Upload
others
View
4
Download
0
Embed Size (px)
Citation preview
4 Statistical Ensembles : Worked Examples
(4.1) Consider a system of N identical but distinguishable particles, each of which has a nondegenerate groundstate with energy zero, and a g-fold degenerate excited state with energy Îľ > 0.
(a) Let the total energy of the system be fixed at E = Mξ, where M is the number of particles in an excited state.What is the total number of states Ί(E,N)?
(b) What is the entropy S(E,N)? Assume the system is thermodynamically large. You may find it convenientto define ν ⥠M/N , which is the fraction of particles in an excited state.
(c) Find the temperature T (ν). Invert this relation to find ν(T ).
(d) Show that there is a region where the temperature is negative.
(e) What happens when a system at negative temperature is placed in thermal contact with a heat bath atpositive temperature?
Solution :
(a) Since each excited particle can be in any of g degenerate energy states, we have
Ί(E,N) =
(N
M
)gM =
N ! gM
M ! (N âM)!.
(b) Using Stirlingâs approximation, we have
S(E,N) = kBlnΊ(E,N) = âNk
B
ν ln ν + (1â ν) ln(1â ν)â ν ln g
,
where ν = M/N = E/Nξ.
(c) The inverse temperature is
1
T=
(âS
âE
)
N
=1
NÎľ
(âS
âν
)
N
=k
B
ξ¡ln
(1â ν
ν
)+ ln g
,
hencek
BT =
Îľ
ln(1âνν
)+ ln g
.
Inverting,
ν(T ) =g eâÎľ/k
BT
1 + g eâÎľ/kBT
.
(d) The temperature diverges when the denominator in the above expression for T (ν) vanishes. This occurs atν = νâ ⥠g/(g + 1). For ν â (νâ, 1), the temperature is negative! This is technically correct, and a consequenceof the fact that the energy is bounded for this system: E â [0, NÎľ]. The entropy as a function of ν therefore has amaximum at ν = νâ. The model is unphysical though in that it neglects various excitations such as kinetic energy(e.g. lattice vibrations) for which the energy can be arbitrarily large.
(e) When a system at negative temperature is placed in contact with a heat bath at positive temperature, heat flowsfrom the system to the bath. The energy of the system therefore decreases, and since âS
âE < 0, this results in a net
1
Figure 1: Bottom: dimensionless temperature θ(ν) ⥠kBT/Îľ versus dimensionless energy density ν = E/NÎľ forproblem 1, shown here for g = 3. Note that T â ââ for ν â νâÂą0+, where νâ = g/(g+1) is the energy density atwhich the entropy is maximum. Top: dimensionless entropy s(ν) ⥠S/NkB versus dimensionless energy densityν. Note the maximum at νâ = g/(g + 1), where g is the degeneracy of the excited level.
entropy increase, which is what is demanded by the Second Law of Thermodynamics. More precisely, let dQ bethe heat added to the system from the bath. The first law then says dE = dQ. The total entropy change due tosuch a differential heat transfer is
dStot = dS + dSb =
(1
Tâ 1
Tb
)dE ,
where dS = dSsys is the entropy change of the system and T is the system temperature; Tb > 0 is the temperature
of the bath. We see that the Second Law, dStot ⼠0, requires that dE ⤠0. For dQ = dE < 0, the total entropyincreases. Note that the heat capacity of the system is
C =âE
âT= NÎľ
âν
âT=
NÎľ2
kBT 2
g eâÎľ/kBT
(1 + g eâÎľ/k
BT)2
and that C ⼠0. Even though the temperature T can be negative, we always have C(T ) ⼠0; this is necessary forthermodynamic stability. We conclude that the systemâs temperature changes by dT = dE/C, so if dE < 0 wehave dT < 0 and the system cools.
All should be clear upon examination of Fig. 1. When ν > νâ, the system temperature is negative. Placing
the system in contact with a bath at temperature Tb > 0 will cause heat to flow from the system to the bath:dQ = dE < 0. This means dν = dE/NÎľ < 0, hence ν decreases and approaches νâ from above, at which pointT = ââ. At this point, a further differential transfer âdQ > 0 from the system to the bath continues to result in
an increase of total entropy, with dStot = âdQ/Tb at ν = νâ. Thus, ν crosses νâ, and the temperature flips fromT = ââ to T = +â. At this point, we can appeal to our normal intuition. The system is much hotter than thebath, and heat continues to flow to the bath. This has the (familiar) effect of lowering the system temperature,
2
which will then approach Tb from above. Ultimately, both system and bath will be at temperature Tb, as requiredfor thermodynamic equilibrium.
3
(4.2) Solve for the model in problem 1 using the ordinary canonical ensemble. The Hamiltonian is
H = Îľ
Nâ
i=1
(1â δĎ
i,1
),
where Ďi â 1, . . . , g + 1.
(a) Find the partition function Z(T,N) and the Helmholtz free energy F (T,N).
(b) Show that M = âHâÎľ counts the number of particles in an excited state. Evaluate the thermodynamic average
ν(T ) = ăMă/N .
(c) Show that the entropy S = â(âFâT
)N
agrees with your result from problem 1.
Solution :
(a) We have
Z(T,N) = Tr eâβH =(1 + g eâÎľ/k
BT)N
.
The free energy is
F (T,N) = âkBT lnF (T,N) = âNk
BT ln
(1 + g eâÎľ/k
BT)
.
(b) We have
M =âH
âÎľ=
Nâ
i=1
(1â δĎ
i,1
).
