On 08/25/2003 09:47 AM, I posed a riddle:
>
> What can it possibly mean to
> talk about the low-frequency dielectric constant of
> water (or other materials with nonzero conductivity)?
I believe the question is not 100% answerable in
principle, but reasonable answers can be found in
many practical situations.
The basic physics looks like this: The charge-versus-voltage
characteristic of water has multiple contributions:
a) a little piece from the polarizability of the
atoms themselves. This can be modelled as a mass on
a spring, where the mass is the mass of an electron
and the "spring" is the electrical force attaching it
to the nucleus. This contribution is independent of
temperature.
b) a piece due to the alignment of the permanent dipole
of the molecule. The mass is much higher (since nuclei
have to move, not just electrons). The restoring force
is due to thermal agitation.
c) a piece due to DC conductivity.
The bottom line is the sum of all the various contributions.
We can get a unified view by treating each of the
contributions as a damped harmonic oscillator.
The equation for each DHO has three terms:
-- a mass term (force proportional X dot dot)
-- a spring term (force proportional to X), and
-- a damping term (force proportional to X dot)
where X is some sort of displacement variable and dot
means time-derivative.
For each DHO, let's apply a smallish sinusoidal voltage
and see how the charge responds. By Floquet's theorem,
the response will also sinusoidal (since we are keeping
the excitation small enough so that things are linear).
For each oscillator
-- At very high frequencies, the mass term will dominate.
The dielectric constant will be inversely proportional
to the square of the frequency, and the response will
be 180 degrees out of phase with the excitation.
-- At very low frequencies the spring term (if any) will
dominate. The dielectric constant will be independent
of frequency. The response will be in phase with the
excitation.
-- At intermediate frequencies, the damping term will dominate.
The dielectric constant will be inversely proportional
to the frequency, and the response will be 90 degrees
out of phase with the excitation.
Note that for the conductivity-related contribution, the
spring term is non-existent! At low frequencies, the
damping term is the leading-order term.
So this is the answer to the puzzle as posed: At sufficiently
low frequency, contribution (a) [atomic polarizability] and
contribution (b) [permanent dipole] are both in their
low-frequency regimes, and give a contribution that is in
phase with the excitation. Contribution (c) [conductivity]
is 90 degrees out of phase with the excitation and be
somewhat arbitrarily (see footnote) designated as not part
of "the" dielectric constant.
The operational implication is that if you want to measure
the dielectric constant of water, you reeeeally want to
use a phase-sensitive detector (aka "lock-in amplifier").
Pedagogically speaking, this has a great deal of upside
potential. Students ought to get experience using lockins,
the more the better. In a typical real-world physics lab,
you can't swing your arms without hitting a lockin.
Physicists depend on lockins the way carpenters depend on
hammers. As I recall, my thesis apparatus had half a
dozen lockins bolted into the racks, including one operating
at about 35 Hz, one operating at a few kHz, and a home-made
one operating at more than 1 GHz.
Phase-sensitive instruments are also very handy for
analyzing analog electronics (feedback loops and such).
You can use this to motivate your students who don't
happen to be physics majors.
==============================
Footnote:
I consider it somewhat arbitrary to exclude the conductivity
contribution from "the" dielectric constant. IMHO one
should report the imaginary part (out-of-phase) as well
as the real part (in-phase). Also note that things get
really weird if you have water trapped in a porous medium
(below the percolation threshold). Then the ions will move
a mesoscopic distance and stop. So the very-low-frequency
charge-versus-voltage response will be huuuuge but finite,
and in phase with the excitation.