Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let a body $B$ be a compact set in $\mathbb{R}^3$ with a piecewise smooth boundary. Some pieces/patches of the boundary are perfect mirrors; others perfect matte, colored surfaces. Imagine the view of $B$ from infinity in some direction $u$. A light ray from the "eye" at infinity travels parallel to $u$ until it hits $B$. It then reflects from perfect mirror patches and eventually hits a colored patch, or shoots off (in some direction) to infinity. What is seen in direction $u$ is an array of the colors each ray hits, or transparency when a ray runs to infinity.

More precisely, let $u$ be along the $z$-direction of a Cartesian coordinate system. Each ray parallel to $u$ may be identified by $(x,y)$-coordinates. Each point $(x,y)$ is assigned a color, a positive integer corresponding to the color hit by that ray, or 0 if the ray (ultimately) goes to infinity. The image of $B$ in the $z$-direction is the coloring of $\mathbb{R}^2$. Let $R(u)$ be a region of $\mathbb{R}^2$ large enough to include all the nonzero points of the image from any direction $u$.

As the directions $u$ vary over $\mathbb{S}^2$, the image $R(u)$ changes, generally continuously (say, under the Hausdorff metric). I am interested in bodies $B$ that change discontinuously:

Q1. Does there exist a $B$ whose image changes discontinuously with respect to $u$, preferrably with rather dramatic differences, and perhaps several or many such changes?

Such a $B$ could be called a chameleon body, for its appearance changes depending on the viewpoint. Ultimately I would like to control the changes. For example:

Q2. Is there a $B$ whose view changes between these two images? :-)

[This question was inspired by fascinating work by Alexander Plakhov and Vera Roshchina: "Invisibility in billiards" (Nonlinearity 24(3)) and "Bodies invisible from one point" (arXiv:1112.6167). They arrange that a in-ray along line $L$, after ricocheting around inside $B$, emerges and continues along $L$. Orchestrating this for all rays from a point makes $B$ invisible from that point.]

share|cite|improve this question
A cube with five mirror faces and one matte face is a "chameleon": as you rotate the cube, you go from seeing nothing (i.e. seeing infinity) to seeing the matte face. Is that an example of what you're looking for? – André Henriques Dec 31 '11 at 16:57
@André: Nice! But the matte face first appears as an $\epsilon$-sliver. I was hoping for an abrupt appearance... – Joseph O'Rourke Dec 31 '11 at 19:46
Your hope for abruptness suggests to me a "broadcasting" radar dish, where the image rays come from some focal point and hit a parabolic reflector, so that the image can be seen only from a small subset of the horizon. You may want to consult your optics friends for this problem. Gerhard "Ask Me About System Design" Paseman, 2011.12.31 – Gerhard Paseman Dec 31 '11 at 23:50
Off topic, but I think others will approve and let me speak for them: I appreciate your contributions and demeanor on this forum. A good part of the beauty of MathOverflow is represented in your pictures, and the playful, curious, and good natured spirit is often captured in your words, both in your questions and your responses. I wish you and yours good health for all of the year 2012, and look forward to more contributions from you and your students and colleagues. Gerhard "And Happy Valentine's Day, Too!" Paseman, 2011.12.31 – Gerhard Paseman Dec 31 '11 at 23:55
I haven't the time to check this out thoroughly, so I'm writing this as a comment for now. What you describe reminds me of "digital sundials" described by Ian Stewart (Ian Stewart, "What in heaven is a digital sundial?", Scientific American, pages 104-106, 1991). More detail can also be found at ("The present invention embodies a digital sundial, a device that displays the current time in digits, words, or pictures that change with the direction of the sunlight.") – Joel Reyes Noche Jan 1 '12 at 2:09

2 Answers 2

(Too long for a comment.)

Following on from Gerhard's parabolic mirrors suggestion: take a parabolic mirror surface, cut off by a plane perpendicular to the axis of symmetry, so that the resulting surface is still rotationally symmetric. Let's require that the cutoff plane is further out than the focus, so that the focus is not visible from the side of the mirror. Then put a single coloured point at the focus. From exactly one direction, this looks like a solid disk of colour. From any other direction, we either get a single point of colour, or nothing at all. I think that multiple such configurations arranged in a plane, all facing almost the same direction (near perpendicular to the plane) should not interfere with each other (here the focus being hidden from the side is important).

With this set-up, the "2011"/"2012" image could be done with an array of circular "pixels", where half of them show for the 1, and half for the 2. Viewed with less than perfect (ignoring null sets) vision, the single points of the incorrect colour would vanish, and the result would be a (less than) 50% density halftone "newspaper style" image.

I think we can improve on the 50% figure as well. Let's assume that the image we "render" from a given direction is centered on the origin in $\mathbb{R}^3$. We have desired directions for viewing "2011" and "2012", which are very close to each other on $S^2$. No matter how close they are however, we can set "2011" and "2012" back far enough from the origin, in the directions that they are to be viewed in, so that the parabolic mirror pixels for the two images need not be interleaved. Then the density can be made the same as that for circle packing in $\mathbb{R}^2$.

Can it be done with polygonal shapes, so as to get full density of colour? Can it be done so that the images are viewable from subsets of $S^2$ containing open sets?

share|cite|improve this answer
This strikes me as theoreticians trying to recreate a TV screen. I like the idea, and think it could work, but I am more struck by the notion that this is an engineering question, and that even better answers might come from flat panel screen designers and the like. Perhaps some of those are reading this forum and will chime in. Gerhard "MathOverflow Isn't Just For Mathematicians" Paseman, 2012.01.01 – Gerhard Paseman Jan 1 '12 at 8:11
@Henry: Very clever to arrange the sudden change from just a point of color to a solid disk of color! – Joseph O'Rourke Jan 1 '12 at 14:27

For question 2, it is not too difficult to construct such an object under your idealized assumptions of perfect mirrors and non-diffracting light.

  1. Put a black object that looks like "201" in front of everything, then put a mirror that is shaped like a section of a circular paraboloid in the place of the last digit somewhat further from the observer. This mirror should have focus directly behind one of the black "201" regions, and reflect light from the focus to the observer. The rest of the body will be small, and lie directly behind the black region (thus invisible to the observer).

  2. At the focal point, place the edge of a reflective cube. I don't think it is reasonable to define what the observer sees through the mirror at this precise point, since the edge of a cube is a sort of idealization. However, under small perturbations of the observer's position, the observer sees whatever is reflected off one or the other faces of the cube adjacent to the edge. (There are also positions where the observer gathers light from both adjacent faces, due to the failure of the mirror to be a perfect focusing device away from the initial direction, but we may ignore them.)

  3. Arrange small objects near the cube so that one of the faces of the cube adjacent to the edge should reflect the figure "1", and the other face should reflect "2".

share|cite|improve this answer
@Scott: Great idea to place an edge supporting two mirrors at the focus! – Joseph O'Rourke Jan 1 '12 at 14:28

Your Answer


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.