Re: funny capacitor

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

At 02:18 PM 3/8/01 -0500, Ludwik Kowalski wrote:
> I was impressed that Dr. Laplace took
> care of the charge conservation, and of the
> Cij=Cji automatically, I did not have to ask
> for it, it was the consequence of solving the
> equation.

Yes, that's the right way to look at it.
And yes, that's an important observation.

We can, using a little theory, get some insight as to _why_ this was
guaranteed to happen in any non-buggy Laplace-equation solver.

A) We observe (not assume!) the sum across each row of Cij is zero.  That
is easy to explain.  It is a direct consequence of Maxwell's equations,
because:
        del^2 (phi + const) = del^2 (phi).
and in accordance with convention, this is called "gauge invariance".

B) We observe (not assume!) the sum down each column of Cij is zero.  That
is almost as easy to explain:
   -- This can be described as conservation of charge,
     -- which is a consequence of conservation of electric flux as
        it propagates from conductor to conductor
       -- which is a consequence of applying Gauss's law to
           Maxwell's equations

C) We observe (not assume!) that Cij=Cji.  The explanation for this will be
given below.


> 4) But how can one justify Cij=Cji? John
> justified it be referring to Dr. Maxwell. Can
> this be done on the basis of what is explained
> in a typical introductory physics textbook? The
> question is "why does the coefficient of influence
> of object i on object j is the same as that of
> object j on object i? That may be obvious for
> two identical spherical conductors but not for
> two conductors whose sizes and shapes are very
> different.

Well, I have not heretofore done a very good job of justifying it.  It's
more complicated than charge conservation or gauge invariance.

1) Easy case:  Observe that if the full Cij matrix is 2x2 (i.e. the case of
only two conductors) then charge conservation and gauge invariance are
sufficient to force the Cij matrix to be symmetric.
                a    b
        Cij =                   (equation 1)
                c    d

        a+b = 0 = a+c           (equation 2)

but don't get lulled into complacency by this easy case.

2) Meanwhile, for three or more conductors, the symmetry is deeper than
anything you can get from charge conservation and/or gauge invariance.  For
example, consider this alleged Cij matrix
               1   0  -1
        Cij = -1   1   0                (equation? 3)
               0  -1   1
which sums to zero along rows and along columns, but is not symmetric.

BTW, any matrix M can be written as Mij = M(ij) + M[ij] where the symmetric
part of M is given by
        M(ij) = 1/2  (Mij + Mji)        (equation 4a)
and the antisymmetric part is given by
        M[ij] = 1/2  (Mij - Mji)        (equation 4b)

So for our alleged capacitance matrix we have the symmetric part:
                 1    -.5  -.5
        C(ij) = -.5    1   -.5          (equation 5)
                -.5   -.5   1

and the antisymmetric part:
                 0    .5   -.5
        C[ij] = -.5    0    .5          (equation? 6)
                 .5   -.5   0

If we are going to rule out Cij matrices with nonzero antisymmetric parts,
we cannot do it based on Gauss's law, nor based on del^2(const) = 0.  We
will have to go back to Laplace's equation and find some other properties
that we can use.

WLoG consider the following three objects:

         11111
        1111111
        1111111            2222
         11111            222222        (figure 1)
                          222222
                           2222
                33
               3333
                33

and in accordance with the usual prescription we will calculate the full
capacitance matrix according to
        Cij = (delta Qi) / (delta Vj)           [all Vk const except Vj)

That is, we hold N-1 of the objects at constant potential, wiggle the
voltage on the remaining one, and find out what happens to the charge on
each and every object by turning the crank on Laplace's equation.

During this procedure, WLoG we can assume there will come a time when there
is 1 volt on object 1 and zero volts on the others.  We can then map out
the "field lines".  More formally, we can find **flux tubes**, examples of
which are shown in the following:

         11111
        1111111
        1111111yyyyyy      2222
         11111      yyyyyy222222        (figure 2)
              x           222222
               x      zzzz 2222
                33 zzzz
               3333
                33

where there are three flux tubes, denoted by xx...xx, yy...yy, and zz...zz
respectively.  Each flux tube has the property that
        E dot dA = 0
everywhere along the side-walls of the tube (but not the end-caps), where E
is the electric field and dA is the element of side-wall area.

Later during the procedure, there will come a time when there is 1 volt on
object 2, and zero on the others.  We can find another set of flux tubes.

Similarly there will come a time when there is 1 volt on object 3, and zero
on the others.  We can find a third set of flux tubes.

Now let's consider the superposition of these three cases;  that is, 1 volt
on all three objects on the same time.  The charge-distributions will
simply be additive, because everything is linear.  Also the
flux-distributions will be additive.

The alleged Cij matrix given in equation (3) corresponds to flux flowing
around in circles:
        -- from object 1 to object 2, then
        -- from object 2 to object 3, then
        -- from object 3 back to object 1.

That flux distribution is perfectly consistent with Gauss's law.  If you
don't like this flux distribution, you will need something other than
Gauss's law to disprove it.

You could more-or-less model this flux distribution by using three "D
cells" arranged in a circle:  Flux lines go from the + end of one battery
to the - end of the next...   We let the V of each object denote the
voltage at the - end, and let Q denote the total charge...  And it
basically works.  However, this obviously violates the requirement that our
objects be conductors, i.e. equipotentials.

It's interesting that we were able to prove charge conservation without
invoking the requirement that the objects must be equipotentials.  But to
prove Cij=Cji, we will have to invoke it.

So let's invoke that, and also invoke the assumption that there are no
time-varying magnetic fields in the problem.  Then we can see that the
flux-flowing-in-circles scenario violates Maxwell's equations because it
would have circulation, i.e. del cross E nonzero.

We conclude that the Cij in equation (3) does not describe a genuine,
physical capacitance matrix.

To summarize:
  *) No flux flowing in circles implies for any genuine physical
situation, the antisymmetric part of the full capacitance matrix
vanishes:  C[ij] = 0.
  *) Therefore Cij must be symmetric.
  *) This symmetry is a consequence of a Maxwell equation, while charge
conservation is a consequence of a different Maxwell equation.