MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Physics Department
8.044 Statistical Physics I
Spring Term 2013
Solutions, Exam #2
Problem 1 (30 points) Entropy of a Surface Film
γ and CA are given in terms of T and A so it is reasonable to choose T and A as the variables
in which to expand the entropy.
∂S
∂S
dS =
dT +
dA
∂T A
∂A T
∂S
∂S
CA
N kB N kB T
d/Q CA ≡
=T
=
=
+
⇒
dt A
∂T A
∂T A
T
T
T02
To find (∂S/∂A)T use a Maxwell Relation. You may either use the magic square or derive
the required relation as follows.
F ≡ U − TS
dF = −SdT + γdA
cross derivatives of the prefactors of the differentials are equal
∂S
∂A
= −
T
∂γ
∂T
=
A
N kB
A − bN
1
Substituting in these results gives
dS =
N kB N kB T
+
T
T02
dT +
N kB
A − bN
dA
2
1
T
S = N kB ln T + N kB
+ f (A)
2
T0
∂S
N kB
= f 0 (A) =
⇒ f (A) = N kB ln(A − bN ) + c
∂A T
A − bN
2
1
T
S(T, A) = N kB ln T + N kB
+ N kB ln(A − bN ) + c
2
T0
[Note: One can make the arguments of the logs dimensionless by distributing part of the
additive constant c among the various other terms.]
1
S(T, A) = N kB ln(T /T1 ) + N kB
2
2
T
T0
2
+ N kB ln((A − bN )/A1 ) + c0
Problem 2 (40 points) Crystal Field Splitting
a)
Z1 = 1 + 2 exp[−∆/kB T ]
X
<> =
states
2∆ exp[−∆/kB T ]
1
state p(state) =
= 2∆
1 + 2 exp[−∆/kB T ]
exp[∆/kB T ] + 2
U (T, N ) = N < >= 2∆N
1
exp[∆/kB T ] + 2
b) At T = 0 only the non-degenerate ground state is occupied. S(T = 0, N ) = kB N ln(1) = 0.
As T → ∞, all three states are equally probable. S(T, N ) → kB N ln(3).
c)
CV
=
∂U
∂T
V
d
= 2∆N
dT
1
exp[∆/kB T ] + 2
CV
(∆/kB T 2 ) exp[∆/kB T ]
= 2∆N
(exp[∆/kB T ] + 2)2
2
∆
exp[∆/kB T ]
= 2N kB
kB T
(exp[∆/kB T ] + 2)2
energy gap behavior
1/T2
d)
F (T, N ) = −kB T ln Z = −N kB T ln Z1 = −N kB T ln(1 + 2 exp[−∆/kB T ])
P (T, N ) =
=
=
=
∂F
d∆
∆
∂F
−
=−
=γ
∂∆ T dV
V
∂∆ T
T
∆ 1
−2
−N kB T γ
exp[−∆/kB T ]
V Z1 kB T
∆
exp[−∆/kB T ]
2N γ
V
1 + 2 exp[−∆/kB T ]
γ
1
U
2N ∆
=γ
V
exp[∆/kB T ] + 2
V
∂F
∂V
3
T
Problem 3 (30 points) Heating a Shell
a) For the shell,
P in = 4πr2 σTH4
P out = 4πR2 σTS4
P out = P in ⇒ r
b)
2
TH4
=R
2
TS4
r
→ TS = TH
r
R
1
e(ω, T )heater = (1)
c u(ω, TH )
4
c Z ω0 k T B H
2
P in = (4πr )
ω 2 dω
2
3
4 0
π c
Z
r2 kB TH ω0 2
kB ω03 2
=
ω
dω
=
r TH
πc2
3πc2
0
Note that the power is coming from the central object (not from the shell) and from its
surface (not volume). Thus this result is proportional to r2 .
c)
1
e(ω, T )shell = α(ω)
c u(ω, TS )
4
c Z ω0 k T kB ω03 2
B S
2
2
P out = (4πR )
R TS
ω
dω
=
4 0
π 2 c3
3πc2
r 2
P out = P in ⇒ TS = TH
R
This is an example of a poor absorber being a poor emitter (Kirchoff’s law, on the information
sheet). The shell does not absorb beyond ω0 , thus it does not radiate beyond ω0 .
4
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8.044 Statistical Physics I
Spring 2013
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