statistical physics (macfarlane a.)

7/27/2019 Statistical Physics (Macfarlane a.)

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Statistical Physics

Dr. A. J. Macfarlane1

Lent 1998

1LATEXed by Paul Metcalfe comments to [email protected].


2/43

Revision: 1.10

Date: 1998-06-03 20:13:39+01

The following people have maintained these notes.

date Paul Metcalfe


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Contents

Introduction v

1 Quantum Statistical Mechanics 1

1.1 Introduction to Quantum Statistical Mechanics . . . . . . . . . . . . 1

1.2 Canonical ensemble . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.3 Temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.4 Towards thermodynamic variables . . . . . . . . . . . . . . . . . . . 5

1.5 Towards applications . . . . . . . . . . . . . . . . . . . . . . . . . . 71.5.1 Nparticle partition function . . . . . . . . . . . . . . . . . . 71.5.2 Extensive and intensive variables . . . . . . . . . . . . . . . 8

1.5.3 Density of states . . . . . . . . . . . . . . . . . . . . . . . . 8

1.5.4 Gas of spinless particles . . . . . . . . . . . . . . . . . . . . 9

1.5.5 Entropy and the Gibbs paradox . . . . . . . . . . . . . . . . . 9

1.6 Harmonic oscillator model . . . . . . . . . . . . . . . . . . . . . . . 10

2 Thermodynamics 11

2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.2 Applications ofdE= TdS PdV . . . . . . . . . . . . . . . . . . 122.2.1 Integrability conditions . . . . . . . . . . . . . . . . . . . . . 12

2.2.2 Specific heats . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.2.3 Adiabatic changes . . . . . . . . . . . . . . . . . . . . . . . 14

2.2.4 Entropy ofnmoles of ideal gas . . . . . . . . . . . . . . . . 142.2.5 van der Waals equation . . . . . . . . . . . . . . . . . . . . 15

2.2.6 The Joule effect . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.3 Some thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.3.1 The second law . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.4 Heat flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3 Grand ensemble methods 19

3.1 The formalism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3.2 Systems of non-interacting identical particles . . . . . . . . . . . . . 21

3.2.1 A little quantum mechanics . . . . . . . . . . . . . . . . . . 21

3.2.2 The partition functions . . . . . . . . . . . . . . . . . . . . . 223.2.3 Classical limit . . . . . . . . . . . . . . . . . . . . . . . . . 24

3.3 Black body radiation . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.4 Degenerate Fermi gas . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3.4.1 Heat capacity at low temperature . . . . . . . . . . . . . . . . 27

3.5 Bose-Einstein condensation . . . . . . . . . . . . . . . . . . . . . . . 28

iii


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iv CONTENTS

3.6 White dwarf stars . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

4 Classical statistical mechanics 31

4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

4.2 Diatomic gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

4.3 Paramagnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

4.4 Specific heats . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 334.5 Weak interparticle forces . . . . . . . . . . . . . . . . . . . . . . . . 34

4.6 The Maxwell distribution . . . . . . . . . . . . . . . . . . . . . . . . 35

Approximations 37


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Introduction

These notes are based on the course Statistical Physics given by Dr. A. J. Macfarlane

in Cambridge in the Lent Term 1998. These typeset notes are totally unconnected

with Dr. Macfarlane. The recommended books for this course are discussed in the

bibliography.

Other sets of notes are available for different courses. At the time of typing these

courses were:

Probability Discrete Mathematics

Analysis Further AnalysisMethods Quantum Mechanics

Fluid Dynamics 1 Quadratic Mathematics

Geometry Dynamics of D.E.s

Foundations of QM Electrodynamics

Methods of Math. Phys Fluid Dynamics 2

Waves (etc.) Statistical Physics

General Relativity Dynamical Systems

They may be downloaded from

http://home.arachsys.com/pdm/ or

http://www.cam.ac.uk/CambUniv/Societies/archim/notes.htm

or you can email [email protected] to get a copy of thesets you require.

v


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Chapter 1

Quantum Statistical Mechanics

1.1 Introduction to Quantum Statistical Mechanics

Statistical mechanics deals with macroscopic systems of many particles. Consider an

isolated systemS

of gas in a vessel whose walls neither let heat in or out and is subject

to no mechanical action. No matter how it is prepared, it is an experimental fact that

S reaches a steady state (a state of thermodynamic equilibrium) in which a set of(rather few) thermodynamic variables are constant. includes the pressure P, volumeV, temperature T, total energy Eand entropy S.

Any state ofS (before or after thermodynamic equilibrium is reached) containsparticles with positions and momenta changing in time. We cannot possibly analyse

this in detail even if we knew the forces and initial conditions, and could solve the

resulting system there is no hope that we could organise usefully the vast amount of

data.

At best, we aim to treat the possible states of motion ofS by some averagingor statistical procedure that allows us to predict values of the variables in that areconstant in states of thermodynamic equilibrium ofS.

We use quantum mechanical ideas to approach the subject. Consider

microstatesofS, the stationary quantum states |i. macrostates, which correspond to states of thermodynamic equilibrium.

The latter are not states in the quantum mechanical sense, but do involve the vast

numbers of microstates.

The ergodic hypothesis (which is provable for some systems) is thatS passes intime through all its possible states compatible with its state of thermodynamic equilib-

rium. This allows us to replace time averages for a single system Swith averages at afixed time over a suitable ensemble Eof systems, each identical to Sin its macroscopicproperties. The most probable state ofE reveals which microstates ofScontribute to itsmacroscopic states of thermodynamic equilibrium and yield excellent approximationsto the values of its thermodynamic variables.

LetE,Nbe the total energy and total number of particles ofSrespectively. Onemeets three types of ensemble.

microcanonical: each member ofEhas the same values ofEand N.

1


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2 CHAPTER 1. QUANTUM STATISTICAL MECHANICS

canonical: each member ofE hasNparticles butE is not fixed, although theaverage total energy is time independent.

grand: neitherEnor Nis fixed, although the averages are.

For the latter two, fluctuations about the average values are found to be very small

and all three types of ensemble give the same thermal physics. This course studies

mainly canonical ensembles. We will do systems with identical bosons/fermions via

the grand ensemble for technical simplicity.

The fact that averaging procedures give reliable estimates of values of physical

variables and apparently differentprocedures yield equivalent results is due to the effect

of the very large number of particles involved. For instance, consider the probability

pmof obtainingN/2 + mheads fromNcoin tosses.

pm = 2N

NN2 + m

2

Ne

2m2

N ,

using Stirlings formula1 forN m 0 . IfN = 1023 andm = 1017 then theexponential terme

1011

is effectively zero.

