solutions manual for equilibrium and non-equilibrium statistical thermodynamics
TRANSCRIPT
1
ERRATApage 205, equation(4.82)
G(~r, ~r ′) =δM(~r)
∂B(~r ′)or G−1(~r, ~r ′) =
δ2L[M ]
δM(~r)δM(~r ′)
page 209, equation (4.92)
G−1ij (~r, ~r ′) =
δL[M ]
δMi(~r)δMj(~r ′)
∣∣∣M1=v,M2=0
page 244, equation following (4.179)
D − 4
(D
2− 1
)= 4 −D
page 245, first equation of the page
Z−1 =d
dq2G−1(~q)
∣∣∣q2=0
page 359, equation following (6.81)Tαβ(~u = 0) = T ′
αβ = Pαβ
page 366, first equation
θ(x, t) = −θ0 +θ0√πDt
0∫
−∞
dx′ exp
(− (x− x′)2
4Dt
)
page 512, read: An example of the increase of the H-function is given by Jaynes [60].ignore: but the argument is incomplete.
page 565, equation (9.179)
χ′′AB(t) =
1
2〈[A(t),B†(0)]〉
page 577, equation (9.193)
∂t n+1
m~∇ · ~gL = 0
equation following (9.196)~∇× (~∇ × ~V ) = ~∇(~∇ · ~V ) − ~∇2~V
page 584, first line of question 11. By examining the moments ∆X, ∆P , (∆P )2, (∆X)2 and ∆P∆X...
2
Solutions for selected exercises and problems
Solutions for chapter 1
Exercise 1.6.2 Internal variable at equilibrium
1. Considering the energy as a function of S and y, one obtains from (1.36)
∂E
∂y
∣∣∣∣S
∂y
∂S
∣∣∣∣E
∂S
∂E
∣∣∣∣y
= −1
At equilibrium (∂E/∂S)y = T , and
∂E
∂y
∣∣∣∣(eq)
S
= −T ∂S
∂y
∣∣∣∣(eq)
E
= 0
because one must have by definition (∂S/∂y)E = 0 at equilibrium.
Let us also consider a function1 f(S(E, y), y) :
∂f
∂y
∣∣∣∣E
=∂S
∂y
∣∣∣∣E
∂f
∂S
∣∣∣∣y
+∂f
∂y
∣∣∣∣S
=∂f
∂y
∣∣∣∣S
the last equality being valid at equilibrium, and let us use this result in (1.64)
∂2S
∂y2
∣∣∣∣(eq)
E
= − ∂
∂y
∣∣∣∣S
(∂E
∂y
∣∣∣∣S
∂S
∂E
∣∣∣∣y
)
=∂E
∂y
∣∣∣∣S
∂2S
∂E∂y− 1
T
∂2E
∂y2
∣∣∣∣S
The first term vanishes at equilibrium, which demonstrates the second equation of (1.63), taking (1.62)into account.
2. The condition of maximum entropy reads
dS =
(1
T1dE1 +
1
T2dE2
)+
(P1
T1dV1 +
P2
T2dV2
)
Since dE1 = −P1dV1, the condition dS = 0 is automatic.
3. The condition of minimum energy at fixed entropy, volume and particle number, is
dE = (T1 − T2)dS1 − (P1 − P2)dV1 + (µ1 − µ2)dN1 = 0
Because the superfluid does not transport entropy, there cannot be entropy transfer between the twocompartments: dS1 = 0. Furthermore dV1 = 0 because the volumes are fixed. The only condition isthat of equality of the chemical potentials: µ1 = µ2. Assume that the temperature of compartment 1is increased by ∆T . As the chemical potential remains unchanged, the condition dµ1 = 0 added to theGibbs-Duhem relation (1.41)
dµ = − S
NdT +
V
NdP
1The following equality is a particular case of a general identity which we shall encounter many times. Let f(xi) be afunction of several variables, which are themselves functions of another set of variables: xi(yj). One easily shows that
∂f
∂yj
˛
˛
˛
˛
yj 6=k
=X
i
∂f
∂xi
˛
˛
˛
˛
xl 6=i
∂xi
∂yj
˛
˛
˛
˛
yj 6=k
3
shows that the pressure must increase by ∆P = (S1/V1)∆T .
Problem 1.7.1 Reversible and irreversible free expansion of an ideal gas
1. One starts from the expression of (∂E/∂V )T
∂E
∂V
∣∣∣∣T
= T∂S
∂V
∣∣∣∣T
− P = T∂P
∂T
∣∣∣∣V
− P
where we have used the Maxwell relation (1.27). By assumption, this expression must vanish, which givesanother proof of (1.61). taking into account
∂P
∂T
∣∣∣∣V
=1
Vφ′(T )
One gets Tφ′(T ) = φ(T ) and by integration φ(T ) = aT .
2. In the case of an ideal gas∂S
∂V
∣∣∣∣T
=∂P
∂T
∣∣∣∣V
=R
V
The expression giving S is
S(T, V ) = S(T0, V0) +
T∫
T0
dTCV
T+
V∫
V0
dVR
V
that is
S(T, V ) = S(T0, V0) + CV lnT
T0+R ln
V
V0
In order to express S as a function of the pressure, and not of the volume, one uses
V
V0=
T
T0
P0
P
and, with CP = CV +R
S(T, P ) = S(P0, T0) + CP lnT
T0+R ln
P0
P
3.
S(T, V ) = S(T0, V0) +R ln
(T l/2V
Tl/20 V
)
A transformation at constant entropy thus corresponds to
T l/2V = const ou T 1/(γ−1)V = const
with γ = (l + 2)/l. If we use the pressure instead of the volume
TP (1−γ)/γ = const Tf = Ti
(Pf
Pi
)(γ−1)/γ
One can compute explicitly the work provided to the gas by integrating the formula for infinitesimal work
Wi→f = −∫ f
i
P dV
for a reversible adiabatic transformation: PV γ =const, but it is simpler to remark that the increase inwork is equal to that in internal energy, as there is no heat exchange
Wi→f = Ei→f = CV (Tf − Ti)
4
the last equality being valid only for an ideal gas, and CV independent of T . It is a good exerciseto check that this result coincides with that obtained from calculating the work; one will notice thatCV = R/(γ − 1)
4. The work supplied to the gas is determined by the external pressure, which is equal to the finalpressure Pf : W ′
i→f = −Pf (V ′f − Vi). As the transformation is adiabatic and as the gas is ideal, we have
once moreW ′
i→f = Ei→f = CV (T ′f − Ti)
Using the ideal gas law, V ′f = RT ′
f/Pf , one determines T ′f
T ′f =
Ti
γ
(1 + (γ − 1)
Pf
Pi
)
The final temperature in the irreversible expansion is higher than that of the reversible expansion: thiscan be seen by using the concavity of the curve giving Tf as a function of x = Pf/Pi.
5. In the case of the reversible transformation, the entropy variation ∆S vanishes, while in the irreversiblecase ∆S > 0. Since the final pressures are identical, one may use in the comparison the result of question 2
S(Tf , Pf ) = S(Ti, Pi) + Cp lnTf
Ti+R ln
Pi
Pf
whence
CP lnT ′
f
Ti> CP ln
Tf
Ti
and thus T ′f > Tf . The variation of internal energy is minus the work supplied to the external medium
W = CV (Ti − Tf ). As T ′f > Tf , Wirr < Wrev. In order to convince oneself that the external gas does not
play any role in the entropy balance, one remarks that it can be replaced by a mass M on the piston, withPfA = Mg. The external medium is then entirely mechanical, and does not contribute to the entropybalance.
If one puts again the mass on the piston, P ′′f = Pi, and the expression of the entropy shows that T ′′
f > Ti
because one must have ∆S > 0 after completion of both operations. An explicit computation gives
T ′′f =
Ti
γ2
(1 + (γ − 1)
Pi
Pf
)(1 + (γ − 1)
Pf
Pi
)
It is easy to check in this expression that T ′′f > Ti by setting x = Pf/Pi and by remarking that (x+1/x) ≥
2.
6. Since the expansion is adiabatic and does not supply any work to the system, and since the gas is ideal,the internal energy and therefore the temperature remains unchanged. From the results of question 2,dS = R ln[(V + dV )/V ] ≃ RdV/V > 0. Since d−Q = 0, the transformation cannot be quasi-static:d−Q = TdS does not hold.
7. Since the internal energy does not vary (adiabatic wall and no work supply), Tf = Ti, and ∆S =R ln(Vf/Vi), although no heat was supplied to the gas.
Problem 1.7.3 Equation of state for a solid
1. The knowledge of the equation of state allows us to express the temperature
T =∂E
∂S
∣∣∣∣V
= f(V )g′(S) = S1/3 ×[f(V )eS/3R
3R(S + 4R)
](1)
The quantity between square brackets staying finite when S → 0, one sees in the preceding expressionthat S and T vanish simultaneously. This behaviour is in agreement with the third law.
2. Expression (1) allows one to display the dominant behaviours of S, and thus of CV , in the two limitingcases
5
1. low temperature: T ∼ S1/3 → S ∼ T 3 → CV ∼ T 3
2. high temperature T ∼ eS/3R → S ∼ 3R lnT → CV ∼ 3R
3. From (1.68), one gets
P = − ∂E
∂V
∣∣∣∣S
= −f ′(V )g(S) = −2Ab(V − V0)eb(V −V0)
2
g(S) (2)
The physical interpretation of V0 is clearly displayed in (2): V0 is the volume at zero pressure, namelythe maximum volume of the solid. For V > V0, the atoms which form the solid are organized in anotherphase of matter described by another equation of state. One can point out an inconsistency of the model:it predicts a finite pressure for a vanishing volume! At zero pressure, the dilatation coefficient vanishes,since V0 is temperature independent.
4. Let us apply (1.36) to the variables (S, V, T )
∂S
∂V
∣∣∣∣T
= − ∂T
∂V
∣∣∣∣S
(∂T
∂S
∣∣∣∣V
)−1
= −f′(V )g′(S)
f(V )g′′(S)
There are many possible proofs of relation (1.69), we propose below the proof which seems to us theshortest. Examination of the relevant variables for the calculation of the dilatation coefficient lead us towork with the Gibbs potential G(T, P ) = E − TS + PV
dG = −SdT + V dP
One may then use the Maxwell relation
∂V
∂T
∣∣∣∣P
= −(∂P
∂S
∣∣∣∣T
)−1
Relation (2) provides us with the expression for P in terms of the variables (S, V ). Using a procedurewhich should be familiar by now, we display explicitly the partial derivatives of P with respect to thesevariables
∂P
∂S
∣∣∣∣T
=∂P
∂S
∣∣∣∣V
+∂P
∂V
∣∣∣∣S
∂V
∂S
∣∣∣∣T
= − (f ′(V ))2(g′(S))2 − f(V )f ′′(V )g(S)g′′(S)
f ′(V )g′(S)
= − 1
CV
(f ′(V ))2(g′(S))2 − f(V )f ′′(V )g(S)g′′(S)
f ′(V )g′′(S)
The last line is obtained by noticing that
CV = T∂S
∂T
∣∣∣∣V
= T
(∂T
∂S
∣∣∣∣V
)−1
=g′(S)
g′′(S)
We invite the reader to work out another demonstration starting from
∂V
∂T
∣∣∣∣P
∂T
∂P
∣∣∣∣V
∂P
∂V
∣∣∣∣T
= −1
Problem 1.7.5 Surface tension of a soap film
1. The TdS equation reads in this case
dE = TdS + σℓdx
6
Let us also write the differential of the free energy
dF = −SdT + σℓdx
and the Maxwell relation which follows from it2
∂S
∂x
∣∣∣∣T
= −ℓ ∂σ∂T
∣∣∣∣x
2. Using the differentiation rule for implicit functions and the preceding Maxwell relation, one showsthat
∂E
∂T
∣∣∣∣x
=∂E
∂S
∣∣∣∣x
∂S
∂T
∣∣∣∣x
= T∂S
∂T
∣∣∣∣x
= Cx
∂E
∂x
∣∣∣∣T
=∂E
∂x
∣∣∣∣S
+∂E
∂S
∣∣∣∣x
∂S
∂x
∣∣∣∣T
=
(σ − T
∂σ
∂T
∣∣∣∣x
)ℓ
3. At a constant temperature
dE = −T ℓ ∂σ∂T
∣∣∣∣x
dx+ σℓ dx
The second term is identified with the work: d−W = σℓ dx, and thus the energy which is exchanged inthe from of heat is
d−Q = −T ℓ ∂σ∂T
∣∣∣∣x
dx = aσ0T ℓ dx > 0
One must supply heat to the film in order to stretch it.
4. Let us show that (∂Cx/∂x)T = 0
∂Cx
∂x
∣∣∣∣T
= T∂
∂T
[∂S
∂x
∣∣∣∣T
]
x
= −T ℓ ∂2σ
∂T 2
∣∣∣∣x
= 0
5. The surface tension, and thus f , are independent of the film area, and thus of x, whence
∂f
∂x
∣∣∣∣T
= 0 ⇒ κT = −∞
One recognizes a well-known, and easily observable, property: any fluctuation destroys the soap film.
6. Let us express the TdS equation in terms of the variables (T, x)
TdS = Cx dT + aσ0T ℓ dx
When a transformation is done at constant entropy
dT
T= −aσ0ℓ
Cxdx
Numerical application: dT = −4.3 102 dx (dx in m).
2One can derive directly this relation from (1.27) with the substitutions −V → x et P → σℓ.
7
Solutions for chapter 2
Exercise 2.7.1 Density operator for spin-1/2
1. The second degree equation giving the eigenvalues of D is
λ2 − λ+ a(1 − a) − |c|2 = 0
and the product of the roots is given by
λ+λ− = a(1 − a) − |c|2
The positivity condition on D implies λ+λ− ≥ 0 and the condition TrD = 1, or λ+ + λ− = 1, so that0 ≤ λ+λ− ≤ 1/4, whence (2.105). In the case of a pure state one of the eigenvalues vanishes (the otherbeing equal to one) and a(1 − a) = |c|2. The density matrix of the state |ψ〉 is
D =
(|α|2 αβ∗
α∗β |β|2)
and one can indeed check that |α|2|β|2 = (αβ∗)(α∗β).
2. One writes a = (1 + bz)/2, c = (bx − iby)/2, whence (2.107) for D. As Tr σi = 0, one deduces from(2.107)
Trσiσj = 2 δij
and (2.108). One observes that a(1−a) = (1− b2z)/4 and that |c|2 = (b2x + b2y)/4. The inequalities (2.105)
are equivalent to 0 ≤ ~b 2 ≤ 1, the pure case corresponding to ~b 2 = 1.
3. With ~B parallel to Oz, the Hamiltonian reads
H = −1
2γσz
and the equation of evolution of the density matrix is
dD
dt=
1
i~[H,D] = −1
2γB(bxσy − byσx)
which is equivalent todbxdt
= γBbydbydt
= −γBbxdbzdt
= 0
that isd~b
dt= −γ ~B ×~b
This is nothing other than the evolution equation of a classical magnetic moment in a constant magneticfield.
4. The non-zero matrix elements of D are
D+−;+− = D−+;−+ =1
2D+−;−+ = D−+;+− = −1
2
On taking the partial trace, we obtain the density matrix of the first spin D(1)
D(1)++ = D
(1)−− =
1
2D
(1)+− = D
(1)−+ = 0
D(1) then describes a state with zero polarization. One has of course D(1) = D(2), and D(1)⊗D(2), whichis a diagonal matrix with all elements equal to 1/4, is not equal to D, which would be the case were D atensor product.
8
Exercise 2.7.7 Galilean transformation
The occupation probability Pr of a level r is unchanged under a Galilean transformation, otherwiseone could detect a translational motion without looking outside the system.This implies D′ = D andinvariance of the entropy. Under a Galilean transformation, the Hamiltonian becomes
H ′ = H − uP +1
2Mu2
and the equality of the operator parts of D and D′ implies
βH − λP = β′(H − uP)
that is β = β′ and βu = λ. One also gets
Z = Z ′ exp
(1
2βMu2
)= Z ′ exp
(M
2βλ2
)
From (2.66), this gives 〈P 〉〈P 〉 =
∂ lnZ
∂λ=Mλ
β= Mu
which is the expected value. Concerning the energy one gets
E = −∂ lnZ
∂β= −∂ lnZ ′
∂β+Mλ2
2β2= E′ +
1
2Mu2
The total energy is the sum of the rest energy E′ and the kinetic energy Mu2/2 of the moving mass.
Exercise 2.7.8 Fluctuation-response theorem
1. Let us compute dK/dx
dK(x)
dx= ex(A+B)(A+B)e−xA − ex(A+B)A e−xA = ex(A+B)B e−xA
On integrating this equation from x = 0 to x = 1 one gets
K(1) −K(0) = e(A+B)e−A − 1 =
1∫
0
dx ex(A+B)Be−xA
which leads to (2.118) if one multiplies the RHS by expA. Applying this equation to exp(A+ λB) leadsto
eA+λB = eA +
1∫
0
dx ex(A+λB) (λB) e(1−x)A
One may replace exp(x(A + λB)) by exp(xA) with an error on the order of λ2. Using the invariance ofthe trace under circular permutation leads to (2.119)
Tr e(A+λB) ≃ Tr eA + λTr (eAB)
By definition
eA(λ+dλ) = eA(λ)+dλA′(λ) = eA(λ) + dλd eA(λ)
dλ
The substitution λ → dλ in (2.119) leads to (2.120). Using once more the invariance of the trace undercircular permutation allows one to recover (2.57) in a particular case
d
dλTr eA(λ) = Tr
(dA(λ)
dλeA(λ)
)(3)
9
2. From (2.66), or (2.124)
Ai =1
Z
∂
∂λiTr[exp(∑
k
λkAk
)]=
1
ZTr[Ai exp
(∑
k
λkAk
)]= Tr (DAi)
In order to differentiate Ai with respect to λj , one must use (2.120)
∂
∂λjTr[Ai exp
(∑
k
λkAk
)]= Tr
[Ai
1∫
0
dx exp[x∑
k
λkAk
]Aj exp
[(1 − x)
(∑
k
λkAk
)]]
= Z Tr[ 1∫
0
dxAiDxAjD
(1−x)]
from which one deduces
∂Ai
∂λj=
1
Z
∂
∂λjTr[Ai exp
(∑
k
λkAk
)]−AiAj
=
1∫
0
dxTr[(Ai − Ai)D
x(Aj −Aj)D(1−x)
]
Let us set B =∑
i ai(Ai −Ai)
∑
ij
aiajCij =
1∫
0
dxTr (DxBD(1−x)
B) ≥ 0
because, in a basis where D is diagonal
Tr (DxBD(1−x)
B) =∑
n,m
DxnnBnmD
(1−x)mm Bmn =
∑
n,m
DxnnD
(1−x)mm |Bnm|2 ≥ 0
using the hermiticity of B and the fact that the diagonal elements of D are positive.
Exercise 2.7.9 Phase space volume for N free particles
The condition for the total energy to be ≤ E is
p21
2m+ . . .+
p2N
2m≤ E
and thus
ΦN (E) =V N
N !hN
∫
~p 21 +...+~p 2
N≤2mE
d3p1 . . . d3pN
One expects the average energy per particle to be ∼ E/N , which suggest to introduce the dimensionlessvariable ~u defined as
~p =
√2mE
N~u ~u2
1 + . . .+ ~u2N ≤ N
since the average value of u is on the order of 1. This change of variables leads immediately to (2.122),with
C(N) =
∫
~u21+...+~u2
N≤N
d3u1 . . .d3uN
In order to compute C(N) one calculates in two different ways the integral In
In =
+∞∫
−∞
dx1 . . .
+∞∫
−∞
dxn e−(x21+...+x2
N )
10
This integral is the product of n independent Gaussian integrals and its value is πn/2. One can alsocompute this integral in polar coordinates in a n-dimensional space
In = An
∫ ∞
0
dr e−r2
rn−1
where An is the area of the unit sphere in a n-dimensional space. The change of variables v = r2 castsIn in the form
In =1
2An
∫ ∞
0
dv e−vvn2−1 =
1
2AnΓ
(n2
)
which leads to
An =2 πn/2
Γ(n/2)
One computes C(N) in polar coordinates in a 3N -dimensional space (r = (~u 21 + . . .+ ~u 2
N)1/2)
C(N) = A3N
√N∫
0
dr r3N−1 =2π3N/2(N)3N/2
3NΓ(3N2 )
=π3N/2(N)3N/2
Γ(3N2 + 1)
The final form of C(N) is obtained by using Stirling approximation
Γ
(3N
2+ 1
)≃(
3N
2e
)3N/2
One deduces ΦN (E)
ΦN (E) =V N
N !
(4πemE
3Nh2
)3N/2
(4)
One remarks that the result is well approximated by
1
N !
[Φ
(E
N
)]N
where Φ ≡ Φ1 is the number of levels calculated in (2.22) for a single particle. The density of states ρ(E)is the derivative of ΦN (E) with respect to E, ρ(E) = Φ′
N (E), and Ω(E) is given by Φ′N (E)∆E.
Exercise 2.7.10 Entropy of mixing and osmotic pressure
1. The number of possible configurations available to place molecule B is proportional to the number ofA molecules, NA, as each molecule defines a possible localization for B. The entropy is then k lnNA+ afunction of (T, P ). If one adds NB molecules, this can be done in
CNB
NA≃ (NA)NB
NB!
ways, whence the expression of δS. This entropy is partly an entropy of mixing (see (2.93)).
2. The terms δE and PδV do not depend on NA because the energy and volume variations depend onlyon the interaction of molecule B with its neighbours A, and not on the number of A molecules: δE andPδV are local terms. If one adds NB molecules of type B, because of the condition NB ≪ NA, each Bmolecule interacts only with its neighbours of type A, and one can add the terms δE and PδV of each ofthe B molecules. To the Gibbs potential NAµ0(T, P ) of the solvent one must finally add NBf(T, P ) andthe term −TδS, whence (2.123). In a scale transformation, ln(eNA/NB) is unchanged, and G → λG.The chemical potentials follow by differentiation
µA =∂G
∂NA= µ0(T, P ) − NBkT
NA
µB =∂G
∂NB= f(T, P ) + kT ln
NB
NA
11
One checks that G = µANA + µBNB.
3. The chemical potentials of the type A molecules must be the same in both compartments
µ0(T, P0) = µ0(T, P1) −NBkT
NA
By performing a Taylor expansion
µ0(T, P0) = µ0(T, P0) + (P1 − P0)∂µ0
∂P
∣∣∣T− NBkT
NA
= µ0(T, P0) + (P1 − P0)V
NA− NBkT
NA
namely
P1 − P0 ≃ NBkT
V
The osmotic pressure is identical to that of an ideal gas with NB molecules in a volume V 3. However thevolume V is typical of that of a liquid, on the order of 10−3 of that of a gas, and the osmotic pressurecan be very large.
