Problemas Resueltos Del Reichl.tmp

Problems for the course

Statistical Physics

(FYS3130)

Prepared by Yuri Galperin

Spring 2004

Contents

1 General Comments 5

2 Introduction to Thermodynamics 72.1 Additional Problems: Fluctuations. . . . . . . . . . . . . . . . . . . . . . . . . 282.2 Mini-tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

2.2.1 A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302.2.2 B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3 The Thermodynamics of Phase Transitions 333.1 Mini-tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

3.1.1 A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 443.1.2 B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

4 Elementary Probability Theory . . . 49

5 Stochastic dynamics ... 59

6 The Foundations of Statistical Mechanics 77

7 Equilibrium Statistical Mechanics 79

8 Tests and training 103

A Additional information 107A.1 Thermodynamics. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

A.1.1 Thermodynamic potentials. . . . . . . . . . . . . . . . . . . . . . . . . 107A.1.2 Variable transformation. . . . . . . . . . . . . . . . . . . . . . . . . . . 107A.1.3 Derivatives from the equation of state. . . . . . . . . . . . . . . . . . . 107

A.2 Main distributions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109A.3 The Dirac delta-function. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110A.4 Fourier Series and Transforms. . . . . . . . . . . . . . . . . . . . . . . . . . . 111

B Maple Printouts 115

3

4 CONTENTS

Chapter 1

General Comments

This year the course will be delivered along the lines of the book [1]. The problems will also beselected from this book. It iscrucially importantfor students to solve problems independently.

The problems will be placed on the course homepage. The same page will contain solutionsof the problems. So if a student is not able to solve the problem without assistance, then she/heshould come through the solution. In any case, student have to able to present solutions, obtainedindependently, or with help of the course homepage.

FYS 3130 (former FYS 203) is a complicated course, which requires basic knowledge ofclassical and quantum mechanics, electrodynamics, as well as basics of mathematics.

Using a simple test below please check if your knowledge is sufficient. Answers can be foundeither in theMaple file.

A simple test in mathematics

Elementary functions

Problem 1.1: Function is given by the definition

f (x) = x4 +ax3 +bx2 .

(a) Show that by a proper rescaling it can be expressed as

f (x) = a4Fη(ξ) where Fη(ξ) = ξ4 +ξ3 +ηξ2 ,

ξ≡ x/a, η≡ b/a2. FunctionFη(ξ) is very important in theory of phase transitions.

(b) InvestigateFη(ξ).

– How many extrema it has?

– When it has only 1 minimum? When it has 2 minima? At what value ofη it has aninflection point?

5

d:/Documents/Lectures/Statphys/Reichl/test1.mws

6 CHAPTER 1. GENERAL COMMENTS

– Plot Fη(ξ) for this value ofη.

Problem 1.2: Logarithmic functions are very important in statistical physics. Check you mem-ory by the following exercises:

• Plot function

f1(x) = ln1−x1+x

for |x| ≤ 1.

Discuss properties of this function at|x|> 1.

• Plot functionf2(x) = ln(tanπx) .

• Simplify the expressione4lnx− (x2 +1)2 +2x2 +1.

Problem 1.3: Do you remember trigonometry? Test it!

• Simplifysin−2x− tan−2x−1.

• Calculate infinite sums

C(α,β) =∞

∑n=0

e−αncosβn, α > 0.

S(α,β) =∞

∑n=0

e−αnsinβn, α > 0.

Hint: take into account thatcosx = Re(eix

), sinx = Im

(eix

).

Basic integrals

Problem 1.4: Calculate interglars:

•In(γ) =

∫ ∞

0xne−γxdx, γ > 0.

• ∫dx

x2±a2 ,∫

dxx(1−x)

.

•G(γ) =

∫ ∞

0e−γx2/2dx, γ > 0.

Chapter 2

Introduction to Thermodynamics

Quick access:1 2 3 4 5 6 7 8 9 10111213141516171819202122

Problem 2.1: Test the following differentials for exactness. For those cases in which the dif-ferential is exact, find the functionu(x,y).

(a) dua = −ydxx2+y2 + xdy

x2+y2 .

(b) dub = (y−x2)dx+(x+y2)dy .

(c) duc = (2y2−3x)dx−4xydy .

Solution 2.1:

(a) The differential is exact,ua(x,u) =−arctan(x/y).

Here one point worth discussion. The functionua(x,u) = −arctan(x/y) has a singularityatx = 0,y = 0. As a result, any close path embedding this point contributes2π to the vari-ation of the quantityu(x,y). Consequently, this function cannot serve as a thermodynamicpotential if both positve and negative values ofx andy have physical meaning.

(b) The differential is exact,ub(x,y) = yx+(y3−x3)/3.

(c) The differential is not exact.

The functionu(x,y) is reconstructed in the following way. For an exact differential,du= uxdx+uydy,

ux =∂u(x,y)

∂x, uy =

∂u(x,y)∂y

.

If we introduce

u1(x,y) =∫ x

0ux(ξ,y)dξ ,

7

8 CHAPTER 2. INTRODUCTION TO THERMODYNAMICS

then the differencef ≡ u(x,y)−u1(x,y) is a function only ofy. Consequently,

d fdy

=∂u(x,y)

∂y− ∂u1(x,y)

∂y= uy(x,y)−

∫ x

0

∂ux(x,y)∂y

dx.

As a result,

f (y) =∫ y

0dηuy(x,η)−

∫ y

0

∫ x

0dξdη

∂ux(ξ,η)∂η

.

Finally,

u(x,y) =∫ x

0dξux(ξ,y)+

∫ y

0dηuy(x,η)−

∫ y

0

∫ x

0dξdη

∂ux(ξ,η)∂η

.

For calculation see the Maple fileit1.mws.

Problem 2.2: Consider the two differentials

1. du1 = (2xy+x2)dx+x2dy, and

2. du2 = y(x−2y)dx−x2dy.

For both differentials, find the changeu(x,y) between two points,(a,b) and(x,y). Compute thechange in two different ways:

(a) Integrate along the path(a,b)→ (x,b)→ (x,y),

(b) Integrate along the path(a,b)→ (a,y)→ (x,y).

Discuss the meaning of your results.

Solution 2.2: The calculations are shown in the Maplefile. In the case (b) the results aredifferent because the differential is not exact.

Problem 2.3: Electromagnetic radiation in an evacuated vessel of volumeV at equilibriumwith the walls at temperatureT (black body radiation) behaves like a gas of photons havinginternal energyU = aVT4 and pressureP = (1/3)aT4, wherea is Stefan’s constant.

(a) Plot the closed curve in theP−V plane for a Carnot cycle using black body radiation.

(b) Derive explicitly the efficiency of Carnot engine which uses black body radiation as itsworking substance.

d:/Documents/Lectures/Statphys/Reichl/it1.mws

9

1 2

34

V o l u m e

Pres

sure T h

T c

Figure 2.1:On the Carnot cycle with black-body radiation.

Solution 2.3: We will follow example shown in Exercise 2.2. Let us start withisotherms. Atthe isotherms the pressure isV-independent, thus isotherms arehorizontal, see Fig.2.1 Alongthe first isothermal path,

∆Q1→2 = ∆U +P∆V = (4/3)aT4h (V2−V1) =

43

aT4h V1

[V2

V1−1

]. (2.1)

In a similar way,

∆Q3→4 =43

aT4c V4

[1−V3

V4

]. (2.2)

Now let us consider an adiabatic path. Along an adiabatic path,

dQ= 0 = dU +PdV = 4aVT3dT +aT4dV +(1/3)aT4dV = 4aVT3dT +(4/3)aT4dV .

Consequently,dTT

=−13

dVV

→ VT3 = const, PV4/3 = const.

Let us start form the point 2 characterized by the valuesP1,V2 and adiabatically expand thegas to the point 3 characterized by the volumeV3. We haveV2T3

h = V3T3c . In a similar way,

V4T3c = V1T3

h . Combining these equalities, we get:

V2T3h = V3T3

c , V4T3c = V1T3

h → V2

V3=

V1

V4=

(Tc

Th

)3

(2.3)

Combining Eqs. (2.1), (2.2) and (2.3), we find

∆W = ∆Q1→2 +∆Q3→4 =43

aT4h V1

(V3

V4−1

)(1− Tc

Th

).


Remember: along a closed path∆U = 0 and the total heat consumption is equal to mechanicalwork.

On the

∆Q1→2 =43

aT4h V1

(V2

V1−1

).

As a result,

η =∆W

∆Q1→2=

Th−Tc

Th

as it should be.

Problem 2.4: A Carnot engine uses a paramagnetic substance as its working substance. Theequation of state is

M =nDH

TwhereM is magnetization,H is the magnetic field,n is the number of moles,D is a constantdetermined by the type of substance, andT is is the temperature.

(a) show that the internal energyU , and therefore the heat capacityCM, can only depend onthe temperature and not the magnetization.

(b) Let us assume thatCM = C = constant. Sketch a typical Carnot cycle in theM−H plane.

(c) Compute the total heat absorbed and the total work done by the Carnot engine.

(d) Compute the efficiency of the Carnot engine.

Solution 2.4:

(a) By definition [see Eq. [1](2.23)],

dU = T dS+H dM .

Thus atM = const the internal energy is independent of the magnetization.

(b) SinceC = const,U0 = NcMT, wherecM is the specific heat per one particle whileN is thenumber of particles. Introducing molar quantities we getU0 = ncT.

A Carnot cycle is shown in Fig.2.2. We have:

(c) For an isothermal process atT = Tc,

Q1→2 =−∫ H2

H1

H(M)dM =Tc

2nD(M2

1−M22) .

In a similar way,

Q3→4 =Th

2nD(M2

3−M24) .

11

M

H

1

2

34

T

T h

c

Figure 2.2:Sketch of the Carnot cycle.

The total work is thenW = Q1→2 +Q3→4, and the efficiency is

η =W

Q3→4= 1+

Tc

Th

M21−M2

2

M23−M2

4

.

Now let us discuss adiabatic paths. We have at each path,

0 = dQ= dU−H dM = ncdT−MTnD

dM .

Immediately we getdTT

=1

n2cDM dM .

Integrating this equality from point 2 to point 3 we obtain,

2n2cD lnTh

Tc= M3

3−M22 .

In a similar way, integrating from 4 to 1 we obtain

2n2cD lnTc

Th= M3

1−M24 .

As a result,M2

3−M22 = M2

4−M21 → M2

3−M24 =−(M1−M2

2) .

(d) Using this expression we obtain the efficiency

η =Q1→2 +Q3→4

Q3→4=

Th−Tc

Th.


Coming back to the item (c) we find

W = Q3→4η =M2

3−M24

2nDTh−Tc

T2h

.

Problem 2.5: Find the efficiency of the engine shown in Fig.2.3 ([1] -Fig.2.18). Assume that

3

1

2

4

P

V

a d i a b a t i c

Figure 2.3:Sketch of the cycle.

the operating substance is an ideal monoatomic gas. Express your answer in terms ofV1 andV2.(The processes1→ 2 and3→ 4 are adiabatic. The processes4→ 1 and2→ 3 occur at constantvolume).

v

Solution 2.5: Let us start with the processes at constant volume. The mechanical work duringtheses processes does not take place. Consequently,

Q2→3 = (3/2)nR(T3−T2) ,

Q4→1 = (3/2)nR(T1−T4) ,

W = (3/2)nR(T1 +T3−T2−T4) .

The efficiency is given by the expression

η =T1 +T3−T2−T4

T1−T4= 1− T2−T3

T1−T4. (2.4)

Now let us consider the adiabatic processes whereTV2/3 = const (monoatomic ideal gas!). Thus,

T1

T2=

T4

T3=

(V2

V1

)2/3

≡ α .

13

Substituting this expression into Eq. (2.4) we obtain:

η = 1− 1α

= 1−(

V1

V2

)2/3

.

Problem 2.6: One kilogram of water is compressed isothermally at20 C from 1 atm to 20atm.

(a) How much work is required?

(b) How much heat is rejected?

Assume that the average isothermal compressibility of water during this process isκT = 0.5×10−4 (atm)−1 and the average thermal expansivity of water during this process isαP = 2×10−4

(C)−1.

Solution 2.6: Since for an isothermal processdQ= T dSwe have

Q = T∫ P2

P1

(∂S∂P

)

TdP.

Using the Maxwell relation for the Gibbs free [see Eq. [1]-(2.112)] energy we obtain(

∂S∂P

)

T=−

(∂V∂T

)

P=−VαP .

Thus

Q =−T∫ P2

P1

V(P)αT(P)dP

Now let us assume thatαT it P-independent, and

V(P) = V0 [1−κT(P−P0)] = (M/ρ) [1−κT(P−P0)] .

As a result,Q =−(M/ρ)TαT(P2−P1) [1−κT(P−P0)/2] .

We see that since the compressibility of water is very low one can neglect the correction due tochange in the volume and assumeV ≈M/ρ. The mechanical work is

W =−∫ P2

P1

PdV =−∫ P2

P1

P

(∂V∂P

)

TdP= (M/2ρ)κT(P2

2 −P21) .


2 J 0

2 L 0

J 0

L 0

a

b c

Figure 2.4:

Problem 2.7: Compute the efficiency of the heat engine shown in Fig.2.4(Fig. [1]-2.19). Theengine uses a rubber band whose equation of state is

J = αLT , (2.5)

whereα is a constant,J is the tension,L is the length per unit mass, andT is the temperature inKelvins. The specific heat (heat capacity per unit mass) is a constant,cL = c.

Solution 2.7: From Fig.2.4 we see that the patha→ b is isothermal. Indeed, sinceJ ∝ L, itfollow from Eq. (2.5) thatT = const. Then, from the same equation we get,

Ta = Tb = J0/αL0 , Tc = J0/2αL0 = Ta/2.

As a result,

Qb→a = MαTa

∫ 2L0

L0

LdL = (3/2)MαL20Ta = (3/2)MJ0L0 ,

Qa→c = Mc(Tc−Ta) =−(1/2)McTa ,

Qc→b = Mc(Tb−Tc)−MJ0L0 = (1/2)McTa−MJ0L0 .

Hence, the total work is

W = Qb→a +Qa→c +Qc→b = (1/2)MJ0L0

and

η =W

Qb→a=

13

.

This result is also clear from geometrical point of view.

15

Problem 2.8: Experimentally one finds that for a rubber band

(∂J∂L

)

T=

aTL0

[1+2

(L0

L

)3]

,

(∂J∂T

)

L=

aLL0

[1−

(L0

L

)3]

,

whereJ is the tension,a = 1.0×103 dyne/K, andL0 = 0.5 m is the length of the band when notension is applied. The mass of the rubber band is held fixed.

(a) Compute(∂L/∂T)J and discuss its physical meaning.

(b) Find the equation of state and show thatdJ is an exact differential.

(c) Assume that heat capacity at constant length isCL = 1.0 J/K. Find the work necessary tostretch the band reversibly and adiabatically to a length 1 m. Assume that when no tensionis applied, the temperature of the band isT = 290K. What is the change in temperature?

Solution 2.8:

(a) We have (∂L∂T

)

J·(

∂T∂J

)

L·(

∂J∂L

)

T=−1.

Consequently,

(∂L∂T

)

J=−

[(∂T∂J

)

L·(

∂J∂L

)

T

]−1

=−

(∂J∂T

)L(

∂J∂L

)T

=−LT

1−(

L0L

)3

1+2(

L0L

)3 .

Physical meaning -

αJ =1L

(∂L∂T

)

J

is the thermal expansion coefficient at given tension.

(b) The equation of state has the form

J =aTLL0

[1−

(L0

L

)3]

.

The proof of the exactness is straightforward.


(c) Consider adiabatic expansion of the band,

0 = dQ= CL dT +J(L,T)dL.

Consequently,dTdL

=−J(L,T)CL

=− aTLCLL0

[1−

(L0

L

)3]

.

Measuring length in units ofL0 asL = ` ·L0 and introducingβ ≡ aL0/CL we obtain thefollowing differential equation

dT/T =−β`(1− `−3)d` .

Its solution is

lnTf

T0=−β

2

(L f

L0

)2[

1+2

(L0

L f

)3]

.

HereL f andTf are final values of the length and temperature, respectively. This is theequation for adiabatic process which provides the change in the temperature. The mechan-ical work is then

W = CL(Tf −T0) .

Problem 2.9: Blackbody radiation in a box of volumeV and at temperatureT has internalenergyU = aVT4 and pressureP = (1/3)aT4, wherea is the Stefan-Boltzmann constant.

(a) What is the fundamental equation for the blackbody radiation (the entropy)?

(b) Compute the chemical potential.

Solution 2.9:

(a) Let us first find the Helmholtz free energy. From the relationP =−(

∂A∂V

)T

we get

A =−PV =−13

aVT4 =−13U .

Consequently,

S=−(

∂A∂T

)

V=

43

aVT3 =43

UT

.

As a result,(

∂S∂U

)

V=

∂(S,V)∂(U,V)

=∂(S,V)/∂(T,V)∂(U,V)/∂(T,V)

=(∂S/∂T)V

(∂U/∂T)V=

1T

.

(b) SinceA is N-independent,µ= 0.

17

Problem 2.10: Two vessels, insulated from the outside world, one of volumeV1 and the otherof volumeV2, contain equal numbersN of the same ideal gas. The gas in each vessel is originallyat temperatureTi . The vessels are then connected and allowed to reach equilibrium in such a waythat the combined vessel is also insulated from the outside world. The final volume isV =V1+V2.What is the maximum work,δWf ree, that can be obtained by connecting these insulated vessels?Express your answer in terms ofTi , V1, V2, andN.

Solution 2.10: The Gibbs free energy of an ideal gas is given by the equation

G = NkT lnP+Nχ(T)

whereχ(T) is some function of the gas excitation spectrum.1

Consequently,

S=−(

∂G∂T

)

P=−NklnP−Nχ′(T) .

Before the vessels are connected,

Si =−Nkln(P1P2)−2Nχ′(T) .

After the vessels are connected the temperature remains the same, as it follows from the conser-vation law, the entropy being

Sf =−2NklnP−2Nχ′(T) .

Consequently,∆S=−Nkln(P2/P1P2). On the other hand,

1P

=V1 +V2

2NkTi,

1Pi

=Vi

NkTi→ P2

P1P2=

4V1V2

(V1 +V2)2 .

1 For an ideal gas, the energy of the particle can be written as a sum of the kinetic energy,εp = p2/2m, and theenergy of internal excitations,εα (characterized by some quantum numbersα),

εpα = εp + εα .

As we will see later, the free energy of the ideal gas can be constructed as

A = −kTN!

ln

(∑pα

e−εpα/kT

)N

≈−NkT ln

[eVN

(mkT

2πh2

)3/2

∑α

e−εα/kT

]

= −NT ln(eV/N)+N f(T) ,

f (T) = −kT ln

[(mkT

2πh2

)3/2

∑α

e−εα/kT

].

The Gibbs free energy is then

G = A+PV = NkT lnP+Nχ(T) , χ(T)≡ f (T)−kT lnkT .


As a result, the maximum work is

∆Wf ree = Ti∆S= NkTi ln(V1 +V2)2

4V1V2.

Problem 2.11: For a low-density gas the virial expansion can be terminated at first order in thedensity and the equation of state is

P =NkTV

[1+

NV

B2(T)]

,

whereB2(T) is the second virial coefficent. The heat capacity will have corrections to its idealgas value. We can write it in the form

CV,N =32

Nk− N2kV

f (T) .

(a) Find the form thatf (T) must have in order for the two equations to be thermodynamicallyconsistent.

(b) FindCP,N.

(c) Find the entropy and internal energy.

Solution 2.11:

(c) The equation of state under consideration can be obtained from the Helmholtz free energy2

A = Aideal+kTB2(T)N2

V.

Then

S=−(

∂A∂T

)

V= Sideal+δS, δS≡−k

N2

V

[B2(T)+TB′2(T)

].

Since we know both entropy and Helmholtz free energy, we find the internal energy as

U = A+TS= Uideal+kTB2(T)N2

V−kT

N2

V

[B2(T)+TB′2(T)

]

= Uideal+δU , δU ≡−kT2B′2(T)N2

V.

2Remember thatP =−(

∂A∂V

)T.

19

(a) As a result,

CV = T

(∂S∂T

)

V=

(∂U∂T

)

V= Cideal

V +δCV , δCV ≡−kTN2

V

[2B′2(T)+TB′′2(T)

].

We find in this way,

f (T) = 2TB′2(T)+T2B′′2(T) .

(b) Let us express the equation of state asV(P,T). Since the density is assumed to be small inthe correction one can use equation for the ideal gas to find the the volume. We have,

V =NkT

P+NB2(T) .

Consequently, the entropy can be expressed as

S= Sideal+δS1 , δS1≡−kNB2 +TB′2kT/P+B2

.

Now

CP = CidealP +T

(∂δS1

∂T

)

P

= −kN(B2 +TB′2)

′(kT/P+B2)− (k/P+B′2)(B2 +TB′2)(kT/P+B2)2

= CidealP −k

N2

V

[T(2B′2 +TB′′2)− (B2 +TB′2)

]

= CidealP +δCV +k

N2

V(B2 +TB′2) .

Here in all corrections we used equation of state for an ideal gas. As a result,

CP−CV = (CP−CV)ideal+kN2

V(B2 +TB′2) .

Problem 2.12: Prove that

CY,N =(

∂H∂T

)

Y,Nand

(∂H∂Y

)

T,N= T

(∂X∂T

)

Y,N−X .


Solution 2.12: Let us first recall definitions forX, Eq. ([1]-2.66),U = ST+YX+ ∑J jµ′jdNj .The enthalpy is defined as

H = U−XY = ST+∑J

µ′jdNj .

SincedU = T dS+Y dXwe getdH = T dS−X dY.

Consequently,

CY,N = T

(∂S∂T

)

Y,N=

(∂H∂T

)

Y,N.

To prove the second relation we do the following(

∂H∂Y

)

T,N=

(∂H∂Y

)

S,N+

(∂H∂S

)

Y,N

(∂S∂Y

)

T.

Now, (∂H∂Y

)

S,N=−X ,

(∂H∂S

)

Y,N= T .

Now we have to use the Maxwell relation, which emerges for theGibbsfree energyG= H−TS.From

dG=−SdT−X dY

we get (∂S∂Y

)

T=

(∂X∂T

)

Y,N.

Thus we obtain the desired result.

Problem 2.13: Compute the entropy, enthalpy, Helmholtz free energy, and Gibbs free energyfor a paramagnetic substance and write them explicitly in terms of their natural variables if pos-sible. Assume that mechanical equation of state ism= (D/T)H and the the molar heat capacityat constant magnetization iscm = c, wherem is the molar magnetization,H is the magnetic field,D is a constant,c is a constant, andT is the temperature.

Solution 2.13: Let us start with the internal energyu(T,m) per one mole. We have the mag-netic contributionumag=

∫ m0 H(m)dm. SinceH(m) = (T/D)m we getumag= (T/2D)m2. The

“thermal” contribution iscT. As a result,

u(T,m) = T(c+m2/2D) , T(u,m) =u

c+m2/2D.

The molar entropys is then derived from the definition(

∂s∂u

)

m=

1T

=c+m2/2D

u→ s(u,m) = (c+m2/2D) ln(u/u0) .

21

Hereu0 is a constant. As a result, in “natural” variables

u(s,m) = u0exp

(s

c+m2/2D

).

To find other thermodynamic potentials we needs(T,m). We can rewrite the above expressionfor the entropy as

s(T,m) = (c+m2/2D) lnT(c+m2/2D)

u0.

In particular, the Helmholtz free energy is

a(T,m) = u−Ts= T(c+m2/2D)[1− ln

T(c+m2/2D)u0

].