1.2 Canonical ensemble

We study a system SofNparticles in a volume V with Nand V fixed. The microstatesofSare the (complete orthogonal) set|i with energy eigenvaluesEi. We assume thespectrum is discrete but with possible degeneracy.

We associate withS a canonical ensembleE. This is a very large numberA ofdistinguishable replicas ofS,S1, . . . , SA. Suppose that in a possible state ofE thereareai members ofEin the microstate |i. Then

i

ai= A and i

aiEi = AE. (1.1)

The average energy of the members ofEis thus the fixed valueE.Given a set {ai} (aconfigurationofE) there areW(a) = A!

iai!ways of realising

it.

Proof. We can distributea1 systems in the microstate|1overE inAa1

ways. Then

we can distribute thea2 states in |2 inAa1a2

ways (and so on). Thus

W(a) =

A

a1

A a1

a2

= A!

i ai!.

We assign equal a priori probability to each way of realising each possible con-

figuration ofEand thus there is a probability proportional to W(a) of realising theconfiguration {ai}.

1logn! n logn n+ 12log 2n


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1.2. CANONICAL ENSEMBLE 3

We will associate the state of thermodynamic equilibrium ofS(for the fixed valuesN, V, E) with the configuration ofE that is most probable in the presence of theconstraints (1.1).

To compute this supposeA0 such thata i0 (negligible probability attachesto configurations where this fails). For large n, Stirlings formula allows log n!n (log n

1)and so

log W A log A A i

ai(log ai 1) = A log A i

ailog ai.

We seek to maximise this subject to the constraints (1.1). We use two Lagrange

multipliersandand solve

0 =

aj

log W

i

ai i

aiEi

and solog aj+ 1 + + Ej = 0. Thus

aj =e1Ej . (1.2)

We eliminate via A =

i ai = e1Z, defining the canonical partition func-

tion

Z=i

eEi Ej

(Ej)eEj , (1.3)

where(Ej)is the degeneracy of the energy level Ej .The fraction of members ofEin the microstate |i in the macrostate of thermody-

namic equilibrium is

i =aiA

= 1

ZeEi. (1.4)

This is the Boltzmann distribution, and may be thought of as the probability of

finding |i in the state of thermodynamic equilibrium.We define the average X of a physical variable Xtaking the value Xiin the state

|i by

X =i

iXi. (1.5)

For instance

E

= 1

A i aiEi = E(reassuringly).Zis very important. It leads directly from quantum mechanical data to calculation

of the thermodynamic variables for Sin thermodynamic equilibrium. For instance,

E=log Z

. (1.6)


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Here, holdingVfixed corresponds to keeping all theEi fixed.ForA large (as in all cases of interest) the most probable state is overwhelmingly

so. It gives in effect the average over all possible states of the ensemble. The idea

of associating average values with actual physical predictions depends on the possible

variances being negligible.

We can calculate the variance in the energy in a similar way. Note that E=

1Z

Z

and so

E

= 1

Z

2Z

2 +

1

Z2

Z

2= E2 + E2 = (E)2 .

For typical large systems E N, and as E depends smoothly on we expectE

Nas well. Thus

|E|E

N

N =N

12 ,

which is very small for very largeN.

1.3 Temperature

Given two systems Sa and Sb with fixed volumesVa,Vb and numbers of particlesNa,Nb respectively we place them in contact to form a composite systemSab such thatenergy can pass from one to the other and a state of thermodynamic equilibrium is

reached. We make a canonical ensembleEab forSab by distributingA replicas ofSaandAreplicas ofSb independently across theA members ofEab so that each memberofEab has a component of type Sab.

Suppose that Sahas microstates |i of energy Eai, Sbhas microstates | of energyEb and Sab has microstates |i, of energy Eai+ Eb.2

Suppose the|ioccursai times,| occursbi times and|i, occursci times inEab. Then we have

i

ai= A

b = A (1.7)

ai =

ci b =i

cii

ci = A (1.8)

We demand that the average energy or Eab be fixed at

AE=i

aiEai+

bEb (1.9)

AE=i

ci(Eai+ Eb) . (1.10)

Now the configuration {ci} arises in2If the interaction of particles ofSawith those ofSb is negligible, except insofar as necessary to allow a

state of thermal equilibrium to be reached.


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1.4. TOWARDS THERMODYNAMIC VARIABLES 5

W(c) A!ici!

(1.11)

ways and the configuration {ai}{b} arises in

W(a)W(b) A!i ai!

A!b!

(1.12)

ways. We can calculate the most probable distribution of{c i} by maximising(1.11) subject to (1.8) and (1.10). We get

ci =e1ab(Eai+Eb) i =

ciA

.

We can also maximise (1.12) subject to (1.7) and (1.9) to get

ai= e1aEai i= aiA

b = e1bEb =

bA

.

Defining

Zab() =i

e(Eai+Eb) Za() =i

eEai Zb() =

eEb

we see that Zab() = Za()Zb() and i = i. A common value forcharacterises the state of thermal equilibrium ofSab and it is natural to assume that is some function of temperature Tand that the average energy (theenergy) of a system

is a function ofN,V andT.We defineT by= 1

kT wherek is a constant.

As this argument applies to any two systems, k should be a universal constant(Boltzmanns constant) the same for all systems when a universal definition ofT isused. We will define T for a standard system (a thermometer) and for other systems byputting them into a state of thermal equilibrium with our standard system.

Recall the ideal gas law, P V =N kTfor a sample ofNmolecules.3 kis the samebecause equal volumes of all gases at fixed temperature and pressure are observed to

have the same number of molecules. Tis in degrees Kelvin.

1.4 Towards thermodynamic variables

Recall that in quantum mechanics, spinless particles in a cube of side L have energiesEn =

2

2mL2|n|2, where n N3. Note that E V23 and we generalise this to

Ei = Ei(V). Consider Sgasin thermodynamic equilibrium in a container of volume Vand changeVby a small amount by slowly moving one of its walls.

3This applies to all gases at sufficiently low density.


12/43


TheEi will change according toEiEi+ EiV Vand changes in which the ido not change will be examined first.

The change in energy is supplied by the work done on Sby a piston applied tothe right hand wall. For slow motion the system passes through successive positions

of thermal equilibrium and then the applied force just balances the pressure forceP A.Thus the work done to the system is

PAl =

P V. E = W =

P V. Now

E=

i iEi and so

P =i

iEiV

= 1

V log Z. (1.13)

Consider next changes in which the i are allowed to vary. ThenE=P V +i Eii, and note that

i i= 0.

