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Another interesting infinite lattice problem I found while watching a physics documentary.

Imagine an infinite square lattice of point masses, subject to gravity. The masses involved are all $m$ and the length of each square of the lattice is $l$.

Due to the symmetries of the problem the system should be in (unstable) balance.

What happens if a mass is removed to the system? Intuition says that the other masses would be repelled by the hole in a sort of "anti-gravity".

  • Is my intuition correct?
  • Is it possible to derive analytically a formula for this apparent repulsion force?
  • If so, is the "anti-gravity" force expressed by $F=-\frac{Gm^2}{r^2}$, where $r$ is the radial distance of a point mass from the hole?

Edit:

as of 2017/02 the Video is here (start at 13min): https://www.youtube.com/watch?v=mYmANRB7HsI

Helder Velez
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Sklivvz
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    Interesting question! :) – Noldorin Dec 23 '10 at 16:17
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    I think it should be correct because of superposition principle, but the sign of force might be positive. – unsym Dec 23 '10 at 16:28
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    The question is if the vector sum of all forces converge to 0 or diverge, if it diverges the question is meaningless. – TROLLHUNTER Dec 23 '10 at 16:35
  • @kalle43: the force on a point goes to zero as $o(r^{-2})$ as r->infinity, so the sum should converge – Sklivvz Dec 23 '10 at 16:41
  • @Sklivvz: That's not considering it rigourously enough. This is quite a mathematically challenging question, and involves the Euler-Maclaurin formula for sums. – Noldorin Dec 23 '10 at 16:46
  • @Noldorin: it was not a proof, hence the should – Sklivvz Dec 23 '10 at 17:09
  • @Sklivvz: Intuitively, yes perhaps! I wouldn't even like to guess, mathematically. – Noldorin Dec 23 '10 at 17:19
  • You can regularize the sum by assembling the lattice in a particular way, and not have to worry about the formal divergences. – Zo the Relativist Dec 26 '10 at 00:00
  • To me it's fairly simple to get over the contradiction. The contribution to the force at (x,y) is exactly the opposite of the one at (-x,-y). I know that mathematically is not very sound... but any other value would also not conserve energy. A non-zero net force means acceleration (and this would apply to every point mass equally). The whole system would be accelerating out of nowhere. – Sklivvz Dec 26 '10 at 10:36
  • @Sklivvz: The first part of what you said is basically right--if you are careful about how you take the limit to an infinite sum, you can sometimes get a finite number when you subtract two infinite quantities. You point out a symmetry of the problem that enables one to do this to this sum. The second contraction is wrong, however--systems can expand and contract without having a center--look at expanding (or contracting) cosmologies. – Zo the Relativist Dec 26 '10 at 14:21
  • @Jerry Schirmer: Regarding the first part, actually: if you take a series of open circular intervals that tends to infinity and take the sum over the enclosed point masses, the force at the center is always zero, so the limit is zero. Regarding the second part: I never said the system would contract or shrink. I meant that since all points are equivalent, if there is any acceleration it would be an accelerating translation of all points at the same time (which is the contradiction) – Sklivvz Dec 26 '10 at 23:16
  • The remaining masses are not "repelled" from the hole - they're attracted to where the hole isn't. – Bob Jarvis - Слава Україні Aug 28 '14 at 03:07

4 Answers4

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This is not correct, but Newton believed this. The infinite system limit of a finite mass density leads to an ill defined problem in Newtonian gravity because $1/r^2$ falloff is balanced by density contributions of size $r^2\rho$, and there is no well defined infinite constant-mass-density system. The reason is that there is no equilibrium of infinite masses in Newtonian gravity--- you need an expanding/contracting Newtonian big-bang.

This is subtle, because symmetry leads you to believe that it is possible. This is not so, because any way you take the limit, the result does not stay put. This was only understood in Newtonian Gravity after the much more intricate General Relativistic cosmology was worked out.

