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Problem 1: Heat Capacity at Constant Pressure in a Simple Fluid
For a simple fluid show that CP =(∂U/∂T )P +αV P . Since the thermal expansion coefficient α can be either positive or
negative, CP could be either less than or greater than (∂U/∂T )P . [Hint: Use the first law to find an expression for
d/Q,then expand in terms of the variables T and P .]
Problem 2: Heat Supplied to a Gas
An ideal gas for which CV = 5Nk is taken from point a to point bin the figure along three
2
paths: acb, adb, and ab. Here P2 = 2P1 and V2 = 2V1. Assume that (∂U/∂V )T = 0.
a) Compute the heat supplied to the gas (in terms of N , k, and T1) in each of the three processes. [Hint: You may
wish to find CP first.]
b) What is the “heat capacity” of the gas for the process ab?
Problems
Statistical Physics

Problem 3: Thermodynamics of a Curie Law Paramagnet
Simple magnetic systems can be described by two independent variables. State variables of interest include the
magnetic field H, the magnetization M , the temperature T , and the internal energy U . Four quantities that are often
measured experimentally are
T
χ ≡
∂M
∂H
, the isothermal magnetic susceptibility,
T
∂M
∂T
, the temperature coefficient,
H
C M ≡
d
/Q
, the heat capacity at constant M , and
dT M
d
/Q
H
C ≡ , the heat capacity at constant H.
dT H
A particular example of a simple magnetic system is the Curie law paramagnet defined by an equation of state of the
form M = aH/T where a is a constant.
For such a system one can show that (∂U/∂M )T = 0 and we shall assume that CM = bT
where bis a constant.
a) Use T and M as independent variables and consider an arbitrary simple magnetic system (that is, not necessarily
the Curie law paramagnet). Express CM as a derivative of the internal energy. Find an expresson for C H − CM in
terms of a derivative of the internal energy, H, and the temperature coefficient. Write an expression for dU (T, M )
where the coefficients of the differentials dT and dM are expressed in terms of measured quantities and H(T, M ).
b) Find explicit expressions for C H (T, M ) and U (T, M ) for the Curie law paramagnet. You may assume that U (T =
0, M = 0) = 0.

c) Consider again an arbitrary simple magnetic system, but now use H and M as the independent variables. Write an
expression for dU (H, M ) where the coefficients of the differentials dH and dM are expressed in terms of the
measured quantities and H. Is the coefficient of the dM term the same as in part a)?
d) Find explicit expressions for the coefficients in c) in the case of the Curie law paramagnet. You will need your
result from b) for C H . Convert the coefficients to functions of H and M , that is, eliminate T . Integrate dU (H, M )
to find U(H, M ). Compare your result with that found in b).
e) Using T and M as the independent variables, find the general constraint on an adiabatic change; that is, find
(∂T/∂M )∆ Q = 0 in terms of a derivative of the internal energy, H(M, T ), and CM (M, T ).
f) Evaluate (∂T/∂M )∆ Q = 0 for the C urie law paramagnet and integrate the result to find the equation of an
adiabatic path in the T, M plane through the point T0, M0.
g) For the Curie law paramagnet, draw an isothermal path on a plot of M verses H. Pick a point on that path; show
that the slope of an adiabatic path going through that point is less than the slope of the isothermal path.
Problem 4: Classical Magnetic Moments
Consider a system made up of N independent classical magnetic dipole moments located on fixed lattice sites. Each
moment µ
→i has the same length µ, but is free to rotate in 3 dimensions. When a magnetic field of strength H is applied in
the positive z direction, the energy of the ith moment is given by ϵi = −mi H where mi is the z component of µ
→i (that
is, µ
→
i · zˆ= mi).
The magnetization M and the total energy E are given by
Σ Σ
N N
M = m i i
E = ϵ = − M H
i= 1 i= 1
a) What are the physically allowed ranges of values associated with mi, M , and E?
b) How many microscopic variables are necessary to completely specify the state of the system?

