SECOND EDITIONSTATISTICALMECHANICSKERSON HUANGMassachusetts Institute of TechnologyJOHN WILEY & SONSNew York Chichester Brisbane Toronto SingaporeCopyright 1963,1987, by John Wiley & Sons, Inc.All rights reserved. Published simultaneously in Canada.Reproduction or translation of any part ofthis work beyond that permitte9 by Sections107 and 108 of the 1976 United States CopyrightAct without the permission of the copyrightowner is unlawful. Requests for permissionor further information should be addressed tothe Permissions Department, John Wiley & Sons.Library 0/ Congress Cataloging-in-Publication Data:Huang, Kerson, 1928-Statistical mechanics.Bibliography: p.Includes indexes.1. Statistical mechanics. I. Title.QCI74.8.H83 1987 530.1'3 86-32466ISBN 0-471-81518-7Printed and bound in the United States of Americaby Braun-Brumfield, Inc.2019 18 17 16 15 14 13 12 11To RosemaryFROM THE PREFACE TO THE FIRST EDITIONThis book is an outgrowth of a year course in statistical mechanics that I havebeen giving at the Massachusetts Institute of Technology. It is directed mainly tograduate students in physics.In this book the starting point of statistical mechanics is taken to be certainphenomenological postulates, whose relation to quantum mechanics I try to stateas clearly as I can, and whose physical consequences I try to derive as Simply anddirectly as I can.Before the subject of statistical mechanics proper is presented, a brief butself-contained discussion of thermodynamics and the classical kinetic theory ofgases is given. The order of this development is imperative, from a pedagogicalpoint of view, for two reasons. First, thermodynamics has successfully describeda large part of macroscopic experience, which is the concern of statisticalmechanics. It has done so not on the basis of molecular dynamics but on thebasis of a few simple and intuitive postulates stated in everyday terms. If we firstfamiliarize ourselves with thermodynamics, the task of statistical mechanicsreduces to the explanation of thermodynamics. Second, the classical kinetictheory of gases is the only known special case in which thermodynamics can bederived nearly from first principles, i.e., molecular dynamics. A study of thisspecial case will help us understand why statistical mechanics works.A large part of this book is devoted to selected applications of statisticalmechanics. The selection is guided by the interest of the topic to physicists, itsvalue as an illustration of calculating techniques, and my personal taste.To read the first half of the book the reader needs a good knowledge ofclassical mechanics and some intuitive feeling for thermodynamics and kinetictheory. To read the second half of the book he needs to have a workingknowledge of quantum mechanics. The mathematical knowledge required of thereader does not exceed what he should have acquired in his study of classicalmechanics and quantum mechanics.KERSON HUANGCambridge, MassachusettsFebruary 1963viiPREFACEThis book is intended for use as a textbook for a one-year graduate course instatistical mechanics, and as a reference book for workers in the field.Significant changes and additions have been made at various places in thetext and in the problems to make the book more instructive. The generalapproach remains the same as that stated in the preface to the first edition.The most substantive change is the addition of the last three chapters on theLandau-Wilson approach to critical phenomena. I hope that these chapters willprovide the uninitiated reader with an introduction to this fascinating andimportant subject, which has developed since the first edition of this book.To make room for the additions, I have omitted or abridged some of theoriginal material, notably the sections on many-body theory and "rigorous"statistical mechanics, which by now have become separate fields.All the material in this book probably cannot be covered in a one-yearcourse (even if that were deemed desirable). It might be helpful, therefore, to listthe chapters that form the "hard core" of the book. They are the following:Chapters 3, 4, 6, 7, 8, 11, 12 (excepting 3.5, 7.6, 7.7, 7.8).KERSON HUANGMarblehead, MassachusettsixCONTENTSPART A THERMODYNAMICS AND KINETIC THEORYCHAPTER 1 THE LAWS OF THERMODYNAMICS 31.1 Preliminaries 31.2 The First Law of Thermodynamics 71.3 The Second Law of Thermodynamics 91.4 Entropy 141.5 Some Immediate Consequences of the Second Law 191.6 Thermodynamic Potentials 221.7 The Third Law of Thermodynamics 25CHAPTER 2 SOME APPLICATIONS OF THERMODYNAMICS 312.1 Thermodynamic Description of Phase Transitions 312.2 Surface Effects in Condensation 352.3 Van der Waals Equation of State 382.4 Osmotic Pressure 432.5 The Limit of Thermodynamics 48CHAPTER 3 THE PROBLEM OF KINETIC THEORY 523.1 Formulation of the Problem 523.2 Binary Collisions 563.3 The Boltzmann Transport Equation 603.4 The Gibbsian Ensemble 623.5 The BBGKY Hierarchy 65CHAPTER 4 THE EOUILIBRIUM STATE OF A DILUTE GAS 734.1 Boltzmann's HTheorem 734.2 The Maxwell-Boltzmann Distribution 754.3 The Method of the Most Probable Distribution 794.4 Analysis of the H Theorem 854.5 The Poincare Cycle 90CHAPTER 5 TRANSPORT PHENOMENA 935.1 The Mean Free Path 935.2 Effusion 955.3 The Conservation Laws 965.4 The Zero-Order Approximation 1005.5 The First-Order Approximation 104vixii CONTENTS5.6 Viscosity 1085.7 Viscous Hydrodynamics 1115.8 The Navier-Stokes Equation 1135.9 Examples in Hydrodynamics 117PART B STATISTICAL MECHANICS 125CHAPTER 6 CLASSICAL STATISTICAL MECHANICS 1276.1 The Postulate of Classical Statistical Mechanics 1276.2 Microcanonical Ensemble 1306.3 Derivation of Thermodynamics 1356.4 Equipartition Theorem 1366.5 Classical Ideal Gas 1386.6 Gibbs Paradox 140CHAPTER 7 CANONICAL ENSEMBLE ANDGRAND CANONICAL ENSEMBLE 1437.1 Canonical Ensemble 1437.2 Energy Fluctuations in the Canonical Ensemble 1457.3 Grand Canonical Ensemble 1497.4 Density Fluctuations in the Grand Canonical Ensemble 1527.5 The Chemical Potential 1547.6 Equivalence of the Canonical Ensemble and the GrandCanonical Ensemble 1577.7 Behavior of W(N) 1617.8 The Meaning of the Maxwell Construction 163CHAPTER 8 QUANTUM STATISTICAL MECHANICS 1718.1 The Postulates of Quantum Statistical Mechanics 1718.2 Density Matrix 1748.3 Ensembles in Quantum Statistical, Mechanics 1768.4 The Third Law of Thermodynamics 1788.5 The Ideal Gases: Microcanonical Ensemble 1798.6 The Ideal Gases: Grand Canonical Ensemble 1858.7 Foundations of Statistical Mechanics 189CHAPTER 9 GENERAL PROPERTIES OF THE PARTITION FUNCTION 1939.1 The Darwin-Fowler Method 1939.2 Classical Limit of the Partition Function 1999.3 Singularities and Phase Transitions 2069.4 The Lee-Yang Circle Theorem 210CONTENTSCHAPTER 10 APPROXIMATE METHODS 21310.1 Classical Cluster Expansion 21310.2 Quantum Cluster Expansion 22010.3 The Second Vi rial Coefficient 22410.4 Variational Principles 22810.5 Imperfect Gases at Low Temperatures 230CHAPTER 11 FERMI SYSTEMS 24111.1 The Equation of State of an Ideal Fermi Gas 24111.2 The Theory of White Dwarf Stars 24711.3 Landau Diamagnetism 25311.4 The De Haas-Van Alphen Effect 26011.5 The Quantized Hall Effect 26111.6 Pauli Paramagnetism 26711.7 Magnetic Properties of an Imperfect Gas 272CHAPTER 12 BOSE SYSTEMS 27812.1 Photons 27812.2 Phonqns in Solids 28312.3 Bose-Einstein Condensation 28612.4 An Imperfect Bose Gas 29412.5 The Superfluid Order Parameter 298PART c SPECIAL TOPICS IN STATISTICAL MECHANICS 305CHAPTER 13 SUPERFLUIDS 30713.1 Liquid Helium 30713.2 Tisza's Two-Fluid Model 31113.3 The Bose-Einstein Condensate 31313.4 Landau's Theory 31513.5 Superfluid Velocity 31713.6 Superfluid Flow 32113.7 The Phonon Wave Function 32513.8 Dilute Bose Gas 329CHAPTER 14 THE ISING MODEL 34114.1 Definition of the Ising Model 34114.2 Equivalence of the Ising Model to Other Models 34414.3 Spontaneous Magnetization 34814.4 The Bragg-Williams Approximation 35214.5 The Bethe-Peierls Approximation 35714.6 The One-Dimensional Ising Model 361xiiixlv CONTENTSCHAPTER 15 THE ONSAGER SOLUTION 36815.1 Formulation of the Two-Dimensional Ising Model 36815.2 Mathematical Digression 37415.3 The Solution 378CHAPTER 16 CRITICAL PHENOMENA 39216.1 The Order Parameter 39216.2 The Correlation Function and the Fluctuation-Dissipation Theorem 39416.3 Critical Exponents 39616.4 The Scaling Hypothesis 39916.5 Scale Invariance 40316.6 Goldstone Excitations 40616.7 The Importance of Dimensionality 407CHAPTER 17 THE LANDAU APPROACH 41617.1 The Landau Free Energy 41617.2 Mathematical Digression 41817.3 Derivation in Simple Models 4201 7.4 Mean-Field Theory 42217.5 The Van der Waals Equation of State 42617.6 The Tricritical Point 42817.7 The Gaussian Model 43417.8 The Ginzburg Criterion 43717.9 Anomalous Dimensions 438CHAPTER 18 RENORMALIZATION GROUP 44118.1 Block Spins 44118.2 The One-Dimensional Ising Model 44318.3 Renormalization-Group Transformation 44618.4 Fixed Points and Scaling Fields 44918.5 Momentum-Space Formulation 45218.6 The Gaussian Model 45518.7 The Landau-Wilson Model 458APPENDIX N-BODY SYSTEM OF IDENTICAL PARTICLES 468A.1 The Two Kinds of Statistics 468A.2 N-Body Wave Functions 470A.3 Method of Quantized Fields 477A.4 Longitudinal Sum Rules 484INDEX 487THERMODYNAMICSANDKINETIC THEORYRTHE LAWS OFTHERMODYNAMICS1.1 PRELIMINARIESThermodynamics is a phenomenological theory of matter. As such, it draws itsconcepts directly from experiments. The following is a list of some workingconcepts which the physicist, through experience, has found it convenient tointroduce. We shall be extremely brief, as the reader is assumed to be familiarwith these concepts.( a) A thermodynamic system is any macroscopic system.(b) Thermodynamic parameters are measurable macroscopic quantities as-sociated with the system, such as the pressure P, the volume V, thetemperature T, and the magnetic field H. They are defined experimen-tally.