Re: Solving field equations in Excel

["John S. Denker" <jsd@MONMOUTH.COM>]

At 11:20 AM 2/11/01 -0500, Ludwik Kowalski wrote:

> Should I assume, however, that the problem .... can be solved
> numerically to any degree of accuracy without dealing with
> space charges?

I guess so.

1) The relaxation algorithm necessarily computes the potential in all the
"space" cells in the model universe.

2a) However, nothing requires you, personally, to look at these cells.  The
computer need not print them out.

2b) The computer need not evaluate the Laplacian in these regions (although
the operator it does evaluate has practically the same form as the
Laplacian).

3) Depending on what Ludwik means by "any degree of accuracy", it might be
good programming practice to evaluate the space charge, to make sure that
the algorithm is successfully driving it to zero.  But good programming
practice is not required.

4) Similarly, it might be good programming practice to sum the charge in
the entire universe;  if it's nonzero then the code has room for improvement.

I'm not making this up;  early versions of my spreadsheet (in the
rotationally-invariant case) were only 99% accurate in conserving
charge.  That really irked me;  it should have been better.  After sleeping
on it, I realized that the nonlinearity in the Laplacian (in polar
coordinates) was killing me.  I had been approximating the term (1/r)(d/dr)
using the centered long-span difference ... now things are much better
because I am using the average of the left difference and the right
difference; for details, see
  http://www.monmouth.com/~jsd/physics/laplace.html

But again, good programming practice is not required.  If you're smart
enough, you can get the right answer without checking the sum rules.  Or if
you want, you can get the wrong answer without checking the sum
rules.  It's a free country.

====

To summarize:  If you don't want to look at the space charge, you don't
have to look.  I don't have a problem with that.  OTOH I do have a problem
when somebody says the model doesn't conserve charge.  It does.  It has
to.  There is a big difference between
  a) choosing not to see the conservation, and
  b) seeing non-conservation.