10.6 Bose–Einstein Statistics
575
Thus, following the introduction of the dimensionless variable x ≡ βhν, the internal
energy for the blackbody radiation is given by
U =
8πV (k B T ) 4
c 3 h 3
∞
n=1
∞
0
x
3 e
−nx dx =
48πV
c 3 h 3 (k B T )
4 ζ(4) ,
(10.6.39)
in which ζ(4) is the zeta function
ζ(4) ≡
∞
n=1
1
n 4 =
π 4
90
.
(10.6.40)
In this way, we obtain the internal energy U of a photon gas as
U =
8π 5 k 4
B V
15c 3 h 3 T
4
=
4σ V
c
T
4 ,
(10.6.41)
with σ = 2π 5 k 4
B /(15c 2 h 3 ) 5.67037×10 −8 J m −2 s −1 K −4 the Stefan-Boltzmann
constant of thermodynamics. An expression for the heat capacity at constant
volume associated with thermal radiation may be obtained directly from expression
(10.6.41) as
C V =
∂U
∂T
V
=
16σ V
c
T
3 .
(10.6.42)
Let us now examine the radiation pressure P , which can be obtained through the
equation of state
P V
k B T
= −
i
ln(1 − e
−βhν i )
= −
∞
0
(ν) ln(1 − e
−βhν ) dν
= −
8πV
c 3
∞
0
ν
2 ln(1 − e
−βhν ) dν .
(10.6.43)
This expression can be simplified by carrying out an integration by parts, to give
P =
8πh
3c 3
∞
0
ν 3
e βhν − 1
dν =
U
3V
.
(10.6.44)
Radiation thus exerts a pressure given by
P =
4σ
3c
T
4 .
(10.6.45)