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Forum index » Science and Technology » Math » Recreational
Multiple Reflections (Kaleidoscope) problem
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Christopher J. Henrich
science forum Guru Wannabe


Joined: 03 May 2005
Posts: 107

PostPosted: Fri Jun 09, 2006 1:03 am    Post subject: Re: Multiple Reflections (Kaleidoscope) problem Reply with quote

In article <e6ad8q$8d6$1@reader2.panix.com>, Paul Ciszek
<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.
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Mike Amling
science forum Guru


Joined: 05 May 2005
Posts: 525

PostPosted: Thu Jun 08, 2006 11:52 pm    Post subject: Multiple Reflections (Kaleidoscope) problem Reply with 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
by an equilateral triangle of mirrors self-consistent?

--
Please reply to: | "Any sufficiently advanced incompetence is
pciszek at panix dot com | indistinguishable from malice."
Autoreply is disabled |
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