To get enthalpy we have to subtract fromu the quantityHm= (D/T)H2 and to expressm throughH asm= (D/T)H. As a result, we obtain:

h(T,H) = u−Hm= T(c+DH2/2T2)− (D/T)H2 = T(c−DH2/2T2) .

Finally, Gibbs free energy isg = a−Hm, which has to be expressed throughT andH. We get

g = a−Hm= T(c+DH2/2T2)[1− ln

T(c+DH2/2T2)u0

]− (D/T)H2

= T(c−DH2/2T2)−T(c+DH2/2T2) lnT(c+DH2/2T2)

u0.

Problem 2.14: Compute the Helmholtz free energy for a van der Waals gas. The equation ofstate is (

P+αn2

V2

)(V−nb) = nRT,

whereα andb are constants which depend on the type of gas andn is the number of moles.Assume that heat capacity isCV,n = (3/2)nR.

Is this a reasonable choice for the heat capacity? Should it depend on volume?

Solution 2.14: Let us express pressure through the volume,

P =nRT

V−nb− αn2

V2 .

SinceP =−∂A/∂V we obtain

A =−∫

P(V)dV =−nRTln(V−nb)− (αn2/V)+A(T) .


HereA(T) is the integration constant, which can be found from the given specific heat. Indeed,

S=∫ T CV,ndT′

T ′= (3/2)nRln(T/T0) .

Consequently,

A(T) =−∫ T

S(T ′)dT′ = (3/2)nRT[1− ln(T/T0)] .

Here we omit temperature-independent constant.The suggestion regarding specific heat is OK since the difference between the entropies of

van der Waals gas and the ideal gas istemperature independent. (Check!)

Problem 2.15: Prove that

(a) κT(CP−CV) = TVα2P

(b) CP/CV = κT/κS.

Solution 2.15: We use the method of Jacobians:

(a)

CV = T (∂S/∂T)V = T∂(S,V)/∂(T,V) =∂(S,V)/∂(T,P)∂(T,V)/∂(T,P)

= T(∂S/∂T)P(∂V/∂P)T − (∂S/∂P)T (∂V/∂T)P

(∂V/∂P)T

= CP−T(∂S/∂P)T (∂V/∂T)P

(∂V/∂P)T.

Now, from the Maxwell relations(∂S/∂P)T =−(∂V/∂T)P. Thus,

CP−CV =−T[(∂V/∂T)P]2

(∂V/∂P)T= TV

α2P

κT.

The first relation follows from this in a straightforward way from definitions.

(b) Let us first calculate the adiabatic compressibility(∂V/∂P)S as(

∂V∂P

)

S=

∂(V,S)∂(P,S)

=∂(V,S)/∂(V,T)∂(P,S)/∂(P,T)

· ∂(V,T)∂(P,T)

=(∂S/∂T)V

(∂S/∂T)P·(

∂V∂P

)

T.

Consequently,CP

CV=

(∂V/∂P)T

(∂V/∂P)S=

κT

κS.

23

Problem 2.16: Show that

Tds= cx(∂T/∂Y)x dY+cY (∂T/∂x)Y dx,

wherex = X/n is the amount of extensive variable,X, per mole,cx is the heat capacity per moleat constantx, andcY is the heat capacity per mole at constantY.

Solution 2.16: Let us substitute the definitions

cx = T (∂S/∂T)x , cY = T (∂S/∂T)Y .

Now the combinationcx(∂T/∂Y)x dY+cY (∂T/∂x)Y dx can be rewritten as

T (∂S/∂T)x(∂T/∂Y)x dY+T (∂S/∂T)Y (∂T/∂x)Y dx

= T (∂S/∂Y)x dY+T (∂S/∂x)Y dx= T ds.

Problem 2.17: Compute the molar heat capacitycP, the compressibilities,κT and κS, andthe thermal expansivityαP of a monoatomic van der Waals gas. Start from the fact that themechanical equation of state is

P =RT

v−b− α

v2 .

and the molar heat capacity iscv = 3R/2, wherev = V/n is the molar volume.

Solution 2.17: Let us start with ther specific heat. Using the method similar to the Problem2.15 we can derive the relation

cP−cv =−T[(∂P/∂T)v]

2

(∂P/∂v)T=

R1−2α(v−b)2/RTv3

.

Now let us compute the compressibility

κT =−1v

(∂v∂P

)

T=−1

v

(∂P∂v

)−1

T=

(v−b)2

vRT1

1−2α(v−b)2/RTv3.

Givencv, other quantities can be calculated using results of the Problem 2.15.

Problem 2.18: Compute the heat capacity at constant magnetic fieldCH,n, the susceptibilitiesχT,n andχS,n, and the thermal expansivityαH,n for a magnetic system, given that the mechanicalequation of state isM = nDH/T and the heat capacityCM,n = nc, whereM is the magnetization,H is the magnetic field,n is the number of moles,c is the molar heat capacity, andT is thetemperature.


Solution 2.18: Let us start with susceptibilities. By definition,

χT,n =(

∂M∂H

)

T,n=

nDT

.

Now, (∂M∂H

)

S=

∂(M,S)∂(H,S)

=∂(M,S)/∂(M,T)∂(H,S)/∂(H,T)

· ∂(M,T))∂(H,T)

= χT,nCM,n

CH,n.

Thus we have found one relation between susceptibilities and heat capacities,

χS,n

χT,n=

CM,n

CH,n.

Now let us findCH,n. At constantH, M becomes dependent only on temperature. Then,

dM =dMdT

dT =−nDHT2 dT .

Consequently, the contribution to the internal energy is

dU =−H dM =nDH2

T2 dT .

As a result,

CH,n−CM,n =nDH2

T2 .

GivenCH,n = ncwe easily computeCH,n andχS,n.According to Eq. (R2.149),αH is defined as

αH =(

∂M∂T

)

H=−nDH

T2 .

Problem 2.19: A material is found to have a thermal expansivityαP = v−1(R/P+a/RT2) andan isothermal compressibilityκT = v−1 [T f(P)+b/P]. Herev = V/n is the molar volume.

(a) Find f (P).

(b) Find the equation of state.

(c) Under what condition this materials is stable?

25

Solution 2.19:

(a) By definition, we have

∂v∂T

=RP

+a

RT2 .

∂v∂P

= −T f(P)− bP

.

To makedvan exact differential we need:

− RP2 =− f (p) .

Thusf (p) = R/P2 .

(b) We can reconstruct the equation of state as:

v =∫ P ∂v

∂PdP=

∫ PdP

[−RT

P2 −bP

]

=RTP−blnP+g(T) .

Hereg(T) is some function of the temperature. Now,

∂v∂T

=RP

+g′(T)≡ RP

+a

RT2 .

Thusg(T) =−a/RT+const. As a result, we can express the equation of state as

v−v0 =RTP

+blnP0

P− a

RT.

(c) Since the compressibility must be positive, we have the stability condition

T f(P)+bP

> 0 → TRP2 +

bP

> 0.

Consequently, the stability condition is

P/T < R/b.

Problem 2.20: Compute the efficiency of the reversible two heat engines in Fig.2.5 (R2.20).Which engine is the most effective? (Note that these are not Carnot cycles. The efficiency of aheat engineisη = ∆Wtotal/∆Qabsorbed.


T

T 1

T 2

S1S 2S

T

T 1

T 2

S1S 2S

a ab

bc c

( a ) ( b )

Figure 2.5:

Solution 2.20: SincedQ= T dS, we immediately get for any closed path in theT−Splane:

∆Wtotal =∮

T dS.

This is just the area of the triangle,

∆Wtotal = (1/2)(T2−T1)(S2−S1) .

The heat absorbed in the case(a) is

∆Qabsorbed= T2(S2−S1) .

Thus,

ηa =T2−T1

2T2.

In the case(b), it easy to show that

∆Qabsorbed= (1/2)(T2 +T1)(S2−S1) .

Thus

ηb =T2−T1

T2 +T1> ηa .

Problem 2.21: It is found for a gas thatκT = Tv f(P) andαP = Rv/P+ Av/T2, whereT isthe temperature,v is the molar volume,P is the pressure,A is a constant, andf (P) is unknownfunction.

(a) What is f (P)?

(b) Findv = v(P,T).

27

Solution 2.21: The solution is similar to the problem2.19. We have:

∂v∂T

= v2(

RP

+AT2

),

∂v∂P

= −v2T f(P) .

Let us introduceγ(P,T)≡ [v(P,T)]−1. We get,

∂γ∂T

= −RP− A

T2 ,

∂γ∂P

= T f(P) .

Again, from the Maxwell relation we getRP2 = f (p). Then we can expressγ as

γ =∫ P ∂γ

∂PdP= TR

∫ P dPP2 =−RT

P+g(T) .

Then,g′(T) =−A/T2, or g(T) = A/T +const. As a result,

γ = γ0 +AT− RT

P.

Consequently,

v(P,T) =1

γ0 +A/T−RT/P.

Problem 2.22: A monomolecular liquid at volumeVL and pressurePL is separated from a gasof the same substance by a rigid wall which is permeable to the molecules, but does not allowliquid to pass.The volume of the gas is held fixed atVG, but the volume of the liquid cam bevaried by moving a piston. If the pressure of the liquid is increased by pushing in on the piston,by how much does the pressure of the gas change? [Assume the liquid in incompressible (itsmolar volumeis independent of the pressure) and describe the gas by the ideal gas equation ofstate. The entire process occurs at fixed temperatureT].

Solution 2.22: Let us consider the part of the system, which contains both liquid and gasparticles. In this part the chemical potentials must be equal,µL = µG. On the other hand, thechemical potentials and pressures of gas in both parts must be equal. Thus we arrive at theequation,

µL(PL,T) = µG(PG,T) .

If one changes the pressure of liquid byδPL, then(

∂µL

∂PL

)

TδPL =

(∂µG

∂PG

)

TδPG .


For an ideal gas,(∂µ/∂P)T = kT/PG = VG/NG ≡ vL. the quantityvL has a physical meaning ofthevolume per particle. For a liquid, the relation is the same,

(∂µ∂P

)

T=

∂∂P

(∂G∂N

)

P,T=

∂∂N

(∂G∂P

)

N,T=

(∂V∂N

)

T= vL .

The last relation is a consequence ofincompressiblecharacter of the liquid. As a result,

δPG

δPL=

vL

vG.

2.1 Additional Problems: Fluctuations

Quick access:2324252627

Problem 2.23: Find the mean square fluctuation of the internal energy (usingV andT as inde-pendent variables). What is the mean square fluctuation of the internal energy for a monoatomicideal gas?

Solution 2.23: We have

∆U =(

∂U∂V

)

T∆V +

(∂U∂T

)

V∆T =

[T

(∂P∂T

)

V−P

]∆V +CV ∆T .

Here we use Maxwell relation which can be obtained from Helmholtz free energy. Squaring andaveraging we obtain (note that〈∆V ∆T〉= 0)

〈(∆U)2〉=[T

(∂P∂T

)

V−P

]2

〈(∆V)2〉+C2V〈(∆T)2〉 .

Now,

〈(∆T)2〉= kT2/CV , 〈(∆V)2〉=−kT

(∂V∂P

)

T.

Thus

〈(∆U)2〉=−kT

[T

(∂P∂T

)

V−P

]2(∂V∂P

)

T+CVkT2 .

For the ideal gas,P = NkT/V , CV = (3/2)Nk.

Thus,〈(∆U)2〉= (3/2)N(kT)2 .

2.1. ADDITIONAL PROBLEMS: FLUCTUATIONS 29

Problem 2.24: Find 〈∆T ∆P〉 (with variablesV andT) in general case and for a monoatomicideal gas.


∆P =(

∂P∂V

)

T∆V +

(∂P∂T

)

V∆T .

Squaring and averaging we obtain (note that〈∆V ∆T〉= 0)

〈∆T ∆P〉=(

∂P∂T

)

V〈(∆T)2〉=

(kT2

CV

) (∂P∂T

)

V.

For an ideal gas,

〈∆T ∆P〉=23

kT2

V.

Problem 2.25: Find 〈∆V ∆P〉 with variable (V andT).

Solution 2.25: Using results of the previous problem we get

〈∆V ∆P〉=(

∂P∂V

)

T〈(∆V)2〉=−kT .

Problem 2.26: Using the same method show that

〈∆S∆V〉= kT

(∂V∂T

)

P, 〈∆S∆T〉= kT .

Solution 2.26: Starightforward.

Problem 2.27: Find a mean square fluctuation deviation of a simple pendulum with the length` suspended vertically.

Solution 2.27: Let m be the pendulum mass, andφ is the angle of deviation from the vertical.The minimal work is just the mechanical work done against the gravity force. For smallφ,

Wmin = mg· `(1−cosφ)≈mg φ2/2.

Thus,〈φ2〉= kT/mg .


2.2 Mini-tests

2.2.1 A

The Helmholtz free energy of the gas is given by the expression

A = Nε0−NkT ln(eV/N)−NcT lnT−NζT .

Heree= 2.718. . . is the base of natural logarithms,N is the number of particles,V is the volume,T is the temperature in the energy units, whileε0, c andζ are constants.

(a) Find the entropy as function ofV andT.

(b) Find internal energyU as a function of the temperatureT and number of particlesN.

(c) Show thatc is the heat capacity per particle at given volume,cv.

(d) Find equation of state.

(e) Find Gibbs free energy, enthalpy. Find the entropy as a function ofP andT.

(f) Using these expression find the heat capacity at constant pressure,cP.

(g) Show that for an adiabatic process

TγP1−γ = constant, whereγ = cP/cv .

Solution

By definition,

S=−(

∂A∂T

)

V,N= N ln

eVN

+Nc(1+ lnT)+NζT .

Now,

U = A+TS= Nε0 +NcT,

P = −(

∂A∂V

)

T,N=

NTV

,

G = A+PV = A+NT = Nε0−NT lnVN−NcT lT+NζT

= Nε0 +NT lnP−NT(1+c) lnT−NζT ,

W = U +PV = U +NT = Nε0 +N(c+1)T ,

S = −N lnP+N(1+c) lnT +N(ζ+1+c) ,

cP = T

(∂S∂T

)

P,N= 1+c.

2.2. MINI-TESTS 31

Immediately, to keep entropy constant we get

−N lnP+NcP lnT = const → TcP/P = const.

SincecP−cV = cP−c = 1, we obtain

TcPPcP−cV = const, → TγP1−γ = const.

2.2.2 B

Problem 1

Discuss entropy variation for(a) adiabatic process,(b) isothermic process,(c) isochoric process,(d) isobaric process.

Problem 2

(a) Discuss the difference between Gibbs and Helmholtz free energy.(b) Prove the relation

U =−T2(

∂∂T

AT

).

(c) A body with constant specific heatCV is heated under constant volume fromT1 to T2. Howmuch entropy it gains?(d) Discuss the heating if the same body is in contact with a thermostat atT2. In the last case theheating is irreversible. Show that the total entropy change is positive.(e) Two similar bodies with temperaturesT1 andT2 brought into contact. Find the final tempera-ture and the change in entropy.

Solution

Problem 1

(a) Constant(b)

S= Q/T .

(c)

S=∫ T2

T1

CV(T)T

dT = CV lnT2

T1.


(d)

S=∫ T2

T1

CP(T)T

dT = CP lnT2

T1.

Problem 2

Mechanical work underisothermic processis given by

dW = dU−dQ= dU−T dS= d(U−TS) .

The function of stateA = U−TS

is calledHelmholtzfree energy. We have

dA= dU−T dS−SdT= (T dS−PdV)−T dS−SdT=−SdT−PdV.

Thus,

S=−(

∂A∂T

)

V, P =−

(∂A∂V

)

T.

Thus,A is thethermodynamic potentialwith respect toV andT.The thermodynamic potential with respect toP andT is calledGibbsfree energy. We have

G = U−TS−PV→ dG=−SdT+V dP.

(b) Substituting

S=−(

∂A∂T

)

V

into definition ofA we get the result.(c) We have

S=∫ T2

T1

CV

TdT = CV ln

(T2

T1

).

(d) Use 2nd law of thermodynamics(e) Energy conservation law yields

CV(T2−TB) = CV(TB−T1) → TB = (T1 +T2)/2.

for the entropy change we have

δS= CV lnTB

T1+CV ln

TB

T2= 2CV ln

(T1 +T2

2√

T1T2

)≥ 0.

Chapter 3

The Thermodynamics of Phase Transitions

Quick access:1 2 3 4 5 6 7 8 9 10111213

Problem 3.1: A condensible vapor has a molar entropy

s= s0 +Rln

[C(v−b)

(u+

av

)5/2]

,

wherec ands0 are constants.

(a) Compute equation of state.

(b) Compute the molar heat capacities,cc andcP.

(c) Compute the latent heat between liquid and vapor phases temperatureT in terms of thetemperatureT, the gas constantR and gas molar volumesvl andvg. How can you findexplicit values ofvl andvg if you need to?

Solution 3.1:

a) Let us first find the temperature as function ofu ands. We have

1T

=(

∂s∂u

)

v=

5R2

(u+

av

)−1.

Thus,

s = s0 +Rln[C(5RT/2)5/2(v−b)

],

u =52

RT− av

,

a = u−Ts=52

RT− av−T

s0 +Rln

[C(v−b)(5RT/2)5/2

].

33

34 CHAPTER 3. THE THERMODYNAMICS OF PHASE TRANSITIONS

As a result,

P =−(

∂a∂v

)

T=

RTv−b

− av2 .

This is the van der Waals equation.

(b) We have,

cv = T

(∂s∂T

)

v=

52

RT.

The result forcP can be obtained in the same way as in the problem2.17.

(c) The latent heatq is just

q = T(sg−sl ) = RT lnvg−b

vl −b.

Since the pressure should be constant along the equilibrium line,Pl = PG, we have

RTvg−b

− av2

g=

RTvl −b

− a

v2l

.

Another equation is the equality of chemical potentials,µl = µg. We know thatµ =(∂a/∂n)T,V , so everything could be done. Another way is to plot the isotherm and findthe volumes using the Maxwell rule.

Problem 3.2: Find the coefficient of thermal expansion,αcoex= v−1(∂v/∂T)coex, for a gasmaintained in equilibrium with its liquid phase. Find an approximate explicit expression forαcoex, using the ideal gas equation of state. Discuss its behavior.

Solution 3.2: It is implicitly assumed that the total volume of the system is kept constant. Wehave,

∂v∂T

=(

∂v∂T

)

P+

(∂v∂P

)

T

dPdT

.

For an ideal gas,v = RT/P. Thus(

∂v∂T

)

P= R/P,

(∂v∂P

)

T=−RT/P2 .

Consequently,

αcoex=1v

(∂v∂T

)

coex=

1T

[1− T

P

(∂P∂T

)

coex

].

According to the Clapeyron-Clausius formula,(

∂P∂T

)

coex=

qT(vg−vl )

≈ qTvg

,

35

whereq = (∆h)lg, we can rewrite this expression as

αcoex≈ 1T

(1− q

RT

).

The coefficient of thermal expansion is less because with the increase of the temperature undergiven pressure the heat is extracted from the system.

Problem 3.3: Prove that the slope of the sublimation curve of a pure substance at the triplepoint must be greater than that of the vaporization curve at the triple point.

Solution 3.3: The triple point is defined by the following equations for the 2 phases:

P1 = P2 = P3 , T1 = T2 = T3 , µ1 = µ2 = µ3 .

The definitions of the sublimation and vaporization curves are given in Fig. 3.4 of the text-book [1]. Consequently, the slopes of the vaporization and sublimation curves are given by therelations (

∂P∂T

)

lg=

sg−sl

vg−vl≈ sg−sl

vg,

(∂P∂T

)

sg=

sg−ss

vg−vs≈ sg−ss

vg.

Since solid state is more ordered,ss < sl , and(

∂P∂T

)

lg−

(∂P∂T

)

sg=

ss−sl

vg< 0.

Problem 3.4: Consider a monoatomic fluid along its fluid-gas coexistence curve. Compute therate of change of chemical potential along the coexistence curve,(∂µ/∂T)coex, whereµ is thechemical potential andT is the temperature. Express your answer in terms ofsl ,vl andsg,vg

which are the molar entropy and molar volume of the liquid and gas, respectively.

Solution 3.4: We haveµ(P,T) = µl (P,T) = µg(P,T). Thus,

dµdT

=(

∂µg

∂T

)

P+

(∂µg

∂P

)

T

dPdT

.

Consequently, (dµdT

)

coex=−sg +vg

sg−sl

vg−vl=

sgvl −vgsl

vg−vl.


Problem 3.5: A system in its solid phase has a Helmholtz free energy per mole,as = B/Tv3,and in its liquid phase it has a Helmholtz free energy per mole,al = A/Tv2, whereA andB areconstants,v is the volume per mole, andT is the temperature.

(a) Compute the Gibbs free energy density of the liquid and solid phases.

(b) How are the molar volumes,v, of the liquid and solid related at the liquid-solid phasetransition?

(c) What is the slope of the coexistence curve in theP−T plane?

Solution 3.5:

(a) By definition,g = a+Pv, andP =−(∂a/∂v)T . Thus,

g = a−v(∂a/∂v)T , gs = 4B/Tv3s , gl = 3A/Tv2

l .

(b) SinceP =−(∂a/∂v)T we have,

Ps = 3B/Tv4s , Pl = 2A/Tv3

l .

Since at the phase transitionPs = Pl = P, we obtain

v3l

v4s

=2A3B

.

(c) Now we can express the Gibbs free energies per mole in terms ofP,

gs =43(3B)1/4P3/4T−1/4 , gl =

32(2A)1/3P2/3T−1/3 .

Since at the transition pointgl = gs, we have

PT

=

(37/4

28/3

A1/3

B1/4

)12

=321

224

A4

B3 .

Problem 3.6: Deduce the Maxwell construction using stability properties of the Helmholtzfree energy rather than the Gibbs free energy.

Solution 3.6: Since the system in the equilibrium, the maximum work extracted during theprocess of the phase transition is∆A−P0∆V. Let us look at Fig.3.1.

37

P

V

ab c

v l v g

P 0

Figure 3.1:We see thatP0∆V = P0(Vg−Vl ) is just the area of the dashed rectangle. To find∆A let is followisotherm. SinceT = const,

∆A =∫ c

aPdV,

which is just the areabelowthe isotherma−b−c. Subtracting areas we reconstruct the Maxwellrule.

Problem 3.7: For a van der Waals gas, plot the isotherms in the in theP− V plane (P andVare the reduced pressure and volume) for reduced temperaturesT = 0.5, T = 1.0, andT = 1.5.For T = 0.5, is P = 0.1 the equilibrium pressure of the liquid gas coexistence region?

Solution 3.7: We use Maple to plot the curves. The graphs have the form We see that forT = 0.5 there is no stavle region at all. To illustrate the situation we plot the curves in thevicinity of T = 1.

Problem 3.8: Consider a binary mixture composed of two types of particles,A andB. For thissystem the fundamental equation fog the Gibbs energy is

G = nAµA +nBµB , (3.1)

the combined first and second laws are

dG=−SdT+V dP+µAdnA +µBdnB (3.2)

(S is the total entropy andV is the total volume of the system), and the chqemical potentialsµA

andµB are intensive so that

µA = µA(P,T,xA) and µB(P,T,xB) .


–4

–2

0

2

4

6

8

10

12

0.6 0.7 0.8 0.9 1 1.1 1.2v

Figure 3.2:The upper curve correspondsT = 1.5, the lower one - toT = 0.5.

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

P

1 2 3 4 5v

Figure 3.3:The upper curve correspondsT = 1.05, the lower one - toT = 0.85.

