Statistical Mechanics. Atomistic view of Materials

Transcription

1 Statistical Mechanics Atomistic view of Materials

2 What is statistical mechanics? Microscopic (atoms, electrons, etc.) Statistical mechanics Macroscopic (Thermodynamics) Sample with constrains Fixed thermodynamics variables Boundary-conditions du = TdS-pdV+μdN+ E, V, N, etc. (extensive variable, proportional to amount of material)

3 Micro Macro Relate microscopic phenomena to macroscopic properties Given a series of microscopic states, what is the corresponding macroscopic state? Given a thermodynamic state of a material, what are the probabilities of finding the system in the various possible microscopic states?

4 Thermodynamical ensembles Microcanonical ensemble: E(N,V,E) Isolated box Canonical ensemble: E(T,V,N) Box in a heat bath Grand canonical ensemble: E(T,V,μ) Open system

5 Review of probability theory i: space of possible states of the system Pi: ΣPi Pi Pi =

6 Review of probability theory i: space of possible states of the system Pi: probability of finding system in state i ΣPi = 1: sum of probability is 1 (consider discrete and finite states) Pi 0 Pi = Ni/N = number trials get i / total number of trials (limit large N)

7 Review of probability theory Quantity Fi associated with state i Coin: F(head) = +1 and F(tail) = -1 Average <F> =

8 Review of probability theory Quantity Fi associated with state i Coin: F(head) = +1 and F(tail) = -1 Average <F> = ΣPiFi (quantity weighted by the probability)

9 Microscopic probability: isolated system N,V,E Consider N atoms in a rigid container of volume V with constant energy E What is the probability of finding the system in a given microscopic state? {Ri,Pi} (count states) k m P H = P 2 2m kx2 X

10 Microscopic probability: Phase-space isolated system Number of states with energy E:

11 Microscopic probability: isolated system Discretize phase-space X P ~ ~ Number of states with energy E:

12 Microscopic probability: Phase-space isolated system X P ~ ~ Number of states with energy E: (E) = 1 ~ Z dxdp (E H(X, P))

13 Microscopic probability: isolated system N,V,E Consider N atoms in a rigid container of volume V with constant energy E What is the probability of finding the system in a given microscopic state? {Ri,Pi} (count states) Number of different microscopic states Z Z 1 (E,V,N)= N~ 3N d 3N R d 3N P (E H(R i,p i ))

14 Microscopic probability: isolated system N,V,E Consider N atoms in a rigid container of volume V with constant energy E Postulate: probability of the material being in any of the omega state are the same (equally likely)

15 Microscopic probability: isolated system N,V,E Consider N atoms in a rigid container of volume V with constant energy E Postulate: probability of the material being in any of the omega state are the same (equally likely) P ({R i,p i })= ( 1 (N,V,E) if H({R i,p i })=E 0 otherwise

16 Statistical mechanics 1 2 Consider a fictitious separation N1,V1,E1 N2,V2,E2=E-E1 that divides the system into 2 subsystems Energy can be exchanged between subsystem 1 & 2 What is the probability of subsystem 1 having the energy E1? P (E 1,E E 1 )= number of micro states with E 1 in subsystem 1 (N,V,E)

17 Statistical mechanics 1 2 Consider a fictitious separation N1,V1,E1 N2,V2,E2=E-E1 that divides the system into 2 subsystems Energy can be exchanged between subsystem 1 & 2 What is the probability of subsystem 1 having the energy E1? P (E 1,E E 1 )= number of micro states with E 1 in subsystem 1 (N,V,E) P (E 1,E E 1 )= 1(N 1,V 1,E 1 ). 2 (N 1,V 2,E 2 ) (N,V,E)

18 Statistical mechanics 1 2 N1 = N2 = 3 qtot = q1 + q2 = 6 7 microstates with q1 = 0, 1, 2,,6 What is the probability of subsystem 1 having the energy q1 P (E 1,E E 1 )= 1(N 1,V 1,E 1 ). 2 (N 1,V 2,E 2 ) (N,V,E) For each microstate 1 there are omega 2 micro states accessible by system 2

19 Statistical mechanics 1 2 N1 = N2 = 3 qtot = q1 + q2 = 6 7 microstates with q1 = 0, 1, 2,,6 Number of microstates 1 having energy 1?

