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
Answer
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)
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