Use these facts to derive the relations

sdT−vdP+ ∑α=A,B

xαµα = 0 (3.3)

and

∑α=A,B

xα (dµα +sα dT−vα dP) = 0, (3.4)

wheres= S/n, n= nA+nB, sα = (∂S/∂nα)P,T,nβ6=α, andvα = (∂V/∂nα)P,T,nβ 6=α

with α = A,B and

β = A,B.

Solution 3.8: Let us expressnα asnα = nxα and divide Eq. (3.1) by n. We get

dg= d

(∑α

xαµα

)= ∑

α(xα dµα +µα dxα) (3.5)

39

The let us divide Eq. (3.2) by n to obtain

dg=−sdT+vdP+∑α

µα dxα . (3.6)

Equating the right hand sides of these equation we prove Eq. (3.3). Further, the entropy is anextensive variable. Consequently,is should be ahomogeneousfunction ofnα.

S= ∑α

(∂S/∂nα)P,T,nβ6=αnα = n∑

αsαxα .

Substituting these expressions into Eq. (3.3) we prove Eq. (3.4).

Problem 3.9: Consider liquid mixture (l ) of particlesA andB coexisting in equilibrium withvapor mixture (g) of particlesA andB. Show that the generalization of of the Clausius-Clapeyronequation for the coexistence curve between the liquid and vapor phases when the mole fractionof of A in the liquid phase is fixed is given by

(∂P∂T

)

xlA

=xg

A(sgA−sl

A)+xgB(sg

B−slB)

xgA(vg

A−vlA)+xg

B(vgB−vl

B), (3.7)

wheresα = (∂S/∂nα)P,T,nβ 6=α, andvα = (∂V/∂nα)P,T,nβ 6=α

with α = A,B and β = A,B. [Hint:Equation (b) of the Problem (3.8) is useful.]

Solution 3.9: Let us apply Eq. (3.4) to thegasphase. We get

∑α

xgα(dµg

α +sgα dT−vg

α dPg)

= 0.

Dividing this equation bydT we obtain

∑α

xgα

(dµg

αdT

+sgα−vg

αdPg

dT

)= 0.

Now, let us take into account that at the coexistence curveµgα = µl

α. Thus

dµgα

dT=

dµlα

dT=−sl

α +vlα

dPl

dT.

Since at the equilibriumPg = Pl , we prove Eq. (3.7).

Problem 3.10: A PVT system has a line of continuous phase transitions (a lambda line) sepa-rating two phases, I and II, of the system. The molar heat capacitycP and the thermal expansivityαP are different in the two phases. Compute the slope(dP/dT)coexof theλ line in terms of thetemperatureT, the molar volumev, ∆cP = cI

P−cIIP , and∆αP = αI

P−αIIP .


Solution 3.10: At a continuous phase transition the entropy is continuous,∆s= 0. At the sametime, s= s(P,T) and along the coexistence curveP is a function of the temperature. For eachphase, we get

d sdT

=∂s∂T

+∂s∂P

dPdT

.

We know that(∂S/∂T)P = cP/T and from the Maxwell relation

(∂s/∂P)T =−(∂v/∂T)P .

Since∆s= 0,(∆cP)/T = (dP/dT)∆(∂v/∂T)P = (dP/dT)v(∆αP) .

The answer is (dPdT

)

coex=

∆cP

T ∆αP.

Problem 3.11: Water has a latent heat of vaporization,∆h = 540cal/gr. One mole of steam iskept at its condensation point under pressure atT1 = 373K. The temperature is then lowered toT2 = 336K. What fraction of the steam condenses into water? (Treat the steam as an ideal gasand neglect the volume of the water.)

Solution 3.11: Let us use the Clausius-Clapeyron equation for the case of liquid-vapor mixture.Sincevg¿ vl we get (

dPdT

)

coex=

qTvg .

Hereq is the latent heatper mole. Considering vapor as an ideal gas we get

dPdT

=qP

RT2 → P = P0e−q/RT .

As a consequence,P1

P2= exp

[q(T1−T2)

RT1T2

].

Since the volume is kept constant,

ng1

ng2

=P1

P2= exp

[q(T1−T2)

RT1T2

]= exp

[18∆h(T1−T2)

RT1T2

].

Here we have taken into account that the olecular weight of H2O is 18. The relative number ofcondensed gas is then

n1−n2

n1= 1−exp

[−18∆h(T1−T2)

RT1T2

].

41

Problem 3.12: A liquid crystal is composed of molecules which are elongated (and often haveflat segments). I behaves like a liquid because the locations of the center of mass of the moleculeshave no long-range order. It behaves like crystal because the orientation of the molecules doeshave long range order. The order parameter of the liquid crystal is given by the diatic

S= η [nn− (1/3)I ] ,

wheren is a unit vector (called the director), which gives the average direction of alignment ofthe molecules. The free energy of the liquid crystal can be written as

Φ = Φ0 +12

ASi j Si j − 13

BSi j SjkSki +14CSi j SjkSklSkl , (3.8)

whereA = A0(T−T∗), A0, B andC are constants,I is the unit tensor so

xi · I · x j = δi j , Si j = xi ·S· x j ,

andthe summation is over repeated indices. The quantitiesxi are the unit vectors

x1 = x , x2 = y , x3 = z.

(a) Perform the summations in the expression forΦ and writeΦ in terms ofη, A, B, C.

(b) Compute the critical temperatureTc, at which the transition from isotropic liquid to liquidcrystal takes place.

(c) Compute the difference between the entropies between the isotropic liquid (η = 0) and theliquid crystal at the critical temperature.

Solution 3.12:

(a) First, for brevity, let us express the matrixSasη(s− I/3), where

s= nn =

n21 n1n2 n1n3

n2n1 n22 n2n3

n3n1 n3n2 n23

.

The Eq. (3.8) can be expressed as

Φ = Φ0 +12

Aa2η2− 13

Ba3η3 +14Ca4η4 ,

a2 = Tr

(s− 1

3I)2

, a3 = Tr

(s− 1

3I)3

, a4 =

[Tr

(s− 1

3I)2

]2

. (3.9)

It is easy to calculatebm = Tr(s− 1

3I)m

. Indeed,

Tr

(s− 1

3I

)m

=m

∑k=0

m!k!(m−k)!

(−1)k

3k Tr(

skIm−k)

.


It is straightforward that form> 1 andk 6= 0 we get

Tr(

skIm−k)

= Tr(

skI)

= Tr(

sk)

= 1, Tr Im = Tr I = 3.

Here we have used the properties of trace and ofsoperators. Since trace is independent ofthe presentation, let us direct the axis1 along the vectorn. Then

s=

1 0 00 0 00 0 0

→ Tr

(sk

)= Tr s= 1.

Consequently,

bm =(

1− 13

)m

+2

(−13

)m

, a2 = b2 =23

, a3 = b3 =29

, a4 = b22 =

49

.

Finally, we get

Φ = Φ0 +13

Aη2− 227

Bη3 +19Cη4 . (3.10)

(b) At the transition point the dependenceΦ(η) has two equal minima. To find this point, letus shift the origin ofη by someη0 in order to killodd in η−η0 items. It we putη = η0+ξwe have

Φ−Φ0 =13

A(ξ+η0)2− 227

B(ξ+η0)3 +19C(ξ+η0)4 . (3.11)

In order to kill the term proportional toξ3 one has to put (see the Maple file)η0 = B/6C.Substituting this value into the expression for the coefficientξ,

(2/3)Aη0− (2/9)Bη20 +(4/9)Cη3

0 ,

and equating the result to 0 we getA = B2/27C. SinceA = A0(T−T∗) we have

Tc = T∗+B2

27A0C.

(c) SinceS=−∂Φ/∂T =−(1/3)A0η2 we get

∆S=−(1/3)A0(η+−η−)(η+ +η−) =−(2/3)A0η0(ξ+−ξ−) ,

whereη± are the roots of the equation∂Φ/∂η = 0 while ξ± are the roots of the equation∂Φ/∂ξ = 0. Note that the potentialΦ(ξ) is symmetric, thusξ+ + ξ− = 0. The importantpartΦ(ξ) at the critical point can be obtained substitutingη0 andA into Eq. (3.11). It hasthe form,

δΦ(ξ) =C9

(ξ4− B2

18C2ξ2)

.

43

Thus,

ξ± =± B6C

, ξ+−ξ− =B3C

.

As a result,

∆S=− 127

A0B2

C2 .

Problem 3.13: The equation of state of a gas is given by the Berthelot equation(

P+a

Tv2

)(v−b) = RT.

(a) Find values of the critical temperatureTc, the critical molar volumevc, and the criticalpressure,Pc in terms ofa, b, andR.

(b) Does the Berthelot equation satisfy the law of corresponding states.?

• Find the critical exponentsβ, δ, andγ from the Berthelot equation.

Solution 3.13:

(a) Let us express the pressure as a function of volume,

P =RT

v−b− a

Tv2 .

For the critical temperature we have(

∂P∂V

)

T= − RTc

(vc−b)2 +2a

Tcv3c

= 0,

(∂2P∂V2

)

T=

2RTc

(vc−b)3 −6a

Tcv4c

= 0.

From this we obtain:

vc = 3b, Tc =

√8a

27Rb, Pc =

RTc

8b.

(b) We putP = Pc · P, T = Tc · T , v = vc · v

to get: (P+

3Tv2

)(3v−1) = 8T .


(c) First let us simplify the equation of state near the critical point. Putting

P = 1+ p, v = 1+ν , T = 1+ τ

and expanding inp, ν, andτ up to lowest non-trivial order (see Maple file), we get

p =−32

ν3−12ντ+7τ . (3.12)

The derivative(∂p/∂ν)τ is (∂p∂ν

)

τ=−9

2ν2−12τ . (3.13)

Since this function isevenin ν the Maxwell construction

∫ νg

νl

νdp=∫ νg

νl

ν(

∂p∂ν

)

τdν = 0

requires−νl = νg. Since along the coexistence curvep(νg,τ) = p(−νg,τ) we get

νg =√−8τ → β = 1/2.

Substituting this value ofνg into Eq. (3.13) we observe that

(∂p∂ν

)

τ= 24τ → γ = 1.

To obtain the critical exponentδ let us introduceρ = 1/v, the equation of state being

P(ρ, T) =8Tρ3−ρ

− 3ρ2

T.

Expanding near the critical pointP = T = ρ = 1 we get

P = 1+(3/2)(ρ−1)3 → δ = 3.

3.1 Mini-tests

3.1.1 A

Properties of Van der Waals (VdW) liquid

(i) Find the critical temperatureTc at which the Van der Waals isotherm has an inflectionpoint. Determine the pressurePc and volume,Vc, for a system ofN particles atT = Tc.

3.1. MINI-TESTS 45

(ii) Express the VdW equation in units ofTc, Pc, andVc. Show that it has the form(

P′+3

V ′2

)(3V ′−1) = 8T ′ (3.14)

whereP′ ≡ P/Pc, V ′ ≡V/Vc, andT ′ ≡ T/Tc

(iii) Analyze the equation of state (3.14) near the critical point. Assume that

P′ = 1+ p, T ′ = 1+ τ V ′ = 1−n

and show that for smallp, τ andn the equation of state has the approximate form

p = 4τ+6τn+(3/2)n3 . (3.15)

(iv) Plot p(n,τ) versusn for τ =±0.05and discuss the plots.

(v) Using the above equation find the stability region. Show this region in the plot.

(vi) Show that the Maxwell relation can be expressed as∫ nr

nl

n(∂p/∂n)τ dn= 0 (3.16)

along the equilibrium liquid-gas line. Using this relation and the equation of state findnl

andnr .

(vii) Discuss why the stability condition and Maxwell relation lead to different stability criteria?Illustrate discussion using the plot.

Solution

(i) Writing the VdW equation as

P =NkT

V−Nb− N2a

V2

and requiring (∂P∂V

)

Tc

= 0,

(∂2P∂V2

)

Tc

= 0

we obtain

Tc =827

abk

, Vc = 3Nb, Pc =127

ab2 .

(ii) The result is straightforward in one substitutesP = PcP′, V = VcV ′ andT = TcT ′.

(iii) Straightforward.

(iv) Straightforward.


(v) The stability region is determined by the equation(∂p/∂n)τ = 0. We have

(∂p/∂n)τ = 6τ+(9/2)n2 .

Consequentlynst =±2

√−τ/3.

(vi) Since in the main approximation(∂p/∂n)τ is an odd function ofn, nl = −nr . UsingEq. (3.15) we get (see the Maple file)

nr =√−4τ .

3.1.2 B

Curie-Weiss theory of a magnet

The simplest equation for a non-ideal magnet has the form

m= tanh[β(Jm+h)] (3.17)

whereβ = 1/kT, J is the effective interaction constant, whileh is the magnetic field measured inproper units.

(i) Consider the caseh = 0 and analyze graphically possible solutions of this equation.Hint: rewrite equation in terms of an auxiliary dimensional variablem≡ βJm. Show thata spontaneous magnetization appears atT < Tc = J/k.

(ii) Simplify Eq. (3.17) at h = 0 near the critical temperature and analyze the spontaneousmagnetization as a function of temperature.Hint: PutT = Tc(1+ τ) and consider solutions for smallm andτ.

(iii) Analyze the magnetization curvem(h) nearTc. Hint: express Eq. (3.17) in terms of thevariablesm andh≡ βh≈ h/J and ploth as a function ofm.

(iv) Plot and analyze the magnetization curves forT/Tc = 0.6 andT/Tc = 1.6.

Solution

(i) At h = 0 the equation (3.17) has the form:

(βJ)−1m= tanhm.

From the graph of the functionF(m)≡ (βJ)−1m− tanhm (see the Maple file) one observesthat only trivial solutionm= 0 exists atβJ < 11, or atT ≥ Tc = J/k. At T < Tc there exists2 non-trivial solutions corresponding spontaneous magnetization. Thus(βJ)−1 = T/Tc

3.1. MINI-TESTS 47

(ii) The dimensionless functionF(m)≡ (T/Tc)m− tanhm, nearTc acquires the formF(m) =τm+ m3/3. Thus the solutions have the from

m= 0,±√−3τ .

Since nearTc the ratioTc/T can be substituted by 1, the same result is true form.

(iii) In general, Eq. (3.17) can be rewritten as

(T/Tc)m= tanh(m+ h) → m+ h = artanh[(1+ τ)m] .

Consequently,h = artanh[(1+ τ)m]− m

To plot the magnetization curve one has to come back to the initial variables.

Chapter 4

Elementary Probability Theory and LimitTheorems

Quick access:1 2 3 4 5 6 7 8 9 101112131415

Problem 4.1: A bus has 9 seats facing forward and 8 seats facing backward. In how manyways can 7 passengers be seated if 2 refuse to ride facing forward and 3 refuse to ride facingbackward?

Solution 4.1: Three people refuse ride facing backward, and they should be definitely placedto 9 seats facing forward. The number of ways to do that is9·8·7. In a similar way, 2 people canoccupy 8 seats facing backward in8·7 ways. Now17−5 = 12seats left. They can be occupiedby 7−3−2 = 2 nice people in12·11ways. As a result, we get

9·8·7×8·7×12·11= 3725568

ways.

Problem 4.2: Find the number of ways in which 8 persons can be assigned to 2 rooms (A andB) if each room must have at least 3 persons in it.

Solution 4.2: The number of persons in the room is between 3 and 5. Let us start with thesituation where room A has 3 persons. The number of ways to do that is8 · 7 · 6. The totalnumber of ways is then

8·7·6+8·7·6·5+8·7·6·5·4 = 8736.

49

50 CHAPTER 4. ELEMENTARY PROBABILITY THEORY . . .

Problem 4.3: Find the number of permutations of the letters in the word MONOTONOUS. Inhow many ways are 4 O’s together? In how many ways are (only) 3 O’s together?

Solution 4.3: The total number of the letters is 10, we have 4 O’s, 2 N’s, and other letters do notrepeat. Thus the total number of permutations is10!/(4!2!) = 75600. To get the number of waysto have 4 O’s together we have to use model word MONTNUS which has 7 letters with 2 repeatedN’s. Thus the number of ways is7!/2! = 2520. To have three O’s together we have 6 places toinsert a separate O to the word MONTNUS, thus the number of ways is6×2520= 15120.

Problem 4.4: In how many ways can 5 red balls, 4 blue balls, and 4 white balls be placed in arow so that the balls at the ends of the row are of the same color?

Solution 4.4: Let us start with 5 red balls and choose 2 of them to be at the ends. The rest5+4+4−2= 11balls are in the middle. They have to be permuted, but permutation of the ballsof the color must be excluded. Among them the have 3 rest red balls, 4 blue balls and 4 whiteballs. Thus we have11!/(3!4!4!) = 11550ways. Now let us do the same with white balls. Nowwe have 2 rest white balls, but 5 blue balls to permute. We get11!/(2!5!4!) = 6930ways. Thesame is true for blue balls. The total number of ways is138600+2×83160= 25410ways.

Problem 4.5: Three coins are tossed.

(a) Find the probability of gettingnoheads.

(b) Find the probability of gettingat least one head.

(c) Show that the event “heads on the first coin” and the events “tails on the last coin” areindependent.

(d) Show that the event “only two coins heads” and the event “three coin heads” are dependentand mutually exclusive.

Solution 4.5:

(a) The answer is(1/2)3 = 1/8.

(b) The sum of the probabilities is 1. Since the probability to get no heads is 1/8, the answeris 1−1/8 = 7/8.

(c) We assume that result of coin tossing is fully random.

(d) The event “3 heads” contradicts the requirement “only 2”.

51

Problem 4.6: Various 6 digit numbers can be formed by permuting the digits 666655. Allarrangements are equivalently likely. Given that the number is even, what is the probability thattwo fives are together?

Hint: You must find a conditional probability.

Solution 4.6: Since the number must be even, the last number is 6, and there are 4 ways tochoose this number. There are 5 digits left, and the total number of their permutations give 5!.We have to divide it by 3! of permutations of 6 and 2! permutations between 5. We end at thenumber of different numbers is4 ·5!/3!2! = 40. If we want to have two fives together, it is thesame as permute 4 units, the number of permutations4·4!/3! = 16. Thus the probability is

P = 16/40= 2/5.

Problem 4.7: Fifteen boys go hiking. 5 get lost, 8 get sunburned, and 6 return home withoutproblems.

(a) What is the probability that a sunburned boy got lost?

(b) What is the probability that a lost boy got sunburned?

Solution 4.7: See. pp. 176-177 of the book [1].Let us define the event to get sunburned asA, and the event to get lost -B. The probability to

get sunburned isP(A) = 8/15, while the probability to get lost isP(B) = 5/15= 1/3. ThusWe know that15− 6 = 9 boys had problems. Thus, the joint probability to be either sun-

burned or lost is,P(A∪B) = 9/15. Thus the probability to be both sunburned and lost is

P(A∩B) = P(A)+P(B)−P(A∪B) = (8+5−9)/15= 4/15.

Consequently, conditional probabilities are

P(A|B) = P(A∩B)/P(A) = 1/2 (a), P(B|A) = P(A∩B)/P(B) = 4/5 (b),

respectively.

Problem 4.8: A stochastic variableX can have valuesx = 1 andx = 3. A stochastic variableY can have valuesy = 2 andy = 4. Denote the joint probability density

PX,Y(x,y) = ∑i=1,3

∑j=2,4

pi, j δ(x− i)δ(y− j) .

Compute the covariance ofX andY for the following 2 cases:

(a) p1,2 = p1,4 = p3,2 = p3,4 = 1/4;

(b) p1,2 = p3,4 = 0 andp1,4 = p3,2 = 1/2.

For each case decide ifX andY are independent.


Solution 4.8:

(a) First let us calculate

〈XY〉= (1·2+1·4+3·2+3·4)/4 = 6.

At the same time〈X〉= 2 and〈Y〉= 3. Since〈XY〉= 〈X〉 · 〈Y〉 the variables are indepen-dent.

(b) We have,〈XY〉= (1·4+3·2)/2 = 5.

It is not equal to〈X〉 · 〈Y〉, so the variables are dependent.

Problem 4.9: The stochastic variablesX andY are independent and Gaussian distributed withfirst moment〈x〉= 〈y〉= 0 and standard deviationσX = σY = 1. Find the characteristic functionfor the random variableZ = X2 +Y2, and compute the moments〈z〉, 〈z2〉 and〈z3〉. Find the first3 cumulants.

Solution 4.9: First let us find the distribution function forZ. Let us do the problem in twoways.Simple way:

Let us transform the problem to the cylindrical coordinates,x,y→ ρ,φ. We have

P(x)P(y) =12π

e−(x2+y2) =12π

e−ρ2, dxdy= ρdρdφ, .

Thus,

PZ(z) =12π

∫ρdρdφe−ρ2/2δ(z−ρ2) =

12

e−z/2 .

General way:

PZ(z) =∫

dxPX(x)∫

dyPY(y)δ(x2 +y2−z)

=∫ √

z

−√z

dx

2√

z−x2PX(x)∑

±PY

(±

√z−x2

)

=1π

∫ √z

0

dx√z−x2

exp

(−x2

2− z−x2

2

)=

12

e−z/2 .

Now, the characteristic function is

fZ(k) = 〈eikz〉=12

∫ ∞

0dze(ik−1/2)z =

11−2ik

.

53

Then,

〈zn〉= (−i)n dn fZ(k)dkn

∣∣∣∣k→0

.

Performing calculations (see the Maple file), we obtain:

〈z〉= 2, 〈z2〉= 8, 〈z3〉= 48.

Thus,

C1(Z) = 〈z〉= 2, , C2(Z) = 〈z2〉−〈z〉2 = 4,

C3(Z) = 〈z3〉−3〈z〉〈z2〉+2〈z〉3 = 48−3·2·8+3·23 = 24.

Problem 4.10: A die is loaded so that even numbers occur 3 times as often as odd numbers.

(a) If the die is thrownN = 12 times, what is the probability that odd numbers occur 3 times?If is is thrownN = 120times, what is the probability that odd numbers occur 30 times?Use the binomial distribution.

(b) Compute the same quantities as in part (a), but use the Gaussian distribution

Note: For parts (a) and (b) compute your answers to four places.

(c) Compare answers (a) and (b). Plot the binomial and Gaussian distributions for the caseN = 12.

Solution 4.10:

(a) For the random process, the parial probability for an odd numberp = 1/4 and for an evennumber isq = 1− p = 3/4. The probability to havek odd numbers, according to binimialdistribution is

PbN(k) =

N!k!(N−k)!

pkqN−k .

Consequently (see Maple file),

Pb12(3) =

12!3!9!

39

412 = 0.2881.

In a similar way,Pb120(30) = 0.08385.

(b) The number of odd numbers is〈k〉= N/4 and the dispersion isσ = Np·q = 3N/16. Thus

PGN (k) =

1√3Nπ/8

exp

(−(k−N/4)2

3N/8

).

As a result,PG

12(3) = 0.2660, PG120(30) = 0.0841.

(c) Plots are shown in the Maple file.


Problem 4.11: A book with 700 misprints contains 1400 pages.

(a) What is the probability that one page contains no mistakes?

(b) What is the probability that one pages contains 2 mistakes?

Solution 4.11: n mistakes can be distributed betweeng pages by

wg(n) =(g+n−1)!(g−1)!n!

.

If one has a page without mistakes, it means thatn mistakes are distributed betweeng−1 pages.The probabiliy of that event is

P(0) =(g+n−2)!(g−2)!n!

n!(g−1)!(g+n−1)!

=g−1

g+n−1.

In our caseg = 1400andn = 700, thusP(0) = 1399/2099≈ 0.67. In a similar way, if we havetwo misprints at a page, thenn− 2 misprints are distributed betweeng− 1 pages. Thus, thenumber of ways to do that iswg−1(n−2). Consequently,

P(2) =wg−1(n−2)

wg(n)=

(g+n−4)!(g−2)!(n−2)!

n!(g−1)!(g+n−1)!