Ron Maimon
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I think that your initial intutiion is right--before the point particle is removed, you had (an infinite set of) two $\frac{G\,m\,m}{r^{2}}$ forces balancing each other, and then you remove one of them in one element of the set. So initially, every point particle will feel a force of $\frac{G\,m\,m}{r^{2}}$ away from the hole, where $r$ is the distance to the hole. An instant after that, however, all of the particles will move, and in fact, will move in such a way that the particles closest to the hole will be closer together than the particles farther from the hole. The consequence is that the particles would start to clump in a complicated way (that I would expect to depend on the initial spacing, since that determines how much initial potential energy density there is in the system)

Zo the Relativist
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  • This is a qualitative answer, but really just states the obvious. Doesn't help a huge amount. – Noldorin Dec 23 '10 at 16:36
  • @Noldorin: Well, yes, but time-evolving this problem isn't going to be simple at all, and most likely won't be analytically solveable. The many-body problem is highly nonlinear. A simple analysis of the particles with a pairwise consideration of the particles shows that you instantaneously have a hole in the system. After the initial timestep, however, that is gone, and the system will behave in an unpredictable manner. – Zo the Relativist Dec 25 '10 at 19:20
  • On second look, this problem is rather intractable analytically, so your intuitive answer is actually growing on me. KennyTM provides a good reason why the quantities blow up, but I suppose there's still more to the story. – Noldorin Dec 26 '10 at 14:55
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    This is incorrect--- but Newton made the same mistake, so no downvote. The constant density over all space Newtonian gravitating system is not allowed, except in an expanding or contracting Newtonian cosmology, and there is no antigravity for the missing particle. The symmetry arguments are misleading, because the limit is subtle. – Ron Maimon Sep 16 '11 at 18:55
  • @Jerry You followed your intuition in the last part of your answer and you are expecting a "clump in a complicated way". But the clumping will be along the outer shell of a growing 'void' as I show in my answer. Either analytically either by simulation it is what really happens. – Helder Velez Sep 19 '11 at 07:53
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    Just a quick note on another aspect: I suspect homogeneity considerations will make the dynamics independent of the initial lattice spacing. – Emilio Pisanty Jan 13 '13 at 01:21
  • I agree with @EmilioPisanty - the only dimensionful quantities in the problem are $G$, $m$, and $l$, and there's no way to combine those values into a dimensionless ratio. So regardless of their values, we can choose units in which all three equal 1, so this is actually a uniquely determined problem. – tparker Feb 01 '17 at 07:55
  • The density of matter is very low, of the order of one molecule per 10 m³ and therefore can be treated without collisions and with temperature 0ºK. – Helder Velez Sep 16 '18 at 10:47
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I assume by square lattice you mean a 3D cubic lattice because there's no translational symmetry along the $z$-axis for a 2D square lattice.

Suppose the masses are located at $(n_x, n_y, n_z)$ where $n_{x,y,z}\in\mathbb Z$. Let's also define the unit of mass and length so that $m=l=1$.

Consider the total force acted on the mass point at (0, 0, 0) just due to the 1st octant $(x>0,y>0,z>0)$: $$\begin{aligned} \mathbf F_{+++} &= -G \sum_{n_x=1}^\infty \sum_{n_y=1}^\infty \sum_{n_z=1}^\infty \frac{n_x \hat{\mathbf x} + n_y \hat{\mathbf y} + n_z \hat{\mathbf z}}{(n_x^2+n_y^2+n_z^2)^{3/2}} \\ &= -G \left( \hat{\mathbf x} \sum_{n_x=1}^\infty \sum_{n_y=1}^\infty \sum_{n_z=1}^\infty \frac{n_x}{(n_x^2+n_y^2+n_z^2)^{3/2}} + \dotsb \right), \end{aligned}$$ however, the sum actually diverges, since, $$\begin{aligned}\sum_{n_x=1}^\infty \sum_{n_y=1}^\infty \sum_{n_z=1}^\infty \frac{n_x}{(n_x^2+n_y^2+n_z^2)^{3/2}} &\ge \int_1^\infty \int_1^\infty \int_1^\infty \frac{n_x}{(n_x^2+n_y^2+n_z^2)^{3/2}} dn_x dn_y dn_z \\ &= \int_1^\infty \int_1^\infty \frac1{\sqrt{1+n_y^2+n_z^2}} dn_y dn_z \\ &= \infty, \end{aligned} $$ so while symmetry suggests that the force at center is 0, mathematically it is not well defined.