20 Statistical mechanics 1 2 N1 = N2 = 3 qtot = q1 + q2 = 6 7 microstates with q1 = 0, 1, 2,,6 Number of microstates 1 having energy 1? =C q+n 1 q

21 Statistical mechanics 1 2 N1 = N2 = 3 qtot = q1 + q2 = 6 7 microstates with q1 = 0, 1, 2,,6 E1 Omega1 E2 Omega2 Omega Pi Total number of micro states: 462

22 Statistical mechanics N1 = 300, N2 = 200, qtot = 100 N1= 300 N2= 200 q1 Omega1 q2 Omega2 Omega Pi E E E E E E E E E E E E E E E E E E " 0.07" 0.06" 0.05" 0.04" 0.03" 0.02" 0.01" Séries1" E E E E E E E E- 25 0" 0" 20" 40" 60" 80" 100" 120" E E E E E E E E- 23

23 Statistical mechanics 1 2 Consider a fictitious separation N1,V1,E1 N2,V2,E2=E-E1 that divides the system into 2 subsystems Energy can be exchanged between subsystem 1 & 2 What is the probability of subsystem 1 having the energy E1? P (E 1,E E 1 )= 1(N 1,V 1,E 1 ). 2 (N 1,V 2,E 2 ) (N,V,E) Additive measure of number of states logp(e 1,E E 1 )=log 1 (N 1,V 1,E 1 )+log 2 (N 2,V 2,E 2 )+C

24 Statistical mechanics Equilibrium 0.08" 0.07" 0.06" " N1,V1,E1 N2,V2,E2=E-E1 0.04" 0.03" Séries1" 0.02" 0.01" 0" 0" 20" 40" 60" 80" 100" 120" Equilibrium: subsystems have the most likely energies: maximum of Pi

25 Statistical mechanics Equilibrium 0.08" 0.07" 0.06" " N1,V1,E1 N2,V2,E2=E-E1 0.04" 0.03" Séries1" 0.02" 0.01" 0" 0" 20" 40" 60" 80" 100" 120" Equilibrium: subsystems have the most likely energies: maximum of 1,E E 1 ) 1(N 1,V 1,E 1 ) 2(N 2,V 2,E 2 ) 1 Since E 2 = E E 2 In 1 (N 1,V 1,E 1 1 2(N 2,V 2,E 2 2

26 Statistical mechanics Equilibrium 0.08" 0.07" 0.06" " N1,V1,E1 N2,V2,E2=E-E1 0.04" 0.03" Séries1" 0.02" 0.01" 0" 0" 20" 40" 60" 80" 100" 120" Equilibrium: subsystems have the most likely energies: maximum of 1 (N 1,V 1,E 1 1 2(N 2,V 2,E (N 1,V 1,E 1 )= 2 (N 2,V 2,E 2 )

27 Entropy Statistical mechanics: microcanonical ensemble Temperature Pressure S = klog (N,V,E) Chemical = = (N,V,E) = µ = = 1 Ludwig Boltzmann ( ) (Image from wikipedia) kt

28 Micro-canonical ensemble N,V,E Equal probability postulate ( 1 (N,V,E) if H({R i,p i })=E P ({R i,p i })= 0 otherwise Relationship between macroscopic and microscopic world S = klog (N,V,E) (E) = 1 Z dxdp (E H(X, P)) ~

29 Canonical ensemble (Ale) E + Ebath = Etot = Constant system bath Probability of the system being in a microscopic state ({Ri},{Pi})? P ({R i }, {P i })= bath(e tot H({R i }, {P i })) tot Since E << Etot we expand log Ω(bath) around Etot logp({r i }, {P i })=log bath (E tot E=E tot E = log bath (E tot ) E kt P ({R i }, {P i })= e H({R i },{P i }) P micro states e H({R i },{P i }) Maxwell-Boltzmann distribution

30 Canonical ensemble (Ale) Maxwell-Boltzmann distribution P ({R i }, {P i })= Partition function Z(N,V,T) = micro X states e H({R i },{P i }) P micro states e H({R i },{P i }) e H({R i },{P i }) Lot of states with same energy (N,V,E)e E kt P(E)% E kt e Ω ( N, V, E) