Re: Entropy and states of matter (VERY long)

[David Bowman <David_Bowman@GEORGETOWNCOLLEGE.EDU>]

Regarding John Denker's comments:

> ...  Entropy and energy and temperature are related, but one should
> not be too glib about seeming to define one in terms of the others.

I'm not so sure about the glibness factor, but thermodynamic
temperature *is* defined as the reciprocal of the rate at which the
equilibrium entropy function increases with increasing values of the
system's internal energy under rigid conditions that prevent
macroscopic parameters on the system's Hamiltonian from changing so
that no work is done on the system while the change takes place.  The
(macroscopically) infinitesimal change (due to heating) is taken to
be quasi-static so that the system doesn't fall out of equilibrium
while the change takes place which would otherwise prevent the
equilibrium entropy function (of the macrostate parameters) from
actually describing the system's entropy if it had fallen out of
equilibrium during the change.  (For more details consult just about
any textbook on stat mech.)  

From the point of view of stat mech, not all thermodynamic
quantities are equally fundamental.  The quantities entropy and
energy are quite fundamental.  The entropy is probably most
fundamental. In thermo our macroscopic system's exact microscopic
state is not known and is unknowable.  At best the system has its
macrostate (knowable state) defined in terms of a suitable set of
macro-level parameters that determine the macrostate once they
are all specified.  The average further minimal amount of extra
information necessary to determine, with certainty, the system's
exact microstate given only this macro-level description *is* the
system's entropy.

Also quite fundamental is the system's energy.  The energy is the
value of the quantity (Hamiltonian) that generates a canonical
transformation of the system that results in the system being
subject to an infinitesimal virtual displacement in time.  The
internal energy is the system's energy as evaluated in the
frame (i.e. rest frame) in which its total momentum is zero.  (Here
the total momentum happens to be the value of the particular
quantity that generates a canonical transformation for an
infinitesimal virtual displacement of the system in space.)  In
thermodynamics the internal energy used is a coarse-grained
macroscopic value that is averaged over the microscopic fluctuations
that are too tiny to be detected at the macro-level.

But temperature is a *derived* concept/quantity that describes the
rate at which these two fundamental quantities vary w.r.t. each other
in equilibrium when no macro-work is being done on the system during
the variation.  A thermodynamic system that is 'hot' is one that
would have its rate of entropy increase be very low if it absorbed a
tiny amount of heat in a work-free quasi-static manner.  OTOH, a
macroscopic system is considered to be 'cold' if a little heat would
tend to greatly further microscopically 'disorder' it.

> > The kinetic energy of the particles in the transition is unaffected
> > because the transition happens at a fixed temperature.
>
> I'm not convinced.  Can you prove it?  I suspect it's
> not true.

Well I can't prove a first order transition always must happen at a
constant temperature (since it is possible to it to take place
adiabatically and that can result in a changing temperature as the
transition proceeds, or the system can be constrained to follow along
a line of two-phase coexistence and then change the relative amounts
of the two phases as it goes--changing its temperature as it goes).
Nevertheless a first order transition *is* discontinuous in at least
some of its extensive macroscopic parameters as a function of the
temperature (and/or pressure, chemical potential, or some other
relevant intensive parameter(s)).

The reason the overall kinetic energy of the particles is fixed when
the temperature is fixed is because 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, and the kinetic energies of
these classical particles *do* contribute to the Hamiltonian
quadratically because the system has a Newtonian relationship between
a momentum and its corresponding kinetic energy.  In essence, the
equipartition theorem holds for classical quadratic degrees of
freedom, and all the individual momentum components each give a
quadratic contribution to the kinetic energy part of the Hamiltonian.
Since we are assuming that classical mechanics is sufficiently
accurate when doing out stat mech for the system, this means that we
are implicitly assuming that our system is hot enough so that the
relevant momentum degrees of freedom describing the relative motions
of the particles are accurately described by classical mechanics--
which for the motions of whole molecules means that the temperature
is much hotter than both the discrete energy spacing of the
energy levels for the various relevant microscopic modes that might
arise from Van der Waals bindings between molecules, *and* that the
temperature is high enough so that the so-called thermal wavelength
is much smaller than the typical interparticle spacing.  Any
canonical (p,q) pair in the system's Hamiltonian for which classical
mechanics is not a valid approximation requires that quantum stat
mech be applied to it rather than classical stat mech.  Since the
equipartition theorem is only valid for sufficiently classically-
behaving degrees of freedom we realize that any such mode for which
QM is an essential part of the description in the temperature range
of interest will not be subject to the equipartition theorem.

