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Question about gravity and the shell theorem

  1. Dec 18, 2014 #1
    Consider a point in the universe I'll call X and a test particle some distance R away. For the purposes of calculating the gravitational acceleration of the test particle I'll consider all the matter within radius R of point X to be a point mass attracting the particle to X. Furthermore, all the gravity from all the matter outside this sphere should cancel out per the shell theorem. So, I conclude that the test particle accelerates towards X.

    But that can't be right. in a homogeneous and isotropic universe the test particle wont prefer to accelerate in any particular direction because they are all the same. What is wrong with the reasoning in my first paragraph?
     
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  3. Dec 18, 2014 #2

    Bandersnatch

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    The gravitational force from the matter outside R only cancels out if said matter is in uniform spherical distribution. Can you say that about matter distribution in a homogeneous and isotropic universe?
     
  4. Dec 18, 2014 #3

    ShayanJ

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    Shell theorem is a consequence of inverse square law of gravity which is true only in the weak field and low speed limit of General Relativity.
    (I also give a vote for Bandersnatch's objection! In fact that's the main thing.)
     
  5. Dec 18, 2014 #4

    PeterDonis

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    The first obvious question to ask is, what if the radius R is zero? I.e., what if the test particle itself is at point X? In that case all the gravity from the matter everywhere else cancels out and the test particle remains at point X. (Yes, the shell theorem is valid in this case--see below.)

    The second question is then: why should changing where we pick point X to be change the behavior of our test particle? Obviously it shouldn't. So we must be making some error in our analysis of the case where R is nonzero. It isn't the shell theorem (again, see below). So it must be the other part, where we calculate the motion of the test particle relative to point X when R is nonzero.

    Try this way of looking at it: put a second test particle at point X, so we have two test particles a distance R apart, both in free fall. Are they at rest relative to each other? No; they are not. Both test particles are following "comoving" worldlines (by hypothesis), and comoving worldlines are moving apart, because the universe is expanding. But the claim that the gravity of the "point mass" at X would make the test particle at radius R from X fall towards it was, implicitly, based on an assumption that the test particle starts out at rest relative to point X. In fact, it isn't; it's moving outward. So the gravity of the "point mass" at X will not cause an acceleration of inward motion of the test particle; it will cause a deceleration of the outward motion of the test particle.

    And, in fact, if we look at what happens in a matter-dominated universe (which we have also been implicitly assuming, based on how we expect the gravity of the point mass at X to act), this is exactly what happens! The matter in the universe causes the expansion of the universe to decelerate. And the scenario we've just described explains why, from a local viewpoint; in other words, it explains why an observer at X would compute that the test particle at R should decelerate its outward motion, relative to him.

    Yes. A homogeneous and isotropic universe is spherically symmetric about any point. So this escape route won't work.

    The Newtonian version of the shell theorem is a consequence of the inverse square law of gravity. The GR version of the shell theorem is true in any spherically symmetric spacetime. So this escape route won't work either. But fortunately you don't need an escape route; see above.
     
  6. Dec 18, 2014 #5

    Bandersnatch

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    I was hoping to elicit this answer from the OP. Once this is realised, it's easy to exclude the influence of the mass that's supposed to be attracting the test particle simply by choosing a different mass for which the previous one constitutes the shell.
     
  7. Dec 18, 2014 #6

    PeterDonis

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    I'm not sure exactly what you mean by "choosing a different mass", unless it's just to make the first point I made, that if we put the test particle at point X all the mass in the universe is in the "shell".

    But in any case, as the rest of my post shows, the intuition that, if we put the test particle somewhere other than point X, the mass closer to X can be treated as a point mass at X, is correct. That mass does attract the test particle. The incorrect implicit assumption in the OP's post was that the test particle starts at rest relative to X. In fact, it's moving outward, so the effect of the "point mass" at X is to decelerate it.
     
  8. Dec 18, 2014 #7

    Bandersnatch

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    Yes. No disagreement with your posts, at least none intended. Although I'd rather let the OP get there himself.
     
  9. Dec 18, 2014 #8

    marcus

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    I think Bandersnatch, Peter, and Shyan have addressed this and there probably isn't anything anyone needs to add. I just want to mention something related in an odd way, to your question about the shell theorem. This is about Newtonian gravity in ordinary Euclidean space. Something fun to consider.

    Imagine a infinite block of material with uniform density (say Euclidean R3 filled with material the average density of the Earth.)
    Imagine a spherical bubble the size of the Earth removed from it. You and your friends get to explore inside that hollow sphere. What is the gravity like?

    It is the same as normal Earth gravity, only reversed, so it is outwards. You walk on the inside of the sphere and your weight feels the same.

    (I am neglecting the fact that the Earth is not a perfect sphere and its sealevel gravity varies with latitude due to rotation etc etc. Let's idealize and speak approximately : ^)

    At first this seems funny because wouldn't the complement of a round ball be LIKE a very thick shell? Why wouldn't there be zero gravity inside the bubble?

    But it has to be that way because ball that we imagined being REMOVED was the size and density of the Earth. So its gravity would have to exactly cancel the gravity inside the hollow ball you are exploring. Gravity inside the hollow ball has to be equal and opposite to that on Earth.
     
    Last edited: Dec 18, 2014
  10. Dec 18, 2014 #9

    PeterDonis

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    Isn't this impossible in Newtonian gravity? The gravitational potential energy would be minus infinity at every point.
     
