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A Classical gravity on a torus

  1. Nov 15, 2016 #1
    Some time ago I tried to define classical inverse-square gravity on a 3-dimensional (cubical) torus T3: the quotient space obtained by identifying opposite faces of a unit cube. (Or more rigorously, the quotient space

    T3 = R3/Z3

    of R3 by its subgroup Z3 of integer points.)

    I assumed there was a unit point mass (call it mass A) at (0, 0, 0) and tried to define the force on another unit point mass (call it mass B) at an arbitrary point (x, y, z). My method — admittedly naïve — was to sum up the inverse-square forces from a copy of mass A at every integer point

    (K, L, M) ∈ Z3

    acting on mass B. I hoped that — if summed cleverly — the forces from all these copies of mass A would mostly cancel, and the triply-infinite sum would converge.

    That is, I was trying to sum the force terms

    -(x-K, y-L, z-M) / ((x-K)2 + (y-L)2 + (z-M)2)3/2

    over the array of all integer points

    (K, L, M) ∈ Z3

    But I tried every method I could think of: spherical shells, cubical shells, etc. . . . and clearly none of these summation schemes converged.

    QUESTION: Is there a smart way to arrange for these forces to converge? Or perhaps this force method is just wrong and I should be using a potential function method instead?

    Or perhaps it is known that there is no way to define classical gravity on a 3-torus? (And what about a 2-dimensional torus — and what about an n-torus for a dimension n greater than 3 ?)
     
    Last edited: Nov 15, 2016
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  3. Nov 17, 2016 #2

    mfb

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    Crystals have a similar problem, where the crystal structure replaces the spacetime wrapping. This leads to the concept of the Madelung constant, where you also have to be careful with the integration order, but you can get convergence.
     
  4. Nov 17, 2016 #3
    The trouble with the Madelung stuff is that it's based on adding up both positive and negative charges — which have a much better chance to cancel so as to converge, which they indeed do. But gravity doesn't have that advantage.
     
  5. Nov 17, 2016 #4

    mfb

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    You have the different directions which perform a similar role as you calculate forces not potentials (the potential would diverge for sure).

    Heuristic argument: The number of masses in a shell grows with r2, but the forces scale with r-2. If you would add their magnitude, every shell would contribute the same, but the contributions should cancel at least as good as their square root, leaving you with an overall contribution of r-1 from that shell. As different shells will have a different random direction, giving more cancellation, the overall integral should converge.
     
  6. Nov 17, 2016 #5
    Having tried that kind of thing I was not able to achieve convergence. If you do a numerical simulation and get convergence, or find a proof that it must converge — when it is summed up in a certain order — please let me know!
     
  7. Nov 18, 2016 #6
    "You have the different directions which perform a similar role as you calculate forces not potentials (the potential would diverge for sure)."

    But I'm not sure the potential should be summed over anything. Maybe just assume that near the point mass at the origin it has to look like

    P(v) = -1/|v|​

    locally, and that P has to be a harmonic function on the torus.
     
  8. Nov 18, 2016 #7

    mfb

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    Let's assume we can extend classical gravity naturally to your space. Both a local minimum and a local maximum of the field (apart from the mass points) violate Laplace's equation. We want minima at the mass points to get something interesting, the potential should be continuous and our space is compact, which means there has to be a maximum somewhere. Contradiction. You cannot reproduce classical gravity on your space.
     
    Last edited: Nov 18, 2016
  9. Nov 18, 2016 #8
    I agree that that would be the case if the potential or force from a point mass were a continuous function on the space, the 3-torus.

    But as regards say the potential: I would expect it to be of infinite magnitude at the location of the point mass. This would avoid having an extremum at a point where the function is defined. Which as far as I can tell would avoid a contradiction.

    If this is wrong, please correct me!
     
  10. Nov 18, 2016 #9

    mfb

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    It doesn't avoid the contradiction: if the potential is not flat (boring), you need a maximum and a minimum, only avoidable via singularities, but your mass point is the only singularity and you want the potential to be low there.
     
  11. Nov 18, 2016 #10
    Good point. Maybe that's the whole story. But it also seems possible that the heuristic of [trying to add up the forces to get the effective force] might have some nonempty discrete set of point where it just can't be summed in any manner. (This set should *not* include the antipode of the origin: the point (1/2, 1/2, 1/2) that is farther from (0, 0, 0) than any other point. At this point the force should, by symmetry, ideally be zero. But it is also possible that this point is included.) This might lead to a result that is non-contradictory.
     
  12. Nov 19, 2016 #11

    Orodruin

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  13. Nov 19, 2016 #12

    mfb

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    You can use Yukawa potentials, for example, they have an exponential suppression which makes the sum convergent. For large distance parameters, you still get the 1/r^2 for all nearby instances of your mass.
     
  14. Nov 19, 2016 #13
    The situation is similar to an infinite homogeneous mass distribution which has no finit potential and no defined forces for a test mass. Integration of the forces from all mass points in the distribution results an integrational constant which cannot be determined. But you can at least calculate the tidal forces between two test masses.
     
  15. Nov 19, 2016 #14

    mfb

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    This is also a point where general relativity works better.
     
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