浙江大学物理系本科《统计力学》讲义-chapter2.pdf

Thermodynamics and Statistical Physics B Zheng 1 Zhejiang Institute of Modern Physics Zhejiang University Hangzhou 310027 P R China 1e mail zheng tel 0086 571 87952753 Fax 0086 571 87952754 1 Chapter 1 Thermodynamics 2 Chapter 2 Classical statistical mechanics From thermodynamics one does not understand every thing E g what is pressure and especially what are temperature and entropy Kinetic theory of molecules off ers certain understanding but not suffi ciently gen eral Thermodynamics does not look fundamental and systematic and all the laws seem isolated each other 2 1Postulates Statistical physics equilibrium non equilibrium Statistical mechanics here is mainly concerned with the equilibrium state Calculate macroscopic parameters from microstruc tures and interactions 3 Derive thermodynamics e g defi ne and calculate Temperature Internal energy Entropy Prove that the laws in thermodynamics Statistical mechanics goes beyond thermodynamics The system a classical system isolated in the sense that the energy is conserved or nearly conserved is com posed of a large number N of elements typically N 1023molecules Thermodynamic limit N V but V N v a constant Here V is measured in a microscopic unit e g V 1023molecular volumes The thermodynamic properties of the system may be tackled from three levels Fundamental level Solve the microscopic equations of motion such as New ton Heisenberg Eqs Diffi culties too many degrees of freedom 4 microscopic initial conditions and boundary conditions irregular disturbance from environments Here the time t is microscopic Quasi fundamental level Do not trace the motion of each molecule and con sider only the probability distribution pi qi t with pi qi t Y i d3pid3qi being the number of molecules at time t and lying within a volumeQid3pid3qiof the coordinate qiand mo mentum pi Equations of motion Liouville s eq Boltzmann eq These eqs can be solved only in some simple cases such as dilute gases The time t is mesoscopic Statistical mechanics Do not solve any eqs of motion but assume a form of pi qi t in the equilibrium state i e pi qi p q it can be tested by experiments it can be derived from eqs of motion in some special cases More strictly this is the so called ensemble theory The time t is macroscopic space the phase space spanned by p q each point in space represents a microscopic state of the system 5 An ensemble A collection of systems identical in composition and macroscopic conditions but existing in diff erent microscopic states without interactions each other A system can be represented by a point in space then p q dpdq pi qi Y i dpidqi is the number of systems at the volume element dpdq Physically it is reasonable to do this just because a macroscopic state may correspond to many microscopic states The ensemble is introduced to replace the dynamic evolution of the microscopic states Postulate of equal a priori probability When an isolated system is in thermodynamic equi librium its state is equally like to be any state satisfying the macroscopic conditions i e p q const if E H p q E 0otherwise here EH is the Hamitonian and the ensemble described by this distribution is the so called microcanomical ensemble why is introduced Theoretically it is convenient Practically isolation is not strict why is p q a const for possible states 6 One may understand from ergodicity which may be achieved by the measure of non ergodic states is negligibly small disturbance boundary conditions interactions If there is ergodicity it is natural that the practical path of a system should be a simple loop in space The ensemble average of a measurable property f p q is defi ned as R dpdqf p q p q R dpdq p q In the thermodynamic limit it is usually assumed 2 2 1 We may defi ne the most probable value of f p q as the value with the maximum probability P f0 Here we should note that P f is diff erent from p q as shown in Fig 2 1 In the thermodynamic limit the ensemble average and the most probable value should be nearly the same Otherwise statistical mechanics should be questioned Question in what case Eq is not valid Answer strongly correlated systems 7 Figure 2 1 8 Figure 2 2 9 2 2Microcanomical ensemble H is naturally defi ned as the internal energy So it is important how to defi ne the entropy and temperature Let us denote the volume in space of the micro scopic ensemble E Z E H p q E dpdq E is understood to be dependent of N V and also For example it is shown in Fig 2 2 for H p q p2 q2 The entropy is defi ned by S E V k log E where k is a universal constant eventually shown to be Boltzmann s constant To justify the defi nition of S we should show a S is extensive b S satisfi es the properties of the entropy as re quired