3This remark must be taken as a mnemonic rule: as emphasized in our derivation, the osmotic pressure is of entropicorigin .
12
Solutions for chapter 3
Exercise 3.7.5 Solid and liquid vapour pressures
1. The chemical potentials will coincide when the two phases are in mutual equilibrium
µs = µg
only if the zeroes of energy have been chosen in an identical way in the two phases. One possibleconvention is to decide that an atom in the gaseous phase, located at infinity with a vanishing kineticenergy, has zero energy. On must then give the solid an internal energy at zero temperature whichcompensates exactly the ionization energy, so that if the solid is destroyed with all atoms sent to infinity,the total energy vanishes.
E(T = 0) = −Nu0
2. The solid being incompressible, the specific heats at constant volume and at constant pressure coincideCP = CV = C = Nc; the energy of the solid becomes
E(T ) = −Nu0 +N
T∫
0
dT ′ c(T ′)
and its entropy
S(T ) = N
T∫
0
dT ′ c(T′)
T ′
This allows one to deduce the expression for the Gibbs potential
G = E(T ) − TS(T ) + PNvs = Nµs(T )
= N
Pvs − u0 +
T∫
0
dT ′(
1 − T
T ′
)c(T ′)
whence (3.149). The equality of the Gibbs potentials µs = µg leads to an implicit equation for P (T ) =kT/v. In the limit where vs ≪ v one gets
P (T ) =
(2πm
h2
)3/2
(kT )5/2 exp
[µs(T )
kT
]
3. The partition function for the N molecules of the liquid is
ZN =1
N !ζNl ζl =
Nv0h3
∫d3p e−βp2/2meβu0 =
Nv0λ3
eβu0
from which one derives the liquid Gibbs potential Gl
Gl = FN + PV = N(Pv0 − u0 − kT ln
(ev0λ3
))
and µl = G/N . Since V = Nv0, N and V are not independent variables, one should not differentiatewith respect to N at fixed V . The equality of the chemical potentials µg et µl and the ideal gas lawPv = kT give the relation
u0
kT+ ln
(ev0λ3
)= ln
(kT
Pλ3
)+v0v
13
and taking the exponential of this relation
eu0/kT ev0 =kT
PePv0/kT
Since Pv0/kT = v0/v, this quantity ≪ 1 if the specific volume of the liquid is very small with respectto that of the gas, an excellent approximation in the case of boiling water under standard conditionsv0/v ≃ 10−3. Within this approximation
P ≃ kT
ev0e−u0/kT
4. The entropies per molecules in the gas and in the liquid phases are deduced from (3.17) or from
σg = −∂µg
∂T
∣∣∣P
= k lnv
λ3+
5
2k
σl = −∂µl
∂T
∣∣∣P
= k lnv0λ3
+5
2k
so thatℓ = T (σg − σl) = kT ln
v
v0
On the other handdP
dT=
k
ev0
(1 +
u0
kT
)e−u0/kT =
P
T
(1 +
u0
kT
)
which allows one to check that the Clapeyron formula holds true, as must be the case because our startingpoint was the equality of chemical potentials. In the case (i) one finds u0/kT ≃ 12.1 and in the case (ii)u0/kT ≃ 6.4. This difference by a factor of two illustrates the rough character of the model.
Problem 3.8.4 Models of a boundary surface
A. 1. The length of the interface is obtained from Pythagoras’ theorem
L =
L∫
0
dx
(1 +
(dy
dx
)2)1/2
≃ L+1
2
L∫
0
dx
(dy
dx
)2
(5)
2. On differentiating (3.158) with respect to x one gets
dy
dx=
∞∑
n=1
(πnL
)An cos
(πnxL
)
whence the expression of the Hamiltonian
H =α
2
∫ L
0
∞∑
n,m=1
(πnL
)(πmL
)AnAm cos
(πnxL
)cos(πmx
L
)
=απ2
4L2
∞∑
n,m=1
nmAnAm
∫ L
0
dx
[cos
((n+m)πx
L
)+ cos
((n−m)πx
L
)]
The result (3.159) for H is obtained thanks to
∫ L
0
dx
[cos
((n+m)πx
L
)+ cos
((n−m)πx
L
)]= Lδnm
This result is nothing other than the decomposition of H into independent normal modes An.
14
3. The probability law for the Ans is given by the Boltzmann weight exp[−βH ]. Since the normal modesAn are uncoupled
P (An) ∝ exp
[− απ2
4LkTn2A2
n
]P (An, Am) ∝ exp
[− απ2
4LkT
(n2A2
n +m2A2m
)]
and average values follow from the equipartition theorem (3.53)
〈A2n〉 =
2LkT
απ2n2〈AnAm〉 =
2LkT
απ2n2δmn
This expression allows one to derive 〈y(x)y(x′)〉
〈y(x)y(x′)〉 =2LkT
απ2
∞∑
n=1
1
n2sin(πnxL
)sin
(πnx′
L
)
=2LkT
απ2
∞∑
n=1
1
n2
[cos
(πn(x′ − x)
L
)− cos
(πn(x′ + x)
L
)]
Using the identity (3.160)
∞∑
n=1
1
n2
[cos
(πn(x′ − x
L
)− cos
(πn(x′ + x
L
)]=π2
L2x(L − x′)
we compute the average value
〈y(x)y(x′)〉 =kT
αLx(L − x′)
as well as
〈(∆y)2〉 = 〈(y(x) − y(x′))2〉 =kT
αL|x′ − x|(L − |x′ − x|)
When |x′ − x| ≪ L, the fluctuation 〈(∆y)2〉 is
〈(∆y)2〉 ≃ kT
α|x′ − x| (6)
One can also derive these last two results without appealing to normal modes, by discretizing the integralwhich gives L and by using the identity (A.45) on multiple Gaussian integrals. Indeed, the problem isequivalent to that of a random walk on a straight line x(t) (exercise 6.2), with the following correspon-dence: x → y, t → x et D → kT/2α, where D is the diffusion coefficient. Relation (6) is the analogueof: 〈(∆x)2〉 = 2D|t′ − t|. The fluctuation is largest for |x′ − x| = L/2, and in that case 〈(∆y)2〉1/2 ∝
√L.
The interface width grows as√L, and is not a constant.
B. 1. The energy associated with a plane interface, corresponding to the ground state, is 2JN . Let usobserve that in the preceding situation N is the length of the interface, in units of the lattice spacing. Ina general configuration where the interface is not a straight line, the interface length is
L =
N∑
p=1
(1 + |yp|)
and the energy
H = 2JL = 2JN + 2J
N∑
p=1
|yp|
2. Let us define hp = 2J(1+ |yp|) and H =N∑
p=1hp: the steps are independent variables and the partition
function factorizes
Z =N∏
p=1
Tr e−βhp = ζN
15
where ζ represents the partition function of a single step
ζ =
+∞∑
y=−∞e−2βJ(1+|y|) = e−2βJ cothβJ
From the knowledge of ζ (or of Z), one gets ε and f using (3.2) and (3.6)
ε = 2J
(1 +
1
sinh 2βJ
)f = 2J − 1
βln cothβJ
To each of the extreme regimes corresponds a very different behaviour of the interface
1. βJ ≪ 1: the steps behave as independent random variables and the interface is very irregular
ε→ 2J + kT f → 2J − kT ln(kT/J)
One then finds a kind of ‘equipartition theorem’ with ν = 2. In this ‘classical limit’
ζ =
+∞∫
−∞
dy e−2βJ(1+|y|) = e−2βJ 1
βJ
2. βJ ≫ 1: the interface is in its ground state, meaning that it is a straight line.
ε→ 2J f → 2J
3. The constraint is obeyed in two situations y = ±p, so that
P(|y| = |p|) =2e−2βJ(1+|p|)
ζ(7)
Setting ζ′ = e2βJζ = cothβJ and K = βJ in (7), one easily shows the two following results
〈|y|〉 = −1
2
1
ζ′∂ζ′
∂K=
1
sinh 2βJ
〈y2〉 =1
4
1
ζ′∂2ζ′
∂K2=
1
2(sinhβJ)2
The first result allows one to check ε = 2J(1 + 〈|y|〉.4. The central limit theorem tells us that σ2
∆y = qσ2y. Moreover σ2
y = 〈y2〉 because 〈y〉 = 0, whence
〈(∆y)2〉 = q〈y2〉 =q
2(sinh 2βJ)2
As in the continuum model (part A), one finds that 〈(∆y)2〉 ∝ q ∝ |x′ − x|.
Problem 3.8.5 Debye-Huckel approximation
1. The densities n±(~r) are proportional to the probabilities of finding a charge ±q located at position~r as measured from the ion at the origin of coordinates. They are thus proportional to the Boltzmannweight
n±(~r) ∝ exp
[∓qΦ(~r)
kT
]
since the potential energy of an ion with charge ±q is ±qΦ(~r). The normalisation is fixed thanks to thecondition that, if r → ∞, Φ(~r) → 0 and n(~r) → n. The condition of weak deformations, in other wordsthe fact that there are only small deviations with respect to the ideal gas case, corresponds to a regime
16
where the modulus of the potential energy, |qΦ| is very small with respect to the kinetic energy, ∼ kT .Therefore4
n±(~r) = ne∓βqΦ(~r) ≃ n (1 ∓ βqΦ(~r)) (8)
Let us note that in the general definition (3.75) of the density, the origin of coordinates can be takenarbitrarily, in contrast to the present situation where the position ~r which appears in the densities n±(~r)is measured from a specific ion. In question 4, where we shall give the expression for the pair density, weshall see very clearly that n±(~r) must be interpreted in terms of conditional probabilities.
2. The second form of the Poisson equation illustrates very clearly the meaning of the ionic cloud: acentral charge, qδ(~r), which modifies locally the charge distribution around it q(n+(~r) − n−(~r)). Using(8), the Poisson equation becomes
−∇2Φ(~r) ≃ q
ε0
[−2nq
kTΦ(~r) + δ(~r)
]
and its Fourier transform reads
p2Φ = − 2nq2
ε0kTΦ +
q
ε0= − 1
b2Φ +
q
ε0
so that
Φ =q
ε0
1
p2 + b−2
Using (3.161), the inverse Fourier transform of the preceding equation gives
Φ(~r) =q
4πε0 re−r/b
whence
n+(~r) − n−(~r) = − 1
4πb2re−r/b
3. The charge Q(R) contained in a sphere with radius R is
Q(R) = q
(1 −
∫
0≤r≤R
d3r (n+(~r) − n−(~r))
)= q
1 − 1
b2
R∫
0
dr re−r/b
= qe−R/b(1 +R/b)
When R → ∞ one expects that this charge tends to zero because the solution is electrically neutral. Onechecks this to hold true for R ≫ b, which allows us to interpret the Debye length b as measuring theextension of the ionic cloud: far beyond the Debye length, the central charge is completely screened.
4. The expression (3.162) of the average potential energy is obtained directly from the general relation(3.84), by identifying separately the contributions of the densities associated with the two kinds of ion,n+ and n−. For the sake of simplicity, let us consider one of the two pair densities: n+
2 (~r ′, ~r′′
)d3r′d3r′′
is interpreted as the mean number of pairs of identical charges in d3r′d3r′′ around ~r ′ and ~r′′
. However,assuming that an ion is located at ~r = ~0, then n+(~r)d3r represents the mean number of pairs of ions withidentical charge which one may build around the position ~r with the central ion. Hence
n±2 (~r ′, ~r
′′
) = nn±(~r = ~r′′ − ~r ′) = n2 e∓βqΦ(~r)
The last equality gives the pair correlation function (3.77) of the problem.
4The expansion of the exponential is clearly not valid for r → 0. However we only need it to be valid on the scale ofthe average distance between ions: qΦ(r) ≪ kT , with r ∼ n−1/3. As we shall see in question 5, the approximation is validprovided the density of potential energy is small with respect to the density of kinetic energy. If that were not the casen±(~r) would not be given simply by a local Boltzmann weight!
17
The expression (3.162) of the potential energy is clearly of the form (3.84), once the result is multipliedby a factor of two in order to take into account the fact that two signs are possible for the central charge
Epot =q2
4πε0
∫
V
d3r′d3r′′1
|~r ′′ − ~r ′|(n+
2 (~r ′, ~r′′
) − n−2 (~r ′, ~r
′′
))
= nVq2
4πε0
∫
V
d3r1
r
(n+(~r) − n−(~r)
)
= −nV q2
4πε0b2
+∞∫
0
dr e−r/b = −nV q2
4πε0b
5. The average potential energy between two ions (taken with the same sign to simplify the discussion)is
q2
4πε0 r≃ q2
4πε0 n−1/3
The condition of weak deformations also reads
q2
4πε0 n−1/3≪ kT
so that1
n2/3≪ 4πε0 kT
nq2= 8πb2
which leads correctly to the condition r ≪ b; this last condition implies that the present model is onlyvalid if the volume b3 contains a large number of charges. In an equivalent way, one may observe thatthe density of potential energy nq2/(4πε0b) is very small with respect to the density of potential energy(3/2)nkT provided b≫ r.
Problem 3.8.7 Beyond ideal gas: first term of virial expansion
1. One deduces (3.166) from the expression (3.86) of the pressure
P =1
β
∂ lnZ
∂V=
1
β
∂ lnZK
∂V+
1
β
∂ lnZU
∂V
and from ∂/∂V = (1/N)(∂/∂v).
2. Low values of the temperature enhance the effect of the attractive part of the potential, while highvalues enhance the repulsive part. The average number of molecules in a sphere of radius r0 is 〈N1〉 ∼nr30 ≪ 1 if one chooses n to be arbitrarily small. Since the molecules are assumed to be independent,the probability P (N1) of finding N1 molecules in a sphere with radius r0 is a Poisson distribution. Theprobability of finding two molecules in the sphere is then ≃ 〈N1〉2/2 ≪ N1. If one draws around eachmolecule a sphere with radius r0, one sees that the average number of molecules whose distance is lessthan r0 is Nnr30/2. This number is negligible if Nnr30 = n2V r30 ≪ 1, namely if one considers a sufficientlysmall volume of gas at a fixed density. This condition being satisfied, let us assume that the molecules 1and 2 are located at a distance less than r0.The probability of finding in the gas another pair of moleculeswhose distance is less than r0 is then ∼ (Nnr30)
2 and thus negligible. The integrand in (3.167) is differentfrom zero only if |~r1 − ~r2| < r0, while for all the other pairs one can take this distance to be larger thanr0, and the integral can be evaluated as
I ≃ V (N−2)
∫d3r1 d3r2
(e−βU(|~r1−~r2|) − 1
)≃ V (N−1)
∫d3r
(e−βU(|~r|) − 1
)
Since the number of pairs is N(N − 1)/2 ≃ N2/2, one deduces (3.168) and (3.169) with
B(T ) = −2π
∫ ∞
0
dr(e−βU(|~r|) − 1
)r2 (9)
18
using polar coordinates. As B(T ) ∼ r30 , one has (N2/V )B(T ) ≪ 1 and
lnZU ≃ −N2
VB(T ) = −Nf(v, T )
One checks the extensivity of the free energy and obtains Pv/kT from lnZU by using (3.166).
3. If the gas is sufficiently diluted, v ≫ b and
P =kT
v − b− a
v2≃ kT
v
(1 +
b
v
)− a
v2
so that by identification with (3.170)
B(T ) = b− a
kT
4. In the case of a dilute gas, the probability of finding a molecule at a distance r from another moleculeis proportional to the Boltzmann weight exp[−βU(r)], because the potential energy of the set of thetwo molecules is U(r). This is not true for a dense gas, because this potential energy is modified bythe presence of neighbouring molecules, but this effect is negligible for a dilute gas. The normalisationlimr→∞ g(r) = 1 demands that g(r) ≃ exp[−βU(r)]. Putting this value of g(r) in (3.172) and using polarcoordinates, one finds
P = nkT − 2πn2
3
∫ ∞
0
r3dr U ′(r)e−βU(r)
One then integrates by parts, taking as a primitive of U ′(r) exp[−βU(r)] the function
− 1
β
(e−βU(r) − 1
)
so that the pressure is given by a convergent integral. The final result is
P = nkT − 2πn2
∫ ∞
0
r2dr(e−βU(r) − 1
)
which coincides with (3.170).
5.
1. Hard sphere gas
B(T ) = −2π
∫ σ
0
r2dr =2π
3σ3
The first virial coefficient is positive and independent of temperature. A naive reasoning relyingon excluded volume considerations would lead to a coefficient 4π/3 in the van der Waals equation,instead of the correct value 2π/3; actually it is the number of pairs of colliding molecules which isthe relevant quantity.
2. In this case the virial coefficient is given y
B(T ) =2π
3
[σ3 − (eβu0 − 1)(r30 − σ3)
]
When T → 0, exp(βu0) is large and the attractive part of the potential dominates: B(T ) < 0. Onthe contrary, when T → ∞, the repulsive part of the potential dominates and B(T ) ≃ (2πσ3)/3;the potential energy is then negligible with respect to the the kinetic energy, except if the repulsivepart is concerned. One can then predict a change of sign of B(T ).
19
3. Lennard-Jones potential
B(T ) = −2π
∫ ∞
0
dr r2(
exp
[−4βu0
[(σr
)12
−(σr
)6]]
− 1
)
= −2πσ3
∫ ∞
0
dxx2
(exp
[−4
u0
kT
[(1
x
)12
−(
1
x
)6]]
− 1
)
= σ3f( u0
kT
)= σ3f
(1
θ′
)
If one draws B(T )/σ3 as a function of θ′, one should obtain a universal curve, independent of the typeof molecule. However, at low temperatures, and for the lightest gases, hydrogen and helium, quantumeffects may not be neglected and one observes deviations with respect to the universal curve which relieson the classical theory.
Problem 3.8.8 Theory of nucleation
1. Using the results of subsection 3.5.4, one easily shows that
∆(E − T0S − µ0N) ≤ 0 (10)
2. If the system exchanges work with the external medium, the expression of energy conservation becomes
∆E = Q+W + µ0∆N
so that using once more Q ≤ T0∆S
W ≥ ∆(E − T0S − µ0N)
In the case of a reversible transformation we have
∆Stot = ∆(E − T0S − µ0N) −W = 0
and one checksWmin = ∆(E − T0S − µ0N)
2. The entropy is an increasing function of the energy, as is clearly illustrated by the curve Stot(Etot).
3. in the vicinity of point D (or C), one may use a linear approximation
∆Stot ≃ − dStot
dEtot
∣∣∣∣D
Wmin = − 1
T0Wmin
When going from D to B, one goes from a situation where the system is in equilibrium with the reservoir,and where the number of configurations is Ωeq, to an off-equilibrium situation where the number ofconfigurations is Ω. Therefore the probability of such a fluctuation is
P ∝ Ω
Ωq= e(Stot−Sq
tot)/k = e∆Stot/k ≃ e−Wmin/kT0
4. One works in a container of volume V . Before the droplet is formed, the gas occupies the wholevolume, and thus (see (3.133))
Jin = −P2V
After formation of a droplet of volume V1,
Jfin = −P2(V − V1) − P1V1
20
The transformation gas-droplet is made at constant temperature and chemical potential. Indeed thesupersaturated vapour plays the role of the reservoir (figure 3.24), and the droplet that of the system A.The minimum work which is necessary in order to form the droplet is thus given by (3.173), when thesurface energy is taken into account
Wmin = ∆J + σA = V1(P2 − P1) + σA
5. In the case of a spherical droplet, V = (4/3)πR3 et A = 4πR2, one has
dWmin
dR= 4πR2(P2 − P1) + 8πRσ
and R∗ is given by
R∗ =2σ
P1 − P2(11)
One observes on figure 3.36 that a droplet disappears if it is formed with a radius less than the criticalradius R∗. One must overcome a potential barrier of value Wmin(R
∗) in order to obtain the dropletnucleation. This is the meaning of the expression ‘activation energy’ which is given to the critical valueW ∗
min = Wmin(R∗)
W ∗min =
16π
3
σ3
(P1 − P2)2(12)
R∗
3σ/(P1 − P2)
R
Wmin
16π
3
σ3
(P1 − P2)2
Figure 1: Curve giving Wmin(R).
7. Clearly the probability of nucleation is given by
P ∗ ∝ e−W∗min/kT
Taking (12) into account, one understands that droplet formation is more and more likely as the pressuredifference between the liquid and gaseous phases increases.
8. One must necessarily have µ1(P1, T ) = µ2(P2, T ) otherwise one would observe a flux of molecules:at the transition µ1(P , T ) = µ2(P , T ). Therefore, at fixed temperature
µ1(P1) − µ1(P ) = µ2(P2) − µ2(P )
µ1(P + δP1) − µ1(P ) = µ2(P + δP2) − µ2(P )
v1δP1 = v2δP2
21
where we have used (∂µ/∂P )T = v which is deduced from the Gibbs-Duhem relation (3.105). We cannow rewrite the denominator of (11) as
P1 − P2 = δP1 − δP2 =δP2
v1(v2 − v1)
=δP2
v1
(∂µ2
∂P
∣∣∣∣T
− ∂µ1
∂P
∣∣∣∣T
)=
1
v1(µ2(P2, T ) + µ1(P2, T ))
and one then obtains (3.174), namely an expression for the critical radius which does not involve theknowledge of the pressure inside the droplet.
22
Solutions for chapter 4
Exercise 4.6.1 High temperature expansion and the Kramers-Wannier duality
1. One first proves the identity
exp (KSiSj) = coshK + SiSj sinhK
by noticing that there are only two possibilities, SiSj = 1 and SiSj = −1, and that the identity is validin both cases. Then, expanding the expression of the partition function, one readily finds
Z(K) = (coshK)L∑
C
∏
〈i,j〉(1 + SiSj tanhK) (13)
Let us represent graphically the product of Ising spins
Sn1
1 · · ·Sni
i · · ·SnN
N
by drawing a line on the lattice between two Sis if they are nearest neighbours. The contribution of thespin product to the partition function will be non zero only if there is an even number of lines (0, 2 or 4)leaving from all sites. If a term in the expansion corresponds to an odd number of lines leaving any site,then it gives a vanishing contribution. Thus the contribution of the preceding spin product will be nonzero only if the lines form a closed polygon on the lattice.
2. Let us count the number of polygons for the first values of b, without paying attention to the boundaryconditions.
1. b = 4. The polygon is a square with perimeter 4 (in units of the lattice spacing). Since this squarecan be placed in N differents ways on the lattice, ν(4) = N .
2. b = 6. The polygons are rectangles with perimeter 6, whose largest side can be either horizontal(‘horizontal rectangle’) or vertical (‘vertical rectangle’). Then ν(6) = 2N .
3. b = 8. The leading term in N is obtained by placing two squares with b = 4 on the lattice. Thiscan be done in N2/2 ways. Other possibilities (i.e. rectangles with perimeter b = 8) give only afactor proportional to N . Thus ν(8) = N2/2. It is left to the reader to show that the exact result,using periodic boundary conditions, is ν(8) = N(N − 5)/2.