( c) A thermodynamic state is specified by a set of values of all thethermodynamic parameters necessary for the description of the system.( d) Thermodynamic equilibrium prevails when the thermodynamic state ofthe system does not change with time.( e) The equation of state is a functional relationship among the thermody-namic parameters for a system in equilibrium. If P, V, and T are thethermodynamic parameters of the system, the equation of state takesthe formf(P,v,T)=Owhich reduces the number of independent variables of the system fromthree to two. The function f is assumed to be given as part of thespecification of the system. It is customary to represent the state ofsuch a system by a point in the three-dimensional P-V-T space. Theequation of state then defines a surface in this space, as shown in Fig.1.1. Any point lying on this surface represents a state in equilibrium. In34TTHERMODYNAMICS AND KINETIC THEORYAn equilibriumstateSurface ofequationof statepr - - - - - = - - ~ vFig. 1.1 Geometrical representation of theequation of state.(f)(g)(h)thermodynamics a state automatically means a state in equilibriumunless otherwise specified.A thermodynamic transformation is a change of state. If the initial stateis an equilibrium state, the transformation can be brought about onlyby changes in the external condition of the system. The transformationis quasi-static if the external condition changes so slowly that at anymoment the system is approximately in equilibrium. It is reversible ifthe transformation retraces its history in time when the externalcondition retraces its history in time. A reversible transformation isquasi-static, but the converge is not necessarily true. For example, agas that freely expands into successive infinitesimal volume elementsundergoes a quasi-static transformation but not a reversible one.The P- V diagram of a system is the projection of the surface of theequation of state onto the P- V plane. Every point on the P- V diagramtherefore represents an equilibrium state. A reversible transformationis a continuous path on the P- V diagram. Reversible transformationsof specific types give rise to paths with specific names, such asisotherms, adiabatics, etc. A transformation that is not reversiblecannot be so represented.The concept of work is taken over from mechanics. For example, for asystem whose parameters are P, V, and T, the work dW done by asystem in an infinitesimal transformation in which the volume in-creases by dV is given bydW=PdV(i) Heat is what is absorbed by a homogeneous system if its temperatureincreases while no work is done. If !1Q is a small amount of the heatabsorbed, and !1T is the small change in temperature accompanyingthe absorption of heat, the heat capacity C is defined by!1Q = C!1TThe heat capacity depends on the detailed nature of the system and isgiven as a part of the specification of the system. It is an experimentalfact that, for the same !1T, !1Q is different for different ways of heatingTHE LAWS OF THERMODYNAMICS 5up the system. Correspondingly, the heat capacity depends on themanner of heating. Commonly considered heat capacities are Cv andCp , which respectively correspond to heating at constants V and P.Heat capacities per unit mass or per mole of a substance are calledspecific heats.(j) A heat reservoir, or simply reservoir, is a system so large that the gainor loss of any finite amount of heat does not change its temperature.(k) A system is thermally isolated if no heat exchange can take placebetween it and the external world. Thermal isolation may be achievedby surrounding a system with an adiabatic wall. Any transformationthe system can undergo in thermal isolation is said to take placeadiabatically.(I) A thermodynamic quantity is said to be extensive if it is proportionalto the amount of substance in the system under consideration and issaid to be intensive if it is independent of the amount of substance inthe system under consideration. It is an important empirical fact thatto a good approximation thermodynamic quantities are either exten-sive or intensive.(m) The ideal gas is an important idealized thermodynamic system. Experi-mentally all gases behave in a universal way when they are sufficientlydilute. The ideal gas is an idealization of this limiting behavior. Theparameters for an ideal gas are pressure P, volume V, temperature T,and number of molecules N. The equation of state is given by Boyle'slaw:PV- = constantN (for constant temperature)The value of this constant depends on the experimental scale oftemperature used.(n) The equation of state of an ideal gas in fact defines a temperaturescale, the ideal-gas temperature T:PV=NkTwherek =1.38 x10-16 erg/degwhich is called Boltzmann's constant. Its value is determined by theconventional choice of temperature intervals, namely, the Centigradedegree. This scale has a universal character because the ideal gas has auniversal character. The origin T = 0 is here arbitrarily chosen. Laterwe see that it actually has an absolute meaning according to the secondlaw of thermodynamics.To construct the ideal-gas temperature scale we may proceed as follows.Measure PV/Nk of an ideal gas at the temperature at which water boils and at6 THERMODYNAMICS AND KINETIC THEORYoPV/NkFreezingpoint ofwater100 divisionsBoilingpoint ofwater Fig. 1.2 The ideal-gas temperature scale.which water freezes. Plot these two points and draw a straight line through them,as shown in Fig. 1.2. The intercept of this line with the abscissa is chosen to bethe origin of the scale. The intervals of the temperature scale are so chosen thatthere are 100 equal divisions between the boiling and the freezing points of water.The resulting scale is the Kelvin scale (K). To use the scale, bring anything whosetemperature is to be measured into thermal contact with an ideal gas (e.g., heliumgas at sufficiently low density), measure PV/Nk of the ideal gas, and read off thetemperature from Fig. 1.2. An equivalent form of the equation of state of an idealgas isPV= nRTwhere n is the number of moles of the gas and R is the gas constant:R = 8.315 joule/deg= 1.986 caljdeg= 0.0821liter-atm/degIts value follows from the value of Boltzmann's constant and Avogadro'snumber:Avogadro's number = 6.023 X 1023atoms/molMost of these concepts are properly understood only in molecular terms.Here we have to be satisfied with empirical definitions.In the following we introduce thermodynamic laws, which may be regardedas mathematical axioms defining a mathematical model. It is possible to deducerigorous consequences of these axioms, but it is important to remember that thismodel does not rigorously correspond to the physical world, for it ignores theatomic structure of matter, and will thus inevitably fail in the atomic domain. Inthe macroscopic domain, however, thermodynamics is both powerful and useful.It enables one to draw rather precise and far-reaching conclusions from a fewseemingly commonplace observations. This power comes from the implicit as-sumption that the equation of state is a regular function, for which the thermody-namic laws, which appear to be simple and naive at first sight, are in factenormously restrictive.THE LAWS OF THERMODYNAMICS1.2 THE FIRST LAW OF THERMODYNAMICS7In an arbitrary thermodynamic transformation let I1Q denote the net amount ofheat absorbed by the system and I1W the net amount of work done by thesystem. The first law of thermodynamics states that the quantity I1U, defined byI1U = I1Q - I1W (1.1)is the same for all transformations leading from a given initial state to a givenfinal state.This immediately defines a state function U, called the internal energy. Itsvalue for any state may be found as follows. Choose an arbitrary fixed state asreference. Then the internal energy of any state is I1Q - I1W in any transforma-tion which leads from the reference state to the state in question. It is definedonly up to an arbitrary additive constant. Empirically U is an extensive quantity.This follows from the saturation property of molecular forces, namely, that theenergy of a substance is doubled if its mass is doubled.The experimental foundation of the first law is Joule's demonstration of theequivalence between heat and mechanical energy-the feasiblity of convertingmechanical work completely into heat. The inclusion of heat as a form of energyleads naturally to the inclusion of heat in the statement of the conservation ofenergy. The first law is precisely such a statement.In an infinitesimal transformation, the first law reduces to the statement thatthe differentialdU= dQ - dW (1.2)is exact. That is, there exists a function U whose differential is dU; or, theintegral f dU is independent of the path of the integration and depends only onthe limits of integration. This property is obviously not shared by dQ or dW.Given a differential of the form df = g(A, B) dA + h(A, B) dB, the condi-tion that df be exact is BgjBB = BhjBA. Let us explore some of the conse-quences of the exactness of dO. Consider a system whose parameters are P, V, T.Any pair of these three parameters may be chosen to be the independentvariables that completely specify the state of the system. The other parameter isthen determined by the equation of state. We may, for example, considerU = U(P, V). Then*dU = ( ~ ; L dP + ( ~ ~ L dV (1.3)The requirement that dU be exact immediately leads to the resultB ~ [ ( ~ ~ L L = B: [( ~ ~ L L (1.4)The following equations, expressing the heat absorbed by a system during aninfinitesimal reversible transformation (in which dW = PdV), are easily ob-*The symbol (iJU/iJP)v denotes the partial derivative of U with respect to P, with V heldconstant.8 THERMODYNAMICS AND KINETIC THEORYtained by successively choosing as independent variables the pairs (P, V), (P, T),and (V, T):dQ = ( : ~ L dP + [( : ~ L + p] dV (1.5)dQ = [( ~ ~ L + p( ~ ~ ) JdT+ [( ~ ~ ) T + p( ~ ; L ] dP (1.6)dQ = ( ~ ~ L dT+ [( ~ ~ L + p] dV (1.7)Called dQ equations, these are of little practical use in their present form,because the partial derivatives that appear are usually unknown and inaccessibleto direct measurement. They will be transformed to more useful forms when wecome to the second law of thermodynamics.It can be immediated deduced from the dQ equations that(1.8)(1.9)where H = U + PV is called the enthalpy of the system.We consider the following examples of the application of the first law.