When I claimed that the kinetic energy of the particles in our
system was constant when the temperature was constant I had assumed
(as is typically the actual case for normal freezing of fluids, etc.)
that the degrees of freedom for those kinetic energies we were
discussing where those of the translation of the centers of mass of
the molecules at hand and their internal rotations.  Also it was
assumed that any intermolecular Van der Waals binding forces between
neighboring molecules in the solid and liquid phases were weak enough
so that the relevant temperature was plenty high enough so that the
relevant momenta (contributing quadratically to the Hamiltonian) *did*
behave classically near the transition temperature.  This is usually
a good assumption for typical systems found under typical terrestrial
conditions.  Also at these temperatures the angular momentum degrees
of freedom of the molecules about their centers of mass are also
quite classical in their behavior.  Even the lightest molecules
(worst case being) of Hydrogen have their rotational angular momenta
behaving quite classically at temperatures in the vicinity of room
temperature.  The angular momenta, too, contribute to their kinetic
energies quadratically in the Hamiltonian.

When I mentioned the kinetic energy being determined by the

temperature and it being constant when the temperature was constant,
I did not mean to necessarily mean the kinetic energies associated
with various *intra*-molecular motions of the various degrees of
freedom describing the various internal parts of the molecules.  For
instance I didn't mean to include the internal kinetic energies of
the electrons and quarks of the molecules.  Nor did I mean to include
the kinetic energies associated with the canonical momenta for
whatever vibrational modes that happen to occur as the relative
motion between the various atomic nuclei that compose each molecule
when the energy level spacing for these vibrational modes was not
tiny compared to the temperature range of interest.  At temperatures
in the vicinity of room temperature the temperature is typically
small compared to the excitation energy for these modes, and for all
practical purposes these modes stay in their ground state in the
temperature range of interest.  Thus, these various relatively stiff
internal degrees of freedom tend to be frozen out, and I did not mean
to include the kinetic energies associated with them when I discussed
'the system's kinetic energy'.

> The usual classical statement is that there is 1/2 kT
> per degree of freedom, but that leaves us wondering
> what a degree of freedom is.

As I mentioned above, its supposed to be (1/2)*k*T for each degree of
freedom that contributes quadratically to the system's Hamiltonian
*and* for which classical mechanics is a good approximation for the
motion of the relevant (q,p) pairs.  If the contribution to the
Hamiltonian is some other function of the degree of freedom than a
quadratic one, *or* if the degree of freedom is not sufficiently
classical, then the (1/2)*k*T association breaks down.

> (Classical thermo
> requires us to count degrees of freedom, but doesn't
> tell us how to do it.

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.

> In stat mech, I know how to
> count states, which makes a lot more sense.)

In stat mech individual degrees of freedom *are* a concern, and the
macroscopic behavior of the system is found by averaging over the
statistics of the microscopic degrees of freedom.  When these
microscopic degrees of freedom behave according to classical
mechanics and when they contribute quadratically to the Hamiltonian,
*then* each one of them contributes (on average) the same energy of
(1/2)*k*T to the total thermal energy.  If some of the quadratic
degrees of freedom happen to be canonical position coordinates then
their contributions are to the system's *potential* energy, and if
the degrees of freedom happen to be canonical momenta, then their
contributions are to the system's kinetic energy.  If they behave
classically but do not contribute quadratically to the system's
Hamiltonian then their average contribution to the system's
energy is some other function than (1/2)*k*T.  For a sufficiently
complicated Van der Waals potential energy function, the average
of the total of the potential energies of the relative
intermolecular positions to the system's energy can be a
discontinuous function of the system's temperature.  Such a
discontinuity in the overall potential energy contributions of the
intermolecular positions (and orientations for molecules that are
not spherically symmetric) *are* the latent heat of transition for
a first order transition that might happen to occur at the
temperature for which these degrees of freedom are classically-
behaving.

