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Christopher J. Henrich
science forum Guru Wannabe

Joined: 03 May 2005
Posts: 107

Posted: Fri Jun 09, 2006 1:03 am    Post subject: Re: Multiple Reflections (Kaleidoscope) problem

<nospam@nospam.com> wrote:

 Quote: My apollogies if this appears more than once; I am having trouble posting it. There must be a theorem of group theory somewhere that addresses this issue. Someone is trying to model an infinite 2-D grid of identical features by modeling one of the features, and "reflecting" it about each of the sides of a square. This is not quite the same thing as periodic boundary conditions, but we hope it will yield similar results. I realized though, that not all 2-D grids fare equally well under this treatment. Imagine this: You draw a lower-case "b" on a piece of paper and then surround it with four mirrors arranged in a square. You will see something that looks like this: b d b d b p q p q p b d b d b p q p q p b d b d b This grid is self consistent; whether you look at the relection of the north mirror in the west mirror, or the reflection of the west mirror in the north mirror, you will see a "q" in the square to the northwest of the original "b". Next, I looked at creating a hexagonal grid by the same method. I think that a hexagon of mirrors creates a hexagonal grid of reflections, but I'm not absolutely sure. I do know, though, that the grid would not be self-consistent--you can get to some cells by either an even or an odd number of reflections, depending on the "path" taken. Is there a simple way to determine whether a given "kaleidoscope", i.e., a prism without ends constructed of mirrors, will result in a self-consistent infinite grid or not? Is the grid produced Yes, there is.

In the plane, two mirrors (represented by half-lines) at an angle of
Pi/m will generate a "dihedral group" of order 2m. This is the symmetry
of many flowers and flwer-like figures.

Three mirrors, as the sides of a triangle, will generate a discrete
group of transformations if the angles between them are Pi/2, Pi/3,
Pi/4, or Pi/6. There are three possibilities: equilateral triangle,
right isosceles triangle, and half-of-an-equilateral triangle.

four mirrors, surrounding a rectangle, generate a discrete group.

A very good reference for this theory (which goes on into more
dimensions) is James E. Humphreys, _Reflection_ _Groups_ _and_
_Coxeter_ _Groups_ , Cambridge University Press.

 Quote: by an equilateral triangle of mirrors self-consistent?

--
Chris Henrich
http://www.mathinteract.com
God just doesn't fit inside a single religion.
Mike Amling
science forum Guru

Joined: 05 May 2005
Posts: 525

Posted: Thu Jun 08, 2006 11:52 pm    Post subject: Multiple Reflections (Kaleidoscope) problem

My apollogies if this appears more than once; I am having
trouble posting it.

There must be a theorem of group theory somewhere that addresses
this issue.

Someone is trying to model an infinite 2-D grid of identical
features by modeling one of the features, and "reflecting" it
about each of the sides of a square. This is not quite the
same thing as periodic boundary conditions, but we hope it
will yield similar results. I realized though, that not all
2-D grids fare equally well under this treatment.

Imagine this: You draw a lower-case "b" on a piece of paper
and then surround it with four mirrors arranged in a square.
You will see something that looks like this:

b d b d b
p q p q p
b d b d b
p q p q p
b d b d b

This grid is self consistent; whether you look at the relection
of the north mirror in the west mirror, or the reflection of the
west mirror in the north mirror, you will see a "q" in the square
to the northwest of the original "b". Next, I looked at creating
a hexagonal grid by the same method. I think that a hexagon of
mirrors creates a hexagonal grid of reflections, but I'm not
absolutely sure. I do know, though, that the grid would not
be self-consistent--you can get to some cells by either an even
or an odd number of reflections, depending on the "path" taken.

Is there a simple way to determine whether a given "kaleidoscope",
i.e., a prism without ends constructed of mirrors, will result
in a self-consistent infinite grid or not? Is the grid produced
by an equilateral triangle of mirrors self-consistent?

--
pciszek at panix dot com | indistinguishable from malice."

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