(a) Analysis of Joule's Free-Expansion Experiment. The experiment in ques-tion concerns the free expansion of an ideal gas into a vacuum. Theinitial and final situations are illustrated in Fig. 1.3.Experimental Finding. T1 = T2.Deductions. LlW= 0, since the gas performs no work on its externalsurrounding. LlQ = 0, since LlT = 0. Therefore, LlU = by the first law.Thermometer _ Tl~Water bathBeforeGas occupies volume VI.AfterGas occupies volume V2 >VI'Fig. 1.3 Joule's free-expansion experiment.THE LAWS OF THERMODYNAMICS 9Thus two states with the same temperature but different volumes havethe same internal energy. Since temperature and volume may be takento be the independent parameters, and since U is a state function, weconclude that for an ideal gas U is a function of the temperature alone.This conclusions can also be reached theoretically, without reference toa specific experiment, with the help of the second law of thermody-namics.(b) Internal Energy of Ideal Gas. Since U depends only on T, (1.8) yields(au) dUCv = aT v = dTAssuming Cv to be independent of the temperature, we obtainU = CvT +constantThe additive constant may be arbitrarily set equal to zero.(c) The Quantity Cp - Cv for an Ideal Gas. The enthalpy of an ideal gas isa function of T only:H= U+ PV= (Cv + Nk)THence, from (1.9),Cp = (aH) = dH =Cv+NkaT p dTorThus it is more efficient to heat an ideal gas by keeping the volume constant thanto heat it by keeping the pressure constant. This is intuitively obvious; atconstant volume no work is done, so all the heat energy goes into increasing theinternal energy.1.3 THE SECOND LAW OF THERMODYNAMICSStatement of the Second LawFrom experience we know that there are processes that satisfy the law ofconservation of energy yet never occur. For example, a piece of stone resting onthe floor is never seen to cool itself spontaneously and jump up to the ceiling,thereby converting the heat energy given off into potential energy. The purpose ofthe second law of thermodynamics is to incorporate such experimental facts intothermodynamics. Its experimental foundation is common sense, as the followingequivalent statements of the second law will testify.Kelvin Statement. There exists no thermodynamic transformation whose soleeffect is to extract a quantity of heat from a given heat reservoir and toconvert it entirely into work.10 THERMODYNAMICS AND KINETIC THEORYClausius Statement. There exists no thermodynamic transformation whosesole effect is to extract a quantity of heat from a colder reservoir and todeliver it to a hotter reservoir.In both statements the key word is "sole." An example suffices to illustratethe point. If an ideal gas is expanded reversibly and isothermally, work is doneby the gas. Since !:J.U = 0 in this process, the work done is equal to the heatabsorbed by the gas during the expansion. Hence a certain quantity of heat isconverted entirely into work. This is not the sole effect of the transformation,however, because the gas occupies a larger volume in the final state. This processis allowed by the second law.The Kelvin statement K and the Clausius statement C are equivalent. Toprove this we prove that if the Kelvin statement is false, the Clausius statement isfalse, and vice versa.Proof that K False => C False Suppose K is false. Then we can extract heatfrom a reservoir at temperature T1 and convert it entirely into work, with noother effect. Now we can convert this work into heat and deliver it to a reservoirat temperature T2 > T1 with no other effect. (A practical way of carrying out thisparticular step is illustrated by Joule's experiment on the equivalence of heat andenergy.) The net result of this two-step process is the transfer of an amount ofheat from a colder reservoir to a hotter one with no other effect. Hence C isfalse. Proof that C False => K False First define an engine to be a thermodynamicsystem that can undergo a cyclic transformation (i.e., a transformation whosefinal state is identical with the initial state), in which system does the followingthings, and only the following things:(a) absorbs an amount of heat Q2 > 0 from reservoir T2;(b) rejects an amount of heat Ql > 0 to reservoir T1, with T1< T2;(c) performs an amount of work W> O.Suppose C is false. Extract Q2 from reservoir T1 and deliver it to reservoirT2, with T2 > T1. Operate an engine between T2 and T1 for one cycle, andarrange the engine so that the amount of heat extracted by the engine from T2 isexactly Q20 The net result is that an amount of heat is extracted from T1 andentirely converted into work, with no other effect. Hence K is false. The Carnot EngineAn engine that does all the things required by the definition in a reversible way iscalled a Carnot engine. A Carnot engine consists of any substance that is made togo through the reversible cyclic transformation illustrated in the P- V diagram ofFig. 1.4, where ab is isothermal at temperature T2, during which the systemabsorbs heat Q2; be is adiabatic; cd is isothermal at temperature T1, withTHE LAWS OF THERMODYNAMICSTl\\\\\\\"" .....' - - - - - - - - - - - - ~ V11Fig. 1.4 The Carnot engine.T1 < T2, during which time the system rejects heat Ql; and da is adiabatic. Itmay also be represented schematically as in the lower part of Fig. 1.4. The workdone by the system in one cycle is, according to the first law,W= Q2 - Qlsince !:J.U = 0 in any cyclic transformation. The efficiency of the engine is definedto beWe show that if W> 0, then Ql > 0 and Q2 > 0. The proof is as follows. Itis obvious that Ql *" 0, for otherwise we have an immediate violation of theKelvin statement. Suppose Ql < 0. This means that the engine absorbs theamount of heat Q2 from T2 and the amount of heat - Ql from T1 and convertsthe net amount of heat Q2 - Ql into work. Now we may convert this amount ofwork, which by assumption is positive, into heat and deliver it to the reservoir atT2, with no other effect. The net result is the transfer of the positive amount ofheat - Ql from T1 to T2 with no other effect. Since T2 > T1 by assumption, thisis impossible by the Clausius statement. Therefore Ql > 0. From W = Q2 - Qland W> 0 it follows immediately that Q2 > 0. In the same way we can show that if W < and Ql < 0, then Q2 < 0. Inthis case the engine operates in reverse and becomes a "refrigerator."The importance of the Carnot engine lies in the following theorem.12 THERMODYNAMICS AND KINETIC THEORYFig. 1.5 Construction for the proof ofCarnot's theorem.-w'Q(--w---L.f-'----------Lj-J---- Tl---rl.-----------r-t-r---- T2CARNOT'S THEOREMNo engine operating between two given temperatures is more efficient than aCarnot engine.Proof Operate a Carnot engine C and an arbitrary engine X between thereservoirs T2 and T1 (T2 > T1), as shown in Fig. 1.5. We have, by the first law,W= Q2 - QlW' = Q2 - Q{LetQ2 N'Q2 Nwhere N' and N are two integers. This equality can be satisfied to any degree ofaccuracy by making N', N sufficiently large. Now operate the C engine N cyclesin reverse, and the X engine N' cycles. At the end of this operation we have~ o t a l = N'W' - NW(Q2)total = N'Q2 - NQ2 = 0(Ql)total = N'Q{ - NQlOn the other hand we can also write~ o t a l = (Q2)total - (Ql)total = -(Ql)totalThe net result of the operation 'is thus a violation of the Kelvin statement, unless~ o t a l S 0which implies thatIn other words we must haveN'Q{ - NQl ~ 0Q2Q{ - Q2Ql ~ 0Ql Q{- O. This is easily proved by using an ideal gas to form aCarnot engine. From now on we do not distinguish between the two anddenote the absolute temperature by T.1.4 ENTROPYThe second law of thermodynamics enables us to define a state function S, theentropy, which we shall find useful. We owe this possibility to the followingtheorem.CLAUSIUS' THEOREMIn any cyclic transformation throughout which the temperature is defined,the following inequality holds:dQA.._ off relative to i. (b) The individual momenta of the collidingpartners PI and pz may be constructed from P and 2p, as shown.The points A and B will coincide at 0, the midpoint of P, if thecolliding molecules have equal mass.The dynamical aspects of the collision are contained in the differential crosssection da/ dQ, which is defined experimentally as follows. Consider a beam ofparticle 2 incident on particle 1, regarded as the target. The incident flux I isdefined as the number of incident particles crossing unit area per second, fromthe viewpoint of the target:1= nlv1 - vzl (3.18)where n is the density of particles in the incident beam. The differential crosssection da/ dQ is defined by the statementI(da/dQ) dQ == Number of incident molecules scattered per second (319)into the solid angle element dQ about the direction Q .The differential cross section has the dimension of area. The number of moleculesscattered into dQ per second is equal to the number of molecules in the incidentbeam crossing an area da/dQ per second. The total cross section is the numberof molecules scattered per second, regardless of scattering angle:f daatot = dQ dQ (3.20)In classical mechanics the differential cross section can be calculated fromthe intermolecular potential as follows. First we transform the coordinate systemto the center-of-mass system, in which the total momentum is zero. Since we areconsidering only the nonre1ativistic domain, this involves a trivial translation of58pTHERMODYNAMICS AND KINETIC THEORYp'p' o(3.22)Fig. 3.5 Classical scattering of a molecule by a fixed center of force O.all velocities