Solution

PHYC - 505: Statistical Mechanics
Homework Assignment 2
Solutions
Due February 11, 2014
1. Work out fully an expression for the Debye specific heat for a 1-dimensional solid at temperature T .
Show all your steps so it will be clear you have understood the Debye derivation. Show what limits the
expression you get has at high and low temperatures. What does high and low mean here precisely?
In the Debye model, the atoms in the lattice are interacting with each other. These interactions cause
the existence of quantized lattice vibrations, known as phonons. Let us consider a simple model for
these interactions. Suppose we have a 1-d ring of N particles, each of mass, m, and coupled to its
nearest neighbors with spring forces as in the figure below.
The equation of motion of the nth particle is
m¨
xn = −k (xn − xn+1 + xn − xn−1 )
x
¨n = −ω02 (2xn − xn+1 − xn−1 )
where ω0 is the natural frequency of the coupling. Here it is necessary to change the space from one
in which the particles are interacting (n-space) to one in which they are not (q-space). The equations
of motion can be decoupled by performing a discrete Fourier transform defined as
X
xq ≡
eiqn xn
n
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Applying this transform we have
X
iqn
2
e
x
¨n = −ω0 (2xn − xn+1 − xn−1 )
n
!
q
x
¨ = −ω
2
−iq
q
2x − e
X
e
iq(n+1)
iq
xn+1 − e
n
X

= −ω 2 2xq − e−iq
X
eiqα xα − eiq
α
= −ω
xn−1
n

2
e
iq(n−1)
−iq q
q
2x − e
X
eiqβ xβ 
β
iq q
x −e x
2 q
= −ω x (2 − 2 cos(q))
q
= −4ω 2 xq sin
2
= −ωq2 xq
where
q ωq = 2ω0 sin
2
This q-space equation of motion takes on the form of a simple harmonic oscillator with frequency ωq .
This is the q th normal mode.
Notice that, for a ring of N spring-connected masses, the periodic boundary conditions are such that
eiqn = eiq(n+N )
From this it is seen that the allowed values of q can be calculated as
eiqn = eiqn eiqN
1 = eiqN
2π
q=
ν
N
where ν takes on integer values between 0 and N − 1, inclusively, and so q takes on the values from 0 to
2π. Thus we have converted N coupled oscillators in n-space into N decoupled oscillators in q-space.
The price we pay is that each oscillator oscillates at a different frequency. The partition function for
a single mode with frequency ωq is calculated (using the fact that the energy of a quantum harmonic
oscillator, with frequency ωq , is En = n + 21 ~ωq as follows
Z=
∞
X
e−β~ωq (n+ 2 )
1
n=0
= e−β~ωq /2
∞
X
e−β~ωq
n
n=0
−β~ωq /2
=
e
1 − e−β~ωq
and the average energy for this single mode can be found as
∂
ln Z
∂β
∂
~ωq
=−
−β
− ln 1 − e−β~ωq
∂β
2
hEi = −
~ωq
~ωq e−β~ωq
+
2
1 − e−β~ωq
~ωq
~ωq
=
+ β~ωq
2
e
−1
=
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~ω
(Note that, without the ground state energy, 2 q , there is a resemblance to Bose-Einstein statistics.)
The average energy for the entire 1-d solid would then be found by summing over all the modes in
q-space as
X ~ωq
~ωq
+ β~ωq
hEi =
2
e
−1
q modes
Ignoring the ground state energy we have
X
hEi =
q modes
~ωq
−1
eβ~ωq
If we approximate N >> 1, we can convert this sum over the q-modes into an integral with respect to
q as follows: first we multiply by 1, that is, ∆ν = 1, since ν is an integer between 0 and N − 1.