Clearly this counts all the excited particles, since the expression 1 â δĎi,1 vanishes if i = 1, which is the ground
state, and yields 1 if i 6= 1, i.e. if particle i is in any of the g excited states. The thermodynamic average of M is
ăMă =(âFâÎľ
)T,N
, hence
ν =ăMăN
=g eâÎľ/k
BT
1 + g eâÎľ/kBT
,
which agrees with the result in problem 1c.
(c) The entropy is
S = â(âF
âT
)
N
= NkBln(1 + g eâÎľ/k
BT)+
NÎľ
T
g eâÎľ/kBT
1 + g eâÎľ/kBT
.
Working with our result for ν(T ), we derive
1 + g eâÎľ/kBT =
1
1â ν
Îľ
kBT
= ln
(g(1â ν)
ν
).
Inserting these results into the above expression for S, we verify
S = âNkBln(1â ν) +Nk
Bν ln
(g(1â ν)
ν
)
= âNkB
ν ln ν + (1â ν) ln(1â ν)â ν ln g
,
as we found in problem 1b.
4
(4.3) Consider a system of noninteracting spin trimers, each of which is described by the Hamiltonian
H = âJ(Ď1Ď2 + Ď2Ď3 + Ď3Ď1
)â Âľ0H
(Ď1 + Ď2 + Ď3
).
The individual spin polarizations Ďi are two-state Ising variables, with Ďi = Âą1.
(a) Find the single trimer partition function Îś.
(b) Find the magnetization per trimer m = Âľ0 ăĎ1 + Ď2 + Ď3ă.
(c) Suppose there are N trimers in a volume V . The magnetization density is M = Nm/V . Find the zero fieldsusceptibility Ď(T ) = (âM/âH)H=0.
(d) Find the entropy S(T,H,N).
(e) Interpret your results for parts (b), (c), and (d) physically for the limits J â +â, J â 0, and J â ââ.
Solution :
The eight trimer configurations and their corresponding energies are listed in the table below.
|Ď1Ď2Ď3 ă E |Ď1Ď2Ď3 ă E
|âââ ă â3J â 3Âľ0H |âââ ă â3J + 3Âľ0H|âââ ă +J â Âľ0H |âââ ă +J + Âľ0H|âââ ă +J â Âľ0H |âââ ă +J + Âľ0H|âââ ă +J â Âľ0H |âââ ă +J + Âľ0H
Table 1: Spin configurations and their corresponding energies.
(a) The single trimer partition function is then
Îś =â
Îą
eâβEÎą = 2 e3βJ cosh(3β¾0H) + 6 eâβJ cosh(β¾0H) .
(b) The magnetization is
m =1
βΜ
âÎś
âH= 3Âľ0 ¡
(e3βJ sinh(3β¾0H) + eâβJ sinh(β¾0H)
e3βJ cosh(3β¾0J) + 3 eâβJ cosh(β¾0H)
)
(c) Expanding m(T,H) to lowest order in H , we have
m = 3β¾20 H ¡
(3 e3βJ + eâβJ
e3βJ + 3 eâβJ
)+O(H3) .
Thus,
Ď(T ) =NV
¡ 3¾20
kBT
¡(3 e3J/kB
T + eâJ/kBT
e3J/kBT + 3 eâJ/k
BT
).
(d) Note that
F =1
βlnZ , E =
â lnZ
âβ.
5
Thus,
S =E â F
T= k
B
(lnZ â β
â lnZ
âβ
)= Nk
B
(ln Îś â β
â ln Îś
âβ
).
So the entropy is
S(T,H,N) = NkBln(2 e3βJ cosh(3β¾0H) + 6 eâβJ cosh(β¾0H)
)
â 6NβJkB¡(
e3βJ cosh(3β¾0H)â eâβJ cosh(β¾0H)
2 e3βJ cosh(3β¾0H) + 6 eâβJ cosh(β¾0H)
)
â 6Nβ¾0HkB¡(
e3βJ sinh(3β¾0H) + eâβJ sinh(β¾0H)
2 e3βJ cosh(3β¾0H) + 6 eâβJ cosh(β¾0H)
).
Setting H = 0 we have
S(T,H = 0, N) = NkBln 2 +Nk
Bln(1 + 3 eâ4J/k
BT)+
NJ
T¡(
12 eâ4J/kBT
1 + 3 eâ4J/kBT
)
= NkBln 6 +Nk
Bln(1 + 1
3 e4J/k
BT)â
NJ
T¡(
4 e4J/kBT
3 + e4J/kBT
).
(e) Note that for J = 0 we have m = 3Âľ20H/k
BT , corresponding to three independent Ising spins. The H = 0
entropy is then NkBln 8 = 3Nk
Bln 2, as expected. As J â +â we have m = 9Âľ2
0H/kBT = (3Âľ0)
2H/kBT ,
and each trimer acts as a single Z2 Ising spin, but with moment 3Âľ0. The zero field entropy in this limit tendsto Nk
Bln 2, again corresponding to a single Z2 Ising degree of freedom per trimer. For J â ââ, we have
m = Âľ20 H/k
BT and S = Nk
Bln 6. This is because the only allowed (i.e. finite energy) states of each trimer are the
three states with magnetization +Âľ0 and the three states with magnetization âÂľ0, all of which are degenerate atH = 0.
6
(4.4) In §4.9.4 of the lecture notes, we considered a simple model for the elasticity of wool in which each of Nmonomers was in one of two states A or B, with energies ξ
A,Band lengths â
A,B. Consider now the case where the A
state is doubly degenerate due to a magnetic degree of freedom which does not affect the energy or the length ofthe A
Âą monomers.
