0198506961.pdf

F.2 Bose–Einstein condensation 317

Hence,except for the ground-state population,μcan be neglected and
fBEbecomes the same as the distribution for photons:

f(ε)

1

eβε− 1

. (F.8)

From what has been said above you may wonder how neglecting the
chemical potential can be consistent with conservation of particle num-
berineqnF.4. Thisequationcanbeexpressedasanintegraloff(ε)
times the density of statesfor particles,D(ε),

N=N 0 +

∫∞

0

f(ε)D(ε)dε. (F.9)

The number in the ground state,N 0 , has to be put in explicitly because
the integral does not properly count these atoms. Effectively, we have
replacedμas a parameter byN 0 (these are related by eqn F.7). The two
terms in eqn F.9 give the number of particles in the two parts, or sub-
systems, that make up the whole. From this perspective we regard the
N−N 0 particles in the excited states (ε>ε 0 ) as a sub-system that ex-
changes particles with the condensate (atoms in the ground state). Thus
atoms in the excited states behave as if there is no number conservation:
N−N 0 →0whenT→0, as for photons.
The integral in eqn F.9 contains the distribution function from eqn F.8
times the density of states for particles given by

D(ε)=AV ε^1 /^2 dε, (F.10)

whereAis a constant.^6 With the substitutionx=βε, eqn F.9 becomes^6 D(ω) differs fundamentally from
Dph(ω) in eqn F.1 because a particle’s
energy is proportional to the square of
its wavevector,ε∝k^2 , i.e.ε=p^2 / 2 M
with momentump=
k.

N 0 =N−AV(kBT)^3 /^2 ζ, (F.11)

whereζrepresents the value of the integral given in statistical mechanics
texts as

ζ=

∫∞

0

x^1 /^2

ex− 1

dx=2. 6 ×

√

π

2

. (F.12)

The ground-state occupation goes to zero,N 0 =0,at the critical tem-
peratureTCgiven by

N

V

=A(kBTC)^3 /^2 ζ. (F.13)

WithA=2π(2M)^3 /^2 /h^3 and eqn F.12 forζ, this gives eqn 10.14. The
discussion here supposes that there is a large population in the lowest
level (the Bose–Einstein condensate) and determines the temperature at
whichN 0 goes to zero. (A different perspective adopted in many treat-
ments is to consider what happens as atoms are cooled down towards
TC.) Dividing eqn F.11 by F.13 gives the fraction of particles in the
ground state for a Bose gas in a box as

N 0

N

=1−

(

T

TC

) 3 / 2

. (F.14)