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Re: Entropy and states of matter

I usually agree 100% with what David Bowman writes, but
this time I'd like to push back on a couple of points.
He wrote:

... for classical degrees of freedom
that contribute to the Hamiltonian quadratically there is an average
contribution of (1/2)kT to the system's internal thermal energy
relative to the ground state energy,

OK, that's carefully said. Call that proposition [A].

In the context of my statement:
(Classical thermo
requires us to count degrees of freedom, but doesn't
tell us how to do it.

He replied:
No it doesn't. Classical thermo doesn't even know about any
microscopic degrees of freedom in the first place--let alone count
them. Classical thermo is exclusively concerned with the
thermodyanmic macrostate and the macroscopic quantities describing
it and their inter-relationships.

Call that proposition [B]. I disagree with it. First let
me get some experimental facts on the table, and interpret
them a little; then let's see how all this fits in with
what David is saying.

Let's look at the data for the adiabatic exponent, also
known as the ratio of specific heats, often denoted k or
gamma.

1) Classical thermodynamics makes an iron-clad prediction
that a monatomic gas should have gamma = 4/3. The observed
data is pretty close to this value, but there are measurable
deviations. See the tables in:
http://www.physicsofmatter.com/Book/Chapters/Chapter5/5.html

2) Similarly, classical thermo makes an iron-clad prediction
that a diatomic gas should have gamma = 7/5. It's different
because the diatomic gas has rotational "degrees of freedom"
that the monatomic gas lacks. Again, the data isn't too far
off, but it's definitely off, and there is a clear trend
toward lower gamma at higher temperature.

3) You would think that a triatomic _linear_ molecule
like CO2 would behave just like a diatomic molecule, but
it doesn't. Gamma is markedly lower.

4) OK, you say, let's let the triatomic molecules have a
lower gamma. You think you've spotted a pattern? Consider
a pentatomic molecule, CH4. That should have a really low
gamma, right? Ooops, its larger than CO2. Closer to CO
than to CO2.

IMHO there is no way you can reconcile this data with
a classical notion of 1/2 kT per degree of freedom.
Even if you do a shameless smoke-and-mirrors routine
to figure out _a posteriori_ how many "degrees of
freedom" there must have been, you can't really make
it work.

I had this data very much in mind when I wrote:
Classical thermo
requires us to count degrees of freedom, but doesn't
tell us how to do it.

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

Indeed I suspect that the phase transition in question
(freezing) may be associated with a wholesale loss
of degrees of freedom.

No, I disagree. When a typical liquid freezes under typical
terrestrial conditions (i.e. not cryogenic and or hyperbaric ones)
then the relative positions (and orientations) of the molecules,
and their associated momenta (& angular momenta) tend to be quite
classical in their behavior.

I don't know enough about liquid water to respond specifically
to the "freezing" example. So instead
let's consider sublimation of ice directly to water
vapor. This is not particularly unusual under terrestrial
conditions, especially in the winter. ISTM the H2O molecules
are free to rotate in the liquid, and not free in the solid,
so this is a better illustration of my notion that we can
have a first-order phase transition where the kinetic energy
per molecule is not constant.

If we lose, say, some rotational
degrees of freedom, the kinetic energy per particle
will decrease.

But the relevant angular momenta tend to be already classical at the
transition temperature. After the molecules freeze they *will*
typically have their own rotational motions interfered with by the
steric torques exerted by neighboring molecules via Van der Waals
interactions. Such interaction torques typically may prevent most
molecules from ever executing a complete rotation. But this does not
mean that any rotational degrees of freedom have been 'lost'. Rather
it means that for the particles in close proximity to their
equilibrium lattice sites with their equilibrium lattice orientations
the individual molecules begin to effectively behave as a set of
classically vibrating torsion pendula.

That's a fine theoretical argument, but I don't buy it.
In particular, consider the !!careful!! wording of proposition
[A]. There is a requirement that the DoF in question be
classical. That's crucial! And I strongly suspect that the
rotational DoF in the water vapor is classical, while the
torsion-pendulum DoF in the ice crystal is not. That is,
I suspect that the hydrogen bonding is so strong that the
torsion-pendulum frequency is not small enough compared
to kT/hbar. Even if I'm wrong about this particular example,
I'm sure I can (with a little work) come up with another
example that indubitably illustrates my point.

The point remains that classical thermo requires us to
count degrees of freedom, but doesn't tell us how to do
it. In particular, we might agree on a set of possible
modes, but classical arguments won't tell us which of
the possible modes are "classical" and which are not.