=(g−1)n(n−1)

(g+n−1)(g+n−2)(g+n−3).

We getP(2)≈ 0.07.

Problem 4.12: There old batteries and a resistor,R, are used to construct a circuit. Each batteryhas a probabilityp to generate voltageV = v0 and has a probability1− p to generate voltageV = 0 Neglect internal resistance in the batteries.

Find the average power,〈V2〉/R, dissipated in the resistor if

(a) the batteries are connected in the series and

(b) the batteries are connected in parallel.

In cases (a) and (b), what would be the average power dissipated in all batteries werecertain to generate voltagev0.

(c) How you realize the conditions and results in this problem in the laboratory?

55

Solution 4.12: Let us start with series connection and find the probability to find a given valueof the voltage. We have 4 possible values of the voltage. In unitsv0 they are 0, 1, 2, and 3. Wehave,

P(0) = (1− p)3 , P(1) = 3p(1− p)2 , P(2) = 3p2(1− p) , P(3) = p3 .

Consequently,

〈W〉= 〈V2〉/R= ∗(v20/R)

[3p(1− p)2 +12p2(1− p)+9p3] = (3v2

0/R) p(1+2p) .

Now let us turn to parallel connection. Let us for a while introduce conductanceGi = 1/Ri ofthe batteries, as well as conductanceG = 1/R of the resistor. The voltage at the resistor is givenby the expression

V =∑i=1,2,3EiGi

G+∑i=1,2,3Gi.

HereEi areelectro-motive forcesof the batteries. Now the crucial point is to assume somethingabout the resistances of the batteries. Let us for simplicity assume that they are equal, andGi À G. Actually, it is a very bad assumption for old batteries. Under this assumption,V =(E1 +E2 +E3)/3. The the result is just 9 times less that that for series connection.

To check this kind of phenomena in the lab one has to find many old batteries at a junk-yard,first check distribution of their voltages and then make circuits.

The problem is badly formulated.

Problem 4.13: Consider a random walk in one dimension. In a single step the probability ofdisplacement betweenx andx+dx is given by

P(x)dx=1√

2πσ2exp

(−(x−a)2

2σ2

)dx. (4.1)

After N steps the displacement of the walker isS= X1 + X3 + XN whereXi is the displacementafter ith step. AfterN steps,

(a) what is the probability density for the displacement,S, of the walker, and

(b) what is his standard deviation?

Solution 4.13:

(a) By definition,

PS(s) =∫

dx1P(x1)∫

dx2P(x2) . . .∫

dxN P(xN)δ(s−x1−x2 . . .−xN) .

To calculate this integral we expandδ-function into Fourier integral

δ(x) =∫ ∞

−∞

dk2π

eikx .


In this way we get

PS(s) =∫ ∞

−∞

dk2π

eiks∫

dx1e−ikx1P(x1)∫

dx1e−ikx2P(x2) . . .

∫dxN e−ikxNP(xN)

=∫ ∞

−∞

dk2π

eiks(∫ ∞

−∞dxe−ikxP(x)

)N

=∫ ∞

−∞

dk2π

eiks[P (k)]N . (4.2)

HereP (k) =

∫ ∞

−∞dxe−ikxP(x)

is the Fourier component of the single-step probability. From Eq. (4.1) we obtain

P (k) = exp(−ika−k2σ2/2

) → [P (k)]N = exp(−iNka−k2Nσ2/2

).

Now,

PS(s) =∫ ∞

−∞

dk2π

exp[ik(s−Na)−k2Nσ2/2

]=

1√2πNσ2

exp

(−(s−Na)2

2Nσ2

).

(b) We have (see Maple file),

〈S〉=∫

dssPS(s) = Na, 〈S2〉−〈S〉2 = Nσ2 .

Problem 4.14: Consider a random walk in one dimension for which the walker at each stepis equally likely to take a step with displacement anywhere in the intervald− a≤ x≤ d + a,wherea < d. Each step is independent of others. AfterN steps the displacement of the walker isS= X1 +X3 + · · ·+XN whereXi is the displacement afterith step. AfterN steps,

(a) what is the average displacement,〈S〉, of the walker, and

(b) what is his standard deviation?

Solution 4.14: Starting from the definition, we have

〈S〉 =∫

sPS(s)ds=∫

sds∫

dx1P(x1) · · ·∫

dxN P(xN)δ(s−x1− . . .−xN)

=∫

dx1P(x1) · · ·∫

dxN P(xN)(x1 +x2 + . . .xN) = N〈X〉= Nd.

In a similar way,

〈S2〉=∫

dx1P(x1) · · ·∫

dxN P(xN)(x1 +x2 + . . .xN)2 .

57

We know that (∑i

xi

)2

= ∑i

x2i + ∑

i 6= j

xix j .

The number of pairs with non-equali and j is N(N−1). After averaging the obtain

〈S2〉= N〈X2〉+N(N−1)〈X〉2 .

To calculate〈X2〉 let us specify the normalized probability as

P(x) =12a

1, d−a≤ x≤ d+a0, otherwise

.

Then,

〈X2〉=12a

∫ d+a

d−ax2dx= d2 +a2/3.

Summarizing we get (see Maple file)

〈S2〉−〈S〉2 = N(d2 +a2/3)+N(N−1)d2− (Nd)2 = Na2/3.

Problem 4.15: Consider a random walk in one dimension. In a single step the probability ofdisplacement betweenx andx+dx is given by

P(x)dx=1π

ax2 +a2 dx. (4.3)

Find the probability density for the displacement of the walker afterN steps. Does it satisfy theCentral Limit Theorem? Should it?

Solution 4.15: Using the method of the Problem4.13we get

P (k) = e−a|k| → PS(s) = 2∫ ∞

0

dk2π

e−Nkacosks=1π

Na2

Na2 +s2 .

It does not satisfy the Central Limit Theorem. It should not because〈X2〉 is divergent.

Chapter 5

Stochastic Dynamics and Brownian Motion

Quick access:1 2 3 4 5 6 7 8 9 1011Test 1

Problem 5.1: Urn A has initially 1 red and 1 white marble, and urn B initially has 1 whiteand 3 red marbles. The marbles are repeatedly interchanged. In each step of the process onemarble is selected from each urn at random and the two marbles selected are interchanged. Letthe stochastic variableY denote “configuration of the units”. There configurations are possible.They are shown in Fig.5.1

3

1

2

Figure 5.1:To the problem 5.1

We denote these 3 realizations asy(n) wheren = 1, 2 and 3.

(a) Compute the transition matrixQ and the conditional probability matrixP(s0|s).(b) Compute the probability vector,〈P(s)|, at times, given the initial condition stated above.

What is the probability that there are 2 red marbles in urn A after 2 steps? After manysteps?

(c) Assume that the realization,y(n), equals ton2. Compute the moment,〈y(s)〉, and theautocorrelation function,〈y(0)y(s)〉, for the same initial conditions as in part (b).

59

60 CHAPTER 5. STOCHASTIC DYNAMICS ...

Solution 5.1: The configurations are shown in Fig.5.1. By inspection of the figure we get,

Q11 = 0, Q12 = 1, Q13 = 0,

Q21 =12· 14

=18

, Q22 =12· 14

+12· 34

=12

, Q23 =12· 34

=38

,

Q31 = 0, Q32 =12

, Q33 =12

.

Thus the transition matrixQ is given by the expression

Q =

0 1 01/8 1/2 3/80 1/2 1/2

,

while the initial condition corresponds to the probability vector

〈p(0)|= (0,1,0) .

Below we proceed as in exercise 5.1 of the book [1].The following procedure is clear form the Maple printout. First we find eigenvalues ofQ

which are equal to1,∓1/4. The corresponding “right” eigenvectors are:

[1, 1, [1, 1, 1]], [−14

, 1, [6,−32

, 1

]], [

14, 1, [4, 1,−2]]

Thus we denoteλ1 = 1, λ2 =−1/4, λ3 = 1/4, and

|1〉= (1,1,1) , |2〉= (6,−3/2,1) , |3〉= (1,2,−3) .

The “left” eigenvectors are just the right eigenvalues of the transposed matrix,QT . We have forthem the Maple result

[14, 1, [1, 2,−3]], [

−14

, 1, [1,−2, 1]], [1, 1, [1, 8, 6]]

Thus,〈1|= (1,8,6) , 〈2|= (1,−2,1) , 〈3|= (1,−2,1) .

It is easy to prove that

〈i|k〉= ni δik , where n1 = 15, n2 = 10, n3 = 12.

Now we construct matricesPi = |i〉〈i| to get

P1 =

1 8 61 8 61 8 6

, P2 =

6 −12 6−3/2 3 −3/2

1 −2 1

, P3 =

4 8 −121 2 −3

−2 −4 6

.

61

Now, P(s0|s) = P(s−s0) with

P(s) = ∑i

λsi

niPi

=

115

+35

(−14

)s+13

(14)s 8

15− 6

5(−14

)s+23

(14)s 2

5+

35

(−14

)s− (14)s

115− 3

20(−14

)s+112

(14)s 8

15+

310

(−14

)s+16

(14)s 2

5− 3

20(−14

)s− 14

(14)s

115

+110

(−14

)s− 16

(14)s 8

15− 1

5(−14

)s− 13

(14)s 2

5+

110

(−14

)s+12

(14)s

As it should be,P(0) = I . In this way we solve the part (a).Now we come to the point (b). Straightforward calculation gives

〈P(s)| = 〈p(0)|P(s)

=[

115− 3

20(−14

)s+112

(14)s,

815

+310

(−14

)s+16

(14)s,

25− 3

20(−14

)s− 14

(14)s]

.

We are asked to find the probability of the configuration 3. After 2 steps it is 3/8, and ats→ ∞we get 2/5.

The part (c) is calculated in a straightforward way. We obtain

〈y(s)〉 =295− 3

10(−14

)s− 32

(14)s,

〈y(0)y(s)〉 =7950

(−1)(1+s) 4(−s)− 334

16(−s) +81100

(−1)(1+2s) 16(−s)

+112

(−1)(1+s) 16(−s)− 32

4(−s) +84125

Problem 5.2: Three boys, A,B, and C, stand in a circle and play catch (B stands to the right ofA). Before throwing the ball, each boy flips a coin to decide whether to throw to the boy on hisright or left. If “heads” comes up, the boy throws to his right. If “tails” come up, the boy throwsto his right.

The coin of boy A is “fair” (50% heads and 50% tails), the coin of boy B has head of bothsides, and the coin of boy C is weighted (75% heads and 25% tails).

(a) Compute the transition matrix, its eigenvalues, and its left and right eigenvalues.

(b) If the ball is thrown in regular intervals, approximately what fraction of time each boy havethe ball (assuming they throw the ball many times)?

(c) If boy A has the ball to begin with, what is the chance it has it after 2 throws? What is thechance he will have in afters throws.


Solution 5.2: The problem is similar to the previous one. Let us map the notations

A→ 1, B→ 2, C→ 3,

and note that “right” means counterclockwise. Then

Q12 = Q13 = 1/2, Q23 = 1, Q21 = 0, Q31 = 3/4, Q32 = 1/4.

At the same timeO11 = Q22 = Q33 = 0. As a result, the transition matrix has the form

Q =

0 1/2 1/20 0 1

3/4 1/4 0

.

The following solution is similar to the previous problem (see the Maple printout). Note that theeigenvalues and eigenvectors are complex, but conjugated. It means that the probability is real,but oscillates in time).

The answers are: after many throws the chance for the boy A to hold the ball is 6/19; startingfrom the given initial conditions his chance to hold the ball after 2 throws is 3/8.

Problem 5.3: A trained mouse lives in the house shown in Fig.5.2.

C (3)

B (2)

A (1)

Figure 5.2:

A bell rings at regular intervals (short compared with the mouses lifetime). Each time it rings,the mouse changes rooms. When he changes rooms, he is equally likely to pass through any ofthe doors of the room he is in.

Let the stochastic variableY denotes “mouse in a particular room”. There 3 realizations ofY:

1. “Mouse in room A”

2. “Mouse in room B”

3. “Mouse in room C”

63

which we denote asy(1), y(2), andy(3), respectively.

(a) compute the transition matrix,Q, and the conditional probability matrix,P(s0|s).

(b) . Compute the probability vector,〈P(s)|, at times, given that the mouse starts in room C.Approximately, what fraction of his life does he spend in each room?

(c) Assume that the realization,y(n) equals ton. Compute the moment,〈y(s)〉, and the auto-correlation function,〈y(0)y(s)〉, for the same initial condition as in part (b).

Solution 5.3: The problem is very much similar to the previous ones. We have

Q12 = Q13 = 1/2, Q21 = 1/3, Q23 = Q32 = 2/3, Q31 = 1/3, Qii = 0.

Thus

Q =

0 1/2 1/21/3 0 2/31/3 2/3 0

.

Now we do the same as in previous problems (see Maple printout). The answers are obviousfrom the printout.

Problem 5.4: The doors in the mouse’s house shown in Fig.5.2so thay get periodically largerand smaller. This causes the mouse’s transition between the rooms to become time periodic. Letthe stochastic variableY have the same meaning as in Problem 5.3. The transition matrix is nowgiven by

Q11(s) = Q22(s) = Q33(s) = 0, Q12(s) = cos2(πs/2) , Q12(s) = sin2(πs/2) ,

Q21(s) = 1/4+(1/2)sin2(πs/2) , Q23(s) = 1/4+(1/2)cos2(πs/2)Q31(s) = (1/2)cos2(πs/2) , Q32(s) = (1/2)+(1/2)sin2(πs/2) ,

(a) If initially the mouse is in room A, what is the probability to find it in room A after2sroom changes?

(b) If initially the mouse is in room B, what is the probability to find it in room A after2s roomchanges? In room B?

Solution 5.4: The solution is rather tedious, but obvious from the Maple file.


Problem 5.5: Consider a discrete random walk on a one-dimensionalperiodic lattice with2N+1 sites (label the sites from−N to N). Assume that the walker is equally likely to move onthe lattice site to the left or right at each step. Treat this problem as a Markov chain.

(a) Compute the transition matrix,Q, and the conditional probability,P(s0|s).(b) Compute the probabilityP1(n,s) at times, given the walker starts at siten = 0.

(c) If the lattice has 5 sites(N = 2), compute the probability to find the walker on each siteafters= 2 steps and afters= ∞ steps.Assume that the walker starts at siten = 0.

Hint: Let us understand the wordperiodicas peridically extended, i. e. possessing the property

P[n+(2N+1), t] = P(n, t) .

Solution 5.5: The transition probabilities are given by the expressions

wn−1,n = wn,n−1 = wn+1,n = wn,n+1 = 1/2.

The Master equation has the form

dPn/ds = wn−1,nPn−1 +wn+1,nPn+1−Pn(wn,n−1 +wn,n+1)= (1/2)(Pn−1 +Pn+1−2Pn) .

Let us look for the solution in the form

Pn(s) ∝ eλs+iκn .

Then we getλ =−[1−cos(κ)] =−2sin2(κ/2) .

Now we have to findκ. If we apply cyclic boundary conditions (assuming that the wordperiodicin the problem formulation means that) we have to require:P(n+2N+1) = P(n) at anys. Thus

eiκ(2N+1) = 1 → κ(2N+1) = 2πk,

wherek is the integer number ranging from−N to N. Thus, the general solution is

P(n,s) =N

∑k=−N

Pk exp

[−in

2πk2N+1

−2ssin2 πk2N+1

].

The quantitiesPk should be found from the initial conditions. Ats= 0 we can rewrite the l. h. s.of the requirementP(n,0) = δn,0 in the form

P(n,0) =N

∑k=−N

Pk e−in2πk/(2N+1) .

65

On the other hand,

δn,0 =1

2N+1

N

∑k=−N

e−in2πk/(2N+1) .

From that we immediately getPk = const(k). The proper constant1/N (s) is the normalizationcoefficient found from the condition

N

∑n=−N

P(n,s) = 1.

Finally

P(n,s) =1

N (s)

N

∑k=−N

exp

[−in

2πk2N+1

−2ssin2 πk2N+1

].

Here

N (s) =N

∑n=−N

exp

[−in

2πk2N+1

−2ssin2 πk2N+1

]= 2N+1.

Now let us find the conditional probabilityP(n,s|m,0). Now we have to apply the initial condi-tion P(n,0) = δm,0, or

P(n,0) =N

∑k=−N

Pk e−in2πk/N = δn,m =N

∑k=−N

Pk e−i(n−m)2πk/(2N+1) .

Thus,Pk =∝ eim2πk/(2N+1) .

Consequently,

P(s−s0|0)mn = P(n,s|m,0) =1

2N+1

N

∑k=−N

exp

[−i(n−m)

2πk2N+1

−2ssin2 πk2N+1

].

As s→ ∞ the probability tends to1/(2N+1). The calculation forN = 2 is straightforward, seeMaple file. Finally, the matrixQmn = P(1|0)mn.

Problem 5.6: At time t, a radioactive sample containsn identical undecayed nuclei, each withthe probability for the unit time,λ, of decaying. The probability of decay during the timet →t +∆t is λ∆t. Assume that at timet = 0 there aren0 undecayed nuclei present.

(a) Write down and solve the master equation of the process [findP1(n, t)].

(b) Compute the the mean number of undecayed nuclei and the variance as a function of time.

(c) What is the half-life of this decay process?


Solution 5.6: The master equation has the form

dPdt

=−λnP(n, t)+λ(n+1)P(n+1, t) .

This is a linear master equation, and we solve it by the method of generating functional, p. 262of the book [1]. We introduce

F(z, t) =∞

∑n=−∞

znP1(n, t) .

Then we multiply the master equation byzn and sum overn. We get

∂F∂t

=−λ(z−1)∂F∂z

. (5.1)

Indeed,

∑n

zn(n+1)P(n+1, t) =ddz∑

nzn+1P(n+1, t) =

∂F∂z

.

In a similar way

∑n

znnP(n, t) = z∑n

zn−1nP(n, t) = z∂F∂z

Introducingτ = λt we rewrite Eq.5.1 in the form

∂F∂τ

+(z−1)∂F∂z

= 0. (5.2)

The characteristics of this equation meet the equation

dτ =dz

z−1→ ln(z−1)− τ = const.

Consequently,

F(z, t) = Φ(

e−τ+ln(z−1))

= Φ[(z−1)e−τ] .

SinceP(n,0) = δn,n0 we get

Φ(z−1) = zn0 → Φ(u) = (u+1)n0 .

As result,F(z, t) =

[(z−1)e−τ +1

]n0 .

Now,

〈n(t)〉=(

∂F∂z

)

z=1= n0e−τ .

In a similar way,〈n2(t)〉−〈n(t)〉2 = n0e−τ (

1−e−τ) .

67

To find the probabilities let us express the quantity

[(z−1)e−τ +1

]n0 = e−n0τ [z+(eτ−1)]n0 = e−n0τn0

∑n=0

n0!n!(n0−n)!

zn(eτ−1)n0−n .

Thus,

P1(n,τ) =n0!

n!(n0−n)!e−n0τ (

1−e−τ)n0−n.

Problem 5.7: Consider a random walk on the lattice shown in Fig.5.3. The transition rates are

4

1

2 3

Figure 5.3:A sketch to the problem5.7.

w12 = w13 = 1/2, w21 = w23 = w24 = 1/3, w31 = w32 = w34 = 1/3,

w42 = w43 = 1/2, w14 = w41 = 0.

(a) Write the transition matrix,W, and show that that this system obey detailed balance.

(b) Compute the symmetric matrix,V, and find its eigenvalues and eigenvectors.

(c) Write P1(n, t) for the caseP(n,0) = δn,1. What isP(4, t)?

Solution 5.7: The solution is similar to the Exercise 5.2 in p. 245 of the book [1]. The answersare obvious from the Maple file.

Problem 5.8: Consider a random walk on the lattice shown in Fig.5.4. The siteP absorbs thewalker. The transition rates are

w12 = w13 = 1/2 w21 = w23 = w2P = 1/3,

w31 = w32 = w3P = 1/3, wP1 = wP2 = 1/2.

(a) Write the transition matrix,M , and compute its eigenvalues and eigenvectors.

(b) In the walker starts at siten = 1 at timet = 0, compute the mean first passage time.


P

1

2 3

Figure 5.4:A sketch to the problem5.8.

Solution 5.8: The solution is similar to the Exercise 5.4 of the book [1]. The answers are clearfrom the Maple file.

Problem 5.9: Let us consider onRL electric circuit with resistance,R, and inductance,L,connected in series. Even there is no average electromotive force (EMF) exists across the resistor,because of the discrete character of electrons in the circuit and their random motion, a fluctuatingEMF, ξ(t), exists whose strength is determined by the temperature,T. This, in turn, induces thefluctuating current,I(t), in the circuit. The Langevin equation for the current is

Ld I(t)

dt+RI(t) = ξ(t) . (5.3)

If the EMF is delta-correlated,

〈ξ(t2)ξ(t1)〉= gδ(t2− t1)

and〈ξ(t)〉ξ = 0, computeg and the current correlation function〈〈I(t2)I(t1)〉ξ〉T . Assume thatthe average magnetic energyL〈I2

0〉T/2 = kT/2 and〈I0〉T = 0 whereI(0) = I0.

Solution 5.9: Let us divide Eq. (5.3) by R, denoteL/R≡ τ and measure time in units ofτ andthe current in unitsI1 ≡ R−1√g. Namely, defineθ = t/τ, and j(t) = I(t)/I1. Then the equationfor the current has the form

d j/dθ+ j = η(θ) , 〈η(θ2)η(θ1)〉= δ(θ−θ1) .

Its solution has the form

j(θ) = j0e−θ +∫ θ

0dses−θη(s) .

Multiplying solutions for j(θ2) and for j(θ1) and averaging over realizations ofη we get

〈 j(θ2) j(θ1)〉η = j20e−(θ2+θ1)

+∫ θ2

0ds2

∫ θ1

0ds1δ(s2−s1)es1+s2−θ1−θ2

=(

j20−12

)e−θ1−θ2 +

12

e−|θ2−θ1| .

69

The first item must vanish since in the equilibrium the process must bestationary, i. e. it mustdepend only on the differenceθ2−θ1. Thus,

〈 j20〉T =12

→ 〈I20〉T =

g2R2 .

On the other hand,

L〈I20〉= kT → g = 2R2kT/L .

Consequently,

〈〈 j(θ2) j(θ1)〉η〉T =12

e−|θ2−θ1|, .

In dimensional variables,

〈〈I(t2)I(t1)〉ξ〉T = I21〈〈 j(θ2) j(θ1)〉η〉T =

kTL

e−|t2−t1|/τ .

Problem 5.10: Due to the random motion and discrete nature of electrons, an LRC seriescircuit experience a random electromotive force (EMF),ξ(t). This, in turn, induces a randomvarying charge,Q(t), on the capacitor plates and a random current,I)t) = dQ(t)/dt, through theresistor and inductor. The random charge,Q(t), satisfies the Langevin equation

Ld2Qdt2

+RdQdt

+QC

= ξ(t) .

Assume that EMF is delta-correlated,

〈ξ(t2)ξ(t1)〉= gδ(t2− t1)

and〈ξ(t)〉ξ = 0. Assume that the circuit is at the temperatureT and the average magnetic energyin the inductor and average electric energy in the capacitor satisfy the equipartition theorem,

12

L〈I20〉T =

12C〈Q2

0〉T ,

whereQ(0) = Q0, I(0) = I0. Assume that〈I0〉T = 0, 〈Q0〉T = 0, 〈Q0 I0〉T = 0.