In a certain limiting case, the accessible volume in phase space for the microcanonical ensemble is given by
N M 2
Ω ≈ (2µ) exp[− 2N µ2
].
3
F
c) Use the microcanonical ensemble to find the equation of state, M as a function of H
and T .
d) Is there some condition under which the solution to c) is unphysical for the system under consideration? Explain
your answer. For what values of T is the expression for Ω a good approximation?
e) The probability density p(M ) for the z component of a single magnetic moment can be written as p(m) = ΩJ/Ω
where Ω is given above. What is ΩJ?
f) Find p(m). [Note: For the limit which applies here, an expression for p(m) includ- ing powers of m no higher
than the first is adequate.] Sketch p(m) and check its normalization.
g) Use p(m) to computed < m >. Compare the result with that which one would expect.
Problem 5: A Strange Chain
l
F
A one dimensional chain is made up of N identical elements, each of length l. The angle between successive elements can
be either 00 or 1800, but there is no difference in internal energy between these two possibilities. For the sake of counting,
one can think of each element as either pointing to the right (+) or to the left (-). Then one has
N = n+ + n−
L = l(n+ − n− ) = l(2n+ − N )
a) Use the microcanonical ensemble to find the entropy as a function of N and n+ , S(N, n+).

b) Find an expression for the tension in the chain as a function of T , N , and n+, F(T, N, n+). Notice the
strange fact that there is tension in the chain even though there is no energy required to reorient two neighboring
elements! The “restoring force” in this problem is generated by entropy considerations alone. This is not simply an
academic oddity, however. This system is used as a model for elastic polymers such as rubber.
c) Rearrange the expression from b) to give the length as a function of N , T , and F .
d) Use the result for the high temperature behavior from c) to find an expression for the thermal expansion coefficient α
≡ L− 1 (∂L/∂T )F . Note the sign. Find a stout rubber band. Hang a weight from it so that its length is extended by
about a factor of two. Now heat the rubber band (a hair drier works well here) and see if the weight goes up or
down.
Problem 6: Classical Harmonic Oscillators
Consider a collection of N identical harmonic oscillators with negligible (but non-zero) in- teractions. In a
microcanonical ensemble with energy E, the system is on a surface in phase
space given by
Σ
N
p2
i mω q
2 2
+ i
= E .
2m 2
i= 1
a) Find the volume of phase space enclosed, Φ(E), as follows. Transform to new variables
1
xi = √
2m
pi 1 ≤ i ≤ N
r
mω 2
xi = qi−N
2
N + 1 ≤ i ≤ 2N
Note that in terms of these variables the constant energy surface is a 2N dimensional sphere. Find its volume.
Find the corresponding volume in p-q space.
b) Find the entropy S in terms of N and E .
c) Find T and express E in terms of N and T .
d) Find the joint probability density for the position coordinate qi and the momentum coordinate pi of one of the
oscillators. Sketch p(pi, qi).

Problem 1: Heat Capacity at Constant Pressure in a Simple Fluid
Start with the first law of thermodynamics.
d
/Q = dU + P dV
dU =
∂U
∂T
dT +
P
∂U
dP
∂P
The relation for CP we are looking for involves (∂U/∂T )P so it is natural to try to do our expansion in terms of the
variables T and P . We expand both dU and dV in terms of dT and dP .
T
dV =
∂V
∂T
dT +
P
∂V
∂P
dP
T
∂U ∂V ∂U ∂V
d
/Q = + P dT + + P dP
∂T ∂T ∂P ∂P
P P T T
Since we need the derivative at constant P the second term in the above expression will drop out.
d
/Q
P
C ≡ .
dT P
=
∂U
∂T
+ P
P
∂V
∂T P
=
∂U
∂T
+ αV P
P
Problem 2: Heat Supplied to a Gas
To find CP we proceed as follows. Rearrange the first law to isolate d/Q.
d
/Q = dU + P dV
Expand the differential of the energy in terms of dT and dV .
∂U ∂U
dU = dT +
∂T V ∂V
dV
T
Substitute into the expression for d/Q.
∂U ∂U
d
/Q =
∂T
dT +
V ∂V
+ P dV
T
Solutions