by the second law of thermodynamics Proof a Let the system be divided into two subsystems 1 N1 V1 H1 2 N2 V2 H2 with N N1 N2 V V1 V2 and H p q H1 p1 q1 H2 p2 q2 10 Here it is assumed that the interaction between two subsystems is negligible p1 q1 are the coordinates of particles in the system 1 and p2 q2 are the coordinates of particles in the system 2 E g if the intermolecular potential is fi nite range and the surface to volume ratio of each subsystem is negligibly small If we defi ne S1 E1 V1 k log 1 E1 S2 E2 V2 k log 2 E2 and S E V k log E 2 we should show in the thermodynamic limit S E V S1 E1 V1 S2 E2 V2 If E E1 E2is a decomposition of the energy the volume in space of the whole system is Z E1 E2 H1 H2 E1 E2 2 dp1dp2dq1dq2 Z E1 H1 E1 dp1dq1 Z E2 H2 E2 dp2dq2 1 E1 2 E2 This means that the extensive property of S has been proven if the two subsystems are added up without in teractions The key point is that the possible decomposition of the energy E E1 E2is not unique when the system is divided into two subsystems 11 Let us divide E into E intervals with Ei i i 1 E then E E X i 1 1 Ei 2 E Ei We will show or have to show that only one term in the sum is dominating Reading materials Let the largest term be 1 E1 2 E2 with E1 E2 E then 1 E1 2 E2 E E 1 E1 E2 or S1 E1 V1 S2 E2 V2 S E V S1 E1 V1 S2 E2 V2 k log E In the thermodynamic limit we expect log 1 N1 log 2 N2 E N1 N2 then S E V S1 E1 S2 E2 O log N In summary the extensive property of S is based on that a decomposition E E1 E2of the energy is domi nating when the system is divided into two subsystems 12 with fi xed N1 N2and V1 V2 Such a dominating decom position is expected from Eqs In thermodynamic equilibrium the system is homo geneous therefore N V In other words E N1 N2 N is simply the ex tensive property of E However log 1 N1 V1in indicates already that S1is extensive Therefore derivation of E E1 E2 Sounds not very convincing It tells only that if S1and S2are extensive then S is also extensive End reading materials Understanding Probably alternatively we may think E1and E2are the averaged energy of the two subsystems In the thermodynamic limit the fl uctuation of E1around E1should be much smaller than E1 Let us take then only 1 E1 2 E2 is dominating E E X i 1 1 Ei 2 E Ei Therefore S E V S1 E1 V1 S2 E2 V2 with E E1 E2 This also explains why we can add two systems up in equilibrium Question why is it only an understanding 13 Answer 1 Ei 2 E Ei is just P E1 Ei the prob ability E1takes the value Ei When we assume it indicates already that all other terms are negligible Reading materials Why E E X i 1 1 Ei 2 E Ei e g N 2 H 1 2m 2 X i 1 p2 1 1 2 2 2 X i 1 q2 i H1 1 2mp 2 1 1 2 2q2 1 H2 1 2mp 2 2 1 2 2q2 2 E Z E H E 2 dp1dq1dp2dq2 1 E1 Z E1 H1 E1 dp1dq1 2 E E1 Z E E1 H2 E E2 dp2dq2 If E1 is fi xed E 1 E1 2 E E1 If not E E X i 1 1 Ei 2 E Ei 14 since E1takes value from 0 to E corresponding to dif ferent states End reading materials Actually this implies that E E1 E2maximizes the function 1 E1 2 E2 under the restriction E E1 E2 0 i e 1 E1 2 E2 0 with E1 E2 0 This leads to E1 log 1 E1 fl fl fl fl E1 E1 E2 log 2 E2 fl fl fl fl E2 E2 or S1 E1 E1 fl fl fl fl E E1 S2 E2 E2 fl fl fl fl E2 E2 We defi ne the temperature of any system by S E V E 1 T Then S1 E1 S E2 is simply the zeroth law T1 T2 T defi ned in this way is an intensive parameter and S E 1 T is also one of the Maxwell relations in thermodynamics If S is correctly defi ned T should be also correct 15 2 3Thermodynamics Let us defi ne X E Z H p q E dpdq E P E E Then E E E X E X E It can be proved that up to an additive const of the order O logN the following defi nitions are equivalent S k log E S k log E S k log X E why question to think about Keeping in mind that the energy does not fl uctuate so much it is obvious that P E P E1 P E2 b With the defi nition S E V k log X E it is easy to show that S never decreases i e the second law for a thermally isolated system in thermodynamics Proof For our system considered parameters are N E V By defi nition of an isolated system N and E 16 can not change V can not decrease Therefore the second law here is simply stated as that S is a non decreasing function of V This is obvious for P E is a non decreasing function of V by its defi nition S is really entropy Assuming that the system is changed slowly by cou pling the system to external environments Then it is a quasi static process dS E V S E V dE S V E dV 1 T dE S V E dV Defi ne the pressure of the system to be P T S V E then dS 1 T dE PdV or dE TdS PdV This looks like the fi rst law Question Is it reasonable to defi ne P T S V E Hint P is intensive 17 Exercise prove P E V S If we accept the defi nition of P then S should be en tropy by the fi rst law In other words