(a) Compute the current correlation function〈〈I(t2)I(t1)〉ξ〉T .

(b) Compute the variance of the charge distribution,〈〈(Q(t)−Q0)2〉ξ〉T .


Solution 5.10: The problem is exactly the same as in the Exercise 5.5 of the book [1]. Aftermapping

Q→ x, I → v, L→m, LC→ ω20, R→ γ

we obtain the same equation. Thus,

〈〈I(t + τ)I(t)〉ξ〉T =kTL

e−Γ|τ|C(|τ|) ,

whereC(τ) = coshδτ− (Γ/δ)sinhδτ , Γ = R/L , δ =

√Γ2−ω2

0 .

The variance of charge distribution can be calculated by integrating the formal solution

I(t) = I0e−Γt −Q0ω2

0

δe−Γt sinhδt +

1L

∫ t

0dt′ ξ(t ′)e−Γ(t−t ′)C(t− t ′)

over time:

Q(t) = Q0 + I01−e−Γt

Γ−Q0

ω20

2δ

(1−e−(Γ−δ)t

Γ−δ− 1−e−(Γ+δ)t

Γ+δ

)

+1L

∫ t

0dt′′

∫ t

0dt′ ξ(t ′)e−Γ(t ′′−t ′)C(t ′′− t ′)

Now we do averages overξ andT to obtain (linear terms vanish!)

〈〈Q2(t)〉ξ〉T = 〈Q20〉T

[1− ω2

0

2δ

(1−e−(Γ−δ)t

Γ−δ− 1−e−(Γ+δ)t

Γ+δ

)]2

+ 〈I20〉T

(1−e−Γt

Γ

)2

+gL2

∫ t

0dt1

∫ t

0dt2

∫ t

0dt3

∫ t

0dt4e−Γ(t1−t2+t3−t4)C(t1− t2)C(t3− t4)δ(t2− t4)

Now we substitute〈Q20〉T = CkT, 〈I2

0〉T = kT/L, andg = 4RkT, and then applyδ function. Thelast term becomes

4RkTL2

∫ t

0dt1

∫ t

0dt2

∫ t

0dt3e−Γ(t1−2t2+t3)C(t1− t2)C(t3− t2) .

The rest integrations must be done explicitly which results in lengthy expressions.

Problem 5.11: Consider a Brownian particle of massm in one dimension in the presence ofa constant forcef0 in a fluid with force constantγ and in th4e presence of a delta-correlatedrandom forceξ(t) such that〈ξ(t1)ξ(t2)〉ξ = gδ(t1− t2) and〈ξ(t)〉ξ = 0. Assume that the velocityand displacement of the Brownian particle att = 0 arev0 andx0, respectively.

(a) Compute the velocity correlation function〈v(t1)v(t2)〉ξ.

(b) Compute the displacement variance〈(x(t)−x0)2〉.

71

Solution 5.11: The differential equation for the particle is

d v(t)dt

+γm

v(t)− f0m

=1m

ξ(t) ,dxdt

= v(t) .

Its solution is

v(t) = v0e−(γ/m)t +f0γ

(1−e−(γ/m)t

)+

1m

∫ t

0dt′e(γ/m)(t ′−t)ξ(t ′) .

The quantity

vd(t) =f0γ

(1−e−(γ/m)t

)

is just the drift velocity in the forcef0. Using results of the Sec. 5.E.1 of the book [1] we obtain

〈〈[v(t + τ)−vd(t + τ)] [v(t)−vd(t)]〉ξ〉T =kTm

e−(γ/m)|τ| .

Since〈v(t)〉T = vd(t) we obtain

〈〈v(t + τ)v(t)〉ξ〉T = vd(t + τ)vd(t)+kTm

e−(γ/m)|τ| .

In a similar way, we cam introduce drift displacement

xd(t) =∫ t

0dt′

f0γ

(1−e−(γ/m)t ′

)=

f0γ

[t−m

γ

(1−e−(γ/m)t

)].

Finally we obtain

〈〈[x(t)−x0−xd(t)]2〉ξ〉T =

2kTγ

[t−m

γ

(1−e−(γ/m)t

)].

Tests for training

Problem 5.12: Stochastic processesConsider an electron (spinS= 1/2) bounded to an impiruty center and assume that the tempera-tureT is much less than the energy difference between the levels of orbital motion. Consequently,only spin degrees of freedom are excited.

Let the system be placed in external magnetic fieldB, the additional energy in magnetic fieldis ±βB where± corresponds to the values±1/2 of the spin component along the field. Hereβ = |e|h/2mc is the Bohr magneton,c is the velocity of light since the Gaussian system is used.The magnetic moment of the states withSz =±1/2 is∓β.

(a) Find the stationary probabilities for the states withSz =±1/2, P±.- Find average magnetic moment of the system as function of temperatureT and magnetic fieldB.


(b) Find mean square fluctuation of magnetic moment,〈(∆M)2〉 and the ratio〈(∆M)2〉/〈M〉2.- Show that

〈(∆M2)〉=β2

cosh2(βB/T). (5.4)

- Discuss limiting cases of large and low temperatures. Explain these results qualitatively.

(c) Write down the Master Equation for the probabilityPi(t) to find the system in the statei =±at timet.- Using the result for the stationary case find the relation between the transition probabilitiesW+− andW−+ for the transition from the state “+” to the state “-” and for the reverse transition,respectively. Express the probabilitiesW+− andW−+ through the quantityW = W+−+W+−.- Derive the Master Equation for the population difference,P (t)≡ w+(t)−w+(t).

(d) Write down the Master Equation for theconditional probability, P( f , t|i, t0), to find the systemin the statef at timet under the conditions that at timet0 it was in the statei.- Specify initial conditions to this equation.- Show that solution of this equation with proper initial condition is

P( f , t|i, t0) = wf +(δi f −wf )e−W|t−t0| (5.5)

wherewf is the stationary probability to find the system in the statef .

Hint: It is convenient to express the set of Master Equations for the quantitiesP( f , t|i, t0) atdifferenti and f as an equation for the2×2 matrixP with matrix elementsPf i(t|t0)≡P( f , t|i, t0).The solution of the matrix Master Equation∂P/∂T = WP at t ≥ t0 can then be searched in in thematrix asP ∝ ∑kC(k)e−λk(t−t0). Hereλk are eigenvalues of the matrixW, while C(k) are time-independent matrices. They are determined from the intial conditions and from the fact that att− t0→ ∞ the conditional probabilities tend to the stationary ones.

(e)Using Eq. (5.5) show that the correlation function〈∆M(t)∆M(0)〉 has the form

〈∆M(t)∆M(0)〉= 〈(∆M)2〉e−W|t| (5.6)

where〈(∆M)2〉 is given by Eq. (5.4) while W = W+−+W−+.- Find the fluctuation spectrum.

Solution 5.12: (a) Denotew± the probabilities of the statesSz =±1/2. We have

M =−β(w+−w−) . (5.7)

According to the Gibbs distribution,

w+/w− = e−2βB/T . (5.8)

73

Since only 2 spin levels are involved,

w+ +w− = 1. (5.9)

Combinig Eqs. (5.8) and (5.9) we obtain,

w+ =1

e2βB/T +1, w− =

e2βB/T

e2βB/T +1, w−−w+ = tanh

(βBT

)(5.10)

As a result,

M = β tanh

(βBT

)(5.11)

(b) We get:

〈M2〉 = β2(w+ +w−) = β2

〈M〉2 = β2 tanh2(βB/T) . (5.12)

As a result,

〈(∆M2)〉=β2

cosh2(βB/T),

〈(∆M)2〉〈M〉2 =

1

sinh2(βB/T)(5.13)

At low temperatures,T ¿ βB, fluctuations are exponentially suppressed because spin is alignedto magnetic field.

(c) The equation reads,

∂wi(t)∂t

=− ∑s=±

[Wi→swi(t)−Ws→iws(t)] (5.14)

- In the stationary situation, using the detailed equilibrium principle,

Wi→swi(t) = Ws→iws(t)

we obtain

W+−W−+

=w−w+

= e2βB/T , , W−+ =W

1+e2βB/T, W+− =

W

1+e−2βB/T(5.15)

- Coming back to Eq. (5.14) we get

∂w+(t)∂t

= −W+−w+(t)+W−+w−(t) ,

∂w−(t)∂t

= −W−+w−(t)+W+−w+(t) . (5.16)

Subtracting equations we arrive at the result,

∂P (t)∂t

=−WP (t) (5.17)


(d) We consider astationaryrandom process, andP( f , t|i, t0) is a function oft−t0. Let us assumet0 = 0, t ≥ 0 and denote Let us for brevity denoteP( f , t|i,0)≡ Pf i(t). The Master Equation forPf i(t) has the form

∂Pf i(t)∂t

=−∑s[Wf→sPf i(t)−Ws→ f Psi(t)] . (5.18)

The initial condition to this equation is

Pf i(t) = δi f at t → 0. (5.19)

Let us introduce the matrixP(t) with matrix elementsPi f (t) = P( f , t|i,0). Then Eq. (5.18) canbe rewritten as

∂P(t)∂t

= WP(t) , with W =( −Wf→s Ws→ f

Wf→s −Ws→ f

). (5.20)

It is natural to search the solution of Eq, (5.18) in the formP(t)∑kC(k)e−λkt . Thenλk are theeigenvalues of the matrixW while Ck are the numerical matrices to be determined from intialconditions att = 0 and from the fact thatPf i(t) → wf at t → ∞. Herewf is the stationaryprobability of the statef .

Equating the determinant ofW−λI we get

λ1 = 0, λ2 =−Wf→s+Ws→ f ≡−W .

HereI is the unit matrix Then, in general,

Pf i(t) = C(1)f i +C(2)

f i e−Wt

whereC(k) are time-independent matrices. From the initial conditions (5.19),

C(1)f i +C(2)

f i = δ f i .

On the other hand, att → ∞ the conditions probability tends to the stationary one,wf . Conse-quently,

C(1)f i = wf .

Finally, because of the time reversibility,Pf i(t) depends only on|t|. As result,

Pf i(t) = wf +(δ f i−wf )e−W|t|

This expression coincides with Eq. (5.5).

(e)By definition,

〈∆M(t)∆M(0)〉= 〈M(t)M(0)〉−〈M〉2 = ∑i f

MiM f [P( f , t|i,0)−wf ]wi .

75

Using Eq. (5.5), one can rewrite the above equation as

〈∆M(t)∆M(0)〉= e−W|t|∑i f

MiM f (δi f −wf )wi = 〈(∆M)2〉e−W|t|

This expresssion coincides with Eq. (5.6).The fluctuation spectrum is

(∆M)ω =∫ ∞

∞dt e−ωt〈∆M(t)∆M(0)〉= 〈(∆M)2〉

∫ ∞

∞dt e−ωt−W|t| .

The result reads as,

(∆M)ω = 2〈(∆M)2〉 . Wω2 +W2

Chapter 6

The Foundations of Statistical Mechanics

This chapter will be skipped this year

77

78 CHAPTER 6. THE FOUNDATIONS OF STATISTICAL MECHANICS

Chapter 7

Equilibrium Statistical Mechanics

Quick access:1 2 3 4 5 6 7 8 9 101112131415161718192021222324

Problem 7.1: Compute the structure function forN noninteracting harmonic oscillators, eachwith frequencyω and massm. Assume the system has total energyE. Using this structurefunction and the microcanonical ensemble, compute the entropy and the heat capacity of thesystem.

Solution 7.1: One can use either classical, or quantum Hamiltonian. Let us do the problemclassically. For brevity let us assume the the oscillators are one-dimensional. Then the Hamilto-nian is

H =N

∑i=1

(p2

i

2m+

mω2x2i

2

)= E .

The total number of states is

Ω(E) =∫

Vdx1

∫

Vdx2 . . .

∫

VdxN

∫dp1

∫dp2 . . .

∫dpN Θ

[E−

N

∑i=1

(p2

i

2m+

mω2x2i

2

)].

Here

Θ(x) =

1, x > 00, x < 0

is the Heaviside unit step function. We have

dΘ(x)/dx= δ(x) ,

thus the structure functionΣ(E) = ∂Ω(E)/∂E is given by the equation

Σ(E) =∫

Vdx1

∫

Vdx2 . . .

∫

VdxN

∫dp1

∫dp2 . . .

∫dpN δ

[E−

N

∑i=1

(p2

i

2m+

mω2x2i

2

)].

79

80 CHAPTER 7. EQUILIBRIUM STATISTICAL MECHANICS

Let us use the following trick. Denoteqi = xi√

mω2/2, i = 1..N, qi+N = pi√

1/2mandR2 = E.Then

Ω =(

2ω

)N ∫dq1 . . .

∫dq2N Θ

(R2−

2N

∑j=1

q2j

)≡

(2ω

)N

Ω(R) .

From the dimensionality considerations,

Ω(R) = A2NR2N .

To find the coefficientA2N let us consider the integral

∫ ∞

−∞dq1 . . .

∫ ∞

−∞dq2N exp

(−

2N

∑j=1

q2j

)=

(∫ ∞

−∞dqe−q2

)2N

= πN .

On the other hand, the same integral is

∫ ∞

0dR

∂Ω∂R

e−R2= 2NA2N

∫ ∞

0dRR2N−1e−R2

= NA2NΓ(N)

Thus

A2N =πN

NΓ(N).

Finally,

Ω(E) =1

NΓ(N)

(2π

Eω

)N

, Σ(E) =1

EΓ(N)

(2π

Eω

)N

.

Now, the entropy is

S/k = lnΩ(E)N!hN = ln

(Ehω

)N 1(N!)2 = N ln(E/hω)−2N lnN+2N .

Here we used the Stirling’s formula. Now,

1T

=∂S∂E

=NkE

→ E = NkT, C = kN.

This is the classical answer. The quantum calculation is demonstrated in the Exercise 7.1 of thebook [1].

Problem 7.2: A system consists ofN noninteracting, distinguishable two-level atoms. Eachatom can exist in one of two states,E0 = 0, andE1 = ε. The number of atoms in energy levelE1

is n1. The internal energy of the system isU = n0E0 +n1E1.

(a) Compute the entropy of the system as a function of internal energy.

(b) Compute the heat capacity of a fixed number of atoms,N.

81

Solution 7.2: SinceE0 = 0 the energyU cam be realized inU/ε≡ n1 particles are in the state1, and(N−U/ε) = N−n1 particles are in the state 0. The number of ways to do that is

W =N!

n1!(N−n1)!Consequently,

Sk

= lnN!

(N−n1)!≈ N lnN−N−n1 lnn1 +n1− (N−n1) ln(N−n1)+(N−n1) .

FinallySk≈ N lnN−U

εln

Uε−

(N−U

ε

)ln

(N−U

ε

).

To find temperature we do usual procedure:

1T

=∂S∂U

≈ kε

lnNε−U

U.

As result,

U =Nε

eε/kT +1, C = Nk

( εkT

)2 eε/kT

(eε/kT +1

)2 .

Problem 7.3: A lattice containsN normal lattice sites andN interstitial lattice sites.N identicalatoms sit on the lattice,M on the interstitial sites andN−M on the normal sites (N ÀM À 1.If an atom occupies a normal sites, its energyE = 0. If an atom occupies an interstitial site, itsenergy isE = ε. Compute the internal energy and heat capacity as a function of temperature forthis lattice.

Solution 7.3: The internal energy isU = Mε, so we have to expressM as a function of temper-ature. The number of ways in which one can placeM atoms atN interstitial sites is

Wi =N!

M!(N−M)!.

The restN−M atoms can be placed atN normal states in the same number of ways. Thus

W =(

N!M!(N−M)!

)2

→ Sk

= 2lnN!

M!(N−M)!.

Using the Stiling’s formula we get

S/k ≈ 2[N lnN−N−M lnM +M− (N−M) ln(N−M)+(N−M)]≈ 2[N lnN−M lnM− (N−M) ln(N−M)] ,

T−1 = (2k/ε) ln(N−M)/M , M = U/ε .

Thus,

U =Nε

eε/2kT +1≈ Nεe−ε/kT , C≈ 2kN(ε/2kT)2e−ε/2kT .


Problem 7.4: Consider a lattice withN spin-1 atoms with magnetic momentµ. Each atom canbe in one of 3 spin states:,Sz = −1,0,+1. Let n−1, n0 andn1 denote the respective number ofatoms in each of those spin states. Find the total entropy and the configuration which maximizesthe total entropy. What is the maximum entropy? (Assume that no magnetic field is present,so all atoms have the same energy. Also assume that atoms on different lattice sites cannot beexchanged, so they are distinguishable).


W =N!

n−1!n1!(N−n−1−n1)!,

lnW ≈ N lnN−n−1 lnn−1−n1 lnn1− (N−n−1−n1) ln(N−n−1−n1) .

Differentiating this function byn−1 andn1 and equating the derivatives to 0 we get:

n1 = (N−n1−n−1) = n−1 → n1 = n0 = n−1 = N/3

The maximum entropy isSmax = kNln3.

Problem 7.5: A system has 3 distinguishable molecules at rest, each with a quantized magneticmoment which can havez-components+µ/2 or −µ/2. Find an expression for the distributionfunction, fi (i denotes the configuration), which maximizes entropy subject to the conditions

∑i

fi = 1, ∑i

Mi,z fi = γµ,

whereMi,z is the magnetic moment of the system in thei-th configuration. For the caseγ = 1/2compute the entropy and computefi .

Solution 7.5: Let us measure magnetic moment in unitsµ. We have 4 configurations withmagnetic moments in parentheses

↑↑↑ (3/2), ↑↑↓ (1/2), ↑↓↓ (−1/2), ↓↓↓ (−3/2) .

The configurations with magnetic moments±1/2 are triply degenerated. Denoting the configu-rations as 1,2, 3 and 4, we construct the functional

Sk

= ∑i

gi fi(α+βMi,z− ln fi) .

From this expression we get

α+βMi,z− ln fi−1 = 0 → fi = exp(1−α−βMi,z) .

83

Heregi is the degeneracy factor:g1,4 = 1, g2,3 = 3. We have 2 equations forα andβ:

∑i

gi exp(1−α−βMi,z) = 1,

∑i

giMi,z exp(1−α−βMi,z) = γ .

or

2e1−α(

cosh3β2

+3coshβ2

)= 1

2e1−α(

32

sinh3β2

+32

sinhβ2

)= γ .

For γ = 1/2 that yields:

cosh3β2

+3coshβ2

= 3sinh3β2

+3sinhβ2

Denotingx = eβ/2 we can rewrite this equation as

x6−3x2−4 = 0,

the meaningful solution beingx =

√2, or β = ln2.

Consequently,

α = 1+ ln27√

24

.

As a result, we have

fi = N eMi,z ln2 , N =23/2

27.

The entropy is

Sk

=−∑i

gi fi ln fi =−∑i

gi fi (lnN +Mi,z ln2) =− lnN − 12

ln2 = ln274

.

Problem 7.6: A fluid in equilibrium is contained in the insulated box of volumeV. The fluid isdivided (conceptually) intom cells. Compute the variance of the enthalpy fluctuations,〈(∆Hi)2〉in the i-th cell. (For simplicity assume the fluctuations occur at fixed particle number,Ni).

Hint: UseP andSas independent variables.


Solution 7.6: We have∆H = T ∆S+V ∆P.

Thus〈(∆H)2〉= T2〈(∆S)2〉+V2〈(∆P)2〉 .

From the general theory the fluctuation probability is proportional to

w ∝ eWmin/kT , Wmin = ∆U−T ∆S+P∆V (7.1)

is the minimum work needed to carry out reversibly the given change in the thermodynamicquantities of small part considered (relative to which the remainder of the body acts as a medium).Expanding∆U in the series, we obtain

Wmin =12

[∂2U∂S2 (∆S)2 +2

∂U∂S∂V

∆S∆V +∂2U∂V2 (∆V)2

]

=12

[∆S∆

(∂U∂S

)

V+∆V ∆

(∂U∂V

)

S

]

=12

(∆S∆T−∆P∆V) .

Thus

w ∝ exp

(∆P∆V−∆S∆T

2kT

). (7.2)

One can use this general formula and choose different couples as independent variables. In ourcase the proper pair isSP. We get

∆V =(

∂V∂P

)

S∆P+

(∂V∂S

)

P∆S,

∆T =(

∂T∂P

)

S∆P+

(∂T∂S

)

P∆S

=(

∂T∂P

)

S∆P+

TCP

∆S.

The Maxwell relation following fromdH = T dS+V dP is(

∂V∂S

)

P=

(∂T∂P

)

S.

Thus

∆V =(

∂V∂P

)

S∆P+

(∂T∂P

)

S.

Substituting the above relations into Eq. (7.2) we obtain

w ∝ exp

[1

2kT

(∂V∂P

)

S(∆P)2− 1

2CP(∆S)2

].

85

As a result,〈∆S∆P〉= 0, 〈(∆S)2〉= kCP , 〈(∆P)2〉=−kT (∂P/∂V)S .

Using these relations we get

〈(∆H)2〉= kT2CP−V2kT (∂P/∂V)S .

Problem 7.7: A fluid in equilibrium is contained in the insulated box of volumeV. The fluidis divided (conceptually) intom cells. Compute the variance of the internal energy fluctuations,〈(∆Ui)2〉 in the i-th cell. (For simplicity assume the fluctuations occur at fixed particle number,Ni). What happens to the internal energy fluctuations near the critical point?

Solution 7.7: We have,

∆U =(

∂U∂V

)

T∆V +

(∂U∂T

)

V∆T =

[T

(∂P∂T

)

V−P

]∆V +CV ∆T .

Now we can use Eq. (7.2) and chooseV,T as independent variables. We have:

∆S =(

∂S∂T

)

V∆T +

(∂S∂V

)

T∆T =

CV

T∆T +

(∂P∂T

)

V∆T ,

∆P =(

∂P∂T

)

V∆T +

(∂P∂V

)

T∆V .

The we get

w ∝ exp

[− CV

2kT(∆T)2 +

12kT

(∂P∂V

)

T(∆V)2

],

〈∆T ∆V = 0〉〈(∆T)2〉= kT2/CV , 〈(∆V)2〉=−kT (∂V/∂P)T .

Finally,

〈(∆U)2〉=−kT

[T

(∂P∂T

)

V−P

]2(∂V∂P

)

T+kT2CV .

Near the critical point〈(∆U)2〉 → ∞.

Problem 7.8: What is the partition function for a van der Waals gas withN particles? Notethat the result is phenomenological and might involve some guessing. It is useful to compare itto the partition function for an ideal gas. Remember that the particles are indistinguishable , sowhen using a partition function one must insert a counting factor. Use this partition function tocompute the internal energy,U(N,T,V), the pressure,P(N,T,V), and the entropy,S(N,T,V).


Solution 7.8: The partition function is (β≡ (kT)−1),

ZN =1

N!h3N

∫dr1 . . .drN

∫dp1 . . .dpN e−βH (r ,p) .

First we can integrate out the momenta,

Z(k)N =

1N!h3N

∫dp1 . . .dpN e−β∑i p

2i /2m =

1

N!λ3NT

,

whereλT = h/√

2πmkT. Since the partition function for the ideal gas isZ(0)N = VNZ(k)

N Then weare left with

ZN

Z(0)N

=1

VN

∫dr1 . . .drN e−βV(r i) = 1+

1VN

∫dr1 . . .drN

[e−βV(r i)−1

]≡ 1+zN .

Let us try to consider rarefied gas where only pair interaction is important. Then we can chooseN(N−1)/2 pair to get

zn≈ N2

2V2

∫dr1dr2

[e−βV(r1−r2)−1

].

Introducing the virial coefficient

B(T) =12

∫dr

(1−e−βV(r)

),

wherer = r1− r2, we obtainzN =−N2B(T)/V .