by a constant amount. We only need to follow the trajectory of oneof the particles, which will move along an orbit, as if it were scattered by a fixedcenter of force 0, as illustrated in Fig. 3.5. It approaches 0 with momentum p,the relative momentum, and will recede from 0 with momentum p', the rotatedrelative momentum. The normal distance between the line of approach and 0 iscalled the impact parameter b. By conservation of angular momentum, it is alsothe normal distance between the line of recession and O. This is indicated in Fig.3.5, together with the scattering angles. From the geometry it is clear thatdaI dfl dfl = Ibdbdq, (3.21)We can find the relation between b and the scattering angles from the classicalorbit equation, thereby obtaining da/ dfl as a function of the scattering angles. *The use of classical mechanics to calculate the differential cross section inthis problem is an old tradition (started by Maxwell) predating quantum mechan-ics. To be correct, however, we must use quantum mechanics, notwithstandingthe fact that between collisions we regard the molecules as classical particles. Thereason is that when the molecules collide their wave functions necessarily overlap,and they see each other as plane waves of definite momenta rather than wavepackets of well-defined positions. Furthermore, formulating the scattering prob-lem quantum mechanically makes the kinematics and symmetries of the problemmore obvious.In quantum mechanics the basic quantity in a scattering problem is thetransition matrix (T matrix), whose elements are the matrix elements of a certainoperator T( E) between the initial (i) and final (f) state:Tfi = (l',2'IT(E)ll,2)T(E) =' +'(E -0 + i)-l, +where 0 is the unperturbed Hamiltonian, , the potential, and 0+. A'See any book on classical mechanics, for example, L. D. Landau and E. M. Lifshitz, Mechanics(Pergamon, Oxford, 1960), Chapter IV.THE PROBLEM OF KINETIC THEORY 59(3.23)collision is a transition from the initial state to a set of final states. For finalstates in the infinitesimal momentum-space element d 3Pl d3p5., the rate is*dP12 -">1'2' = Ida = d3pi d3p2 S4(Pr- PJITrd2S4(Pr- PJ == S3(P - P')S(E - E')The transition rate into any region of momentum space can be obtained byintegrating the above over the appropriate region. To obtain the differential crosssection, integrate over the recoil momentum PI (which is fixed by momentumconservation) and the magnitude P5. (which is determined by energy conserva-tion) to obtain(3.24)(3.25)The integrations are trivial, and yield a factor representing the density of finalstates because of the S functions that enforce momentum-energy conservation.For the formal manipulations that we are going to do, however, it is best to leavethe integrations undone.The T matrix is invariant under spatial rotations and reflections and undertime reversal, because all molecular interactions originate in the electromagneticinteraction, which have these invariance properties. Explicitly, we have(Pz,P1ITlpl,P2) = (Rpz, RpliTIRpl' Rp2)(Pz,P1ITlpbP2) = ( - P2, -PIIT!- Pl, -pz)where Rp is the vector obtained from P after performing a spatial rotation aboutan axis, and/or a reflection with respect to a plane. For elastic scattering thedensity of states are the same in the initial and final states. Thus invariances ofthe T matrix directly implies corresponding invariances of the differential crosssection.If the molecules have spin, (3.25) remains valid provided we interpret P toinclude the spin coordinate. A rotation rotates both the momentum and the spin,but a reflection does not affect the spin. Under time reversal both the momentumand the spin change sign.From (3.25) we can deduce that the inverse collision, defined as the collisionwith initial and final states interchanged, has the same T matrix (and hence thesame cross section):or (3.26)To show this we represent the collision by the schematic drawing A of Fig. 3.6,which is self-explanatory. The diagram A' beneath it has the same meaning as*A word about normalization. With the definition (3.22), and with single-particle statesnormalized to one particle per unit volume, there should be a factor (2'IT1i)3 multiplying thethree-dimensional S function for momentum conservation, a factor 2'IT/1i multiplying the S functionfor energy conservation, and a factor multiplying each volume element d3p of momentumspace. Since we are not going to calculate cross sections, these factors are a nuisance to write out. Weredefine the T matrix appropriately to absorb these factors.60ATHERMODYNAMICS AND KINETIC THEORYp P-PI' -P2' RpI I RP2 P2 I PI__- 1 \ - ~ n __* _ ~ n >K-PI -P2 RPI': Rp2' P2: PI'p' p'BCDPIII~ nA' B' C' D'Fig. 3.6 Symmetry operations that take a collision to the inverse collision. InA', B', C', and D', i and f, respectively, denote initial and final relativemomenta.Fig. 3.4. The T matrix for this collision is the same as the time-reversed collisionrepresented by B. Now rotate the coordinate system through 1800about asuitable axis D perpendicular to the total momentum, and then reflect withrespect to a plane pp' perpendicular to D. As a result we obtain the collision D,which is the inverse of the original collision, and which has the same T matrixbecause of (3.25).If collisions were treated classically, the inverse collision could be verydifferent from the original collision. As a concrete example consider the classicalcollision between a sphere and a wedge. * A glance at Fig. 3.7 proves the point.But this is irrelevant for molecular scattering, because molecules are not describ-able as "wedges" or the like. A nonspherically symmetric molecule is one withnonzero spin, and exists in an eigenstate of the spin. The angular orientation,being conjugate to the angular momentum, is completely uncertain. The symme-try between collision and inverse collision remains valid when the spin is takeninto account, as stated previously.3.3 THE BOLTZMANN TRANSPORT EQUATIONTo derive an explicit formula for (a//at)con, we assume that the gas is suffi-ciently dilute that only binary collisions need be taken into account. The effect of*Both made of concrete.CollisionTHE PROBLEM OF KINE!TIC THEORYInverse collision61Before:After:Before:After:Fig. 3.7 Classical collision between macroscopic objects.external forces on collisions are ignored on the assumption that these forces, ifpresent, would vary little over the range of the intermolecular potential.The number of transitions 12 ~ 1'2' in a volume element d 3r at r, owing tocollisions during the time interval St isdN12 dP12-">1'2' Stwhere dN12 is the initial number of colliding pairs (Pi' P2)' We introduce thetwo-particle correlation function F bydN12 = F(r,pl,p2' t) d3rd3Pl d3P2Thus, in the notation of (3.10), we haveR Std3rd3Pl = Std 3rd3pl f d3Pl dP12 -">1'2' F(r,pl' P2' t)Using (3.23), we obtainR = j d 3P2 d3Pl d3p2 84(Pr- P;)ITrd2F(r, Pl,P2' t)Similarly, we find(3.27)(3.28)(3.29)(3.30)The S functions in (3.29) and (3.30) are identical, and Tfi = Tir by (3.25). Hence(af) =R-R= jd3P2d3pld3P2S4(Pr-P;)ITrd2(Fl'2,-F12) (3.31)at coliwhere F12 = F(r, Pi' P2' t). Note that we can integrate over the vector P ~ and themagnitude Pl, so that the differential cross section appears in the integrand of(3.30):(3.32)For formal manipulations, however, it is more convenient to leave it in the form(3.31).62 THERMODYNAMICS AND KINETIC THEORYThe expression we obtained is exact for a sufficiently dilute gas. But itcontains the unknown correlation function F.We now introduce the crucial assumption(3.33)(3.35)This says that the momenta of two particles in the volume element d 3r areuncorrelated, so that the probability of finding them simultaneously is theproduct of the probability of finding each alone. This is known as the" assump-tion of molecular chaos." It is necessary to obtain a closed equation for thedistribution function, but there is otherwise no justification for it at this point.We shall come back to analyze its meaning later.With the assumption of molecular chaos, we have( = f d3p2 - v21(da/d&2)(J{f{ - Id2) (3.34)t coliwhere the following abbreviations have been used:11 ==/(r,pl' t)12 == I (r, P2' t)f{ == t)f{ == t)Substituting (3.34) into (3.8) we obtain the Boltzmann transport equation(:t + \lr + F \lPI )/1 = f d 3P2 d 3pl d 3p2 S4(pf - PJITfd2(J{I{ - 12/1)(3.36)which is a nonlinear integro-differential equation for the distribution function.We have considered only the case of a single species of spinless molecules. Ifwe consider different types of molecules, then we have to introduce a separatedistribution function for each type, and the collision term will couple them ifdifferent types of molecules can scatter each other. If the molecules have spin, orif we consider excitation of the molecules through scattering, then the differentspin states or excited states should be considered as different species of mole-cules.3.4 THE GIBBSIAN ENSEMBLEGibbs introduced the idea of a statistical ensemble to describe a macroscopicsystem, which has proved to be a very important concept. We shall use it here topresent another approach to the Boltzmann transport equation.A state of the gas under consideration can be specified by the 3N canonicalcoordinates ql' .. ' q3N and their conjugate momenta Pl,., P3N. The 6N-dimensional space spanned by {Pi' qi} is called the r space, or phase space, ofTHE PROBLEM OF KINETIC THEORY 63the system. A point in r space represents a state of the entire N-particle system,and is referred to as the representative point. This is in contrast to the I-t spaceintroduced earlier, which refers to only one particle.It is obvious that a very large (in fact, infinite) number of states of the gascorresponds to a given macroscopic condition of the gas. For example, thecondition that the gas is contained in a box of volume 1 cm3is consistent with aninfinite number of ways to distribute the molecules in space. Through