Re: funny capacitor (gauge invariance)

[Ludwik Kowalski <KowalskiL@MAIL.MONTCLAIR.EDU>]

John S. Denker wrote (in part):

> I won't give up my gauge invariance until they pry it from
> my cold, dead fingers.
>
> > We need a method of calculating Cij which does not introduce
> > the ambiguity. Do you agree?
>
> I would rather say that we need a way to calculate the
> physically-meaningful quantities.  ...

Unfortunately the gauge is not a physically meaningful quantity
TO ME. I have no trouble in accepting the "charge invariance"
(the total net charge of the universe remains zero) because charge
is a familiar physical quantity.

But how can I accept the "gauge invariance" without knowing
how to expressed the gauge numerically. We are dealing with
a classical physics problem and I would like to know how "the
amount of gauge ?" is defined in classical physics. Is this a
dimensionless quantity? If not then what is its unit? Couple of
good examples based on familiar problems would be useful.
Ludwik Kowalski




> We can allow V to remain ambiguous, as
> long as our physically-meaningful results don't depend on absolute
> V.  Ambiguity in V is not _per se_ a problem that needs solving.  It's like
> elevation, or like relativity:  any reported velocity is ambiguous, unless
> we know what it is relative to.
>
> Specifically, the Q values are physically meaningful, the delta_V values
> are physically meaningful, but the absolute V values are not physically
> meaningful.  Therefore
>    a) We need a way to calculate Q(delta_V).  We have that.  No problem.
>    b) We need a way to calculate delta_V(Q).  We have that.  No problem.
>    c) We don't need a way to calculate V(Q).  We don't have that.  No problem.
>    d) We don't need a way to calculate Q(V).  Actually we happen to have
> that, even though it's not strictly needed.  No problem.
>
> >  By using this method we would
> > not be forced to use tricks to solve V(Q) problems.
>
> Calculating the absolute V(Q) is not just tricky -- it's impossible.
>
> > What is the
> > correct way of calculating nine (not sixteen) Cij coefficients
> > for the three equations with three unknowns? I suppose the
> > 3 by 3 matrix of such coefficients would not be singular.
>
> It suffices to calculate the 4x4 full capacitance matrix, and then throw
> away the 4th row and 4th column.  This results in the 3x3 diminished
> capacitance matrix, which applies to the (V1, V2, V3, 0) subspace.
>
> There exist other approaches that will arrive at the same results.
>
> The alert reader will notice the possibility of speeding up the calculation
> by a factor of 4/3 by not bothering to calculate the 4th column of the full
> capacitance matrix.  IMHO this optimization is for experts only, and is NOT
> recommended for students, because the 4th column provides a valuable check
> on the performance of the Laplace-equation solver.
>
> In all cases, calculating the 4th row is extremely cheap, so you might as
> well do it, even if you plan to throw it away.  Before throwing it
> away  you should use it as a valuable check on the number-crunching.