We get

A = A0−kT ln(1+zn)≈ A0−kTzN = A0 +kTN2B(T)/V , A0≡−kT lnZ(0)N .

In the spirit of the van der Walls model let us assume that the interaction is large atr ≤ 2r0, andat r > 2r0 the interaction is small. Then

B(T) = 2π∫ 2r0

0r2dr

(1−e−βV(r)

)+2π

∫ ∞

2r0

r2dr(

1−e−βV(r))

≈ 16πr30/3+2πβ

∫ ∞

2r0

r2drV(r)≡ b−a/kT ,

whereb = 16πr30/3 anda = π

∫ ∞2r0

r2drV(r). Finally,

zn =−N2

V

(b− a

kT

).

For an ideal gas,

A0 =−NkT lneVN

(mkT

2πh2

)3/2

= N f(T)−NkT ln(eV/N) .

87

Thus,A = N f(T)−NkT ln(e/N)−NkT(lnV−Nb/V)−N2a/V .

Sinceln(V−Nb)≈−Nb/V + lnV we get

A≈ A0−NkT ln(1−Nb/V)−N2a/V .

From this we get

P = −∂A∂V

=NkTV−b

− N2aV

, S=−∂A∂T

= S0 +kNln

(1− Nb

V

),

U = U0− N2aV

=32

kT− N2aV

.

Problem 7.9: Consider a solid surface to be a two-dimensional lattice withNs sites. AssumethatNa atoms (Na¿ Ns) are adsorbed on the surface, so that each site has either 0 or 1 adsorbedatom. At adsorbed atom has energyE =−ε, whereε > 0. Assume that atoms on the surface donot interact with each other.

(a) If the surface is at temperatureT, compute chemical potential of the adsorbed atoms as afunction ofT, ε, andNa/Ns (use the canonical ensemble).

(b) If the surface is in equilibrium with an ideal gas of similar atoms at temperatureT, computethe ratioNa/Ns as a function of pressure,P of the gas. Assume the gas has number densityn. (Hint: equate chemical potentials of the adsorbed atoms and the gas).

Solution 7.9:

(a) The canonical partition function is

Z =1

Na!

Ns

∑i=1

e−βEi =Ns!

Na!(Ns−Na)!eβNaε .

Hereβ = 1/kT. SinceNa¿ Ns,

Z≈ NNas

Na!eβNaε .

Equating this expression toe−Naβµ we get

µ=−ε−kT ln(Ns/Na) .


(b) For an ideal gas,

µ= kT ln

(Pλ3

T

kT

).

Equating chemical potentials, we obtain

− εkT− ln

Ns

Na= ln

(Pλ3

T

kT

)→ Na

Ns=

Pλ3T

kTeε/kT .

This expression is valid at high temperatures when

nλ3T ¿ e−ε/kT .

Problem 7.10: Consider a two-dimensional lattice lattice lattice in thex−y plane with sides oflengthLx andLy which containN atoms(N is very large) coupled by nearest neighbor harmonicforces.

(a) Compute the Debye frequency for this lattice.

(b) In the limit T → 0, what is the heat capacity?

Solution 7.10: We follow the section 7.E of the book [1]. SinceNÀ 1 the boundary conditionsare not important, and we can apply zero boundary conditions. As in this exercise, we usecontinual approximation and replace

∑k

. . . → LxLy

(2π)2

∫dkxdky . . . .

The the density of states is

g(ω) = ∑k,α

δ(ω−ωk,α) =LxLy

(2π)2 ∑α

∫dk δ(ω−ωk,α) .

Hereα denotes the vibrational mode. In two-dimensional lattice we have two modes, longitudi-nal, l , and transverse,t. In the Debye approximation it is assumed that the dispersion lawωk,α islinear and isotropic:

ωk,α = cαk, ki =πLi

si , si = 1,2, . . . .

wherecα is the sound velocity for each mode. Since in two-dimensional case∫

dk . . . =∫

kdkdφ . . . = 2π∫

kdk. . .

we obtain

g(ω) =LxLy

2πh2

(1

c2l

+1

c2t

)ω .

89

Since the total number of vibrations should be normalized to2N, where 2 is the number of modesandN is number of atoms, we should get

∫ ωD

0g(ω)dω = 2N .

Consequently

2N =LxLy

4πh2

(1

c2l

+1

c2t

)ω2

D

and1

ωD=

√LxLy

8Nπh2

(1

c2l

+1

c2t

).

The density of states can be expressed as

g(ω) =4N

ω2D

ω .

As result,

C =4kN

(hβωD)2

∫ TD/T

0dx

x3ex

(ex−1)2 =

2kN, T → ∞24ζ(3)kN(T/TD)2, T → 0.

Hereζ(x) is the Riemannζ-function,ζ(3)≈ 1.20, TD = hωD/k.

Problem 7.11: A cubic box (with infinitely hard walls) of volumeV = L3 contains an ideal gasof N rigid HCl molecules (assume the effective distance between the H atom and the Cl atomd = 1.3 A).

(a) If L = 1.0 cm, what is the spacing between the translation energy levels?

(b) Write the partition function for this system (include the translational and rotational contri-butions). At what temperature do rotational degrees of freedom become important.

(c) Write expressions for the Helmholtz free energy, the entropy, and the heat capacity ofthe system for temperatures where the rotational degrees of freedom make a significantcontributions.

Solution 7.11: Let us put the origin of the reference frame in the cube’s corner. Since at thewalls the wave function should vanish, for each componentα of the momentump we get:

sin(pαL/h) = 0 → kα = nαπ/L , kα ≡ pα/h.

Herenα is an integer number. Thus the energy levels of translational motion are given by theexpression

εn = ε0(n2x +n2

y +n2z) , εt =

h2π2

2mL2 .


Thus the spacing between the levels is

εni+1− εni = (2ni +1)εt .

The rotational degrees of freedom are determined by the moment if inertia,

I =m1m2

m1 +m2d2 ,

the energy of rotation being

εK = εrK(K +1) , K = 0,1, . . . , εr =h2

2I.

Thus the spacing between the levels is

εK+1− εK = 2(K +1)εr .

It is important that each rotational level is degenerate by the factor(2K +1).ThusZ = Ztr ·Zrot where

Ztr =

(∞

∑nx=1

e−βεtn2x

)3

=

V/λ3T , βεt ¿ 1,

exp(−βεt), βεt À 1,

Zrot =∞

∑K=0

(2K +1)e−βεrK(K+1) =

kT/εr , βεr ¿ 1,

1+3e−2βεr , βεr ¿ 1.

The following is straightforward.

A = −kT lnZtr−kT lnZrot;

S = −(

∂A∂T

)

V, C = T

(∂S∂T

)

V=−T

(∂2A∂T2

)

V.

The calculations are obvious.

Problem 7.12: An ideal gas is composed ofN “red” atoms of massm , N “blue” atoms of massm, andN “green” atoms of massm. Atoms of the same color are indistinguishable. Atoms ofdifferent color are distinguishable.

(a) Use the canonical ensemble to compute the entropy of this gas.

(b) Compute the entropy of ideal gas of3N “red” atoms of massm. Does if differ from that ofthe mixture. If so, by how much.

91

Solution 7.12: Since the atoms of different color distinguishable the partition function is justthe product of partition function. For each component

ZN =

(∑k

e−βp2/2m

)N

=1

N!

(V

λ3T

)N

.

From this expression,

AN =−NkT lneVN

(mkT

2πh2

)3/2

= N f(T)−NkT ln(eV/N) .

Here

f (T) =(

mkT

2πh2

)3/2

.

Thus,Z = Z3N, and

A = 3N f(T)−3NkT ln(eV/N) , S= 3NK ln(eV/N)−3N f ′(T) .

In the case of single-color gas of3N particles we have

S1 = 3Nkln(eV/3N)−3N f ′(T) → S−S1 = 3N ln3.

Problem 7.13: An ideal gas consists of a mixture of “green” and “red” spin-1/2 particles. Allparticles have massm. A magnetic field,B, is applied to the system. The “green” particles havemagnetic momentγG, and the “red” particles have magnetic momentγR, whereγR < γG. Assumethat temperature is high enough that Fermi statistics can be neglected. The system will be inequilibrium if chemical potentials the the “red” and the “green” gases are equal. Compute theratio NR/NG, whereNR is the number of “red” atoms andNG is the number of “red” particles.Use the canonical ensemble (no other ensembles will be accepted).

Solution 7.13: Let us consider one of the gases since they are noninteracting. We haveZ =Ztr ·Zs whereZtr is the usual transport partition function while

Zs =(

e−βγB +eβγB)N

= 2N coshN(βγB) .

Since

µ=−kTN

ln1

N!

(V cosh(βγB)

λT

)N

=−kT ln

(eVcosh(βγB)

NλT

).

Equating chemical potentials for “red” and “green” gases we get

cosh(βγRB)NR

=cosh(βγGB)

NG→ NR

NG=

cosh(βγRB)cosh(βγGB)

.


Problem 7.14: Consider a one-dimensional lattice withN lattice sites and assume thati-thlattice site has spinsi =±1. The Hamiltonian describing this lattice is

H =−εN

∑i=1

sisi+1 .

Assume periodic boundary condition, sosN+1 = s1. Compute the correlation function,〈s1s2〉.How does it behave at very high temperature and at very low temperature?


ZN(T) = ∑s1=±1

. . . ∑sN=±1

exp

(βε∑

isisi+1

).

As it is recommended in the book [1], we introduce the matrix

P =(

eβε e−βε

e−βε eβε

)= eβε I +e−βεσσσ1 .

We have〈si |P|si+1〉= eβsisi+1 .

ThenZN(T) = ∑

s1=±1. . . ∑

sN=±1〈s1|P|s2〉〈s2|P|s3〉 . . .〈sN|P|s1〉= Tr PN .

The matrixP has eigenvalues

λ1 = 2coshβε , λ2 = 2sinhβε .

ThusZN(β) = 2N(coshN βε+sinhN βε) .

We have

〈sisi+1〉=1

Nε∂ lnZN

∂β=

coshN−1(βε)sinh(βε)+sinhN−1(βε)cosh(βε)coshN(βε)+sinhN(βε)

.

At low temperature,βε À 1, cosh(βε) ≈ sinh(βε) and〈sisi+1〉 = 1. Thus we have ferromagnetordering.

At high temperature,

cosh(βε)→ 1, sinh(βε)→ 0 → 〈sisi+1〉= βε¿ 1.

93

Problem 7.15: In the mean field approximation to the Ising lattice, the order parameter,〈s〉,satisfied the equation

〈s〉= tanh

(〈s〉 Tc

T

),

whereTc = νε/2k whereε is the strength of the coupling between lattice sites andν is the numberof nearest neighbors.

(a) Show that〈s〉 is the following temperature dependence

〈s〉 ≈

1−e−2Tc/T , T → 0,√3(1−T/Tc), T → Tc .

(b) Compute the jump in the heat capacity atT = Tc.

(c) Compute the magnetic susceptibility,χ(T,N)B=0, in the neighborhood ofTc both forT >Tc andT < Tc. What is the critical exponent for both cases?

Solution 7.15:

(a) Sincetanhξ≈ 1−2e−2ξ at largeξ, we get

〈s〉 ≈ 1−2e−2〈s〉Tc/T → 〈s〉 ≈ 1−2e−2Tc/T .

At small ξ, tanhξ≈ ξ−ξ3/3, we obtain

〈s〉 ≈ 〈s〉 Tc

T+〈s〉33

(Tc

T

)3

.

Thus,〈s〉 ≈

√3(1−T/Tc) .

(b) We will use notations of Sec. 7.F.2 of the book [1]. The the Hamiltonian is

H =−N

∑i=1

E(ε,B)si , E(ε,B) =νε2〈s〉+µB. (7.3)

ThusU =−NE〈s〉=−NE〈s〉, and

C =−N

(∂E(ε,B)〈s〉

∂T

)

N,B=−N

[E + 〈s〉 ∂E

∂〈s〉](

∂〈s〉∂T

)

N,B.

At T → Tc−0 andB = 0 we get

〈s〉=√

3(1−T/Tc) → C|Tc−0 =32

NνεTc

.

At T > Tc the average spin is zero, thus the jump in specific heat is given byC|Tc−0.


(c) To get magnetic susceptibility we have to differentiate the magnetic momentN = Nµ〈s〉with respect toB:

χ = Nµ

(∂〈s〉∂B

)

N,B→0.

Definingη = (∂〈s〉/∂B)N,B→0 we get the following equation atB→ 0:

η =1

cosh2(〈s〉Tc/T)

[βνε2

η+βµ

].

Its solution is

η =2βµ

2cosh2(〈s〉Tc/T)−Tc/T.

Finally,

χ =Nµ2

νεTc/T

cosh2(〈s〉Tc/T)−Tc/T.

AboveTc we have〈s〉=0, and

χ =Nµ2

νεTc

T−Tc.

BelowTc we get

cosh2(〈s〉Tc/T)≈[

1+〈s〉22

(Tc

T

)2]2

≈ 1+3

(1− T

Tc

).

As a result, belowTc

χ =Nµ2

νεTc

2(Tc−T).

Thus there is a divergence atT → Tc, χ ∝ |T−Tc|−1, with critical exponent equal to -1.

Problem 7.16: The density of states of an ideal Bose-Einstein gas is

g(E) =

αE2, E > 0,0, E < 0,

wherea is a constant. Compute the temperature for Bose-Einstein condensation.

Solution 7.16: We have the equation forβc = 1/kTc:

N =∫ ∞

Emin

g(E)dE

eβcE−1=

αβ3

c

∫ ∞

0

x2dxex−1

=2αζ(3)

β3c

.

Hereζ(x) is the Riemannzeta-function. Finally,

kTc =(

N2αζ(3)

)1/3

.

95

Problem 7.17: An ideal Bose-Einstein gas consists of noninteracting bosons with massmwhich have an internal degree of freedom which can be described by assuming, that the bo-son are two-level atoms. Bosons in the ground state have the energyE0 = p2/2m, while bosonsin the excited state have the energyE1 = p2/2m+ ∆, wherep is the momentum and∆ is theexcitation energy. Assume that∆À kT. Compute the Bose-Einstein condensation temperature,Tc, for this gas of two-level bosons. Does the existence of internal degree of the internal degreeof freedom raise or lower the condensation temperature?

Solution 7.17: We start with calculation of density of states.

g0(E) =∫

dpδ(

E− p2

2m

)=

4πh3

∫p2dpδ

(E− p2

2m

)= a

√E , a≡ m3/2

√2π2h3 .

In a similar way,

g1(E) = a√

E−∆ .

The equation forβc = 1/kTc is:

Nβ3/2c

a=

∫ ∞

0

√xdx

ex−1+

∫ ∞

βc∆

√x−βc∆dxex−1

≈ ζ(3/2)√

π2

+√

π2

e−βc∆ .

We can solve this equation by iterations since∆ is large. Denoting

β0 =(

aζ(3/2)√

π2N

)2/3

we get

βc = β0

(1+

√πa

2Nβ3/20

e−β0∆

)2/3

≈ β0 +2√

πa

6N√

β0e−β0∆ .

Thus the internal degree of freedomdecreasesthe condensation temperature.

Problem 7.18: Compute the Clausius-Clapeyron equation for an ideal Bose-Einstein gas andsketch the coexistence curve. Show that the line of transition points in theP−v plane obeys theequation

pv5/3 =2πh2

m

g5/2(1)

[g5/2(1)]5/3.


Solution 7.18: Let us first recall the notations. The quantityz= eβµ is called thefugacity, thedistribution function is written as

〈nl 〉=1

eβ(εl−µ)−1=

z

eβεl −z.

The functionsgi(z) are introduced as

g5/2(z) = − 4√π

∫ ∞

0x2dx ln

(1−z−x2

)=

∞

∑α=1

zα

α5/2, (7.4)

g3/2(z) = zddz

g5/2(z) =∞

∑α=1

zα

α3/2. (7.5)

Now, the grand canonical potential is

Ω =V(kT)5/2m3/2√

2π2h3

∫ ∞

0dy√

y ln(1−ze−y) =−V

kT

λ3T

g5/2(z) .

Thus the equation of state is

P =kT

λ3T

g5/2(z) .

In equilibrium with the condensate atT = Tc,

Pc =kTc

λ3T

g5/2(1) .

The critical particle density in the condensation point is

〈nc〉=g3/2(1)

λ3T

.

DenotingλT = a/T1/2 wherea =√

2πh2/mkwe get at the critical point:

NV

=g3/2(1)T3/2

c

a3 → Tc =

(Na3

Vg3/2(1)

)2/3

.

Thus,

Pc =kTcg5/2(1)

λ3T

= ka2 g5/2(1)

[vg3/2(1)]5/3=

2πh2

m

g5/2(1)

[vg3/2(1)]5/3.

Problem 7.19: Show that the pressure,P, of an ideal Bose-Einstein gas can be written in theform P = αu, where “u” is the internal energy per unit volume anda is a constant.

(a) what isu?

(b) What isα?

97

Solution 7.19: Let us do integration by parts in Eq.7.5to get

g5/2(z) = − 2√π

∫ ∞

0dy√

y ln(1−ze−y)

=2√π· 23

∫ ∞

0dyy3/2 z

ey−z

Calculation the average energy

U =∫ ∞

0g(ε)εdε

z

eβε−z

and recalling thatg(ε) ∝√

ε we immediately see that

PV =23U → P =

23

u.

Thus,α = 2/3.

Problem 7.20: Electrons in a piece of Cu metal can be assumed to behave as an ideal Fermigas. Cu metal in the solid state has mass densityρ = 9 gr/cm3. Assume that each Cu atomdonates an electron to the Fermi gas. Assume the system is atT = 0 K.

(a) Compute the Fermi energy,εF , of the electron gas.

(b) Compute the Fermi temperature,TF = εF/k.

Solution 7.20: Let us start with the calculation of the density of states:

g(ε) =2V ·4π

h3

∫p2dpδ

(ε− p2

2m

)

=√

2Vm3/2

π2h3

√ε≡Vg0

√ε .

Hereg0 =√

2m3/2/π2h3, we have taken into account that spin degeneracy for electrons 2. Inte-gration expression for densoty of states from 0 toεF and equationg to the number of electronswe obtain

(2/3)g0ε3/2F = ne → εF = (3ne/2g0)2/3, .

Herene = Ne/V = ρ/mCu is the electron density.


Problem 7.21: The density of states of an ideal Fermi-Dirac gas is

f (E) =

D, if E > 0,0, if E < 0,

whereD is a constant.

(a) Compute the Fermi energy.

(b) Compute the specific heat at low temperature.


N = D∫ ∞

0

dE

eβ(E−µ) +1=

Dβ

∫ ∞

−βµ

dxex +1

=Dβ

ln(

1+eβµ)

.

Thus

eβµ = eβN/D−1 → µ=1β

ln(

eβN/D−1)≈ N

D−kTe−N/DkT .

The energy is then

U = D∫ ∞

0

E dE

eβ(E−µ) +1.

Let us use the trick called the Sommerfeld expansion. Consider an auxiliary integral,

I =∫ ∞

0

f (E)dE

e(E−µ)/kT +1=

∫ ∞

−µ/kT

f (µ+kTz)dEez+1

= kT∫ µ/kT

0

f (µ−kTz)dEe−z+1

+kT∫ ∞

0

f (µ+kTz)dEez+1

=∫ µ

0f (E)dE+kT

∫ ∞

0dE

f (µ+kTz)− f (µ−kTz)ez+1

Here we have used the fact that(e−z+1)−1 = 1− (ez+1)−1 and thateµ/kT À 1. Expanding theintegrand, we obtain

I ≈∫ µ

0f (E)dE+2(kT)2 f ′(µ)

∫ ∞

0

zdzez+1

+ . . . =∫ µ

0f (E)dE+

π2

6(kT)2 f ′(µ)+ . . . .

In our case,f (E) = DE, f ′(µ) = D, and we get

UD≈ µ2

2+

π2

6(kT)2 .

The specific heat is thenCD

= µ∂µ∂T

+π2

3k2T .

The first item is exponentially small at low temperatures, and

C≈ π2

3Dk2T ≈ k

π2

3kTµ

.

99

Problem 7.22: Compute the magnetization of an ideal gas of spin-1/2 fermions in the presenceof magnetic field. Assume that fermions each have magnetic momentµe. Find the expression formagnetization in the limit of weak magnetic field andT → 0 K.

Solution 7.22: We will calculate here only spin (Pauli) susceptibility. Since the spin degree offreedom leads to additional energy±µeB we get

Ω(µ) =12

Ω0(µ−µeB)+12

Ω0(µ+µeB)

whereΩ0 is the grand potential in the absence of magnetic field,

Ω0(µ) =−23U =−2A

3

∫ ∞

0

E3/2dE

e(E−µ)/kT +1, A =

√2Vm3/2

π2h3 .

As a result,

−Ω(µ)A

=13

∫ ∞

0

E3/2dE

e(E−µ−µeB)/kT +1+

13

∫ ∞

0

E3/2dE

e(E−µ+µeB)/kT +1.

In small magnetic fields or at large temperature,

Ω(µ)≈Ω0(µ)+(µeB)2

2∂2Ω0(µ)

∂µ2 .

Consequently,

M =−(

∂Ω∂B

)

T,V,µ= µ2

eB∂2Ω0(µ)

∂µ2 .

Since∂Ω∂µ =−N we get

M = µ2eB

(∂N∂µ

)

T,V.

This formula works in the wide region

µeB¿ µ.

In very strong magnetic fields the analysis is more complicated. First the chemical potential isgiven by the expression

NA

=13

[(µ+µeB)3/2 +(µ−µeB)3/2

].

In a similar way,

−Ω(µ)A

=215

[(µ+µeB)5/2 +(µ−µeB)5/2

].

These equations are written assuming thatµeB≤µ, otherwise only one item is present. Physicallyit means that only spins-up-states are occupied. Let us consider the case of very strong magneticfield, µeBÀ µ. Then

Ω(µ) =−2A15

(µ+µeB)5/2 → M =Aµe

3(µ+µeB)3/2 = Nµe.


Problem 7.23: Show that the entropy for an ideal Fermi-Dirac gas (neglecting spin) can bewritten in the form

S=−k∑l

[〈nl 〉 ln〈nl 〉+(1−〈nl 〉) ln(1−〈nl 〉)] ,

where〈nl 〉=(

eβ(εl−µ) +1)−1

.

Solution 7.23: The number of ways to distributeNl identical particles amongGl states withnot more than 1 particle per state is just

Wl =Gl !

Nl !(Gl −Nl )!.

Using the Stirling’s formula we obtain

S= k∑l

ln [Gl lnGl −Nl lnNl − (Gl −Nl ) ln(Gl −Nl )] .

Using the mean occupation number〈nl 〉= Nl/Gl we can rewrite this formula as

S=−k∑l

Gl [〈nl 〉 ln〈nl 〉+(1−〈nl 〉) ln(1−〈nl 〉)] .

If we ignore spin, thenGl = 1.

Problem 7.24: To the lowest order in the density, find the difference in the pressure and theisothermal compressibility between an ideal boson and ideal fermion gas. Assume that thefermions and bosons have the same mass and both are spin-less. (Note: You are now consid-ering fairly high temperature).

Solution 7.24: Assuming that the density of states is given by the expression

g(ε) = g0V ε1/2

we obtain

Ω =−23

g0V∫ ∞

0

ε3/2dεeβ(ε−µ)±1

≈−23

g0V∫ ∞

0ε3/2dεeβ(µ−ε)

(1∓eβ(µ−ε)

).

Introducing the fugacityz= eβµ we get

P =23

g0(kT)5/2z∫ ∞

0ξ3/2dξe−ξ

(1∓ze−ξ

)=√

π2

g0(kT)5/2z

(1∓z

√2

8

).