Define the entropy

S= k

Alog Wmax that is,Wat thermal equilibrium

= k

AA log A i ailog ai

= k

A

A log A

i

ailog i log Ai

ai

=ki

ilog i at thermal equilibrium.

(1.14)

We see that

S= k

i

log i+ i

1

ilog i

=

k i(Ei log Z+ 1)=

1

T

i

Eii

and thus

i Eii = T S. We have defined Ssuch that

E= T S P Vor in terms of exact differentials,dE= TdS PdV. (1.15)

We might regard this as arising from the comparison of ensembles with infinitesi-

mally different states of thermal equilibrium. (1.15) states theFirst Law of Thermody-

namics4, written in the form

dE= Q + W, (1.16)

whereQ is the heat supplied to SandWis the work doneon S.4You cant win...


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1.5. TOWARDS APPLICATIONS 7

(1.15) also shows thatT = ES

andP =EV

.

For a gas of spinless particles, E =

i Eii V23 and so P = 23

EV

, giving

P V = 23E. Combined with the result for a perfect gas,E = 32N kT (see later), we

have Boyles lawP V =N kT.We want to be able to calculate Sfrom the partition function. Now, at thermal

equilibrium,

S= ki

ilog i

= ki

i(Ei log Z)

= 1

TE+ k log Z.

Define thefree energyF=E T S, then

F=1

log Z. (1.17)

Also,

S= 1T

log Z

+ k log Z=

T (kTlog Z) =F

T. (1.18)

We have a very tangible definition ofS, as proportional tolog Wmax, whereWmaxis the number of microstates ofScontributing to the state of thermal equilibrium.

Given the partition function (1.3), we can calculate the thermodynamic variables

in a state of thermal equilibrium, P(by (1.13)),E(by (1.6)),F(by (1.17)) andS(by(1.18)).

1.5 Towards applications

1.5.1 Nparticle partition function

Consider a composite system with Hamiltonian H = H1+H2 andH1 and H2 inde-pendent (they commute). LetH1|a= E1a|a andH2|= E2|. Then asH1 andH2 commute,H|a, = Ea|a, , whereEa = E1a + E2. Then the 2 particlepartition function is

Z=a,

eEa =a

eE1a

eE2 =Z1Z2.

LetZ ZNdescribe the sum over states of a gas ofNvery weakly interactingparticles. If these are supposed independent (distinguishable) thenZ = z N, wherez is the one particle partition function. This allows easy calculations, but fails for

systems of identical (indistinguishable) bosons or fermions. For the latter it seems bestto use grand ensemble methods.

Later we will consider a gas of diatomic molecules, H= HT+ Hrot+ Hvib, whereHT describes the centre of mass motion, Hrot describes rotation about the centre ofmass andHvib describes vibration along the axis. We see that the one particle partitionfunction isz = zTzrotzvib.


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1.5.2 Extensive and intensive variables

Extensive variables are proportional to the amount of matter in a system at fixed tem-

perature, whereas intensive variables are independent of the amount of matter.

In general,N,V,Sand Eare extensive and T andPare intensive.

Think initially of a system made up of two independent subsystems. ThenV =

V1+V2. In a state of thermal equilibrium the systems have the same (and hence sameT), and soTis intensive. In a state of equilibrium there is mechanical equilibrium andso the systems have the same pressure, and so Pis intensive.

AsZ= Z1Z2 we see that E= E1+ E2,S= S1+ S2 and so on. A logical ex-tension of this argument gives thatV,Eand Sare proportional toNand so extensive.This can fail if volume energies do not swamp surface energies or if intermolecular

forces are neither weak nor short range.

1.5.3 Density of states

Consider a spinless particle in a box of side L. Instead of using solutions of theSchrodinger equation such that = 0 on the walls we use periodic boundary condi-

tions,(x + Li) = (x)(and so on). Thus we can usek ekx with= 2

k2

2m .

The periodicity gives that Lk2

Z3. There is one such state per unit volume inn-space and in the continuum limit, there ared3nstates with n in the range nn+dn.There are thusd3nstates with k in the range k k +dk and hence L2 3 d3kstateswith k in the range kk + dk.

We see that

i

L

2

3d3k (1.19)

= V

(2)3 4k2 dk=

V

(2)34

2m

2

12 2md

22

= 2V

2m

h

32

12 d (1.20)

=

dg().

We insert a factorgS= (2S+ 1) to cope with massive particles of spinS, and

g() = gS2V 2m

h2

32

12

gives the density of states. We use (1.20) in isotropic contexts, but (1.19) is needed

in kinetic theory.

In two dimensions we consider a square of side LandA= L 2 enters in the role ofV. We find thatg() =

2mh2

gS2Ais independent of.


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1.5. TOWARDS APPLICATIONS 9

For relativistic particles of rest mass m we have =

m2c4 + k2c2 12 and use

(1.19) to get d3k=

4k2 dk=

4

(c)3

mc2

d

2 m2c4 12 . (1.21)We also supply a factor gS = 2S+ 1 for a particle of spin Swith nonzero restmass. For photons, m= 0,gS= 2 and so

i

2 V(2)3

4

c

3 0

2 d=8V

c3

0

2 d. (1.22)

1.5.4 Gas of spinless particles

Consider Nspinless particles in a cube of side length L= V 13 . Then the single particle

partition function

z= i

ei

0

d g()e,

whereg() = DV 12 andD= 4

2mh2

32 . Settingy = we get

z= DV(kT)32

0

dy y12 ey =V

2mkT

h2

32

. (1.23)

Nowz = zN and solog Z= Nlog z. Thus

E=N

32log = 32NkT.This is the classical result: there is an energy 12kTassociated with each degree of

freedom of each particle. Combining withP V = 2

3

Ewe getP V =N kT.

1.5.5 Entropy and the Gibbs paradox

UsingZ= zN, (1.23) and (1.18) we have

S= N k log V + 32N k log

2mkT

h2

+ 32N k,

which is not extensive, asVN. This is Gibbs paradox.If instead we useZ= z

N

N! , (1.23), (1.18) and Stirlings formula we have

S= N k log VN

+ 32N k log

2mkT

h2

+ 32N k

and we see thatS is extensive.Often an additive constant in Sdoes not matter and (1.23)gives all thermodynamics

otherwise correctly. It reflects a normal classical view of how to treat indistinguishable

particles. Taking Z = zN

N! cures this but has no foundation in classical statistical

physics.


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1.6 Harmonic oscillator model

Model a crystal by placing one atom at each point of some regular lattice with N sites.In 1D, simplify by taking a simple harmonic oscillator of frequency r instead of therth atom.

NowZN =rz(r), and usingEn = (n+ 12)for the simple harmonic oscil-lator with frequency we get

z() = e

2

1 e .