(a) Generalize the results from this section of the lecture notes and show that you can write the Hamiltonian H
and chain length L in terms of spin variables Sj â â1, 0, 1, where Sj = Âą1 if monomer j is in state AÂą,
and Sj = 0 if it is in state B. Construct the appropriate generalization of K â H â ĎL.
(b) Find the equilibrium length L(T, Ď,N) as a function of the temperature, tension, and number of monomers.
(c) Now suppose an external magnetic field is present, so the energies of the AÂą states are split, with Îľ
AÂą =ÎľA â Âľ0H . Find an expression for L(T, Ď,H,N).
Solution :
(a) Take
H =Nâ
j=1
[ÎľB+ (Îľ
Aâ Îľ
B)S2
j
], L =
Nâ
j=1
[âB+ (â
Aâ â
B)S2
j
],
resulting in
K = H â ĎL = N(ÎľBâ Ďâ
B) + â
Nâ
j=1
S2j ,
whereâ = (Îľ
Aâ Îľ
B)â Ď(â
Aâ â
B) .
(b) The partition function is
Y (T, Ď,N) = eâG/kBT = Tr eâK/k
BT
= eâN(ÎľBâĎâB)/kBT(1 + 2 eââ/k
BT)N
.
Thus, the Gibbs free energy is
G(T, Ď,N) = âkBT lnY (T, Ď,N) = N(Îľ
Bâ Ďâ
B)âNk
BT ln
(1 + 2 eââ/k
BT)
.
The equilibrium length is
L = ââG
âĎ= Nâ
B+N(â
Aâ â
B) ¡ 2 eââ/k
BT
1 + 2 eââ/kBT
.
Note that L = NâA
for â â ââ and L = NâB
for â â +â.
(c) Accounting for the splitting of the two A states,
L = NâB+N(â
Aâ â
B) ¡ 2 eââ/k
BT cosh(Âľ0H/k
BT )
1 + 2 eââ/kBT cosh(Âľ0H/k
BT )
.
7
(4.5) Consider a generalization of the situation in §4.4 of the notes where now three reservoirs are in thermalcontact, with any pair of systems able to exchange energy.
(a) Assuming interface energies are negligible, what is the total density of states D(E)? Your answer should beexpressed in terms of the densities of states functions D1,2,3 for the three individual systems.
(b) Find an expression for P (E1, E2), which is the joint probability distribution for system 1 to have energy E1
while system 2 has energy E2 and the total energy of all three systems is E1 + E2 + E3 = E.
(c) Extremize P (E1, E2) with respect to E1,2. Show that this requires the temperatures for all three systemsmust be equal: T1 = T2 = T3. Writing Ej = Eâ
j + δEj , where Eâj is the extremal solution (j = 1, 2), expand
lnP (Eâ1 + δE1 , E
â2 + δE2) to second order in the variations δEj . Remember that
S = kBlnD ,
(âS
âE
)
V,N
=1
T,
(â2S
âE2
)
V,N
= â 1
T 2CV
.
(d) Assuming a Gaussian form for P (E1, E2) as derived in part (c), find the variance of the energy of system 1,
Var(E1) =â¨(E1 â Eâ
1 )2âŠ
.
Solution :
(a) The total density of states is a convolution:
D(E) =
ââŤ
ââ
dE1
ââŤ
ââ
dE2
ââŤ
ââ
dE3 D1(E1)D2(E2)D3(E3) δ(E â E1 â E2 â E3) .
(b) The joint probability density P (E1, E2) is given by
P (E1, E2) =D1(E2)D2(E2)D3(E â E1 â E2)
D(E).
(c) We set the derivatives â lnP/âE1,2 = 0, which gives
â lnP
âE1
=â lnD1
âE1
â âD3
âE3
= 0 ,â lnP
âE2
=â lnD3
âE2
â âD3
âE3
= 0 ,
where E3 = E â E1 â E2 in the argument of D3(E3). Thus, we have
â lnD1
âE1
=â lnD2
âE2
=â lnD3
âE3
⥠1
T.
Expanding lnP (Eâ1 + δE1 , E
â2 + δE2) to second order in the variations δEj , we find the first order terms cancel,
leaving
lnP (Eâ1 + δE1 , E
â2 + δE2) = lnP (Eâ
1 , Eâ2 )â
(δE1)2
2kBT 2C1
â (δE2)2
2kBT 2C2
â (δE1 + δE2)2
2kBT 2C3
+ . . . ,
where â2 lnDj/âE2j = â1/2k
BT 2Cj , with Cj the heat capacity at constant volume and particle number. Thus,
P (E1, E2) =
âdet(Câ1)
2ĎkBT 2
exp(â 1
2kBT 2
Câ1ij δEi δEj
),
8
where the matrix Câ1 is defined as
Câ1 =
(Câ1
1 + Câ13 Câ1
3
Câ13 Câ1
2 + Câ13
).
One findsdet(Câ1) = Câ1
1 Câ12 + Câ1
1 Câ13 + Câ1
2 Câ13 .
The prefactor in the above expression forP (E1, E2) has been fixed by the normalization conditionâŤdE1
âŤdE2P (E1, E2) =
1.
(d) Integrating over E2, we obtain P (E1):
P (E1) =
ââŤ
ââ
dE2 P (E1, E2) =1â
2ĎkBC1T
2
eâ(δE1)2/2k
BC
1T 2
,
where
C1 =Câ1
2 + Câ13
Câ11 Câ1
2 + Câ11 Câ1
3 + Câ12 Câ1
3
.
Thus,
ă(δE1)2ă =
ââŤ
ââ
dE1 (δE1)2 = k
BC1T
2 .
9
(4.6) Show that the Boltzmann entropyS = âkB
ân Pn lnPn agrees with the statistical entropy S(E) = k
BlnD(E, V,N)
in the thermodynamic limit.
