MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Physics Department
8.044 Statistical Physics I
Spring Term 2013
Solutions to Problem Set #4
Problem 1: Heat Capacity at Constant Pressure in a Simple Fluid
Start with the first law of thermodynamics.
d
/Q = dU + P dV
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 .
∂U
∂U
dU =
dT +
dP
∂T P
∂P T
∂V
∂V
dV =
dT +
dP
∂T P
∂P T
∂U
∂V
∂U
∂V
d/Q =
+P
dT +
+P
dP
∂T P
∂T P
∂P T
∂P T
Since we need the derivative at constant P the second term in the above expression will drop
out.
d
/Q CP ≡
dT P
∂U
∂V
=
+P
∂T P
∂T P
∂U
=
+ αV P
∂T 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 +
dV
∂T V
∂V T
Substitute into the expression for d/Q.
∂U
∂U
d/Q =
dT +
+ P dV
∂T V
∂V T
1
Form the derivative with respect to T by dividing by dT and specifying the path as one of
constant P .
d/Q ∂U
∂U
∂V
+
+P
≡ CP =
dT P
∂T V
∂V T
∂T P
| {z }
CV
Now use the given information (∂U/∂ V )T =0, P V = N kT , and CV = (5/2)N k in the
equation above to find that
CP = (7/2)N k.
a) It is easy to find the heat along the two rectangular paths by integration.
Z c
Z b
∆Q(acb) =
CV dT +
CP dT
a
Z
c
P2
=
P1
V1
CV
dP +
Nk
Z
V2
CP
P2
dV
Nk
CV
V2
dP
Nk
V1
= (19/2)N kT1
Z
d
Z
b
CV dT
CP dT +
∆Q(adb) =
a
Z
d
V2
=
V1
P1
CP
dV +
Nk
Z
P2
P1
= (17/2)N kT1
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.
Z
Z Vd
∆W (adb) = −P dV = −
P1 dV = −N kT1
Va
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 .
Z b
Z V2
∆W (ab) = −
P dV = −(P1 /V1 )
V dV
a
V1
= −(1/2)(P1 /V1 )(V22 − V12 ) = −(3/2)P1 V1 = −(3/2)N kT1
2
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 ∂U
=
dT +
+P dV
∂T V
∂V T
| {z }
| {z }
0
CV
Cab ≡
d/Q dV = CV + P
dT ab
dT ab
Along the path ab
P1
N kT
V1
V =
⇒ V2 =
N kT.
V1
V
P1
So along ab we can construct an expression relating dV to dT by taking the derivative of
this expression.
P =
2V dV
V1
N k dT
P1
∂V
V1 N k
⇒
=
∂T ab P1 2V
=
Finally
Cab = CV +
V1 N k P
= CV + (1/2)N k = 3N k.
P1 2 |{z}
V
P1 /V1
As a check we can integrate this heat capacity along the path.
Z b
∆Q(ab) =
Cab dT = 3N k(Tb − Ta ) = 9N kT1
a
This is identical to the result we found above in part a).
Comment: CV = (5/2)N k 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)N k we would have found CP = (5/2)N k, ∆Q(acb) = (13/2)N kT1 , ∆Q(adb) =
(11/2)N kT1 , ∆U = (9/2)N kT1 , ∆Q(ab) = 6N kT1 , and Cab = 2N k.
3
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
∂U
dU =
dT +
dM
∂T M
∂M T
expansion
d
/Q = dU − d/W = dU − H dM
first law
=
CM ≡
CH ≡
CH − CM =
∂U
∂M
=
T
∂U
∂U
dT +
− H dM
∂T M
∂M T
d/Q ∂U
=
dT M
∂T M
d/Q ∂U
∂U
∂M
+
−H
=
dT H
∂T M
∂M T
∂T H
∂U
∂M
−H
∂M T
∂T H
CH − CM
+H
∂M
∂T
substitution
from line above
from d
/Q
by rearrangement
H
Now substitute into the general expansion of dU to arrive at
#
"
CH (T, M ) − CM (T, M )
+ H(T, M ) dM
dU (T, M ) = CM (T, M ) dT +
∂M
∂T
4
H
b) Now find the results specific to the Curie law paramagnet.