4. b = 10. The leading term is given by a square with perimeter 4 and a rectangle with perimeter 6.This rectangle can be either vertical or horizontal, so that ν(10) = 2N2.
5. b = 12. The leading term corresponds to three squares with perimeter 4, giving ν(12) = N3/3!.
The first terms in the high temperature expansion of the partition function are then, setting tanhK = x
ZN(K) ≃ 2N (coshK)L[1 +Nx4 + 2Nx6 +
1
2N2x8
+ 2N2x10 +1
6N3x12 + · · ·
] (14)
The free energy FN is proportional to the logarithm of ZN ; we have, to order x12
FN ∝ lnZN ≃ Nx4 + 2Nx6 +1
2N2x8 + 2N2x10 +
1
6N3x12
− 1
2
(Nx4 + 2Nx6 +
1
2N2x8
)2
+1
3N3x12 + · · ·
= N(x4 + 2x6 + · · · ) +N2x12 + · · ·
23
At least to order x10, we have thus checked the extensivity of the free energy, but our estimates of ν(b)are not precise enough to give the coefficient of N in the x8 term and to eliminate the N2 factor in thex12 term.
3. Let us flip all the spins of the dual lattice inside a given polygon of the original lattice. It is clear thatthere is a one-to-one correspondence between the links on the polygon and the number of broken linkson the dual lattice: each broken link on the dual lattice crosses a side of the polygon once and only once.Starting from a configuration with all spins up, the energy of a configuration with n broken links will be
En = −KL+ 2nK
because each broken link costs an energy 2K, hence the partition function is, in the low temperatureregion
Z∗(K) = 2 exp(KL)[1 +
∑
n=4,6,8···ν(n) exp(−2nK)
](15)
where the factor 2 comes from the fact that we could have chosen to start from a configuation with allspins down. Note that (13) converges for temperatures T > Tc1, while (15) is convergent for T < Tc2.with Tc2 ≤ Tc1: The temperatures Tc1 and Tc2 are defined as those temperatures which limit the radiusof convergence of the expansions (13) and (15) respectively.
4. From the high temperature expansion we have
Z(K)
2N (coshK)L=
1 +
∑
b=4,6,8,...
ν(b)(tanhK)b
while the low temperature expansion gives for a temperature T ∗
Z∗(K∗)
2 exp(K∗L)=[1 +
∑
n=4,6,8···ν(n) exp(−2nK∗)
]
The two ratios are equal if one chooses to relate K and K∗ by
exp(−2K∗) = tanhK
This gives a relation between the partition function at two different temperatures. If there is a unique
critical temperature where both the high and low temperature expansions diverge, Tc1 = Tc2 ≡ Tc, weobtain this critical temperature Kc as
exp(−2Kc) = tanhKc or sinh2Kc = 1
Note that the assumption that there is a unique critical temperature is a strong one: one easily findsmodels where this assumption is not satisfied, see for example problem 7.9.3.
Exercise 4.6.5 Shape and energy of an Ising wall
1. The ‘equation of motion’ for Md2M
dz2= V ′(M)
is immediately deduced from the identities (A.59) and (A.60). In the mechanical analogy, the conservationof kinetic energy is
1
2
(dx
dt
)2
+ U(x) = const
The ‘particle’ starts at t = −∞ with zero velocity from a point of abscissa x = −M0 and arrives att = +∞ with zero velocity at the point x = M0. At the points x = ±M0, the kinetic energy vanishes and
24
the potential energy is U(M0), so that the constant in the preceding equation is U(M0). Transposingthis result to the original problem leads to
1
2
(dM
dz
)2
= V (M) − V (M0) (16)
Of course, one could prove (16) directly from the equation of motion exactly as one proves energyconservation in classical mechanics. One checks that differentiating (16) with respect to z gives back theequations of motion, while the boundary conditions are limz→±∞M = ±M0 and lim|z|→∞(dM/dz) = 0.
2. Dividing equation (16) by |r0|M20 , one gets
1
2|r0|M20
(dM
dz
)2
= − 1
2M20
(M2 −M20 ) +
u0
4!|r0|M20
(M4 −M4
0
)
Now the zero field magnetization is
M20 =
6|r0|u0
whenceu0
4!|r0|M20
=1
4M40
This yields the equation for f
ξ2(
df
dz
)2
= −1
2(f2 − 1) +
1
4(f4 − 1) =
1
4
(1 − f2
)
Taking the square root of the preceding equation and choosing the positive sign (the negative sign wouldalso lead to an equally acceptable solution, corresponding to a decreasing function of z, one gets
df
1 − f2=
dz
2ξ
from which one deduces
f(z) = tanh
(z − z0
2ξ
)(17)
Taking z0 = 0 simply fixes the position of the wall at z = 0.
3. The surface tension is the difference per unit area of the energy in the presence of a wall (first bracketin the following equation) and the energy without any wall (second bracket)
σ =
+∞∫
−∞
dz
([12
(dM
dz
)2
+ V (M)]−[V (M0)
])
Using (16) allows us to cast the expression for σ in the form
σ =
+∞∫
−∞
dz
(dM
dz
)2
= M20
+∞∫
−∞
dz
(df
dz
)2
= M20
+1∫
−1
df
dzdf =
M20
2ξ
+1∫
−1
(1 − f2)df =2
3
M20
ξ
Exercise 4.6.6 The Ginzburg-Landau theory of superconductivity
1. The magnetic field ~B vanishes with the vector potential. Furthermore, decomposing ψ into real andimaginary parts
ψ = ψ1 + iψ2
25
we get|~∇ψ|2 = |~∇ψ1 + i~∇ψ2|2 = (~∇ψ1)
2 + (~∇ψ2)2
so that (Γ − ΓN ) becomes
Γ − ΓN =
∫d3r
(a|ψ|2 + b|ψ|4 +
~2
2m
[(~∇ψ1)
2 + (~∇ψ2)2])
which is exactly (4.88) with a suitable redefinition of the parameters a, b and ψ.
2. The covariant derivative~D = −i~~∇− q ~A
transforms in the following way under a local gauge transformation
exp(−iqΛ(~r ))(i~~∇− q ~A) exp(−iqΛ(~r )) = −i~~∇− q ~A(~r) + q~∇Λ(~r) = −i~~∇− q ~A′(~r)
which shows the invariance of (Γ − ΓN ) under local gauge transformations.
3. Let us give some details on the computation of functional derivatives. We begin with
Γ1 =1
2
∫d3r ~B2 =
1
2
∫d3r (~∇× ~A)2
We use the identity, easily proved from the techniques given at the end of § 10.4.1
~∇ · (~V × ~W ) = ~W · ~∇~V − ~V · ~∇× ~W (18)
to write δΓ1 as
δΓ1 =
∫d3r(~∇× ~A) · (~∇× δ ~A) =
∫d3r ~B · (~∇× δ ~A)
=
∫d3r
[δ ~A · (~∇× ~B) + ~∇ · (δ ~A× ~B)
]
=
∫d3r δ ~A · (~∇× ~B) −
∫δ ~A · (d2S × ~B)
where we have used the divergence (Green’s) theorem in going from the second to the third line. Thesecond term in the last line is a surface term, and neglecting this term for the moment yields
δΓ1
δ ~A(~r)= ~∇× ~B(~r) (19)
Let us now compute the functional derivative with respect to ~A of Γ2
Γ2 =1
2m
∫d3r
∣∣∣(−i~~∇− q ~A
)ψ∣∣∣2
=1
2m
∫d3r
[~
2|~∇ψ|2 + i~q( ~Aψ∗ · ~∇ψ − (~∇ψ)2 · ~Aψ + q2|ψ|2 ~A2]
from which one easily deduces
δΓ2
δ ~A=
i~q
2m
[ψ∗(~∇ψ) − (~∇ψ∗)ψ
]+
q2
2m|ψ|2 ~A2 (20)
One recognizes the familiar expression for the probability current of the wave function in the presence ofa magnetic field. It remains to deal with
Γ3 =
∫d3r ~B · ~H
26
The variation of Γ3 is given by
δΓ3 =
∫d3r ~H · ∇ × δ ~A
=
∫d3r ~∇ · (δ ~A× ~H) +
∫d3r δ ~A · (~∇× ~H)
= −∫
d3 ~A · (d2S × ~H)
We have used Green’s theorem and the fact that ~∇× ~H = 0 inside the superconductor. The surface termderived a few lines above combines with that obtained previously to give the final boundary condition
n× ( ~B − ~H) = 0 (21)
where n is a unit vector normal to the boundary of the superconductor: thus the tangential componentof ( ~H − ~B) is continuous at the boundary between the normal and superconductor phases. The variationwith respect to ψ leads to much easier calculations and we only quote the final results
1. The ‘equations of motion’
0 =1
2m
(−i~~∇− q ~A
)ψ + aψ + 2b|ψ|2ψ (22)
~∇× ~B = − i~qµ0
2m
[ψ∗~∇ψ − (~∇ψ)∗ψ
]− µ0q
2
m~A |ψ|2 (23)
2. The boundary conditions
n× ( ~B − ~H) = 0 n · (−i~~∇− q ~A)ψ = 0 (24)
4. In a uniform situation and zero magnetic field
|ψ|2 = − a
2ba|ψ|2 + b|ψ|4 = −a
2
4b(25)
The difference of Gibbs potentials between the normal phase in the presence of an external field and inthe absence of such a field is
ΓN (H) − ΓN(H = 0) = −H∫
0
dH ′B(H ′) = −H2
2µ0
while ΓS(H) = ΓS(H = 0) since ~B = 0 in the superconductor phase. The phase transition between thenormal and superconductor phases occurs when the Gibbs potentials are equal. This defines the criticalfield Hc(T )
ΓS(H = 0) = ΓN (H = 0) − 1
2µ0H2
c (T )
Now one uses (25)
ΓS(H = 0) − ΓN (H = 0) = −a2
4b
so that
H2c (T ) =
µ0a2
2b
The critical field vanishes linearly at T = Tc since
a(T ) = a(T − Tc)
27
5. Let us assume a semi-infinite geometry where the normal-superconductor boundary is the plane z = 0.Then the order parameter ψ obeys
|ψ(z = 0)| = 0 limz→∞
|ψ(z)| =
√|a|2b
We may choose ψ to be real without loss of generality. As in exercise 4.6.5, we define a dimensionlessfunction f(z)
f(z) =
√b
2|a| ψ(z) f(z = 0) = 0 limz→∞
f(z) = 1
The ‘equation of motion’ for ψ(z) is
− ~2
2m
d2ψ
dz2+ aψ + 2bψ3 = 0
from which one deduces the corresponding equation for f
−2ξ2d2f
dz2− f + f3 = 0
which is integrated to
ξ2(
df
dz
)2
= −1
2f2 +
1
4f4 + const
For z → ∞, df/dz → 0 and f → 1, which means that the constant is 1/4. One finally gets
ξ2(
df
dz
)2
=1
4(1 − f2)2 (26)
which is easily integrated as in the preceding exercise and allows one to obtain the z-dependence of theorder parameter
ψ(z) =
√|a|2b
tanh
(z
2ξ
)(27)
Starting at z = 0 from a vanishing order parameter, the order parameter grows from zero toward its bulkvalue
√|a|/(2b) over a length∼ ξ, the coherence length.
6. We now assume ψ = cst and ~A 6= 0. In other words, we consider the situation deep inside thesuperconductor, at distances ≫ ξ from the boundaries. The electromagnetic currrent ~ in (25) becomes
~(~r) = −q2nS
m~A (28)
so that the ‘equation of motion’ (23) becomes in the Coulomb gauge ~∇ · ~A = 0
∇2 ~A− µ0q2nS
m~A = 0 (29)
This equation defines a second characteristic length of the superconductor, the London penetration
length λ
λ =
[m
µ0q2nS
]1/2
=
[2mb
µ0q2|a|
]1/2
(30)
In order to make explicit the fact that λ is indeed the penetration length, let us first express (29) in termsof the magnetic field by taking the curl of both sides
∇2 ~B − 1
λ2~B = 0
28
In a one-dimensional situation where ~B is perpendicular to the z-axis, ~B = Bu, u · z = 0, one finds
d2B
dz2− 1
λ2B = 0
whose solution isB(z) = B0 exp
(− z
λ
)
Thus the magnetic field vanishes in the bulk of the superconductor.
Exercise 4.6.11 Critical exponents to order ε for n ≫ 1
1. We write
Φ4 =[(Φ2 − 〈Φ2〉) + 〈Φ2〉
]2= 〈Φ2〉2 + 2〈Φ2〉(Φ2 − 〈Φ2〉) + (Φ2 − 〈Φ2〉)2
We may omit 〈Φ2〉2, which adds a constant to H . Furthermore the term
u0
n(Φ2 − 〈Φ2〉)2 ∼ 1
and may be neglected. With these simplifications, the Hamiltonian reads
H =
∫dDr
[1
2(~∇Φ(~r))2 +
1
2r0Φ
2(~r) +u0
2n〈Φ2〉Φ2(~r)
](31)
and is now of Gaussian type. The coefficient of Φ2 is
1
2
(r0 +
u0
n〈Φ2〉
)
and one may write immediately the correlation function from (4.171)
Gij(q) =δij
q2 + r0 +u0
n〈Φ2〉
(32)
On the other handn∑
i=1
〈ϕ2i 〉 =
n∑
i=1
〈ϕi(0)ϕi(0)〉 = n
∫dDq
(2π)DGii(q)
which leads to the consistency equation
〈Φ2〉 = nKD
Λ∫
0
qD−1dq
q2 + r0 +u0
n〈Φ2〉
(33)
where we have useddDq
(2π)D=SDq
D−1dq
(2π)D= KD qD−1dq
We may check in (33) the consistency of our assumption 〈Φ2〉 ∼ n. The first two terms in H are oforder n. As Φ4 is of order n2, one must choose the combination u0/n so that all terms in H are of thesame order in n when n→ ∞.
2. From the fluctuation response theorem, the inverse susceptibility ρ(T ) is proportional to 1/G(q = 0),so that
ρ(T ) = r0(T ) +u0
n〈Φ2〉
= r0(T ) + u0KD
Λ∫
0
qD−1dDq
q2 + ρ(T )(34)
29
The critical temperature Tc corresponds to an infinite susceptibility, and thus to ρ(Tc) = 0. This leads to
0 = ρ(Tc) = r0(Tc) + u0KD
Λ∫
0
qD−1dq
q2(35)
Subtracting (35) from (34) we obtain
ρ(T ) = r0(T − Tc) − u0ρ(T )KD
Λ∫
0
qD−1 dq
q2(q2 + ρ(T ))(36)
Let us make the change of variables q = x√ρ. The equation for ρ(T ) becomes
ρ(T ) = r0(T − Tc) + [ρ(T )](D−2)/2u0KD
Λ/√
ρ∫
0
xD−1 dx
x2(x2 + 1)(37)
where Λ is the ultraviolet cut-off.
3. The integral in (35) is divergent for D ≤ 2, and the theory can only be valid for D > 2. For 2 < D < 4,the integral in (37) is convergent at infinity, and one may write
ρ(T ) = r0(T − Tc) + Cρ(D−2)/2 (38)
whenceρ(T ) ∝ (T − Tc)
2/(D−2)
Indeed, for 2 < D < 4, 2/(D− 2) < 1 and one may neglect ρ(T ) on the LHS (38). The critical exponentγ is then
γ =2
D − 22 < D < 4
For T = Tc, G(q) ∝ 1/q2, so that η = 0. Finally
ξ(T ) ∝ (ρ(T ))−1/2 ∝ (T − Tc)− 1
D−2 =⇒ ν =1
D − 2
To summarize, for 2 < D < 4 and with an error O(1/n)
η = 0 ν =1
D − 2γ =
2
D − 2
which obeys the scaling laws (4.166). Note also that the results agree with the n → ∞ limit of (4.203).For D > 4, one recovers, as one should, the mean-field exponents. This can be seen as follows. Sincethe integral over q is ultraviolet divergent and infrared convergent even when ρ = 0, we may write theintegral in (36)
u0KD
Λ∫
0
qD−1 dq
q2(q2 + ρ(T ))≃ u0KD
Λ∫
0
qD−5dq =u0KD
D − 4ΛD−4 = C(Λ)
The equation for ρ(T ) becomesρ(T ) = r0(T − Tc) − C(Λ)ρ(T )
so that ρ(T ) ∝ (T − Tc), and γ = 1. The only effect of the interaction is to renormalize the value of r0,leaving the critical exponents unchanged.
Exercise 4.6.12 Irrelevant exponents
30
1. From the definition u0 = ξ−εg0 and (4.185) we deduce
u0 = ξ−ε g exp
−
g∫
0
dg′(
ε
β(g′, ε)+
1
g′
) (39)
We now write the integral in (39) as
∫ g
0
dg′(
ε
β(g′, ε)+
1
g′
)=
g∫
0
dg′ε
ω(g′ − g∗)+
∫ g
0
dg′(
ε
β(g′, ε)+
1
g′− ε
ω(g′ − g∗)
)
The second integral is convergent at g′ = g∗, so that it may be expanded in powers of g
A(g) = exp
(∫ g
0
dg′(
ε
β(g′, ε)+
1
g′− ε
ω(g′ − g∗)
))= 1 + c1g + c2g
2 + · · ·
Performing the integrals in (39) yields
u0 ≃ ξ−εg∗A(g∗)∣∣∣
g∗
g∗ − g
∣∣∣ε/ω
(40)
from which one deduces (g∗ − g)
g∗ − g = u−ω/ε0 ξ−ω [g∗A(g∗)]ω/ε
g∗
or
g = g∗(
1 −(
u0ξε
g∗A(g∗)
)−ω/ε)
(41)
namelyg ≃ g∗
(1 + O(ξ−ω)
)
The difference between the value g∗ of g at the fixed point and g is controlled by definition by the highestirrelevant exponent −ω3.
2. We want to keep ε finite and are not allowed to take the limit ε → 0 of the various expressions. Westart from the expression of β(g, ε) given in § 4.5.3
β(g, ε) = −εg +3
2ε
Γ(ε/2)
(4π)D/2g2 + O(g3)
= −εg +3
2B(ε)g2 + O(g3)
with
B(ε) =εΓ(ε/2)
(4π)D/2
This gives the value of g∗
g∗ =2ε
3B(ε)
On the other hand, from the results of § 4.5.3 and
∫ ∞
0
uαdu
(u+ 1)β=
Γ(α+ 1)Γ(β − α− 1)
Γ(β)
we get for (r0 − r0c)
r0 − r0c = ξ−2
(1 + u0
ξεΓ(ε/2)
2(4π)D/2(1 − ε/2
)
31
This leads to the following expression of Z(g)
Z(g) = −1
2ξ2
d
dξ(r0 − r0c)
∣∣∣u0
= 1 +gB(ε)
2ε
from which one deduces
γ(g) = β(g, ε)dZ
dg=
(−εg +
3
2B(ε)g2
)B(ε)
2ε= −1
2gB(ε)
Subsituting the value of g∗ in this expression leads to
γ(g∗) = −1
3ε = −1
3if D = 3
For D = 3 and using (4.196), the critical exponent ν becomes
ν =3
5= 0.6
instead of ν = 7/12 = 0.583 from the ε-expansion.
Exercise 4.6.13 Energy-energy correlations
1. The only temperature dependence in the Landau-Ginzburg Hamiltonian (4.96) is hidden in r0, asr0 = r0(T − T0). We may write the partition function (4.95) in zero magnetic field as
Z =
∫dP [ϕ] exp
(−1
2r0
∫dDrϕ2(~r)
)→∫
dP [ϕ] exp
(1
2β
∫dDr ϕ2(~r)
)(42)
Indeed, since we are only interested in the singularity structure of the physical quantities, we do not payattention to the proportionality factors, so that we may make the replacement
r0 = r0(T − T0) → −β
by noticing that
r0(T − T0) =r0k
(1
β− 1
β0
)≃ r0
(β0 − β
kβ20
)
It is important to notice that the factor dP [ϕ] is temperature independent
∫dP [ϕ] =
∫Dϕ(~r) exp
(−1
2[~∇ϕ]2 +
1
4!u0ϕ
4 − 1
2
r0kβ0
ϕ2
)(43)
The average values 〈•〉 are by definition
〈•〉 =1
Z
∫dP [ϕ] exp
(1
2β
∫dDr ϕ2(~r)
)(•) (44)
Thus the average energy E is
E = −∂ lnZ
∂β= −1
2
∫dDr 〈ϕ2(~r)〉
We obtain the specific heat by differentiation
C =dE
dT= − 1
kT 2
dE
dβ=
1
kT 2
d2 lnZ
dβ2=
1
4kT 2
∫dDr dDr′ 〈ϕ(~r)ϕ(~r ′)〉c (45)
This is of course an example of the fluctuation-response theorem.