> 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.  As such, the equipartition theorem
holds for these canonical momenta on *both* sides of the transition.

> 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.

These torsion pendulua still have a quadratic kinetic rotational
energy just as they had before when the phase was liquid. It's just
that now it also has an angular-dependent potential energy of mutual
interactions between the neighbors that is stronger than it was
before in the liquid when more of the particles were more angularly
free so that more of them could execute a full rotation before
interaction torques reversed their instantaneous rotation direction.
These interactions now tend to provide a restoring torque on the
angularly displaced molecules.  Just because the molecules in the
solid may spend most of their time angularly rocking back and forth
as they translationally vibrate, it is not an indication that any
translational or rotational degrees of freedom have been 'lost'.
If the frozen system's specific heat is lower than that of the
melted one, that merely means that the *potential* energy due to the
*positional and angular* degrees of freedom contributes a more weakly
temperature dependent contribution to the system's energy than they
did in the melt when their potential energy contributions were
different for the different significantly sampled configurations in
that other phase.

*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.

> And why is it allegedly the kinetic energy that is
> determined by the temperature?  Why wouldn't it be
> equally true that the potential energy is determined
> by temperature?  The physics doesn't distinguish.

It's not a matter of the energy being kinetic or potential *per se*,
but a matter of the relevant degree of freedom contributing
*quadratically* to the Hamiltonian.  If, as is the case of a
classical simple harmonic oscillator mode *both* the kinetic energy
*and* the potential energy were quadratic functions of the
relevant degrees of freedom (i.e. (1/2m)*p^2 + (k/2)*q^2) then
both the kinetic *and* the potential energy contribute (1/2)*k*T to
the thermal energy.  In this case the SHO mode contributes a full
(2/2)*k*T to the internal thermal energy.

It just so happens that for systems whose particles do not have a
relativistically fast relative motion, their contributions to the
kinetic energy are Newtonian and *are* quadratic and each contribute
(1/2)*k*T in classical stat mech.  OTOH, the potential energy
contributions of the interparticle interactions tend to be a very
complicated function of the mutual particle positions (and
orientations).  The actual contribution of such a complicated
potential energy function is not characterized by a simple formula.
In this case a messy configurational partition function has to be
worked out to find how the positional and orientational degrees of
freedom contribute to the overall (classical) potential energy.

BTW, if a degree of freedom behaves classically but its contribution
is a nonquadratic power law so that its contribution to the energy
still goes as the power law--but a non-quadratic power p of the
absolute value of the displacement of that degree of freedom from
its minimal energy value, then in thermal equilibrium that degree of
freedom contributes (1/p)*k*T to the overall internal energy budget.
The 2 in the denominator in the coefficient in the usual (1/2)*k*T
is an artifact of the 2nd power nature of a quadratic power law.  It
is *only* for *power law classical* degrees of freedom for which
their average energy in equilibrium is strictly proportional to the
temperature, and for which the specific heat is therefore a constant.

> To illustrate what I mean by the physics not distinguishing,
> consider the thermodynamics of an electrical LC oscillator.
> Designate the energy in the inductor as "kinetic" and the
> energy in the capacitor as "potential".  Then there will
> be 1/2 kT in each.

Correct.  In the case of your example *both* of the canonical
coordinates of the relevant generalized (q,p) pair contribute equally
to the thermal energy because they are both quadratic.  But suppose
that the capacitor in the circuit had a very squishy elastic
dielectric medium between the plates which nonlinearly changed the
capacitance as it was charged and the plates attracted each other.
In this case the electrostatic energy would *not* be a quadratic
function of the charge on the (nonlinear) 'capacitor', and then in
this case it would *not* contribute (1/2)*k*T to the overall thermal
energy in equilibrium at sufficiently (typically unrealistically
extremely) high temperatures that had thermal fluctuations result in
the capacitor being charged to its nonlinear regime in thermal
equilibrium.  In this latter case the system still behaves
classically, but one of the relevant degrees of freedom is some other
function than a quadratic one and its contribution to the thermal
energy is then some other function than (1/2)*k*T.