Form the derivative with respect to T by dividing by dT and specifying the path as one of constant P .
d
/Q ∂U
P
. ≡ C = + + P
dT P ∂T V
∂U
∂V
∂V
∂T
T P
C V
` ˛¸ x
Now use the given information (∂U/∂V )T =0, PV = NkT , and CV = (5/2)Nk in the equation above to find that
CP = (7/2)Nk.
a) It is easy to find the heat along the two rectangular paths by integration.
c b
∆Q(acb) =
a
CV dT + CP dT
c
∫ ∫
∫
P 2
V1
P1
∫
= CV
Nk
dP +
V 2
P2
CP
Nk
dV
V1
= (19/2)NkT1
∆Q(adb) =
d ∫ b
CP dT + CV dT
a
∫
∫
d
V 2
P1
V1
= CP
Nk
dV +
∫ P 2
V2
CV
Nk
dP
P1
= (17/2)NkT1
Before we compute the heat along the diagonal path, it is useful to find the difference in internal energy between b and a.
Since the internal energy is a state function, it does not matter what path we use to find it. We already know the heat
input ∆ Q along the path adb and the work ∆W is easy to find.
V d
Va
∫ ∫
∆W (adb) = −P dV = − P1 dV = −NkT1
Then the path-independent result for ∆U can be computed along this particular path.
∆ U = ∆ Q + ∆ W = (17/2 −2/2)N kT1 = (15/2)N kT1
The work along the diagonal path ab can be calculated by integration using the V dependence of P : P = (P1/V1)V .
∆W (ab) = − P dV = −(P1/V1)
∫ ∫
b V 2
a V1
V dV
2 2
2 1
1 1 1 1
= −(1/2)(P /V )(V −V ) = −(3/2)P V = −(3/2)N kT

Finally we can compute the heat supplied along the diagonal path.
∆ Q = ∆ U −∆ W = 9N kT1
b) Examine the constraint along the diagonal path ab.
d/Q = dU + P dV
∂U
=
∂T V
C
dT +
∂U
V
` ˛¸ x
∂V T
0
+ P dV
` ˛¸ x
d
/Q
ab
C ≡ dV
V
= C + P
dT ab dT ab
. .
Along the path ab
P1 NkT
P = V =
V1 V
2 V1
⇒ V = NkT.
P1
∂V V1 Nk
=
⇒
ab
∂T P1 2V
So along ab we can construct an expression relating dV to dT by taking the derivative of this expression.
2V dV =
V1
N k dT
P1
Finally
ab V
V1 Nk P
C = C + = CV + (1/2)Nk = 3Nk.
1
P 2 V
P1/V1
As a check we can integrate this heat capacity along the path.
`˛¸x
b
∫
∆ Q(ab) = C ab dT = 3N k(Tb −Ta) = 9N kT1
a
This is identical to the result we found above in part a).
Comment: CV = (5/2)Nk is an approximation to a diatomic gas where the rotational degrees of freedom are contributing
to CV but the vibrational degrees of freedom are not (they are frozen out; we will understand why later in the course). If we
had used the monatomic result CV = (3/2)Nk we would have found CP = (5/2)Nk, ∆Q(acb) = (13/2)NkT1, ∆Q(adb)
= (11/2)NkT1, ∆U = (9/2)NkT1, ∆Q(ab) = 6NkT1, and Cab = 2Nk.

