Statistical Physics, Second Revised and Enlarged Edition

98 Bose–Einsteingases

apossibilityina‘friendly’boson system since thereisnoexclusion principle. Hence
we must haveB>1 to describe thefinitenumberNof gas particles.
Thevalue ofBwill befoundto vary withdensity andtemperature, asdidμin
the discussionofFDstatistics orαinthe case ofMB statistics. We can recognize
from (9.1) that a simplification to the MB limit will occur whenB1. Under these
circumstances the−1inthedenominator will becomeinsignificantfor allvalues of
ε,evenfor thegroundstate. Hence (9.1) could be replacedbythesimplefffMB(ε)=
( 1 /B)exp(−ε/kkkBT).
Theobvious approachtodeterminingB,andthe one usedwithsuccessintheMB
and FD cases, is to use the densityof states approximation, and to replace the sum
over all states by an integration overkorε.So we enumerate the states by the usual
function

g(k)dk=V/( 2 π^3 )· 4 πk^2 dk·G (9.2) and (4.5)

Weshall for simplicityconsider spin-0bosonsfor whichthespinfactorG =1.
Making the substitution into (9.1) ofε=^2 k^2 / 2 M,as is appropriate for agas of
particles of massM, we obtain

N=

∑

i

ni

=

∑

i

giiifffi

=

∫∞

0

∫∫

g(k)f(k)dk

=V/( 2 π)^3 · 4 π

∫∞

0

∫∫

k^2 dk/[Bexp(^2 k^2 / 2 MkkkBT)− 1 ]

or,after a little work,

N=Z·F(B) (9.3)

In(9.3)the factorZis the same as the MB partition function for the gas (hence the
notation),i.e.Z=V( 2 πMkkkBT/h^2 )^3 /^2 .ThefunctionF(B)isdefinedbywhatisleft
inthe equation, whichbythesubstitutiony^2 =^2 k^2 / 2 MkkkBTis seen tobe

F(B)=( 4 /

√

π)

√√

∫∞

0

∫∫

y^2 dy/[Bexp(y^2 )− 1 ] (9.4)

Equation (9.3) can equally well be written asF(B)=A,whereAis the usual degen-
eracy parameter (definedasN/Z).Aisgivenbythe(N,V,T)macrostate conditions,
so thatBcanbedeterminedfromatableofvalues ofF(B)againstB,a numerical
task since (9.4) cannot be readily inverted. So is the problem solved?
To takethe goodnewsfirst, thissolution certainlymakes sense athighToratlow
densityN/V,i.e.intheMBlimitA 1 .Inthese circumstances,Bislarge, the− 1 in