566
10 Quantum Statistics
What happens now, if we add another boson to the system? Even if we add the
boson to the ground state population, the number of microstates available to the
system will increase slightly (because the number of ways of distributing the internal
energy of the system amongst the particles normally increases with the addition
of another particle) and consequently, the internal energy of the gas should be
allowed to decrease slightly (by cooling the system) in order that the number of
available microstates not increase, thereby holding the system entropy fixed. The
small change U in U will thus be negative, so that μ will be very small in
magnitude, and negative.
By extending this argument, we can see that the chemical potential of a Bose gas
remains close to zero, and negative, until a temperature T c , which characterizes the
temperature for which the occupancy of the ground state becomes vanishingly small,
is reached. As we have already seen for spinless bosons, this critical temperature is
given by
T c =
1
g
−
3
2
(1)
2
3
h 2
2πmk B
N
g I V
2
3
(10.6.12)
for a gas of bosons with spin I (g I ≡ 2I + 1 is the relevant spin degeneracy). For
temperatures T > T c , there will no longer be any ground state bosons present, so
that if an extra boson is added to the system, the entropy will increase significantly,
as the number of ways of distributing the internal energy among the particles is
now large, giving rise to many more microstates becoming available to the system.
The system must hence be cooled significantly to maintain the entropy at a fixed
value. The chemical potential then becomes rapidly more negative, as can be seen
in Fig. 10.11.
10.6.3 The Strongly Degenerate Bose–Einstein Gas:
Bose–Einstein Condensation
We begin with Eqs. (10.6.4) and (10.6.5), and employ the definition (10.2.16) to
replace the integrals in terms of the Bose–Einstein functions g −
n (λ) to obtain for
P /(k B T ) and ρ the expressions
P
k B T
=
1
3 g
−
5
2
(λ) −
1
V
ln(1 − λ) ,
(10.6.13)
and
ρ =
1
3 g
−
3
2
(λ) +
λ
V (1 − λ)
.
(10.6.14)
10 Quantum Statistics
What happens now, if we add another boson to the system? Even if we add the
boson to the ground state population, the number of microstates available to the
system will increase slightly (because the number of ways of distributing the internal
energy of the system amongst the particles normally increases with the addition
of another particle) and consequently, the internal energy of the gas should be
allowed to decrease slightly (by cooling the system) in order that the number of
available microstates not increase, thereby holding the system entropy fixed. The
small change U in U will thus be negative, so that μ will be very small in
magnitude, and negative.
By extending this argument, we can see that the chemical potential of a Bose gas
remains close to zero, and negative, until a temperature T c , which characterizes the
temperature for which the occupancy of the ground state becomes vanishingly small,
is reached. As we have already seen for spinless bosons, this critical temperature is
given by
T c =
1
g
−
3
2
(1)
2
3
h 2
2πmk B
N
g I V
2
3
(10.6.12)
for a gas of bosons with spin I (g I ≡ 2I + 1 is the relevant spin degeneracy). For
temperatures T > T c , there will no longer be any ground state bosons present, so
that if an extra boson is added to the system, the entropy will increase significantly,
as the number of ways of distributing the internal energy among the particles is
now large, giving rise to many more microstates becoming available to the system.
The system must hence be cooled significantly to maintain the entropy at a fixed
value. The chemical potential then becomes rapidly more negative, as can be seen
in Fig. 10.11.
10.6.3 The Strongly Degenerate Bose–Einstein Gas:
Bose–Einstein Condensation
We begin with Eqs. (10.6.4) and (10.6.5), and employ the definition (10.2.16) to
replace the integrals in terms of the Bose–Einstein functions g −
n (λ) to obtain for
P /(k B T ) and ρ the expressions
P
k B T
=
1
3 g
−
5
2
(λ) −
1
V
ln(1 − λ) ,
(10.6.13)
and
ρ =
1
3 g
−
3
2
(λ) +
λ
V (1 − λ)
.
(10.6.14)