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  4. Is the Universe in a net-zero state, balancing energy against gravity?
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Is the Universe in a net-zero state, balancing energy against gravity?

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Offline paul cotter (OP)

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Is the Universe in a net-zero state, balancing energy against gravity?
« on: 03/06/2022 10:50:03 »
Some years ago I saw on a tv program the late Stephen Hawking claim that all the mass/energy in the universe is balanced by negative gravitational energy and the net sum is zero. I fully understand the principle of the negative gravitational but I don't see how a weak force like gravity could be the causation of a negation of the enormous values deriving from e=m(csq).
« Last Edit: 08/06/2022 16:06:07 by chris »
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Offline Halc

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Re: Is the Universe in a net-zero state, balancing energy against gravity?
« Reply #1 on: 03/06/2022 13:37:06 »
Quote from: paul cotter on 03/06/2022 10:50:03
I don't see how a weak force like gravity could be the causation of a negation of the enormous values deriving from e=m(csq).
That's like saying that raindrops are so small and I'm so big, therefore drowning isn't a possibility.

I tried computing this, and my reaction was how the positive energy of a rock could possibly make a dent in the negative gravitational potential energy of that rock. I got a net-negative, and not by a little bit either, so I wondered how there can be enough positive energy to bring that back up to zero.

Not hard to do: Use the shell theorem. Don't need a calculator.
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Offline paul cotter (OP)

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Re: Is the Universe in a net-zero state, balancing energy against gravity?
« Reply #2 on: 07/06/2022 09:10:03 »
I need some clarification here. Can I define the gravitational negative energy as the work needed from contact  to separation at infinity(to simplify the integration) for a simple universe or is there other factors at play? PS never heard of the shell theorem before, however I would a always assume point masses, if only to simplify the maths.
« Last Edit: 07/06/2022 09:15:02 by paul cotter »
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Offline evan_au

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Re: Is the Universe in a net-zero state, balancing energy against gravity?
« Reply #3 on: 07/06/2022 10:39:20 »
Quote from: paul cotter
never heard of the shell theorem
The Shell theorem was developed by Isaac Newton, and used when calculating the gravitational filed inside or outside a spherical object (eg the Earth or the Sun).
- As far as I know, Newton didn't try applying it to the whole universe, but if you assume that the universe is isotropic, I guess you could...
https://en.wikipedia.org/wiki/Shell_theorem
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Offline Halc

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Re: Is the Universe in a net-zero state, balancing energy against gravity?
« Reply #4 on: 07/06/2022 13:30:54 »
Quote from: paul cotter on 07/06/2022 09:10:03
I need some clarification here. Can I define the gravitational negative energy as the work needed from contact  to separation at infinity(to simplify the integration) for a simple universe or is there other factors at play?
'from contact'?  I didn't get that part, but I get your point. We have some arbitrary 1kg rock here on Earth, and if Earth was in isolation in the universe, the negative PE of the rock is the same as the positive KE it would take to get it off Earth, which is about 11 km/sec.  Problem is, Earth is not in isolation, and if you ejected the rock at that speed, it would indeed indefinitely leave Earth, but it would not leave the solar system, so it still has negative PE.
So your definition above only works for masses in isolation, and the universe isn't a mass in isolation. No matter how far away I get my rock, it is always going to be near some mass or other and will be getting negative PE from it. There is nowhere to aim to get it to PE zero, a location infinitely distant from all mass, except to wait an infinitely long time to let all the mass expand infinitely far from you. If there's no such location, how do we know the negative PE of our rock?

Quote
PS never heard of the shell theorem before, however I would a always assume point masses, if only to simplify the maths.
Actually, the shell theorem (Newton) assumes a homogeneous mass for the gravitating object. The universe is homogeneous at sufficiently large scale, so it works, sort of. Newton used the technique to demonstrate that your weight within a planet (say you at the bottom of a 1000 km hole) depends completely on the material within your radius and not at all on the material at higher altitude than you. He also used it to show that a spherically distributed mass can act as a point mass. I'm not doing that, but I am borrowing the technique. See below.

Quote from: evan_au on 07/06/2022 10:39:20
Quote from: paul cotter
never heard of the shell theorem
- As far as I know, Newton didn't try applying it to the whole universe, but if you assume that the universe is isotropic, I guess you could...
Homogeneous, not isotropic, but yes. The idea follows, but I think it is flawed.

We divide the universe into shells centered on our rock. The mass of each shell contributes to the negative PE of the rock. Each shell is the same thickness, and is sufficiently distant to be effectively a 2D surface. The universe is homogeneous, so the mass of a shell is proportional to its area.
A shell at radius R contributes -X to our potential. (I'm switching to potential here, not PE. To translate, simply multiply the potential by the mass of the rock).  A shell at radius 2R has 4x the mass but twice the distance, so a -2X contribution to our potential. That means that more distant objects collectively contribute more to our potential than do near ones. The total negative potential of here is thus the sum of X+2X+3X... which is ∞² so to speak, which is considerably more negative than the finite positive energy locked up in the rock via E=mc².