X
~ωq
hEi =
∆ν
eβ~ωq − 1
q modes
X
~ωq
N
=
∆
q
eβ~ωq − 1
2π
q modes
=
N
2π
N
≈
2π
X
q modes
2π
Z
0
~ωq
∆q
eβ~ωq − 1
~ωq
dq
−1
eβ~ωq
Next we notice that for small q,
q ωq = 2ω0 sin
→ ω0 q
2
We will therefore approximate the dispersion relation to be
ωq ≈ ω0 q
and can be rewritten in terms of cs is the speed of sound in the medium and a is the interatomic
spacing as
cs
ωq = q
a
Note: This is a good approximation for small q (acoustic phonons), but rather bad for large q. This
yields a differential
dωq = ω0 dq
Combining these two approximations, we have that the average energy can be rewritten as
Z
N 2πω0
dωq
~ωq
hEi ≈
β~ω
q
2π 0
e
− 1 ω0
Z ωmax
N
~ωq
=
dωq
2πω0 0
eβ~ωq − 1
This is typically rewritten by introducing the Debye temperature, TD , defined by the relation
kB TD = ~ωmax = 2πω0
Changing the integration variables above to dimensionless ones, x = β~ωq , we have
Z ωmax
N
~ωq
hEi =
dωq
2πω0 0
eβ~ωq − 1
Z β~ωmax
N
1 β~ωq d (β~ωq )
=
2πω0 0
β eβ~ωq − 1
β~
Z TD /T
N
1
x
=
dx
2
x
2πω0 β ~ 0
e −1
Z
N kB T 2 TD /T x
=
dx
TD
ex − 1
0
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Unfortunately, this integral cannot be evaluated in terms of elementary functions (although an expression exists in terms of special functions), but we can look at some limiting cases. When the temperature
is much greater than the Debye temperature x in the integrand is always small so we can make the
approximation
ex ≈ 1 + x
so we have that the average energy for T >>
2π~ω0
kB
N kB T 2
hEi =
TD
Z
2
Z
≈
N kB T
TD
N kB T 2
=
TD
is
TD /T
ex
0
TD /T
x
dx
1+x−1
TD /T
x
dx
x
0
Z
0
x
dx
−1
Z
N kB T 2 TD /T
dx
=
TD
0
N kB T 2 TD
=
TD
T
= N kB T
and the heat capacity at constant volume of this limit is then found by taking the temperature derivative
of the average energy as
dhEi
CV =
= Nk
dT
in the high temperature limit.
On the other hand, in the limit that the temperature is much smaller than the Debye temperature,
0
that is, T << 2π~ω
kB , the upper limit can be replaced by infinity without much loss of accuracy due to
the fact that the integrand goes like ex for large x. When the integral is extended to infinity, it can be
calculated analytically in terms of the Riemann zeta and gamma functions as
Z
N kB T 2 TD /T x
dx
TD
ex − 1
0
N kB T 2
≈
ζ(2)Γ(2)
TD
π 2 N kB T 2
=
6
TD
hEi =
and the heat capacity at constant volume becomes
CV =
π2
T
N kB
3
TD
Notice that this is proportional to T since we are in one dimension instead of T 3 as in the three
dimensional case.
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2. Calculate the energy density of states for a 2-dimensional system that is otherwise similar to a free
electron of mass m confined to a square of side L but is different in that the energy momentum
relationship is E = Gp3 rather than E ∼ p2 , G being a constant of appropriate dimensions. Calculate
the (temperature dependence of the) specific heat of a collection of N such noninteracting electrons in
thermal equilibrium at temperature T .