Solution :
Letâs first examine the canonical partition function, Z =ââŤ0
dE D(E) eâβE . We compute this integral via the saddle
point method, extremizing the exponent, lnD(E)â βE , with respect to E. The resulting maximum lies at E such
that 1T = âS
âE
âŁâŁE
, where S(E) = kBlnD(E) is the statistical entropy computed in the microcanonical ensemble. The
ordinary canonical partition function is then
Z â D(E) eâβE
ââŤ
ââ
d δE eâ(δE)2/2kBT 2CV
= (2ĎkBT 2CV )
1/2 D(E) eâβE .
Taking the logarithm, we obtain the Helmholtz free energy,
F = âkBT lnZ = âk
BlnD(E) + E â 1
2kBT ln
(2Ďk
BT 2CV
).
Now SOCE
= âkB
ân Pn lnPn, with Pn = 1
Z eâβEn . Therefore
SOCE
(T ) =k
B
Z
ââŤ
0
dE D(E) eâβE(lnZ + βE
)
= kBlnZ +
1
T¡âŤâ0 dE ED(E) eâβE
âŤâ0 dE D(E) eâβE
.
The denominator of the second term is Z , which we have already evaluated. We evaluate the numerator using thesame expansion about E. The only difference is the additional factor of E = E+ δE in the integrand. The δE term
integrates to zero, since the remaining factors in the integrand yield D(E) eâβE eâ(δE)2/2kBT 2CV , which is even in
δE. Thus, the second term in the above equation is simply E/T , and we obtain
SOCE
= kBlnD(E) + 1
2kBln(2Ďk
BT 2CV ) .
The RHS here is dominated by the first term, which is extensive, whereas the second term is of order lnV . Thus,
we conclude that SOCE
(T, V,N) = SÂľCE(E, V,N), where E and T are related by 1
T = âSâE
âŁâŁE
.
10
(4.7) Consider rod-shaped molecules with moment of inertia I , and a dipole moment Âľ. The contribution of therotational degrees of freedom to the Hamiltonian is
Hrot =p2θ2I
+p2Ď
2I sin2θâ ÂľE cos θ ,
where E is the external electric field, and (θ, Ď) are polar and azimuthal angles describing the molecular orienta-tion1.
(a) Calculate the contribution of the rotational degrees of freedom of each dipole to the classical partition func-tion.
(b) Obtain the mean polarization P = ăÂľ cos θă of each dipole.
(c) Find the zero-field isothermal polarizability, Ď(T ) =(âPâE
)E=0
.
(d) Calculate the rotational energy per particle at finite field E, and comment on its high and low-temperaturelimits.
(e) Sketch the rotational heat capacity per dipole as a function of temperature.
Solution :
(a) The rotational contribution to the single particle partition function is
Ξrot =
ââŤ
ââ
dpθ
ââŤ
ââ
dpĎ
ĎâŤ
0
dθ
2ĎâŤ
0
dĎ eâp2
θ/2IkBT eâp2
Ď/2IkBT sin2θ eÂľE cos θ/k
BT
= 2Ď Âˇ (2ĎIkBT )1/2
ĎâŤ
0
dθ e¾E cos θ/kBT
ââŤ
ââ
dpθ eâp2
Ď/2IkBT sin2θ
= 4Ď2IkBT
ĎâŤ
0
dθ sin θ e¾E cos θ/kBT =
8Ď2I(kBT )2
ÂľEsinh
(ÂľE
kBT
).
The translational contribution is Ξtr = V Îťâ3T . The single particle free energy is then
f = âkBT ln
(8Ď2Ik2
BT 2)+ k
BT ln(ÂľE)â k
BT ln sinh
(ÂľE
kBT
)â k
BT ln
(V/Îť3
T
).
(b) The mean polarization of each dipole is
P = â âf
âE= âk
BT
E+ Âľ ctnh
(ÂľE
kBT
).
(c) We expand ctnh (x) = 1x + x
3 + O(x3) in a Laurent series, whence P = Âľ2E/3kBT + O(E3). Then Ď(T ) =
Âľ2/3kBT , which is of the Curie form familiar from magnetic systems.
(d) We have Ξrot = Tr eâβhrot , hence
Îľrot = ăhrotă = ââ ln Ξrotâβ
= â â
âβ
â 2 lnβ + ln sinh(β¾E)
= 2kBT â ÂľE ctnh
(ÂľE
kBT
).
1This is problem 4.12 from vol. 1 of M. Kardar.
11
Figure 2: Rotational heat capacity crot(T ) for problem 7.
At high temperatures T ⍠¾E/kB
, the argument of ctnhx is very small, and using the Laurent expansion we findÎľrot = k
BT . This comports with our understanding from equipartition, since there are only two quadratic degrees
of freedom present (pθ and pĎ). The orientational degree of freedom θ does not enter because ÂľE cos θ ⪠kBT
in this regime. Unlike the rotational kinetic energy, the rotational potential energy is bounded. In the limit T âŞÂľE/k
B, we have that the argument of ctnhx is very large, hence Îľrot â 2k
BT â ÂľE. This can be understood as
follows. If we change variables to pĎ âĄ pĎ/ sin θ, then we have
Ξrot =
ââŤ
ââ
dpθ
ââŤ
ââ
dpĎ
ĎâŤ
0
dθ sin θ
2ĎâŤ
0
dĎ eâp2
θ/2IkBT eâp2
Ď/2IkBT eÂľE cos θ/k
BT
=
ââŤ
ââ
dpθ
ââŤ
ââ
dpĎ
1âŤ
â1
dx
2ĎâŤ
0
dĎ eâp2
θ/2IkBT eâp2
Ď/2IkBT eÂľEx/k
BT ,
where x = cos θ. We see that x appears linearly in the energy, and simple dimensional analysis reveals that anydegree of freedom Îś which appears homogeneously as U(Îś) â Îśr contributes k
BT/r to the average energy. In our
case, we have quadratic contributions to the Hamiltonian from pθ and pĎ, a linear contribution from x = cos θ,
and Ď itself does not appear. Hence Îľ = âÂľE + 2 Ă 12kB
T + kBT = âÂľE + 2k
BT . The âÂľE term is the minimum
value of the potential energy.