> As another illustration, consider cooling a chunk of metal
> from +1C to -1C.  No phase transition is involved, but
> otherwise I can make a analogy between cooling water from
> +1C to -1C.  Each system transfers some entropy delta_S to
> the environment.  In each case the transfer occurs at a
> temperature of roughly T=273K.  In each case this requires
> an energy T*delta_S.  But in the case of the metal I don't
> think you can attribute the energy to potential energy or
> to "large scale reconfiguration".  (Choose a piece of Invar-
> like metal if you want no reconfiguration at all.)

Of course.  In this case there is no phase transition and thus,
there is no large scale reconfiguration of the particles so
there is no discontinuous jump in the extensive energy nor in the
extensive entropy as a function of temperature.  In this case, to the
extent that the temperature *is constant* to that extent the
kinetic energy of the nuclear motions of the metal atoms is constant.
But in this case the temperature is not constant (because there is no
1st order transition).  The temperature changes by 2 parts in 273 and
so does the kinetic energy of the nuclear motions of the metal atoms/
ion cores change by the same proportion.  To the extent that the
overall energy changed by something else, that was mostly caused by
the disproportionate changes in the *quantum-behaving* electrons in
the conduction band (contributing approximately ([gamma]/2)*T^2 to
their energy above their ground state) *and* any possible change in
the potential energy of the nuclei/ion cores changes due to an
increase in their RMS fluctuations about their equilibrium lattice
positions.  (Because the metal is made of an invar-type material the
equilibrium lattice positions of the ion cores do not change, and the
lattice stays the same size.)  But to the extent that the
fluctuations in the positions of the nuclei/ion cores of the metal
are small enough so that the quadratic/harmonic approximation holds
for the expansion of the potential energy about the equilibrium
lattice positions for all of the vibrational modes, then the Dulong-
Petit law would tend to hold for the vibrating lattice and the
potential energy of the lattice would increase by the *same*
proportion of 2 parts in 273 that their kinetic energy changes by.
In such a case the lattice's thermal energy is 1/2 kinetic and 1/2
potential as required by the Virial theorem.

> ...
> >  If the transition to the high energy
> > phase had (at least partially), instead, happened adiabatically then
> > the system's temperature would *decrease* (as in the case of
> > evaporative cooling, or salt-induced melting) since in that case the
> > increased potential energy of the particles comes from a
> > corresponding decrease in the particles' overall kinetic energy.
>
> Yes, but such a process doesn't shed any light on the
> relationship between energy and entropy.

True.  But I was covering my bases here since it is not strictly a
requirement that a first order transition always happen at a fixed
temperature.  What is relevant is that certain extensive bulk
macroscopic parameters (such as the internal energy, entropy,
volume etc.) of the system be discontinuous functions of the
intensive parameters (temperature, pressure, chemical potential,
etc.)

> (I take it "adiabatic" means "isentropic" this case.)

Well, not quite.  Here I meant that the system be effectively
considered to be sealed or insulated from heating or cooling across
the system's boundaries.  As the transition proceeds and the
temperature drops the total internal energy remains constant, the
potential energy increases, the kinetic energy decreases, and the
entropy also increases.  Since the entropy increases the process is
not isentropic.  In this case imagine an insulated sealed container
containing only ice and salt or another insulated sealed container
containing only liquid water in a puddle on the bottom with a huge
previously evacuated volume above it.

> ...
> > > The argument is incomplete at this point, because
> > > we've been measuring the energy, not directly
> > > measuring the entropy.  So we must show the
> > > connection.
> >
> > The connection is via the *temperature* at which the
> > transition takes place.  The temperature is, by definition,
> > the proportionality factor between quasistatic infinitesimal
> > energy increments and quasistatic infinitesimal entropy
> > increments (under conditions for which no macro-work is done
> > on the system at hand).
>
> That definition of temperature bothers me on
> several counts.

Then you may have to take the matter up with the community of
statistical mechanicians.

> 1) For one thing, "quasistatic" is not the key
> issue.