Even ifr = for allr, the atoms are distinguishable as there exists one at eachlattice site. This is a one dimensional version of theEinstein solid. We find that

E

N =log z

= 12+

e 1 .

If kTthis gives the classical result EN 12 = k T, with energy 12kT per

degree of freedom per atom.

If

kT then

1 and E

N

12

e, which tends to zero as

T 0.Quantum statistical mechanics knows when to count the full classical 1

2kT value

per degree of freedom: do it only when Tis large on some scale set by the problem. Inthis case the critical temperature Tc=

k

.


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Chapter 2

Thermodynamics

2.1 Introduction

Consider a volumeVof gas with a fixed numberNof particles. The state of thermalequilibrium of this gas is characterised by T andV.

Our work with quantum statistical mechanics has produced some concepts natu-

rally:

E: the total energy of the system, which arose as the average energy and hasnegligible fluctuations.

S: the entropy. This has a clear significance as the number of microstates con-tributing to the macrostate of thermal equilibrium.

equations of state: for instanceP V =N kT. dE= TdS PdV, which is true for any thermodynamic change between two

infinitesimally close states of thermodynamic equilibrium.

All thermodynamic variables for the sample of gas we are talking about can beregarded as a function of two suitably chosen independent variables. In the case of the

energy, Earises asE(S, V).We also found thatdE= Q+W, whereQ is the heat supplied to the system

andWis the work done onthe system.For finite changes from the initial to the final state conservation of energy gives

E= Q+ W. This is the first law of thermodynamics, equivalent to the conser-vation of energy.

Suppose we change from a state withE1(S1, V2) toE2(S2, V2). In general, weexpect

W V2

V1

PdV

because of wasted effort (against friction or in producing convection or turbulence).

This gives us

Q S2S1

TdS.

11


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12 CHAPTER 2. THERMODYNAMICS

These inequalities hold as equalities for reversiblechanges, that is changes which

arequasi-staticand non-dissipative.

Quasi-static changes are changes which are done so slowly as to pass through states

arbitrarily close to a state of thermodynamic equilibrium, so that all thermodynamic

variables are well-defined throughout the change.

Some (obvious) irreversible processes are fast piston movement, free expansion of

a gas into a vacuum and the mixing of samples of different gases.In induced/allowed changes in real life with Q = 0, S > 0 and so the entropy

increases.

2.2 Applications ofdE= TdS PdV2.2.1 Integrability conditions

The internal energy Eis seen asE(S, V)within

dE= TdS PdV =

E

S

V

dS+

E

V

S

ThusT =ES

V andP =

EV

S. dE is an exact differential, so

2

ESV =

2EVS

and

P

S

V

=

T

V

S

This is aMaxwell relation.

The free energy F =E T Sis a natural function ofT andV, asdF = dE TdS SdT= SdT PdV.

ThusS=FT

V

andP =FV

T

. This gives the Maxwell relation

SV

T

= PT

V

.

H=E+ P Vdefines theenthalpyH(S, P)and G = E T S+P Vdefines theGibbs function G(T, P).

These give four Maxwell relations, but they are interdependent.

E, F, H and G are all extensive. If (exceptionally)G = 0 (true for a gas ofphotons) thenG = N and so the chemical potential can vanish for arbitrary finiteN.

We need some rules for shunting partial derivatives around. Consider z = z (x, y).Then

z = z

xy

x +z

yx

y and hence

1 =

x

z

y

z

x

y

and

0 =

x

y

z

z

x

y

+

z

y

x

.


19/43

2.2. APPLICATIONS OFDE= TDS PDV 13

This can be rewritten in a slightly different form:

x

z

y

z

x

y

= 1

xy

z

yx

z

xz

y =1.

(2.1)

When we wrote down E, F, Hand G we wrote them in terms of the mathematicallynatural independent variables in each case. In practice, we would likeEas a functionofT andV. RecallTdS= dE+ PdV, so that

TdS=

E

T

V

dT+

E

V

T

dV + PdV and so

T

S

T

V

=

E

T

V

and

T SV

T

= EV

T

+ P.

Also, we know that 2S

TV =

2SVT

and soE

V

T

=T

P

T

V

P.

For a perfect gas, P V =N kTand soEV

T

= 0 . ThusEis a function ofTonly.Given the laws of thermodynamics we need two inputs from experiment to specify

all thermodynamic variables completely. For instance, for a perfect gas we need the

equation of state andE= CT.

2.2.2 Specific heats

An amount of gas withNA (Avogadros number) of molecules is called a mole. Theideal gas law tells us that at fixed temperature and pressure, a mole ofany ideal gas

occupies the same volume. Written in terms of moles, the ideal gas law for nmoles ofgas is

P V =nNAkT=nRT. (2.2)

Ris called thegas constantand is the same for all ideal gases. In fact, all gases areessentially ideal at sufficiently low density.

For a sample of1mole of ideal gas we define the specific heat(or heat capacity permole) usingQ = TdSunder suitable conditions.

DefineCV =TSTV, thespecific heat at constant volume andCP =T

STP,

thespecific heat at constant pressure. Now

TdS= dE+ PdV = dH VdP

and so


20/43


CV =T

S

T

V

=

E

T

V

CP =

H

T

P

.

(2.3)

For one mole of ideal gas we know that E= C Tand soC=CV. To findCP weneed to useP andTas the independent variables.

Now

dS=

S

T

V

dT+

S

V

T

dV

=

S

T

V

dT+

S

V

T

V

T

P

dT+

V

P

T

dP

.

ThusST

P

=ST

V

+SV

T

VT

P

and so

CP =CV + TS

VTV

TP =CV + TV

TPP

TV . (2.4)This allows the calculation ofCP CVfrom the equation of state: for an ideal gas

CP CV =R. Define the ratio of specific heats= CPCV . Note that= CV+RCV >1for an ideal gas.

Statistical mechanics gives CV = 32

R, 52

R , . . . (for monatomic, diatomic (etc)gases). Thus= 53 ,

75 , . . . .

2.2.3 Adiabatic changes

These are (defined as) changes which are reversible and satisfyQ = 0. We refer tonmoles of ideal gas usingE= nCVT andP V =nRT. Now

0 = RdE+ RPdV

=RnCVdT+ RPdV

=CV (PdV + VdP) + RPdV

=CPPdV + CVVdP.

Thus0 =dVV

+ dPP

and soP V is constant on adiabatics.Note that adiabaticsP V constant are steeper than isothermals P Vconstant.

2.2.4 Entropy ofn moles of ideal gas

We start (as usual) from

TdS= dE+ PdV

=nCVdT+ nRTdV

V .