(e) The rotational heat capacity per molecule, sketched in Fig. 2, is given by
crot =âÎľrotâT
= 2kBâ k
B
(ÂľE/k
BT
sinh(ÂľE/kBT )
)2.
12
(4.8) Consider a surface containing Ns adsorption sites which is in equilibrium with a two-component nonrela-tivistic ideal gas containing atoms of types A and B . (Their respective masses are mA and mB). Each adsorptionsite can be in one of three possible states: (i) vacant, (ii) occupied by an A atom, with energy ââA, and (ii) occupiedwith a B atom, with energy ââB.
(a) Find the grand partition function for the surface, Îsurf(T, ÂľA, ÂľB, Ns).
(b) Suppose the number densities of the gas atoms are nA and nB. Find the fraction fA(nA, nB, T ) of adsorptionsites with A atoms, and the fraction f0(nA, nB, T ) of adsorption sites which are vacant.
Solution :
(a) The surface grand partition function is
Îsurf(T, ÂľA, ÂľB, Ns) =(1 + e(âA+ÂľA)/kB
T + e(âB+ÂľB)/kBT)Ns
.
(b) From the grand partition function of the gas, we have
nA = Îťâ3T,A eÂľA/kB
T , nB = Îťâ3T,B eÂľB/kB
T ,
with
ÎťT,A =
â2Ď~2
mAkBT
, ÎťT,B =
â2Ď~2
mBkBT
.
Thus,
f0 =1
1 + nA Îť3T,A eâA
/kBT + nB Îť
3T,B eâB
/kBT
fA =nA Îť
3T,A eâA/kB
T
1 + nA Îť3T,A eâA
/kBT + nB Îť
3T,B eâB
/kBT
fB =nB Îť
3T,B eâB/kB
T
1 + nA Îť3T,A eâA
/kBT + nB Îť
3T,B eâB
/kBT
.
Note that f0 + fA + fB = 1.
13
(4.9) Consider a two-dimensional gas of identical classical, noninteracting, massive relativistic particles with
dispersion Îľ(p) =âp2c2 +m2c4.
(a) Compute the free energy F (T, V,N).
(b) Find the entropy S(T, V,N).
(c) Find an equation of state relating the fugacity z = eÂľ/kBT to the temperature T and the pressure p.
Solution :
(a) We have Z = (ÎśA)N/N ! where A is the area and
Îś(T ) =
âŤd2p
h2eâβ
âp2c2+m2c4 =
2Ď
(βhc)2(1 + βmc2
)eâβmc2 .
To obtain this result it is convenient to change variables to u = βâp2c2 +m2c4, in which case p dp = u du/β2c2,
and the lower limit on u is mc2. The free energy is then
F = âkBT lnZ = Nk
BT ln
(2Ď~2c2N
(kBT )2A
)âNk
BT ln
(1 +
mc2
kBT
)+Nmc2 .
where we are taking the thermodynamic limit with N â â.
(b) We have
S = ââF
âT= âNk
Bln
(2Ď~2c2N
(kBT )2A
)+Nk
Bln
(1 +
mc2
kBT
)+Nk
B
(mc2 + 2k
BT
mc2 + kBT
).
(c) The grand partition function is
Î(T, V, Âľ) = eâβΊ = eβpV =
ââ
N=0
ZN(T, V,N) eβ¾N .
We then find Π= exp(ΜA eβ¾
), and
p =(k
BT )3
2Ď(~c)2
(1 +
mc2
kBT
)e(Âľâmc2)/k
BT .
Note that
n =â(βp)
âÂľ=
p
kBT
=â p = nkBT .
14
(4.10) A nonrelativistic gas of spin- 12 particles of mass m at temperature T and pressure p is in equilibrium witha surface. There is no magnetic field in the bulk, but the surface itself is magnetic, so the energy of an adsorbedparticle is âââ Âľ0HĎ, where Ď = Âą1 is the spin polarization and H is the surface magnetic field. The surface hasNS adsorption sites.
(a) Compute the Landau free energy of the gas Ίgas(T, V, ¾). Remember that each particle has two spin polar-ization states.
(b) Compute the Landau free energy of the surface Ίsurf(T,H,NS). Remember that each adsorption site can bein one of three possible states: empty, occupied with Ď = +1, and occupied with Ď = â1.
(c) Find an expression for the fraction f(p, T,â, H) of occupied adsorption sites.
(d) Find the surface magnetization, M = Âľ0
(Nsurf,â âNsurf,â
).
Solution :
(a) We have
Îgas(T, V, Âľ) =
ââ
N=0
eNÂľ/kBT Z(T, V,N) =
ââ
N=0
V N
N !eNÂľ/k
BT 2N Îťâ3N
T
= exp(2V k
BTÎťâ3
T eÂľ/kBT)
,
where ÎťT =â2Ď~2/mk
BT is the thermal wavelength. Thus,
Ίgas = âkBT ln Îgas = â2V k
BTÎťâ3
T eÂľ/kBT .
