Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Imagine there's a point-like particle undergoing radioactive decay at some position $(0,0,0)$ in Euclidean $3$-space. We encapsulate this particle with a spherical detector for the decay products it emits s.t. the walls of the detector are some distance $L$ from the particle in all directions.

Here, we assume that the particle decay products travel strictly along a linear interval from the particle to the detector walls, and that any obstructions in the path of a decay particle will adsorb (equiv. destroy/erase) the particle emission. We also assume isotropy for the emission of decay products, i.e. that a particular decay event will propel a decay product along a trajectory uniformly selected from all possible trajectories.

The catch is that the detector only reports to us the geometric centroid, $C_L$, of the 3-space coordinates corresponding to decay product collision events along its walls. The exact coordinates of these collision events are otherwise unknown. We can, however, vary the distance to the detector walls, $L$, allowing us to collect an arbitrary number of centroids, $(C_{L_1},...,C_{L_N})$ corresponding to these different particle to detector wall distances.


We'd like to play a game with this detector.

We have some shape $P$ of maximum cross-sectional diameter $< L$, which we place somewhere inside the detector chamber s.t. the radioactive particle is within some $\epsilon$ or its outer surface. If a decay product collides with the 3-polytope, i.e. if its trajectory falls along an interval intersecting the polytope, it will fail to be recorded at the detector wall.

We can wait to collect as many particle emissions as I'd like in the context of having $P$ placed inside the detector in the aforementioned manner. Afterwards, the detector will return the geomtric centroid, $C_L$, for the positions where decay products were recorded. We can also repeat this experiment for an arbitrary number of different particle to detector wall distances to collect a set of centroids $(C_{L_1},...,C_{L_N})$.

Is there anything surprising we can learn about $P$ provided a set of geometric centroids corresponding to different particle to detector wall distances, $(C_{L_1},...,C_{L_N})$? What if we assume weak / strict convexity for $P$?

share|improve this question
    
Is this a correct interpretation? The "detector" is the entire sphere of radius $L$. It only reports the centroid as a point in $\mathbb{R}^3$, inside the sphere, of all the radiation it receives. –  Joseph O'Rourke May 11 '13 at 23:38
    
@Joseph O'Rourke Yes, your interpretation is correct. –  LesserOrchard May 12 '13 at 2:38

1 Answer 1

up vote 1 down vote accepted

If I have interpreted your situation correctly, all you can learn from your center of gravities $C_L$ is the angular aperture of $P$ at the origin (from any one $L$), and the orthogonal to the extremes of $P$ (from a sequence of $L$s):
           SphericalDetector
In $\mathbb{R}^3$, you could not distinguish between a vertical and a horizontal segment $P$ (or a thickened segment $P$).

share|improve this answer
    
@Joseph O'Rourke What if we allow for $L < D$, where $D$ is the cross-sectional diameter of our polytope. From a series of $C_{L_i}$ values, can we learn something about its geometry? –  LesserOrchard May 12 '13 at 3:06
    
Perhaps not if $P$ is convex. –  LesserOrchard May 12 '13 at 3:07
    
@Joseph O'Rourke Specifically, if $P$ is non-convex, the apparent angular aperture may change as $L$ is increased to the point where it is greater than the maximum cross-sectional diameter of $P$. –  LesserOrchard May 12 '13 at 3:20
    
@Lesser: I don't see that. Are not the spheres cocentric? That's what I assumed. Of course, if you both increase the radius and move the sphere centers around, yes, you can learn more about $P$, even when it is convex. –  Joseph O'Rourke May 12 '13 at 13:16

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.