  11. Dec 18, 2014 #10

    ShayanJ

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    I see it as redefining vacuum and then that bubble becomes a spherical distribution of negative mass.
     
  12. Dec 18, 2014 #11

    marcus

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    I think you can see how to make the example more like our actual universe :)
    Use a different density, much lower than that of Earth. Closer to the actual average density of matter we measure. If you like, say it is Einstein gravity. If you want, adjust the cosmological constant the way Einstein originally wanted (it is an ideal example) so that space can be flat and doesn't have to be expanding. Or let it be very gradually expanding. It doesn't matter. The idealized example is of an infinite flat universe, filled with uniform density ("homogeneous and isotropic"), with a spherical bubble removed.
    You can make it as close as you want to the familiar infinite LambdaCDM case, homogeneous and isotropic except for a bubble removed. The gravity outwards on the inside of the bubble must be equal and opposite to the gravity inwards on the surface of the ball that was removed.
    That is the puzzle---I don't want to intrude to much in Speedybob's thread, but wanted him to have this puzzle which is related to the one he originally asked about.
     
  13. Dec 18, 2014 #12

    PeterDonis

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    If you intend this as an actual puzzle, I'll refrain from commenting to give the OP a chance to respond. But just for the record, I do not think the gravity on the inside of the "bubble", if the mass inside the bubble were removed, would be as you describe; I think it would be zero, just as the shell theorem says.
     
  14. Dec 18, 2014 #13

    marcus

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    Excellent, Peter! I think it is a good idea to let Speedybob have time to think, let us know if he is still interested, if he has any questions? Sees a flaw in his original problem?
    I hope he's still around, but if he's moved on that leaves the rest of us to hash it out.
     
  15. Dec 19, 2014 #14

    marcus

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    Hi Mr. Speedybob,
    I thought you asked an intriguing question. Hoping you'll come back to it, ask more questions, guess at an answer. We kind of paused to see if you would comment, or draw some conclusions. If not then maybe later today others will try various ways to resolve the puzzle. (Or do you think it has already been resolved? How?)

    I think to make the simplest assumption about context we should think of the standard LCDM cosmic model which is spatially flat infinite universe. And homogenous isotropic of course. That's the model cosmologists most often use and having it infinite makes it simpler to think about shells.

    I was curious about what I highlighted. Can you explain why "gravity from all the matter outside this sphere should cancel out per the shell theorem" ?

    Why can you apply the shell theorem when the shell is, so to speak, infinitely thick? Was it proved for that case? Hoping you are still around and still wondering about this.
     
  16. Dec 19, 2014 #15
    That is an intriguing question. Both your conclusion in post 8 and Shyan's thought in post 10 make sense, and I can't find fault with them. On the other hand gravity is zero inside a shell, so should be zero inside 2 concentric shells, or 3 or 4 or infinite.
    I'm still pondering...
     
  17. Dec 19, 2014 #16

    marcus

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    Could it be the "infinite" that causes trouble?
    You know how infinite series can be summed in various ways to give various different answers?
     
  18. Dec 19, 2014 #17

    marcus

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    Have you looked at how the shell theorem is proved? Maybe it is proved only for shells with a definite thickness? An infinite "shell" is hardly a real shell, isn't it? It has no definite thickness and without that something might go wrong in the proof. I guess that's probably just one way to look at it, and there must be more pictorial intuitive ways.

    Here's a thought. You are imagining filling infinite space with concentric spheres, perfect spheres. Suppose instead you fill it with lopsided spheres all of whose walls have been thickened on the East side, but they fit together perfectly because each one is swollen similarly on that side. Then each misshapen shell exerts a slight net pull to the east. So when you add their effects up you get a (possibly infinite) pull to the east. the fact that space is infinite lets you divide it up different ways.

    Here's a numerical analogy. -1,-1,-1,0,1,1,1 (each -1 is a pull to the west, each 1 is a pull to the east) they add up to zero
    As long as the double sided sequence is finite and symmetrical it will add to zero. The opposite pulls cancel.
    -1,-1,-1,-1,0,1,1,1,1 ? Yes. But what about the infinite double sided series that goes out to infinity in either direction?
    .......-1,-1,-1,-1,0,1,1,1,1......

    Suppose I add it up this way
    .......-1,-1,(-1,(-1,0,1,1),1,1)......
    Each "sphere" or "shell" is giving me a net +1 pull to the east. And I have infinite resources so I can use up faster on the right and never run out.

    What I'm saying is the sum of the doubly infinite series is not defined

    ......(-1,(-1,(-1,(-1,0,1,1),1,1)1,1,)1,1,)...... This already gives a +4 pull to the east.

    But if I arbitrarily decided to add it up in some other way, I could get zero every time. The way we did when it was finite and symmetric.
     
  19. Dec 19, 2014 #18

    marcus

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    the problem is with the infinity. The lack of a definite thickness of the shell. So the "shell theorem" does not apply.
    But there are other ways to calculate the gravity inside the spherical void :w
     
  20. Dec 19, 2014 #19

    PeterDonis

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    This will work if we're considering how to answer the question in GR; but not if we're considering how to answer the question using Newtonian gravity.

    (I should clarify that the answer I gave in an earlier post, that I think gravity inside the "bubble" will be zero, was a GR answer, not a Newtonian answer. I'm still not sure the infinite case can be consistently modeled using Newtonian gravity.)

    (I should also clarify that my "GR answer" was assuming a matter-dominated universe with zero cosmological constant. So it wasn't quite the same model as the standard LCDM model.)
     
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