(b) Each site on the surface is independent, with three possible energy states: E = 0 (vacant), E = ââ â Âľ0H(occupied with Ď = +1), and E = ââ+ Âľ0H (occupied with Ď = â1). Thus,
Îsurf(T,H,NS) =(1 + e(Âľ+â+Âľ
0H)/k
BT + e(Âľ+ââÂľ
0H)/k
BT)NS
.
The surface free energy is
Ίsurf(T,H,NS) = âkBT ln Îsurf = âNSkB
T ln(1 + 2 e(Âľ+â)/k
BT cosh(Âľ0H/k
BT ))
.
(c) The fraction of occupied surface sites is f = ăNsurf/NSă. Thus,
f = â 1
NS
âΊsurf
âÂľ=
2 e(Âľ+â)/kBT cosh(Âľ0H/k
BT )
1 + 2 e(Âľ+â)/kBT cosh(Âľ0H/k
BT )
=2
2 + eâ(Âľ+â)/kBT sech(Âľ0H/k
BT )
.
To find f(p, T,â, H), we must eliminate Âľ in favor of p, the pressure in the gas. This is easy! From Ίgas = âpV ,
we have p = 2kBTÎťâ3
T eÂľ/kBT , hence
eâÂľ/kBT =
2kBT
p Îť3T
.
Thus,
f(p, T,â, H) =p Îť3
T
p Îť3T + k
BT eââ/k
BT sech(Âľ0H/k
BT )
.
15
Note that f â 1 when â â â, when T â 0, when p â â, or when H â â.
(d) The surface magnetization is
M = ââΊsurf
âH= NS Âľ0 ¡
2 e(Âľ+â)/kBT sinh(Âľ0H/k
BT )
1 + 2 e(Âľ+â)/kBT cosh(Âľ0H/k
BT )
=NS Âľ0 p Îť
3T tanh(Âľ0H/k
BT )
p Îť3T + k
BT eââ/k
BT sech(Âľ0H/k
BT )
.
16
(4.11) A classical gas consists of particles of two species: A and B. The dispersions for these species are
ÎľA(p) =
p2
2m, Îľ
B(p) =
p2
4mââ .
In other words, mA= m and m
B= 2m, and there is an additional energy offset ââ associated with the B species.
(a) Find the grand potential Ί(T, V, ¾A, ¾
B).
(b) Find the number densities nA(T, Âľ
A, Âľ
B) and n
B(T, Âľ
A, Âľ
B).
(c) If 2A B is an allowed reaction, what is the relation between nA
and nB
?(Hint : What is the relation between Âľ
Aand Âľ
B?)
(d) Suppose initially that nA= n and n
B= 0. Find n
Ain equilibrium, as a function of T and n and constants.
Solution :
(a) The grand partition function Î is a product of contributions from the A and B species, and the grand potentialis a sum:
Ί = âV kBT Îťâ3
T eÂľA/k
BT â 23/2 V k
BT Îťâ3
T e(ÂľB+â)/k
BT
Here, we have defined the thermal wavelength for the A species as ÎťT ⥠ΝT,A =â2Ď~2/mk
BT . For the B species,
since the mass is twice as great, we have ÎťT,B = 2â1/2 ÎťT,A.
(b) The number densities are
nA= â 1
V¡ âΊ
âÂľA
= V Îťâ3T eÂľA
/kBT
nB= â 1
V¡ âΊ
âÂľB
= 23/2 V Îťâ3T e(ÂľB
+â)/kBT .
If the reaction 2A B is allowed, then the chemical potentials of the A and B species are related by ÂľB= 2Âľ
A⥠2¾.
We then haven
AÎť3T = eÂľ/kB
T , nBÎť3T = 23/2 e(2Âľ+â)/k
BT .
(c) The relation we seek is thereforen
B= 23/2 n2
AÎť3T eâ/k
BT .
(d) If we initially have nA= n and n
B= 0, then in general we must have
nA+ 2n
B= n =â n
B= 1
2
(nâ n
A
).
Thus, eliminating nB
, we have a quadratic equation,
23/2 Îť3T eâ/k
BT n2
A= 1
2 (nâ nA) ,
the solution of which is
nA=
â1 +â1 + 16
â2nÎť3
T eâ/kBT
8â2Îť3
T eâ/kBT
.
17
(4.12) The potential energy density for an isotropic elastic solid is given by
U(x) = ÂľTr Îľ2 + 12Îť (Tr Îľ)
2
= Âľâ
ι,β
Îľ2ιβ(x) +12Îť(â
Îą
ξιι(x))2
,
where Âľ and Îť are the Lame parameters and
ξιβ =1
2
(âuÎą
âxβ+
âuβ
âxÎą
),
with u(x) the local displacement field, is the strain tensor. The Cartesian indices ι and β run over x, y, z. Thekinetic energy density is
T (x) = 12Ď u
2(x) .
(a) Assume periodic boundary conditions, and Fourier transform to wavevector space,
uÎą(x, t) = 1âV
â
k
uÎąk(t) e
ik¡x
uÎąk(t) =
1âV
âŤd3x uÎą(x, t) eâik¡x .
Write the Lagrangian L =âŤd3x(T â U
)in terms of the generalized coordinates uÎą
k(t) and generalized
velocities ËuÎąk(t).
(b) Find the Hamiltonian H in terms of the generalized coordinates uÎąk(t) and generalized momenta ĎÎą
k (t).
(c) Find the thermodynamic average ău(0) ¡ u(x)ă.
(d) Suppose we add in a nonlocal interaction of the strain field of the form
âU = 12
âŤd3x
âŤd3xâ˛
Tr Îľ(x) Tr Îľ(xâ˛) v(xâ xâ˛) .
