# Anti-gravity in an infinite lattice of point masses

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:

Video here (start at 7min): http://www.disclose.tv/action/viewvideo/45729/Stephen_Hawking__The_Story_of_Everything_pt_2_9/

-
Interesting question! :) –  Noldorin Dec 23 '10 at 16:17
I think it should be correct because of superposition principle, but the sign of force might be positive. –  hwlau Dec 23 '10 at 16:28
The question is if the vector sum of all forces converge to 0 or diverge, if it diverges the question is meaningless. –  user1708 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

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)

-
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. –  Jerry Schirmer 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
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 7:53
show 1 more comment

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.

-
but the elapsed time, since the beginning of matter, is finite. –  Helder Velez Sep 16 '11 at 19:22
but the question is asking about Newton's model, where gravity is instantaneous. –  Ron Maimon Sep 17 '11 at 19:12
Why "infinite mass density" ? In my answer I made clear that your statement "there is no equilibrium of infinite masses in Newtonian gravity" is simply wrong. –  Helder Velez Sep 19 '11 at 7:19
@Helder: I was using shorthand for "infinite system/finite density". There is no equilibrium of infinite masses in Newtonian gravity. They always collapse. –  Ron Maimon Sep 22 '11 at 15:34

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.)

-
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
@Marek: Yep. But this is mainly a respond to OP's comment that "the sum should converge". –  KennyTM Dec 23 '10 at 18:57
Are you saying he won't come? =( –  Bruce Connor Dec 23 '10 at 21:32
@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. –  Jerry Schirmer Dec 25 '10 at 19:17
In an infinite lattice you can not restrict to the 1st quadrant. IMO, this answer is simply wrong. –  Helder Velez Sep 19 '11 at 7:09

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:

-
@all . My answer goes against all others and also against some conceptions that are deep inside our believes. If you feel that my answer is wrong I kindly ask you to leave a few words to explain your viewpoint. I'm used to be down-voted and it's your privilege. I know that I have controversial opinions that are not shared by the community but and when I saw unpleasant words like 'crackpot' and not a single argument left is offensive. I'm used to be responsible and justify my sayings and I expect the same from others. –  Helder Velez Sep 19 '11 at 8:27
Hey. -1 because of first sentence. The two other answers are right. –  user1708 Sep 26 '11 at 15:02
@solomon - Thanks for the explanation. I see your point. Can you read the 2nd sentence? –  Helder Velez Sep 26 '11 at 17:38