Chronology Current Month Current Thread Current Date
[Year List] [Month List (current year)] [Date Index] [Thread Index] [Thread Prev] [Thread Next] [Date Prev] [Date Next]

[Phys-l] physics of dissipation (was: T dS versus dQ)

On 01/13/2010 07:35 PM, I wrote:

For example, stirring a gas is a dissipative process.
Sudden movements of a piston stir the gas; the
equations of fluid dynamics, even in their simplest
form, leave no doubt about this.

Here is another way of coming to the same conclusion.

I know there are about 50% high school teachers on
this list, and the following explanation is not one
I would present directly to typical HS students ...
but I'm not talking directly to students today. The
folks on this list may find this interesting, or may
find it sufficient to just file away the qualitative
fact that we are on reeeeally solid grounds when we
say that sudden movements of the piston stir the gas.

Let's forget about all the complexities of the
aforementioned fluid dynamics approach. Instead
let's take the quantum mechanical approach. Let's
simplify the gas down to a single particle, the
familiar particle in a box, and see what happens.

As usual, we assume the box is rigid and thermally
isolated / insulated / whatever. No entropy flows
across the boundary of the box. Also, no energy
flows across the boundary except for the work done
by the piston.

Since we are interested in entropy, it will not
suffice to talk about "the" quantum state of the
particle. The entropy of any particular quantum
state (microstate) is zero. We can however
represent the thermodynamic state (macrostate)
using a density matrix ρ.

For some background on density matrices in the
context of thermodynamics, see

The entropy is
S := -Tr ρ log ρ [1]

which is _the_ gold-standard most-general definition
of entropy; in the classical limit it reduces to
the familiar workhorse expression S = - ∑ P log P.
Reference: op. cit.

For simplicity we consider the case where the initial
state is a pure state, i.e. a single microstate.
That means the initial entropy is zero, as you can
easily verify. Hint: equation [1] is particularly
easy to evaluate in a basis where ρ is diagonal.

Next we perturb our particle-in-a-box by moving one
wall of the box inward. We temporarily assume this
is done in such a way that the particle ends up in
the corresponding microstate, i.e. identical to the
original quantum state except for the shorter
wavelength as required to fit into the smaller box.
It is a straightforward yet useful exercise to show
that this does P dV "work" on the particle. The KE
of the new state is higher than the KE of the old

Now the fun begins. We retract the previous assumption
about the final state; instead we calculate the final
macrostate using perturbation theory. In accordance with
Fermi's golden rule we calculate the _overlap integral_
between the original quantum state (original wavelength)
and each of the possible final quantum states (slightly
shorter wavelength).

In the original basis set, each basis function is
orthogonal to each of the other original basis
functions. In the final basis set, each of the basis
wave functions is orthogonal to each of the other final
basis wave functions, but it is only approximately
orthogonal to the original unperturbed state function.
So the previous assumption that the particle would
wind up in the corresponding state is provably not
quite true; when we do the overlap integrals there
is always some probability of transition to nearby

It is straightforward to show that if the perturbation
is slow and gradual, the corresponding state gets the
lion's share of the probability. Conversely, if the
perturbation is large and sudden, there will be lots
of state transitions. The final state will not be a
pure quantum state. It will be a mixture. The entropy
will be nonzero, i.e. greater than the initial entropy.

The repeated assertion that the compression of the gas
is the same no "matter whether done slowly or quickly"
is completely untenable.

To summarize:
-- slow and gradual ==> isentropic, non dissipative
-- sudden ==> dissipative

So, like I said, we are on reeeally solid grounds
when we say that thermally insulated + gradual is
isentropic, while a sudden motion of the piston is
dissipative. Saying that the system is adiabatic in
the sense of thermally insulated does *not* suffice
to make it adiabatic in the sense of isentropic.

Note that in the quantum mechanics literature the slow
and gradual case is conventionally called the "adiabatic"
approximation in contrast to the "sudden" approximation.
These terms are quite firmly established ... even though
it conflicts with the also well-established convention
in other branches of physics where "adiabatic" means
thermally insulated; see next message.


I file this example of gaseous dissipation under the
same category as Rumford's cannon boring. Actually
I find it to be a better pedagogical example, because
we understand the microscopics of dissipation in a
gas better than we understand the microscopics of
metal-on-metal sliding friction.

However, we can discuss sliding friction using many
of the same ideas. Let's model friction in terms
of _asperities_ on each metal surface. Each of
the asperities sticks and lets go, sticks and
lets go. When it lets go it wiggles and radiates
ultrasound into the bulk of the metal. This
produces in the short term a nonequilibrium state
due to the sound waves, but before long the sound
field dissipates, depositing energy and creating
entropy in the metal.

Again, if you think in terms only of the
(average force) dot (average dx)
you will never understand friction or dissipation.
You need to model many little contributions of the
form (short term force) dot (short term dx) and
then add up all the contributions. This is where
you see the work being done against the radiation

Also: there is a nice introduction to the idea
of "radiation resistance" in Feynman volume I
chapter 32.