PH2b
[15 pts] Problem #2: Thermodynamic Potentials
(a) (4 pts) To show that Φ is minimized with respect to particle and energy variations, we must frst show that its
diFerential vanishes at equilibrium. At equilibrium,
τ
and
μ
are constants, so the diFerential o± Φ is given by
d
Φ
=
dU

τdσ

μdN
=0
.
(20)
We must also show that Φ is indeed a minimum at equilibrium. Knowing that entropy is a maximum at equilibrium,
we seek to write entropy in terms o± Φ. Assuming a system
S
±ree to exchange energy and particles with a bath
R
,
we write the total entropy as
σ
=
σ
R
+
σ
S
=
σ
R
(
U

U
S
,N

N
S
)+
σ
S
(
U
S
S
)
±
σ
R
(
U,N
)

U
S
±
∂σ
R
∂U
R
²
V,N

N
S
±
R
∂N
R
²
V,U
+
σ
S
(
U
S
S
)
.
(21)
Using
±
R
R
²
V,N
≡
1
τ
,
(22)
and
±
R
R
²
V,U
≡
μ
τ
,
(23)
we fnd
σ
=
σ
R
(
)

U
S
τ
+
μN
S
τ
+
σ
S
=
σ
R
(
)

Φ
S
τ
.
(24)
Now
σ
R
(
) is a constant since
U
and
N
are constants, and
σ
is maximized at equilibrium. There±ore, Φ must be
minimized at equilibrium.
(b) (3 pts) The Gibbs sum
Z
is given by
Z
=
³
U,N
g
(
)
e
(
Nμ

U
)
/τ
,
(25)
where the multiplicity
g
(
) gives the number o± states with energy
U
and particle number
N
. Writing the
multiplicity as exp(
σ
), we can write
Z
=
³
U,N
e
(
τσ
+
μN

U
)
/τ
.
(26)
By substituting the expression ±or Φ, we fnd
Z
=
³
U,N
e

Φ(
τ,μ,V
;
U,N
)
/τ
.
(27)
The constant ±actor is unity! (Pretty trivial.)
3