32
2. Using translation invariance, with ~u = ~r − ~r ′ and the scaling properties of 〈ϕ2ϕ2〉c for u <∼ ξ
C =V
4kT 2
∫dDu 〈ϕ2(~u)ϕ2(0)〉c
≃ V
4kT 2
∫
u <∼ ξdDu u−2σ ∝ V
4kT 2ξD−2σ
Note that the last result follows also directly from from dimensional analysis. On the other hand, fromthe very definition of the critical exponent α and the scaling law (4.165)
C ∼ |T − Tc|−α = |T − Tc|−2+νD ∼ ξ2/ν−D
and by identification of the exponents we conclude that
σ = dϕ2 = D − 1
ν(46)
3. In the case T > Tc, where 〈ϕ(~r)〉 = 0, we have for the connected correlation function of ϕ
〈ϕ(~r1)ϕ(~r2)〉c =1
Z
∫dP [ϕ]ϕ(~r1)ϕ(~r2) exp
(1
2β
∫dDr ϕ2(~r)
)
Then, by differentiation with respect to β
∂
∂β〈ϕ(~r1)ϕ(~r2)〉c =
(− 1
Z
∂Z
∂β
)〈ϕ(~r1)ϕ(~r2)〉c
+1
Z
∫dP [ϕ]ϕ(~r1)ϕ(~r2)
[1
2ϕ2(~r3)
]dDr3 exp
(1
2β
∫dDr ϕ2(~r)
)
=
∫dDr3 〈ϕ(~r1)ϕ(~r2)
1
2ϕ2(~r3)〉c
Integrating the preceding equation over r1 and r2 gives
∂
∂β
∫dDr1 dDr2〈ϕ(~r1)ϕ(~r2)〉c =
1
2
∫dDr1 dDr2 dDr3 〈ϕ(~r1)ϕ(~r2)ϕ
2(~r3)〉 (47)
Within a numerical factor, the LHS of (47) is the temperature derivative of the magnetic susceptibility χ,which is the integral of the 〈ϕϕ〉 correlation function
∂χ
∂t∼ ∂
∂tt−ν(2−η) ∼ V ξ2−ηξ1/ν ∼ V ξD−2dϕ+1/ν
Let us evaluate the RHS of (47). From translation invariance
〈ϕ(~r1)ϕ(~r2)ϕ2(~r3)〉 = g(~r1 − ~r2, ~r1 − ~r3) = g(~u,~v)
The integral over ~u and ~v is limited by u, v <∼ ξ. On the other hand, assuming a scaling behaviour ofg(~u,~v) at T = Tc with scaling exponent p
g(λ~u, λ~v) = λpg(~u,~v)
suggests the change of variables ~u = ξ~u′, ~v = ξ~v′
I = V
∫dDu dDv g(~u,~v) = V ξ2D
∫dDu′ dDv′ g(~u′, ~v′)
= V ξ2D+p
∫
u′,v′<∼1
dDu′ dDv′ g(~u′, ~v′) ∼ V ξ2D+p
By identification with the LHS we get
p = −2dϕ −D +1
ν= −(dϕ2 + 2dϕ)
Thus the scaling exponent of 〈ϕϕϕ2〉c is nothing other than 2dϕ +dϕ2 . If we are interested in the averagevalue of N ϕs and M ϕ2s, it is clear that the following scaling law will hold at the critical point
〈ϕ(λ~r1) · · ·ϕ(λ(~rN )ϕ2(λ~r ′1) · · ·ϕ2(~r ′
M )〉c = λ−(Ndϕ+Mdϕ2)〈ϕ(~r1) · · ·ϕ(λ~rN )ϕ2(~r ′1) · · ·ϕ2(~r ′
M )〉c
33
Solutions for chapter 5
Exercise 5.6.4 Non-degenerate Fermi gas
1. We have shown in exercise 2.7.2 that the density of states not only depends on the dispersion lawε(p), but also on the dimension of space. For example in space-dimension two
ρ(ε) =mS
π~2= D
2. With a uniform density of states, one immediately obtains
εF =N
DE =
1
2NεF (48)
The classical approximation is valid when z = eβµ ≪ 1 (section 5.1.2); in such a case
N = z
∞∫
0
dε ρ(ε)e−βε = zDkT
Then the classical limit corresponds to
N ≪ DkT (49)
The order of magnitude of the maximal energy is kT . The quantity DkT gives the number of levelsbetween ε = 0 and ε = kT and provides an estimate of the number of accessible levels. The condition(49) is nothing other than the validity condition of the Maxwell-Boltzmann statistics in the particularcase of a uniform density of states, namely that giving the small probability of occupation of the levels.
3. Let us make the change of variables y = βx. Then, using βµ ≫ 1, we determine immediately thedominant terms which we were looking for
+∞∫
µ
dx1
eβx + 1=
1
β
+∞∫
βµ
dy1
ey + 1∼ 1
βe−βµ (50)
+∞∫
µ
dxx
eβx + 1=
1
β2
+∞∫
βµ
dyy
ey + 1∼ 1
β2e−βµ (51)
In order to demonstrate (5.86), we are going to use the symmetries of the integral. We call f(x) theFermi function written in terms of the variable x = ε− µ and we set
ϕ(x) = ϕS(x) + ϕA(x) ϕS,A(x) =1
2
(ϕ(x) ± ϕ(−x)
)
f(x) = fS(x) + fA(x) fS,A(x) =1
2
(f(x) ± f(−x)
)
From f(x) = 1 − f(−x), we deduce
fS =1
2fA(x) = f(x) − 1
2
As the integration domain is symmetrical with respect to x = 0 we may eliminate the integrands which
34
are antisymmetric in x↔ −x in the calculation of (5.86)
µ∫
−µ
dxϕ(x)f(x) = 2
µ∫
0
dxϕS(x)fS(x) + 2
µ∫
0
dxϕA(x)fA(x)
=
µ∫
0
dx(ϕS − ϕA
)+ 2
µ∫
0
dxϕA(x)f(x)
=
0∫
−µ
dxϕ(x) +
µ∫
0
dxϕ(x) − ϕ(−x)
eβx + 1
4. The chemical potential is given by
N = D
∞∫
−µ
dx1
eβx + 1= D
µ∫
−µ
dx1
eβx + 1+D
∞∫
µ
dx1
eβx + 1
Using (4.86) with ϕ(x) = 1, one obtains
N = Dµ+ O(e−βµ)
while, from (48) one gets
µ = εF + O(e−βµ) (52)
The average energy reads
E = µD
∞∫
−µ
dx1
eβx + 1+D
∞∫
−µ
dxx
eβx + 1
The first integral is identical to that which appears in the calculation of the chemical potential; the secondone features once more (5.86) with ϕ(x) = x, and one uses (51) in order to be able to use (A.51). Onefinally gets for the energy
E =1
2Dµ2 +
π2
6D(kT )2 + O(e−βµ) (53)
5. As a general rule, when one applies the Sommerfeld expansion of the Fermi distribution to a generictest function g(ε) (g(ε) = ρ(ε) for N and g(ε) = ερ(ε) for E), one obtains its Taylor expansion in termsof the (small) parameter (kT/µ)2. The peculiarity of the two dimensional situation, in which the densityof states is uniform, leads to a Taylor expansion which is necessarily finite. Indeed
N = D
µ∫
0
dε + all other terms vanish
E = D
µ∫
0
dε ε+π2
6D(kT )2
∞∫
0
dε δ(ε− µ) + all other terms vanishing
The calculation of the preceding integrals leads indeed to the results (52) and (53). Clearly the correctionsof order e−βµ cannot be taken into account in an expansion in powers of (kT/µ)2. Exercise 5.6.5 gives,in the case of a Bose gas, a still more spectacular illustration of the influence of the space dimension onthe behaviour of a quantum gas. Let us also remark that the techniques developed in this exercise allowone to give an alternative proof of Sommerfelds’s formula (5.29).
Problem 5.7.4 Quark-gluon plasma
35
A. 1. Because the π-meson has three different charge states, the density of states of the π-meson gasreads
ρ(ε) dε = 3V
(2π)34πp2 dp =
3V
2π2ε2 dε
For a boson gas with a vanishing chemical potential, the grand partition function is given by (see (5.7))
lnQ = −∞∫
0
dε ρ(ε) ln(1 − e−βε
)
= − βV
2π2
∞∫
0
dεε3
eβε − 1=π2V
30T 3
(54)
One uses the integral (A.53) to obtain the final result. From the equation of sate of an ultra-relativisticideal gas
PV = T lnQ =1
3E
and using (54), one easily finds the pressure and the energy density
P (T ) =π2
30T 4 (55)
ǫ(T ) =π2
10T 4 (56)
2. cv(T ) et s(T ) are obtained from (56)
cv(T ) =dǫ
dT=
2π2
5T 3
s(T ) =
T∫
0
dT ′ cv(T′)
T ′ =2π2
15T 3
A direct calculation gives the energy density
n(T ) =3
2π2
∞∫
0
dεε2
eβε − 1=
3ζ(3)
π2T 3 (57)
One notes that the proportionality of cv, s and n to T 3 is the consequence of a simple dimensionalargument.
3. From (56) et (57) one can express the mean particle energy as
〈ε(T )〉 =ǫ(T )
n(T )=
π4
30ζ(3)T
Let’s use once more a reasoning which was already developed in the discussion of (5.67). From n and〈ε〉, one may build two length scales. One of them, d ∝ n−1/3 ∼ T−1, is a measure of the mean distancebetween particles, the other one, λ ∝ 〈p−1〉 ∝ 〈ε−1〉 ∼ T−1, provides an order of magnitude of quantumeffects. If one compares these two typical distances, one observes that, whatever the temperature, λ/d ∼ 1.The π-meson gas can never be treated as a classical gas.
B. 1. The reader is referred to subsection 5.1.3.
2. Because massless particles are ultra-relativistic, the density of states of this Fermion gas reads
ρ(ε) dε =V
π2ε2 dε
36
Expression (5.105) is obtained thanks to the change of variables x = βε in
lnQ =
∞∫
0
dε ρ(ε)[ln(1 + e−β(ε−µ)
)+ ln
(1 + e−β(ε+µ)
)]
=βV
3π2
∞∫
0
dε ε3
[1(
1 + e−β(ε−µ)) +
1(1 + e−β(ε+µ)
)] (58)
For µ = 0, one gets (see appendix A.5.2)
lnQ =7π2V
180T 3 P (T ) =
7π2
180T 4 n(T ) =
3ζ(3)
π2T 3
The entropy is deduced from the grand potential J = −T lnQ
s(T ) = − 1
V
∂J∂T
∣∣∣∣V
=7π2
45T 3
3. Let us call respectively I− et I+ the two integrals between square brackets in the second line of (58).One can cast the quantity T lnQ into the form
T lnQ =V
3π2
∞∫
0
dε ε3 [I− + I+]
The condition T = 0 implies that the two integrals are to be calculated in the limit β → ∞. Let usassume µ > 0
limβ→∞
I− = θ(µ− ε) limβ→∞
I+ = 0 if µ > 0
limβ→∞
I− = 0 limβ→∞
I+ = θ(−(µ+ ε)) if µ < 0
From these equation it follows that
T lnQ =V
3π2
µ∫
0
dε ε3 =V
12π2µ4
P (µ) =T lnQV
=1
12π2µ4
ǫ(µ) = 3P (T ) =1
4π2µ4
n(µ) =1
V
∂T lnQ
∂µ
∣∣∣∣T,V
=1
3π2|µ|3
The entropy of a fully degenerate Fermi gas of course vanishes.
4. Let us start from (5.105) and set
I1 =
+∞∫
0
dxx3
e(x−µ′) + 1I2 =
+∞∫
0
dxx3
e(x+µ′) + 1
We do not seek to compute separately I1 and I2, but rather we seek to cast them into a form where their
37
sum has a simple expression
I1 = −µ′∫
0
dy(y − µ′)3
e−y + 1+
+∞∫
0
dy(y + µ′)3
ey + 1
= −µ′∫
0
dy (y − µ′)3(
1 − 1
ey + 1
)+
+∞∫
0
dy(y + µ′)3
ey + 1
I2 = −µ′∫
0
dy(y − µ′)3
ey + 1+
+∞∫
0
dy(y − µ′)3
ey + 1
The sum I1 + I2 is then
I1 + I2 = −µ′∫
0
dy (y − µ′)3 + 2
+∞∫
0
dyy3
ey + 1+ 6µ′2
+∞∫
0
dyy
ey + 1(59)
The first integral in (59) is elementary integral, the other two are given in appendix A.5.2. Inserting (59)in (5.105), one finally gets the expected result (5.106).
C. 1. In order to estimate the contribution of the quark gas to the plasma pressure, one can simply usethe results for the Fermion gas; it is only necessary to take properly into account the degeneracy linkedto the three colour states
Pquark = 2 × 3 × 7π2
180T 4
For the contribution of the gluon gas, one simply uses the expression derived previously for the π-mesongas
Pgluon = (2 × 8) ×(
1
3× π2
30T 4
)
The gluons have spin one, but since they are massless, their spin degeneracy gives only a factor two. Thefactor 1/3 comes from the fact that gluons have only one charge state. On gathering all contributions,one gets the expression (5.107) of the pressure of the quark-gluon plasma.
2. The additional term proportional to the volume in the expression of the free energy leads to a constantterm in the pressure
Pplasma = − ∂F
∂V
∣∣∣∣T
=37π2
90T 4 −B
3. Figure 2 shows the behaviour of the pressure as a function of temperature for the π-meson gas and forthe quark-gluon plasma. The most stable phase is that which minimizes the free energy, or equivalently,that which has the highest pressure. The transition temperature Tc is fixed by the equality of the pressures
37π2
90T 4
c −B =π2
30T 4
c
a condition leading to
Tc =
(45
17π2
)1/4
B1/4 ≃ 144 Mev
This rough estimate of Tc is in qualitative agreement with the results from lattice quantum chromody-namics.
5. The entropy densities of the two phases are different
s(T ) =∂
∂T
(T lnQ
V
)=∂P
∂T=
74π2
45T 3 for the plasma
2π2
15T 3 for the π − meson gas
38
P
plasma
TTc
−B
Pπ(T )
Pplasma(T )
π mesons
Figure 2: Curves giving P (T ) for the π-mesons and the quark-gluon plasma.
At the transition, the entropy variation does not vanish and is given by
∆s =68π2
45T 3
c
The phase transition is thus first order and is accompanied by a latent heat per unit volume ℓ = Tc∆s.
Problem 5.7.5 Bose-Einstein condensates of atomic gases
1. One uses the fact that the potential is slowly varying over distances ∼ ℓ. The total energy is
ε =1
2
(p2
x + p2y + p2
z
)+ U(~r) px =
h
ℓnx etc.
so that (n2
x + n2y + n2
z
)≤ 2mℓ
h2(ε− U(~r))
Counting the number of states in the box is done exactly as in the U = 0 case, making the substitutionε → ε − U(~r). In order to get the total density, it suffices to integrate in the whole space; however oneshould limit the integration to the region where ε ≥ U(~r), because (n2
x + n2y + n2
z) ≥ 0.
In the second method, one starts from the single-particle Hamiltonian
H(~p,~r) =~p 2
2m+ U(~r)
The semi-classical approximation reads
ρ(ε) =
∫d3pd3r
h3δ
(ε− ~p 2
2m− U(~r)
)
=2π
h3
∫d3r
∫d(p2) p δ
(ε− ~p 2
2m− U(~r)
)
=2π (2m)3/2
h3
∫d3r
√ε− U(~r) θ(ε− U(~r))
(60)
In the case U(~r) = 0 in a volume V , U(~r) = +∞ outside V , (60) is in agreement with the familiar result
ρ(ε) =2π (2m)3/2
h3V√ε
39
2. In the case of a harmonic potential, (60) becomes
ρ(ε) =8π2 (2m)3/2
h3
∞∫
0
dr r2√ε− 1
2mω2r2 θ
(ε− 1
2mω2r2
)
=8π2 (2m)3/2
h3
√ε
∞∫
0
dr r2√
1 − mω2r2
2εθ
(2ε
mω2− r2
)
=8π2 (2m)3/2
h3
√ε
q
2ε
mω2∫
0
dr r2
√1 − mω2r2
2ε
On setting u = mω2r2/2ε, one can feature the integral (5.108) and obtain
ρ(ε) =1
2
1
(~ω)3ε2 (61)
3. As in (5.68), one can isolate the contribution of the ground state ε0 = 3~ω/2
N =z
eβε0 − z+
∞∫
ε0
dερ(ε)
z−1eβε − 1= N0 +N1
with
N1 ≤∞∫
ε0
dερ(ε)
eβ(ε−ε0) − 1
Indeed, at the transition, the chemical potential takes its maximum value µ = ε0: the ratio N0/N cantend toward a finite value only if z = exp(βε0). It is more convenient to choose the ground state energyas a reference energy, and thus to redefine the chemical potential as follows
µ′ = µ− ε0 = µ− 3
2~ω z = eβµ′
The critical temperature Tc (βc = 1/kTc) is then fixed by the condition
N = N1 =
∞∫
0
dερ(ε)
eβcε − 1
namely, taking into account (61),
N =
(kTc
~ω
)3
ζ(3) (62)
In the case T ≤ Tc, z = 1 and
N = N0 +
(kT
~ω
)3
ζ(3) (63)
On comparing (62) et (63), one obtains the expression (5.109) of the ratio N0/N .
4. With the data given in the text
~ω
(N
ζ(3)
)1/3
= 1.86 × 10−29
in MKS units, which leads to a critical temperature Tc = 1.35µK.
40
5. At T = Tc particles occupy excited states and
N =
∞∫
0
dερ(ε)
eβε − 1
The potential U(~r) being slowly variable, one can define, by using (60), a local density of states as
n(ε, ~r) =2π (2m)3/2
h3
√ε− U(~r)
eβε − 1θ(ε− U(~r))
(5.110) is obtained by integrating n(ε, ~r) over ε. To establish this formula, we have assumed, as inquestion 3, that the chemical potential is zero when one chooses the ground state energy to be zero. Onenotices that for U(~r) = 0, the density in the center of the trap corresponds to the density (5.70) at thecritical temperature. With the data given in the text and Tc = 1.35µK, one finds for the density
n(~r = 0) =
(2πmkTc
h2
)3/2
ζ(3/2) = 8.6 × 1019 atoms.m−3
.
Problem 5.7.6 Solid-liquid equilibrium for helium-3
A. 1. From the expression (5.64) of the specific heat in the Debye model, one deduces that of the entropyper atom
σ =S
N=
4
5π4k
(T
TD
)3
and that of the energy per atom (5.112) thanks to ∂f/∂T = −σ. For a value of the temperature T = 1 K,kT ≃ 0.1 meV, while (T/TD)3 ≃ 10−3. The last term of (5.112) is then on the order of 6× 10−3 meV andit is negligible compared to the other two terms which are on the order of the meV.
2. The entropy of a system of N disordered spins 1/2 is kN ln 2. One must add to f the term −Tσ =−kT ln 2, because, in the solid, interactions between nuclear moments are negligible with respect tothermal motion, except when T reaches values on order of a mK.
3. To go from the free energy F to the Gibbs potential G, one must add PV ; as G = µN , one must addvsP to f . This expression of µ does not obey the third law σs → k ln 2 for T → 0, and it does not vanishwhen T = 0. In fact the spins begin to be ordered below 2 mK: solid helium-3 becomes anti-ferromagneticbelow this temperature and the entropy vanishes5
B The logarithm of the grand partition function is given by
lnQ =V (2m)3/2
2π2~3
∞∫
0
dε√ε ln (1 + exp(β(µ − ε)))
One integrates by parts
P =1
βVlnQ =
V (2m)3/2
3π2~3
∞∫
0
dεε3/2
exp(β(ε− µ)) + 1
5This nuclear anti-ferromagnetism occurs for an exceptionally high temperature in the case of solid helium-3, because ofan exchange interaction of quantum origin. In general nuclear anti-ferromagnetism, due to a magnetic interaction betweennuclear moments, occurs only at the µK level.
41
At low temperatures, the integral is computed to the order (kT/µ)2 thanks to the Sommerfeld formula(5.29) with the result
P =2
15
(2m)3/2
π2~3µ5/2
(1 +
5π2
8
(kT
µ
)2)
= Bµ5/2
(1 +
5π2
8
(kT
εF (P )
)2)
B =2
15
(2m)3/2
π2~3
The replacement of µ by εF (P ) is correct to this order in T . This substitution allows one to obtain µ asa function of P
µ = AP 2/5
(1 − π2
4
(kT
εF (P )
)2)
A = B−2/5
At T = 0
εF (P ) = AP 2/5 = εF (P0)
(P
P0
)2/5
and one finally obtains (5.114). Let us compute the compressibility6 at T = 0
κ(T = 0) =1
n
∂n
∂P=∂ lnn
∂P=
3
5P
Using the expression (5.22) of the pressure yields
κ(T = 0) =3
5
5m
~2
1
(3π2n)2/3n=
3m
~2
(3π2n)1/3
3π2n2=
1
n2ρ(εF )
C. 1. The expressions of σℓ and vℓ are derived from (5.111)
σℓ = −∂µℓ
∂T= 2kb
(kT
A
)(P
P0
)α
≃ 2kb
(kT
A
)(64)
vℓ =∂µℓ
∂P= α
A
P0
(P
P0
)α−1(1 − b
(kTA
)2)
≃ αA
P0
(P
P0
)α−1
(65)
Differentiating the entropy with respect to T allows one to obtain the specific heat
cv ≃ cP = T∂σℓ
∂T≃ 2kb
(kT
A
)
As nℓ = 1/vℓ ∝ P (1−α) one gets
κ(T = 0) =∂ lnnℓ
∂P=
1 − α
P
2. The transition line is given by the equality of the chemical potentials of the liquid and solid phases
−u0 +9
8~ωD − kT ln 2 + vsP = A
( PP0
)α(
1 − b(kTA
)2)
(66)
Differentiating this equation with respect to T , or equivalently, using Clapeyron’s formula (3.97), onegets
dP
dT=σs − σℓ
vs − vℓ=k ln 2 − 2bk(kT/A)
vs − (4A/5P0)(67)
Writing (dP/dT = 0) leads tob
A=
ln 2
2kTm≃ 12.5 (meV)−1
6Using the Gibbs-Duhem relation gives a more direct proof of (5.117).
42
3. One may cast (67) into the form
dP
dT=k ln 2(1 − T/Tm)
vs − vℓ
hence (vs−vℓ) ≃ 0.25×10−29 m3, in good agreement with the value given in the introduction: (vs−vℓ) ≃0.22 × 10−29 m3.
4. A = 54 vℓP0 ≃ 1.1 meV, A/k ≃ 13 K, while
b =ln 2
2Tm
A
k≃ 14
For T = 1 K, b(kT/A)2 ≃ 0.08. In the case of an ideal Fermi gas with specific volume vℓ, one findsεF ≃ 0.55 meV, and
b
A=
π2
4εF≃ 4.5 (meV)−1
while
A =
(2
15
(2m)3/2
π2~3
)−2/5
P2/50 ≃ 2.4 meV
using the value α = 2/5 of the ideal Fermi gas. The numerical result of the ideal gas and that of theactual liquid differ by a factor ∼ 2-3. The Debye frequency follows from the value of A
9
8~ωD = u0 − vsP0 +A
leading to ~ωD ≃ 0.75 meV, that is TD ≃ 9 K, of the same order of magnitude as the experimental valueof 16K, and a sound velocity
cs =ωD
(6π2ns)1/3≃ 102 m.s−1
5. As (σs − σℓ) ≥ 0 for T ≤ Tm, the quantity of heat which is absorbed in in the solid→liquid transfor-mation is ∆Q = T (σs − σℓ) ≥ 0.
6. Because of the third principle, one must have σs(T = 0) = σℓ(T = 0) = 0, and (dP/dT ) in (67) mustvanish, since vs − vℓ 6= 0. The transformation being adiabatic and quasi-static, the entropy is constantand is represented by an horizontal line on figure 5.21. The method relying on the Pomeranchuk effectdoes not allow one to cool down below a temperature ∼ 1 mK. In the solid, the spins can be identified:they form a paramagnetic system for T >∼ 2 mK, and the entropy per atom of the solid is dominatedby the term k ln 2. In the same range of temperatures, the entropy of the liquid tends to zero, becausePauli’s principle implies that the ground state is unique.
D. 1. One finds numerically that µHeB/k = 0.78×10−2 K, and µHeB/kT = 0.78 for T = 0.01 K, whence
2〈Sz〉
~= tanh
µHeB
kT≃ 0.65
2. As the term k ln 2 is absent in the entropy of the fully polarized solid, σs − σℓ ≤ 0, and from (67),the pressure is an increasing function of T and does go through a minimum as in the unpolarized state.The Pomeranchuk effect is no longer present. As expected, the polarized solid is more organized thanthe polarized solid.