ThusdS= nCVdT

T + nR

dV

V ,

and soS= nCV log T+nR log V +c1. Thermodynamics cannot determine theconstant c1, and does not care that Sis not explicitly extensive.S= nCV log P V

+c

and so (as expected), Sis constant on adiabatics (isentropics).


21/43

2.2. APPLICATIONS OFDE= TDS PDV 15

2.2.5 van der Waals equation

We get a better agreement with experiment by replacing the perfect gas law with

P+

n2A

V2

(V nB) = nRT, (2.5)

whereA,B and Rare strictly positive constants.

Molecules are not treated as point particles, and as P at constantT,VnB, which is the residual volume of all the molecules.

P+ n2AV2

reduces the real gas pressure by an amount n2AV2

, due to intermolecular

attractive forces. If these are short range then the smaller V is the more importantthese become.

2.2.6 The Joule effect

Consider the apparatus shown, with adiathermal walls and containing n moles of gas atvolumeV1, pressureP1 and temperatureT1. Pull back the partition and allow the gasto expand (irreversibly) into the total volumeV2, and then to reach a state of thermalequilibrium specified by P2,V2 and T2.

AsQ = 0 (adiathermal) and W = 0 (no work is done), dE = 0 and so for aperfect gas

T2= T1 and S2 S1 = nR log V2V1 >0 forV2> V1.For a van der Waals gas we still have dE= 0(which is true in general) and so E

stays constant. Now

T

V

E

=EV

T

ET

V

=P TP

T

V

nCV

=n2A

V21

nCVfor van der Waals gas


22/43


2.3 Some thermodynamics

Suppose a sample Sofnmoles of perfect gas is put in thermal contact with a heat bathBand compressed reversibly from volumeV1 to volumeV2. If the gas is perfect thenE= E(T) = nCVTand soEstays the same. Now

Q + W = 0 Q= W= V2V1

PdV =nRTlog V2V1

.

ThusW > 0 for compression (V2 < V1) andQ < 0. Heat is given out toBby S. NowS= nR log V2

V1and so the entropy ofSdecreases. However, because the

isolated universe ofSand Bhas S= 0the entropy ofBmust increase.

2.3.1 The second law

(Kelvin) No process can continuously (by going round a cycle) extract heat from a heat

bath and perform an equal amount of work.

(Clausius) No process can continuously transfer heat from a colder to a hotter body. 1

Take a sample Sof perfect gas around a closed Carnot cycle by means of adiabaticsand isothermals (T2 > T1). OnABthe heat bath atT2 supplies heatQ2 > 0 to Sattemperature T2. On CD the heat bath at T1supplies heatQ1 < 0to Sat temperatureT1.

E= E(T)is unchanged over one complete cycle so that the work done in a cycleby Sis W= Q1+ Q2 by the first law (asQ = 0on adiabatics).

This agrees with the Kelvin statement of the second law, (Q1) > 0of heat iswasted at the low temperature heat bath. The efficiency is defined

=WQ2

= 1 +Q1Q2

0 andQ1 = nRT1logVDVC

=nRT1 log VBVA .

Thus= T2T1T2

, and clearly0<


23/43

2.4. HEAT FLOW 17

2.4 Heat flow

LetS1 be a sample ofn1 moles of ideal gas at temperature T1 in a fixed volumeV1.ThenE1(T1) = n1CV1T1 = x1T1. Similarly for S2. Put S1 and S2 in thermal contactallowing no change inV1+ V2. Suppose the state of thermal equilibrium is reached ata temperature T. Then (by conservation of energy),

x1(T1 T) + x2(T2 T) = 0

and we can solve for T. As for the entropy, S1,2 = x1,2log TT1,2

and S1+S2 0.


24/43



25/43

Chapter 3

Grand ensemble methods

3.1 The formalism

Our approach is a direct extension of the approach we adopted earlier for the canonical

ensemble. Given a systemS

of fixed volume we construct a grand (canonical) ensem-

bleG with a large numberA of distinguishable replicas ofS in microstates|i ofSwithNi particles and energies Ei. Suppose thatai members ofG are in the microstate|i so that

i

ai = Ai

aiEi = AEi

aiNi = AN.

We have thus fixed the average energy and number of particles of members of the

ensemble. Each configuration{ai} ofG defines a state of the ensembleand we assign toeach configuration equala prioriprobability. We associate the state of thermodynamic

equilibrium ofSwith the most probable configuration ofG in the presence of ourconstraints. We find this by maximising the samelog Was before to get

0 =

ai

log W

i

ai i

aiEi i

aiNi

.

This gives us that

ai = e(1+)e(EiNi),

where we have defined thechemical potential by =. We now define thegrand partition function

Z=

i

e(EiNi). (3.1)

The fraction of members ofG in the microstate |i is

i=aiA

=e(EiNi)

Z .

19


26/43

20 CHAPTER 3. GRAND ENSEMBLE METHODS

The grand ensemble average is

O=i

iOi

and E=Eand N=N(which is, on the whole, a good thing). As before we can

use Z to calculate thermodynamic variables:

N= 1

log Z

,V

, (3.2)

E N=

log Z

,V

. (3.3)

As before, from consideration of changes at constant i we find thatE=P Vand

P=i i

Ei

V =

1

log Z

V

,

. (3.4)

More general changes in which i= 0obeyi

i = 0i

Nii = N E = P V +i

Eii.

Defining the entropy

S= kA

log Wmax =ki

ilog i

gives that

S=ki

ilog ii

i1

ii

=ki

i((Ei Ni) log Z) .

This gives the fundamental thermodynamic relation

T S= E+ P V N, (3.5)

yielding the first law dE=Q+Wmech+ Wchem, whereWchem =Nis the

work done in adding Nparticles to the system S. The chemical potential is thereforegiven by

=

E

N

S,V

. (3.6)


27/43

3.2. SYSTEMS OF NON-INTERACTING IDENTICAL PARTICLES 21

Returning to the entropy Swe see that

S= k(E N) + k log Z (3.7)which can be put in the same form as (1.18), namely

S=

k Tlog ZT

,V

. (3.8)

A similar argument to that used to define temperature in 1.3 gives that two systems

placed in thermal and diffusive contact will reach a thermodynamic equilibrium char-

acterised by common values of= 1kT

and.We keep the idea that at fixed NA, NB energy flows from the system with the

higher temperature to that with the lower temperature until thermodynamic equilibrium

is reached. We add that if the contact allows the diffusion of particles then in the state

of thermodynamic equilibrium with constantthere is (on average) no diffusion.We return to the first law, dE= TdSPdV+ dN. We can view Eas a function

ofS,V andN, and when (as for many large systems), E,SandVare extensive weget

E(S,V,N) =E(S,V,N)

and so

E= S

E

S

V,N

+ V

E

V

S,N

+ N

E

N

S,V

=T S P V + N.We define thegrand potentialfor a state of thermodynamic equilibrium by

= E T S N=E N (E N+ kTlog Z)=

kTlog

Z.

