Re: volume polarization vs. surface charge

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

At 01:14 AM 3/5/99 -0500, Bob wrote:
> Donald has made the crucial, clarifying point; let me just expand a little:

This "crucial" point is not entirely correct.  There is a physics mistake,
compounded by confusion over terminology and viewpoints.

> A neutral but polarized (either permanent or induced by an applied field)
> dielectric produces a weak 1/r^3 dipole field.  This dipole field is
> numerically equivalent to the 1/r^2 field produced by a fictitious

According to the "del dot D" viewpoint it is fictitious, but according to
the "del dot E" viewpoint it is not fictitious.  In defence of the "del dot
E" viewpoint, let me point out that if you count *real* electrons and
*real* protons in a small volume encompassing the surface of a polarized
dielectric, there is a mismatch.  There has never been a charge that is
more real, more non-fictitious.  For more on this, see _The Feynman
Lectures on Physics_ volume II section 32-2.

> volume
> density of charge equal to (minus) the divergence of the polarization volume
> density;  this is the mathematical origin of the so-called "bound" charges
> in and on a polarized dielectric.  These fictitious charges are introduced
> only as a convenience for calculation.

I agree that it *may* be convenient to write the Maxwell equations in terms
of "del dot D" rather than "del dot E" but that is a choice of viewpoint,
not a physical requirement, and someone who finds it convenient use "del
dot E" is free to do so.

I am not trying to argue in favor of one viewpoint over the other;  I am
trying to *prevent* this from becoming a war between the little-endians and
the big-endians.

> The neutral dielectric produces no
> monopole field, but its dipole field is calculationally equivalent to the
> monopole field of this distribution of fictitious bound charges.

The absence of a monopole term in the far field is merely evidence of
overall neutrality (i.e. if you integrate over the whole dielectric).  It
is absolutely not evidence as to whether the charges are bound or free.

My physical point remains:  in a polarizing field, the behavior of a metal
(which I think we agree contains an abundance of real, free charges) is not
macroscopically different from an insulator with a high dielectric
constant.  [To forestall nitpickers, let me restate some of the usual
reasonable assumptions:  we are talking about reasonable frequencies,
overall neutrality, no charge injection, ....]

OTOH we agree that when the dielectric constant is not very large, the
behavior can be quite different from the high-dielectric and/or metallic
case.  In particular, consider the following geometry:


    PPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP

    iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
    iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
    iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
    iiiiiiii   iiiiiiiiii   iiiiiiiiii   iiiiiiii
    iiiiiiii   iiiiiiiiii x iiiiiiiiii   iiiiiiii
    iiiiiiii   iiiiiiiiii   iiiiiiiiii   iiiiiiii
    iiiiiiii   iiiiiiiiii   iiiiiiiiii   iiiiiiii
    iiiiiiii   iiiiiiiiii   iiiiiiiiii   iiiiiiii
    iiiiiiii   iiiiiiiiii   iiiiiiiiii   iiiiiiii
    iiiiiiii   iiiiiiiiii   iiiiiiiiii   iiiiiiii
    iiiiiiii   iiiiiiiiii   iiiiiiiiii   iiiiiiii
    iiiiiiii   iiiiiiiiii   iiiiiiiiii   iiiiiiii
    iiiiiiii   iiiiiiiiii   iiiiiiiiii   iiiiiiii
                             y
    QQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQ


where (P) and (Q) are the capacitor plates, and (i) is an insulating
dielectric.  Consider the field at point (x).  In the low-dielectric case,
the field will be very nearly the same as it would be with no dielectric at
all.  In the high-dielectric case, the field will be greatly reduced.
 * according to the "del dot E" viewpoint, we would say the field is
screened by the induced charges on the face of the dielectric near points
such as (y).
 * according to the "del dot D" viewpoint, we would say the E field is less
than the D field because of the polarization terms that appear explicity in
the "del dot D" version of Maxwell's equations.

The answer is the same.  The physics is the same.  Only the viewpoint is
different.

Suggestion:  Let's try to be explicit about which viewpoint is being taken,
and let's try to be tolerant of other viewpoints as long as they get the
right answer.

--- jsd