*But* if the freezing process results in a very stiff fully
covalently bonded crystal, such as a diamond lattice, and *if* the
freezing process occurred at a temperature that was still mundane
(not likely, as we would expect it to instead occur in the multi-
kilokelvin regime) *then* the neighboring interactions in the solid
could possibly be so strong that the energy level spacing of some of
the high energy (so-called optical) modes would be large enough so
that the rotational and translational degrees of freedom would no
longer behave classically when describing how the various position/
momentum and angle/angular momentum conjugate pairs of degrees of
behave. In such a situation we would then effectively have a set
of quantum harmonic oscillators and quantum torsion oscillators that
were not excited enough at the relevant temperatures for the
equipartition theorem to hold for them, and these high energy
quantum oscillator modes then would be in various states of partial
freeze-out.

We agree on the physics. Perhaps we disagree as to how
mundane this freeze-out is. I claim it happens quite commonly.
The gamma of CH4 seems like pretty good support for this
claim.

...
Free Joule expansion into a vacuum is also not quasi-static
because it is not resolvable into a continuous sequence of
mutually equilibrated states. Joule-Thompson throttling
expansion is, again, a steady state situation--not a
quasi-static one.

Really? I thought that was exactly how one analyzed
the Joule-Thompson expansion.

We can't restrict thermodynamics to cases where everything
is always _exactly_ in equilibrium; otherwise there would
never be any heat flow from anywhere to anywhere. The
usual requirement is that things be close enough to
equilibrium that you can transfer energy without "too
much" entropy production, where "too much" is relative
to some relevant scale of the overall problem.

We are supposed to be ignorant of the system's
microscopics in characterizing the macrostate. We are not allowed to
just define the macrostate as the actual microstate. If we could so
define and characterize the system in terms of all its actual
microscopic dynamical variables, we would no longer be doing
stat mech/thermo--we would just be doing ordinary mechanics of the
system and the system would not have any entropy at all since all of
its dynamical variables would be known ...

That's the conventional wisdom.

... and in principle monitored.
Such extensive fine-scale observation for truly macroscopic systems
is forbidden by the uncertainty principle anyway.

"Known" is not the same as "monitored" or "observed". In
particular, consider the spin-echo experiment. If I know
what pulse-sequence to apply, I can turn what otherwise
would have been a high-entropy description into a low-
entropy description.

What we mean by macro-work is a macro-level change in the system's
energy because of a change in one or more macro-level parameters in
the system's Hamiltonian, (e.g. a change in the system's volume,
shape, magnetic moment, etc.). If the system's energy changes
by a noticeably macroscopic amount, and yet all of such energy-
changing macro-parameters are held rigidly constant, then the energy
change is due to heat.

How do you apply that to the spin-echo case?
You can't tell what's thermal and what's not
if I don't tell you what pulse-sequence to try.

If the system's 'macrostate' is exactly defined in terms of its actual
microstate (as in the Szilard engine) then there is no randomness to
average the generalzed force over when tweaking the system's
macrostate, and the supposed (statistically calculable) macro-work
expression just boils down to actual the ordinary mechanical work.

The Szilard engine is quite a bit more interesting than
that, because it has a heat reservoir whose macrostate
is, by hypothesis, not known. The one-particle "working
fluid" is used to extract energy from the heat reservoir.

... in this case the state of the engine is not
written in terms of a temperature *either*.

The state of the heat reservoir is most certainly
written in terms of its temperature.

That system is just not a thermodynamic system.

Completely disagree. A good fraction of what I know
about entropy I learned by thinking about the thermodynamics
of Szilard engines. In particular, there's a paper by
Zurek which lays out the only really-convincing argument
I've ever seen that Shannon entropy really is the same
as Carnot entropy ... and it is based in large part on
an analysis of a Szilard engine. (Another large part
goes back to the physics-of-computation ideas of
Landauer and Bennett).

... when things
*do* behave effectively classically, it would be silly overkill that
only further complicated and confused the situation to go beyond it
to pedantically insist on the more rigorously correct quantum
description.

But my point remains that classical arguments won't tell
you the limits of their own validity. One would think
that the gamma of CH4 is a perfectly well-posed classical
question, but if you try to answer it in classical terms
you've got three strikes against you before you even come
to the plate.

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

To repeat: There's a very great deal of what David is
saying that I agree with, and it seems ungracious to
focus on the places where I have doubts ... but I didn't
want to clutter 700 mailboxes by quoting a long list
of agreed-upon points......