UNIVERSITETET I OSLO
Det matematisk-naturvitenskapelige fakultet
Exam FYS 3130: Statistical Physics
Friday June 8 at 14:30 (3 hours)
Store fysiske lesesal, Fysikkbygningen
The Problem set has 3 pages
Allowed material for the exam:
Clark: Physical and Mathematical Tables
Øgrim: Størrelser og enheter i fysikken
Oliver and Boyd: Science Data Book
Tabeller i fysikk for den videregående skole
Rottmann: Matematisk formelsamling
Standard numeric calculator.
The problem set may be answered in Norwegian, or English
Check that the set of problems is complete before you start answering the questions
Problem 1: Black-body radiation
Consider a photon gas (electromagnetic radiation) in thermal equilibrium with a confining
enclosure at a fixed temperature T .
1a: Assume that the chemical potential of the gas is equal to zero, µ = 0. Replace the
sum over states with an integral using
Z ∞
X
V
ω 2 dω . . .
(1)
... → 2 3
π
c
0
j
and find the partition function of the photon gas.
Solution 1a: Photons are electromagnetic radiation or standing waves, characterized
by their circular frequency, ω, and energy = ~ω. The frequency is related to the
wavenumber q by ω = cq, where c is the speed of light, and the energy is = ~cq.
Photons also have two independent directions of polarization that have the same
energy in vacuum.
At equilibrium photons are absorbed and emitted from the container walls and consequently the number of photons is not fixed. It follows that Helmholtz free energy
A(T, V, N ) = A(T, V ) cannot depend on the number of photons. It follows that
µ = ∂A/∂N = 0.
The electromagnetic field inside a box of volume V = L × L × L, where L is the
length of the sides of the box, may be written
e(x, y, z) = e0 sin(qx x) sin(qy y) sin(qz z) sin(ωt)
1
where qx , qy , and qz are the wave numbers, and x, y, z ∈ [0, L], and A is the field
amplitude. If we assume that E0 = 0 on the walls of the (metallic) box we must
require that the wave numbers have the form
π
jx , where jx = 1, 2, 3, . . .
L
qx =
(2)
and similar expressions for qy and qz . Note that jx = 0, is not included since the
field would vanish altogether for that value. Also negative values of jx , say, brings
nothing new due to the oscillating term sin(ωt), where t is time.
The photon energy is discrete:
j =
~cπ q 2
jx + jy2 + jz2 , where j ↔ (jx , jy , jz )
L
In the occupation number representation we have nj photons of energy j . Photons
are bosons therefore nj is arbitrary: nj = 0, 1, 2, . . . ∞. The partition function is
then
−βA
e
∞
X
= Q(T, V ) =
−β
e
P∞
j
j nj
=
∞ X
∞
Y
−βj nj
e
=
j=0 nj =0
n1 ,n2 ,...nj ...=0
∞ Y
j=0
1
1 − e−βj
(3)
The Helmholtz free energy is
A(T, V ) = kB T
X
ln(1 − e−βj )
(4)
j
For large volumes V we may replace the sum
X
j
... =
XXX
jx
jy
V
=
(2π)3
jz
P
j
3 X X X
L
... =
...
π
q
q
q
x
ZZZ
by an integral
∞
∞
y
(5)
z
V
dqx dqy dqz . . . =
4π
(2π)3
Z
∞
2
q dq . . .
0
In the last integral we may change variables to ω = cq, and multiplying with the
degeneracy of 2 to account for the two polarizations that have the same energy we
conclude that
Z ∞
X
V
... → 2 3
ω 2 dω . . .
(6)
π c 0
j
2
We may now evaluate the Helmholtz free energy
Z ∞
V
A = kB T 2 3
ω 2 ln(1 − e−β~ω )dω
π c 0
Z
(kB T )4 ∞ 3
=V 2 3 3
x ln(1 − e−x )dx
π c~ 0
4 Z ∞
x3 dx
1 (kB T )
= −V 3 2 3 3
π c ~ 0 ex − 1
π 2 (kB T )4
= −V
45(~c)3
(7)
It then follows that the partition function is given by
2
π (kB T )3
−A/kB T
Q(T, V ) = e
= exp V
45(~c)3
1b: Argue why the chemical potential for the photon gas is zero µ = 0 and find Gibbs free
energy.
Solution 1c: The chemical potential is given by
∂A(T, V, N )
µ=
=0
∂N
T,V
since A(T, V, N ) = A(T, V ) is independent of the number of photons N
The pressure of the photon gas is
∂A(T, V )
π 2 (kB T )4
P =−
=
∂V
45(~c)3
T
The Gibbs free energy is
G(T, P ) = A + P V = 0
1c: Show that the total radiation energy E = −3A, where A is Helmholtz free energy.