Repeat parts (b) and (c).
Solution :
To do the mode counting we are placing the system in a box of dimensions Lx Ă Ly Ă Lz and imposing periodicboundary conditions. The allowed wavevectors k are of the form
k =
(2Ďnx
Lx
,2Ďny
Ly
,2Ďnz
Lz
).
We shall repeatedly invoke the orthogonality of the plane waves:
LxâŤ
0
dx
LyâŤ
0
dy
LzâŤ
0
dz ei(kâkâ˛)¡x = V δk,kⲠ,
where V = LxLyLz is the volume. When we Fourier decompose the displacement field, we must take care to note
that uÎąk
is complex, and furthermore that uÎąâk
=(uÎąk
)â, since uÎą(x) is a real function.
(a) We then have
T =
ââŤ
ââ
dx 12Ď u
2(x, t) = 12Ďâ
k
âŁâŁ ËuÎą
k(t)âŁâŁ2
18
and
U =
ââŤ
ââ
dx
[12Âľ
âuÎą
âxβ
âuÎą
âxβ+ 1
2 (Îť+ Âľ) (â¡u)2]
= 12
â
k
(¾ διβ + (Ν+ ¾) kι kβ
)k2 uÎą
k(t) uβâk
(t) .
The Lagrangian is of course L = T â U .
(b) The momentum ĎÎąk conjugate to the generalized coordinate uÎą
k is
ĎÎąk =
âL
â ËuÎą
k
= Ď ËuÎąâk ,
and the Hamiltonian is
H =â
k
ĎÎąkËuÎąk â L
=â
k
âŁâŁĎÎąk
âŁâŁ2
2Ď+ 1
2
[Âľ(διβ â kÎą kβ
)+ (Ν+ 2¾) kι kβ
]k2 uÎą
k uβâk
.
Note that we have added and subtracted a term ¾ kι kβ within the expression for the potential energy. This is
because Pιβ = kÎą kβ and Qιβ = διβ â kÎą kβ are projection operators satisfying P2 = P and Q2 = Q, with P+Q = I,
the identity. P projects any vector onto the direction k, and Q is the projector onto the (two-dimensional) subspace
orthogonal to k.
(c) We can decompose uk
into a longitudinal component parallel to k and a transverse component perpendicular to
k, writing
uk = ik uâk+ iek,1 u
âĽ,1k
+ iek,2 uâĽ,2k
,
where ek,1 , ek,2 , k is a right-handed orthonormal triad for each direction k. A factor of i is included so that
uââk
=(uâk
)â, etc. With this decomposition, the potential energy takes the form
U = 12
â
k
[Âľk2
(âŁâŁuâĽ,1k
âŁâŁ2 +âŁâŁuâĽ,2
k
âŁâŁ2)+ (Îť+ 2Âľ)k2
âŁâŁuâk
âŁâŁ2]
.
Equipartition then means each independent degree of freedom which is quadratic in the potential contributes an
average of 12kB
T to the total energy. Recalling that uâk
and uâĽ,jk
(j = 1, 2) are complex functions, and that they areeach the Fourier transform of a real function (so that k and âk terms in the sum for U are equal), we have
â¨Âľk2
âŁâŁuâĽ,1k
âŁâŁ2âŠ=â¨Âľk2
âŁâŁuâĽ,2k
âŁâŁ2âŠ= 2Ă 1
2kBT
â¨(Îť+ 2Âľ)k2
âŁâŁuâk
âŁâŁ2âŠ= 2Ă 1
2kBT .
Thus,
â¨|uk|2
âŠ= 4Ă 1
2kBT Ă 1
Âľk2+ 2Ă 1
2kBT Ă 1
(Îť+ 2Âľ)k2
=
(2
Âľ+
1
Îť+ 2Âľ
)k
BT
k2.
19
Then
â¨u(0) ¡ u(x)
âŠ=
1
V
â
k
â¨|uk|2
âŠeik¡x
=
âŤd3k
(2Ď)3
(2
Âľ+
1
Îť+ 2Âľ
)k
BT
k2eik¡x
=
(2
Âľ+
1
Îť+ 2Âľ
)k
BT
4Ď|x| .
Recall that in three space dimensions the Fourier transform of 4Ď/k2 is 1/|x|.
(d) The k-space representation of âU is
âU = 12
â
k
k2 v(k) kι kβ uιk uβ
âk,
where v(k) is the Fourier transform of the interaction v(xâ xâ˛):
v(k) =
âŤd3r v(r) eâik¡r .
We see then that the effect of âU is to replace the Lame parameter Îť with the k-dependent quantity,
Îť â Îť(k) ⥠Ν+ v(k) .
With this simple replacement, the results of parts (b) and (c) retain their original forms, mutatis mutandis.
20
(4.13) For polyatomic molecules, the full internal partition function is written as the product
Ξ(T ) =gel ¡ gnucgsym
¡ Ξvib(T ) ¡ Ξrot(T ) ,
where gel is the degeneracy of the lowest electronic state2, gnuc =â
j(2Ij + 1) is the total nuclear spin degeneracy,
Ξvib(T ) is the vibrational partition function, and Ξrot(T ) is the rotational partition function3. The integer gsymis the symmetry factor of the molecule, which is defined to be the number of identical configurations of a givenmolecule which are realized by rotations when the molecule contains identical nuclei. Evaluate gnuc and gsymfor the molecules CH4 (methane), CH3D, CH2D2, CHD3, and CD4. Discuss how the successive deuteration ofmethane will affect the vibrational and rotational partition functions. For the vibrations your discussion can bequalitative, but for the rotations note that all one needs, as we derived in problem (6), is the product I1I2I3 of themoments of inertia, which is the determinant of the inertia tensor Iιβ in a body-fixed center-of-mass frame. Using theparallel axis theorem, one has
Iιβ =â
j
mj
(r2j διβ â rÎąj rβj
)+M
(R2 διβ âRÎąRβ
)
where M =â
j mj and R = Mâ1â
j mjrj . Recall that methane is structurally a tetrahedron of hydrogen atomswith a carbon atom at the center, so we can take r1 = (0, 0, 0) to be the location of the carbon atom and r2,3,4,5 =(1, 1, 1) , (1,â1,â1) , (â1, 1,â1) , (â1,â1, 1) to be the location of the hydrogen atoms, with all distances in unitsof 1â
3times the CâH separation.