macro-scopic measurements we would not be able to distinguish between two gasesexisting in different states (thus corresponding to two distinct representativepoints) but satisfying the same macroscopic conditions. Thus when we speak of agas under certain macroscopic conditions, we are in fact referring not to a singlestate, but to an infinite number of states. In other words, we refer not to a singlesystem, but to a collection of systems, identical in composition and macroscopiccondition but existing in different states. With Gibbs, we call such a collection ofsystems an ensemble, which is geometrically represented by a distribution ofrepresentative points in r space, usually a continuous distribution. It may beconveniently described by a density function p(p, q, t), where (p, q) is anabbreviation for (Pi' ... , P3N; ql' ... , q3N)' so defined that(3.37)is the number of representative points that at time t are contained in theinfinitesimal volume element d 3Npd3Nq of r space centered about the point(p, q). An ensemble is completely specified by p(p, q, t). It is to be emphasizedthat members of an ensemble are mental copies of a system and do not interactwith one another.Given p(p, q, t) at any time t, its subsequent values are determined by thedynamics of molecular motion. Let the Hamiltonian of a system in the ensemblebe .Yf(Pl, ... , P3N; ql'., q3N) The equations of motion for a system are givenby(i = 1, ... ,3N)a.Yfqi= -aPi(i = 1, ... , 3N)(3.38)These will tell us how a representative point moves in r space as time evolves.We assume that the Hamiltonian does not depend on any time derivative of pand q. It is then clear that (3.38) is invariant under time reversal and that (3.38)uniquely determines the motion of a representative point for all times, when theposition of the representative point is given at any time. It follows immediatelyfrom these observations that the locus of a representative point is either a simpleclosed curve or a curve that never intersects itself. Furthermore, the loci of twodistinct representative points never intersect.We now prove the following theorem.64 THERMODYNAMICS AND KINETIC THEORYLIOUVILLE'S THEOREMap 3N ( ap ap )- + L -P+ -q =0at ;=1 api I aq; I(3.39)Proof Since the total number of systems in an ensemble is conserved, thenumber of representative points leaving any volume in r space per second mustbe equal to the rate of decrease of the number of representative points in thesame volume. Let w be an arbitrary volume in r space and let S be its surface. Ifwe denote by v the 6N-dimensional vector whose components arev == (PI' P2,"" P3N; q1' q2,"" q3N)and n the vector locally normal to the surface S, thend-- f dw p = f dS n 0 vpdt w sWith the help of the divergence theorem in 6N-dimensional space, we convertthis to the equationf dW[ap+ \7 o (vp )] = 0w at (3.40)where \7 is the 6N-dimensional gradient operator:\7 == (a:1' a:2"'" a:3N ; a:1' a:2"'" a:3JSince w is an arbitrary volume the integrand of (3.40) must identically vanish.Henceap 3N [ a a]- - = \7 o (vp) = L -(PiP) + -(qiP)at ;=1 api aqi3N ( ap ap ) 3N (ap. aq )L -P + -q. + L p -' + -'; ~ 1 api' aqi' ;=1 api aqiBy the equations of motion (3.38) we haveap; aq;- + - = 0 (i = 1, ... ,3N)api aqiThereforeap--=at Liouville's theorem is equivalent to the statementdp-=0dt (3.41)since by virtue of the equations of motion Pi and q; are functions of the time. ItsTHE PROBLEM OF KINETIC THEORY 65geometrical interpretation is as follows. If we follow the motion of a representa-tive point in r space, we find that the density of representative points in itsneighborhood is constant. Hence the distribution of representative points movesin r space like an incompressible fluid.The observed value of a dynamical quantity of the system, which isgenerally a function of the coordinates and conjugate momenta, is supposed to beits averaged value taken over a suitably chosen ensemble:f d 3Npd3NqO(p, q)p(p, q, t)(0) = --------f d 3Np d 3Nq p( p, q, t)(3.42)This is called the ensemble average of 0. Its time dependence comes from that ofp, which is governed by Liouville's theorem. In principle, then, this tells us how aquantity approaches equilibrium-the central question of kinetic theory. In thenext section we shall derive the Boltzmann transport equation using this ap-proach.Under certain conditions one can prove an ergodic theorem, which says thatif one waits a sufficiently long time, the locus of the representative point of asystem will cover the entire accessible phase space. More precisely, it says thatthe representative point comes arbitrarily close to any point in the accessiblephase space. This would indicate that the ensemble corresponding to thermody-namic equilibrium is one for which p is constant over the accessible phase space.This is actually what we shall assume. *3.5 THE BBGKY HIERARCHYOne can define correlation functions fs' which give theoprobability of finding sparticles having specified positions and momenta, in the systems forming anensemble. The function f1 is the familiar distribution function. The exactequations of motion for fs in classical mechanics can be written down. Theyshow that to find f1 we need to know f2' which in turns depends on a knowledgeof f3' and so on till we come the full N-body correlation function f N' This systemof equations is known as the BBGKyt hierarchy. We shall derive it and showhow the chain of equations can be truncated to yield the Boltzmann transportequation. The "derivation" will not be any more rigorous than the one alreadygiven, but it will give new insight into the nature of the approximations.Consider an ensemble of systems, each being a gas of N molecules enclosedm volume V, with Hamiltonian .Yr. Instead of the general notation {Pi' qi}*See the remarks about the relevance of the ergodic theorem in Section 4.5.tBBGKY stands for Bogoliubov-Bom-Green-Kirkwood-Yvon. For a detailed discussion andreferences see N. N. Bogoliubov in Studies in Statistical Mechanics, J. de Boer and G. E. Uhlenbeck,Eds., Vol. I (North-Holland, Amsterdam, 1962).66 THERMODYNAMICS AND KINETIC THEORY(i = 1, , 3N), we shall denote the coordinates by the Cartesian vectors {Pi' ri}(i = 1, , N), for which we use the abbreviation(3.43)The density function characterizing the ensemble is denoted by p(l, ... , N, t),and assumed to be symmetric in Z1, . ' ZN. Its integral over all phase space is aconstant by Liouville's theorem; hence we can normalized it to unity:f dZ1 . dzN P(1, ... , N, t) = 1 (3.44)Thus the ensemble average of any function 0(1, ... , N) of molecular coordinatescan be written as(0) == f dZ1 . dzN P(l, ... , N, t)O(l, ... , N) (3.45)Using the Hamiltonian equations of motion (3.38), we rewrite Liouville'stheorem in the formNL (\7p,P' \7r,.Yf- \7r,P' \7p,.Yf);=1Assume that the Hamiltonian is of the formN p2 N.Yf= L - + L U; + LVi);=1 2m;=1 ;UJ Pi\7Jt=-p, mN\7 .Yf= - F - " K'"J"rj l.i...l)=1(j*i)whereF;= -\7rP(r;)K;)= -\7r,u(lr;-r))Liouville's theorem can now be cast in the form(3.46)(3.47)(3.48)(3.49)(3.50)THE PROBLEM OF KINETIC THEORYwhereN 1 NhN(l, ... , N) = LSi +"2 L Pi) i, }=1(i*})PiS==-o\7 +F.0\7I m r1 I PIThe single-particle distribution function is defined by67(3.51)f,(p,r, t) = ( - p;)8'(r - r;)) Nf dz, .. , dzN p(l, ... , N, t)(3.52)The factor N in the last form comes from the fact that all terms in the sum in thepreceding term have the same value, owing to the fact that P is symmetric inZI, ... , ZN' Integrating II over ZI yields the correct normalization N, by virtue of(3.44).The general s-particle distribution function, or correlation function, isdefined byN!Is(I, ... , z, t) == ( ) jdzS+1 '" dzNP(I"", N, t) (s = 1, ... , N)N-s!(3.53)The combinatorial factor in front comes from the fact that we do not care whichparticle is at ZI' which is at z2, etc. The equation of motion isa N! a N!atls = (N - s)! f dZs+1 ... dzNatp = - (N _ s)! f dZs+ 1 '" dzNhNP(3.54)We isolate those terms in hN involving only the coordinates Zl,"" zs:s N Is 1 N s NhN (I, ... , N) = LSi + LSi + - L Pi) + - L Pi) + L L Pi) s+l 2 2 i,)=s+1 i=1 (i*j) (i*j)Note thatN=hs(l, ... ,s) + hN-s(s + 1, ... ,N) + L L Piji=l j=s+lj dZs+1 . dzNhN_,(s + 1, .. " N)p(I, ... , N) = 0(3.55)(3.56)68 THERMODYNAMICS AND KINETIC THEORYbecause h N -s consists of gradient terms in P with p-independent coefficients, anda gradient term in r with an r-independent coefficient. Thus the integral evaluatesp on the boundary of phase space, where we assume p to vanish. Substituting(3.55) into (3.54), we obtain

