Re: funny capacitor

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

At 05:23 AM 3/10/01 -0500, Bob Sciamanda wrote:
> A simple, transparent illustration of the traditional interpretation of the
> equation set Qi = Cij Vj and its inverse (the Einstein summation convention
> is implied):

It would be better to call this *one* of the traditional equations that
take the form Qi = Cij Vj, rather than to call it "the" traditional
interpretation.  Bob's equation has certain advantages and
disadvantages.  Other similar-looking equations have other advantages and
disadvantages.

As to the question of whether one member of the Qi = Cij Vj family is "more
traditional" than another:
  a) I care a little (but only a little!) what's traditional and what's not.
   -- On the do-not-care side: If we happen to discover a completely
novel, heretical, unorthodox, unconventional yet physically-correct
equation that gives charge in terms of voltage, "tradition" will not
prevent us from writing down this equation and discussing its consequences.
   -- On the somewhat care side:  We have some responsibility to avoid
unnecessary confusion and "negative transference" w.r.t other,
similar-looking equations.  In the present thread, I have suggested
reducing confusion by distinguishing, when appropriate, between the *full*
capacitance matrix and the associated *diminished* capacitance matrices.
  b) Neither the diminished capacitance matrix nor the full capacitance
matrix is novel or heretical.  Examples of the sensible use of the latter
are easy to find;  for instance
  http://ee.www.ecn.purdue.edu/~chengkok/ee695K/lecture2.pdf
  [beware of a confusing sign convention therein;  the off-diagonal
  matrix elements Cij (capital C) are written as -cij (lower-case c
  with a minus sign) so that the the cij will be positive numbers.
  The cij are not, _per se_, the matrix elements.]

> Consider the isolated system of two concentric, thin conducting spherical
> shells.

Minor point:  One needs to be slightly careful when reading the word
"isolated", because it means different things to different people.  In the
present context, the word is apparently _not_ intended to mean
"insulated".  See below (near the end) for some remarks about the
"insulated" case.

> The smaller sphere has radius a, charge Q1 and potential V1
> relative to infinity.

Ahh, "relative to infinity".  That's an important statement.  We can
express that as an equation
        V3 = 0          (equation 1)
where V3 is the voltage at "infinity".

My main point is that equation (1) is an assumption.  It is not a law of
physics.  Perhaps Bob's point is that equation (1) is
"traditional".  Certainly it is one of the possible traditional
assumptions, but IMHO it would be awfully dogmatic to claim that it is the
only possible assumption.

> The larger sphere has radius b, charge Q2 and
> potential V2 relative to infinity.
> Since we know the fields and potentials of a uniform, spherical shell of
> charge, we can quickly write, using k = 1/(4*PI*epsilon):
>
> V1 = k ( Q1/a + Q2/b )
>
> V2 = k ( Q1/b + Q2/b )                  [(equations 2)]

Collecting the foregoing equations all in one place, we have

        V1 = k/a Q1 + k/b Q2
        V2 = k/b Q1 + k/b Q2            (equations 3)
        V3 = 0

As the saying goes, "you have to measure voltages with respect to
something."  Using V3=0 as a reference is conceptually straightforward:
 -- To measure V1-V3, drag a test charge from location 3 to location 1
    and measure the energy-per-unit-charge that is required.
 -- To measure V2-V3, drag a test charge from location 3 to location 2
    and measure the energy-per-unit-charge that is required.

The fact the location 3 is "infinitely" far away makes this rather
laborious in practice, but conceptually it is OK.

Now suppose we send Joe Schmoe into the lab with only vague instructions,
namely to measure the voltages on the spheres.  There is a fair chance that
Joe will hook the black lead of the voltmeter to the large sphere, and use
the red lead to measure all the voltages in the vicinity.   This will
result in:

        V1 = (k/a - k/b) Q1 +   0 Q2
        V2 = 0                                  (equations 4)
        V3 = -k/b Q1        - k/b Q2

Joe's measurements are inconsistent with the values predicted by equation
(3).  But are Joe's measurements wrong?  IMHO they are not wrong.  They
differ from equation (3) by a gauge transformation.

There is an analogy here to Galileo's principle of relativity:  If I
measure velocities in one frame, and you measure velocities in another
frame, the numbers will come out different, but it doesn't mean that your
velocities are wrong (or that mine are wrong).  To argue that one frame is
"traditional" to the exclusion of all other frames would be little-endian
dogma, not physics.
  http://www.bruchez.org/erik/litt/endianne.html

> These are easily inverted to:
> Q1 = {ab/(k(b-a))} * {V1 - V2}
>
> Q2 = {b/(k(b-a))} * {-aV1 + bV2)                [(equations 5)]

OK, and we can rewrite this in matrix form, to wit:

        Cij =  ab/(k(b-a))   -ab/(k(b-a))       (equations 6)
              -ab/(k(b-a))   b^2/(k(b-a))

which we recognize as the *diminished* capacitance matrix for the (V1,V2,0)
subspace of the (V1,V2,V3) system.

> Note that neither coefficient matrix is singular. (multiply the two matrices
> and you will get the identity matrix.)

We are not surprised that a diminished capacitance matrix is invertible.

OTOH we know that anybody who (accidentally or otherwise) calculates the
*full* capacitance matrix will be faced with a singular matrix... and will
have to project down to one of the diminished capacitance matrices in order
to get something that is invertible.

> Note that the total charge  Q1 + Q2 = (b/k)V2  is not fixed, since V2 is a
> freely adjustable variable.  This allows us to consider all possible charge
> states of the system.  Q1 and Q2 are freely adjustable, without constraint,
> and will determine the V's.  Or, the V's are adjustable and will determine
> the Q's.

That means that the "isolated" (Q1,Q2) system is isolated but not
"insulated" from the rest of the world.  There needs to be a hole in the
boundary so that we can throw charge "overboard" when making a change in Q1+Q2.

If we switch to consider the *insulated* system, a couple of interesting
things happen:
  *) Any change in Q1 must be balanced by an equal-and-opposite
     change in Q2, because we can't throw charge overboard.
  *) We can't measure V3, and we can't measure anything relative to V3,
     because we can't drag a test charge to/from location 3,
     because the insulator gets in the way.

In this case, the equations simplify quite a bit.  The main thing that
survives after simplification is
        delta_V = C Q1                  (equation 7)
where
        delta_V := V1 - V2
and where C is the two-terminal capacitance, which in this example has the
value
        C = (k/a - k/b)

This C can be viewed as a 1x1 matrix which is a diminished capacitance
matrix.  The full capacitance matrix for this two-object system is given by
        C11 = C22 =  (k/a - k/b)
        C21 = C12 = -(k/a - k/b)

As always:
 -- The Q(V) problem has a unique solution
 -- The unrestricted V(Q) problem has NON-unique solutions
 -- The delta_V(Q) problem has a unique solution