Re: There's work, and then there's work

[John Mallinckrodt <ajm@CSUPOMONA.EDU>]

Robert Cohen wrote:

> I would like to respond to David Rutherford's argument that energy
> is associated with both the heat gained by the surroundings (via
> what he calls the transverse component velocity) and the kinetic
> energy associated with the current (which he relates to the
> longitudinal component velocity).
>
> This is actually a common source of confusion for my students and so
> I would like to see it better addressed than it has.  DR is right
> that, at any given time, only part of the energy (associated with
> the kinetic energy of the electrons) will be converted to heat as
> measured by the calorimeter experiment.  However, the key point that
> I think my students (if not DR) miss is that eventually the current
> will stop.  At that point, all of the energy will be dissipated to
> the calorimeter.  Consequently, the calorimeter should be measuring
> *all* of the energy dissipated by the resistor (which, in turn, must
> be equivalent to the initial energy stored by the capacitor since
> the final current in the circuit is zero).

We can do some simple calculations with typical values to address
this question and see that the words "utterly negligible" almost fail
to be strong enough to describe just how incredibly miniscule the
kinetic energy represented by the drift of the conduction electrons
really is.

Suppose we have a C = 1000 microF cap initially charged to Vo = 10 V
that is discharging through a small R = 10 ohm resistor and L = 20 cm
of copper wire with a diameter of d = 1.00 mm.

Using the density of copper (rho = 8.9 g/cm^3), the atomic mass of
copper (M = 64 g/mole),  and the assumption of one conduction
electron per copper atom, we find a conduction electron number
density of n = (rho/M)*N_avog = 8.5 x 10^22 electrons/cm^3 and a
total of N = n(pi*d^2*L/4) = 1.3 x 10^16 conduction electrons.

The initial energy stored by the capacitor is U = 1/2 CV^2 = 50 mJ.

The initial current is I = Vo/R = 1.0 A which means that the
electrons are collectively moving with a drift speed of v = I/e*n =
75 micrometers/s.

The TOTAL kinetic energy associated with the electrical current is,
therefore, K = N(1/2 mv^2) = 3 x 10^-23 J.    (!!!!!)

This maximum (and entirely negligible) kinetic energy vanishes
exponentially over the ensuing few time constants (t ~= 5*RC = 50
ms).  The average rate of dissipation of energy initially stored in
the capacitor during this period is P = U/t ~= 1 J/s.

Thus, no more than about 30 trillionths of 1 picosecond (!!!) would
be required for the electrons to gather the maximum kinetic energy
they will EVER have during the process.  (The actual process takes a
"lot longer" than that--on the order of 1000 picoseconds--because of
complications resulting from the finite propagation speed of the
electromagnetic fields that establish and sustain the discharge
current.)

The picture that emerges is that the energy being lost by the
capacitor appears ENTIRELY and, for all practical purposes
INSTANTANEOUSLY in the form of "heat" (for lack of a better word) in
the wire and the resistor.  The drifting electrons are merely a very
"light weight" intermediary for the transfer and at NO time account
for more than a billionth of a billionth of 1% of the total energy
being dissipated.

--
   A. John Mallinckrodt        http://www.csupomona.edu/~ajm
   Professor of Physics      mailto:ajm@csupomona.edu
   Physics Department         voice:909-869-4054
   Cal Poly Pomona              fax:909-869-5090
   Pomona, CA 91768-4031     office:Building 8, Room 223