Problem 3: Thermodynamics of a Curie Law Paramagnet
All the manipulations we perform here for a simple magnetic system mirror those we carried out in lecture for a simple
hydrostatic system.
a) Heat capacities for the generic magnet.
∂U
∂T
dT +
∂U
∂M
M T
dU =
d
/Q = dU −d
/W = dU −H dM
dM expansion
first law
∂U
=
∂T
dT +
M
∂U
∂M T
—H dM substitution
d
/Q
M
C ≡ .
dT M
=
∂U
∂T
from line above
M
d
/Q
.
∂U
H
C ≡ =
dT H ∂T
+
M
∂U ∂M
—H from d/Q
∂M T ∂T H
∂U
H M
C −C = —H ∂M
∂M ∂T
T H
∂U
∂M
C H −C M
T
by rearrangement
∂M
= + H
∂T H
Now substitute into the general expansion of dU to arrive at
H M
"
∂M
C (T, M ) −C (T, M )
dU(T, M ) = C M (T, M ) dT +
∂T H
+ H (T, M ) dM
#

b) Now find the results specific to the Curie law paramagnet.
M
C = bT,
∂U
∂M
= 0 given
T
∂U ∂U
dU = dT + dM expansion
∂T M
` ˛¸ x
∂M T
C M 0
` ˛¸
U(T, M ) =
∫ T
J J
bT dT = (1/2)bT2
+ f (M )
x
integration
0
∂U
∂M
= f J
(M ) = 0 (given) ⇒ f (M ) = constant = U(T = 0)
T
0 by assumption
U(T ) = (1/2)bT2
+
¸
U(T
x`
= 0
˛
)
∂M
∂T
∂ aH
=
H ∂T T
= − aH
T 2
H
from eq. of state
aH aH2
M 2
a
C H −C M = (0 −H )(−
T 2 ) = = from a)
T 2
C H (T, M ) = bT + M 2
a

c) Here we practice working with an alternative pair of independent variables.
∂U
dU =
∂H
dH +
M
∂U
∂M
dM expansion
H
∂U ∂U
∂H
dH +
M
d
/Q = dU −H dM = ∂M
—H dM first law and substitution
H
d
/Q ∂U ∂H
=
.
M
C ≡ dividing the above by dT
dT M ∂H M ∂T M
∂H
∂T
=
M
C
∂M
−1
∂T ∂M
∂M H ∂ H T
= − ∂T H
chain rule
χ T
1 ∂U ∂M
M = −
T
χ ∂H M ∂T
substitution
H
∂U
∂H
= −
M
∂M
CM χT
∂T H
rearrangement of above
d
/Q ∂U
H
C ≡ =
.
∂M
from d/Q expression
dT H
—H
∂M H ∂T H
∂U
∂M
C H
=
H
+ H rearrangement of above
∂M
∂T H
Now substitute into the general expansion of dU(H, M ) to arrive at
H
C M (H , M ) χ T (H , M ) C (H , M )
"
dU(H, M ) = − dH +
∂M ∂M
∂T H ∂T H
+ H dM
#
Note that the coefficient of the dM term in the expansion of dU (H, M ) found here is different from the coefficient of the
dM term in the expansion of dU(T, M ) found in part a).

d) Now we use the equation of state associated with a Curie law paramagnet.
T
χ ≡
∂H
∂M ∂ aH a
= =
T T
∂H T T
= −
∂M
∂T
aH
T 2
H
bT
dU(H, M ) = −
a
T
2
bT + a H
− aH
T 2
!
x
dH +
`
bT
˛
2
¸
/H
− aH
T 2
T 2
+ H dM
!
`
−bT
˛
2
¸
/M
x
bT2
bT2
H
= dH −
M
dM
ba2H ba2H2
= dH − dM by eliminating T
M 2 M 3
ba2 H2
U(H , M ) = integration
2 M 2 + f (M )
∂U
∂M
ba2H2
= − J
+ f (M )
H M 3
Comparison with the coefficient of dM in the differential form dU(H, M ) above shows that
f J(M ) = 0 which, when integrated, gives f = constant. Thus we can write
ba2 H2
2
U(H, M ) = + constant
M 2
There is no reason to carry around a constant term in the internal energy which never responds to any change in the
independent variables, so we are free to set the constant equal to zero.
ba2 H2
U(H , M ) =
2 M 2
By using the equation of state, M = aH/T , we see that this reduces to the same result obtained in b), that is U =
(b/2)T 2.