101

The normalization condition in this case gives

N = g0V(kT)3/2z∫ ∞

0ξ1/2dξe−ξ

(1∓ze−ξ

)=√

π2

zg0V

(1∓z

√2

4

).

As result,

z0 =2√

πg0(kT)3/2

NV

.

We obtain

P≈ NkTV

(1∓z0

√2

8

)(

1∓z0

√2

4

) ≈ NkTV

(1±z0

√2

8

)≈ NkT

V

(1±

√2

4√

πg0(kT)3/2

NV

).

The sign “+” hold for the Fermi statistics, the sign “-” holds for the Bose statistics. The com-pressibility is calculated in a straightforward way since

V ≈ NkTP

(1±

√2

4√

πg0(kT)5/2P

).

Chapter 8

Tests and training

Here we put some exercises for self-testing.

103

104 CHAPTER 8. TESTS AND TRAINING

Problem 8.1:

(a) Discuss the difference between Gibbs and Helmholtz free energy.

(b) Prove the relation

U =−T2(

∂∂T

AT

),

whereU is the internal energy.

(c) A body with constant specific heatCV is heated under constant volume fromT1 to T2. Howmuch entropy it gains?

(d) Discuss the heating if the same body is in contact with a thermostat atT2. In the last casethe heating is irreversible. Show that the total entropy change is positive.

(e) Two similar bodies with temperaturesT1 andT2 brought into contact. Find the final tem-perature and the change in entropy.

Solution 8.1:

(a) Mechanical work underisothermic processis given by

dW = dU−dQ= dU−T dS= d(U−TS) .

The function of stateF = U−TS

is calledHelmholtzfree energy. We have

dF = dU−T dS−SdT= (T dS−PdV)−T dS−SdT=−SdT−PdV.

Thus,

S=−(

∂F∂T

)

V, P =−

(∂F∂V

)

T.

Thus,F is thethermodynamic potentialwith respect toV andT.

The thermodynamic potential with respect toP andT is calledGibbsfree energy. We have

G = U−TS−PV→ dG=−SdT+V dP.

(b) Substituting

S=−(

∂F∂T

)

V

into definition ofF we get the result.

105

(c) We have

S=∫ T2

T1

CV

TdT = CV ln

(T2

T1

).

(d) Use 2nd law of thermodynamics

(e) Energy conservation law yields

CV(T2−TB) = CV(TB−T1)−> TB = (T1 +T2)/2.

for the entropy change we have

δS= CV lnTB

T1+CV ln

TB

T2= 2CV ln

(T1 +T2

2√

T1T2

)≥ 0.

Problem 8.2: Find fluctuation of the energy per particle for the Boltzmann gas of spinlessparticles at a given temperature . Assume thatεp = p2/2m.

Solution 8.2:

〈(δε)2〉 =⟨(εi−〈ε〉)2⟩ = 〈ε2〉−〈ε〉2 .

Now

〈ε〉 =1Z1

∑i

εie−βεi =− 1

Z1

∂Z1(β)∂β

.

〈ε2〉 =1Z1

∑i

ε2i e−βεi =

1Z1

∂2Z1(β)∂β2 .

Thus

〈(δε)2〉=1Z1

∂2Z1(β)∂β2 −

(1Z1

∂Z1(β)∂β

)2

=∂2 logZ1

∂β2 .

Now we have to specifyZ1 for εi = p2/2m.

Z1 = ∑p

exp(−βεp) =1

(2πh)3

∫ ∞

04πp2dpexp(−βεp) .

It is convenient to introduce thedensity of statesas number of states per energy interval.

g(ε) = ∑p

δ(ε− εi) =4π

(2πh)3

∫ ∞

0p2dp∑

iδ(ε− εi) =

m3/2√

2π2h3

√ε .

106 CHAPTER 8. TESTS AND TRAINING

In this way,

Z1 =m3/2√

2π2h3

∫ ∞

0dεε1/2exp(−βε) =

m3/2√

2π2h3β3/2

∫ ∞

0dxx1/2exp(−x) .

As ∫ ∞

0dxx1/2exp(−x) = Γ(3/2) = (1/2)

√π

we get

Z1 =a

β3/2, a =

√π8

m3/2

π2h3 → logZ1 = loga− 32

logβ .

Thus

〈(δε)2〉=3

2β2 .

Problem 8.3: Two-level systems with random energy spacingsε are distributed with a constantprobability

p(ε) = P0 .

Find specific heat of the system.

Solution 8.3: As we know from the lecture,

n+

n−= e−βε , n+ +n− = 1.

Thus

n± =1

exp(±βε)+1.

The average energy is (we start from the middle of the distance between the levels)

〈ε〉ε =ε2(n+−n−) =

ε2

tanhβε2

.

The contribution to specific heat is

c(ε) =∂〈ε〉ε∂T

==−β2∂〈ε〉ε∂β

=14

1

cosh2(βε/2).

The total specific heat is

C =∫ ∞

0p(ε)dεc(ε) = P0

∫ ∞

0dε

14

1

cosh2(βε/2)=

P0

2β

∫ ∞

0

dx

cosh2x=

P0

2β.

Note that it is proportional toT.

Appendix A

Additional information

A.1 Thermodynamics

A.1.1 Thermodynamic potentials

Thermodynamic potential Notation Independent variables DifferentialInternal energy U S, V, N T dS−PdV+µdN

Heat function (enthalpy) H S, P, N T dS+V dP+µdNHelmholtz free energy F T, V, N −SdT−PdV+µdN

Gibbs free energy G T, P, N −SdT+V dP+µdNLandau free energy Ω T, V, µ −SdT−PdV−N dµ

Table A.1:Thermodynamic potentials (summary)

A.1.2 Variable transformation

Jacobian∂(u,v)/∂(x,y) is defined as the determinant

∂(u,v)∂(x,y)

=∣∣∣∣

∂u/∂x ∂u/∂y∂v/∂x ∂v/∂y

∣∣∣∣ (A.1)

The following relations can be useful:

∂(v,u)∂(x,y)

=−∂(u,v)∂(x,y)

,∂(u,y)∂(x,y)

=(

∂u∂x

)

y,

∂(u,v)∂(x,y)

=∂(u,v)∂(t,s)

· ∂(t,s)∂(x,y)

. (A.2)

A.1.3 Derivatives from the equation of state

Important relations arise from the properties of partial differentiations. Consider 3 quantities,X,Y,Z, related by the equation of stateK (X,Y,Z) = const. Now let us considerX,Y as indepen-

107

108 APPENDIX A. ADDITIONAL INFORMATION

dent variables, whileZ = Z(X,Y). We have(

∂Z∂X

)

YdX+

(∂Z∂Y

)

XdY−dZ = 0. (A.3)

If Y,Z are taken as independent variables, then

−dX+(

∂X∂Y

)

ZdY+

(∂X∂Z

)

YdZ = 0 (A.4)

Now we multiply Eq.A.3 by(

∂X∂Y

)Z

and Eq.A.4 by(

∂Z∂Y

)X

and subtract the results. We obtain

[(∂Z∂X

)

Y

(∂X∂Y

)

Z+

(∂Z∂Y

)

X

]dX+

[−

(∂X∂Y

)

Z−

(∂X∂Z

)

Y

(∂Z∂Y

)

X

]dZ = 0.

SincedX anddY areindependent, this equation is compatible if(

∂Z∂X

)

Y

(∂X∂Y

)

Z+

(∂Z∂Y

)

X= 0

(∂X∂Y

)

Z+

(∂X∂Z

)

Y

(∂Z∂Y

)

X= 0,

or (∂X∂Y

)

Z

(∂Y∂Z

)

X

(∂Z∂X

)

Y=−1, (A.5)

(∂X∂Y

)

Z=

(∂Y∂X

)−1

Z. (A.6)

General scheme for transformation: Consider any quantityF (X,Y), the differential of whichcan be expressed as

dF =(

∂F∂X

)

YdX+

(∂F∂Y

)

XdX .

Then we divide it bydZ and assumeY = const, (dY = 0). We get(

∂F∂Z

)

Y=

(∂F∂X

)

Y

(∂X∂Z

)

Y. (A.7)

Another important relation arises if one divides the expression fordF by xX,(

∂F∂X

)

Z=

(∂F∂X

)

Y+

(∂F∂Y

)

X

(∂Y∂X

)

Z. (A.8)

Equations (A.7), (A.8) together with Eqs. (A.5), (A.6) and the Maxwell relations are usuallyused for transformations and computation of derivatives from the equation of state.

A.2. MAIN DISTRIBUTIONS 109

A.2 Main distributions

Binomial distribution

WN(m) =N!

m!(N−m)!pm(1− p)N−m. (A.9)

Poisson distribution

P(m) =mm

m!e−m. (A.10)

Gaussian distribution

G(x) =1√2πσ

e−(x−x)/2σ . (A.11)

Distribution of displacemnet for 1D random walk with the stepl

G(x) =1

2l√

2πNp(1− p)e−[x−(p−q)l ]2/8Nl2p(1−p) . (A.12)

Gaussian integrals

I1(a) =∫ ∞

−∞dze−az2/2 =

√2π/a. (A.13)

∫ ∞

−∞z2dze−z2/2 = 2

[dI1(a)

da

]

a=1= I1(a)

√2π . (A.14)

Gaussian distribution for more than one variables Define entropy asS(x1, . . . ,xn. Then

S−S0 =−12∑

i,k

βikxixk , βik = βki .

For convenience, let us assume summation over repeated subscripts and rewrite the above equa-tion

S−S0 =−12

βikxixk . (A.15)

Consequently,

w = Ae−12βikxixk .

Let us first calculate the normalization factorA from∫

dwdx1 · · ·dxn = 1.

To calculate the integral let us introduce the linear transform

xi = aikyk


to make the quadratic form (refmgs01diagonal. In order that

βikxixk = yiyi = yiyk δik

the relationβikail akm = δkm (A.16)

should be valid. Denoting determinants of the matricesβ anda asβ anda, respectively, we getthe relationβa2 = 1. The Jacobian of the transformationxi → yi is justa,

J =∣∣∣∣∂(x1, . . . ,xn)∂(y1, . . . ,yn)

∣∣∣∣ = |a|= a.

Consequently,

Aa

[∫dye−y2/2

]n

=A(2π)n/2

√β

= 1.

Thus

w =

√β

(2π)n/2exp

(−1

2βikxixk

). (A.17)

Let define generalized forces as

Xi =−∂S/∂xi = βikxk, (A.18)

and first calculate

〈xiXk〉=

√β

(2π)n/2

∫xiβikxk e−βikxixk/2dx1 · · ·dxn . (A.19)

The easiest way to calculate this integral is to calculate to assume for a while thatxi 6= 0. Then

x =

√β

(2π)n/2

∫xi e

−βik(xi−xi0)(xk−xk0/2dx1 · · ·dxn = xi0 .

Then we can differentiate both sides with respect toxk0 and then putxi0 = xk0 = 0. The l.h.s. isjust 〈xiXk〉 while the r.h.s. isδik. Thus

〈xiXk〉= δik , or βik〈xixk〉= δik , 〈xixk〉=(

β−1)

ik. (A.20)

A.3 The Dirac delta-function

The main property is to single-out one particular valuex−x0 of a variablex. It is defined by thecharacteristic properties

δ(x−x0) =

0 for x 6= x0

∞ for x = x0(A.21)

A.4. FOURIER SERIES AND TRANSFORMS 111

but in such way that for anyε > 0∫ x0+ε

x0−εδ(x−x0)dx= 1. (A.22)

Since delta-function has a very (infinitely) sharp peak atx = x0 and a unit area∫ B

Af (x)δ(x−x0)dx=

f (x0) if A < x0 < B0 otherwise

. (A.23)

Representations for the Diracδ-function

Let introduce a positive parameterγ and at final stage tend it to zero. Physically it means thatγis less than all other involved scales. The main representations are the following: rectangular,

δ(x) =

γ−1 for − γ/2 < x < γ/e0 otherwise

, (A.24)

Lorentzian,

δ(x) =1π

γx2 + γ2 , (A.25)

Gaussian,

δ(x) =1

γ√

2πe−x2/2γ2

, (A.26)

and integral,

δ(x) =12π

∫ ∞

−∞eikxdk (A.27)

We will also need a representation for the Kronekker’s symbol

δn,0 =12π

∫ π

−πe−nφdφ . (A.28)

A.4 Fourier Series and Transforms

Fourier series

Generally the Fourier series is defined as an expansion of a real function in series as

f (x) =a0

2+

∞

∑k=1

(ak sinkx+bk coskx) . (A.29)

The sum of the series is periodic with the period2π.Another form is complex Fourier series

f (x) =∞

∑k=−∞

ck eikx , c±k = (ak∓bk)/2, c0 = a0/2. (A.30)


Since∫ 2π

0sinmxsinnxdx=

πδmn, m 6= 0,0, m= 0,

∫ 2π

0cosmxcosnxdx=

πδmn, m 6= 0,2π, m= n = 0,

∫ 2π

0sinmxcosnxdx=

0, for integer m, n

we have

ak

bk

ck

=

1π

∫ 2π

0dt f(t)

cosktsinkt12 e−ikt

, k 6= 0. (A.31)

At k = 0,

a0 = c0 =12π

∫ 2π

0dt f(t) .

If the function has a discontinuity atx = x0 then the series gives

f (x0) =12

[ f (x0 +0)+ f (x0−0)] .

If one is willing to change the interval2π → 2L than k has to be replaced byπk/L and thenormalization factor is1/L rather than1/π.

Advantages of the Fourier series: it can represent discontinuous functions. Let us take anexample of periodic function with the period2π defined as

F(x) =

1, −π/2 < x < π/20, −π < x <−π/2, π/2 < x < π,

This function is even, thusbk = 0, ak = (2/πk)sinπk/2.

Below we plot the sum of 5 and 50 Fourier harmonics. It is useful to know the Parseval’s identity

1π

∫ π

−π[ f (x)]2 dx=

a20

2+

∞

∑k=1

(a2k +b2

k) .

Fourier transform

We assumeL→ ∞ and replace

∞

∑k=−∞

(. . .)→ L∫ ∞

−∞dk(. . .) .

g(k) =12π

∫ ∞

−∞f (t)e−ikt dt . (A.32)

A.4. FOURIER SERIES AND TRANSFORMS 113

0

0.2

0.4

0.6

0.8

1

–1 –0.8 –0.6 –0.4 –0.2 0.2 0.4 0.6 0.8 1

x

Figure A.1:

The completeness condition

f (x) =12π

∫ ∞

−∞dkeikt

∫ ∞

−∞dx f(x)e−ikx . (A.33)

The Dirac delta-function is represented as

δ(x− t) =12π

∫ ∞

−∞dkeik(t−x) . (A.34)

We have several useful rules for the Fourier transform. Ifg(k) is the Fourier transform off (x)then

f ′(x)≡ d f/dx → ikg(k) ;

f (n)(x)≡ dn f/dxn → (ik)ng(k) ;∫ ∞

−∞dy f1(y) f2(x−y) → g1(k) ·g2(k) (convolution theorem);∫ ∞

−∞f1(t) f2(t)dt →

∫ ∞

−∞g1(k) f2(−k)dk;

∫ ∞

−∞f1(t) f ∗2 (t)dt →

∫ ∞

−∞g1(k) f ∗2 (k)dk the Parseval’s theorem. (A.35)

These properties are very important for solving differential equations, etc.


f (x) = 1√2π

∫ ∞−∞ F(ω)e−iωxdx F(ω) = 1√

2π

∫ ∞−∞ f (x)e−iωxdx

1a2+x2 ,a > 0

√π2

e−a|ω|a

f (x) =

1, |x|< a0, |x|> a

√2π

sin(aω)ω

e−a2x2 1a√

2e−ω2/4a

Table A.2:Table of Fourier transform pairs

Examples

One-dimensional diffusion: The equation has the form

∂n(x, t)∂t

=−D∂2n(x, t)

∂x2 .

Let us assume that we add a particle at timet = 0 at the pointx = 0. Then the equation togetherwith initial condition can be written as

∂n(x, t)∂t

+D∂2n(x, t)

∂x2 = δ(x)δ(t) .

The Fourier transform of this equation has the form:

(Dk2− iω)n(k,ω) = 1 → n(k,ω) =1

(2π)2

1Dk2− iω

.

Now we can come back to real space-time representation:

n(x, t) =∫ ∞

−∞

dk2π

∫ ∞

−∞

dω2π

1Dk2− iω

=∫ ∞

−∞

dk2π

e−Dk2|t| =1

2√

πD|t| e−x2/4D|t| .

Appendix B

Maple Printouts

Quick access:1, 2.1, 2.2, 2.17, 3.7, 3.12, 3.13, 4.9, 4.10, 4.13, 4.14, 5.1, 5.2, 5.3, 5.4, 5.5, 5.7,5.8

Test in mathematics> F:=(xi,eta)->xi^4+xi^3+eta*xi^2:> F1:=(xi,eta)->subs(y=xi,diff(F(y,eta),y)): solve(F1(xi,eta)=0,xi);

0,−38

+18

√9−32η,−3

8− 1

8

√9−32η

One extremum at eta>9/32, otherwise 3 extrema

> x1:=-3/8+1/8*sqrt(9-32*eta): x2:=-3/8-1/8*sqrt(9-32*eta):

> F2:=(xi,eta)->subs(y=xi,diff(F1(y,eta),y)):> g0:=eta->simplify(F2(0,eta)): g1:=eta->simplify(F2(x1,eta)):> g2:=eta->simplify(F2(x2,eta)):

> g0(eta); g1(eta); g2(eta);

2η98− 3

8

√9−32η−4η

98

+38

√9−32η−4η

> plot(g0(eta),g1(eta),g2(eta),eta=-0.1..9/32);

115

116 APPENDIX B. MAPLE PRINTOUTS

0.5

1

1.5

2

2.5

–0.1 –0.05 0.05 0.1 0.15 0.2 0.25eta

x0-> maximum at eta<0, otherwise minimum; x1-> minimum at eta>0, maximum at9/32>eta>0; x2-> minimum at 9/32>eta.

Two specific values: eta=0 and eta=9/32.

> plot(F(xi,0), xi=-1.1..0.45); plot(F(xi,9/32), xi=-0.7..0.25);

117

–0.1

–0.05

0.05

0.1

–1 –0.8 –0.6 –0.4 –0.2 0.2 0.4xi

0.005

0.01

0.015

0.02

0.025

0.03

0.035

–0.6 –0.4 –0.2 0.2xi

Problem 1.2> f1:=x->ln((1-x)/(1+x)): plot(f1(x),x=-1..1);


–6

–4

–2

0

2

4

6

–1 –0.8 –0.6 –0.4 –0.2 0.2 0.4 0.6 0.8 1x

> f2:=x->ln(abs(tan(Pi*x))): plot(f2(x),x=0..1,f2=-5..5);

–4

–2

0

2

4

f2

0.2 0.4 0.6 0.8 1x

119

> simplify(exp(4*ln(x))-(x^2+1)^2+2*x^2+1);

0

Problem 1.3> simplify((sin(x))^(-2)-(tan(x))^(-2)-1,trig);

0> assume(alpha>0): assume(beta>0):> k:=alpha->sum(exp((I*beta-alpha)*n),n=0..infinity):

> a1:=alpha->evalc(Re(k(alpha))): a2:=alpha->evalc(Im(k(alpha))):> a1(alpha); a2(alpha);

− eα˜ (cos(β˜)−eα˜)(cos(β˜)−eα˜)2 +sin(β˜)2

eα˜ sin(β˜)(cos(β˜)−eα˜)2 +sin(β˜)2

Problem 1.4> assume(zeta>0): assume(n,natural):> simplify(int(x^n*exp(-zeta*x),x=0..infinity));

ζ˜(−n˜−1) Γ(n˜+1)> int((x^2+a^2)^(-1),x); int((x^2-a^2)^(-1),x);

arctan(xa)

a12

ln(x−a)a

− 12

ln(a+x)a

> int(exp(-zeta*x^2/2),x=0..infinity);

12

√2√

π√ζ˜

Problem 2.1

(a)> uax:=(x,y)->-y/(x^2+y^2): uay:=(x,y)->x/(x^2+y^2):

> simplify(diff(uax(x,y),y)-diff(uay(x,y),x));

0

The differential is exact> ua:=(x,y)->int(uax(xi,y),xi=0..x): ua(x,y);

−arctan(xy)


(b)> ubx:=(x,y)->(y-x^2): uby:=(x,y)->(x+y^2):

> simplify(diff(ubx(x,y),y)-diff(uby(x,y),x));

0

The differential is exact> ub:=(x,y)->int(ubx(xi,y),xi=0..x)+int(uby(x,eta),eta=0..y)-int(int(di> ff(ubx(xi,eta),eta),eta=0..y),xi=0..x);

ub := (x, y)→∫ x

0ubx(ξ, y)dξ+

∫ y

0uby(x, η)dη−

∫ x

0

∫ y

0

∂∂η ubx(ξ, η)dηdξ

> ub(x,y);

yx− 13

x3 +13

y3

(c)> ucx:=(x,y)->(2*y^2-3*x): ucy:=(x,y)->-4*x*y:

> simplify(diff(ucx(x,y),y)-diff(ucy(x,y),x));

8y

The differential is not exact

Problem 2.2

General definitions> ua:=(x,y)->int(ux(xi,b),xi=a..x)+int(uy(x,eta),eta=b..y);

ua := (x, y)→∫ x

aux(ξ, b)dξ+

∫ y

buy(x, η)dη

> ub:=(x,y)->int(uy(a,eta),eta=b..y)+int(ux(xi,y),xi=a..x);

ub := (x, y)→∫ y

buy(a, η)dη+

∫ x

aux(ξ, y)dξ

> ux1:=(x,y)->2*x*y+x^2: uy1:=(x,y)->x^2:

> ua1:=(x,y)->int(ux1(xi,b),xi=a..x)+int(uy1(x,eta),eta=b..y):

> ub1:=(x,y)->int(uy1(a,eta),eta=b..y)+int(ux1(xi,y),xi=a..x):

> ua1(x,y); ub1(x,y); simplify(ua1(x,y)-ub1(x,y));

b(x2−a2)+13

x3− 13

a3 +x2(y−b)

a2(y−b)+y(x2−a2)+13

x3− 13

a3

0> ux2:=(x,y)->y*(2-2*y): uy2:=(x,y)->-x^2:

121

> ua2:=(x,y)->int(ux2(xi,b),xi=a..x)+int(uy2(x,eta),eta=b..y):

> ub2:=(x,y)->int(uy2(a,eta),eta=b..y)+int(ux2(xi,y),xi=a..x):

> ua2(x,y); ub2(x,y); simplify(ua2(x,y)-ub2(x,y));

b(2−2b)(x−a)−x2(y−b)

−a2(y−b)+y(2−2y)(x−a)

2bx−2ba−2b2x+2b2a−x2y+x2b+a2y−a2b−2yx+2ya+2y2x−2y2a

Why?

> simplify(diff(ux2(x,y),y)-uy2(x,y),x);

2−4y+x2

The differential is not exact.

