PHYSICS 210A : STATISTICAL PHYSICS
HW ASSIGNMENT #2 SOLUTIONS
(1) Compute the density of states D(E, V, N ) for a three-dimensional gas of particles with
P
4
Hamiltonian Ĥ = N
i=1 A |pi | , where A is a constant. Find the entropy S(E, V, N ), the
Helmholtz free energy F (T, V, N ), and the chemical potential µ(T, p).
Solution :
Let’s solve the problem for a general dispersion ε(p) = A|p|α . The density of states is
Z d
Z d
d pN
VN
d p1
α
α
D(E, V, N ) =
·
·
·
δ
E
−
Ap
−
.
.
.
−
Ap
.
1
N
N!
hd
hd
The Laplace transform is
dd p −βApα N
e
hd
N
Z∞
V N Ωd
d−1 −βApα
dp p
e
=
N ! hd
VN
b
D(β,
V, N ) =
N!
N
V
=
N!
Z
0
Ωd Γ(d/α)
αhd Ad/α
N
β −N d/α .
Now we inverse transform, recalling
K(E) =
We then conclude
and
E t−1
Γ(t)
VN
D(E, V, N ) =
N!
⇐⇒
b
K(β)
= β −t .
Ωd Γ(d/α)
αhd Ad/α
N
Nd
E α −1
Γ(N d/α)
S(E, V, N ) = kB ln D(E, V, N )
V
E
d
= N kB ln
+ N kB ln
+ N kB a0 ,
N
α
N
where a0 is a constant, and we take the thermodynamic limit N → ∞ with V /N and E/N
fixed. From this we obtain the differential relation
d N kB
N kB
dV +
dE + s0 dN
dS =
V
α E
1
µ
p
= dV + dE − dN ,
T
T
T
where s0 is a constant. From the coefficients of dV and dE, we conclude
pV = N kB T
d
E = N kB T .
α
1
Note that we have replaced E =
variables’ T , V , and N .
d
α
N kB T in order to express F in terms of its ’natural
V
N
The Helmholtz free energy is
E
d
N kB T ln
− N kB T a0
α
N
d
V
d
d
kB T − N kB T ln
− N kB T a0 .
= N kB T − N kB T ln
α
α
α
N
F = E − T S = E − N kB T ln
−
The chemical potential is
∂F
d
d
µ=T
k T +
= − kB T ln
∂N T,V
α
α B
d
d
k T +
= − kB T ln
α
α B
V
d
kB T − kB T ln
+ (1 − a0 ) kB T
α
N
kB T
d
k T − kB T ln
+ (1 − a0 ) kB T .
α B
p
Suppose we wanted the heat capacities CV and Cp . Setting dN = 0, we have
dQ
¯ = dE + p dV
d
= N kB dT + p dV
α
d
N kB T
= N kB dT + p d
.
α
p
Thus,
dQ
¯
CV =
dT
V
d
= N kB
α
,
dQ
¯
Cp =
dT
=
p
d
1+
N kB .
α
(2) Consider a gas of classical spin- 23 particles, with Hamiltonian
N
X
X
p2i
Ĥ =
− µ0 H
Siz ,
2m
i=1
i
where Siz ∈ − 32 , − 12 , + 12 , + 23 and H is the external magnetic field. Find the Helmholtz
free energy F (T, V, H, N ), the entropy S(T, V, H, N ), and the magnetic susceptibility χ(T, H, n),
where n = N/V is the number density.
Solution :
The partition function is
Z = Tr e−Ĥ/kB T =
N
1 VN 2
cosh(µ
H/2k
T
)
+
2
cosh(3µ
H/2k
T
)
,
0
B
0
B
N ! λdN
T
2
so
V
F = −N kB T ln
−
N
k
T
−
N
k
T
ln
2
cosh(µ
H/2k
T
)
+
2
cosh(3µ
H/2k
T
)
,
B
B
0
B
0
B
N λdT
p
where λT = 2π~2 /mkB T is the thermal wavelength. The entropy is
∂F
V
1
S=−
d
+
1)N
k
+
N
ln
2
cosh(µ
H/2k
T
)
+
2
cosh(3µ
H/2k
T
)
+
(
= N kB ln
B
0
B
0
B
2
∂T V,N,H
N λdT
−
µ0 H sinh(µ0 H/2kB T ) + 3 sinh(3µ0 H/2kB T )
·
.