+ hs)ls = - ( )' j dZs+1 . dZNt t P;jP(l, ... , N)at N s. s N! j dZs+1 P;,s+l (N _ s + I)! j dZs+2 dz N P(1, ... , N)s- L j dZS+1 Pi,s+lfs+l(l, ... , s + 1) (3.57);=1In passing from the first to the second equation we have used the fact that thesum over j gives N - s identical terms. Now substitute Pij from (3.51), and notethat the second term there does not contribute, because it leads to a vanishingsurface term. We then arrive at( :t + hs)/.(1, ... , s) = - j dZs+1 Ki,s+l 0 \7p/s+l(l, ... , s + 1)(s = 1, ... , N) (3.58)which is the BBGKY hierarchy. The left side of each of the equations above is a"streaming term," involving only the s particles under consideration. For s > 1 itincludes the effect of intermolecular scattering among the s particles. Theright-hand side is the "collision integral," which describes the effect of scatteringbetween the particles under consideration with an "outsider," thus coupling Is tofs+l'The first two equations in the hierarchy read(:t +: \7rl + F1 0\7Pl)tl(Zl,t) = - jdZ2K120\7P,!2(Zl,Z2,t) (3.59)[a Pi P2 1 ( )]-a + - 0 \7r + - 0 \7r + F1 0 \7p + F2 \7p + zK12 0 \7p - \7pt m 1 m 2 1 2 12X/2(Zb Z2, t)= - j dZ3(K13 0 \7P1 + K23 0 \7pJ/3(Zl' Z2' Z3, t) (3.60)The terms in the equations above have dimensions Is/time, and different timescales are involved:1K 0 \7p -'Tc1F 0 \7p - (3.61)'TeP 1- 0 \7 -m r'TsTHE PROBLEM OF KINETIC THEORY 69where 'Te is the duration of a collision, 'Te is the time for a molecule to traverse acharacteristic distance over which the external potential varies significantly, and'Ts is the time for a molecule to traverse a characteristic distance over which thecorrelation function varies significantly. The time 'Te is the shortest, and 'Te thelongest.The equation for 11 is unique in the hierarchy, in that "streaming" sets arather slow time scale, for it does not involve intermolecular scattering, (therebeing only one particle present.) The collision integral, which has more rapidvariations, sets the time scale of 11' This is why the equilibrium condition isdetermined by the vanishing of the collision integral.In contrast, the equation for 12 (and higher ones as well) contains a collisionterm of the order 1/'Te on the left side. The collision integral on the right side issmaller by a factor of the order nri (where n is the density, and ro the range ofthe intermolecular potential) because the integration of r3 extends only over avolume of radius roo Now ro ::::: 10-8cm and n ::::: 1019cm-3under standardconditions. Hence nri ::::: 10-5. Thus for 12 (and higher equations too) the timescale is set by the streaming terms instead of the collision integral, which we shallneglect.With neglect of the right side of (3.60), the hierarchy is truncated at 12, andwe have only two coupled equations for 11 and 12:(3.62)[~ + Pl \lr + P2 \lr + }K12 (\lp - \lp )]/2(Zl' Z2' t) = 0 (3.63)at m 1 m 2 1 2where we have set all external forces to zero, for simplicity. We shall also assumefor simplicity that the force K vanishes outside a range roo To remind us of thisfact, we put the subscript roon the integral in the first equation, indicating thatthe spatial part of the integral is subject to Ir1 - r21< rooThe salient qualitative features of (3.62) and (3.63) are that 12 varies in timewith characteristic period 'Te, and in space with characteristic distance ro, while 11varies much less rapidly, by a factor nri. Thus 11 measures space and time withmuch coarser scales than 12' .The correlations in 12 are produced by collisions between particles 1 and 2.When their positions are so far separated as to be out of molecular interactionrange, we expect that there will be no correlation between 1 and 2, and 12 willassume a product form (neglecting, of course, possible correlations produced bycollisions with a third particle):(3.64)To evaluate (a/1/at)eoll' however, we need 12 not in the uncorrelatedregion, but in the region where the two particles are colliding. To look at this70 THERMODYNAMICS AND KINETIC THEORYFig. 3.8 Illustration of behavior of two-particle correlation function. Theseparation between the two particles is r, and the relative momentum p.The two particles are correlated only inside the range of the intermolecularforce, indicated by the sphere of radius '0' Outside the sphere, thecorrelation function is a product of two one-particle distribution functions.In equilibrium there is a steady scattering of beams of particles of allmomenta, from all directions, at all impact parameters.(3.65)2p=P = P2 + Piregion it is convenient to use total and relative coordinates, defined as follows:P2 - PiThen (3.63) becomes( a P P )- + - . \7 + - . \7 + K(r) 'M f (P R P r t) = 0at m R m r v p 2 , , , ,K(r) =- \7,v (, )(3.66)Transform to the center-of-mass system by putting P = O. The above can then berewritten, to first order in dt, as the streaming condition (with P and Rsuppressed for clarity):12(P + K(r) dt,r + : dt, t + dt) = 12(p,r, t) (3.67)It traces the classical trajectories in the force field K centered at 0, as illustratedIn Fig. 3.8. If 12 were peaked at point A initially, then (3.67) says that as timegoes on the peak will move along the trajectory for that particular initialcondition.THE PROBLEM OF KINETIC THEORY 71The equilibrium situation, for which ah./ at = 0, is a steady-state scattering,by the force field K, of a beam of particles consisting of all momenta, at allimpact parameters. Referring to Fig. 3.8, we may describe the steady state asfollows: Outside the sphere of interaction the uncorrelated factorized form (3.64)holds. However, boundary values of the momenta are correlated through the factthat momenta entering the sphere at a specific impact parameter must leave thesphere at the correct scattering angle, and vice versa.To "derive" the Boltzmann transport equation, we assume that, since 12 hasa shorter time scale than 11' it reaches equilibrium earlier than 11' Thus we setaIII at = 0, and assume 12 has attained the equilibrium form described earlier.Similarly, we assume that the range of force '0 is essentially zero from the pointof view of 11' Thus in the factorized form of 12 just before and after a collision,we can put r2 and rl both equal to the same value.With this in mind, we substitute (3.63) into (3.62) to obtain( aall) == -jdZ2K120'VP/2(Zl,Z2,t)t coli TO(3.68)Using the coordinates defined in (3.66), and neglecting the gradient with respectto R, we have(3.69)where the notation is indicated in Fig. 3.8. Now we set12(xl) = Il(Pl)/l(P2)12(X2) = where P'l' pz are the final momenta in the scattering process, when the initialmomenta are PI' P2 and the impact parameter is b. Using the definition (3.21) ofthe classical cross section, we finally have(3.70)which is the same as (3.34).72 THERMODYNAMICS AND KINETIC THEORYPROBLEMS3.1 Give a few numerical examples to show that the condition (3.1) is fulfilled forphysical gases at room temperatures.3.2 Explain qualitatively why all molecular interactions are electromagnetic in origin.3.3 For the collision between perfectly elastic spheres of diameter a,(a) calculate the differential cross section with classical mechanics in the coordinatesystem in which one of the spheres is initially at rest;(b) compare your answer with the quantum mechanical result. Consider both the low-energy and the high-energy limit. (See, e.g., L. I. Schiff, Quantum Mechanics, 2nd ed.(McGraw-Hill, New York, 1955), p. 110).3.4 Consider a mixture of two gases whose molecules have masses m and M, respec-tively, and which are subjected to external forces F and Q, respectively. Denote therespective distribution functions by f and g. Assuming that only binary collisions betweenmolecules are important, derive the Boltzmann transport equation for the system.3.5 This problem illustrates in a trivial case how the ensemble density tends to a uniformdensity over the accessible phase space. Consider an ensemble of systems, each of whichconsists of a single free particle in one dimension with momentum p and coordinate q.The particle is confined to a one-dimensional box with perfectly reflecting walls located atq = -1 and q = 1 (in arbitrary units.) Draw a square box of unit sides in the pq plane(the phase space). Draw a square of sides 1/2 in the upper left comer of this box. Let theinitial ensemble correspond to filling this comer box uniformly with representative points.(a) What is the accessible part of the phase space? (i.e., the region that the representativepoints can reach through dynamical evolution from the initial condition.)(b) Consider how the shape of the distribution of representative point changes at regularsuccessive time intervals. How does the distribution look after a long time?Suggestion: When a particle is being reflected at a wall, its momentum changes sign.Represent what happens in phase space by continuing the locus of the representative pointto a fictitious adjacent box in pq space, as if the wall were absent. "Fold" the adjacentbox onto the original box properly to get the actual trajectory of the representative point.After a long time, you need many such adjacent boxes. The" folding back" will then giveyou a picture of the distribution.THE EQUILIBRIUM STATEOF A DILUTE GAS4.1 BOLTZMANN'S H THEOREMWe define the equilibrium distribution function as the solution of the Boltzmanntransport equation that is independent of time. We shall see that it is also thelimiting form of the distribution function as the time tends to infinity. Assumethat there is no external force. It is then consistent to assume further that thedistribution function is independent of r and hence can be denoted by I(p, t).The equilibrium distribution function, denoted by lo(p), is the solution to theequation al(p, t)/at = O. According to the Boltzmann transport equation (3.36),lo(p) satisfies the integral equationwhere PI is a given momentum.A sufficient condition for lo(p) to solve (4.1) is10(pz)fo(pD - 10(pz)fo(PI) = 0 (4.2)where {PI' pz} ~ { p ~ , pz} is any possible collision (i.e., one with nonvanishingcross section). We show that this condition is also necessary, and we thus arriveat the interesting conclusion that lo( p) is independent of da/ dD., as long as thelatter is nonzero.To show the necessity of (4.2) we define with Boltzmann the functionalH(t) ;: j d3vl(p, t)log/(p, t)where I(p, t) is the distribution function at time t, satisfyingal(P1,t)_j33/3/4( ) Z(II )at - d pz d PI d pz I) Pc - Pi ITed Iz/1 - Iz/173(4.3)(4.4)74 THERMODYNAMICS AND KINETIC THEORYDifferentiation of (4.3) yieldsdH(t) a/(p, t)-- = fd 3v [1 + log/(p, t)]dt at (4.5)Therefore at/at = 0 implies dH/dt = O. This means that a necessary conditionfor at/at = 0 is dH/dt = O. We now show that the statementdH- = 0 (4.6)dtis the same as (4.2). It would then follow that (4.2) is also a necessary conditionfor the solution of (4.1). To this end we prove the following theorem.BOLTZMANN'S HTHEOREMIf / satisfies the Boltzmann transport equation, thendH(t)-- < 0 (4.7)dt -Proof Substituting (4.4) into the integrand of (4.5) we have*dHdt = f d 3p2 d 3 p ~ d 3p2 l)4(Pr - PJITrd2(f{/{ - /2/1)(1 + log/I) (4.8)Interchanging PI and P2 in this integrand leaves the integral invariant because Triis invariant under such an interchange. Making this change of variables ofintegration and taking one-half of the sum of the new expression