Problem 5.7.7 Superfluidity for hardcore bosons
1. Expressing the ais and the a†i s in terms of the spin operators ~Si gives for the Hamiltonian
H = −t∑
〈ij〉
(S
+i S
−j + S
−i S
+j
)− µ
∑
i
(S
zi +
1
2
)
= −µN2
− t∑
〈ij〉
(S
+i S
−j + S
−i S
+j
)− µ
∑
i
Szi (68)
43
2. The occupation number operator at site i is
ni = a†iai = S
zi +
1
2
(we have used a system of units where ~ = 1). We have for the spin system
Szi |−〉 = −1
2|−〉i S
zi |+〉 =
1
2|+〉i
so that the action of the occupation number operator ni on the states |±〉i is
ni|−〉i = 0 ni|+〉i = |+〉i
Since S+i |+〉i = 0, the only possible eigenvalues of ni are ni = 0 or ni = 1. If site i is in the state |−〉i,
then the site is not occupied, and it is occupied if the state is |+〉i. From these results, the total numberof bosons in the ground state |0〉 vanishes
Nb|0〉 = Nb
∏
i
|−〉i = 0
3. Because of the relation between the occupation number ni and the z-component of the spin operatorSi
Szi = ni −
1
2
the average value of∑
i Szi given by
∑
i
〈Szi 〉 =
∑
i
ni −N
2= Nb −
N
2
Let us compute the average value of∑
i Szi in the state (5.123). We remark (see footnote 34, chap. 5)
that for φi = 0
|ψ〉i = ei(θ−π)Syi |−〉i (69)
The operator exp(−iθ′Syi ) rotates spin i by an angle θ′ around Oy
eiθ′S
yi S
zi e−iθ′Sy
i = Szi cos θ′ − S
xi sin θ′ (70)
Thus, with θ′ = θ − π
〈Szi 〉 = 〈−|
(eiθ′
Syi S
zi e−iθ′Sy
i
)|−〉
= 〈−|Szi cos θ′ − S
xi sin θ′|−〉 = −1
2cos θ′ =
1
2cos θ
where we have used 〈−|Sx|−〉 = 0. We thus obtain the average value∑
i〈Szi 〉 in the state |Ψ〉
∑
i
〈Szi 〉 =
N
2cos θ (71)
from which follows the relation between Nb and N
N
2cos θ = Nb −
N
2
In terms of the density ρ = Nb/N this relation reads
sin2 θ = 4ρ(1 − ρ) (72)
44
4. The number of bosons in the condensate is
N0 = 〈Ψ|a†(~k = 0)a(~k = 0)|Ψ〉 =1
N
∑
ij
〈Ψ|a†jai|Ψ〉
=1
N
∑
ij
〈0|[∏
l
eiθ′S
y
l
]S
+j S
−i
[∏
q
e−iθ′S
yq
]|0〉
This can be simplified by noting that operators referring to different sites commute, so that
〈0|[∏
l
eiθ′S
yl
]S
+j S
−i
[∏
q
e−iθ′S
yq
]|0〉 = 〈0|eiθ′
Syi eiθ′
Syj S
+j S
−i e−iθ′
Syi e−iθ′
Syj |0〉
We use once more the fact that exp(−iθ′Sy) is a rotation operator
eiθ′S
y
S± e−iθ′
Syl = S
x cos θ′ + Sz sin θ′ ± iSy
and the property 〈−|Sx|−〉 = 〈−|Sy|−〉 = 0 to derive at once
〈−|eiθ′S
y
S± e−iθ′
Syl |−〉 = −1
2sin θ′ =
1
2sin θ
so that
N0 =N
4sin2 θ =
N
4[4ρ(1 − ρ)] (73)
or in terms of the superfuid densityρ0 = ρ(1 − ρ)
In general ρ0 < ρ, unlike the ideal gas case where ρ0 = ρ (remember that we are working at zerotemperature). When ρ ≪ 1, the bosons rarely collide, so that the system is approximately free and weregain ρ0 = ρ.
6. The grand potential is given from (5.132) and the expression (68) by
J = −µN2
− µ∑
i
〈Ψ|Szi |Ψ〉 − t
∑
〈ij〉〈Ψ|
(S
+i S
−j + S
−i S
+j
)|Ψ〉 (74)
The second term in (74) is evaluated from the results of question 3 and the third from those of question 4
J = −µN2
− µN
2cos θ − tN sin2 θ (75)
7. The minimization equation dJ /dθ = 0 gives at once the value of θ
cos θ =µ
4t(76)
from which we derive the mean field values of ρ, ρ0 and J
ρ =1
2+µ
8tρ0 =
1
4−( µ
8t
)2
J = −N(t+
µ2
16t
)(77)
8. The state |Ψ〉T is obtained from the ground state |0〉 by a site dependent rotation of (θ−π) around Oyfollowed by a rotation of −φ around Oz. Clearly, the φ rotation around Oz does not affect the calculationof the average value of S
zi , so that we may use the result of question 3
T 〈Ψ|∑
i
Szi |Ψ〉T =
N
2cos θ
45
On the other hand, from the transformation properties of Sx and S
y under a rotation around Oz (or froma simple matrix multiplication)
e−iφSz
S± eiφS
z
= e±iφS±
and we get
T 〈Ψ|S+i S
−j |Ψ〉T =
1
4sin2 θ ei(φi−φj)
T 〈Ψ|S+j S
−i |Ψ〉T =
1
4sin2 θ e−i(φi−φj)
This leads to ∑
〈ij〉T 〈Ψ|S+
i S−j + S
−i S
+j |Ψ〉T =
N
2sin2 θ
(1 + cos
2π
L
)
where we have used the fact that there are N nearest neighbours in the x direction as well as in the ydirection. Gathering all the preceding results gives the grand potential
JT = −µN2
− µN
2cos θ − tN
2(1 + cos δφ) sin2 θ (78)
The minimization with respect to θ gives the following value of θ
cos θ =µ
2t(1 + cos δφ)
9. In the limit δφ≪ 1 the preceding value of θ becomes
cos θ ≃ µ
4t
(1 +
1
4(δφ)2
)
It is then straightforward to compute the difference (JT − J ) per site
1
N(JT − J ) =
∆JN
= tρ(1 − ρ)(δφ)2
But ∆J can also be written in terms of the superfluid density ρs and the velocity v as
∆JN
=1
2mρsv
2
Comparing the two expressions of ∆J /N and using v = 2tδφ leads to
ρs = ρ(1 − ρ) (79)
10. The condensate density ρ0 and the superfluid density ρs are thus identical within the mean fieldapproximation
ρ0|MF = ρs|MF
This equality does not persist if one includes the effect of spin waves. If one takes spin waves into account,the analytical results are in excellent agreement with those of numerical simulations.
46
Solutions for chapter 6
Exercise 6.4.5 Lord Kelvin’s model of earth cooling
1. In the region x ≥ 0, the temperature θ(x, t) = 0, while in the region x < 0, the temperature θ(x, t)obeys the diffusion equation
∂θ
∂t= D
∂2θ
∂t2(80)
The most general solution of (80) is (without taking into account boundary conditions)
θ(x, t) = A+B√
4πDt
∫ +∞
−∞dx′ g(x′) exp
(− (x− x′)2
4Dt
)(81)
in terms of an arbitrary function g(x) and of two constants A and B. This result is easily proved byshowing that (81) obeys (80) thanks to
∂2
∂x2=
∂2
∂(x− x′)2
and the fact that1√
4πDtexp
(− x2
4Dt
)
obeys (80). For t = 0 we have
x ≥ 0 θ(x, t = 0) = 0
x < 0 θ(x, t = 0) = A+B g(x) = θ0
These conditions imply that one may choose g(x) = 1 and A + B = θ0 for x < 0, and they allow us tocompute θ(x = 0, t)
θ(x = 0, t) = A+B
2= 0
whenceA = −θ0 B = 2θ0
The explicit expression for θ(x, t) (for x ≤ 0) follows from (81)
θ(x, t) = −θ0 +θ0√πDt
0∫
−∞
dx′ exp
(− (x− x′)2
4Dt
)(82)
2. The gradient is easily computed by making the change of variable u = x− x′
θ(x, t) = −θ0 +θ0√πDt
∞∫
x
du exp
(− u2
4Dt
)(83)
with the result for x = 0∂θ
∂x
∣∣∣x=0
= − θ0√πDt
(84)
3. The numerical result for the age of the Earth is
t = 8.5 × 1015 s = 2.7 × 108 y
4. In three dimensions, we may use the Laplacian in spherical coordinates
∇2 f(r) =1
r
d2
dr2[rf(r)]
47
valid for a function f(r) which is regular at the origin. The function Θ(r, t) = rθ(r, t) obeys the equation
∂Θ
∂t= D
∂2Θ
∂r2(85)
As in Question 1, the general solution of (85) may be written in terms of three constants A, A′ and B as
Θ(r, t) = Ar +A′ +B√
4πDt
+∞∫
−∞
dr′ exp
(− (r − r′)2
4Dt
)(86)
from which we derive A′ = 0 (because θ must be finite at r = 0) and g(r) = r so that
θ(r, t) = A+B
r√
4πDt
+R∫
−R
r′dr′ exp
(− (r − r′)2
4Dt
)(87)
The symmetric integration from −R to +R ensures that θ(r, t) be finite at r = 0. Furthermore we have
θ(r = 0, t) = A+B(1 + O(e−R2/(Dt))
)
so that A+B = θ0. Let us now compute θ(R, t), with the change of variable u = r − r′
θ(R, t) = A+B
R√
4πDt
2R∫
0
du (R− u) exp
(− u2
4Dt
)
= A+B
2
(1 + O(e−R2/(Dt))
)
so that A+B/2 = 0. As in question 1, A = −θ0 and B = 2θ0. The final result is then
θ(r, t) = −θ0 +θ0
r√πDt
r+R∫
r−R
du (r − u) exp
(− u2
4Dt
)
from which we derive the expression of the gradient
∂θ
∂r
∣∣∣r=R
=θ0√πDt
1 + e−R2/(Dt) +
2R∫
0
duu
R2exp
(− u2
4Dt
)
For R ≫√Dt, the second term in the square bracket is exponentially negligible, while the third one gives
a relative error O(Dt/R2).
The reason for the symmetric integration from −R to +R, which may seem surprising at first sight, maybe understood by going to Fourier space. Assume that θ(r, t = 0) = 1 for r ≤ R. Then the Fouriertransform θ is
θ(~k, t = 0) =
∫
r′≤R
dr′ e−i~k·~r ′
=4π
k
R∫
0
r′dr′ sin kr′
Let us give some intermediate steps
θ(~r, t) = 4π
∫d3k
(2π)3e−i~k·~r e−Dk2t 1
k
R∫
0
r′dr′ sin kr′
=1
πr
R∫
0
r′dr′∞∫
0
dk [cos k(r − r′) − cos k(r + r′)] e−Dk2t
=1
r√
4πDt
R∫
0
r′dr′[exp
(− (r − r′)2
4Dt
)− exp
(− (r + r′)2
4Dt
)]
48
The symmetric integration is recovered by making the change of variable r′ → −r′ in the second term ofthe square bracket.
Problem 6.5.2 Hydrodynamics of the perfect fluid
1. Let us assume that P depends only on z and consider in the fluid a parallelepiped with edges parallelto the axes, the area of the horizontal faces located at z and z + dz z et z + dz being dS. The force onthe parallepiped is
dS [−P(z + dz) + P(z)] ≃ −dS dPdz
dz = −dVdPdz
In the general case, the force is given by
−∫
Pd ~S = −∫~∇PdV
where we have used Green’s theorem7. The z-component of the force is indeed −~∇P , and Euler’s equationis the consequence of Newton’s law applied to a fluid element.
2. From ǫ′ = ǫn, h′ = nh, s = ns and from (3.93) one derives
nds = ds− sdn
and then
dǫ′ = Tds+ (h′ − T s)dn
= Tds+1
n(h′ − Ts)dn
= Tds+1
ρ(h′ − Ts)dρ
3. Using the summation convention on repeated indices
∂
∂t
(1
2ρu2
α
)=
1
2
∂ρ
∂tu2
α + ρuα∂uα
∂t
= −1
2u2
α∂β(ρuβ) − ρuαuβ(∂βuα) − uβ∂βP
= −∂β
(1
2ρu2
αuβ + Puβ
)+ P(∂βuβ)
which gives (6.104).
3. One obtains from (6.103)∂ǫ′
∂t= T
∂s
∂t+
(h′ − Ts
ρ
)∂ρ
∂t
Adding this equation to (6.104) and using the entropy conservation law (6.105) and that of the mass(6.64) yields
∂
∂t(k + ǫ′) + ∂β
[jKβ + Puβ
]= P(∂βuβ) − T∂β(suβ) −
(h′ − Ts
ρ
)∂β(ρuβ)
= −ǫ′(∂βuβ) − Tuβ(∂βs) −(h′ − Ts
ρ
)uβ(∂βρ)
One rewrites ǫ′(∂βuβ)ǫ′(∂βuβ) = ∂β(ǫ′uβ) − uβ∂βǫ
′
7If ~n is an arbitrary fixed vector, Green’s theorem givesR
f(~r)~n · d ~S =R
dV ~n · ~∇f(~r).
49
and one uses once more (6.103) to compute ∂βǫ′ and to derive the final result (6.107). This equation is
that of energy conservation, and it shows that the energy current is
~E =
(1
2ρ~u 2 + h′
)~u = (ǫ+ P)~u
This expression is a particular case of (6.83) when Pαβ = Pδαβ and when the heat current ~E′ is zero.
The fact that this result features the enthalpy density can be explained as follows: the energy flux througha closed surface S is ∫
~E · d~S =
∫ǫ~u · d ~S +
∫P~u · d ~S
The first term is a convection term, while the second term corresponds to the work done by the pressureon the fluid element inside S.
Problem 6.5.4 Isomerization reactions
1. Since the number of molecules is conserved, the balance equation for the number of molecules of agiven form reads
dNi
dt= Ni =
∑
j 6=i
(kijNj − kjiNi)
2. Let us consider an ideal gas made of N = N1 +N2 molecules of two different kinds, at a fixed pressure.From (2.90b), the gas entropy is
S = kN1 lnT 5/2
P1+ kN2 ln
T 5/2
P2+ const
where P1 = (N1/N)P (P2 = (N2/N)P ) represents the partial pressure of the molecules of type 1 (2).The entropy can be cast into a form which features explicitly the mixing term
S = kN lnNT 5/2
P− kN1 lnN1 − kN2 lnN2 + const
Only the mixing term depends on the values of N1 and N2 and, at equilibrium,
S0 = kN lnNT 5/2
P− kN0
1 lnN01 − kN0
2 lnN02 + const
whenceS − S0 = −k
(N1 lnN1 +N2 lnN2 −N0
1 lnN01 −N0
2 lnN02
)
The relation (6.112) is derived in the same way.
3. On expanding to second order the logarithms in (6.112), one establishes the following inequality
S − S0 = −k2
∑
i
x2i
N0i
(88)
One checks on this expression that S−S0 ≤ 0, where the equality only holds at equilibrium. The affinitiesare easily deduced from (88) :
Γi = −k xi
N0i
Because ji = xi = Ni, transport equations are obtained by making explicit the affinities in the kineticequations
ji =∑
j 6=i
(kijNj − kjiNi
)=∑
j 6=i
(kijxj − kjixi
)
=1
k
∑
j 6=i
N0i kji Γi −
1
k
∑
j 6=i
N0j kij Γj
50
One then identifies the response coefficients
Lii =1
k
∑
j 6=i
N0i kji Lij = −1
kN0
j kij
and one checks Onsager’s relations
Lij = −1
kN0
j kij = −1
kN0
i kji = Lji
Because of the conservation of the total number of molecules, the three transport equations are notindependent. Assuming, to fix ideas , N2 and N3 to be independent, the reader will easily express thetransport equations relating j2 − j1 and j3 − j1 to the affinities.
51
Solutions for chapter 7To be completed soon
52
Solutions for chapter 8
Exercise 8.5.4 Calculation of the collision time
1. Let us evaluate the collision time of a target particle with momentum ~p1 by computing the numberof collisions per unit time in a phase space volume d3r d3p1. The flux of incoming particles with velocity~v2 is |~v2 − ~v1|f(~p2)d
3p2, and, from the definition (8.7) of the cross-section, the number of collisions perunit of time is [
f(~p1)d3r1 d3p1
] ∫d3p2
∫dΩ
′
σ(|~v2 − ~v1|,Ω′
)|~v2 − ~v1|f(~p2)
The number of collisions per unit of time suffered by a particle with momentum ~p1 is the inverse of thecollision time
1
τ⋆(~p1)=
∫d3p2 |~v2 − ~v1|f(~p2)
∫dΩ
′
σ(|~v2 − ~v1|,Ω′
)
=
∫d3p2 |~v2 − ~v1|f(~p2)σtot(|~v2 − ~v1|) (89)
The average collision time τ⋆ is obtained from averaging over ~p1 (one must remember that one calculatesthe mean number of collisions, and the average which must be taken is that of 1/τ⋆(~p1))
1
τ⋆=⟨ 1
τ⋆(~p1)
⟩=
1
n
∫d3p1
1
τ⋆(~p1)f(~p1) (90)
whence (8.150).
2. This distribution function is indeed correctly normalized according to (8.1), because∫
d3p f(~p) =
∫p2dp
∫dΩ~p f(~p) = n
As τ⋆(~p1) does not depend upon the direction of ~p2, one may choose ~p1 ‖ Oz. If θ denotes the anglebetween Oz and ~p2
(~v2 − ~v1)2 = 2v2
0(1 − cos θ) = 4v20 sin2 θ
2
assuming the cross-section to be velocity independent. By substituting the values of |~v1 −~v2| and of f in(8.150)
1
τ⋆(~p2)= σtot
n
4π2v0
∫dΩ~p1
sinθ
2
= σtotnv0
1∫
−1
d(cos θ) sinθ
2
=4
3σtotnv0
The collision time and the mean free path are given by
τ⋆ =3
4nv0σtotℓ =
3
4nσtot(91)
3. Let us introduce the center-of-mass momentum ~P = ~p1 + ~p2 of the two colliding particles, and let usmake the change of variables
~p1 =1
2~P + ~p ~p2 =
1
2~P − ~p
where ~p is the reduced momentum. We get from (6.117)
1
τ⋆=
1
nσtot
∫d3P d3p f
(1
2~P + ~p
)f
(1
2~P − ~p
)2|~p|m
53
and then
f
(1
2~P + ~p
)f
(1
2~P − ~p
)= n2
(1
2πmkT
)3
exp
(−
~P 2
4mkT
)exp
(− ~p 2
mkT
)
The first exponential features the total mass 2m and the second one the reduced mass m/2. The integrals
are performed thanks to the change of variables ~P =√
2 ~P ′ et ~p = ~p ′/√
2 and by using the definition ofthe mean velocity for a Maxwell distribution,
〈v〉 =
(1
2πmkT
)3/2 ∫d3p
|~p|m
exp
(− ~p 2
2mkT
)
Problem 8.6.2 Electron gas in the Boltzmann-Lorentz model
A. Since the cross-section does not depend on the angles, the collision term (8.35) is expressed as afunction of τ⋆(v) = 1/(ndvσ(v)) as follows
C[f ] = ndv
∫dΩ′ σ(v)
4π
[f(~r, p′,Ω′) − f(~r, ~p)
]= − 1
τ⋆(v)f(~r, ~p)
because, by assumption (~p ′ = (p′,Ω′))
∫dΩ′ f(~r, p′,Ω′) = 0
One then derives the expression for f
f = −τ⋆(v)(~v · ~∇)f0
and the expressions for the currents from (8.49). When σ does not depend on v, one can extract fromthe integrals in (8.61) and (8.62) the factor vτ⋆(v) = ℓ = 1/(ndσ), which is the mean free path of theelectrons determined by the collisions with the scattering centers. This leads to the expressions for thecurrents
~N = − ℓ
3
∫d3p v~∇f0 ~E = − ℓ
3
∫d3p εv~∇f0
B. 1. The expressions for the currents and for the coefficients Lij are obtained as in section 8.2.5 by
writing ~∇f0 as a function of ~∇(1/T ) and of ~∇(−µ/T ). Let us make in LNN the change of variablesε = p2/2m, mdε = pdp
LNN =8π
3
ℓm
k
∫ ∞
0
ε dε f0
so that ν = 1 for LNN . For LEN one finds an analogous expression with ν = 2, while ν = 3 for LEE.Using the definitions given in the text, one computes the integrals in the form
LNN =τ⋆
mnT
LNE = LEN =2τ⋆
mnkT 2 = 2kTLNN
LEE =6τ⋆
mk2T 3 = 6(kT )2LNN
2. From these expressions, the diffusion coefficient is derived thanks to (6.27)
D =k
nLNN =
τ⋆
mkT
54
and the thermal conductivity coefficient thanks to (6.50)
κ =1
T 2LNN
(LEELNN − L2
EN
)= 2
τ⋆
mnk2T
3. See in section 6.2.2 the discussions which follow equations (6.57) and (6.61). From (6.57)
σel =q2
TLNN =
nq2τ⋆
m
and the ratio κ/σel isκ
σel= 2
k2
q2T
4. When an external force is present8, ~∇f0 is given by
~∇f0 =[~∇α− (ε+ V ) ~∇β − β~∇V
]f0
and f by (~F = −~∇V in (8.32))
f = − ℓ
v
[~v · ~∇~rf0 − (~∇~rV ) · ~∇~pf0
]
taking into account~∇~pf0 = −β~vf0
one finds
f = − ℓ
v~v ·[~∇α− (ε+ V ) ~∇β
]f0
This equation shows that the results in the presence of an external force are deduced from those withoutexternal force through the substitution ε → ε + V in the expression for f . There is no change in theexpression for LNN : LNN = L′
NN , while
LEN =ℓ
3km
∫d3p p(ε+ V )f0 = L′
NE + V L′NN
LEE =ℓ
3km
∫d3p p(ε+ V )2f0 = L′
EE + 2V L′EN + V 2L′
NN
5. One remarks that ~∇~pf0 = ~v(df0/dε), whence
~el = q
∫d3p~vf = −q2ℓ
∫d3p
~v
v( ~E · ~v)df0
dε= − q
2ℓ
3m~E
∫d3p p
df0dε
The change of variables p→ ε leads to
σel = −8πq2ℓm
3
∫dε ε
df0dε
=8πq2ℓm
3kT
∫dε εf0(ε) =
q2
TLNN
for a Maxwell-Boltzmann distribution where (df0/dε) = −βf0.