2T
cosh(µ0 H/2kB T ) + cosh(3µ0 H/2kB T )
The magnetization is
∂F
sinh(µ0 H/2kB T ) + 3 sinh(3µ0 H/2kB T )
M =−
.
= 12 N µ0 ·
∂H T,V,N
cosh(µ0 H/2kB T ) + cosh(3µ0 H/2kB T )
The magnetic susceptibility is
χ(T, H, n) =
where
1
V
∂M
∂H
=
T,V,N
nµ20
f (µ0 H/2kB T )
4kB T
d sinh x + 3 sinh(3x)
.
f (x) =
dx cosh x + cosh(3x)
In the limit H → 0, we have f (0) = 5, so χ = 4nµ20 /4kB T at high temperatures. This is a
version of Curie’s law.
(3) Compute the RMS volume fluctuations in the T − p − N ensemble.
Solution :
Averages within the T − p − N ensemble are computed by
hAi =
Let Y = Tr
−β(Ĥ+pV )
∂G
∂p
Tr e−β(Ĥ+pV )
.
= e−βG . Then
hV 2 i =
and since
Tr A e−β(Ĥ +pV )
2
1 ∂ 2Y
−2 βG ∂
=
β
e
e−βG
β 2 Y ∂p2
∂p2
∂G 2
1 ∂ 2G
+
,
=−
β ∂p2
∂p
= V , we have
hV 2 i − hV i2 = −kB T
3
∂ 2G
.
∂p2
For the case of a nonrelativistic ideal gas, we have
, Z∞
Z∞
−βpV
k
hV i = dV e
dV e−βpV Z(T, V, N )
Z(T, V, N ) V
k
0
0
, Z∞
Z∞
(N + k)! kB T k
−βpV
N
−βpV
N +k
,
= dV e
dV e
V =
V
N!
p
0
0
since Z(T, V, N ) =
1
N
N ! (V /λT ) .
hV i = (N + 1)
Thus,
kB T
p
k T 2
hV 2 i = (N + 1)(N + 2) B
p
,
and therefore
k T 2
2
Vrms
= hV 2 i − hV i2 = (N + 1) B
p
⇒
Vrms = N 1/2
kB T
.
p
Thus Vrms /hV i = N −1/2 ≪ 1. This is, once again, the Central Limit Theorem in action.
(4) For the system described in problem (1), compute the distribution of speeds f¯(v). Find
the most probable speed, the mean speed, and the RMS speed.
Solution :
Again, we solve for the general case ε(p) = Apα . The momentum distribution is
α
g(p) = C e−βAp ,
R
where C is a normalization constant, defined so that ddp g(p) = 1. Changing variables to
t ≡ βApα , we find
d
α (βA) α
.
C=
Ωd Γ αd
The velocity v is given by
v=
∂ε
= αApα−1 p̂ .
∂p
Thus, the speed distribution is given by
Z
α
¯
f (v) = C ddp e−βAp δ v − αApα−1 .
Now
δ v − αAp
α−1
δ p − (v/αA)1/(α−1)
.
=
α(α − 1)Apα−2
4
We therefore have
f¯(v) =
C
α
pd−α+1 e−βAp
α(α − 1)A
We can now calculate
Z
.
p=(v/αA)1/(α−1)
α
hv i = C ddp e−βAp αApα−1
r
and so
α−1
kvkr = hv r i1/r = αA
1−α−1
(kB T )
Γ
r
,
d−r
α +
Γ αd
r
!1/α
.
To find the most probable speed, we extremize f¯(v). We obtain
βApα =
d−α+1
,
α
which means
−1
d − α + 1 1−α
−1
−1
−1
v = αA
= (αA)α (d − α + 1)1−α (kB T )1−α .
αβA
5