Of course, if we assume the net force can be well-defined as 0 (e.g. the gravity actually decays faster than $1/r^3$!), then Points 1 and 3 are correct. When we remove a particle from the lattice, the contribution $-\frac{GMm\hat{\mathbf r}}{r^\alpha}$ will be subtracted from it, so it is as if there is a particle of negative mass $-m$ put to that point. This is because force is additive and gravity is proportional to mass.

(Yeah this is stating the obvious.)

kennytm
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    Let me play devil's advocate here: one can easily fix the problem mathematically by considering faster decaying force: $F \sim r^{\alpha}$ for alpha suitably small. Then the nature of the problem doesn't change but the math starts to converge (and would give the same answer as the intuitive one). So in this case it's obvious that computing integrals is useless math masturbation; pure physical intuition is all this problem requires. Actually this is very similar to regularizations in QFT and why mathematicians argue that QFT doesn't work. Needless to say, experiments prove them wrong :-) – Marek Dec 23 '10 at 18:31
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    @Marek: Yep. But this is mainly a respond to OP's comment that "the sum should converge". – kennytm Dec 23 '10 at 18:57
  • @Marek: I agree with you, and I have nothing against QFT, but (just to be fair) experiments can't prove them wrong because experiments can't prove QFT works. Experiments can only prove a theory is wrong or doesn't work. For all we know, all of the successful predictions could be a series of bizarre coincidences. ;-) – Malabarba Dec 23 '10 at 21:13
  • @Bruce: sure, I thought it was understood (among scientists at least) that all statements about science are purely probabilistic. We don't know our theories are correct. We are only convinced by so many experiments that probability they are incorrect is by now hugely smaller than a probability of a Santa coming tomorrow and bringing you that Bugatti Veyron you always wanted :-) – Marek Dec 23 '10 at 21:19
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    Are you saying he won't come? =( – Malabarba Dec 23 '10 at 21:32
  • I like this answer. It actually addresses the question properly. – Noldorin Dec 23 '10 at 22:23
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    @KennyTM: but you can regularize that divergence by considering the particles pairwise. Start with the central mass, and add two opposing partices at a time. The Newtonian force only depends on pairwise interactions, so it doesn't cause a problem doing it this way. Once you have built the lattice, then it clearly has translational symmetry, so any point is the same as any other point, and the solution is static. – Zo the Relativist Dec 25 '10 at 19:17
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    In an infinite lattice you can not restrict to the 1st quadrant. IMO, this answer is simply wrong. – Helder Velez Sep 19 '11 at 07:09
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Your intuition is OK. This apparent «anti-gravity» is just the usual attractive gravity. The problem is treated analytically, in a light way (*), starting in the post O nascimento de uma Bolha (The birth of a Void). For convenience I grabbed a few images from the blog (CC license) and I adapted some legends.

Image A-top - The gravitational field is null everywhere in an uniform and isotropic distribution.
Image A-middle - due to fluctuations particles acquire motion, and a local defect of density, here represented by a removed particle, originates a field that is symetric of the field of the missing particle.
Image A-bottom - The acceleration of particles surrounding the point of depletion leads them to exceed the following ones; Now, every particle of its boundary will be subject to a field corresponding to all matter that is missing inside. The "Void" grows rapidly because the field grows as the "Void" grows.
Image B -- void + sphere = isodense
Image C -- The field of a void
Image D -- The field of an homogeneous sphere, mass M and radius R.
( graphs C+D = 0 )
Image E -- The fields considering the outer shell
Image F -- The acceleration of the Void.

I already did a simulation of a lattice respecting this conditions (tricky because it is an infinite universe ;-) and the result is the same as I show in this answer.

(*) a complete mathematical treatment is known to me since 1992/Oct and it will be the subject of a paper by my friend Alfredo, as the author states in the last page of his recent paper A self-similar model of the Universe unveils the nature of dark energy (no peer reviewed). It will be shown that the evolution of the initial homogeneous and uniform universe, with T=0, will match the observed large-scale structure of the universe (without DM).

The birth of a Void:

The birth of a Void

Helder Velez
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