What I have in mind by the term 'quasi-static' is a process that
takes the system from one equilibrium state to another
(macroscopically) nearby one which happens so slowly that during the
change the system never 'notices' that the change is taking place and
doesn't fall out of equilibrium.  In essence the change takes place
on a time scale that is very long compared to the equilibration time
scale.  (Here I'm using the concept of 'equilibrium' loosely enough
so as to effectively include various possible metastable macrostates
that may take much too long to come to true equilibrium over
experimental time scales.  Thus, we can think of a diamond in a ring
under constant temperature and pressure conditions as being in
equilibrium, when to be pedantic, true equilibrium will really not
occur until the diamond spontaneously turns to graphite which would
be on a time scale many orders of magnitude longer than the current
age of the universe.)

> I can have a process that is quasistatic
> but clearly irreversible;  examples include:
> -- heat flow down a copper rod, with one hot end
> and one cold end.

I would not consider a nonequilibrium steady state as being
quasi-static according to how I used the term.  In such a steady
state the system does not have a well defined temperature. One
end of the rod is hotter than the other one.

> -- Joule-Thompson expansion.

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.

> I think the key issue is _reversibility_.  I will
> assume that's what was intended, and move on; if
> something else was intended, please explain.

OK. If the process is quasi-static then it is asymptotically
reversible.  Such processes typically take an infinite amount of time
to accomplish.

> 2) The alleged definition also leaves us wondering what
> "macro-work" is.

True.  The definition of temperature does not in itself also define
macro-work.  Usually definitions define one term at a time.

> Suppose we want to analyze a Szilard
> engine (basically a "steam engine" containing only one
> molecule of steam).  Then either all the work is micro-
> work, or all of it is macro-work;  there's no useful
> distinction.  And there is a perfectly clear concept
> of temperature.  So it would be nice to define
> temperature without getting tangled up in macro-work
> versus micro-work.

Why would you want such a distinction in the definition of
temperature?  It is not unreasonable to define a quantity in such a
way that some other term needs to be defined first.

Usually we get our understanding of what macrowork is for a system
once we agree on how to define the system's macrostate in terms of
whatever macroscopic variables are to be used to characterize that
state.  Because we are doing thermo we must define the macrostate as
a strictly coarser-grained description than the actual microstate of
the system.  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 and in principle monitored.
Such extensive fine-scale observation for truly macroscopic systems
is forbidden by the uncertainty principle anyway.

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.  The way we can understand how much macro-work
(d-bar)W is done on the system when macroparameter x is changed by an
amount dx is according to:  (d-bar)W = X*dx where X is the (negative
of) the generalized thermodynamic force conjugate to parameter x.
Here X is the thermal average (over the distribution of the
microscopic states) of the partial derivative of the system's
Hamiltonian w.r.t. parameter x.  In classical mechanics this partial
derivative (like the Hamiltonian itself) is a function of the
microstate that is averaged over, and in quantum mechanics it ends up
being a self-adjoint operator whose product with the density matrix
is traced over to get the thermal average needed for quantity X.

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.

> In the case of the Szilard steam engine of precisely one water
> molecule then if the trajectory of tha molecule is determined by the
> description of the system that is supposed to characterize it, then
> we are not doing stat mech--we are just doing the mechanics of the
> molecule and any work that it does or has done to it is just the
> mechanical work involved for the mechanical system.

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

> Since the
> system's description doesn't involve a coarser-grained description
> than the microscopic one the very concept of macro-work is not
> relevant.

Not for that problem.  That system is just not a thermodynamic
system.  It is like calculating the exact scattering angle for
the Rutherford scattering of an alpha particle off of a gold nucleus
as a function of the impact parameter and incident energy.  The
answer is uniquely determined by the unique trajectory determined by
the initial conditions.  Such a calculation has nothing to do with
stat mech or thermo, and there is nothing *thermal* about it.  Such
a system does not have a temperature (and its entropy is strictly
zero) when described so exactly.  But the concept of macro-work *is*
useful when we *do* do stat mech and thermo, however (as is the
concept of temperature).