Consider the Hamiltonian
H = Gp3
In the semi-classical limit, the partition function for one particle in such a system, accounting for the
spin, can be calculated as follows
Z
Z
2
2
Z1 = 2
d x d2 pe−βE
h
Z L Z L Z
3
2
= 2
dx
dy d2 pe−βGp
h 0
0
Z
3
2L2
d2 pe−βGp
= 2
h
Now considering a circle of radius p, the above integral becomes
Z
3
2L2 ∞
Z1 = 2
2πpdpe−βGp
h
0
Z
3
4πL2 ∞
pdpe−βGp
=
h2 0
see http://en.wikipedia.org/wiki/N-sphere. Now using the substitution that
3
E = Gp
→
p=
E
G
1/3
dE = 3Gp2 dp = 3G1/3 E 2/3 dp
we have that
Z1 =
4πL2
h2
Z
∞
pdpe−βGp
3
0
∞ 1/3
4πL
1
E
=
dEe−βE
h2 0
G
3G1/3 E 2/3
Z ∞
4πL2
= 2 2/3
e−βE E −1/3 dE
3h G
0
2
Z
Here we see that this takes on the form
Z
Z=
∞
g(E)e−βE dE
0
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where g(E) is the energy density of states, and so we have that the density of states is
g(E) =
4πL2
E −1/3
3h2 G2/3
Now, in order to calculate the heat capacity, we find the average energy of the system, by first finding
the partition function of the system as follows
Z=
=
=
=
=
Z1N
N!
N
Z ∞
1
4πL2
−βE −1/3
e
E
dE
N ! 3h2 G2/3 0
N
Z
1
4πL2 −2/3 ∞ −βE
−1/3
e
(βE)
d(βE)
β
N ! 3h2 G2/3
0
N
Z
1
4πL2 −2/3 ∞ −x −1/3
e
x
dx
β
N ! 3h2 G2/3
0
N
2
1
4πL
2
−2/3
β
Γ
N ! 3h2 G2/3
3
The average energy is then found as
hEi = −
∂
1 ∂Z
ln Z = −
∂β
Z ∂β
N!
1
=−
N N !
4πL2
2
β −2/3 Γ 3
3h2 G2/3
2
= N β −1
3
2
= N kB T
3
4πL2
Γ
3h2 G2/3
and the heat capacity at constant volume is
CV =
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2
N kB
3
N 2
2N −2N/3−1
β
−
3
3
3. Starting from the expression −kB T ln Z for the free energy, derive an INTERESTING expression for
the entropy from what you know about the relation between the entropy, energy and free energy. The
emphasis is on the word INTERESTING so be creative.
Knowing that the free energy is given by the relation
F = −kB T ln Z
and that the entropy is given by the relation
S=−
∂F
∂T
N,V
we have that the entropy can be written as
∂
(−kB T ln Z)N,V
∂T
kB T ∂Z
= kB ln Z +
Z ∂T
kB T ∂Z ∂β
= kB ln Z +
Z ∂β ∂T
kB T 1 ∂Z
= kB ln Z −
kB T 2 Z ∂β
∂
= kB ln Z − kB β
ln Z
∂β
= kB (ln Z + βhEi)
S=−
Now we know that the probability that the system occupies a certain microstate s, is defined in terms
of the partition function, Z, as
1
Ps = e−βEs
Z
The average energy of the system, hEi, is defined in terms of, Ps as
X
hEi =
Es Ps
s
We also know that the sum over all the probabilities is equal to unity, that is,
X
Ps = 1
s
Now taking our result for the entropy and maneuvering things a bit we have that
S = kB (ln Z + βhEi)
!
X
1 −βEs
= kB ln
e
+β
Es Ps
Ps
s
!
X
= kB −βEs − ln Ps + β
Es Ps
s
!
= kB
(−βEs − ln Ps )
X
Ps + β
s
X
Es Ps
s
!
= kB
−
X
Ps ln Ps − β
s
= −kB
X
X
s
Ps ln Ps
s
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Ps E s + β
X
s
Es Ps
This is an interesting expression for entropy in that it can be reduced to the well-known expression
written on Boltzmann’s gravestone
Z = kB ln Ω
for the micro-canonical ensemble. This is done through the relation that
Ps =
1
Ω
where Ω is the number of ways of forming the observed system, or the number of states having energy
hEi. This is also known as the multiplicity. We have then, using the above properties in combination
with our definition, that
X
S = −kB
Ps ln Ps
s
= −kB (1) ln
1
Ω
= kB ln Ω
The combination Ps ln Ps is very interesting and you will encounter it in class when the H-theorem is
discussed starting from the Boltzmann equation.
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