C. 1. On using (5.19), εF = µ for T = 0 and pF = (2mµ)1/2
n =8π
3h3(2mµ)3/2
8By convention ~∇ = ~∇~r unless otherwise stated.
55
2. The number of possible microscopic states in d3r′ d3p′ is
2
h3d3r′ d3p′
where the factor 2 comes from the spin degrees of freedom, and one has f d3r′ d3p′ states. The probabilityoccupation of a state is then h3f/2. The probability for the state ~p ′ to be accessible is (1−h3f ′/2). Theintegral in the collision term will be proportional to
f
(1 − h3
2f ′)− f ′
(1 − h3
2f
)= f − f ′
and the collision term remains unchanged9.
3. Given the Sommerfeld formula (5.29)
∂f0∂α
=∂µ
∂α
∂f0∂µ
≃ 2
h3
[1
βδ(ε− µ) +
π2
6β3δ′′(ε− µ)
]
∂f0∂β
=ε
β
∂f0∂ε
≃ 2
h3
[− ε
βδ(ε− µ) − π2ε
6β3δ′′(ε− µ)
]
that is
~∇f0 = −2T
h3
[~∇(− µ
T
)+ ~∇
(1
T
)][δ(ε− µ) +
π2
6β2δ′′(ε− µ)
]
whence c1 = 2T/h3 and c2 = π2/(6β2). From here on, the computation is identical to that of part B.
4. For LNN , s = 1 and∫
dε ε δ′′(ε− µ) = 0. There only remains the contribution of δ(ε− µ)
LNN =16πℓTm
3h3
∞∫
0
dε ε δ(ε− µ) =16πℓTmµ
3h3=ℓTn
pF=nTτ⋆
F
m
We have used the relation between the chemical potential and the density, as well as ℓ = τ⋆F vF . Only
the electrons located close to the Fermi surface p = pF contribute to transport, because the electronsinside the Fermi sea (p < pF ) cannot be thermally excited. The reasoning is analogous to that used insection 5.2.2 for the specific heat.
5. The diffusion coefficient is given by (6.27)
D =1
T
(∂µ
∂n
)
T
LNN =2
3
µτ⋆F
m
and the electrical conductivity by (6.57)
σel =q2
TLNN =
q2
mnτ⋆
F
This last result is identical to that of the classical gas, provided one replaces τ⋆ by τ⋆F , the collision time
on the Fermi surface; on the contrary, the result for D is quite different from the classical one.
6. The electronic density is n = 8.4 × 1028 m−3, which implies τ⋆F = 2.1 × 10−14 s.
7. The coefficient LEN is given by
LEN =16πℓTm
3h3
∞∫
0
dε ε2[δ(ε− µ) +
π2
6β2δ′′(ε− µ)
]
= µLNN
(1 +
π2
3β2µ2
)
9On the contrary, the collision term in the Boltzmann equation would be modified, leading to the so-called Landau-Uhlenbeck equation.
56
while LEE is
LEE =16πℓTm
3h3
∞∫
0
dε ε3[δ(ε− µ) +
π2
6β2δ′′(ε− µ)
]
= µ2LNN
(1 +
π2
β2µ2
)
These results allow us to compute the thermal conductivity coefficient
κ =µ2
T 2LNN
[(1 +
π2
β2µ2
)−(
1 +π2
3β2µ2
)2]
=1
3π2k2LNN =
1
3mπ2k2nTτ⋆
F
Comparison with the electrical conductivity σel = q2LNN/T leads to the Franz-Wiedemann law
κ
σel=π2
3
k2
q2T
The factor 2 in question B.3 has become π2/3. The method followed in question B.5 leads once more tothe electrical conductivity using
df0dε
≃ − 2
h3δ(ε− µ)
because
σel =2
h3
8πq2ℓm
3
∫dε εδ(ε− µ) =
16πq2ℓm
3h3µ =
q2
TLNN
Problem 8.6.5 Electrical conductivity in a magnetic field and the quantumHall effect
A. Given the assumptions in this problem, the Boltzmann-Lorentz equation reads
(~F · ~∇p)f = C[f ] = −ndvσtot(v) f (92)
1. In the case where ~B = ~0, and neglecting terms of second order with respect to f0, (92) yields
f = −qτ⋆(p)( ~E · ~∇~p)f0
We remark that
(~∇f0)α = ∂pαf0 = ∂pα
ε∂f0∂ε
= vα∂f0∂ε
from which follows
f = −qτ⋆(p)(~v · ~E)∂f0∂ε
On the other hand, the electrical current density is given by
~el = q~N = q
∫d3p~vf (93)
that is
~el = −q2∫
d3p τ⋆(p)~v(~v · ~E)∂f0∂ε
= −4π
3
q2
m2
∫dp p4∂f0
∂ε~E
At T = 0 the equilibrium distribution for an electron gas is
f0(ε) =2
h3θ(εF − ε)
57
so that∂f0∂ε
= − 2
h3δ(εF − ε) = − 2
h3
m
pFδ(p− pF ) (94)
One finally obtains for the electrical conductivity
σel =
[2
h3
4π
3p3
F
]q2
mτ⋆F
The term between square brackets is nothing other than the electron density.
2. In the presence of a magnetic field, the full Lorentz force (8.167) must be taken into account in (92).
However, if ~B and ~E are collinear, the magnetic force will average to zero, since, on average, the electronsmove in the direction of ~E: the conductivity will remain unchanged. In the general case, the magneticforce appears as a correction term in (92). Indeed, since ~∇f0 ∝ ~v, the term (~v × ~B) · ~∇f0 = 0. It is thusthe perturbative term f which appears in the form (8.169) of the Boltzmann-Lorentz equation.
Even without knowing its explicit general form, the values taken by ~C in two limiting cases ~B = ~0 and~E = ~0 are easily obtained. Indeed, if ~B = ~0, one reverts to the preceding case, and thus ~C = qτ⋆(p) ~E; if~E = ~0, the electrons suffer only the Laplace force q(~v× ~B) which averages to zero because of the isotropy
of the scattering, and thus ~C = ~0.
3. By making ~C explicit in (8.169), one gets
q~v · ~E ∂f0∂ε
− q(~v × ~B) · ~∇~p
(~v · ~C ∂f0
∂ε
)=
1
τ⋆~v · ~C ∂f0
∂ε
Let us examine the gradient term
~∇~p
(~v · ~C ∂f0
∂ε
)= ~v · ~C ∂2f0
∂ε2~v +
1
m
∂f0∂ε
~C
The first term is eliminated on taking the scalar product with ~v × ~B. Using the invariance of the mixedproduct under circular permutations, one finally obtains
∂f0∂ε
~v ·[q ~E − q ~B
m× ~C
]=∂f0∂ε
~v ·[
~C
τ⋆(p)
]
Equation (8.170) follows because this relation must be satisfied whatever the value of ~v.
From the two vectors ~E and ~B, one builds the natural basis ( ~E, ~B, ~B × ~E) on which one can decompose
the vector ~C. The coefficients of this decomposition are given by (8.170)
(α− qτ⋆ − γ
qτ⋆
mB2
)~E +
(δ + γ
qτ⋆
m~B · ~E
)~B +
(γ + α
qτ⋆
m
)( ~B × ~E) = ~0
where we have used ~B × ( ~B × ~E) = ( ~B · ~E) ~B −B2 ~E. In the case under study, the electric and magneticfields are perpendicular. The preceding system leads to
α =qτ⋆
1 + ω2τ⋆2δ = 0 γ = −qτ
⋆
m
qτ⋆
1 + ω2τ⋆2
Knowing the expression of ~C, and thus of f , one goes easily from (8.170) to (8.169).
4. Taking (94) into account, relation (8.171) can also be written as
f =2
h3
qτ⋆
1 + ω2τ⋆2
[~E + τ⋆(~ω × ~E)
]· ~p
pFδ(p− pF )
58
Putting this result in the expression for ~el (93), one gets
~el =nq2τ⋆
F
m
1
1 + ω2τ⋆2F
( ~E + τ⋆F ~ω × ~E)
=σel
1 + ω2τ⋆2F
( ~E + τ⋆F ~ω × ~E)
The only non vanishing components of the conductivity tensor are
σxx = σyy =σel
1 + ω2τ⋆2F
σxy = −σyx = − σel
1 + ω2τ⋆2F
ωτ⋆F =
σel
1 + ω2τ⋆2F
τ⋆F qB
m
(95)
The minus sign in σxy = −σyx follows from the Onsager relations in the presence of a magnetic field.
B 1. Equation (8.173) is nothing other than Newton’s law, in which, in addition to the Lorentz force,there also appears a friction force which is introduced in order to take collisions into account. This frictionforce governs the dynamics for short times, and vanishes exponentially with a characteristic time scalegiven by the parameter τ⋆.
In this model, the electrical current densities are simply given by jelα = qn〈vα〉. One thus obtains thesame conductivity tensor as above, with the only difference that the Boltzmann-Lorentz model was ableto fix the value of the characteristic time
2. Using (8.172) and (95), the condition jely = 0 gives
EH = −σyx
σyyEx = −ωτ⋆
F Ex
In this situation, one easily checks that the conductivity does not depend on the Laplace force
jelx =
[σxx − σxyσyx
σyy
]Ex
= σxx
[1 +
σ2xy
σ2xx
]Ex
= σelEx
One gets for jelx as a function of EH
jelx = − σel
ωτ⋆F
EH =nq
BEH
and thus
RH =EH l
jelx l d=
B
nd q(96)
The fact that the Hall resistance does not depend on the relaxation time shows that this resistance arisesbecause a stationary situation is reached in the transverse direction. As a consequence of the Laplaceforce, charges accumulate on the surface of the metal. The potential difference which follows from thisaccumulation of charges generates a transverse electric field, which, after a time on the order of τ⋆
F ,reaches an equilibrium value EH . In the framework of the Drude model, this means that the frictionforce vanishes, and that 〈Fy〉 = 0, that is
qEy − q〈vx〉B = 0
so thatqnEy − jelx B = 0
One recovers the preceding expression for RH .
59
C. 2. In two dimensions, the level density is uniform (see exercise 2.7.2)
ρ(ε) =2πmS
h2
2. Since the level density is uniform, the number of states in an interval ~ω,which corresponds to thegap between two consecutive Landau levels, is given by
g = ρ(ε)~ω =mS
hω = −S qB
h
3. Taking the degeneracy into account, there are νg electrons up to level ν with a surface density
nS = −νqBh
Setting nd = nS in relation (96), one gets for the ‘quantum Hall coefficient’
RH = − h
νq2
The preceding reasoning is not enough to justify the existence of plateaus in the Hall resistance. A generalargument relying on gauge invariance allows one to to show rigourously the existence of those plateaus(see Laughlin, [71]).
Problem 8.6.6 Specific heat and two-fluid model for helium II
A. 1. For T ≪ TD the calculation of the internal energy features the integral (A.53)
Ephonon =V
(2π~)3
∫d3p
pc
eβpc − 1=
V
2π2
(kT )4
(~c)3
∞∫
0
dxx3
ex − 1
=π2V
30
(kT )4
(~c)3
One deduces the specific heat by differentiating Ephonon with respect to T
CphononV =
1
V
∂Ephonon
∂T
∣∣∣∣V
=2π2k4
15~3c3T 3
2. For T <∼ 1 K, βε(p) >∼ 8.5, so that exp(βε(p)) >∼ 5000. Hence, replacing the Bose distribution by theMaxwell distribution is quite justified. Figure 8.10 shows that rotons can only exist in a limited range ofvalues of p centered around p0, p0 − ξ <∼ p <∼ p0 + ξ. The integral which occurs in the calculation of theroton contribution to the internal energy depends only upon a limited integration range
Eroton =V
2π2~3e−β∆
p0+ξ∫
p0−ξ
dp p2
(∆ +
(p− p0)2
2µ
)exp
(−β (p− p0)
2
2µ
)
However, since βp20/2µ ≃ 100, the Gaussian factor in the integral exhibits a narrow peak around its mean
value, which allows one to replace p2 by p20 (the error being on the order of µ/(βp2
0) ∼ 10−2) and to
60
integrate from −∞ +∞ on p :
Eroton ≃ V
2π2~3e−β∆
∞∫
−∞
dp p20
(∆ +
(p− p0)2
2µ
)exp
(−β (p− p0)
2
2µ
)
≃ V p20
2π2~3e−β∆
+∞∫
−∞
dp′(
∆ +p′2
2µ
)exp
(−β p
′2
2µ
)
≃ V
2π2~3∆p2
0 exp
(− ∆
kT
)√2µkT
+∞∫
−∞
dx
(1 +
kT
∆x2
)exp
(−x2
)
≃ V
2π2~3∆p2
0 exp
(− ∆
kT
)√2πµkT
(1 +
1
2
kT
∆
)
The third line is obtained by setting x =√β/2µ(p− p0). The Gaussian integrations are performed with
the help of (A.37). In the temperature range which concerns us here, the term ∝ kT/∆ ∼ 1/10 in thelast bracket could be neglected.The specific heat is obtained by differentiation of the internal energy. It reads at the leading order in T
CrotonV ≃ k
2π2~3p20 exp
(− ∆
kT
)√2πµkT
(∆
kT
)2[1 +
kT
∆+
3
4
(kT
∆
)2]
For T → 0, CphononV which varies as T 3 dominates over Croton
V which vanishes exponentially. However,
for T = 1 K, CphononV ∼ 3.5 kJ.K−1 is smaller by a factor ∼ 3 than Croton
V ∼ 11 kJ.K−1.
B. 1. Each atom of the fluid has in the reference frame R′ a momentum ~qi′ = ~qi −m~v
H ′ =
N∑
i=1
(~qi −m~v)2
2m+
1
2
∑
i6=j
U(~ri − ~rj)
=
N∑
i=1
q2i2m
−(
N∑
i=1
~qi
)· ~v +
1
2
N∑
i=1
mv2 +1
2
∑
i6=j
U(~ri − ~rj)
= H − ~P · ~v +1
2Mv2
(97)
If the fluid is in its ground state with energy E = E0 and momentum ~P = 0 in the reference frame R,going form R to R′ leads from (97) to an energy E′ in R′
E′ = E0 +1
2Mv2 (98)
Assume that an elementary excitation is created in the fluid with momentum ~p and energy ε(~p) in R,going from R to R′ implies, again from (97)
E′ = E0 + ε(~p) − ~p · ~v +1
2Mv2 (99)
because the fluid energy in R is E0 + ε(~p) and its momentum ~P = ~p. Subtracting (99) from (98) gives(8.178)
ε′(~p) = ε(~p) − ~p · ~v
As the gas of elementary excitations is in equilibrium with the walls of the container, one must use theenergy in the reference frame R′ in order to write the equilibrium distribution
n(~p,~v) =1
eβε′(~p) − 1= n(ε− ~p · ~v)
61
In the reference frame R, the momentum density of quasiparticles is given by
~g =
∫d3p
(2π~)3~p n(ε− ~p · ~v)
If the flow velocity is small, namely if ~p · ~v ≪ ε, then
n(ε− ~p · ~v) ≃ n(ε) − ~p · ~v dn(ε)
dε
The first term on the RHS does not contribute to the momentum density and
~g = −∫
d3p
(2π~)3~p (~p · ~v)dn(ε)
dε(100)
The final result (8.179)
~g = −1
3
∫d3p
(2π~)3p2 dn
dε~v
is obtained thanks to the symmetry argument leading to (8.60). From ~g = ρn~v and using (8.179), onededuces
ρn = −1
3
∫d3p
(2π~)3p2 dn
dε
For a phonon gas where ε = cp, one integrates by parts the previous equation with the result
ρn =4
3c24π
(2π~c)3
∞∫
0
dε ε3n(ε) (101)
The energy density in R reads, to lowest order in ~v
ǫ =
∫d3p
(2π~)3εn(ε) =
4π
(2π~c)3
∞∫
0
dε ε3n(ε) (102)
On comparing (101) and (102), one gets
ρn =4
3c2ǫ
The dispersion relation of the phonon gas is identical with that of a photon gas, in which case we haveshown that the pressure is one third of the energy density (see (2.29) or (4.35))
P =1
3ǫ
The phonon equation of state is thus
P =c2ρn
4
Each phonon propagates with a velocity c along a direction p which depends on its momentum: p = ~p/p.The expression for the energy current is
~jE =
∫d3p
(2π~)3ε cp n(ε− ~p · ~v) ≃ −c2
∫d3p
(2π~)3p2(p · ~v)dn
dεp
= −c2∫
d3p
(2π~)3p2(p · ~v)dn
dεp = −c2 1
3
∫d3p
(2π~)3p2 dn(ε)
dε~v = c2~g
Taking into account ~∇ · ~E = ~∇ · (c2~g) = c2~∇ · (ρn~v), the continuity equation for the energy reads
∂ǫ
∂t+ c2~∇ · (ρn~v) = 0
62
Differentiating this equation with respect to time yields a wave equation for the energy density with speedc/√
3. Indeed
∂2ǫ
∂t2+ c2~∇ ·
(∂
∂t(ρn~v)
)= 0
∂2ǫ
∂t2− c2
3∇2ǫ = 0 (103)
where we have used Euler’s equation (6.70) which becomes for v → 0
∂
∂t(ρn~v) = −~∇P = −1
3~∇ǫ
Equation (103) shows that since ǫ ∝ T 4, any local heat excess will propagate with velocity c/√
3: Thesecond sound waves may thus be interpreted as temperature waves. We have just exhibited one of theremarkable properties of Helium II, namely that it can transport heat not through diffusion but throughwave propagation, a mechanism which is much more efficient than diffusion. Second sound is a soundwave in the gas of elementary excitations. The previous considerations can be immediately transposedto the case of a photon gas, and in fact to any gas of massless particles at thermal equilibrium. If onereplaces the sound velocity by the light velocity, the preceding calculation shows that the sound velocityin any ultrarelativistic dilute gas is c/
√3.
Problem 8.6.7 The Landau theory of Fermi liquids
I Static properties
1. At zero temperature, the chemical potential µ is by definition
µ = E(N + 1) − E(N)
where N is the number of particles and E(N) the ground state energy for N particles. Then µ is theenergy needed to add one particle in the lowest available momentum state, namely pF , and µ = ε(pF ).In other words, µ is the extra energy of the system if one adds a quasiparticle on the Fermi surface.
2. The interaction term between two particles can be non zero only if the distance between these twoparticles is less than the range a of the interaction. The probability for this to happen is ∼ a3/V . Thenthe expression of ε(~p) is, as it should be, independent of V . As |δf(~p)| ≪ 1, the order of magnitude ofthe second term in the expression for ε(~p) is ρ(µ)λξ. It is independent of V , because ρ(µ) ∝ V , and is oforder ξ. If δf(~p) is isotropic, we can take the z-axis along ~p. If (θ′, φ′) denote the polar and azimuthalangles of ~p ′, we have
∑
l
∫d3p′ αℓPℓ(cos θ′)δf(p′) =
∑
l
∫p′
2dp′ d cos(θ′) dφ′αℓPℓ(cos θ′)δf(p′)
= α0
∫d3p′ δf(p′)
since the only non zero contribution comes from the ℓ = 0 term. We remind the reader of the standardresult for Legendre polynomials
1∫
−1
d(cos θ)Pℓ(cos θ)Pℓ′ (cos θ) =2
2ℓ+ 1δℓℓ′ P0(cos θ) = 1 P1(cos θ) = cos θ
On the other hand
δN =2V
h3
∫d3p′ δf(p′)
and comparing the previous two equations, one derives
ε(p) = ε(p) + α0δN
63
A higher order term in the expansion of ε(~p) would be of the form
(2V
h3
)2 ∫d3p1 d3p2 λ
(2)(~p; ~p1, ~p2)δf(~p1)δf(~p2) ∼ ρ2(µ)λ(2)ξ2
and thus of higher order in ξ.
3. From the definition of δf(~p) we may write, using the fact that [ε(~p) − ε(p)] is of order ξ
δf(~p) = f(~p) − f0((ε(~p) − ε(p)) + ε(p))
≃ f(~p) − f0(ε(p)) − (ε(~p) − ε(p))∂f0∂ε(p)
= δf(~p) + δ(ε(p) − µ)2V
h3
∫d3p′ λ(~p, ~p ′)δf(~p ′)
4. The excitation at temperature T , kT ≪ µ, is
δf(~p) = f0(~p, T )− f0(~p, T = 0) =1
1 + exp[(ε(~p) − µ)/kT ]− θ(µ− ε(~p))
= −π2
6(kT )2δ′(ε(~p) − µ)
where we have used the Sommerfeld approximation (5.29); here ξ must be identified with kT , since theFermi distribution is rounded off around µ on a distance of order kT ≪ µ (see figure 5.2). We haveneglected the variation of µ with T , as this assumption will be shown later on to be self-consistent. Notethat one can write the zero temperature equilibrium distribution either as f0(~p, T = 0) = θ(µ − ε(~p))or f0(~p, T = 0) = θ(µ − ε(p)), since there are no quasiparticles at equilibrium and zero temperature:δf(~p)|T=0 = 0. The interaction term in the quasiparticle energy is proportional to δE given by
δE =∑
ℓ
∫d(cos θ′) dφ′αℓPℓ(cos θ′)
∞∫
0
p′2dp′ δ′(ε(~p ′) − µ)(kT )2
≃∑
ℓ
∫d(cos θ′) dφ′αℓPℓ(cos θ′)
∞∫
0
p′2dp′ δ′(ε(p′) − µ)(kT )2
∝ (kT )2ρ′(µ)
We have made use of the fact that the replacement of ε(~p ′) by ε(p) in the δ′-function is consistent withour approximations: |ε−µ| is on the order of kT , which plays the role of ξ. Hence we may replace ε by εin the calculation of thermodynamic quantities if we neglect terms of higher order in T . For example wemay compute the term proportional to T of the specific heat. Let us first obtain the explicit expressionfor the density of states
ρ(µ) =2V
h3
∫d3p δ(ε(p) − µ)
=8πV
h3
∞∫
0
p2dp δ(ε(p) − µ) =8πV
h3
dp
dε
∣∣∣p=pF
p2F
=8πV
h3
m∗
pFp2
F =V m∗pF
π2~3
The term proportional to T in the specific heat is obtained following § 5.2.2: one has only to replace thefree mass m by the effective mass m∗ and one finds
CV =π2k2T
3ρ(µ) =
V k2T
3~2m∗pF
64
Let us show the consistency of our approach. Following again § 5.2.2, we easily show that
∂µ
∂T∝ k2T
µ
so that it is consistent to neglect the T -dependence of the chemical potential if we limit ourselves to termsof order zero and one in T .