Thus Z=e. Nowalso equals P Vand soP V =kTlog Z (3.9)

allows the calculation of the equation of state from the grand partition function.

The state of thermodynamic equilibrium corresponds to the most probable state

inG. Some thermodynamic variables arise from averages over microstates weightedby the probability of finding them in the state of thermodynamic equilibrium. The

averages are effectively averages over all possible states of the ensemble because the

macrostate of thermodynamic equilibrium dominates overwhelmingly. They are also

very sharp (for the same reason).

3.2 Systems of non-interacting identical particles

3.2.1 A little quantum mechanics

We will treat such systems with only one type of particle. When interactions are ne-

glected then the wavefunction of each stationary state of the system (x 1,x2, . . . )is


28/43


obtained by either symmetrization (for bosons) or antisymmetrization (fermions) of the

product 1(x1)2(x2) . . . of one particle wavefunctions.Consider spin0 bosons in a cube of sideL. Then the one particle wavefunctions

are

n(x) = 2L

32

sin n1xL

sin n2xL

sin n3zL

with corresponding energies

En = 2

2mL2|n|2 .

For identical bosons we cannot use wavefunctions 1(x1)1(x2)2(x3)which cor-respond to particles1and2in state1n1and particle3in state2 n2, but we mustuse instead the symmetrized wavefunction

=

1

3!(1(x1)1(x2)2(x3) + 1(x1)2(x2)1(x3) + 2(x1)1(x2)1(x3)) .

All we can say about this is that there are two particles in state |1 and one particlein state |2. is fully determined by the fact that1 is used twice and2 used once.

For spin 12 fermions we have

(x)n(x)orn(x),

whereis spin up andis spin down withHandEnindependent ofand. We can use determinants to build the antisymmetric wavefunctions.

For instance, given1(x, ) and 2(x, ) (where1 n1,2 n2 and = or) we get the antisymmetrized version

(x1, 2,x2, 2) =

1

2!

1(x1, 1) 1(x2, 2)2(x1, 1) 2(x2, 2)

.

Similarly for three particles,

=

1

3!

1(x1, 1) 1(x2, 2) 1(x3, 3)2(x1, 1) 2(x2, 2) 2(x3, 3)3(x1, 1) 3(x2, 2) 3(x3, 3)

.reflects thePauli exclusion principlewhich forbids any two of the i or any two

sets(x, )from being the same.Eachis determined fully by the number of times each one particle wavefunction

is used in the product term that we (anti)symmetrize.

3.2.2 The partition functions

Let the one particle wavefunctions of the particles of Sbe r(x)with energyr. Sup-pose that in the microstate |i,nr of the particles have wavefunction r(x).

Then

Ni=r

nr and Ei =r

rnr.


29/43

3.2. SYSTEMS OF NON-INTERACTING IDENTICAL PARTICLES 23

We see that the microstate |i ofSis fully determined byi {nr}. We obtain thefull set of microstates ofSby letting thenr vary without restriction over their allowedrange of values. (ThusNi andEi cannot themselves be restricted this is why weuse the grand ensemble method.)

We can now write down the grand partition function

Z=i

e(EiNi)

=

n1,n2,...

e(n11+n22+n1n2... )

=r

nr

enr(r).

For fermions we have thatnr can only be zero or one, so that

Z=r

1 + e(r)

. (3.10)

We use this to find

N= 1

log Z

,V

=r

1

e(r) + 1.

Now

nr =

n1,n2,...

nre(r)

Z

= 1

log Z

r

,,s

s=r

= 1

e(r) + 1 .

We see that N =

rnr . nr is the average occupation numberof the rth one

particle state at thermal equilibrium. This is the Fermi-Diracdistribution.

For bosons,0 nr < and

Z=r

1

1 e(r) .

This gives that

nr = 1

e(r)

1

. (3.11)

This is the Bose-Einstein distribution.

If there aregr one particle states of energyr we can write the average number ofparticles with energy r as

n(r) = gr

e(r) 1 .


30/43


forNlarge we pass to the continuum limit,

r

dg(),

whereg() is the density of states factor, DV 12 as in section 1.5.3, where D =

22mh2

32 gSandgS= 2S+ 1 is the spin degeneracy.

The average number of particles with energy in the range + dis

n()d= g()d

e() 1 .

Thus in the continuum limit, the grand partition function Zis given by

log Z=0

g() log(1 e()) d.

We can now use the results of section 3.1 to find things like NandE,

N=0

g() de() 1 E=

0

g() de() 1 .

In three dimensions,g() = DV 12 and so (integrating by parts) we see that

log Z= DV0

2

3

32

e() 1=

2

3.

Combining this with (3.9) we find that P V = 23E. The 23 comes from the

12 factor

ing().

3.2.3 Classical limit

For a volume V ofNparticles with energy E, we useg() = DV 12 and so

N=DV

0

12 d

e() 1 and E= DV0

32d

e() 1 .

Puttingz = we findN = DV(kT)32 I1

2()and E =DV(kT) 52 I3

2(),

with

In(y) =

0

zn dz

ez+y

1

.

Ifey 1 we can neglect the1 in the denominator and find thatn() eg().We can approximate the integrals (expand the integrand) to get

E 32N kT

1 e

4

2+ . . .

.


31/43

3.3. BLACK BODY RADIATION 25

In the lowest approximation E= 32N kT andN=DV(kT)32

2 .

The condition e 1 is thus

2mkT

h2

32 V

N1 ifgS= 1.

Thisis a classical limit it holds when h is very small on a scale defined by mand the mean energy per particle 3

2kT. This condition is satisfied at low density V

N and

at high temperature.

It is true for all real gases except helium at very low temperature and very high

density. Most real gases liquefy before quantum mechanical effects set in.

It fails to holds for electrons in solids due to the fact that their effective mass inside

the solid is much less than the free electron mass.

We can use our formulae for EandNto get the lowest order quantum correctionto the equation of state recall thatP V = 23E, so

P V =N kT N P(kT)

32

1

D

8.

3.3 Black body radiation

Consider a cubic cavity of side L = V 13 inside a perfectly black body, the walls of

which are maintained at a fixed temperatureT. The atoms of the perfectly black bodyabsorb all photons incident on them and independently emit photons such that even

in thermodynamic equilibrium the number of photons varies significantly. Thus the

total number of photons is not conserved and so no constraint can be applied to them.

