Statistical Physics, Second Revised and Enlarged Edition

100 Bose–Einsteingases

Equation (9. 6 ) has radicallydifferent solutions above and belowTTTB.Abovethis

temperature, the second term (corresponding to the vast majority of available states)

can containallthe particles. WehaveB> 1 andhencen 0 ≈ 0 from equation (9.7).

Equation (9. 6 ) becomesN=N(th), withN(th)given precisely by(9.3). Nothinghas

changed from the previous section.

BelowTTTB, however, the first term in (9. 6 ) comes into play. We haveB≈1, so

that the secondtermbecomesN(th)=Z× 2. 612 =N(T/TTTB)^3 /^2 ,usingthedefinition

(9. 5 ). The first term is given by the difference betweenNandN(th), i.e.

n 0 =N[ 1 −(T/TTTB)^3 /^2 ] (9.8)

This accommodation canbemadewiththe merest perceptiblechangeinB,Bbeing
ofamagnitude 1 +O( 1 /N)throughout the low temperature region. Graphs ofn 0 and
ofBare giveninFigs. 9.1 and9.2.
The properties ofthe ‘condensate’, theground-state particles, are ratherinteresting.
AtT= 0 ,asexpected, we haven 0 =Nand all the particles are in the ground state.
However, even at any non-zero temperaturebelowTTTBwehaveasignificantfraction

T

N

n 0 n^0 = N [1 – (TTT/TTTB)

3 / (^2) ]
TTTB
n 0 ~ 0
Fig. 9. 1 Theground-state occupationn 0 for anidealBEgas as afunction oftemperature. Theground
state is not heavilyoccupied above the condensation temperatureTTTB,but belowTTTBit contains a significant
fractionoftheNbosons.
T
B
TTTB
1
~T3/2in
classical limit
Fig. 9. 2 The variation withTof the normalizationparameterBfor an ideal BEgas. At and belowTTTB,B
becomes hooked up at a value (justgreater than) unity.