and (4.8), weobtaindH 1- = -f d 3p d 3p' d 3p' l)4(p - P)IT.1 2dt 2 2 1 2 r 1 ilX(f{/{ - fd1)[2 + log (fd2)] (4.9)(4.10)(4.11)The integrand of the integral in (4.11) is never positive.This integral is invariant under the interchange of {PI' P2} and { p ~ , P2} becausefor every collision there is an inverse collision with the same T matrix. HencedH 1dt = - 2f d 3p2 d 3 p ~ d 3p2 84(Pr - pJI1fd2(f{/{ - /d1)X [2 + log (f{f{)]Taking half the sum of (4.9) and (4.10) we obtaindH 1dt = '4 f d 3p2 d 3 p ~ d 3p2 l)4(Pr - PJITrd2(f{/{ - /d1)X [log (fd2) -log(f{f{)]*Note that the use of (4.4) presupposes that the state of the system under consideration satisfiesthe assumption of molecular chaos.THE EQUILIBRIUM STATE OF A DILUTE GAS 75As a by-product of the proof, we deduce from (4.11) that dHjdt = 0 if andonly if the integrand of (4.11) identically vanishes. This proves that the statement(4.6) is identical with (4.2). It also shows that under an arbitrary initial conditionI(p, t) ~ lo(p)1-004.2 THE MAXWELL-BOLTZMANN DISTRIBUTIONIt has been shown that the equilibrium distribution function lo(p) is a solution of(4.2). It will be called the Maxwell-Boltzmann distribution. To find it, let us takethe logarithm of both sides of (4.2):log 10 (PI) + log 10 (P2) = log 10 ( p ~ ) + log 10 (P2) (4.12)Since {PI' P2} and { p ~ , P2} are, respectively, the initial and final velocities of anypossible collision, (4.12) has the form of a conservation law. If X(p) is anyquantity associated with a molecule of velocity P, such that X(PI) + X(P2) isconserved in a collision between molecules PI and P2' a solution of (4.12) islog/o(p) = X(P)The most general solution of (4.12) islog/o(p) = XI(P) + X2(P) + ...where the list Xl' X2' ... exhausts all independently conserved quantities. Forspinless molecules, these are the energy and the momentum of a molecule, and, ofcourse, a constant. Hence log I is a linear combination of p2and the threecomponents of P plus an arbitrary constant:log/o(p) = -A(p - PO)2 + logCorlo(p) = Ce-A(p-po)2 (4.13)where C, A, and the three components of Po are five arbitrary constants. We candetermine these constants in terms of observed properties of the system.Applying the condition (3.5), and denoting the particle density NjV by nwe have('TT )3/2n = Cf d 3pe-A(p-po)2 = Cf d 3pe-Ap2= C Afrom which we conclude that A > 0 and(A )3/2C= - n'TTLet the average momentum (p) of a gas molecule be defined byf d 3p p/o(p)(p)= ----f d 3pfo(p}(4.14)(4.15)76ThenTHERMODYNAMICS AND KINETIC THEORY(4.16)Thus we must take Po = 0, if the gas has no translational motion as a whole.Next we calculate the average energy f of a molecule, defined byf d3p (p2j2m)fo(p)f == ---------f d3pfo(p)We have, setting Po = 0,C p2 2 2'1TC 100 2 3f = --fd3p -e-Ap= -- dpp4e-Ap =2nm 2m nm 0 4AmThe constant A is therefore related to the average energy by3A=-4fmSubstituting this into (4.14) we obtain for the constant C the expressionC = n(_3)3/24'1Tfm(4.17)(4.18)(4.19)To relate the average energy f to a directly measurable quantity, let us findthe equation of state corresponding to the equilibrium distribution function. Wedo this by calculating the pressure, which is defined as the average force per unitarea exerted by the gas on one face of a perfectly reflecting plane exposed to thegas. Let the disk shown in Fig. 4.1 represent such a unit area, and let us call theaxis normal to it the x axis. A molecule can hit this disk only if the x componentof its momentum Px is positive. Then it loses an amount of momentum 2px uponreflection from this disk. The number of molecules reflected by the disk persecond is the number of molecules contained in the cylinder shown in Fig. 4.1with vx > O. This number is vxfo(p) d3p, with vx > O. Therefore the pressure is,i-vx~ I/-- - - - - - - - - - - -(FiSkof unit area/ [/?-...!2-I x\ Pzp,-------------Fig. 4.1 Il1ustration for the calculation of the pressure.THE EQUILIBRIUM STATE OF A DILUTE GASfor a gas with zero average velocity77(4.20)(4.21)Cf 2 Cf 2= - d 3pp;e-Ap= - d 3pp2e-Apm 3mwhere the last step comes about because fo(p) depends only on Ipi so that theaverage values of p;, ii;, and pi are all equal to one-third of the average ofp2 = i1 + ii; + p;. Finally we notice thatp2 2P = ICfd3p -e-Ap= lnf3 2m 3This is the equation of state. Experimentally we define the temperature T byP = nkT, where k is Boltzmann's constant. Hencef = 1kT (4.22)In terms of the temperature T, the average momentum Po, and the particledensity n the equilibrium distribution function for a dilute gas in the absence ofexternal force isnI" (p) e-(P-Po)2/2mkT (4.23)Jo = (27TmkT)3/2This is the Maxwell-Boltzmann distribution, the probability of finding a moleculewith momentum P in the gas, under equilibrium conditions. *If a perfectly reflecting wall is introduced into the gas, fo(p) will remainunchanged because fo(p) depends only on the magnitude of P, which is un-changed by reflection from the wall.For a gas with Po = 0 it is customary to define the most probable speed i5 ofa molecule by the value of v at which 47Tp 2f(p) attains a maximum. We easilyfind p = V2mkT. The most probable speed is thereforeV- -- J2mkT(4.24)The root mean square speed vrms is defined by(4.25)At room temperatures these speeds for an O2,gas are of the order of magnitudeof 105cm/s.*We have assumed, in accordance with experimental facts, that the temperature T is indepen-dent of the average momentum Po.78 THERMODYNAMICS AND KINETIC THEORYFig. 4.2 The Maxwell-Boltzmann distribution.(4.26)A plot of 4'ITP%(P) against v = p/m is shown in Fig. 4.2. We notice thatlo(p) does not vanish, as it should, when v exceeds the velocity of light e. This isbecause we have used Newtonian dynamics for the molecules instead of the morecorrect relativistic dynamics. The error is negligible at room temperatures,because v e. The temperature above which relativistic dynamics must be usedcan be roughly estimated by putting v= e, from which we obtain kT "" me2.Hence T"" 1013 K for H2Let us now consider the equilibrium distribution for a dilute gas in thepresence of an external conservative force field given byF = -V'ep(r)We assert that the equilibrium distribution function is nowI(r, p) = lo(p)e-(r)/kT (4.27)where lo(p) is given by (4.23). To prove this we show that (4.27) satisfiesBoltzmann's equation. We see immediately that ai/at = 0 because (4.27) isindependent of the time. Furthermore (al/at)coll = 0 because ep(r) is indepen-dent of p:( ~ L O l l = e-2(r)/kTf d 3p2 d 3Pl d 3p2 84(Pf - PJITfd2(J{f{ - 12/1) = 0Hence it is only necessary to verify that(: V'r + F V'p)/(r,p) = 0and this is trivial. We may absorb the factor exp( -ep/kT) in (4.27) into thedensity n and writewhereI(r,p) = n(r) e-(p-po)'/2mkT(2'ITmkT)3/2 (4.28)(4.29)Finally we derive the thermodynamics of a dilute gas. We have defined thetemperature by (4.22) and we have obtained the equation of state. By the veryTHE EQUILIBRIUM STATE OF A DILUTE GAS 79definition of the pressure, the work done by the gas when its volume increases bydV is P dV. The internal energy is defined byU(T) = N = iNkT (4.30)which is obviously a function of the temperature alone.The analog of the first law of thermodynamics now takes the form of adefinition for the heat absorbed by the system:dQ = dU + PdV (4.31)It tells us that heat added to the system goes into the mechanical work P dV andthe energy of molecular motion dUo From (4.31) and (4.30) we obtain for theheat capacity at constant volumeCv = iNk (4.32)The analog of the second law of thermodynamics is Boltzmann's H theorem,where we identify H with the negative of the entropy per unit volume divided byBoltzmann's constant:SH=--Vk (4.33)Thus the H theorem states that for a fixed volume (i.e., for an isolated gas) theentropy never decreases, which is a statement of the second law.To justify (4.33) we calculate H in equilibrium:Ho= f d3plo log 10 = n{IOg [ n C ? T ~ k T f/2l - ~ }Using the equation of state we can rewrite this as-kVHo= iNk log (PVSI 3) + constant (4.34)We recognize that the right-hand side is the entropy of an ideal gas in thermody-namics. It follows from (4.34), (4.33), and (4.31) that dS = dQ/T.Thus we have derived all of classical thermodynamics for a dilute gas; andmoreover, we were able to calculate the equation of state and the specific heat.The third law of thermodynamics cannot be derived here because we have usedclassical mechanics and thus are obliged to confine our considerations to hightemperatures.4.3 THE METHOD OF THE MOSTPROBABLE DISTRIBUTIONWe have noted the interesting fact that the Maxwell-Boltzmann distribution isindependent of the detailed form of molecular interactions, as long as they exist.This fact endows the Maxwell-Boltzmann distribution with universality. Wemight suspect that as long as we are interested only in the equilibrium behavior of80 THERMODYNAMICS AND KINETIC THEORYpFig. 4.3 The microcanonical ensemble correspond-ing to a gas contained in a finite volume withl.....