Re: Capacitor problem

[David Bowman <dbowman@tiger.gtc.georgetown.ky.us>]

Donald Simanek wrote:

> Some versions of the capacitor problem do not specify resistanceless wire,
> and I've seen them give the answer "The energy goes into heating the wire"
> without ever considering the energy loss resulting from oscillation and
> radiation. But even if both mechanisms of energy loss are present, the
> final equilibrium situation has half the energy one started with. Now if
> the capacitors weren't of identical capacitance, the result wouldn't be
> exactly one half. What would it be, and would that depend on the size of
> the resistance?

Suppose that the initially charged capacitor had a capacitance of C_1 and
the second one had a capacitance of C_2 then the final equilibrium
electrostatic energy of the combined connected system is C_1/(C_1 + C_2).
This value does not depend on the resistance (which includes any radiation
resistance for the circuit, but assumes that the leakage resistance is
infinite; otherwise the final electrostatic energy is zero).

> In these problems, when the two things are identical, one initially has
> all the energy and then the energy is distributed between both, is the
> resulting energy *always* half the initial energy, no matter what objects
> we are talking about, and what methods of energy transfer? If so, why?
> Is there a more general and universal theorem applicable here?

The loss of half of the initial energy is a consequence of a generalization
of Hook's law.  In such a generalization the potential energy of a system
has a minimum as a function of some dynamical parameter.  When such a minimum
is a *quadratic* minimum then there is a restoring (generalized) force which
tends to drive the displaced system back to its potential energy minimum
value which is proportional to the amount of the displacement of the
dynamical parameter from its minimal potential energy value.  This
proportionality is correct in a sufficiently small neighborhood of the
minimal potential energy point.  This is because any quadratic minimum of a
smooth function can be approximated by a parabola in a neighborhood of the
minimum.  The (negative of the) slope of this parabola gives the restoring
force which is linear in that neighborhood.  As long as the system is
displaced by an amount which is within this neighborhood where the quadratic
potential energy approximation is valid then we can write:
U = A*x^2 where x represents the displacement of the dynamical parameter from
the minimal potential energy configuration, U is the potential energy of the
system measured from a zero level which has the minimal potential energy
state chosen as the state of zero potential energy, and A is a positive
constant characterizing the system.  (For the capacitor problem A = 1/(2*C)
and x = Q. For the tank problem A = g*(fluid density)*(tank cross section
area)/2) and x =fluid level in the tank.)  Suppose we connect two similar
such systems together such that the total potential energy is given by
U = A_1*(x_1)^2 + A_2*(x_2)^2.  Now assume that the initial state is one with
x_1 = X and x_2 = 0.  Assume also that the systems are connected in such a
way that there is a conservation constraint connecting x_1 and x_2 such that
x_1 + x_2 must be a constant.  Including this constraint (x_1 + x_2 = X) into
the equation for U gives: U = A_1*(x_1)^2 + A_2*(X - x_1)^2.  If the system
conserves energy then the system will have its x_1 value oscillate in time if
the system's Lagrangian includes a kinetic inertial term for the x_1 degree
of freedom.  If the system is not conservative, but rather dissipative, then  
the parameter x_1 will continue to move in such a way as to generate entropy 
in both the system and its surroundings (as per the 2nd law).  This
neccessitates that the system's energy will spread out among as many
degrees of freedom as possible.  This will eventually result in the system's
macroscopic potential energy being dissipated as much as possible (consistent
with the assumed x_1 + x_2 conservation law/constraint) into all the
microscopic degrees of freedom that interact with the system's
macroscopically displaced state and each other.  The final equilibrium will
effectively minimize the macroscopic U function above.  Now it is
straightforward to minimize the above formula for U over the possible values
of x_1 by setting dU/dx_1 = 0.  The resulting equilibrium (minimal U) value
for x_1 is: x_1 = X*A_2/(A_1 + A_2).  Substitution of this value into the
formula for U gives a final value of U_final = (A_1)*(A_2)*(X^2)/(A_1 + A_2).
The initial value for U was U_initial = A_1*X^2.  Thus, the fraction of the
initial U value which remains in the final equilibrium is
U_final/U_initial = (A_2)/(A_1 + A_2).  If both halves of the system are
identical so that A_1 = A_2, then we see that in this special case
U_final/U_initial = 1/2.                                                 QED

David Bowman
dbowman@gtc.georgetown.ky.us