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

Suppose $S=\mathbb{S}^d$ is a unit sphere in $(d-1)$ dimensional space, with $d=3$ of special interest. The surface of $S$ is a perfect (internal) mirror. You stand at point $x$ (not the sphere center $c$) inside $S$ and emit a single laser light ray in direction $u$. What happens? I believe that the light ray will remain within the plane containing the three points $\{ x, x+u, c \}$.

Now suppose instead that from $x$ you shine a flashlight, a cone with angular extent $\pm \epsilon$. Does this fill the sphere with constant-density energy for any $\epsilon > 0$? Are there are no dark points within $S$?

A somewhat related question is: What would the flashlight-holder see from $x$? What would the visual image be, say in a graphics ray-tracing system (in $d$ dimensions!)?

I've asked enough questions for one MO posting, but ellipsoids in $\mathbb{R}^d$ are the obvious extension. Are they integrable or chaotic?

share|improve this question
1  
I think the question of what you would see inside a perfectly mirrored sphere, i.e. what the radiance would be along any ray, is not well-posed unless you specify other, non-mirrored objects inside the sphere to act as boundary conditions. There exist many radiance distributions that would satisfy the rendering equation (en.wikipedia.org/wiki/Rendering_equation); for example, choose a function f and assign all rays at a distance r from the center a radiance f(r). Practically, a ray tracer would keep chasing reflected rays and never terminate. –  Rahul Aug 21 '10 at 14:16
1  
...although, if the flashlight remains on, the rendering equation says the illumination will go to infinity because you keep pumping energy into the system which is never absorbed. –  Rahul Aug 21 '10 at 14:43

2 Answers 2

Your first question about the ray remaining in a planar region is trivial considering reflection angles.

In a sphere a ray of light that is at distance $r$ from the center will continue to stay at that distance even after multiple reflections and so for a cone of light to illuminate the sphere one of its rays must pass through the origin. If the cone's distance to the origin is $r$ (assuming that the cone sheds light both ways) then all of the sphere minus a smaller concentric sphere of radius $r$ intersected with the figure you get after rotating a plane on the axis $XC$ so as to cover all of the rays from the flashlight will be illuminated. This is easy to show since you will have a continuum of angles that subtend the chords formed by the rays and so most of them will not be rational multiples of $\pi$ and will not have periodic orbits.

So basically there are two obstructions, the distance of the cone from the center of the sphere $C$ and the planes determined by the line $XC$ and the rays of light. You can see that both these obstructions disappear if one of the interior rays passes through $C$.

share|improve this answer
    
@Gjergji: "will continue to stay at that distance [$r$] even after multiple reflections"---Great observation! Which makes the illumination of the spherical shell clear. Thanks! –  Joseph O'Rourke Aug 21 '10 at 0:20
    
If the cone does not contain the origin, I'm not sure all of the sphere at a distance greater than r will be illuminated. View the configuration along the line xc, so that for any light ray in the direction u, the plane {x, x+u, c} that it illuminates appears as a line. Because the range of u is limited, these lines do not cover the view plane. –  Rahul Aug 22 '10 at 0:13
    
You are right, the answer is in fact the intersection of the spherical shell with the region obtained by rotating a plane along the axis xc so as to cover the cone. I will edit that. I guess the main point I was trying to say is that the entire sphere is illuminated only if the center is contained in the interior of the cone. –  Gjergji Zaimi Aug 22 '10 at 0:21

Billiards inside an ellipsis all are integrable in dimension $2$, as is beautifully shown in the small book by Tabachnikov. I do not know if this property extends to higher dimensions, but the proof I know certainly doesn't (it relies to the metric bifocal caracterisation of ellipses).

Let me give a more precise statement and mention an important conjecture in this domain. Given an elliptic domain, there is an open neighborhood of its boundary that is foliated by curves (which are in fact confocal ellipses) in such a way that any billiard trajectory starting close enough to the boundary (in position and direction; here the proximity condition is in fact that the first segment does not go between the focal points), has each of its segments tangent to one and the same of these curves. The conjecture is that ellipses are the only convex 2-dimensional billiards having this property.

share|improve this answer
2  
Serge Tabachnikov, Geometry and Billiards [AMS, 2005], Chapter 5: "Billiards inside Conics and Quadrics," p.68: "As a consequence, the billiard ball map inside an (n-1)-dimensional ellipsoid is also completely integrable: the billiard trajectory remains tangent to n-2 confocal quadrics." Thanks for the pointer! –  Joseph O'Rourke Aug 21 '10 at 12:36

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.