You may find the following integral may be useful
Z ∞
x3
π4
dx
=
ex − 1
15
0
3
Solution 1c: The entropy is given by
∂A
S=−
= −4A/T
∂T V
It follows that
E = A + T S = −3A
Problem 2: Ideal Gas Thermodynamics
2a: The partition function of an ideal gas is consisting of N particles of mass m at temperature T in a volume V is
−βA
e
VN
= Q(N, V, T ) =
N !h3N
2πm
β
3N/2
(8)
where β = kB T .
Show that the Helmholtz free energy, A(T, V ),may be written in the form
h2
3
A = −N kB T ln(eV /N ) + N f (T ) where f (T ) = 2 kB T ln
.
2πmkB T
(9)
Find the pressure P .
Find the Gibbs free energy G(T, P )
Show that the entropy may be written in the form
S(T, P ) = −N kB ln(P ) − N g 0 (T )
(10)
where g 0 (T ) = ∂[f (T ) − kB T ln(kB T )]/∂T .
Show that the energy of the ideal gas is E = 23 N kB T .
Solution 2a:
Using Stirling’s approximation ln N ! = N ln N − N in equation 8 we
4
find
A(T, V ) = −N kB T ln
eV
(2πmkB T )3/2
N h3
= −N kB T ln(eV /N ) + N
3
k T
2 B
(11)
ln
h2
2πmkB T
3
k T
2 B
h2
2πmkB T
= −N kB T ln(eV /N ) + N f (T ) , where f (T ) =
ln
eV
∂f (T )
∂A
= N kB log
−N
S(T, V ) = −
∂T V
N
∂T
V
eV
= N kB log
− N f 0 (T )
N
∂A
N kB T
P =−
=N
∂V T
V
G(T, P ) = A + P V = −N kB T ln(eV /N ) + N f (T ) + P V
ekB T
= −N kB T ln
+ N f (T ) + N kB T
P
= N kB T ln(P ) + N [f (T ) − kB T ln(kB T )]
= N kB T ln(P ) + N g(T ) where g(T ) = f (T ) − kB T ln(kB T )
∂G
∂g(T )
S(T, P ) = −
= −N kB ln(P ) − N
∂T P
∂T
P
0
= −N kB ln(P ) − N g (T )
E(T, V ) = A + T S
= −N kB T ln(eV /N ) + N f (T ) + N kB T ln(eV /N ) − N T f 0 (T )
= N [f (T ) − T f 0 (T )] = 32 N kB T
∂ ln Q
3N
=
hE i =
= 32 N kB T
∂(−β) V
2β
(12)
(13)
(14)
(15)
(16)
(17)
2b: Two vessels (volume V1 and V2 respectively) contain two identical ideal gases at the
same temperature T and with equal numbers of particles N but with different pressures P1 and P2 . The vessels are then connected. Find the change in entropy. The
vessels are thermally isolated from the surroundings.
Discuss the change in entropy for the case P1 = P2
Solution 2b: Before the vessels are connected the entropy is the sum of the entropy
of the two gases:
S0 = −N kB ln(P1 ) − N g 0 − N ln(P2 ) − N g 0 = −N kB ln(P1 P2 ) − 2N g 0 (T )
5
Since the total volume of the system is fixed, and the system is thermally insulated
it follows that the energy of the system is fixed. The energy of an ideal gas depends
only on temperature, therefore it follows that the temperature is also T after the
vessels are connected.
The pressure, P after the vessels are connected is given by
1
V1 + V2
1 1
1
=
=
+
P
2N kB T
2 P1 P2
The entropy is now
S = 2N kB ln
P1 + P2
2P1 P2
− 2N g 0 (T )
The change in entropy is therefore
∆S = N kB ln
(P1 + P2 )2
4P1 P2
In the situation that P1 = P2 = P we find
∆S = N kB ln(1) = 0
Classically one would have an increase in entropy since each of the gases would expand
to occupy the other volume. However, the statistics underlying the expression for A
is based on quantum mechanics. In quantum mechanics one may not say which of
identical particles is in what vessel; it follows that ∆S = 0 for this case.
Problem 3: Rubber elasticity in 2-dimensions
Rubber may be considered to consist of polymers that are linear chains of monomers. A
polymer consisting of N monomers, with a monomer–monomer distance a, can have many
configurations. Here we consider an ideal polymer that has completely flexible links, that
is, the directions of the links are uncorrelated even for neighboring links. In this simple
model a configuration of a polymer is a random walk. We discuss a 2-dimensional version
and insist that distance between the first- and last monomer is x in the x-direction, and 0
in the y-direction. We consider x to be a macroscopic variable analogous to the volume.
We simplify the model further in assuming that the segments (bonds) connecting
monomers may only be in the x- or y-direction. Consider such a chain of N segments
of length a each. Let R, L, U , and D, be the number of segments pointing to the right,
left, up, and down respectively. The state of the polymer is then characterized bye the
four numbers R, L, U , and D. The number of segments is then N = R + L + U + D.
