PHYSICS 140A : STATISTICAL PHYSICS
HW ASSIGNMENT #8 SOLUTIONS
(1) For the Dieterici equation of state,
p(V − N b) = N kB T e−N a/V kB T ,
find the virial coefficients B2 (T ) and B3 (T ).
Solution :
We first write the equation of state as p = (n, T ) where n = N/V :
p=
nkB T −an/k T
B
e
.
1 − bn
Next, we expand in powers of the density n:
p = nkB T 1 + bn + b2 n2 + . . . 1 − βan + 12 β 2 a2 n2 + . . .
h
i
= nkB T 1 + b − βa n + b2 − βab + 12 β 2 a2 n2 + . . .
h
i
= nkB T 1 + B2 n + B3 n2 + . . . ,
where β = 1/kB T . We can now read off the virial coefficients:
B2 (T ) = b −
a
kB T
B3 = b2 −
,
ab
a2
+ 2 2 .
kB T
2kB T
(2) Consider a gas of particles with dispersion ε(k) = ε0 |kℓ|5/2 , where ε0 is an energy scale
and ℓ is a length scale.
(a) Find the density of states g(ε) in d = 2 and d = 3 dimensions.
(b) Find the virial coefficients B2 (T ) and B3 (T ) in d = 2 and d = 3 dimensions.
(c) Find the heat capacity CV (T ) in d = 3 dimensions for photon statistics.
Solution :
(a) For ε(k) = ε0 |kℓ|α we have
g(ε) =
Z
Ω
ddk
δ
ε
−
ε(k)
= d
(2π)d
(2π)
Z∞
1/α /ℓ
d−1 δ k − (ε/ε0 )
dk k
αε0 ℓα kα−1
0
Ωd
1
=
d
(2π) αε0 ℓd
1
ε
ε0
d −1
α
Θ(ε) .
Thus, for α = 52 ,
1
gd=2 (ε) =
5πε0 ℓ2
ε
ε0
−1/5
Θ(ε)
1
gd=3 (ε) =
5πε0 ℓ3
,
ε
ε0
1/5
Θ(ε) .
(b) We must compute the coefficients
Z∞ αd −1
Z∞
Ωd
ε
1
−jε/kB T
e−jε/kB T
=
dε
Cj = dε g(ε) e
(2π)d αε0 ℓd
ε0
0
−∞
=
Ωd Γ(d/α) 1
(2π)d
αℓd
kB T
jε0
d/α
≡ j −d/α λ−d
T ,
where
2πℓ
λT ≡ 1/d
Ωd Γ αd /α
Then
ε0
kB T
1/α
.
d
C2
+1
− α
λdT
= ∓2
B2 (T ) = ∓
2 C12
d
d
C22 2 C3
−α
−α
2
− 3 ·3
B3 (T ) = 4 − 3 = 4
λ2d
T .
C1
C1
We have α = 25 , so
d
α
=
4
5
for d = 2 and
6
5
for d = 3.
(c) For photon statistics, the energy is
Z∞
E(T, V ) = V dε g(ε) ε
1
eε/kB T
V Ωd ε0
=
Γ
(2πℓ)d α
−1
d
α
−∞
Thus,
V Ωd kB
∂E
=
Γ
CV =
∂T
(2πℓ)d α
d
α
+2 ζ
d
α
+1 ζ
d
α
d
kB T α +1
+1
ε0
d
kB T α
.
+1
ε0
(3) At atmospheric pressure, what would the temperature T have to be in order that the
electromagnetic energy density should be identical to the energy density of a monatomic
ideal gas?
Solution :
The pressure is p = 1.0 atm ≃ 105 Pa. We set
E
=
V
3
2
p=
2π 2 (kB T )4
,
30 (~c)3
2
and solve for T :
#
"
3 1/4
1
J
m
45
T =
·
· (105 Pa) · 1970 eV Å · 1.602 × 10−19
· 10−10
1.38 × 10−23 J/K
2π 2
eV
Å
= 1.19 × 105 K .
(4) Find the internal energy and heat capacity for a two-dimensional crystalline insulator,
according to the Debye model.
Solution :
We have
"
#
Z∞
~ω
Ω(T, V ) = N kB T dω g(ω) ln 2 sinh
.
2kB T
0
The internal energy is given by
∂(βΩ)
=
E(T, V ) =
∂β
1
2
Z∞
~ω
N dω g(ω) ~ω ctnh
.
2kB T
0
In the three-dimensional Debye model, the phonon density of states per unit cell is
g(ω) =
9ω 2
Θ(ωD − ω) ,
ωD3
where ωD is the Debye frequency. Thus,
9N ~
E(T ) =
2ωD3
ZωD
~ω
3
dω ω ctnh
2kB T
0
72N
=
(k T )4
(~ωD )3 B
~ωD
2k T
B
Z
ds s3 ctnh (s) .
0
In d = 2 dimensions, we must replace the phonon density of states with
g(ω) =
4ω
Θ(ωD − ω) .
ωD2
3
This guarantees that the integrated phonon density of states per unit cell is 2, which is the
number of acoustic phonon modes in two dimensions. We then have
ZωD
~ω
2~
2
E(T ) = 2 N dω ω ctnh
ωD
2kB T
0
16N
(k T )3
=
(~ωD )2 B
~ωD
2k T
B
Z
ds s2 ctnh (s) .
0
The heat capacity is
∂E
N ~2
CV =
=
∂T
kB T 2 ωD2
= 16N kB
kB T
~ωD
ZωD
~ω
dω ω 3 csch 2
2kB T
0
~ωD
2k T
B
2 Z
ds s2 csch 2 (s) .
0
One can check that limT →∞ CV (T ) = 2N kB , which is the appropriate Dulong-Petit limit.
4