Solution :
The total partition function is given by
Z(T, V,N) =V N
N !
(2Ď~2
MkBT
)3N/2
ΞNint(T ) ,
The Gibbs free energy per particle is
Âľ(T, p) =G(T, p,N)
N= k
BT ln
(p Îťd
T
kBT
)â k
BT ln Ξ(T )
= kBT ln
(p Îťd
T
kBT
)â k
BT ln
(gel ¡ gnucgsym
)
+ kBTâ
a
ln(2 sinh(Îa/2T )
)â k
BT ln
[(2k
BT
~2
)3/2âĎI1I2I3
].
The electronic degeneracy is gel = 1 for all stages of deuteration. The nuclear spin of the proton is I = 12 and
that of the deuteron is I = 1. Thus there is a nuclear degeneracy of 2Ip + 1 = 2 for each hydrogen nucleus and2Id + 1 = 3 for each deuterium nucleus. The symmetry factor is analyzed as follows. For methane CH4, there arefour threefold symmetry axes, resulting in gsym = 12. The same result holds for CD4. For CH3D or CHD3, thereis a single threefold axis, hence gsym = 3. For CH2D2, the two hydrogen nuclei lie in a plane together with thecarbon, and the two deuterium nuclei lie in a second plane together with the carbon. The intersection of these twoplanes provides a twofold symmetry axis, about which a 180 rotation will rotate one hydrogen into the other andone deuterium into the other. Thus gsym = 2.
To analyze the rotational partition function, we need the product I1I2I3 of the principal moments of inertia, whichis to say the determinant of the inertia tensor det I . We work here in units of amu for mass and 1â
3times the CâH
2We assume the temperature is low enough that we can ignore electronic excitations.3Note that for linear polyatomic molecules such as CO
2and HCN, we must treat the molecule as a rotor, i.e. we use eqn. 4.261 of the notes.
21
separation for distance. The inertia tensor is
Iιβ =â
j
mj
(r2j διβ â rÎąj rβj
)+M
(R2 διβ âRÎąRβ
)
where
M =â
j
mj
R = Mâ1â
j
mjrj .
The locations of the four hydrogen/deuterium ions are:
L1 : (+1,+1,+1)
L2 : (+1,â1,â1)
L3 : (â1,+1,â1)
L4 : (â1,â1,+1) .
For CH4 we have M = 16 and R = 0. The inertia tensor is
ICH4=
8 0 00 8 00 0 8
.
Similarly, for CD4 we have
ICD4
=
16 0 00 16 00 0 16
.
For CH3D, there is an extra mass unit located at L1 relative to methane, so M = 17. The CM is at R =117 (+1,+1,+1). According to the general formula above for Iιβ , thie results in two changes to the inertia ten-sor, relative to ICH
4
. We find
âI =
2 â1 â1â1 2 â1â1 â1 2
+1
17
2 â1 â1â1 2 â1â1 â1 2
,
where the first term accounts for changes in I in the frame centered at the carbon atom, and the second term shiftsto the center-of-mass frame. Thus,
ICH3D =
10 + 217 â 18
17 â 1817
â 1817 10 + 2
17 â 1817
â 1817 â 18
17 10 + 217
.
For CHD3, we regard the system as CD4 with a missing mass unit at L1, hence M = 19. The CM is now atR = 1
17 (â1,â1,â1). The change in the inertia tensor relative to ICD4
is then
âI = â
2 â1 â1â1 2 â1â1 â1 2
+
1
19
2 â1 â1â1 2 â1â1 â1 2
.
Thus,
22
mass M degeneracy symmetry det Imolecule (amu) factor gnuc factor gsym (amu) ¡ a2/3
CH4 16 24 = 16 4Ă 3 = 12 83
CH3D 17 23 ¡ 3 = 24 1à 3 = 3 8 ¡(11 + 3
17
)2
CH2D2 18 22 ¡ 32 = 36 1à 2 = 2 12 ¡(8 + 2
9
)¡(16 + 2
9
)
CHD3 19 2 ¡ 33 = 54 1à 3 = 3 16 ¡(13 + 3
19
)2
CD4 20 34 = 81 4Ă 3 = 12 163
Table 2: Nuclear degeneracy, symmetry factor, and I1I2I3 product for successively deuterated methane.
ICHD3
=
14 + 219
1819
1819
1819 14 + 2
191819
1819
1819 14 + 2
19
.
Finally, for CH2D2. we start with methane and put extra masses at L1 and L2, so M = 18 and R = 19 (+1, 0, 0).
Then
âI = â
4 0 00 4 â20 â2 4
+2
9
0 0 00 1 00 0 1
and
ICH2D
2
=
12 0 0
0 12 + 29 â2
0 â2 12 + 29
.
For the vibrations, absent a specific model for the small oscillations problem the best we can do is to say that
adding mass tends to lower the normal mode frequencies since Ď âźâk/M .
23