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Is the packing of the plane by disks of radius 1/2 centered at the points of ${\bf Z} \times {\bf Z}$ "locally rigid" in the sense that no finite subcollection of the disks admits any joint infinitesimal deformations subject to the locations of the other disks?

Certainly it is clear that if you fix the locations of all but 1 of the disks, the 1 "free" disk cannot move in any direction. But it is less clear to me that for all $n \geq 1$, if you fix the locations of all but $n$ of the disks, there is no way for those $n$ disks to move in some joint fashion. Indeed, we do know that when $n$ is large enough, and the $n$ free disks form a sub-cluster of the packing of a suitable kind, there are lots of ways to rearrange those $n$ disks, if we either remove the disks from the plane and then replace them, or if we allow the disks to temporarily intersect the fixed disks on the way to their respective destinations.

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up vote 9 down vote accepted

Yes, every finite subset of a square lattice packing is locked in place if you hold the rest fixed. See Finite and Uniform Stability of Sphere Packings by A. Bezdek, K. Bezdek, and R. Connelly.

However, this jamming is really delicate. For example, see slides 26 through 33 here, which show an apparent unjamming. It's not real, since you can't carry it out without slightly moving the other disks, but the amount you have to move them is very small. The paper I mentioned above has more details. What this means is that the local rigidity is not relevant in the real world, since it disappears if you relax things just slightly.

Of course you can also deform the whole packing by shearing it, but that's not relevant to the local rigidity you asked about.

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