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# Re: visualizing fields near charged objects

• From: Leigh Palmer <palmer@SFU.CA>
• Date: Fri, 9 Feb 2001 14:45:50 -0800

I addressed John Mallinckrodt in an earlier note:

> John
Denker seems to believe that total charge should be constant
during iteration. Have you any reason to believe that should be
true?

John Denker responded to my question to John Mallinckrodt:

I have the following reason:

Notation:
a
b w c
d

Consider a grid (called the potential grid) containing zeros on the two
rows or columns nearest each edge, and arbitrary values in the interior.
Now convolve that with the operator (a+b+c+d)-4w to form the charge grid.

I have no charge grid; it is unnecessary to produce one, so I
don't. I infer surface charge density by calculating the
normal component of the potential gradient at the surface. A
surface charge is calculated only after the relaxation has
converged sufficiently to make that a reasonable calculation.

Every nonzero potential cell contributes to the convolution grid five
times: once as a, once as b, once as c, once as d, and once (weighted by
-4) as w. If you add those five contributions, you get zero every time.
*) A small discrepancy might be attributable to roundoff errors.
*) A large discrepancy cannot have a good reason. Non-good reasons include
-- serious bugs in the algorithm or the implementation
-- perhaps not summing the charge over
all places where charge might be found.

No. Consider the potential relaxation problem as I perform it.
I define the outer boundary to have zero potential everywhere.
I call the potential on my singly connected conductor 100 for
reasons related to simplicity of numerical display. I set the
initial value of the potential in every cell in free space to
zero as well. This initial condition does not satisfy Laplace's
equation. Note that the total charge on the remote boundary
will be zero initially. The charge on the central object will
be finite and, it turns out, much larger than the relaxed final
value of this charge. Note that the solution must have equal
and opposite charges on these two boundaries. Surely in one
case or the other charge cannot be conserved!

Your contention that charge must be conserved is obscure to me.
Conservation of charge is, of course, a requirement in any
physical process, but a relaxation calculation does not in any
way represent a physical process; it is purely a mathematical
exercise based upon a geometrical interpretation of the meaning
of Laplace's equation. I suggest that it is not a good test for
soundness of procedure and I offer my sound procedure as a
Gegenbeispiel. It seems to me that, whatever your procedure is,
you build in charge conservation. Unlike John M., I was unable
to read your code. I'm not an Excel guru, as I said. My
approach does yield a very satisfying result, however, and I
sent you a picture of it. When I can get more quantitative I
will reopen this discussion.

A really good test would be to examine quantitative convergence
on a known analytical solution, something I can do easily with
my procedure. I will take a cylinder of diameter two cells and
length two cells and place it at the origin of a universe
bounded by a cylinder of diameter 512 cells and height 512
cells. I will expect to see some semblance of an inverse first
power of r potential approached when I use my Laplacian cell
algorithm. An unweighted average algorithm should yield a
potential field which resembles that of a long cylindrical
conductor of radius R perpendicular to the plane,

V = Vo(1 + ln r/R)

The difference should be noticeable.

Leigh