Properties ofan ideal Bose–Einstein gas 99thedenominator ofequation (9.4)isnegligible, andthefunctionF(B)reduces simply
to 1/B. (This may be checked using theIII 2 integral of Appendix C.) This reproduces
the correct MB normalization,^1 /B=N/Z(Chapter 6). Furthermore, asArises
towardsunity, nothingobviousgoes wrong,andtheabove treatmentgives plausible
and calculable deviations from MB behaviour. However, as the gas becomes highly
degenerate (((A>1), a nonsense appears, as we shallnowdemonstrate.
9 .1.2 The Bose–Einstein condensation
Ifwelower the temperaturefar enough,then an apparent contradiction occurs. The
difficultyisthat, althoughweknow that theminimum value possibleforBisB=1,
thefunctionF(B)remainsfinite asB→ 1. In fact it takes the valueF( 1 )= 2 .612....
HenceifA>2.612, then equation (9.3) has no acceptable solution. This condition
correspondstoT<TTTB,whereTTTB,the Bose temperature,isdefinedbyA(TTTB)=
2.612, i.e. by
TTTB=(h^2 / 2 πMkkkB)(N/ 2 .612V)^2 /^3 (9.5)So, although our treatment makes sense whenT>TTTB, nevertheless whenT<TTTB
we apparentlyare not able to accommodate enoughparticles. Theintegral(9.3)is
less thanNeven ifB=1.
The source of this difficulty lies in the total reliance on the density of states approxi-
mation. In particular, the approximationgives zero weighttothegroundstate atk=0,
which is effectivelyremoved bythek^2 factor inside the integrand of (9.3). That is
entirely valid ifg(k)andf(k)are slowly varying functions ofk. But we have already
seen thatintheregion ofB≈1, thisisanythingbut the case. Infactf( 0 )diverges as
B→1.
Asimplified(but essentially correct) solution to the problemisasfollows. We
group the statesinto two classes: (i)aground-stategroup, consistingofthelowestfew
states – it makes no real difference exactly how many are taken.Assumeg 0 suchstates
atε= 0 ,(ii)the remaining states ofenergyε> 0 .Sincef(ε)will be a smoothfunction
over allε= 0, thisgroup maybedescribedbythedensityofstates approximation.
Suppose that in thermal equilibrium there aren 0 bosons in the ground-state group,
andN(th)inthehigher states. Then the number condition (9.3)is replacedby
N=n 0 +N(th) (9. 6 )inwhichtheground-state occupationis
n 0 =g 0 /(B− 1 ) (9.7)andinwhichthehigher state occupations continue tobegivenbythe sameintegral
as in (9.3) (the lower limit of the integral is in principle changed from 0 to a small
finite number,but thishas no effect since, as notedabove, theintegrandvanishes at
k=0).