Statistical Physics, Second Revised and Enlarged Edition

16 Distinguishable particles

by differentiatinglntandsettingthe result equalto zero. Usingthefact thatNis
constant, and noting the cancellation of two out of the three terms arising from the
sum in (2.6), this gives simply

d(lnt)= 0 −

∑

j

dnnj(lnnnj+nnj/nnj− 1 )

=−

∑

j

lnnn∗jdnnj=0(2.7)

where thednnjsrepresent anyallowablechangesinthedistribution numbersfrom
the required distribution{nnj∗}. Of course not all changes are allowed. Onlychanges
which maintain the correct values ofN(2.1)and ofU(2.2) may be countenanced.
Thislackofindependence ofthednnjsgives two restrictive conditions, obtainedby
differentiating(2.1) and (2.2)

d(N)=

∑

j

dnnj= 0 (2.8)

d(U)=

∑

j

εεjdnnj=0( 2 .9)

Aconvenient way of dealing with the mathematics of a restricted maximum of this
typeis to use the Lagrange methodofundeterminedmultipliers. The argumentin
our casegoes asfollows. First we note that we can self-evidentlyaddanyarbitrary
multiples of (2.8) and (2.9) to (2.7), and still achieve the result zero. Thus

∑

j

(−lnnn∗j+α+βεεj)dnnj= 0 (2.10)

foranyvalues ofthe constantsαandβ.The secondandclever stepisthen to recognize
that it will always be possible to write the solution in such a waythat each individual
term in the sum of (2.10) equals zero, so long as specific values ofαandβare chosen.
In other wordsthe mostprobabledistribution{nnj∗}will begivenby

(−lnnn∗j+α+βεεj)=0( 2 .11)

withαandβeachhaving a specific(but as yet undetermined)value. This equation
can thenbewritten

nnj∗=exp(α+βεεj) (2.12)

and this is theBoltzmanndistribution.We maynote at once thatithas the exponen-
tial form suggested for the thermal equilibrium distribution by the little example in
Chapter 1. Butbefore we can appreciate thesignificance ofthis centralresult, we
need to explore the meanings of these constantsαandβ.(((A note of caution:αandβ