Download Homework 8 Solutions Problem 1: Kittel 10

Homework 8 Solutions
Problem 1: Kittel 10-4 (a) The partition function of a single oscillator
that can move in three dimensions is given by:
Zs
X
=
nx ,ny ,nz
X
e−nh̄ω/τ )3 =
e(−(nx +ny +nz )h̄ω)/τ ) = eτ (
n
e/τ
(1 − e−h̄ω/τ )3
where − is the ground state energy of an atom in the solid lattice. Here, the
zero of the energy is taken to be the energy of a free atom at rest (which is
the ground state energy of an atom in the gas). We may estimate the chemical
potential of the atoms in the solid:
µs
=
Gs = Fs + pvs ≈ Fs
⇒ λs = eµs /τ ≈ eFs /τ = e− log Zs =
1
= e−/τ (1 − e−h̄ω/τ )3
Zs
Gs , Fs , and vs are the Gibbs energy, Helmholtz energy, and volume per particle
for the atoms in the solid. p is the vapor pressure of the gas-solid system. We
have neglected the term pvs , which is presumably very small for a solid. We
know the chemical potential for the gas phase, which we treat as an ideal gas:
n
µg = τ log(
)
nQ
p
p 2πh̄2 3/2
n
=
=
⇒ λg = eµg /τ =
nQ
τ nQ
τ Mτ
Because the system is in diffusive equilibrium, we have that λg = λs . From this,
we may obtain the vapor pressure:
p 2πh̄2 3/2
= e−/τ (1 − e−h̄ω/τ )3
τ Mτ
p
M 3/2 τ 5/2
(1 − e−h̄ω/τ )3 e−/τ
2π
h̄3
M 3/2 τ 5/2 h̄3 ω 3 e−/τ
≈
2π
τ3
h̄3
M 3/2 ω 3
=
e−/τ
2π
τ 1/2
=
where we have taken a high temperature limit (τ >> h̄ω)
(b) We start with the Clausius-Clapeyron equation (K&K equation 10-15):
L
L
Lp
dp
=
≈
= 2
dτ
τ (vg − vs )
τ vg
τ
where the last step involved the ideal gas law. Next, we differentiate our expression for the vapor pressure:
−1
M 3/2
dp
+
=
ω 3 e−/τ
dτ
2π
2τ 3/2
τ 5/2
1
M 3/2 ω 3
τ
1
e−/τ 2 −
1/2
2π
τ
2
τ
p
τ
=
−
τ2
2
=
Problem 2 – Kittel 10-5 (a) The total free energy of the system is:
F
= Fs + Fg = Us + Ug + p(Vs + Vg ) ≈ Us + (Ug + pV ) = Us + Fg
= −Ns o + Ng τ [log(Ng /V nQ ) − 1]
In this derivation, we have assumed Vg ≈ V and Vs ≈ 0.
(b) If the total particle number N = Ns + Ng is constant, then we may
write:
F (Ng ) = (Ng − N )o + Ng τ [log(Ng /V nQ ) − 1]
Minimizing this with respect to Ng gives the condition:
dF
dNg
=
0 = o + τ [log(Ng /V nQ ) − 1] + τ
⇒ Ng = nQ V exp(−o /τ )
(c) The equilibrium vapor pressure follows from the ideal gas law:
p=
Ng τ
= nQ τ exp(−o /τ )
V
Problem 3 - Kittel 10-6 (a) On Kittel page 256, we have an expression
for the difference in free energy between the normal and superconducting phases
of a material:
[FN (τ ) − FS (τ )]/V = Bc2 (τ )/2µo
Using the relation: σ = −(∂F/∂T )V , we see immediately that:
∂ 1 2 Bc dBc
[σS (τ ) − σN (τ )]/V =
B (τ ) =
∂τ 2µo c
µo dτ
As τ → 0, both entropies go to zero by the third law. This implies that at τ = 0,
(dBc /dτ ) = 0. We are also given that Bc decreases with increasing temperature.
The figure shows a curve consistent with these facts (not the only possibility).
