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Re: funny capacitor

At 10:59 AM 3/8/01 -0500, Ludwik Kowalski wrote:
Last week the following problem was formulated.
Given several conducting objects which are not
connected to anything (electrically floating) find
their potentials (V1, V2, V3, etc.) when the net
charges on the objects are given (Q1, Q2, Q3, etc.).

Yes, that's an interesting question Call that the V(Q) question.

It is related in a deep and NONtrivial way to the other interesting
question we discussed, namely finding the Qi given the Vj. Call that the
Q(V) question.

I believe that mother nature has one, and only one,
reproducible solution.

It will turn out that there is more than one solution. But let's try to
find some solution(s), and see how far we get.

In trying to find that solution,
for a particular example (funny capacitor), two steps
were clearly identified. The first step consists of
calculating the influence coefficients, Cij, defined by:

Q1=C11*V1 + C12*V2 + C13*V3 + C14*V4
Q2=C21*V1 + C22*V2 + C23*V3 + C24*V4 Eqns (2)
Q3=C31*V1 + C32*V2 + C33*V3 + C34*V4
Q4=C41*V1 + C42*V2 + C43*V3 + C44*V4

OK, that's the Q(V) question.

In this example objects 1, 2 and 3 were metallic plates
while object 4 was the enclosure (infinity).

Or not infinity. Or not even a full enclosure. The key concept is that
the list of objects (four objects in this case) include all the objects
with which we exchange charge. If we are exchanging charge with the
enclosure, that case is a routine corollary of the general formalism.

For the chosen
geometrical arrangement the Cij coefficients, calculated
according to JohnD's prescription, turned out to be:

+2.86, -0.46, -1.15, -1.25
-0.46, +2.86, -1.15, -1.25
-1.15, -1.15, +3.26, -0.96
-1.25, -1.25, -0.96, +3.46

Fine.

BTW note that the Cij values calculated according to the prescription are
gauge-invariant. They depend on geometry and not on the choice of gauge.

Be warned that in the literature there is some difference of opinion as to
what constitutes "the" Cij matrix. To avoid any possibility of confusion,
let's call this the *full* Cij matrix, in contrast with various species of
*diminished* Cij matrices.

Knowing these coefficients we can calculate charges for
any given set of potentials. But that is not our problem.

OK, we are now returning to the V(Q) question.

We want to calculate potentials when charges are known.
This is step 2. Many textbooks tell us that this simply a
matter of solving n equations with n unknowns, for example,

+2.86*V1 - 0.46*V2 - 1.15*V3 - 1.25*V4 = -189
-0.46*V1 + 2.86*V2 - 1.15*V3 - 1.25*V4 = +143
-1.15*V1 - 1.15*V2 + 3.26*V3 - 0.96*V4 = +65.2
-1.25*V1 - 1.25*V2 - 0.96*V3 + 3.46*V4 = -19.3

where numbers on the right side are charges Q1,
Q2, Q3 and Q4 (in arbitrary units) calculated at
the same time as Cij coefficients. These are charges
to produce: V1=-50, V2=+50, V3=+20 and V4=0.
John convinced us that the above equation can not
be solved because the determinant of the matrix of
coefficients is always zero (due to gauge invariance
and to the law of charge conservation).

Either charge conservation OR gauge invariance suffices to guarantee it is
singular. So this is doubly guaranteed singular.

Trying to solve the above equation is like trying
to find out the intersection point of two parallel
lines, wrote John. The two parallel lines do not
intersect.

Actually they *do* intersect. They intersect along their whole
length. They coincide. It's just that they don't have a _unique_ point of
intersection.

Does it it mean our problem does not have
a single unique solution?

Here's the deal: We have four independent voltages. We can set the
voltages to anything we want, and turn the crank on Laplace's
equation. That's the Q(V) problem. When we do that, we observe that Dr.
Laplace does *not* give us four independent charge values. Given three of
them, we can always infer the fourth without looking. We are not ASSUMING
charge conservation; we are merely observing that if we sum over all
objects with which we exchange charge, the results of the Q(V) problem
always conserve charge.

This is not a trivial or obvious result; for instance, a buggy
implementation of the Laplace-equation solver might not conserve
charge. But let us restrict our consideration to non-buggy implementations.