> 3) Suppose we are transferring entropy and energy from
> system A to system B, and we want to match up the
> entropy-transfer with the energy-transfer.
> -- System B could receive some extra entropy,
> because the process is dissipative.  For purposes of
> analysis we can (sometimes) rule this out by restricting
> our attention to nondissipative transfers.
> -- System B could receive some extra energy,
> more than is required by the second law, if we do a
> little "macro-work".
>
> So we see that without careful side-conditions, the
> magnitude of the energy-transfer divided by the
> entropy-transfer is neither an upper bound nor a
> lower bound on the temperature.

That quotient is *not supposed* to be the temperature.  Also there
*are* "careful side-conditions" in the definition of the temperature.
The temperature is a partial derivative relationship under precisely
controlled conditions chosen so that the result is well-defined.
recall the side conditions are chosen so that a) the transfer is
quasi-static (meaning it takes place in the asymptotic limit of
being so slow that the change is a continuous sequence of
*equilibrium* states), and b) it is done under conditions where
there is *no* macro-work done on or by the system.

Of course once we have the definition of temperature in hand it is
possible using various provable thermodynamic identity relationships
(from the properties of differential multivariable calculus) to come
up with other expressions for the temperature that *do* involve
macro-work and other extensive quantities.  For example, we observe
that the temperature is also the partial derivative of the system's
enthalpy w.r.t. its entropy under conditions for which the pressure
is held fixed.  If our system has a finite nonzero bulk modulus then
the conditions under which this derivative are done are ones that
*are* accompanied by macro-work.

> I suppose we can
> make this work by searching for the _minimum_ energy
> required to transfer a given amount of entropy _from
> system A_ (not "to B", that's not the same thing).
>
> OTOH a search algorithm strikes me as somewhat messy.

Sometimes definitions of physical concepts *are* messy involving a
strictly non-realizable asymptotic limit of a sequence of evermore
difficult realizable approximations.  Sometimes life is just hard.

> 4) What we are doing here is starting down the path
> toward deriving a classical theory of thermodynamics.

No we are not.  I just used a theorem of classical statistical
mechanics to analyze a freezing process under those conditions for
which it is a good approximation to consider that the relevant
microscopic variables did behave classically.  Besides, classical
thermodynamics was already derived a very long time ago already.

> That doesn't seem like a good idea, because I don't
> believe there is any self-consistent classical theory
> of thermodynamics.

I'm beginning to think you may be accidentally using the phrase
'classical thermodynamics' to mean 'classical statistical mechanics'.
In any case I'm fully aware that the world is really quantum
mechanical, and that QM does fix up some embarrassing conceptual
problems with classical stat mech (e.g. properly giving the 3rd law,
setting a finite unambiguous zero for the entropy function, making
the accessible microscopic states countable, etc.)  But 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.

> *) One of the rules I like to live by is that thou shalt
> not shoot down a theory unless you've got something better
> to replace it with.  In this case, the replacement is
> obvious:  statistical mechanics.

Excuse me, but we have been discussing stat mech all along.

> Counting states isn't
> completely cut-and-dried, but it makes a lot more sense
> than trying to count "degrees of freedom" whatever that means.

In both classical *and* quantum stat mech one is concerned with
counting states.  In classical state mech it is done via
integration over a quasi-continuous domain of classical phase space.
And in quantum stat mech it is done by making discrete sums over
the quantum numbers for the orthonormal basis states of a Hilbert
space.

In the context in which they are used in stat mech a degree of
freedom is one of the microscopic dynamical variables in the system.
For each one of them there is a dimension in classical phase space.
For each (p,q) *pair* of them there is an argument of the wave
function describing the microstate in QM.

> Then we assume (or argue from experiment) that the occupation
> of a state is an exponentially declining function of the
> energy of the state.

This is a *consequence* of the previous definition of temperature
and the requirement that for the system of interest that the system
is in *thermal contact* with *and* equilibrated with its surroundings
which allow the system to exchange thermal energy fluctuations with
them in such a way that the *average* of the system's energy remains
fixed even though the momentary microscopic energy of the system
fluctuates via tiny energy exchanges with the environment.  This
setup is called the canonical ensemble and the distribution is called
the canonical distribution.  It is a straightforward excercise in
stat mech to show the above result *and* to show that the system and
its environment in this case have a common temperature given in terms
of the coefficient of the exponent of the exponential probability that
the system occupies a given particular microstate.