e) We are looking for the adiabatic constraint on dT and dM.
∂U ∂U
d
/Q = dT + —H dM = 0 for an adiabatic path
M
C dT = −
∂T
` ˛¸ M
C M
x
∂M T
∂U
∂M T
—H dM by rearrangement
dT . ∂U
⇒ = −
.
∂M T —H
dM ∆ Q =0 C M
This gives the slope of an adiabatic path for any magnetic system.
f) Next we specialize to the case of a Curie law paramagnet.
∂U
= 0, C M = bT Curie law paramagnet
∂M
dT
T
.
dM ∆Q=0
= =
= − (0 −H ) H M
bT bT ab
using the general result from e)
dT = M dM
1
ab
after rearrangement
1
2ab
0
(T −T ) = 2 2
0
(M −M ) integration
1
(T −T0) =
2ab
(M −M0)(M + M0)

g) An isothermal path in the H M plane is easy to picture from the given equation of state, M = aH/T . It is a straight
line going through the origin with slope (∂M/∂H)T = a/T . This is shown in the figure accompanying the statement of the
problem. In part f) we found the relation which must hold between dT and dM along an adiabatic path: dT = (1/ab)M
dM . In order to explore an adiabatic path in the H M plane we must express dT in terms of dM and dH.
aH
M
T = equation of state
a aH
dT =
M
dH −
M 2 dM differential of above
Substitute this general expression for dT into the adiabatic path derivative and separate the
dH and dM terms. a aH
(1/ab)M dM = dH − dM
M M 2
dH = 2
1 H 2
M T
a2b
M T M 2
+ dM = +
M a2b a
dM =
a
1 + dM
abT
dM a 1
=
.
a
<
dH ∆ Q =0 T
This allows us to find the slope of an adiabatic line in the H M plane in terms of the quantity
a/T which is the slope of an isotherm.
!
1 + M 2
abT T
Note that the slope of the adiabatic path is less than that of the isothermal path going through the same point.

Problem 4: Classical Magnetic Moments
a)
−µ ≤
−µN ≤
−N (µH ) ≤
mi ≤ µ
M ≤ µN
E ≤ N (µH )
b) There are 2N microscopic variables necessary to specify the state of the system. Some possibilities include the x and z
component of each spin, the z component and the angle in the xy plane for each spin, or the polar angles θ and φ for
each spin.
c) H
T
= − ∂S 1 ∂Ω
= −k using S = −k ln Ω
∂M N Ω ∂M
1 2M
= −k −
Ω (2/3)µ2N
N
M
Ω = k
(1/3)Nµ2
2
N µ H
⇒ M (H , T ) = the Curie law result
3k T
d) The expression found in c) allows M to grow without bound as T → 0. But |M|is bounded by µN . Thus the
expression can only be trusted as an approximation when
M (H , T ) < < µN
Nµ2
< < µN
⇒ kT > > (1/3)µH
3kT
This result says that the “thermal energy”, kT , must be much greater than the maximum energy allowable for a
single spin.
e) To find ΩJ reduce N by 1 and reduce M by m.
(M −m)2
ΩJ
= (2µ)N − 1
exp −
(2/3)(N −1)µ2

f)
(M − m) 2
(2µ)N
h
ΩJ
(2µ)N − 1 exp −
i
p(m) =
Ω
= h
(2/3)(N − 1)µ 2
exp − M 2
(2/3)Nµ 2
i
2
M −2M m + m 2
1
exp −
2µ (2/3)(N −1)µ2
= exp
M 2
(2/3)Nµ2
1
2µ
3mM
≈ exp
Nµ2
small
`
sin
˛
c
¸
eM
x
<<µN
1
≈
3M
1 + ( ) m —µ ≤ m ≤ µ
2µ Nµ2
Now check the normalization.
∫ µ
p(m) dm =
− µ
∫ µ
1 ∫ µ
3M
dm + m dm
2µ 2Nµ3
− µ − µ
= 1 + 0 = 1
g)
< m > =
∫
p(m) m dm
=
∫ µ
1 µ
3M
m dm +
∫
m2
dm
− µ 2µ
`
=
˛
¸
0
x
3
µ
3M 1
=
− µ 2N µ
h 3 M
m =
2Nµ3 − µ 3 N
This tells us that the total magnetization M is N times the average moment of an individual dipole, the result that one
would expect on physical grounds.