Why this analysis is flawed: It assumes a sort of Minkowskian spacetime, an inertial frame upon which the shell theorem is valid. But spacetime isn't Minkowskian. Distances are not infinite relative to an inertial frame, so the size of the universe is quite finite. Relative to an inertial frame, the universe is anything but homogeneous, it being more dense at greater distances. It has an edge, so the infinite series I specified has a finite number of terms.

You can do the above trick using cosmic coordinates, the coordinate system used when stating that the visible universe is R=~48 BLY, but the gravitational potential being an inverse function of distance measured that way hasn't been derived. Newton's laws of motion don't apply, so energy conservation derived from those laws does not necessarily apply in such a frame. A fast moving rock will slow over time in the absence of any force acting on it. Light loses energy over time (evidenced by the CMB having cooled to microwave wavelengths by now). The potential also tends toward 'zero', wherever that is. It might balance, but you can't say if one cannot meaningfully express the negative potential of a typical location in space.
« Last Edit: 07/06/2022 13:37:43 by Halc »
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Offline paul cotter (OP)

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Re: Is the Universe in a net-zero state, balancing energy against gravity?
« Reply #5 on: 07/06/2022 14:10:54 »
Thank you, halc, I will TRY to digest this.
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Offline Eternal Student

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Re: Is the Universe in a net-zero state, balancing energy against gravity?
« Reply #6 on: 07/06/2022 19:43:09 »
Hi.

   I didn't see the TV program but it sounds like something mentioned in one of his (Stephen Hawkins') books - which I also only have a vague recollection of.

    The basic idea requires you to stop thinking about mass as something that is different from energy.  Recall that these words from Hawkins were prior to the discovery of the Higgs Boson and even with the Higgs Boson we still believe that most of what we call mass and attribute to a particle is actually just energy.   For example, you  ( @paul cotter ) recently mentioned protons as an example of something with mass >> sum of it's constituent quarks since most of it's mass is binding energy due to the close proximity of those quarks.

     Anyway the basic idea is that there should be a conservation of energy in the universe, lets say the total energy in the universe = some fixed constant  and then we might just as well set that constant to  ~0, it's a fairly arbitrary decision anyway.   We allow negative energies,  gravitational potential is usually considered as a negative energy.   There was several paragraphs in the book about why it's so important that gravitational potential is considered to be negative.  Along with some brief mention that negative mass is not allowed (or not observed at the very least), so that mass-energy is always positive.   I can't recall all of that - but let's just go with it.

    So, we have most of what Hawkins was stating:    The net sum of energy in the universe (at any time) = 0.

    The rest of the discussion in his book was then about the idea that gravity isn't just a destroyer of structure but also it's a creator of mass-energy and hence of material structures in the universe.    If something (some mass or energy) moves closer to a source of gravitation   (e.g. a black hole at the centre of galaxy) then there is a reduction in the gravitational potential energy,  it becomes more negative.  As such some energy somewhere must grow, it must become more positive,  (since the sum of all energy in the universe = 0)    All positive energy is equivalent to mass  (by  E=mc2 as usual),  so ultimately there is some increase in the mass of the universe.

   Anyway, it's only gravitational potential and similar things like electric potential that can give you negative energy values.   Mass and kinetic energy only gives you positive values,  kinetic energy is just more mass anyway  (if you're old fashioned and remember about relativistic mass or even if you're new and recognise that if you had a compound particle comprised of constituent smaller particles whizzing around inside it then it acts as a compound particle with mass that very much includes the kinetic energy of those fast moving constituents).   We can ignore electric (or magnetic) potential for various reasons - but one of the easiest to see is that there are + and - charges and assuming the universe is homogeneous then these are equally spread out and the net potential energy of a test charge anywhere in the universe is 0.    We can ignore other forces like the strong nuclear force because it's so short range,  only gravity matters on these universe-wide scales.
    So,  for a very simple discussion,  fit for a quick TV interview or his popular but easily accessible books like a Brief History of time we have a set of balancing scales....
     With gravitational potential energy on one side because that's the only negative energy we can have.    Meanwhile there's all the positive energy that can exist on the other side, that energy is equivalent to mass.

   Well, that was the gist of it, to the best of my recollection.   In the books Hawkins used this idea of balancing grav. p.e. against  mass-energy   for various purposes.   For example, it appeared when discussing the big bang and some philosophy for the plausibility of a spontaneous appearance of the universe:  The idea there was that you can start from a situation where there was 0 energy (and 0 mass) but because gravitational potential energy can be negative and can evolve to become more negative, we can (and must) have some positive energy (some mass) appear to balance this.

    It's a rough argument.   It assumes things like conservation of energy, which Hawkins would have known is in dispute on cosmological scales.   I've always assumed it was just a passing discussion or plausibility argument.    You should read the books if you want more detail because I honestly can't remember it well.    Anyway, it still works out in some way if you consider the expanding universe and the cosmological models we have today:   Eventually the universe should expand to the point where everything is infinitely separated and then the gravitational potential of the universe = 0.  We also believe that matter decays and ultimately breaks down to photons which stretch and lose energy until they are of impossibly low energy.  Ultimately you have 0 mass-energy and 0 gravitational potential energy in the universe again.
    So, to re-phrase all of that and end on a more upbeat statement, perhaps we should be grateful that there are some black holes and other sources of gravitation in the universe, they aren't destroying the universe, they are what allowed some mass to exist.

    Best Wishes.
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