Statistical Physics, Second Revised and Enlarged Edition

An assembly atfixed temperature 115

temperature. This canbe readilyshown toimplythat thefree energyofthesystem

must not increase, i.e. that it is a minimum in equilibrium.

Thisthen suggests another approachto the statisticalphysicsfor an assemblyin

away(N,V,T)macrostate. We shouldmake(allowable) adjustmentsinthedistri-

butionnumbers{nnj}untilFis a minimum;and this minimum inFwill describe the

thermodynamic state, withthe corresponding{nnj∗}being the equilibriumdistribution.

In practice, this methodis straightforwardfor the sort ofsystems wehavediscussed

previously, since we can calculateFas afunctionofthedistributionfrom

F=U−TS=U({nnj})−kkkBTlnt({nnj}) (10.7)

An examplefollowsinthe next section. Butinaddition the new methodenables some
newproblems to be attacked.

10.2.1 Distinguishable particles revisited

The new statisticalmethodbasedon minimizingF(equation (10.7)) maybeillustrated

by re-deriving the Boltzmann distribution for the situation of Chapter 2. We haveN

weaklyinteractinglocalizedparticles at a temperatureT.The states ofone particle

are asbeforelabelledbyj.

The free energyFis given as in (10.7) by

F=U−TS

=

∑

j

nnjεεj−kkkBTlnt({nnj}) (10.8)

We require to minimizeFinthis equation, subject now to onlyonecondition, namely

that

∑

nnj=N.(There is no energycondition now sinceUis not restricted.)The math-

ematical steps are similar to those in section 2.1. The number of microstatest({nnj}),

givenby(2.3),issubstitutedinto (10.8). This expressionforFisthendifferentiated,

usingStirling’s approximation as in (2. 6 ), togive

dF=

∑

j

[εεj+kkkBTlnnnj]dnnj (10.9)

Usingthe Lagrange method to find the minimum, we set dF−αdN= 0 ,to take

account of the number condition

∑

nnj=N.(Wehave chosen to write themultiplier

as−αsimply inorder to ensurebelow thatαhasits usualphysicalidentification!)

SubstitutingdFfrom (10.9) and removing the summation sign, the whole point of

Lagrange’s method,weobtainfor the equilibriumdistribution

εεj+kkkBTlnnn∗j−α= 0

i.e.

nn∗j =exp(α−εεj/kkkBT) (10.10)