∂U
CM = bT,
=0
∂M T
∂U
∂U
dU =
dT +
dM
∂T M
∂M T
| {z }
| {z }
expansion
0
CM
Z
given
T
U (T, M ) =
bT 0 dT 0 = (1/2)bT 2 + f (M )
integration
0
∂U
∂M
= f 0 (M ) = 0 (given) ⇒ f (M ) = constant = U (T = 0)
T
0 by assumption
z }| {
U (T ) = (1/2)bT 2 +
U (T = 0)
∂M
∂T
H
∂
=
∂T
aH
T
CH − CM = (0 − H)(−
CH (T, M ) = bT +
=−
H
aH
T2
from eq. of state
aH
aH 2
M2
)
=
=
T2
T2
a
M2
a
5
from a)
c) Here we practice working with an alternative pair of independent variables.
∂U
∂U
dU =
dH +
dM
expansion
∂H M
∂M H
∂U
∂U
d
/Q = dU − H dM =
dH +
− H dM
first law and substitution
∂H M
∂M H
d/Q ∂U
∂H
CM ≡
dividing the above by dT
=
dT M
∂H M ∂T M
∂H
∂T
=
M
CM
∂U
∂H
∂U
∂M
1
= −
χT
= −
M
∂U
∂H
M
∂M
∂T
substitution
H
CH
∂M
∂T H
rearrangement of above
H
d/Q ≡
=
dT H
=
chain rule
χT
C M χT
∂M
∂T
H
=−
∂M
∂H T
CH
∂T
∂M H
∂M
∂T H
−1
∂U
∂M
−H
H
∂M
∂T
from d
/Q expression
H
+H
rearrangement of above
Now substitute into the general expansion of dU (H, M ) to arrive at
"
#
CM (H, M ) χT (H, M )
CH (H, M )
dU (H, M ) = −
dH +
+ H dM
∂M
∂M
∂T
∂T
H
H
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).
6
d) Now we use the equation of state associated with a Curie law paramagnet.
∂M
∂
aH
a
χT ≡
=
=
∂H T
∂H
T T
T
∂M
aH
= − 2
∂T H
T
!
!
2
bT Ta
bT + a H
2
T
dU (H, M ) =
− aH dH +
+ H dM
− T2
− aH
T2
| {z }
|
{z
}
bT 2 /H
=
bT 2
bT 2
dH −
dM
H
M
=
ba2 H
ba2 H 2
dH
−
dM
M2
M3
U (H, M ) =
∂U
∂M
ba2 H 2
+ f (M )
2 M2
= −
H
−bT 2 /M
by eliminating T
integration
ba2 H 2
+ f 0 (M )
M3
Comparison with the coefficient of dM in the differential form dU (H, M ) above shows that
f 0 (M ) = 0 which, when integrated, gives f = constant. Thus we can write
U (H, M ) =
ba2 H 2
+ constant
2 M2
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 H 2
U (H, M ) =
2 M2
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 .
7
e) We are looking for the adiabatic constraint on dT and dM .
∂U
∂U
d
/Q =
dT +
− H dM = 0
∂T M
∂M T
| {z }
for an adiabatic path
CM
CM dT
= −
∂U
∂M
−H
dM
by rearrangement
T
dT ⇒
=−
dM ∆Q=0
∂U
∂M T
−H
CM
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,
CM = bT
Curie law paramagnet
∂M T
dT (0 − H)
H
M
= −
=
=
using the general result from e)
dM ∆Q=0
bT
bT
ab
dT =
1
M dM
ab
(T − T0 ) =
1
(M 2 − M02 )
2ab
(T − T0 ) =
1
(M − M0 )(M + M0 )
2ab
after rearrangement
integration
8
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.
T =
dT =
aH
M
equation of state
a
aH
dH − 2 dM
M
M
differential of above
Substitute this general expression for dT into the adiabatic path derivative and separate the
dH and dM terms.
a
aH
dH − 2 dM
M
M
2
1
H
M
T
T
M2
2
dH =
M +
dM =
+
dM =
1+
dM
a2 b
M
a2 b
a
a
abT
(1/ab)M dM =
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.
!
dM a
1
a
=
<
2
M
dH ∆Q=0 T 1 + abT
T
Note that the slope of the adiabatic path is less than that of the isothermal path going
through the same point.