5. At zero temperature, when the chemical potential varies from µ to µ+ dµ, the excitation δf(~p) is bydefinition
δf(~p) = θ(µ+ dµ− ε(~p)) − θ(µ− ε(~p)) ≃ dµ δ(ε(~p) − µ)
It is crucial to understand that in going from µ to µ+dµ, we have added quasiparticles and we must useε(~p) and not ε(p). On the other hand we also have from question 2
δf(~p) = δf(~p) + α0δN δ(ε(~p) − µ)
since the error made in replacing ε(~p) by ε(~p) in the δ-function is of higher order in ξ. Integrating overphase space leads to
ρ(µ)dµ = δN + ρ(µ)α0δN = δN(1 + Λ0)
or, equivalently∂N
∂µ=
ρ(µ)
1 + Λ0
The T = 0 compressibility κ is obtained form (1.42) in terms of the density n = N/V
κ =1
n2
∂n
∂µ
∣∣∣T=0
=1
n2V
∂N
∂µ
∣∣∣T=0
Finally one uses the expression for ρ(µ) in terms of the effective mass m∗ of the preceding question toderive the final form of the result
κ =1
n2
m∗pF
π2~3(1 + Λ0)
II Boltzmann equation
6. From the conservation of the number of quasiparticles and Liouville’s theorem, we obtain in theabsence of collisions
∂f
∂t+ ~∇~p ε · ~∇~rf − (~∇~r ε) · ~∇~pf = 0
7. Let us evaluate the various terms in the above collisionless Boltzmann equation. First
~∇~r ε(~r, ~p, t) =2V
h3~∇~r
∫d3p′ λ(~p, ~p ′)δf(~r, ~p ′, t)
while
~∇~p f ≃ ~∇~pf0 = ~∇~p θ(µ− ε(~p))
= − ~p
m∗ δ(ε(p) − µ) = −~v~p δ(ε(p) − µ)
Inserting these results in the Boltzmann equation and adding the collision term, we get
∂f
∂t+ ~v~p · ~∇~rδf − 2V
h3
[~∇~r
∫d3p′ λ(~p, ~p ′)δf(~r, ~p ′, t)
][−~v~p δ(ε(p) − µ)
]= C[δf ]
Note that we must take f0 as the ~r-independent equilibrium distribution, and not a ~r-dependent localequilibrium distribution, because we want to have
~∇~rf = ~∇~r(f0 + δf) = ~∇~r δf
65
We can now combine the last two terms in the right hand side by using the definition of δf(~p) in question 3
and obtain∂ δf
∂t+ ~v · ~∇~r δf = C[δf ]
8. The difference δn = n − n0 between the equilibrium density n0 and n is obtained by integrating δfover ~p
δn(~r, t) =2
h3
∫d3p δf(~r, ~p, t)
Thus, integrating the Boltzmann equation from the previous question over ~p and using the conservationof the number of particles in a collision, which implies
∫d3p C[δf ] = 0
we get the continuity equation∂δn
∂t+ ~∇~r · ~ = 0
provided the current ~ is defined as
~ =2
h3
∫d3p~v~p δf
The second term in ~~p is due to the fact that the moving quasiparticle drags along with it particles of thesurrounding medium. Another expression for the current is obtained if one expresses δf in terms of δf
~(~r, t) =2
h3
∫d3p
[~v~p δf(~r, ~p, t) + ~v~p δ(ε(p) − µ)
2V
h3
∫d3p′ λ(~p, ~p ′)δf(~r, ~p ′, t)
]
The second term in the preceding equation is rewritten by exchanging the integration variables ~p and ~p ′
2
h3
2V
h3
∫d3p δf(~r, ~p)
∫d3p′ ~v~p ′δ(ε(p′) − µ)λ(~p, ~p ′)
and we may write
~(~r, t) =2
h3
∫d3p~~p δf(~r, ~p, t)
with
~~p = ~v~p +2V
h3
∫d3p′ ~v~p ′δ(ε(~p ′) − µ)λ(~p, ~p ′)
9. In order to compute the integral ~Ip
~Ip =
∫d3p′ ~v~p ′δ(ε(p′) − µ)λ(~p, ~p ′)
we take the z-axis along ~p and denote by (θ′, φ′) the polar angles of ~p ′ (or ~v~p ′)
~v~p ′ = vp(sin θ′ cosφ′, sin θ′ sinφ′, cos θ′)
Only the component along ~p survives the φ′ integration
~Ip = p
∫p′
2dp′ d(cos θ′)dφ′
[∑
l
αℓPℓ(cos θ′)
]cos θ′ δ(ε(p) − µ)
This equation is simplified thanks to the identity for Legendre polynomials (see question 2)
1∫
−1
d(cos θ′)Pl(cos θ′) cos θ′ =2
3δl1
with the final result
~Ip =1
3~vpα1
∫d3p′ δ(ε(p′) − µ) =
1
3~vpα1
h3
2Vρ(µ)
We thus have two possible expressions for the current
66
• ~(~r, t) =2
h3
∫d3p
~p
m∗
(1 +
α1
3ρ(µ)
)δf(~r, ~p, t)
• ~(~r, t) =2
h3
∫d3p
~p
mδf(~r, ~p, t)
and comparison of the two expressions leads to the relation between the mass of the free particle m andthe effective mass m∗
1
m=
1
m∗
(1 +
1
3Λ1
)Λ1 = α1ρ(µ)
Problem 8.6.8 Calculation of the coefficient of thermal conductivity
1. Using the relation between pressure and density for an ideal gas
n(z) =P
kT (z)= Pβ(z)
allows one to write the local equilibrium distribution (8.115) in the form
f0(z) = P [β(z)]5/2
(2πm)3/2exp
(−β(z)p2
2m
)
Let us compute the drift term of the Boltzmann equation (ε(p) = p2/(2m))
∂f0∂z
=
(5
2βf0 − ε(p)f0
)∂β
∂z= − β
T (z)
(5
2kT − ε(p)
)f0∂T
∂z
so that the drift term is, with cP = 5k/2
Df = vz∂f
∂z= − β
T (z)(cPkT − ε) f0
∂T
∂z
2. We multiply the linearized Boltzmann equation
(ε(p) − cPT )pz =mT
β
∫ 4∏
i=2
d3piWf02∆Φ
by f01Φ(~p1) and integrate over ~p1 to get
∫d3p ε(p)pzf0(p)Φ(~p) =
Tm
4β
∫ 4∏
i=1
d3piWf01f02(∆Φ)2
where we have used the condition that the fluid is at rest∫
d3p pzf = −∫
d3pf0(p)Φ(~p)pz = n〈pz〉 = 0
and the symmetry properties of the Boltzmann equation.
3. The energy current jEz is
jEz =
∫d3p fεvz = −
∫d3pΦ(~p)vzεf0(p)
∂T
∂z
From the definition (6.18) of the coefficient of thermal conductivity, we can identify κ
κ =1
m
∫d3p ε(p)pzf0(p)Φ(~p)
67
With
|X〉 = (ε(p) − cPT )pz = L|Φ〉
and from the definition of the scalar product 〈F |G〉 we can write
κ = 〈Φ|X〉 = 〈Φ|L|Φ〉
We define the positive semi-definite scalar product (Ψ,Φ)
(Ψ,Φ) = (Ψ|L|Φ)=T
4β
∫ 4∏
i=1
d3pi ∆Ψf01Wf02∆Φ
Then, restricting ourselves to functions orthogonal to the zero modes, we define κ[Ψ]
κ[Ψ] =(Ψ,Φ)2
(Ψ,Ψ)≤ (Ψ,Ψ)(Φ,Φ)
(Ψ,Ψ)= (Φ,Φ) = κ
4. From the symmetry properties of the problem, we may write Φ(~p) = A(p)pz. In order for the averagevelocity of the gas to vanish, we have seen previously that the function A(p) must obey
∫d3p f0(p)Φ(~p)pz =
∫d3p p2
zA(p)f0(p) = 0
This condition also ensures that Ψ(~p) is orthogonal to the five zero modes. Writing A(p) = A(1 − γp2)we determine γ from the condition
0 =
∫d3p p2
z(1 − γp2)f0(p) ∝∞∫
0
p4dp (1 − γp2) exp
(−βp
2
2m
)
∝ 23/2Γ
(5
2
)(m
β
)5/2
− 25/2Γ
(7
2
)(m
β
)7/2
= 23/2Γ
(5
2
)[1 − 5
2γm
β
]
with the result γ = β/(5m).
5. Starting from
|X〉 = (ε(p) − cPT )pz = L|Φ〉
we obtain
〈Ψ|X〉 =1
m
∫d3p (ε(p) − cPT )pzf0(p)A(1 − γp2)pz
=1
m
∫d3p ε(p)p2
zf0(p)A(1 − γp2)
where we have used the result of the preceding question. The explicit calculation of 〈Ψ|X〉 gives
〈Ψ|X〉 =An
2m2
(β
2πm
)3/2 ∫d3p p2p2
z(1 − γp2) exp
(−βp
2
2m
)
=4π
3
An
2m2
(β
2πm
)3/2[25/2Γ
(7
2
)(m
β
)7/2
− γ 27/2Γ
(9
2
)(m
β
)9/2]
= −Anβ2
68
Let us finally compute 〈Ψ|L|Ψ〉. We first evaluate ∆Ψ
∆Ψ = (1 − γp21)p1z + (1 − γp2
2)p2z − (1 − γp23)p3z − (1 − γp2
4)p4z
= −γ[p21p1z + p2
2p2z − p23p3z − p2
4p4z
]
where we have taken momentum conservation into acount. Using the center-of-mass variables ~p and ~P ,we have for example
p21p1z =
(~p 2 + ~p · ~P +
1
4~P 2
)(pz +
1
2Pz
)
so that∆Ψ = 2
[(~p · ~P )pz − (~p ′ · ~P )p′z
]
and
〈Ψ|L|Ψ〉 =A2Tγ2
β
∫ 4∏
i=1
d3piWf01f02
[(~p · ~P )pz − (~p ′ · ~P )p′z
]2
Now we can write the square backet of the preceding equation as
[. . .]2 = (piPi)(pjPj)p2z + (p′iPi)(p
′jPj)p
′z2 − 2(piPj)(p
′iPj)pzp
′z
Let us take the angular average over the direction of ~P at fixed ~p, ~p ′
PiPj → 1
3P 2δij
and
〈[. . .]2〉ang,P =1
3
[p2p2
z + p′2p′z
2 − 2(~p · ~p ′)pzp′z
]
It remains to take an angular average over the directions of ~p and ~p ′ when the angle θ between ~p and ~p ′
is fixed
〈[. . .]2〉ang,P,p,p ′ =2
9P 2p4(1 − cos2 θ)
Gathering all preceding results, using the change of variables (~p1, ~p2) → (~p, ~P ) and
∫d3p3 d3p4W → 2p
m
∫dΩσ(p,Ω)
leads to
〈Ψ|L|Ψ〉 =4A2Tγ2n2
9βm
(β
2πm
)3 ∫d3P P 2 exp
(−βP
2
4m
)d3p p5 exp
(−βp
2
m
)σ(p,Ω)(1 − cos2 θ)
=32A2Tn2
25√π
(m
β3
)(β
m
)1/2
σtr
where in obtaining the last line we have assumed the cross-section to be independent of p. We finallyobtain
κ =|〈Ψ|X〉|2〈Ψ|L|Ψ〉 =
25√π
32
k
m
√mkT
σtr
69
Solutions for Chapter 9
Exercise 9.7.1 Linear response: The forced harmonic oscillator
1. From our convention for the time-Fourier transform
f(ω) =
∫dt eiωt f(t)
we have ∂t → −iω, and the equation of motion for the forced oscillator becomes in Fourier space
(−ω2 − iωγ + ω2
0
)x(ω) =
f(ω)
m
The susceptibility is by definition
χ(ω) =x(ω)
f(ω)=
1
m(−ω2 − iωγ + ω20)
The poles of χ(ω) are located as follows
1. Underdamped case: γ < 2ω0
ω± = −iγ
2
[1 ± i
(4ω2
0
γ2− 1
)1/2]
2. Overdamped case: γ > 2ω0
ω± = −iγ
2
[1 ±
(1 − 4ω2
0
γ2
)1/2]
2. In the strong friction limit ω0/γ ≪ 1, The susceptibility χ(ω) is large only for
ω ∼ ω20
γ=
(ω0
γ
)ω0 ≪ ω0
Thus we may neglect ω2 with respect to ω20 and write
χ(ω) =1
m
1
ω20 − iωγ
=χ
1 − iωττ =
γ
ω20
χ =1
mω20
We also have
χ′′(ω) =1
m
ωγ
(ω2 − ω20)
2 + ω2γ2→ ωτ−1
ω2 + τ−2χ
where the second expression holds in the strong friction limit.
3. Let us write the response as
x(t) =
∫dt′ χ(t′)f(t− t′)
and the average work
⟨dW
dt
⟩
T=
1
T
T∫
0
dt f(t)x(t) = − 1
T
T∫
0
dt f(t)x(t)
The boundary terms from the integration by parts may be neglected as they tend to zero in the limitT → ∞. We thus have
⟨dW
dt
⟩
T=
1
4T
T∫
0
iω(fωe−iωt − f∗
ωeiωt)dt
∞∫
−∞
χ(t′)(fωe−iω(t−t′) + f∗
ωeiω(t−t′))
dt′
70
On performing the t-integration and taking the large T limit, only the terms which are ω-independentsurvive, and the result of the t-integration is
1
4|fω|2
(e−iωt′ − eiωt′
)= − i
2|fω|2 sinωt′
Then, using χ(t) = 2iθ(t)χ′′(t), we obtain
⟨dW
dt
⟩
T=
ω|fω|22
∞∫
0
dt χ(t) sinωt =iω|fω|2
2
∞∫
0
dt χ′′(t) sinωt
=ω|fω|2
2
∞∫
0
dt χ′′(t)(eiωt − e−iωt
)=
1
2ω|fω|2χ′′(ω)
where we have used the symmetry property χ′′(t) = −χ′′(−t) in obtaining the last line. One can also usethe method of exercise 9.7.6.
Exercise 9.7.6 Dissipation
We writeχ(ω) = χ′(ω) + iχ′′(ω)
where χ′ and χ′′ are in general, but not always, real functions. In all cases they are related through aHilbert transfrom
χ′(ω) =P
π
∫dω′χ′′(ω′)
ω′ − ω
which shows that the parities of χ and χ′′ are opposite
χ′′(ω) = ∓χ′′(−ω) ⇐⇒ χ′(ω) = ±χ′(−ω) (104)
In the usual case χ′′ is an odd function of ω, so that the upper sign holds in the preceding equation.Using the summation convention over repeated indices, we write
δAi(t) =
∫dt′χij(t
′)fj(t− t′)
and ⟨dW
dt
⟩= − 1
T
∫ T
0
dt fi(t)δAi(t)
to obtain
⟨dW
dt
⟩=
1
4T
T∫
0
iω[fω
i e−iωt − fω∗i eiωt
]dt
∫χij(t
′)[fω
i e−iω(t−t′) + fω∗i eiω(t−t′)
]dt′
After performing the t-integration, we are left with
⟨dW
dt
⟩=
iω
4
∫ [fω
i fω∗j e−iωt′ − fω
j fω∗i eiωt′
]χij(t
′)dt′
=iω
4
[fω
i fω∗j χij(−ω) − fω
j fω∗i χij(ω)
]
= − iω
4fω∗
i fωj [χij(ω) − χji(−ω)]
Now, time translation invariance combined with hermiticity yields the symmetry properties
χ′′ij(ω) = −χ′′
ji(−ω) χ′ij(ω) = χ′
ji(−ω)
71
so that we are left with ⟨dW
dt
⟩= − iω
4fω∗
i
[2iχ′′
ij(ω)]fω
j
and the final result in the form ⟨dW
dt
⟩
T=
1
2
∑
i,j
fωi∗ωχ′′
ij(ω)fωj
Exercise 9.7.9 Strong friction limit: the harmonic oscillator
1. We first take the time-Fourier transform of the equation of motion which gives the relation betweenX(ω) and F (ω)
(−ω2 − iγω + ω2
0
)X(ω) =
F (ω)
m(105)
and we then define the function XT (t) as
XT (t) = X(t) if |t| ≤ T
2XT (t) = 0 if |t| > T
2
From the Wiener-Kinchin theorem (9.170) we get
〈XT (ω)X∗T (ω)〉 = TCxx(ω)
〈FT (ω)F ∗T (ω)〉 = TCFF (ω)
Given the relation (105) between X(ω) and F (ω), we deduce that
Cxx(ω) =1
m2
CFF (ω)
(ω2 − ω20)
2 + γ2ω2
If the autocorrelation of the force is given by (9.125), then
CFF (ω) = 2A
and one obtains
Cxx(ω) =1
m2
2A
(ω2 − ω20)
2 + γ2ω2
2. In the strong friction approximation, the autocorrelation function
Cxx(ω) ≃ 2A/m2
ω40 + γ2ω2
(106)
is peaked around ω = 0. One notes that
Cxx(ω = 0) =1
ω40
Cxx
(ω2
0
γ
)≃ 1
2ω40
so that Cxx(ω) is peaked at ω = 0 with a width ∼ ω20/γ. Then the width τx of Cxx(t) is of order of
τx ∼ γ
ω20
One should note the following hierarchy in the strong friction limit
ω20
γ≪ ω0 ≪ γ
3. From P = m2X, we deduceCpp(ω) = mω2Cxx(ω)
72
and
Cpp(ω) =2Aω2
(ω2 − ω20)
2 + γ2ω2
The autocorrelation function Cpp(ω) has maxima around ω = ±ω0 and we note the following remarkablevalues
Cpp(ω0) =1
γ2Cpp
(ω2
0
γ
)=
1
2γ2Cpp(γ) ≃
1
2γ2
Thus the width of Cpp(ω) is of order γ and the autocorrelation time τp ≃ 1/γ. In the strong friction limit
τx ≃ γ
ω20
≫ τp ≃ 1
γ
This result means that the particle acquires its limit velocity over a short time on the order of τp, andthen it evolves much more slowly under the combined influence of the harmonic force and friction overa time on the order of τx. The momentum distribution is almost Maxwellian for t ≫ τp, because theparticle has enough time to reach equilibrium in momentum space.
4. Neglecting the inertial term in the equation of motion leads to
(−iγω + ω2
0
)X(ω) =
F (ω)
m
and one gets directly the strong friction limit (106) of the autocorrelation function. The correspondingequation of motion in t-space is
X = −ω20
γ+F (t)
mγ
One can read directly on this equation that τx = γ/ω20. The x-space viscosity is γ = ω2
0/γ.
Problem 9.7.1 Inelastic light scattering from a suspension of particles
1. Let us start from the standard Maxwell equation in the presence of a polarized material
~∇× ~B = c2~∇× ~∇× ~A =1
ε0
∂ ~P
∂t+∂ ~E
∂t
From the decomposition of the polarization
~P = ε0(ε− 1) ~E(~r, t) + δ ~P (~r, t)
we get
1
ε0
∂ ~P
∂t+∂ ~E
∂t= ε
∂ ~E
∂t+
1
ε0
∂δ ~P
∂t
and the equation for ~A becomes
c2[~∇ · (~∇ · ~A) −∇2 ~A
]= ε
∂ ~E
∂t+
1
ε0
∂δ ~P
∂t
= −ε∂2 ~A
∂t2− ε~∇∂ϕ
∂t+
1
ε0
∂δ ~P
∂t
where we have used
~E = −~∇ϕ− ∂ ~A
∂t
Choosing the gaugeε
c2∂ϕ
∂t+ ~∇ · ~A = 0 (107)
73
leads to the equation for the vector potential
∇2 ~A− ε
c2∂2 ~A
∂t2= − 1
ε0c2∂δ ~P
∂t(108)
The equation for the scalar potential ϕ is obtained from
~∇ · ~E = − 1
ε0~∇ · ~P
or
∇2ϕ+ ~∇ · ∂~A
∂t=
1
εε0c2∂δ ~P
∂t
Using the gauge condition (107), we easily obtain the equation for ϕ
∇2ϕ− ε
c2∂2ϕ
∂t2=
1
εε0~∇ · (δ ~P )
2. One checks at once that the expressions for ϕ and ~A
ϕ = −~∇ · ~Z ~A =ε
c2∂ ~Z
∂t
are consistent with the gauge condition (107). The following partial differential equation for ~Z
∇2 ~Z − ε
c2∂2 ~Z
∂t2= − 1
εε0δ ~P (109)
gives back the equations for ϕ and ~A. The expression for ~E as a function of ~Z is
~E = −εc
∂2 ~Z
∂t2+ ~∇ · (~∇ · ~Z) (110)
3. Let us take the time-Fourier transform of ~Z(~r, t)
~Z(~r, ω) =
∫dt eiωt ~Z(~r, t)
Then equation (109) reads, with k =√εω/c
(∇2 + k2
)~Z(~r, ω) = − 1
ε0εδ ~P (~r, ω) (111)
The varying polarization is located in a small region in space, while we are interested in measuringthe scattered electromagnetic field at large distances. We shall use a technique which is familiar fromscattering theory, by introducing the Green functions G±(~r)
(∇2 + k2
)G±(~r) = −4πδ(~r)
It is easily checked that the functions
G±(~r) =e±ikr
r
are solutions of the preceding equation. An easy proof is as follows. The Laplacian of a sphericallysymmetric function is
∇2f(r) =1
r
d2
dr2[rf(r)]
74
provided f(r) is regular at r = 0. We thus write
∇2 e±ikr
r= ∇2 e±ikr − 1
r−∇2 1
r
The first function is regular at r = 0
∇2 e±ikr − 1
r=
1
r
d2
dr2(e±ikr − 1
)= −k2 e±ikr
r
while
∇2 1
r= −4πδ(~r)
Equation (111) is solved by using the Green function with outgoing waves
~Z(~r, ω) =1
4πεε0
∫d3r′
eik|~r−~r ′|
|~r − ~r ′| δ~P (~r ′, ω) (112)
Now, since δ ~P is confined to a small region in space, r ≫ r′, we may use the approximation
|~r − ~r ′| ≃ r − r · ~r ′
so that (112) becomes
~Z(~r, ω) ≃ 1
4πε0ε
eikr
r
∫d3r′ e−i~k·~r′
δ ~P (~r ′, ω) (113)
In order to compute ~E, we write δ ~P = δP e0, where e0 is the polarization of the incident electric field.We observe that
~∇ ·(
eikr
re0
)= (e0 · r)
[ikeikr
r− eikr
r2
]
and the second term may be neglected since kr ≫ 1. Similarly
~∇ · e0 · r)ikeikr
r≃ −k2(e0 · r)r
eikr
r
so that the electric field is proportional to
(ω2ε
c2e0 − k2re0 · r)
)=ω2ε
c2e0 − k′e0 · k′)) =
ω2ε
c2e0 × e0 × k′)
We finally get the expression for ~E
~E(~r, ω) ≃ ω2
4πε0c2eikr
r
∫d3r′ e−i~k·~r ′
k × (δ ~P (~r ′, ω) × k)
4. In order to prove the Wiener-Kinchin theorem, we start from the result of exercise 9.7.8
I(T ) =
T/2∫
−T/2
dt1
T/2∫
−T/2
dt2 g(t) = T
+T∫
−T
dt g(t)
(1 − |t|
T
)
We define
XT (ω) =
∫dt eiωtXT (t) =
T/2∫
−T/2
dt eiωtX(t)
75
Then
〈XT (ω)X∗T (ω)〉 =
T/2∫
−T/2
dt1
T/2∫
−T/2
dt2〈X(t1)X∗(t2)〉
= T
T/2∫
−T/2
dt eiωt
(1 − |t|
T
)〈XT (t)X∗
T (0)〉
≃ T
T/2∫
−T/2
dt eiωt 〈XT (t)X∗T (0)〉 = TST (ω)
5. The electromagnetic power is given by the flux of the Poynting vector
ε0 ~E × ~B = ε0c| ~E|2k
But δ ~P being parallel to e0, one can write δ ~P = e0δ ~P and use
(k × (e0 × k))2 = 1 − (e0 · k)2 = (e0 × k)2 = sin2 θ
where π/2 − θ is the scattering angle, cos θ = e0 · k. From the Wiener-Kinchin theorem
〈| ~E|2〉 =1
T
∫dω
2π〈 ~E(~r, ω) · ~E∗(~r,−ω)〉
while
〈 ~E(~r, ω) · ~E∗(~r,−ω)〉 =1
4πε0
(ωc
)4 1
r2(e0 × k)2
∫d3r′ d3r′′ e−i~k·(~r′−~r′′)〈δ ~P (~r′ω) · δ ~P ∗(~r′′,−ω)〉
=1
4πε0
(ωc
)4 1
r2(e0 × k)2〈δ ~P (~k, ω) · δ ~P ∗(~k, ω)〉
dP/(dΩ dω) is the flux of the Poynting vector per unit frequency ω and through a surface r2dΩ, so that
dPdΩ dω
=ε0c
(4πε0)2
(ωc
)4 1
2πT(e0 × k)2〈δ ~P (~k, ω) · δ ~P ∗(~k, ω)〉
6.