Therefore = 0.The number of photons depends on the temperature. We must consider a gas of

photons in a volumeVat a temperatureT. These quanta of the electromagnetic fieldare relativistic bosons of rest mass 0.

Our previous work applies, setting = 0. Thus

nr = 1

er 1 and E=r

rer 1 .

We pass to the continuum limit using (1.22) to get

N= 8V

c3

0

2 d

eh 1 and E= 8V

c3

0

h3 d

eh 1 ,

where = h. This is Plancks law. We expect classical behaviour at highT orlow, wheree

hkT 1 h

kT and

dE=

8V

c3

2

dkT =kT g() dV,

which agrees with the classical result of energykTper normal mode of radiation.This is the classical equipartition of energy. Using this result, dEd 2 gives a diver-gent energy, which is called the ultraviolet catastrophe.

Using the full formula for the energy we find that


32/43


e=E

V =

8

(hc)3(kT)4

0

z3 dz

ez 1=T4.

This is Stefans law, and = 85k415h3c3 isStefans constant. We see that the energydensity is a function ofTonly.

Using (3.7) with = 0we see that

S= k log Z+ ET

and we can evaluate k log Z = 4E3T. The entropy density SV = 43T3 0 asT 0. The energy

E= V T4 =

3S

4

43 1

(V )13

V 13 at constant entropy.

The pressure Pcan now be calculated from P = EVS = 13 EV = 13T4.Finally the density of photons, n= N

VT3 (see example sheet 3).

3.4 Degenerate Fermi gas

This is the extreme quantum limit. It occurs for

electrons in solids, electrons in white dwarf stars,

neutrons in neutron stars and

nucleons in nuclear matter.

ForNelectrons in a volumeVat a temperatureTwe have

N=

0

g()F() d and E=

0

g()F() d,

whereg() = DV

,D = 42mh3

32 and

F() = 1

e() + 1.

The range0e


33/43

3.4. DEGENERATE FERMI GAS 27

Now

limT0

e() =

0 < EF

> EF,

so that

limT0

F() =

1 < EF

0 > EF.

Thus atT = 0 F() = ( EF). AtT = 0it is energetically most favourable forelectrons to fill up the one particle energy eigenstates (two electrons at a time, one spin

up and one spin down) with increasing energy according to the Pauli principle, until

the Fermi energy is reached.EFis the highest energy occupied at T = 0.We can now perform the integrals forNandEto get

N=DV 23E32

F (3.12)

E= DV 23E52F =

35N EF. (3.13)

We can solve (3.12) forEF to get

EF =

3N

2DV

23

V 23 .

The equation of stateP V = 23EbecomesP V 53 = const. Classical kinetic theory

gives P 0 as T 0. The Pauli principle requires particles to have non-zeromomentum and so there is pressure even atT= 0.

In fact

P=2

3 32

3N

V 53 h2

2m m1

and so atT = 0the pressure is much bigger for lighter particles.

3.4.1 Heat capacity at low temperature

We have

N=

0

g()F() dand

E=

0

g()F() d.

We want to findC=ET

V

(which is proportional to the usual CV). We take T

of both of the above to find

C=

0

g()F

T( EF) d.


34/43


At low temperature FT

is very like a delta function and so the region EFdominates the integral.

We approximate g() g(EF)andEF. This gives us

C

g(EF)

kT2

0

d(

EF)

e(EF)e(EF) + 1

2 ( EF)2

=g(EF)k2T

EF

z2ez dz

(ez + 1)2 .

We know that EFis finite and positive, so that EF asT 0. We thereforeapproximate the lower limit of the integral by .

C g(EF)k2T

z2ez dz

(ez + 1)2 .

This leaves a convergent integral whose value is 2

3 , so that

C 22

N k TTF

,

whereTF (the Fermi temperature) is defined byEF = kTF. This approximationis expected to be good for T TF.

3.5 Bose-Einstein condensation

Recall the result

N=DV

0

12 d

e() 1=DV(kT)32

0

z12dz

ez 1=DV(kT)32

2 f(),

where

f() =

n=1

en

n32

.

f() is convergent iff 0. Assuming this, the Bose-Einstein denominatordoes not vanish for any point in the range of integration.

It is easy to see that fhas a maximum value (of2.612) at = 0 and decreasesmonotonically as decreases through negative values from = 0.

Now suppose that NV

is fixed. Our equation for N is OK as T decreases sincef()can increase by getting less negative. When reaches0 problems occurasfcannot increase any more. This will be at T =TB , given by

NV

= D(kTB)32 D= Df(0)

2 .

Our equation forNappears to fail beyond this. Why?More care is called for in the passage to the continuum limit. The

12 factor ing()

assigns0 weight to the particles in the = 0 state. This would be no problem for


35/43

3.6. WHITE DWARF STARS 29

fermions (which have the Pauli principle), but no law stops bosons from condensing

into the = 0state if need be. We write N = N0+NC, whereN0 is the number ofparticles in the = 0state andNCis the number of particles in >0 states. Then forT > TB

N

V

=D

2 f(

).

ForT =TB, NV

= D(kTB)32 , definingTB . ForT < TB then

NCV

= D(kT)32 , giving

NCN

=

T

TB

32

.

3.6 White dwarf stars

White dwarfs are abnormally faint: they are stars in which the hydrogen supply has run

out so that they are composed mainly of helium.

Typically,

T Tsun= 107KM Msun, and

107sun.

We regard them as being a mass of helium at T 10 7Kand under extreme com-pression. ForT 107K, k T 103eV is much greater than the binding energy ofelectrons to helium, so that thermal collisions completely ionize all the atoms, produc-

ing an electron gas.

To a good approximation this may be viewed as a Fermi gas in the degenerate

(negligible T) limit. This may be seen by computing TF, the natural temperature scaleof the problem (the Fermi temperature). Now

kTF =EF = h2

2m

3N

8V

23

2 106eV.

Since NV

is so largeTF 1011KT.The helium nuclei neutralise the charge of the star and produce the gravitational

attraction which counteracts the extreme zero-point pressure of the electron gas (which

dominates the zero-point pressure from the nuclei).

We can calculate the total energy E= Eelec(R) + Egrav(R)and find the radiusR0of the star by minimising this.

Treating the electrons relativistically we have

N

V =

8

3c3 EFmc2 d(

2

m2

c

4

)

12

.

Let

2 m2c4 =mc2x, so that the integral for NV

can be done to get

N

V =

8

3c3(mc2)3

x3F3

.