Problem 5: A Strange Chain
a) The number of ways of choosing n+ elements from a total of N is N !/(N −n+)!n+!. It follows that
N !
Ω(N, n+) =
+ +
(N −n )!n !
S (N , n+ ) = k ln Ω
≈ k { N ln N −(N −n+ ) ln(N −n+ ) −n+ ln n+ −N +
(N −n+ ) + n+ }
`
=
˛
¸
0
x
= k{ N ln N −(N −n+ ) ln(N −n+ ) −n+ ln n+ }
b)
F ∂S
∂L
= −
= −
T N , E
∂S ∂n+
1/2l
∂n+ ∂L
`˛¸x
k N −n + n +
= −
2l N −n +
+
n +
+ ln(N −n ) − − ln n +
2lF
− +
N −n
= ln
kT n+
kT
+
F (N , T, n ) = − ln
2l
N −n +
n +

c) Now rearrange the last result, take the exponential of both sides and solve for n+.
N −n+
exp[−2lF /kT ] =
n+ =
n+
1
N
1 + exp[−2lF /kT ]
Next, use the expression for n+ to find L.
+
L = l(2n −N ) = N l 2 1 + exp[]
1 + exp[]
1 + exp[]
−
=
1 −exp[−2lF /kT ]
= Nl
1 + exp[−2lF /kT ]
N l
exp[lF /kT ] −exp[−lF /kT ] exp[lF /kT
] + exp[−lF /kT ]
= N l tanh(lF /kT )
For high temperatures, where kT > > lF, tanh x → x for small x, so
L ≈ N l 2
F .
kT
The fact that the length L is proportional to the tension F shows that Hooke’s law applies to this system, at least for
high temperatures.
d)
α ≡ L − 1 ∂L
∂T F
1 1
= — (L )
L T
1
= −
T

Problem 6: Classical Harmonic Oscillators
a) In x space 2N
Σ
E = x 2
i
i=1
is a sphere in 2N dimensions with radius
√
E. Its volume is π N E N / N !. The corresponding volume in pqspace is
N
2
Ω(E , N ) =
√
( 2m )N
πN
N !
r !
E N
mω2
=
N
2π 1
E N
ω N !
b)
N
S (E , N ) = k ln Ω(E , N ) = k ln
(
E N
2π 1
ω N !
)
c)
1
T
∂S
∂E
= = k =
N { }
N E − 1{ } N k E
⇒ E = NkT CN = Nk

d) Let ΩJ be the volume in a phase space for N −1 oscillators of total energy E −ϵ where
ϵ = (1/2m)p2 + (mω2/2)q2. Since the oscillators are all similar, < ϵ > = E / N = kT .
i i
p(pi , qi ) = ΩJ
/Ω
N − 1
2π
ω
ΩJ
= 1
(N −1)!
N − 1
(E −ϵ)
ΩJ
Ω
=
ω
− 1 N
2π N ! E −ϵ 1
(N −1)! E E −ϵ
1 −
ϵ N
=
≈ < є > − 1
ω N
2π `
E˛
−ϵ
x ` ˛¸E x
¸
p(pi , qi ) =
≈ exp[− є /<є >]
1 exp[−ϵ/ < ϵ >]
(2π/ω) < ϵ >
1
(2π/ω)kT
= 2
i
exp[−p /2mkT ]exp[ 2 2
i
−(mω /2kT )q ]
1 1
= √ exp[ 2
i
−p /2mkT ] √ i
2 2
exp[−q /2(kT/mω )]
2πmkT 2π(kT/mω2)
!
= p(pi ) × p(qi ) ⇒ pi and qi are S.I.

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