- ~ q energy between E and E + ~ .a gas there is a way to derive the Maxwell-Boltzmann distribution withoutexplicitly mentioning molecular interactions. Such a derivation is now supplied.Through it we shall understand better the meaning of the Maxwell-Boltzmanndistribution. The conclusion we reach will be the following. If we choose a stateof the gas at random from among all its possible states consistent with certainmacroscopic conditions, the probability that we shall choose a Maxwell-Boltzmann distribution is overwhelmingly greater than that for any other distri-bution.We shall use the approach of the Gibbsian ensemble described in Sec. 3.4.We assume that in equilibrium the system is equally likely to be found in anystate consistent with the macroscopic conditions. That is, the density function is aconstant over the accessible portion of r space.Specifically we consider a gas of N molecules enclosed in a box of volume Vwith perfectly reflecting walls. Let the energy of the gas lie between E andE + IJ., with IJ. E. The ensemble then consists of a uniform distribution ofpoints in a region of r space bounded by the energy surfaces of energies E andE + IJ., and the surfaces corresponding to the boundaries of the containing box,as illustrated schematically in Fig. 4.3. Since the walls are perfectly reflecting,energy is conserved, and a representative point never leaves this region. ByLiouville's theorem the distribution of representative points moves likes anincompressible fluid, and hence maintains a constant density at all times. Thisensemble is called a microcanonical ensemble.Next consider an arbitrary distribution function of a gas. A molecule in thegas is confined to a finite region of p. space because the values of p and q arerestricted by the macroscopic conditions. Cover this finite region of p. space withvolume elements of volume w = d 3p d 3q, and number them from 1 to K, whereK is a very large number which eventually will be made to approach infinity. Werefer to these volume elements as cells. An arbitrary distribution function isdefined if we specify the number of molecules ni found in the i th cell. These arecalled occupation numbers, and they satisfy the conditionsKL ni = N (4.35)i=1KL f.ini = E (4.36)i=lTHE EQUILIBRIUM STATE OF A DILUTE GAS 81where f.; is the energy of a molecule in the ith cell:p;f..=-I 2mwhere p; is the momentum of the ith cell. It is in (4.36), and only in (4.36), thatthe assumption of a dilute gas enters. An arbitrary set of integers {n;} satisfying(4.35) and (4.36) defines an arbitrary distribution function. The value of thedistribution function in the ith cell, denoted by /;, isn/; = --.!.- (4.37)wThis is the distribution function for one member in the ensemble. The equi-librium distribution function is the above averaged over the microcanonicalensemble, which assigns equal weight to all systems satisfying (4.35) and (4.36):- (n;)1= -I WThis is the same definition as (3.52) except that we have replaced the infinitesimalelement d 3r d 3p by a finite cell of volume w. Unfortunately this ensembleaverage is difficult to calculate. So we shall adopt a somewhat different approach,which will yield the same result for a sufficiently large system.It is clear that if the state of the gas is given, then f is uniquely determined;but if f is given, the state of the gas is not uniquely determined. For example,interchanging the positions of two molecules in the gas leads to a new state of thegas, and hence moves the representative in r space; but that does not change thedistribution function. Thus a given distribution function f corresponds not to apoint, but to a volume in r space, which we call the volume occupied by f. Weshall assume that the equilibrium distribution function is the most probabledistribution function, i.e., that which occupies the largest volume in r space.The procedure is then as follows:(a) Choose an arbitrary distribution function by choosing an arbitrary setof allowed occupations numbers. Calculate the volumes it occupies bycounting the number of systems in the ensemble that have theseoccupation numbers.(b) Vary the distribution function to maximize the volume.Let uS denote by Q{n;} the volume in r space occupied by the distributionfunction corresponding to the occupation numbers {n;}. It is proportional to thenumber of ways of distributing N distinguishable molecules among K cells sothat there are n; of them in the ith cell (i = 1,2, ... , K). ThereforeN!Q{n;} ex g ~ l g;2 ... g;K (4.38)n1!n2 !n3! .,. nK !where g; is a number that we will put equal to unity at the end of the calculationbut that is introduced here for mathematical convenience. Taking the logarithm82 THERMODYNAMICS AND KINETIC THEORYof (4.38) we obtainK Klog 0 {n i } = log N! - L log nil + L nilog g; + constant ;=1Now assume that each n; is a very large integer, so we can use Stirling'sapproximation, log n;! =:: n; log n; - 1. We then haveK Klog 0 {n;} = N log N - L n; log nil + L nilog g; + constant (4.39);=1 To find the equilibrium distribution we vary the set of integers {n;} subjectto the conditions (4.35) and (4.36) until log 0 attains a maximum. Let {ii;}denote the set of occupation numbers that maximizes log O. By the well-knownmethod Lagrange multipliers we have =0 (ni=iiJ (4.40)where a, ,8 are Lagrange's multipliers. Now the n; can be considered independentof one another. Substituting (4.39) into (4.40) we obtainKL [- (log n; + 1) + log g; - a - ,8d = 0;=1(4.41)ii = g.e-a-{l 00(5.134)Taking the divergence of both sides of the first equation of (5.133), weobtain"V 2p = 0 (5.135)Thus the pressure, whatever it is, must be a linear superpoSItIOn of solidharmonics. A systematic way to proceed would be to write P as the most generalsuperposition of solid harmonics and to determine the coefficient by requiringthat (5.133) be satisfied. We take a short cut, however, and guess that P is, apartfrom an additive constant, a pure solid harmonic of order 1:cos ()P = Po + J-tPl -2- (5.136)rwhere Po and Pi are constants to be determined later. With this, the problemreduces to solving the inhomogeneous Laplace equationcos ()"V 2U = P1"V -- (5.137)r2subject to the conditions"Vu=o[u(r)] r=a = 0u(r) ~ Uor-> 00(5.138)A particular solution of (5.137) isu1 = - Pi r2"V cos () = _ Pi ( ~ - 3 r ~ ) (5.139)6 r26 r r3where denotes the unit vector along the z axis, which lies along uo. It is easilyverified that (5.139) solves (5.137), if we note that l/r and z/r3are both solidharmonics. Thus,"V 2U1 = - :1 [- 3"V 2( :: )] = P1"V (:3) = P1"V c:s2() (5.140)The complete solution is obtained by adding an appropriate homogeneoussolution to (5.139) to satisfy (5.138). By inspection we see that the completesolution is(5.141)where we have setTRANSPORT PHENOMENA 121(5.142)(5.143)to have 'V u = o.We now calculate the force acting on the sphere by the fluid. By definitionthe force per unit area acting on a surface whose normal point along the x) axis is- T) of (5.107). It follows that the force per unit area acting on a surface elementof the sphere is(X Y z) ....f = - -T1 + -T2 + -T3 = -f Pr r rwhere f is the unit vector in the radial direction and Pis given by (5.118). Thetotal force experienced by the sphere isF'=jdSf (5.144)where dS is a surface element of the sphere and the integral extends over theentire surface of the sphere. Thus it is sufficient to calculate f for r = a.The vector f Phas the componentsA.... 1 1 [ ( au) aui)](rP).= -x.p.. = -x i)p-II. - +-I r J JI r J JI r ax. ax.I JXi }.t [ a a ]-P - - -(xu) - u + x-ur r ax J J I J ax. II JHence(5.145)where P is given by (5.136) and (5.142), and u is given by (5.141). Since u = 0when r = a, we only need to consider the first and the last terms in the bracket.The first term is zero at r = a by a straightforward calculation. At r = a thesecond term is found to be1 (au)-[(r 'V)u] r ~ a = -r ar r-a3 Uo-- -2 a3 cos ()-fu --2 0 a (5.146)When this is substituted into (5.145), the second term exactly cancels the dipolepart of fP, and we obtainThe constant Po is unknown, but it does not contribute to the force on thesphere. From (5.144) we obtain(5.147)which is Stokes' law.122 THERMODYNAMICS AND KINETIC THEORYThe validity of (5.141) depends on the smallness of the material derivative ofu as compared to /L'V 2U. Both these quantities can be computed from (5.141). Itis then clear that we must require(5.148)Thus Stokes' law holds only for small velocities and small radii of the sphere. Amore elaborate treatment shows that a more accurate formula for F' is(3 puoa )F' = 67T/LaUo 1 + 8" -/L- + ... (5.149)The pure number puoa//L is called the Reynolds number. When the Reynoldsnumber becomes large, turbulence sets in and streamline motion completelybreaks down.PROBLEMS5.1 Make order-of-magnitude estimates for the mean free path and the collision time for(a) Hz molecules in a hydrogen gas in standard condition (diameter of Hz = 2.9 A);(b) protons in a plasma (gas of totally ionized Hz) at T = 3 X 105K, n = 1015protons/cm3, 0 = 'lTrz,where r = e2/kT;(c) protons in a plasma at the same density as (b) but at T = 107K, where thermo-nuclear reactions occur;(d) protons in the sun's corona, which is a plasma at T = 106K, n = 106protons/cc;(e) slow neutrons of energy 0.5 MeV in 238U (0"" 'lTrz, r"" 10-13cm).5.2 A box made of perfectly reflecting walls is divided by a perfectly reflecting partitioninto compartments 1 and 2. Initially a gas at temperature T1 was confined in compartment1, and compartment 2 was empty. A small hole of dimension much less than the mean freepath of the gas is opened in the partition for a short time to allow a small fraction of thegas to escape into compartment 2. The hole is then sealed off and the new gas incompartment 2 comes to equilibrium.(a) During the time when the hole was open, what was the flux dI of molecules crossinginto compartment 2 with speed between v and v + dv?(b) During the same time, what was the average energy per particle of the moleculescrossing into compartment 2?(c) After final equilibrium has been established, what is the temperature Tz in compart-ment 2?Answer. Tz = ~ T l .5.3 (a) Explain why it is meaningless to speak of a sound wave in a gas of strictlynoninteracting molecules.(b) In view of (a), explain the meaning of a sound wave in an ideal gas.TRANSPORT PHENOMENA 1235.4 Show that the velocity of sound in a real substauce is to a good approximation givenby c = 1/ .jP"s , where p is the mass density and "s the adiabatic compressibility, by thefollowing steps.(a) Show that in a sound wave the density oscillates adiabatically ifK CApCvwhere K = coefficient of thermal conductivityA = wavelength of sound wavep = mass densityCv = specific heatC = velocity of sound(b) Show by numerical examples, that the criterion stated in (a) is well satisfied in mostpractical situations.5.5 A flat disk of unit area is pla