δ ~P (~k, ω) = αe0
∫dt d3r′ e−i(~k·~r′−ωt) ei(~k0·~r−ω0t) δn(~r, t)
= αε0 ~E0δn(~k − ~k0, ω − ω0) = αε0δn(~k ′, ω′)
so thatdP
dΩ dω=
ε0c
(4π)2α2E2
0
(ωc
)4 (e0 × k)2
2πT〈δn(~k ′, ω′)δn(−~k ′,−ω′)〉
where we have used the fact that δn(~r, t) is a real function. Using once more the Wiener-Kinchin theorem
〈δn(~k ′, ω′)δn(−~k ′,−ω′)〉 = V TSnn(~k ′, ω′)
We thus get the power per unit volume of target
dPdΩ dω
=ε0c
32π3
(ωc
)4
(e0 × k)2 α2 ~E20 Snn(~k ′, ω′)
76
7. The expression for Snn(~k ′, ω′) stems from a hydrodynamic approximation relying on the continuityequation and Fick’s law. In this case one knows from (9.40) that
χ′′(k′, ω′) = χ(k′)ω′Dk′2
ω′2 + (Dk′2)2
and from the classical form of the fluctuation-dissipation theorem (9.28) , we get
Snn(~k ′, ω′) =2
βχ(~k ′)
Dk′2
ω′2 + (Dk′2)2
8. One can write the emitted power in two different forms
dPdΩ dω
=ε0c
(4π)2
(ωc
)4
α2E20 sin2 θ
2
β
Dk′2
ω′2 + (Dk′2)2χ(~k ′)
and
dPdΩ dω
=ε0cr
2
2πT〈 ~E(kr, ω) · ~E∗(kr, ω)〉
=ε0c
2πSEE(~k′, ω)
The coefficient A is thus
A =1
8π
(ωc
)4
α2E20 sin2 θ
1
βχ(k′) ≃ A(ω0)
Taking into acount ∫dω
2π
e−iωt
ω2 + γ2=
1
2γe−γ|t|
we obtain for the autocorrelation function of ~E
〈~E(~r, t) · ~E(~r, o)〉 ≃ e−iω0t e−Dk′2tA(ω0)
9. Since ~E is a Gaussian random function, its moments of order 4 can be expressed in terms of itsmoments of order 2. Taking into account that ~E is parallel to a fixed direction and that 〈~E(t) · ~E(t)〉 = 0
〈[~E(t) · ~E∗(t)] [~E(0) · ~E∗(0)]〉 = 〈~E(t) · ~E∗(t)〉〈~E(0) · ~E∗(0)〉 + 〈~E(t) · ~E∗(0)〉〈~E(0) · ~E∗(t)〉
This leads to
〈I(t)I(0)〉 = A2(1 + e−Dk′2t)
and taking the Fourier transform
A2
[2πδ(ω) +
4Dk′2
ω2 + (2Dk′2)2
]
it is much easier to measure a deviation ω on the order of 500 Hz with respect to zero, than to measurethe same deviation with respect to 1014 Hz.
Problem 9.8.2 Light scattering from a simple fluid
1. One sets in equation (6.91) ~g = mn~u, ρ = mn and neglects the advection term (~u · ~∇)~u which isquadratic in ~u. Furthermore, let us examine the deviation δ(ρ~u) from equilibrium
δ(ρ~u) = ~uδρ+ ρδ~u
77
The first term in this equation is of second order in the deviations with respect to equilibrium and maybe neglected. A similar reasoning holds for the spatial derivatives, so that we obtain the Navier-Stokesequation, valid to first order in the deviations from equilibrium
∂t~g + ~∇P − 1
mn
(ζ +
η
3
)~∇(~∇ · ~g) − η
mn∇2~g = 0
In the expression (6.83) for ~E , one may replace Pαβ by Pδαβ since the neglected terms give contributionswhich are quadratic in ~u, and one may also replace in ǫ~u the energy density ǫ by the energy density inthe rest frame of the fluid, which is denoted by ε in the present problem.
2. Equation for ~gL: from0 = ~∇× (~∇× ~gL) = ~∇(~∇ · ~gL) −∇2~gL
we immediately get (9.195). We write the expression for the energy current
~E =ε+ P
mn~g − κ~∇T
from which we derive the continuity equation
∂tε+ε+ P
mn(~∇ · ~gL) − κ∇2T = 0
Combining with (9.193) we get (9.197).
3. The derivation follows that of (9.35), with D → η/(mn). Writing the response function of themomentum density
χgαgβ= kαkβχL(~k) − (δαβ − kαkβ)χT (~k)
we get for χT (see 9.40)
χ′′T (~k, ω) =
2
β
ωηk2/(mn)
ω2 +η2
(mn)2k4
χT (~k)
4. We start from
~∇P =∂P
∂n
∣∣∣ε
~∇n+∂P
∂ε
∣∣∣n
~∇ε
Let us first examine the second term on the right hand side of the preceding equation. From
∂P
∂ε
∣∣∣n
=∂P
∂S
∣∣∣n
∂S
∂ε
∣∣∣n
and∂ε
∂S
∣∣∣V
=1
V
∂E
∂S
∣∣∣V
=T
V
We get∂P
∂ε
∣∣∣n
=V
T
∂P
∂S
∣∣∣n
To transform the second term of ~∇P we start from
V dε = TdS − (ε+ P )dV
For dε = 0 we get
0 = T∂S
∂P
∣∣∣V
dP + T∂S
∂V
∣∣∣PdV − (ε+ P )dV
whence∂P
∂V
∣∣∣ε
=ε+ P
T
∂P
∂S
∣∣∣n
+∂P
∂V
∣∣∣S
78
where we have made use of (1.36) in the last step. Finally we use
∂
∂n= −V
n
∂
∂V
to obtain∂P
∂n
∣∣∣ε
= −V (ε+ P )
nT
∂P
∂S
∣∣∣n
+∂P
∂n
∣∣∣S
Collecting all contributions we get the final expression for ~∇P
~∇P =∂P
∂n
∣∣∣S
~∇n+V
T
∂P
∂S
∣∣∣n
~∇q
5. Let us give the interpretation of δq. At fixed N we have
δε = δ
(E
V
)=
1
VδE − δV
V 2E δn = − n
VδV
so that
δε− ε+ P
nδn =
1
VδE − E
V 2δV +
ε+ P
n
n
VδV
=1
V(δE + PδV ) =
T
VδS
We see that δq is indeed related to the entropy variation per unit volume.
6. The determinant of the matrix M is
detM =
∣∣∣∣∣∣∣∣∣∣
z − k
m0
−kmc2 z + ik2DL −VT
∂P
∂S
∣∣∣nk
ik2κ∂T
∂n
∣∣∣S
0 z + ik2 κ
mncV
∣∣∣∣∣∣∣∣∣∣
=
∣∣∣∣∣∣
z ak 0bk z + idk2 fk
igk2 0 z + ihk2
∣∣∣∣∣∣
where we have simplified the notations in order to compute the determinant by setting
a = − 1
mb = −mc2 d = DL f = −V
T
∂P
∂S
∣∣∣n
g = κ∂T
∂n
∣∣∣S
h =κ
mncV
7. The expression for the determinant reads
detM =[z + ihk2
] [z(z + idk2) − c2k2
]+ iafgk4
In the low temperature approximation we may neglect the last term in the preceding expression. Thezeroes of the determinant are then obvious
1. z = O(k2) or more precisely
z = −ihk2 = − iκ
mncVk2 = −iDTk
2
Because CP ≃ CV in the low temperature limit, we may write
DT =κ
mncP
79
2. z = ±ck + δ, |δ| ≪ ck One solves the equation for z to first order in δ
z = ±ck − i
2DLk
2
In the general case, one must evaluate, recalling that the sound velocity c2 = (∂P/∂ρ)S , ρ = mn
afg
c2= −
(∂n
∂P
∣∣∣S
)(κ∂T
∂n
∣∣∣S
)(−VT
∂P
∂S
∣∣∣n
)
=κV
T
∂T
∂P
∣∣∣S
∂P
∂S
∣∣∣n
Remember that we are working at constant N , which means that constant n implies constant V . Thenthe Maxwell relation
∂P
∂S
∣∣∣V
= − ∂T
∂V
∣∣∣S⇒ ∂P
∂S
∣∣∣n
= − ∂T
∂V
∣∣∣S
impliesafg
c2= −κV
T
∂T
∂P
∣∣∣S
∂T
∂V
∣∣∣S
while∂T
∂P
∣∣∣S
= − ∂S
∂P
∣∣∣T
∂T
∂S
∣∣∣P
=T
CP
∂V
∂T
∣∣∣P
where we have used (1.28). From an analogous calculation
∂T
∂V
∣∣∣S
=T
CV
∂P
∂T
∣∣∣V
so thatafg
c2=
κV T
CPCV
∂P
∂T
∣∣∣V
∂V
∂T
∣∣∣P
We finally use
CP − CV = T∂P
∂T
∣∣∣V
∂V
∂T
∣∣∣P
to obtainafg
c2=
κV
CPCV(CP − CV ) =
κ(cP − cV )
mncP cV
where CV,P = mnV cV,P , mcV,P being the specific heat per particle. We may now revert to the calculationof the zeroes of the determinant.
1. Solution z = O(k2). Neglecting terms of order k4 we get
z ≃ −i
(h− afg
c2
)k2 = − iκ
mncPk2
because
h− afg
c2=
κ
mncP
2. Solution z = ±ck + δ. One solves to first order in δ
2δ ≃ −idk2 − iafg
c2k2
or
δ = − i
2k2
[DL +
κ
mncP
(cPcV
− 1
)]
80
Problem 9.8.4 Ito versus Stratonovitch dilemma
1. Preliminary result. In the evaluation of the Chapman-Kolmogorov equation, one must compute thefollowing integral
I =
∫dy f(y)δ(x− y − g(y))
The argument of the delta-function vanishes for y = y0 given by
x = [y + g(y)]y=y0
Differentiating this equation with respect to x gives
1 =
[dy
dx(1 + g′(y))
]
y=y0
Using the standard formula
δ(x− y − g(y)) =1
|1 + g′(y)|δ(y − y0) =dy
dx
∣∣∣y=y0
δ(y − y0)
and writing y0 = f(x) we getδ(x− y − g(y)) = f ′(x)δ(y − f(x))
The value of y0 is given by the equation
y0 = x− εa(y0) −Bε
√D(y0)
We solve this equation by iteration to order ε and B2ε (remember that Bε is formally of order
√ε)
y0 ≃ x− εa(x) −BεD1/2(x−BεD
1/2(x))
≃ x− εa(x) −BεD1/2(x) +
1
2B2
εD′(x) = f(x)
Since
f ′(x) = 1 − εa′(x) − 1
2Bε
D′(x)
D1/2(x)+
1
2B2
εD′′(x)
the Chapman-Kolmogorov equation (9.144) leads to
J =
∫dy P (y, t|x0)
(1 − εa′(x) − 1
2Bε
D′(x)
D1/2(x)+
1
2B2
εD′′(x)
)
×[δ(y − x) +
(εa(x) +BεD
1/2(x))− 1
2B2
εD′(x))δ′(y − x) +
1
2B2
εDδ′′(y − x)
]
Using the average values Bε = 0 and B2ε = 2ε we get for the average over all realizations of b(t)
J = P (1 − εa′(x) + εD′′(x)) − εa(x)P ′(x) + 2εD′(x)P ′(x) + εD(x)P ′′(x)
= P (x) − ε∂
∂x[a(x)P (x)] + ε
∂2
∂x2[D(x)P (x)]
where P (x) = P (x, t|x0). This leads at once to the F-P equation
∂P
∂t= − ∂
∂x[a(x)P ] +
∂2
∂x2[D(x)P ]
2. We now have
X(t+ ε) − y = εa(y) +D1/2 [y + (1 − q)(X(t+ ε))]
∫ t+ε
t
b(t′)dt′
≃ εa(y) +
[D1/2(y) +
(1 − q)D′(y)
2D1/2(y)(X(t+ ε) − y)
]∫ t+ε
t
b(t′)dt′
81
Expanding to order ε and taking the average value over all realizations of b(t) yields
X(t+ ε) − y = εa(y) + (1 − q)D′(y)
∫ t+ε
t
dt′∫ t+ε
t
dt′′ b(t′)b(t′′)
= ε(a(y) + (1 − q)D′(y))
In other words, the only modification with respect to question 1 is in the drift term
a(y) → a(y) + (1 − q)D(y)
so that the F-P equation becomes
∂P
∂t= − ∂
∂x[a(x)P (x)] − (1 − q)
∂
∂x[D′(x)P (x)] +
∂2
∂x2[D(x)P (x)]
=∂
∂x[a(x)P (x)] +
∂
∂x
[D1−q(x)
∂
∂x(Dq(x)P (x))
]
Problem 9.8.5 Kramers equation
1. Let us compute the relevant moments of p and x.
1. ∆p
p(t+ ε) − p(t) = (F (x) − γp)ε+
∫ t+ε
t
f(x, t′)dt′
Taking an average over all values of f gives
1
ε∆p = lim
ε→0
1
ε[p(t+ ε) − p(t)] = F (x) − γp
2. ∆x
x(t+ ε) − x(t) =1
m
∫ t+ε
t
p(t′)dt′ =1
m
∫ t+ε
t
dt′[p+
∫ t′
t
p(t′′)dt′′]
=p
mε+
1
m
∫ t+ε
t
dt′[p+
∫ t′
t
dt′′f(x, t′′)
]+ O(ε2)
so that1
ε∆x =
p
m
3. ∆p2
(p(t+ ε) − p(t))2
= (F (x) − γp)2ε2 + 2ε(F (x) − γp)
∫ t+ε
t
f(x, t′)dt′
+
∫ t+ε
t
∫ t+ε
t
dt′ dt′′ f(x, t′)f(x, t′′)
Taking the average, the first term is of order ε2, the second one vanishes and the third one is equalto 2Aε so that
1
ε∆p2 = 2A
4. ∆x∆p.
It is easy to see that this moment is of order ε2.
82
5. ∆x2
It is also easy to see that this moment is at least of order ε2.
Taking into account the results of exercise 9.7.11 with x1 = p and x2 = x, we have in the notations ofthis exercise
A1 = F (x) − γ(x)p A2 =p
mA11 = A(x)
from which the F-P equation follows
∂P
∂t= − ∂
∂p([F (x) − γp]P ) − ∂
∂x
( pmP)
+∂2
∂p2(A(x)P )
We have taken from the beginning x-dependent γ and A. One remarks that this choice does not implyany modification in the F-P equation since the p-derivative does not act on an x-dependent function.With
kT (x) =A(x)
mγ
we arrive at [∂
∂t+
p
m
∂
∂x+ F (x)
∂
∂p
]P = γ(x)
[∂
∂p(pP ) +mkT
∂2P
∂p2
]
2. Let us differentiate ρ(x, t) with respect to time
∂ρ
∂t=
∫dp
∂P
∂t=
∫dp
[− p
m
∂P
∂x− F (x)
∂P
∂p+ γ(x)
∂
∂p(pP ) +mkT (x)
∂2P
∂p2
]
An integration by parts shows that many terms vanish, for example
∫dpF (x)
∂P
∂p= [F (x)P (p)]∞−∞ = 0
because the distribution P (p) vanishes at infinity. One is thus left with
∂ρ
∂t= −
∫dp
p
m
∂P
∂x= − ∂j
∂x
In the strong friction limit, the particle reaches almost instantaneously its limiting momentum p givenby p = 0
p(x) =F (x)
γ(x)
The momentum follows a Maxwell distribution centered at p = p(x), whence
P (x, p; t) ≃ ρ(x, t)
√1
2πmkT (x)exp
(− [p− p(x)]2
2mkT (x)
)
3. Let us multiply both sides of the F-P equation by p and integrate over p
∫dp
p2
m
∂P
∂x+
∫dpF (x)p
∂P
∂t= γ(x)
∫dp p
∂(pP )
∂p+ γ(x)
∫dpmkT (x)
∂2P
∂p2
Once more some terms vanish after an integration by parts and we are left with
m∂
∂tj(x, t) +
∂
∂xK(x, t) − F (x)ρ(x, t) = −γ(x)mj(x, t)
where we have used ∫dp p
∂(pP )
∂p= −
∫dp (pP )
∂p
∂p= −
∫dp (pP ) = −mj
83
K is of course related to the average kinetic energy of the particle
1
2K =
p2
2m
Setting σ2 = mkT we have
1√2πσ2
∫dp p2 exp
(− (p− p)2
2σ2
)=
1√2πσ2
∫dp′ (p′ + p)2 exp
(− p′2
2σ2
)
= p2(x) + σ2(x) = p2(x) +mkT (x)
so that
K(x, t) ≃ ρ(x, t)
[kT (x) +
p2(x)
m
]
In the strong friction limit, the particle reaches its limiting speed in a time on the order of the velocitycorrelation time ∼ 1/γ, and then it evolves slowly with a characteristic time 1/γ ≫ 1/γ
∣∣∣∂j
∂t
∣∣∣ ∼ γ|j| ≪ γ|j|
Neglecting ∂j/∂t, we obtain an approximate expression for j
j(x, t) =1
mγ(x)
[F (x)ρ(x, t) − ∂K
∂x
]
and if kT (x) ≫ p2(x)/m
∂
∂xK(x, t) =
∂
∂x[kT (x)ρ(x, t)] =
∂
∂x[mγ(x)D(x)ρ(x, t)]
where we have used Einstein’s relation which gives the x-dependent diffusion coefficient in x-space
D(x) =kT (x)
mγ(x)
The final expression for the current is then
j(x, t) =1
mγ(x)
[F (x)ρ(x, t) − ∂
∂x
[mγ(x)D(x)ρ(x, t)
]]
and the F-P equation for ρ(x, t) reads∂ρ
∂t+∂j
∂x= 0
4. If F (x) = 0 and γ is x-independent, the expression for the current reads
j(x, t) = − 1
m2γ2
∂
∂x[A(x)ρ(x, t)]
and the F-P equation is∂ρ
∂t=
1
m2γ2
∂2
∂x2[A(x)ρ(x, t)]
The corresponding Langevin equation is
x =
√A(x)
mγη(t) η(t)η(t′) = 2δ(t− t′)
with the Ito prescription.
84
On the contrary, if A is x-independent the expression for the current reads
j(x, t) = −√A
mγ(x)
∂
∂x
[ √A
mγ(x)ρ(x, t)
]
and the F-P equation is
∂ρ
∂t=
∂
∂x
√A
mγ(x)
∂
∂x
[ √A
mγ(x)ρ(x, t)
]
The corresponding Langevin equation is
x =
√A
mγ(x)η(t) η(t)η(t′) = 2δ(t− t′)
with the Stratonovitch prescription.
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2
0.9
0.95
1
1.05
1.1<E
>
Quantum SHO: λ=0, ω=2, m=1, β=20Line is a fit: Intercept=1.036, exact=1.037
# $% &
? % = , - #
−4 −3 −2 −1 0 1 2 3 4x
0
0.002
0.004
0.006
0.008
P(x)
Quantum SHO: λ=0, ω=2, m=1Solid: β=20, dashed (both): β=0.33, long dashed: β=0.25
' () * & * * & !
? & 1 <# , = 1 <
−2 −1 0 1 2x
0
0.002
0.004
0.006
0.008
0.01P(
x)
Quantum AHO: ω=2, m=1, β=20, Lt=400Bottom to top: λ=0, 1, 2, 3, 4
() $!+% ' $ %
= /
$ ! * + $
*
/
/
0 < ) ( $ ! * *+$
0 2 4 6 8 λ
0.5
0.6
0.7
0.8
Kine
tic e
nerg
y
Kinetic energy: Virial (circles) and direct (star)
, - "
= ! * +%
,,, ,, , , +
,, , ,
,, , ,
,,
8 * +* ?$ $
,, ,$ 0 $ $ $ # ! * ++$
,,, .
,
? * ! ? , ( 1 < # ( = < .+ 1 #
0 1 2 3 4τ
10−5
10−4
10−3
10−2
10−1
100
G(τ
)
Top to bottom: λ=0,1,2,3,4: G(τ) ~ exp(−τ[E1−E0])ω=2, λ=0,1,2,3,4. ∆E= 1.98, 2.156, 2.289, 2.374, 2.47
λ E1
0 2.981 3.1982 3.36683 3.48664 3.6126
. $% / ( " $ %
0 < % ! = , ' , < + ? * % $ $ $ , $ # ? #
< # , $ $ ' ( $ ( 0 ( # ? $ #
−4 −3 −2 −1 0 1 2 3 4x
0
0.005
0.01
0.015
0.02
P(x)
Quantum AHO: ω=2, m=1, λ=8, Lt=400Bottom to top: β=20, 5, 1, 0.1 and 0.1
0 1 0 ( " + " "