the fi rst law should be also an as sumption in statistical mechanics 2 4Equipartition Theorem Reading materials H H p q H pj qj j 1 3N e g H 1 2m X p2 i or H 1 2m X p2 i 1 2 2 X q2 i Let xibe either pior qi xi H xj 1 E Z E H E dpdq xi H xj E 1 Z H E dpdq Z H E dpdq xi H xj E E Z H E dpdq xi H xj 18 Z H E dpdq xi H xj Z H E dpdq xi xj H E Z H E dpdq xj xi H E ij Z H E dpdq H E the fi rst term Z H E dpdq xi H E 0 xi H xj ij E E Z H E dpdp E H ij E Z H E dpdq 1 Z E H E dpdq E H the second term is negligible ij E X E ij 1 E log P E ij k S E ijkT Ifi j xi H xi kT If H X i AiP 2 i X i BiQ2 i Pi Qiare canonical conjugate variables Then X i Pi H Pi Qi H Qi 2H If f of the const Aiand Biare non zero 1 2fkT 19 exerciseProve the equipartition theorem for H 1 2mp 2 1 2 2q2 by explicitly calculating the ensemble average 2 5Classical ideal gas H 1 2m N X i 1 p2 i X E Z H E d3p1 d3pNd3q1 d3qN V N Z H E d3p1 d3pN Let R 2mE X E V N 3N R n R is the volume of an n sphere of radius R n R CnRn Cn 2 n 2 n 2 1 Z is the gamma function logCn n n 2 log n 2 log n 2 n 2 20 hence X E C3N h V 2mE 3 2 iN S E V k log X E Nk log V 4 mE 3N 3 2 3 2Nk U S V E 3N 4 mV 2 3 exp 2S 3Nk 1 T U S V 2U 3Nk U 3 2NkT CV 3 2Nk P U V S NkT V 2 6Maxwell Boltzmann distribution N identical molecules volume V H p q E The system may be described by a microcanonical en semble For quasi independent systems H X i Hi pi qi where Hi pi qi represents the Hamiltonian of each molecule There could be a number of atoms in a molecule but 21 the interactions between two molecules are negligible E g the simplest case is H 1 2m X i p2 i space the phase space spanned by the single molecule coordinates p q A microscopic state of a single molecule can be rep resented by a point in the space A microscopic state of the system is described by a set of the points Since the energy of a molecule is bounded by E the points are confi ned to a fi nite region of space We divide the region into K elements of volume d3pd3q and denote the number of the molecules in a element by nl then K X l 1 nl N K X l 1 lnl E Here the molecules are assumed to be quasi independent lis the energy level of a single particle It is important to note that a microscopic state of the system may be described by a set of nl but a set of nl corresponds to not only one microscopic state e g interchange of two molecules leads to new states That is a given set of nl corresponds to a volume in space which is called the volume occupied by nl To describe a macroscopic state we need to aver age over the microcanonical ensemble i e all possible microscopic states In other words we should obtain then calculate all macroscopic quantities 22 However it is diffi cult to perform this average We assume that the equilibrium state is described by the most probable distribution nl which occupies the largest volume in space Why This is some what similar to the case when we cal culate E 1 E1 2 E2 by dividing the system into two subsystems For a fi xed l if the relative fl uctuation 2 2 is still not small enough we increase the total num ber N to reduce it Finally should be equal to the most probable nl The procedure a Calculate the volume of nl b Maximize it to obtain the most probable nl The volume of nl nl N QK l 1nl K Y k 1 gnk k where gk is introduced for convenience and will fi nally be put to 1 Understanding There are N ways of distributing N distinguishable molecules to N positions However N positions form K groups with the distribution nl Inside a group there are nl ways of distributing nlmolecules 23 For a large nl lognl nl lognl 1 i e nl nnl l log nl N logN 1 K X l 1 nl lognl 1 K X l 1 nlloggl const Now we vary nl under the condition of PK l 1nl N and PK l 1 lnl E to fi nd the most probable nl We introduce the Lagrange multipliers and and calculate log nl K X l 1 nl K X l 1 lnl 0 Now we consider all nlare independent each other K X l 1 lognl loggl l nl 0 log nl l nl e l Finally and can be determined by the conservation of the total particles and the total energy To prove that nl maximizes nl we can simply calculate the second variation K X l 1 1 nl nl 2 0 24 Note that nlis only the function of l does not generally de pends on p q This is important in the equilibrium state The molecules tends to gather in the lower energy states 2 7Boltzmann statistical theory N X l nl X l e l E X l l nl X l le l Let us assume l p q y y yk represent macroscopic external parameters Defi ne the partition function of a single particle Z y X l e l Z E dpdqe p q y then N e Z y or log Z y N E N logZ y Suppose the system is changed very slowly 25 dE X l nld l X l ld nl The fi rst term represents the interaction with the ex ternal environment X l nld l X l k nl l yk dyk X k X l nl l yk dyk Question why