(b) (1) Since Bc (τc ) = 0, we have that FN (τc ) = FS (τc ). Because the
entropies are also equal, we also know that FN0 (τc ) = FS0 (τc ). Therefore, the
curves are tangent at τ = τc . (2) By defintion,F = U − τ σ. Because at
τ = τc , FN = FS and σN = σS ,which together imply that UN = US . (3) The
latent heat per volume of a transition between states N and S is defined as
c dBc
L = τ (σN − σS ) = B
µ0 dτ . At τ = τc , Bc = 0 so therefore L = 0.
(c)
∆C = CS − CN = τ
∂(σS /V )
∂(σN /V )
τ d2 (Bc2 )
−τ
=
∂τ
∂τ
2µ0 dτ 2
2
Bc
T
Figure 1: Cartoon of Bc vs τ .
For τ << τc , we are told that ∆C is dominated by CN . Also, we are told that
CN = γτ + O(τ 2 ) and C2 = O(τ 2 ). Therefore,
∆C = −γτ + O(τ 2 ) =
τ h dBc 2
d2 Bc i
τ d h dBc i
Bc
=
+ Bc
µ0 dτ
dτ
µ0
dτ
dτ 2
−γ = O(τ ) +
d2 Bc i
1 h dBc 2
+ Bc
µ0
dτ
dτ 2
Since the above holds for any low temperature, we know that in particular, it
should hold for τ = 0. Using the fact that (dBc /dτ ) = 0 at τ = 0, we have:
γ=−
d2 B 1
c
Bc
µ0
dτ 2 τ =0
Problem 4 - Kittel 10-8 (a) We seek the free energy of a gas of phonons in
the Debye approximation. A fast way to obtain this is to note that for a photon
gas, the free energy density is Fphoton /V = −π 2 τ 4 /(45h̄3 c3 ). The Debye theory
follows the same derivation where the sound speed v is used instead of c and the
number of polarizations is 3 instead of 2. Following the photon derivation (see
Homework 3, problem 5), we see that we need to multiply the photon expression
by 3/2:
Fphonon /V =
π2 τ 4
3
Fphoton /V = −
2
30h̄3 v 3
(b) Whether the α or β configuration is favored depends on which state has
the lower free energy. When τ = τc , the two free energies are equal. The free
energy of state α (per volume) is the sum of the free energy (per volume) of the
ground state crystal, Uα (τ = 0), plus the free energy density of the phonons,
which are excited when the crystal experiences a finite temperature. Likewise
3
for β.
Fα = Fβ ⇒ Uα (0) −
π 2 τc4
π 2 τc4
=
U
(0)
−
β
30h̄3 vα3
30h̄3 vβ3
τc4 = (30h̄3 /π 2 )[Uβ (0) − Uα (0)]/(vβ−3 − vα−3 )
(c) The latent heat (per volume) is:
L
dF
dFα dτ
dτ τ =τc
π2 τ 3
π2 τ 4
π 2 τc3 π 2 τc4 c
c
= τc 4
=4
3 3 −4
3 3
3 3 −
30h̄ vβ
30h̄ vα
30h̄ vβ
30h̄3 vα3
= τc (σα (τc ) − σβ (τc )) = τc
β
−
= 4[Uβ (0) − Uα (0)]
Problem 5
The equation for the isotherm is given on page 23-2 of the notes:
8τ
3
p(V ) =
− 2
3V − 1 V
where p is given in units of pc , τ in units of τc , and V in units of Vc . For
the case of τ = 0.9, the isotherm is shown in the accompanying figure. Also
drawn are the lines p = pmax and p = pmin which define the pressure range over
which the volume is triple-valued. We find the pressure range using the “fsolve”
function of Maple. The numbers turn out to be pmin ≈ 0.420 and pmax ≈ 0.724.
The corresponding volumes are vmin,l = 0.719, vmin,g = 4.61, vmax,l = 0.591,
and vmax,g = 1.53. In order to determine the pressure where we get liquid-gas
coexistence, we use the procedure outlined in the hint.