Problem 2.17> P:=(T,v)->R*T/(v-b)-alpha/v^2:

> num:=(diff(P(T,v),T))^2: den:=diff(P(T,v),v);

den:=− RT(v−b)2 +

2αv3

> simplify(-T*num/den);

− v3R2T−RT v3 +2αv2−4αvb+2αb2

> factor(2*alpha*v^2-4*alpha*v*b+2*alpha*b^2);

2α(−v+b)2

Problem 3.7

> P:=(v,T)->(8*T/(3*v-1))-3/v^2;

P := (v, T)→ 8T

3v−1− 3

v2

> plot(P(v,1.5),P(v,1),P(v,0.5)> ,v=0.5..1.2,thickness=3,color=black);


–4

–2

0

2

4

6

8

10

12

0.6 0.7 0.8 0.9 1 1.1 1.2v

For T=0.5 there is no stable region at all. The phse transition can exist only nes T=1.> plot(P(v,1.05),P(v,1),P(v,0.85)> ,v=0.45..5.3,P=0..2,thickness=3,color=black);

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

P

1 2 3 4 5v

> ?plot,color

Problem 3.12> phi:=xi->(A/3)*(xi+eta)^2-(2*B/27)*(xi+eta)^3+(C/9)*(xi+eta)^4:> series(phi(xi),xi=0,5);

(13

Aη2− 227

Bη3 +19

Cη4)+(23

Aη− 29

Bη2 +49

Cη3)ξ+(13

A− 29

Bη+23

Cη2)ξ2+

(− 227

B+49

Cη)ξ3 +19

Cξ4

> c3:=eta->-2/27*B+4/9*C*eta: solve(c3(eta)=0,eta);

16

BC

> c1:=(A,eta)->2/3*A*eta-2/9*B*eta^2+4/9*C*eta^3;

c1 := (A, η)→ 23

Aη− 29

Bη2 +49

Cη3

> solve(c1(A,B/(6*C)),A);

127

B2

C> c2:=subs(eta=B/(6*C),A=B^2/(27*C),1/3*A-2/9*B*eta+2/3*C*eta^2);

c2 :=− 1162

B2

C> delta:=xi->(C/9)*xi^4+c2*xi^2:> delta1:=xik->subs(y=xi,diff(delta(y),y)): delta1(xi);

49

Cξ3− 181

B2ξC

> solve(4/9*C*xi^2-1/81*B^2/C,xi);

−16

BC

,16

BC

Problem 3.13> P:=(v,T)->8*T/(3*v-1)-3/(T*v^2);

P := (v, T)→ 8T

3v−1− 3

T v2

> p1:=(nu,epsilon)=P(1+nu,1+epsilon)-1;

p1 := (ν, ε) = 81+ ε

2+3ν− 3

(1+ ε)(1+ν)2 −1

> aux:=series(8*(1+epsilon)/(2+3*nu)-3/(1+epsilon)/(1+nu)^2-1,nu=0,4):a> ux;

(3+4ε− 31+ ε

)+(−6−6ε+6

1+ ε)ν+(−9

11+ ε

+9+9ε)ν2 +(121

1+ ε− 27

2− 27

2ε)

ν3 +O(ν4)> aux1:=(epsilon)->3+4*epsilon-3/(1+epsilon)+(-6-6*epsilon+6/(1+epsilon> ))*nu+(-9*1/(1+epsilon)+9+9*epsilon)*nu^2> +(12*1/(1+epsilon)-27/2-27/2*epsilon)*nu^3;

aux1:= ε→ 3+4ε− 31+ ε

+(−6−6ε+6

1+ ε)ν+(−9

11+ ε

+9+9ε)ν2

+(121

1+ ε− 27

2− 27

2ε)ν3

> aux2:=series(aux1(epsilon),epsilon=0,2);

aux2:=−32

ν3 +(7− 512

ν3 +18ν2−12ν)ε+O(ε2)

Simplified equation of state> p:=(nu,epsilon)->-3/2*nu^3+(7-12*nu)*epsilon;

p := (ν, ε)→−32

ν3 +(7−12ν)ε

Critical point - epsilon=0;> d1:=(nu,epsilon)->subs(y=nu,diff(p(y,epsilon),y)): d1(nu,epsilon);

−92

ν2−12ε

The curve is symmetric. That leads to vg=-vl=v> solve(p(v,epsilon)= p(-v,epsilon),v);;

0, 2√−2ε,−2

√−2ε

Critical exponent beta=1/2

Along the coexitence curve, vˆ2=-8epsilon. exponent gamma=1

Function of pressure> P1:=(rho,T)->8*T*rho/(3-rho)-3*rho^2/T:

Near the critical point

> P1(1+xi,1);

81+ξ2−ξ

−3(1+ξ)2

> series(8*(1+xi)/(2-xi)-3*(1+xi)^2,xi=0);

1+32

ξ3 +34

ξ4 +38

ξ5 +O(ξ6)

the exponent delta=3

Problem 4.9> assume(z>0): int(1/sqrt(z-x^2),x=0..sqrt(z));

12

π

> f:=k->1/(1-2*I*k): d1:=k->subs(y=k,diff(f(y),y)):> d2:=k->subs(y=k,diff(d1(y),y)): d3:=k->subs(y=k,diff(d2(y),y)):

> d1(0); d2(0); d3(0);

2I−8−48I

Problem 4.10> P:=(N,k)->(N!/(k!*(N-k)!))*(1/4)^k*(3/4)^(N-k):

> evalf(P(12,3));

.2581036091

> evalf(P(120,30));

.08385171464

> G:=(N,k)->(1/sqrt(3*N*Pi/8))*exp(-(k-N/4)^2*8/(3*N));

G := (N, k)→ e(−8/3 (k−1/4N)2N )

√38

Nπ

> evalf(G(12,3));

.2659615201

> evalf(G(120,30));

.08410441740

> plot(P(12,k),G(12,k),k=0..12);

0

0.05

0.1

0.15

0.2

0.25

2 4 6 8 10 12k

> plot(P(120,k),G(120,k),k=0..120);

0

0.02

0.04

0.06

0.08

20 40 60 80 100 120k

Problem 4.13

beta = sigmaˆ2> assume(beta>0): assume(a,real): assume(N>0):> P:=k->(2*Pi*beta)^(-1/2)*int(exp(-(x-a)^2/(2*beta)-I*k*x),x=-infinity.> .infinity): simplify(P(k));

e(1/2I k (−2a˜+I kβ˜))

> P1:=s->(2*Pi)^(-1)*int(exp(I*(s-N*a)*k-k^2*N*beta/2),k=-infinity..inf> inity): P1(s);

12

e(−1/2 (−s+Nã˜)2N˜β˜ )√2√

π√

N˜β˜> int(s*P1(s),s=-infinity..infinity);

Nã˜> int(s^2*P1(s),s=-infinity..infinity)-N^2*a^2;

N˜β˜

Problem 4.14> assume(a>0): assume(d>a): assume(N>0):

> simplify((2*a)^(-1)*int(x^2,x=d-a..d+a));

d˜2 +13

a˜2

> simplify(N*(d^2+1/3*a^2)+N*(N-1)*d^2-N^2*d^2);

13

Nã˜2

Problem 5.1> with(linalg):

> Q:=array([[0,1,0],[1/8,1/2,3/8],[0,1/2,1/2]]);

Q :=

0 1 018

12

38

012

12

> QT:=transpose(Q):

> eigenvectors(Q);

[1, 1, [1, 1, 1]], [−14

, 1, [6,−32

, 1

]], [

14, 1, [4, 1,−2]]

> eigenvectors(QT);

[14, 1, [1, 2,−3]], [

−14

, 1, [1,−2, 1]], [1, 1, [1, 8, 6]]

> r1:=vector([1, 1, 1]); r2:=vector([6, -3/2, 1]); r3:=vector([4, 1,> -2]);

r1 := [1, 1, 1]

r2 :=[6,−32

, 1

]

r3 := [4, 1,−2]

> l1:=vector([1, 8, 6]); l2:=vector([1, -2, 1]); l3:=vector([1, 2,> -3]);

l1 := [1, 8, 6]

l2 := [1,−2, 1]

l3 := [1, 2,−3]

> n1:=15: n2:=10: n3:=12:

> P1:=array([[l1[1]*r1[1], l1[2]*r1[1],l1[3]*r1[1]], [l1[1]*r1[2],> l1[2]*r1[2],l1[3]*r1[2]],[l1[1]*r1[3], l1[2]*r1[3],l1[3]*r1[3]]]);

P1 :=

1 8 61 8 61 8 6


P2 :=

6 −12 6−32

3−32

1 −2 1


P3 :=

4 8 −121 2 −3

−2 −4 6

> P:=s->(1/n1)*P1+(1/n2)*(-1/4)^s*P2+(1/n3)*(1/4)^s*P3;> Question (a)

P := s→ P1n1

+(−14

)sP2

n2+

(14)sP3

n3

> Pr:=s->evalm(P(s)): Pr(s); Pr(0);

115

+35

(−14

)s+13

(14)s 8

15− 6

5(−14

)s+23

(14)s 2

5+

35

(−14

)s− (14)s

115− 3

20(−14

)s+112

(14)s 8

15+

310

(−14

)s+16

(14)s 2

5− 3

20(−14

)s− 14

(14)s

115

+110

(−14

)s− 16

(14)s 8

15− 1

5(−14

)s− 13

(14)s 2

5+

110

(−14

)s+12

(14)s

1 0 00 1 00 0 1

> p0:=vector([0,1,0]);

p0 := [0, 1, 0]> PS:=s->evalm(p0&*P(s)): PS(s);

Question (b)

[115− 3

20(−14

)s+112

(14)s,

815

+310

(−14

)s+16

(14)s,

25− 3

20(−14

)s− 14

(14)s

]

> PS(2); PS(infinity);> Answer: after 2 steps 3/8, after many steps 2/5[

116

,916

,38

]

[115

,815

,25

]

> av:=s->1*PS(s)[1]+4*PS(s)[2]+9*PS(s)[3]: av(s);> Moment

295− 3

10(−14

)s− 32

(14)s

> av:=sum(m^2*PS(s)[m],m=1..3);

295− 3

10(−14

)s− 32

(14)s

> corr:=s->simplify(sum(n^2*sum(m^2*PS(s)[m]*Pr(s)[m,n],> m=1..3),n=1..3)): corr(s);

7950

(−1)(1+s) 4(−s)− 334

16(−s) +81100

(−1)(1+2s) 16(−s) +112

(−1)(1+s) 16(−s)− 32

4(−s)

+84125

> corr(0);

16

> corr(infinity);

84125


Warning, the protected names norm and trace have been redefined andunprotected

> Q:=array([[0,1/2,1/2],[0,0,1],[3/4,1/4,0]]);

Q :=

012

12

0 0 134

14

0

> QT:=transpose(Q):

> r:=eigenvectors(Q):

> l:=eigenvectors(QT):

> r[1][1]; r[2][1]; r[3][1];

1

−12

+14

I√

2

−12− 1

4I√

2

> l[1][1]; l[2][1]; l[3][1];

1

−12

+14

I√

2

−12− 1

4I√

2

> r[1][3]; r[2][3]; r[3][3];

[1, 1, 1]

[−1

6− 1

3I√

2, 1,−12

+14

I√

2

]

[−1

6+

13

I√

2, 1,−12− 1

4I√

2

]

> l[1][3]; l[2][3]; l[3][3];

[65, 1,

85

]

[1,−1

3− 1

3I√

2,−23

+13

I√

2

]

[1,−1

3+

13

I√

2,−23− 1

3I√

2

]

> r1:=vector([1, 1, 1]): r2:=vector([-1/6-1/3*I*sqrt(2), 1,> -1/2+1/4*I*sqrt(2)]): r3:=vector([-1/6+1/3*I*sqrt(2), 1,> -1/2-1/4*I*sqrt(2)]): l1:=vector([6/5, 1, 8/5]): l2:=vector([1,> -1/3-1/3*I*sqrt(2), -2/3+1/3*I*sqrt(2)]): l3:=vector([1,> -1/3+1/3*I*sqrt(2), -2/3-1/3*I*sqrt(2)]):> n1:=evalm(l1&*r1): n2:=evalc(evalm(l2&*r2)):> n3:=evalc(evalm(l3&*r3)):

Probability matrices.> P1:=array([[l1[1]*r1[1], l1[2]*r1[1],l1[3]*r1[1]], [l1[1]*r1[2],> l1[2]*r1[2],l1[3]*r1[2]],[l1[1]*r1[3], l1[2]*r1[3],l1[3]*r1[3]]]);

P1 :=

65

185

65

185

65

185


P2 :=

−16− 1

3I√

2 (−13− 1

3I√

2)(−16− 1

3I√

2) (−23

+13

I√

2)(−16− 1

3I√

2)

1 −13− 1

3I√

2 −23

+13

I√

2

−12

+14

I√

2 (−13− 1

3I√

2)(−12

+14

I√

2) (−23

+13

I√

2)(−12

+14

I√

2)


P3 :=

−16

+13

I√

2 (−13

+13

I√

2)(−16

+13

I√

2) (−23− 1

3I√

2)(−16

+13

I√

2)

1 −13

+13

I√

2 −23− 1

3I√

2

−12− 1

4I√

2 (−13

+13

I√

2)(−12− 1

4I√

2) (−23− 1

3I√

2)(−12− 1

4I√

2)

> P:=s->(1/n1)*P1+(1/n2)*(r[2][1])^s*P2+(1/n3)*(r[3][1])^s*P3;> Question (a)

P := s→ P1n1

+r21

sP2n2

+r31

sP3n3

> Pr:=s->evalm(P(s)):

> p0:=vector([1,0,0]);

p0 := [1, 0, 0]> PS:=s->evalm(p0&*P(s)): PS(s)[1];

619

+(−1

2+

14

I√

2)s(−16− 1

3I√

2)

−13− I

√2

+(−1

2− 1

4I√

2)s(−16

+13

I√

2)

−13

+ I√

2

> evalc(PS(infinity)[1]); evalc(PS(2)[1]);

61938



> Q:=array([[0,1/2,1/2],[1/3,0,2/3],[1/3,2/3,0]]);

Q :=

012

12

13

023

13

23

0

> QT:=transpose(Q):



> r[1][1]; r[2][1]; r[3][1];

−13−231

> l[1][1]; l[2][1]; l[3][1];

−131

−23

> r[3][3]; r[1][3]; r[2][3];

[1, 1, 1][−3, 1, 1][0,−1, 1]

> l[2][3]; l[1][3]; l[3][3];

[1,

32,

32

]

[−2, 1, 1][0,−1, 1]

> r1:=vector([1, 1, 1]): r2:=vector([-3, 1, 1]): r3:=vector([0, -1,> 1]): l1:=vector([1, 3/2, 3/2]): l2:=vector([-2, 1, 1]): l3:=vector([0,> -1, 1]):> n1:=evalm(l1&*r1): n2:=evalc(evalm(l2&*r2)):> n3:=evalc(evalm(l3&*r3)):

Probability matrices.> P1:=array([[l1[1]*r1[1], l1[2]*r1[1],l1[3]*r1[1]], [l1[1]*r1[2],> l1[2]*r1[2],l1[3]*r1[2]],[l1[1]*r1[3], l1[2]*r1[3],l1[3]*r1[3]]]);

P1 :=

132

32

132

32

132

32


P2 :=

6 −3 −3−2 1 1−2 1 1


P3 :=

0 0 00 1 −10 −1 1

> P:=s->(1/n1)*P1+(1/n2)*(r[3][1])^s*P2+(1/n3)*(r[2][1])^s*P3;> Question (a)

P := s→ P1n1

+r31

sP2n2

+r21

sP3n3

> Pr:=s->evalm(P(s)): Pr(s);> Conditional probability

1 0 0

012

+12

(−23

)s 12− 1

2(−23

)s

012− 1

2(−23

)s 12

+12

(−23

)s

> p0:=vector([0,0,1]):

> PS:=s->evalm(p0&*P(s)): PS(s);

Probability vector

[14− 1

4(−23

)s,−18

+18

(−23

)s,78

+18

(−23

)s

]

> PS(infinity); PS(0); [0,

12,

12

]

[0, 0, 1]

> moment:=1*PS(s)[1]+2*PS(s)[2]+3*PS(s)[3]: moment;

52

+12

(−23

)s

> corr:=s->simplify(sum(n*sum(m*PS(s)[m]*Pr(s)[m,n], m=1..3),n=1..3)):> corr(s);

254

+54

3(−2s) (−1)(2s) 2(2s) +(−1)s3(1−s) 2(−1+s)

> corr(0); corr(infinity);

9254



> Q0:=s->array([[0,(cos(Pi*s/2))^2,(sin(Pi*s/2))^2],[1/4> +1/2*(sin(Pi*s/2))^2,0,1/4+1/2*(cos(Pi*s/2))^2],[1/2*(cos(Pi*s/2))^2,1> /2+1/2*(sin(Pi*s/2))^2,0]]): Q0(s);

0 cos(12

πs)2 sin(12

πs)2

14

+12

sin(12

πs)2 014

+12

cos(12

πs)2

12

cos(12

πs)2 12

+12

sin(12

πs)2 0

> simplify(Q0(2*s));

0 cos(πs)2 1−cos(πs)2

34− 1

2cos(πs)2 0

14

+12

cos(πs)2

12

cos(πs)2 1− 12

cos(πs)2 0

Since [cos(Pi*s)]ˆ2=1, we get

> Q:=array([[0,1,0],[1/4,0,3/4],[1/2,1/2,0]]);

Q :=

0 1 014

034

12

12

0

> QT:=transpose(Q):



> r[1][1]; r[2][1]; r[3][1];

1

−12

+14

I√

2

−12− 1

4I√

2

> l[1][1]; l[2][1]; l[3][1];

−12

+14

I√

2

−12− 1

4I√

2

1> r[3][3]; r[1][3]; r[2][3];

[1,−1

2− 1

4I√

2,−16

+13

I√

2

]

[1, 1, 1]

[1,−1

2+

14

I√

2,−16− 1

3I√

2

]

> l[3][3]; l[1][3]; l[2][3];

[1,

85,

65

]

[−1

3− 1

3I√

2,−23

+13

I√

2, 1

]

[−1

3+

13

I√

2,−23− 1

3I√

2, 1

]

> r1:=vector([1, 1, 1]): r2:=vector([1, -1/2+1/4*I*sqrt(2),> -1/6-1/3*I*sqrt(2)]): r3:=vector([1, -1/2-1/4*I*sqrt(2),> -1/6+1/3*I*sqrt(2)]): l1:=vector([1, 8/5, 6/5]):> l2:=vector([-1/3-1/3*I*sqrt(2), -2/3+1/3*I*sqrt(2), 1]):> l3:=vector([-1/3+1/3*I*sqrt(2), -2/3-1/3*I*sqrt(2), 1]):

> n1:=evalm(l1&*r1): n2:=evalc(evalm(l2&*r2)):> n3:=evalc(evalm(l3&*r3)):

Probability matrices.


P1 :=

185

65

185

65

185

65


P2 :=

−13− 1

3I√

2 −23

+13

I√

2 1

(−13− 1

3I√

2)(−12

+14

I√

2) (−23

+13

I√

2)(−12

+14

I√

2) −12

+14

I√

2

(−13− 1

3I√

2)(−16− 1

3I√

2) (−23

+13

I√

2)(−16− 1

3I√

2) −16− 1

3I√

2


P3 :=

−13

+13

I√

2 −23− 1

3I√

2 1

(−13

+13

I√

2)(−12− 1

4I√

2) (−23− 1

3I√

2)(−12− 1

4I√

2) −12− 1

4I√

2

(−13

+13

I√

2)(−16

+13

I√

2) (−23− 1

3I√

2)(−16

+13

I√

2) −16

+13

I√

2

> P:=s->(1/n1)*P1+(1/n2)*(r[1][1])^(2*s)*P2+(1/n3)*(r[2][1])^(2*s)*P3;> Question (a)

P := s→ P1n1

+r11

(2s) P2n2

+r21

(2s) P3n3

> Pr:=s->evalm(P(s)):> Conditional probability> p01:=vector([1,0,0]): p02:=vector([0,1,0]):

> PS1:=s->evalm(p01&*P(s)): PS2:=s->evalm(p02&*P(s)):

> PS1(s)[1];

519

+−1

3− 1

3I√

2

−13− I

√2

+(−1

2+

14

I√

2)(2s) (−13

+13

I√

2)

−13

+ I√

2

> PS1(s)[2];

819

+−2

3+

13

I√

2

−13− I

√2

+(−1

2+

14

I√

2)(2s) (−23− 1

3I√

2)

−13

+ I√

2

> PS2(s)[1];

519

+(−1

3− 1

3I√

2)(−12

+14

I√

2)

−13− I

√2

+(−1

2+

14

I√

2)(2s) (−13

+13

I√

2)(−12− 1

4I√

2)

−13

+ I√

2

> PS2(s)[2];

819

+(−2

3+

13

I√

2)(−12

+14

I√

2)

−13− I

√2

+(−1

2+

14

I√

2)(2s) (−23− 1

3I√

2)(−12− 1

4I√

2)

−13

+ I√

2

The results can be transformed to polar form by the command polar.

Problem 5.5> assume(N,natural): assume(k,real): assume(s>0); assume(n,integer):> P:=(n,s,N)->(1/(2*N+1))*(1+2*sum(cos(2*Pi*k*n/(2*N+1))*exp(-2*s*(sin(> Pi*k/(2*N+1)))^2),k=1..N)):

> P(n,s,N);

1+2

(N˜

∑k˜=1

cos(2πk˜n˜

2N˜+1)e(−2s sin( πk˜

2N˜+1)2)

)

2N˜+1> Pm:=[seq(P(i,2,2),i=-2..2)]: evalf(Pm);

[.1220644066, .2223516301, .3111679263, .2223516301, .1220644066]



> w:=matrix(3,3,[0,1/2,1/2,1/3,0,1/3,1/3,1/3,0]);

w :=

012

12

13

013

13

13

0

> M:=matrix(3,3,[-1,1/2,1/2,1/3,-1,1/3,1/3,1/3,-1]);

M :=

−112

12

13

−113

13

13

−1

> MT:=transpose(M):

> eigenvectors(M);

[−56

+16

√13, 1,

[−1

2+

12

√13, 1, 1

]], [−5

6− 1

6

√13, 1,

[−1

2− 1

2

√13, 1, 1

]],

[−43

, 1, [0,−1, 1]]> eigenvectors(MT);

[−43

, 1, [0,−1, 1]], [−56

+16

√13, 1,

[−1

3+

13

√13, 1, 1

]],

[−56− 1

6

√13, 1,

[−1

3− 1

3

√13, 1, 1

]]

> lambda:=vector([-5/6+1/6*sqrt(13),-5/6-1/6*sqrt(13),-4/3]):

> r1:=vector([-1/2+1/2*sqrt(13), 1, 1]); r2:=vector([-1/2-1/2*sqrt(13),> 1, 1]); r3:=vector([0, -1, 1]);

r1 :=[−1

2+

12

√13, 1, 1

]

r2 :=[−1

2− 1

2

√13, 1, 1

]

r3 := [0,−1, 1]> l1:=vector([-1/3+1/3*sqrt(13), 1, 1]); l2:=vector([-1/3-1/3*sqrt(13),> 1, 1]); l3:=vector([0, -1, 1]);

l1 :=[−1

3+

13

√13, 1, 1

]

l2 :=[−1

3− 1

3

√13, 1, 1

]

l3 := [0,−1, 1]1

> n1:=evalm(l1&*r1): n2:=evalm(l2&*r2): n3:=evalm(l3&*r3):

> Q1:=-sum((1/n1)*r1[1]*l1[k]/lambda[1],k=1..3):

> Q2:=-sum((1/n2)*r1[1]*l1[k]/lambda[2],k=1..3):

> Q3:=-sum((1/n3)*r1[1]*l1[k]/lambda[1],k=1..3):> evalf(Q1+Q2+Q3);> First passage time

13.64536110

> :

Bibliography

[1] L. E. ReichlA modern course in statistical physics, 1998.

141

Problemas Resueltos Del Reichl.tmp

Documents

Transcript of Problemas Resueltos Del Reichl.tmp