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Theory of invisible solids


Here, on this page, you can read about the mathematics of invisible solids. If you have never heard about invisible solids, I suggest you read this informal introduction first.

This page is a work in progress.


I discovered the invisible solids in april 2003. I searched for previous results, but could not find any. In april 2005 I finally got about to making a web page about them.

As a reader pointed out, the invisible solids has most probably already been studied by the military and by magicians - but they of course would never publish anything about them.

The math is classical geometry so even Archimedes might have known about the invisible solids.


A light ray is represented by either
1) a list with only one element: a straight line
or
2) a list whith these elements: first a half line, followed by a number (possibly 0) of line segments, possibly followed by a half line; all arranged so that the last point in an element is identical to the first point in the next element.

In our representation the light-ray comes from the first part (from infinity) and goes to the last part.

The first option represents a light-ray that is not reflected.

The other option represents a light-ray that is reflected, which of course happens in the end points of the half lines and line segments.

(1) Notation
Given a light-ray r then λ(n,r) denotes the (n+1)'th element in the list representing the light-ray;
and λ(*,r) denotes the last element in the list representing the light-ray.

In more prosaic terms λ(n,r) is the linear part after the n'th reflection.

Notice that a light-ray r that is not reflected will have λ(0,r) = λ(*,r).

(2) Definiton
A light-ray r is resuming if and only if
there exists a straight line l so that λ(0,r) ⊆ l and λ(*,r) ⊆ l
and
for all points p1 and p2
if p1 ∈ λ(0,r) and p2 ∈ λ(*,r)
then 2·p1-p2 ∈ λ(0,r) and 2·p2-p1 ∈ λ(*,r)

Informally a resuming light ray will, after all reflections, continue as if it had not been reflected at all.

Notice that light-rays that are not reflected are trivially resuming.

(3) Observation
A resuming light-ray can not have just one reflection.

(4) Observation
A resuming light-ray can not have just two reflections.

When a light-ray hits a non-regular surface (e.g. an edge), reflection is not well defined, and its journey stops there. IOW the non-regular surface will be hit by the λ(*,r) part, which will be a line segment - not a half line.

Notice that a light-ray that hits a non-regular surface can not be resuming.

(5) Definiton
Given a vector v ≠0 then
a light-ray r has direction v if and only if
for any point p
if p ∈ λ(0,r) then p-v ∈ λ(0,r)

So v is the direction that the light-ray has when it starts out.


(6) Definiton
A sterling solid L in Rn is defined by
• L ≠∅
• and
• L is the set closure of the interior of L
• and
• the surface of L is S = SF ∪ SE so that
• • SF (the faces) is regular
• • SE (the edges) has area =0 (n-1 dimensional volume)

Notice that a sterling solid don't have to be connected.

(7) Notation
Given a sterling solid L
then F(L) denotes the SF-surface (faces) of L
and E(L) denotes the SE-surface (edges) of L.

I think that sterling solids in R3 are a fair representation of real-world solid objects.


(8) Definiton
An invisible solid (L,v) is defined by
• L is a sterling solid in Rn
• v is a vector in Rn and has |v|=1
• for all light-rays r with direction v
• if r does not hit E(L) it is resuming

Think of L as a solid that is partly mirror coated.

Notice that if (L,v) is an invisible solid, then so is (L,-v).

Notice that n >1 in Rn for there to be any invisible solids.

(9) Definiton
Given an invisible solid then MR denotes the max number of reflections for the resuming light-rays with direction v.

MR can never be 0 because an invisible solid has volume >0.

The smallest MR can be is 3. The fractal-like type 2 solid shows us there is no upper bound for MR.

MR=3 type 3
MR=4 type 1 and type 2
MR=8 1. level fractal-like type 2
MR=12 2. level fractal-like type 2
MR=16 3. level fractal-like type 2
: :

(10) Terminology
A linear sterling solid L is a sterling solid with F(L) consisting of linear surfaces.

The surface of a linear sterling solid in Rn consists of
faces that are n-1 dimensional plane tiles (excluding their borders), and
edges that are n-2 dimensional plane tiles (including borders).

(11) Lemma
Given a linear sterling solid and a resuming light-ray r. Then there exists an ε ∈ R such that all light-rays with the same direction as r and with a distance <ε to r will hit the same plane tiles as r, in the same order as r and will be resuming.

Take two light-rays with the same direction. If they hit the same plane tile (excluding its border) they will be parallel after the reflection. And they will be the same distance apart before and after the reflection.

Take a resuming light-ray r. It only hits the interior of the plane tiles. Because reflection is not well defined on the edges.

Take the cylinder of all light-rays that start out parallel to r with distance less than some arbitrary radius. Look at the first face that r hits. If all light-rays in the cylinder hits that face thats fine. But if some doesn't hit that face then reduce the radius until the whole cylinder hits the same face. Continue, one by one, with the other faces that r hits, each time reducing the radius as necessary.

Now all light-rays in the cylinder will hit the same faces as r and in the same order as r.

All the light-rays will be resuming because they will all be parallel to the resuming light-ray after the last reflection. QED

The border (surface) of a linear sterling solid in R2 consists of
faces that are line segments (excluding end points), and
edges that are points.

(12) Lemma
For a linear sterling solid in R2 all resuming light-rays will be reflected an even number of times.

Take a light-ray r1 reflected by a line segment. Take a light-ray r2 with the same direction as r1 and which hits the same line segment (excluding end points). Then if r2 is to the left of r1 before the reflection, it will be to the right after the reflection. And vice versa.

From this, and that 0 is even, follows the Lemma. QED


(13) Corollary
There are no linear invisible solids in R2 with an odd MR.

TODO/wishlist


I am interested in feedback. I am especially interested to hear if you


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© 1997-2005 Bjørn Hee, mailto:webmaster@h33.dk