However, the V(Q) problem does have four independent variables, namely
three charges and one gauge.

John eliminated one row and one column
from the Cij matrix. This produced a matrix

2.86 -0.46 -1.25
-1.15 -1.15 -0.96 (equation X)
-1.25 -1.25 3.46

In fact it was row 2 and column 3 that were dropped. It was an arbitrary
choice.

Equation (X) is an example of a *diminished* Cij matrix.

whose inverse is:

0.30 -0.18 0.06
-0.30 -0.48 -0.24 (equation Y)
0.00 -0.24 0.22

Fine.

My question is this. How can we use the elements of
this 3 by 3 matrix to solve the problem? Presumably,
the one, and only one, solution is:

V1=B11*Q1 + B12*Q2 + B13*Q3 + B14*Q4
V2=B21*Q1 + B22*Q2 + B23*Q3 + B24*Q4 Eqns (2)
V3=B31*Q1 + B32*Q2 + B33*Q3 + B34*Q4
V4=B41*Q1 + B42*Q2 + B43*Q3 + B44*Q4

There is no unique solution. We *can* find some solution(s), because our
parallel lines coincide. But in general there will be a multiplicity of
solutions for Vk (differing from one another by a change of gauge).

Remember: the V(Q) problem has four independent variables, namely three
charges and one gauge.

There are many ways of dealing with this situation. Here is one way (not
to be construed as the only way).

We can pick (arbitrarily!) one of the charge-nodes and write it in terms of
the others. For instance, we could write
Q2 = -(Q1 + Q3 + Q4) (equation 3)

Since this is a linear equation, and since equation (2) is a linear
equation, we can use equation (3) to eliminate one of the columns from
equation (2). The simplest thing is to eliminate the Q3 column:

V1=(B11-B12)*Q1 + 0*Q2 + (B13-B12)*Q3 + (B14-B12)*Q4
V2=(B21-B22)*Q1 + 0*Q2 + (B23-B22)*Q3 + (B24-B22)*Q4
V3=(B31-B32)*Q1 + 0*Q2 + (B33-B32)*Q3 + (B34-B32)*Q4 (equation 4)
V4=(B41-B42)*Q1 + 0*Q2 + (B43-B42)*Q3 + (B44-B42)*Q4

If you drop the Q2 column, this can be considered a 4x3 matrix that
operates on vectors of the form [Q1 Q3 Q4]. Or you can leave it as a 4x4
matrix with a column of zeros.

Also note that because we are solving for the "intersection" of two
coincident lines, equation (4) cannot possibly represent the general
solution. There will be a whole family of Vk values; family members will
differ by a change of gauge. Any given set of Bki values can only specify
one member of this family. Therefore, the problem as specified above
cannot possibly lead to a UNIQUE set of Bki values.

If we want a unique solution of some kind, we can ask a slightly different
question. We can ask where the solution-family crosses some plane of
interest. For instance, we can (again arbitrarily!) choose the plane
V3=0. Subject to that arbitrary choice, we (finally!) have a well posed
problem, namely finding Aki such that

V1= A11*Q1 + A13*Q3 + A14*Q4
V2= A21*Q1 + A23*Q3 + A24*Q4 (equation 5)
V4= A41*Q1 + A43*Q3 + A44*Q4

This notation is getting a bit ugly, because if we consider A as a 3x3
matrix its matrix elements aren't numbered properly... because the elements
of the vectors on which it operates aren't numbered properly either. Sigh.

We need sixteen coefficient Bij to calculate V's
from Q's. What are the values of these coefficients
for our specific example?

We only need nine Aki values, and they are given by equation Y, above.

============

Starting from the particular solution-point given by equation (5), we can
exhibit the entire solution-set by shifting the gauge. Explicitly, the
general solution is:

V1 = g + A11*Q1 + A13*Q3 + A14*Q4
V2 = g + A21*Q1 + A23*Q3 + A24*Q4 (equation 6)
V3 = g
V4 = g + A41*Q1 + A43*Q3 + A44*Q4

for arbitrary g.

Equation (6) describes *the* solution-set for the V(Q) problem. Other
methods may lead to the same solution-set, but no other solution-set is
possible.

Less-than-general methods may pick out some element of the solution set,
but it would be wrong to claim uniqueness for such an element.