# Characterizing polyhedron from Brownian particle collisions with a boundary

Please imagine that we have an ordinary 2-sphere, of radius $r_{sphere}$, and some three-dimensional polygon that has all of its points fixed at positions strictly internal to the sphere's surface. Also confined in the sphere is point-like particle (with diffusion constant, $D_{particle}$) undergoing Brownian motion. The surface of the 2-sphere, as well as the surface of the polygon internal to the 2-sphere, are perfect reflecting boundaries for the particle.

Working in discrete time, we track the point-like particle for $N$ finite time units (we'll call them seconds), $(t_1, t_2, ..., t_k, ..., t_N) \in T$. However, during this time the only information we are allowed to record is:

1. If a collision between the probe and the surface of the 2-sphere occurs at a given time point, $t_k$.

And if there is at least one such collision during $t_k$...

1. The coordinates of a collision event on the surface of sphere, randomly selected from all collisions that occur during $t_k$.

Beyond, perhaps, the volume of the polygon in the sphere (and I'm not entirely sure this is learnable), how (if at all possible) can we use the information specified above to characterize the polygon in some additional manner?

Update - If we apply a further restriction that the polyhedron is convex, at the limit of large $N$ will there be enough information from the collisions to reconstruct the convex polyhedra?

-
Trivial observation: the symmetry group of a convex polytope about the origin (if nontrivial) will be manifested in the density function of the collisions. –  Steve Huntsman Aug 23 '10 at 16:35
I assume by "three-dimensional polygon" you mean a polyhedron, a closed surface composed of flat faces? –  Joseph O'Rourke Aug 23 '10 at 17:31
You might look at the MO question "Algorithm for finding the volume of a convex polytope," mathoverflow.net/questions/979/… , for most algorithms use random walks inside the polyhedron to estimate the volume. –  Joseph O'Rourke Aug 23 '10 at 17:34
Joseph, yes, I do mean a polyhedron, but not necessarily a convex polyhedron. –  Rob Grey Aug 23 '10 at 18:09
@Rob: Ah, then the MO question on convex polytopes may not be so relevant... –  Joseph O'Rourke Aug 23 '10 at 18:14