6
3a: Find the number of configurations, W (R, L, U, D), in the approximation ln N ! '
N ln N − N .
Solution 3a:
N!
R!L!U !D!
ln W ' (R + L + U + D) ln(R + l + U + D) − R ln R − L ln L − U ln U − D ln D
W =
3b: Find the maximum of ln W , subject to the constraints
R+L+U +D =N
R − L = x/a
U −D =0
(18)
In order to maximize ln W you may introduce Lagrange parameters α, λ, and µ, so
that instead of requiring δ ln W = 0, you maximize
δ{ln W + α(R + L + U + D) + λ(R − L) + µ(U − D)} = 0
(19)
Show that this leads to the equations
ln(R/N ) = α + λ ,
ln(U/N ) = α + µ ,
ln(L/N ) = α − λ
ln(D/N ) = α − µ
(20)
(21)
Show that we may write
ln Wmax = −αN − λ(R − L) − µ(U − D) .
(22)
Solution 3b:
∂ ln W
∂ ln W
δR + . . .
δD
∂R
∂D
= − ln(R/N )δR − ln(L/N )δL − ln(U/N )δU − ln(D/N )δD
δ ln W =
Therefore we find for equation 19
(− ln(R/N ) + α + λ)δR + (− ln(L/N ) + α − λ)δL
+(− ln(U/N ) + α + µ)δU + (− ln(D/N ) + α − µ)δL = 0
This equation must be satisfied for all possible variations δR, . . . ,δD, and it follows
that each of the four (. . . ) terms must vanish:
ln(R/N ) = α + λ ,
ln(U/N ) = α + µ ,
7
ln(L/N ) = α − λ
ln(D/N ) = α − µ
3c: Determine the Lagrange multipliers by inserting equations 20 and 21 into the constraints 18, and show that µ = 0
eα =
and that
1
2 + + e−λ
eλ
eλ − e−λ
x
=
Na
2 + eλ − e−λ
Also show that
eλ =
1+
1−
x
Na
x
Na
Solution 3c:
N eα (eλ + e−λ + eµ + e−µ )) = N
N eα (eλ − e−λ ) = xa
N eα (eµ − e−µ ) = 0
It follows from the last equation that µ = 0. From the first equation it then follows
that
1
eα =
λ
2 + e + e−λ
with these results the second equation becomes
q=
eλ − e−λ
x
=
Na
2 + eλ − e−λ
where we for convenience have introduced the notation q for x/(N a). This equation
is q quadratic equation for eλ :
(eλ )2 − 2
with the solution
eλ =
q λ 1+q
e −
=0
1−q
1−q
1+
1+q
=
1−q
1−
x
Na
x
Na
We then find that
eα = (2 + eλ + e−λ )−1 =
8
1 − q2
1
=
1 − ( Nxa )2
4
4
3d: Show that
ln Wmax = N ln 4 − N ln 1 −
x 2
Na
x
a
− ln
1+
1−
x
Na
x
Na
Solution 3d: Insert the results from questions 3c into the expression 22 in order to
obtain the desired result.
3e: Show that the Helmholtz free energy of the polymer is
A = −kB T ln Wmax
Solution 3e: The Helmholtz free-energy is A = E − T S. The energy of the ideal
polymer is independent of configuration and may be set equal to zero. The configurational entropy of the polymer depends only on the state-variable x and is given by
S = kB ln Wmax . Therefore we find
A = −kB T ln Wmax
3f: Find the force f (x) = − ∂A
conjugate to the position of the polymer end-point,
∂x T
given that it starts at x = 0. Find the force for Nxa 1. Is this the force from the
polymer on the surroundings, or the external force from the surroundings to make
the polymer end at x?
Solution 3f: With q =
x
Na
as before we have
1+q
2
A = −kB T ln Wmax = −kB T N ln 4 − ln(1 − q ) − q ln
1−q
The force is
f =−
But we have
∂q
∂x
=
∂A
∂x
T
1+q
∂q
2
− ln(1 − q ) − q ln
1−q
∂x
∂
= +kB T N
∂q
1
Na
therefore
kB T ∂
1+q
kB T
1+q
2
f =−
ln(1 − q ) + q ln
=−
ln
a ∂q
1−q
a
1−q
for q 1 we find
f = −kB T
2x
= −kHooke x
Na
9
The polymer behaves as a Hookean spring with spring constant
kHooke =
2kB T
Na
that is proportional to temperature and inversely proportional to the length, N a, of
the polymer.
The force, f , is a function of state and depends on x that describes the macro-state
of the system, therefore f acts on the surroundings, it is analogous to the pressure
of a gas for a given volume.
10