36/43


The integral for the energy density is

E

V =

8

3c3(mc2)4

xF0

x2 dx

1 + x2

and so E

N = 34mc

2 xF+

1xF + O(x

2F )

. We know thatV R3

and soxFR1. Hence

E= a

R+ bR

R,

wherea,b andare positive constants. This only has a minimum if a, whichleads to the Chandrasekhar upper limit on the mass of a white dwarf.


37/43

Chapter 4

Classical statistical mechanics

4.1 Introduction

Recall the resultZ = zN for a system of distinguishable non-interacting particles. Itcan be shown that in the classical limit

z=r

er =r

r|e H|r

gives

z = h3

d3p d3q eH(p,q), (4.1)

where H(p,q)is the quantum mechanical Hamiltonian and H(p, q)is the classicalHamiltonian as a function of classical variablespandq.

The average for the system of a physical variable f(p,q)is

f(p,q)= d3p d3q f(p,q)eH(p,q)

d3p d3q eH(p,q) .

We can see that H(p,q)=

log z

V

agrees with the result

E=

log Z

V

= N

log z

V

.

Now

Z=

Ni=1

d3p d3q

h3

e

Ni=1H(pi,qi),

whereN

i=1 H(pi, qi)is the Hamiltonian of theN-particle system.

For a monatomic gas withH= p2

2m we have

z = V

h3

dp e

p2

2m

3=

V

h3

2m

32

.

This result shows thatE= 32N kT, which is the common classical result of energy12

kTper degree of freedom per particle.

31


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32 CHAPTER 4. CLASSICAL STATISTICAL MECHANICS

4.2 Diatomic gases

We will study therigid dumbbell model as shown.

We want to write down the partition functionZ=z N. To apply (4.1) we need theHamiltonianH. To find the Hamiltonian we first write down the Lagrangian

L = 12mx2 + 12I

2 + sin2 2

.

(V0 in this rigid case.) Note that the Lagrangian splits neatly into translationalmotion ofGand rotational motion about the centre of mass.

We define the generalised momentap= Lq

:

pi = mxi, p = Iand p = Isin

2

.

Now

H=

qp L = p2

2m+

p22I

+p2

2Isin2 ,

and

z= 1

h5

d2p d3qdpd dpd e

H =ztzr,

where

zt = 1h3

d3p d3q e

12mp

2=V

2mh2

32

and

zr = 1

h2

dpd dpd e

2I

p2+p2

sin2

= 82IkT

h2 .

We evaluate the above integral by first doing the p , p and integrals and thendoing the integral.

Thusz = ztzr = V2mkT

h2

32 82IkT

h2 . This givesE= 52N kT.

1

4.3 Paramagnetism

Each molecule of an Nmolecule solid acts as a little magnet fixed at its own latticesite and free only to rotate about it. Each molecule has a dipole moment m and gives

1Which ought to be expected!


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4.4. SPECIFIC HEATS 33

a contribution m B to the energy when in an applied magnetic field B= (0, 0, B).Now

H= p2

2m+

p22I

+p2

2Isin2 mB cos

and so

z=2I

h22

2 sinh y

y

wherey = mB. We can calculate the magnetisation of the solid:

M= (0, 0, M) = NM.Only the third component m cos is non-zero:

m cos = 1

log z

B

=m

coth y y1 .For smally (highT) we thus have Curies law, M= 13N my=

Nm2B3kT .

4.4 Specific heats

The resultE = 52N kTfor a diatomic gas is found to be accurate at sufficiently hightemperatures.Eis found to be 32N kTat lower temperatures. It is as if the rotationaldegrees of freedom arefrozen out. This is explained by quantum mechanics.

We have Z = zN and z = zrzt. We will look atzr, using Hr = 12IL

2 (the

quantum mechanical angular momentum operator). Now

Hr|l m = l(l+ 1)2

2I |l m

form= l, l + 1, . . . , l 1, lfor each ofl= 0, 1, 2, . . . . Thus

zr =

l=0

(2l+ 1)el(l+1)Tr

T (4.2)

withTr =

2

2Ik .Tr is typically about50Kand is experimentally accessible.ForTr T(most gases at normal temperatures) we can turn this sum into an

integral to get

zr = T

Tr=

82

h2

kT

as before. ForTr Tall of the terms withl= 0in (4.2) are exponentially smalland we take only thel = 0term to get zr 1 there is no rotational contribution tothe energy (or heat capacity). The contribution to the energy from rotational motion is

thereforeN kT ifT

Tr or 0ifT

Tr.

For high temperatures,Erises to 72N kTdue to vibrational motion along the axisof the dumbbell. There is an extra term in the Hamiltonian

Hv = 1

mP2 +

m2Q2

4

and an extra factorzv in the one particle partition function.


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4.5 Weak interparticle forces

Consider a classical gas ofNmolecules with Hamiltonian

H=N

r=1p2r2m

+ r


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4.6. THE MAXWELL DISTRIBUTION 35

Consider theshown.

Then the contribution tof(T)from region 1 where is infinite is d0

4y2 dy=2b.

For large Tand weak attractive forces the contribution from region 2 isd

4y2 dy ((y)) = 2a.

aandbare positive constants. Nowf(T) =

2(b

a)and (4.3) becomes

P+N2a

V2

V =N kT

1 +

N b

V

.

For NbV

small we can (approximately) take1 + NbV

to the left hand side of the

equation to get P+

N2a

V2

(V N b) = N kT

and the correspondence with (2.5) is completed by setting A= N2aandB = N b.Note thatB is the volume of all the molecules.

4.6 The Maxwell distribution

Consider a gas with (one particle) Hamiltonian p2

2m .

The number of molecules in the region pp + dp is

n(p) d3p= N eH

eHd3p=Ce

p2

2m .

Now

N=

n(p) d3p= m3C

4v2 dve

12mv

2

=N

f(v) dv,

which defines theMaxwell distribution of speeds,

f(v) = const v2e12mv2.

f(v)is the probability of finding a particle with speed in vv + dv. The constantis (of course) chosen to make 0

f(v) dv= 1. We can define averages in the obviousway, that is

g(v)=0

f(v)g(v) dv.


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43/43

Approximations

Stirlings formula

We derive the approximation for Stirlings formula from thefunction using Laplacesmethod.

Recall that

(z) =0 t

z

1

et

dt

and(n + 1) =n!. Then

n! =

0

tnet dt=0

en log tt dt.

Letv = tn

, so that

n! =nn+10

en(log vv) dv.

Now for large n,en(log vv) en

1(v1)2

2

, so

n! nn+1en

enu2

2 du=

2nennn.

This is Stirlings formula.

statistical physics (macfarlane a.)

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