> Then the definition of temperature
> is 100% clear:  it just specifies the base of the exponent,
> i.e. it specifies the energy-scale in the exponential.  And
> then we write down the partition function and use it to
> calculate system entropy, system energy, and everything else.

This only works in an ensemble that allows the system to be in
thermal contact with a fixed temperature surroundings, i.e. heat
bath, where in the system's energy is allowed to fluctuate.  It is
not correct for systems that are strictly energetically isolated from
their environments, i.e. the case of the microcanonical ensemble.  In
this latter case the occupation probability of the system is strictly
zero for all microstates whose energies are other than the particular
energy it initially started out with.  (This is simply due to the
conservation of energy).  Also, of those particular microstates that
do have the correct total energy, all of those that are mutually
accessible will have equal probability in equilibrium in that case.

Since the canonical ensemble case you describe above is only
sometimes correct it is not a good idea to define temperature via it.

> > The relationship between energy
> > changes (under no-work conditions) and entropy changes is
> > quite analogous to that between energy changes (under no heat
> > conditions) and volume changes.  In the first case the
> > appropriate factor relating the changes is the thermodynamic
> > temperature, and in the later case the appropriate factor
> > relating them is the (negative of) the pressure.
>
> That's true if you want it to be true.  That is, you can
> arrange it so you write things in terms of
>        PdV + TdS
>
> but with a simple change of variables you could
> equally well wind up with
>        PdV + SdT
> or
>        VdP + TdS
> et cetera.......  So we should not imagine that there
> is any super-deep correspondence between dS and dV.

I think you missed my point.  Of course you can make various Legendre
transformations to various other thermodynamic potentials whose
natural variables are not V & S.  But these other functions (e.g. the
Hemholtz free energy for V & T, and the enthalpy for P & S, etc.)
are no longer the system's internal energy.  They are different
functions with different values.  My example was for the first law
involving internal energy changes expressed in terms of entropy
changes and in terms of volume changes.  I did not mean to imply that
one couldn't make a Legendre transform to other useful functions.
It's just that I happened to be explicitly discussing the *energy*,
the entropy and the partial derivative relationship between them
(under constant volume, i.e. no work conditions) that serves as the
definition of the thermodynamic temperature, and compared the
relationship to the analogous case where the pressure is expressed in
terms of the partial derivative relationship between the internal
energy and the volume (under equilibrium conditions where the entropy
is held constant by an insulated adiabatic constraint).

> Also:  Even if we restrict ourselves to using dS and dV
> as the independent variables, we should keep in mind that
> there is a conservation law (or rather a local law of
> nondecrease) that applies to S, which makes it pretty
> special, unlike V

True.  But clearly, in *any* analogy not all points are supposed to
hold for the analogical map.  If they did we would not have an
analogy, but an identity instead.  The purpose of the analogy was not
to establish some deep connection between V & S but just to see that
T is related to U & S similarly to how -p is related to U & V, and
that's *only* as far as it is supposed to go.

> > We can understand the what the value of the transition temperature is
> > in physical terms by observing that the transition temperature is
> > simply the quotient of the latent heat of the transition divided by
> > its entropy change.  So *if* we can quantify how much extra inter-
> > particle potential energy the particles acquire because of their
> > redistribution in space when the transition occurs, and *if* we can
> > also quantify how much extra information is required to specify the
> > system's microstate (because the particles redistributed themselves
> > more randomly with now more available phase space per particle) then
> > the quotient of these two quantities *determines* the temperature of
> > the transition.
>
> That doesn't seem like a very practical way to calculate
> transition temperatures.  Also (as I said above) this seems
> to go too far in distinguishing potential thermal energy versus
> kinetic thermal energy.

True but it's not supposed to be a practical method of calculation.
But rather it's supposed to be a conceptual link that shows how the
latent heat of the transition in terms of an average change in the
potential energy, the entropy jump in the transition in terms of
the (logarithm of the ) number of newly allowed possible microstates
that become available in the transition is related to the temperature
at which that transition takes place.

I'm sorry this post ended up being so long.  My hat is off to anyone
who has read this all the way through to here.

David Bowman