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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



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 2D 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
2D grids fare equally well under this treatment.
Imagine this: You draw a lowercase "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 selfconsistentyou 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 selfconsistent infinite grid or not? Is the grid produced
Yes, there is. 
In the plane, two mirrors (represented by halflines) at an angle of
Pi/m will generate a "dihedral group" of order 2m. This is the symmetry
of many flowers and flwerlike 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 halfofanequilateral 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 selfconsistent?


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

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 2D 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
2D grids fare equally well under this treatment.
Imagine this: You draw a lowercase "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 selfconsistentyou 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 selfconsistent infinite grid or not? Is the grid produced
by an equilateral triangle of mirrors selfconsistent?

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