9
Problem 4: Classical Magnetic Moments
a)
−µ ≤ mi ≤ µ
−µN ≤ M ≤ µN
−N (µH) ≤ 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 x y plane for each spin, or the polar angles θ and φ for each spin.
c)
H
T
∂S
1 ∂Ω
= −k
= −
∂M N
Ω ∂M N
1
2M
M
= −k
−
Ω=k
Ω
(2/3)µ2 N
(1/3)N µ2
N µ2 H
⇒ M (H, T ) =
3k
T
using S = −k ln Ω
the Curie law result
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
N µ2
<< µN
3kT
⇒ kT >> (1/3)µH
This result says that the “thermal energy”, kT , must be much greater than the maximum
energy allowable for a single spin.
e) To find Ω0 reduce N by 1 and reduce M by m.
(M − m)2
0
N −1
Ω = (2µ)
exp −
(2/3)(N − 1)µ2
10
f)
p(m) =
h
i
(M −m)2
exp − (2/3)(N −1)µ2
h
i
M2
exp − (2/3)N
2
µ
Ω0
(2µ)N −1
=
Ω
(2µ)N
1
M 2 − 2M m + m2
M2
=
exp −
exp
2µ
(2/3)(N − 1)µ2
(2/3)N µ2
3mM
N µ2
| {z }
small since M <<µN
1
≈
exp
2µ
1
≈
2µ
3M
1+(
)m
N µ2
Now check the normalization.
Z µ
Z
p(m) dm =
−µ
µ
−µ
−µ≤m≤µ
1
dm +
2µ
Z
µ
−µ
3M
m dm
2N µ3
= 1+0=1
g)
Z
<m> =
p(m) m dm
Z
µ
=
| −µ
=
1
m dm +
2µ
{z
}
Z
µ
−µ
3M
m2 dm
2N µ3
=0
3M hµ 1 3
M
m =
3
2N µ −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.
11
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
Ω(N, 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+ }
|
{z
}
=0
= k{N ln N − (N − n+ ) ln(N − n+ ) − n+ ln n+ }
b)
F
T
= −
= −
∂S
∂L
N,E
∂S ∂n+
∂n+ |∂{zL}
1/2l
k N − n+
n+
= −
+ ln(N − n+ ) −
− ln n+
2l N − n+
n+
2lF
N − n+
−
= ln
kT
n+
kT
N − n+
F(N, T, n+ ) = −
ln
2l
n+
12
c) Now rearrange the last result, take the exponential of both sides and solve for n+ .
N − n+
n+
exp[−2lF /kT ] =
n+ = N
1
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[]
= Nl
1 − exp[−2lF /kT ]
1 + exp[−2lF/kT ]
= Nl
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
2
Nl
L≈
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
13
Problem 6: Classical Harmonic Oscillators
a) In x space
2N
X
E=
is a sphere in 2N dimensions with radius
volume in pq space is
√
x2i
i=1
E. Its volume is π N E N /N !. The corresponding
!N
r
πN √
2
N
Ω(E, N ) =
( 2m )
EN
2
N!
mω
N
2π
1 N
=
E
N!
ω
b)
(
S(E, N ) = k ln Ω(E, N ) = k ln
2π
ω
N
1 N
E
N!
c)
1
T
=
∂S
∂E
N
N E −1 {}
Nk
=k
=
{}
E
⇒ E = N kT
14
CN = N k
)
d) Let Ω0 be the volume in a phase space for N − 1 oscillators of total energy E − where
= (1/2m)p2i + (mω 2 /2)qi2 . Since the oscillators are all similar, < >= E/N = kT .
p(pi , qi ) = Ω0 /Ω
0
Ω
Ω0
Ω
N −1
1
(E − )N −1
(N − 1)!
−1
N
2π
N!
E−
1
=
ω
(N − 1)!
E
E−
ω
N
N
=
1−
− } | {zE }
2π |E{z
=
2π
ω
≈<>−1
p(pi , qi ) =
≈exp[−/<>]
1
exp[−/ < >]
(2π/ω) < >
1
exp[−p2i /2mkT ] exp[−(mω 2 /2kT )qi2 ]
(2π/ω)kT
!
1
1
√
p
exp[−p2i /2mkT ]
exp[−qi2 /2(kT /mω 2 )]
=
2
2πmkT
2π(kT /mω )
=
= p(pi ) × p(qi ) ⇒ pi and qi are S.I.
15
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8.044 Statistical Physics I
Spring 2013
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