in mathematically precise terms, admit a time-like Killing vector field. The universe in its entirety is approximately described by an FLRW spacetime, which does not fulfil this crucial condition. So the concept of ‘gravitational potential of the universe’ is meaningless, which is why you weren’t able to find anything on this topic. 

Thanks Markus!  +1

You seem to be the only responder that at least understood the question.  So I'm going to ramble a bit, part of which probably isn't worded as precisely as it should.

'Gravitation potential' is a local concept then, relative to say a system that is stationary like Earth, where one can speak meaningfully of the relative potential difference between sea level, geosync orbit, and the hole we've hollowed out in the center, but all potential is relative to some reference (sea level say) and involves whatever components (Earth, sun, galaxy) we decide to include in our calculation.

A universe described in its entirety by Minkowski spacetime (infinite uniform distribution of superclusters, not receding from each other, but evolving in place, so gravity wells, but not on the largest scale) would admit this time-like Killing vector field since the matter distribution wouldn't change significantly over time. The density in particular would be fixed, and in such a universe, my argument might be meaningful, but not in this universe.

[Curious_dog]

On 2/28/2021 at 1:57 PM, Halc said:

...A universe described in its entirety by Minkowski spacetime (infinite uniform distribution of superclusters, not receding from each other, but evolving in place, so gravity wells, but not on the largest scale) would admit this time-like Killing vector field since the matter distribution wouldn't change significantly over time. The density in particular would be fixed, and in such a universe, my argument might be meaningful, but not in this universe.

Can't we make amends? Can the average gravitational potential inside our Universe within some finite time frame be computed on an assumption the Universe is frozen within this time frame and the Killing vector field can be (approximately) defined? If the Universe were not expanding, and we could treat it as having finite dimension and uniform matter distribution on macro scale, we would get the simple expression for the potential relative to the Universe boundaries (or "center"). 

I wonder if the assumption of the infinite Universe leads to the accelerated expansion as to get density reduction with the distance, resulting in the finite gravitational energy within the Universe (which will likely be constant throughout it on macro scale, so no force).

[Mordred]

As Markus mentioned the question is rather meaningless. We can only measure differences in potential from one geometric location to another.

 There is technically no limit to the amount of  difference involved. Even if you took the entire mass of the universe and calculated that mass within a single infinitesimal and compared that to a region of zero mass density you still would not have a determination of an upper limit.