Re: Basic Statistical Mechanics Questions

whopkins_at_csd.uwm.edu
Date: 01/30/05 

Date: Sun, 30 Jan 2005 19:01:01 +0000 (UTC)

lost.and.lonely.physicist@gmail.com wrote:
> 1) Can thermodynamics be completely derived from statistical
mechanics?
> For instance, the first law of thermodynamics states
>
> dE = T dS - P dV + mu dN
>
> where E is energy, T temperature with the Boltzmann's constant = 1, S
> entropy, P pressure, V volume, mu the chemical potential, and N the
> number of particles.

The first law states only that dE = DQ - DW, where dE is the total
energy gained by a system, DW is the work is did in a process, and DQ
is the heat it gained in that process.

The DQ = T dS comes out of the second law -- for those processes that
stay at or near equilibrium (i.e. quasi-static); otherwise it's only an
inequality: DQ/T < dS.

So, the essence of the problem is to prove that (dE + DW), though
generally an inexact differential form when averaged macroscopically,
still admits a multiplier that makes it exact.

The mechanical work is defined through a differential form:
DW = sum_a (-Q_a dx^a)
where the microscopic force Q_a is defined by the partial derivative
Q_a = dE/dx^a. Macroscopically, you're dealing with the average <E>,
and in a quasi-static process, you have d<E>/dx^a = <dE/dx^a>, so that
the same partial differential can be used to define the average <Q_a>.

So, it all comes down to showing that d<E> - sum_a (d<E>/dx^a dx^a)
admits a multiplier to make it exact. The total differential for E is
a sum of the form:
dE = sum_a (dE/dx^a dx^a) + sum_A (dE/dz^A dz^A)
where the latter variables, the z's, are the microscopic degrees of
freedom. Therefore, in a quasi-static process, you have:
d<E> - sum_a (d<E>/dx^a dx^a)
= <dE - sum_a dE/dx^a dx^a>
= <sum_A dE/dz^A dz^A>
and the goal is to show that this has a multiplier that makes it an
exact form.

For the canonical distribution, averaging is done evenly over all
states at a given energy level, and weighted at one energy level vs.
another by an exponential exp(-beta E). So, the above amounts to a
weighted sum (or integral) of the form:
sum { exp(-beta E) sum_A dE/dz^A dz^A }
---------------------------------------
sum { exp(-beta E) }
the numerator being reducible to
(-1/beta) sum_A d/dz^A sum { exp(-beta E)) } dz^A
which combined with the denominator gives you
(-1/beta) sum_A d/dz^A ln(sum { exp(-beta E)) }) dz^A
which is a total differential apart from the multiplier, (-1/beta).

Defining the temperature T, through Boltzmann's constant k, by T = 1/(k
beta), one gets
DQ = T dS
where
S = -k ln(sum { exp(-beta E) }.

A differential form that becomes exact with a multiplier selects out
certain pathways along which it is exact. In particular, for constant
T, DQ becomes exact. Therefore, the collection of all possible
processes becomes layered into families of processes, with each layer
having the property that any process confined to it yields an exact DQ.
These are the REVERSIBLE processes, the layer defines an equivalence
class of systems in equilibrium. The parameter T, therefore, is a
measure of equilibrium.

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