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Re: Flat conductors (was I need help).



Regarding John Denker's comments:

...
Why not do the experiment? Take a piece of resistor paper and
divide it into four cells. Put the same pattern of electrodes
in each cell, with mirror symmetry:

<snip diagram of 2 x 2 cells of dipoles>

where P denotes a positive electrode and N denotes a negative
electrode. Wire all the Ps together and all the Ns together so that
they are at the same potential. Make all the contacts the same size
and shape. The "dividing lines" in the diagram are purely imaginary
at this stage.

I predict that you will find that the potential in each cell in this
configuration is the same as you would get if you actually cut the
paper into four pieces.

I think we all predict this.

The method of images enforces the same
boundary conditions at the cell-boundary as the actual cut would.

Of course it does.

Of course you can experiment with more than a 2x2 array of cells. Any
MxN will work.

I'm sure you are correct about this.

Note that the correctness of all this depends on having a real
boundary at the edge of the 4-cell piece of paper. In other words, if
you put four cells (and nothing else) in the middle of a large sheet,
the potential would not quite agree with a single cut-out cell. If
you put lots and lots of cells in the middle of a large sheet, the
middle cells would almost agree with a single cut-out cell, but not
exactly.

As far as I can tell John's method *does* effectively have a finite
number of cells in the middle of an infinitely large sheet. And that
is why I expressed some doubt concerning its rate of convergence with
an increasing numbers of cells.

...
The right answer can be obtained easily. No hedging required. Just
impose the appropriate boundary condition at the edge of the universe.
If the universe contains an MxN array of cells, with both M and N
even, then you can use periodic boundary conditions. You can use
hall-of-mirrors boundary conditions for any M or N (even or odd).

If your method can impose the correct boundary conditions to
begin with you don't need to use any image cells at all. The original
finite rectangular cell is sufficient if the correct boundary
conditions are imposed on it all the way around.

Techniques for coding such boundary conditions were discussed
previously in connection with the spreadsheets for solving Laplace's
equation. Look at
http://www.monmouth.com/~jsd/physics/laplace.html
and search for "periodic boundary conditions" and "hall of mirrors".

Of course this method doesn't actually solve Laplace's equation at
all. It solves a finite system of difference equations that is
supposed to approximate Laplace's equation on a finite grid that is
chosen fine enough so that the discretization error is acceptably
small. Such methods trade errors in approximating the correct
boundary conditions for discretization errors in putting a finite
grid on the (actually continuous) problem. It is sort of a question
of picking one's poison whether one wants to exactly solve a
continuum problem that approximates the original problem but doesn't
quite get the boundary conditions fully correct, or approximating
the original continuum problem with a finite grid problem that
*does* properly enforce all the boundary conditions. Either way one
is not quite actually solving the actual problem at hand.

Of course a big advantage of finite difference and finite element
methods is that they are amenable to problems with complicated
geometries with complicated irregular and mixed boundary conditions
for which the method of images is just plain inappropriate. Such
discrete problems tend to work well on regions where the potential is
very slowly varying from grid point to neighboring grid point.

These methods do not do so well for problems where the potential
wildly changes from one grid point to it neighbor. Since problems
with isolated point charges tend to have divergent potentials at
those point charges no finite discretization of the problem will be
fine enough in the immediate neighborhood of an isolated charge to
well represent the potential in such a singular region.

For Ludwik's problem this means we need to have a grid spacing near
each silver dot that is small compared to the radius of each dot if
we are going to use some bare discrete finite element or finite
difference method to solve it.

David Bowman
David_Bowman@georgetowncollege.edu