Z
Z V2
dp
dV
V dp =
V
dV
V1
3V − 1 1
1
1
1
2
+
= −2.4(ln
−
) + 6( − )
3V1 − 1
3V1 − 1 3V2 − 1
V1
V2
= f (V1 , V2 )
Here V1 and V2 are the liquid and gas volumes respectively corresponding to
a particular value of the pressure. Coexistence occurs at the pressure which
causes f (V1 , V2 ) = 0, for reasons discussed more fully in the lecture notes. By
numerical iteration, we find this happens for a pressure p ≈ 0.65.
Problem 6 On page 302 of Kittel, the model of ferromagnetism is discussed.
The energy per unit volume of this system is given by U/V = −λM 2 /2 =
−nτc m2 /2, where the variable are defined in the problem set. We also know
that m = tanh(m/t) where t = τ /τc . Armed with this information, we may
calculate the heat capacity (per volume). First, we need to find dm/dτ :
m 1 dm dm
= sech2 (m/t) − 2 +
dt
t
t dt
2
1 msech (m/t)
dm
=
⇒
dτ
τc tsech2 (m/t) − t2
4
2
1.5
1
0.5
1
2
3
4
5
6
V
Figure 2: Plot of P/Pc vs V /Vc where p = pmax and p = pmin are also shown.
Then:
C=
d(U/V )
dm
nm2 sech2 (m/t)
= nτc m
= 2
dτ
dτ
t − tsech2 (m/t)
The following figure is a plot of C vs t for t < 1 as requested. For t > 1, m = 0
is the only solution to the self-consistent equation m = tanh(m/t). This implies
that C = 0 above the critical temperature. The jump discontinuity is finite.
Problem 7 (a) Suppose a system can be in a number of states and the
probability of it being in state i is given by pi . We wantPto calculate the average
value of some property A of the system, < A >=
i Ai pi . Formally, this
involves summing an infinite number of terms. Monte-Carlo simulation is a way
to get an approximate answer by summing a finite number of terms. The basic
idea is to randomly choose some states {i} and calculate the average of the {Ai }.
The Metropolis algorithm is a way to ensure that, if we take enough states, then
the probability that a given state i will appear in our random sequence is pi .
Therefore, our Monte-Carlo average will well-approximate the actual average.
Here is the idea of Metropolis: Suppose our system begins in some state i.
We randomly choose another state j. If pi < pj , then we accept the choice and
our system is in state j. If pi > pj , then we accept the choice with probability
pj /pi . Depending on the outcome of the random number generator, our system
will stay in state i or move to state j. The important point is that either way
we update our running average for A.
(b) Suppose we have two states 0 and 1 which occur with probabilities p
and 1 − p. Then if our random number generator gives a number x < p, then
5
1.6
’h8_p6.txt’
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Figure 3: Plot of C vs t = τ /τc .
we say our system is in state 0. Otherwise, it is in state 1.
(c) If our system is in state 0 after n steps, then after the (n + 1)st step,
we know that it is either in state 0 (with probability t00 ) or in state 1 (with
probability t10 ). Thus, t00 + t10 = 1. Likewise, t01 + t11 = 1.
(d) Suppose we impose the condition that if our inputs are canonical , then
the outputs must be canonical. This means:
1
e−E/τ
=
t00
1 − t00
1 − t11
t11
1
e−E/τ
which means that:
1
= t00 + (1 − t11 )e−E/τ
⇒ t00 = 1 − e−E/τ (1 − t11 )
(e) The transition matrix for Metropolis is given by:
TMetro =
1 − e−E/τ
e−E/τ
1
0
To see this is Metropolis, we note that if we start in state 1, then it must
transition to the lower state 0, which implies that t01 = 1 and t00 = 0. On the
other hand, if the system is in state 0, then it will transition to state 1 with a
transition probability which is the ratio of the probabilities p1 /p0 = e−E/τ .
6
(f ) If we want canonical outputs regardless of the inputs, then we need the
following matrix. You may explicitly verify that it behaves as required:
TCanon =
1
1
1 + e−E/τ e−E/τ
7
1
e−E/τ