Statistical Physics, Second Revised and Enlarged Edition

What areαandβ? 17

aredefinedwiththe opposite signin some other works. Lookingat afamiliar result
(e.g. the Boltzmann distribution) easily makes this ambiguity clear.)

2.2 What areαandβ?

Here (as ever?) physics andmathematicsgohandinhand.Inthe mathematics,α
was introduced as a multiplier for the number condition (2.1), andβfor the energy
condition(2.2). So it will follow thatαis determinedfromthefixednumberNof
particles, andcanbethoughtofas a ‘potentialfor particle number’. Similarlyβis
determined byensuringthat the distribution describes an assemblywith the correct
energyU, and it can be interpreted as a ‘potential for energy’.

2. 
   
   
   
   2. 1 αandnumber
   
   
   
   

Sinceαenters the Boltzmann distribution in such a simple way, this section will be
short!Wedetermineαbyapplyingthe condition (2.1) whichcauseditsintroduction
in the first place. Substituting(2.12) back into (2.1) we obtain

N=

∑

j

nnj=exp(α)

∑

j

exp(βεεj) (2.13)

since exp(α)(=A, say) is a factor in each term of the distribution. In other words,Ais
anormalization constantfor thedistribution, chosen so that thedistributiondescribes
thethermalproperties ofthe correct numberNofparticles. Another wayofwriting
(2.13)is:A=N/Z,with the ‘partition function’,Z, defined byZ=

∑

j

∑

exp(βεεj).
We may then write theBoltzmanndistribution (2.12) as

nnj=Aexp(βεεj)=(N/Z)exp(βεεj) (2.14)

Weleave anyfullerdiscussion ofthe partitionfunction until laterinthechapter, when
thenatureofβhas been clarified.

2. 
   
   
   
   2. 2 βandenergy
   
   
   
   

In contrast, the way in whichβenters the Boltzmann distribution is more subtle. Nev-
ertheless theformalstatements are easilymade. We substitute thethermaldistribution
(2.14)backinto therelevant condition (2.2) to obtain

U=

∑

j

nnjεεj=(N/Z)

∑

j

εεjexp(βεεj)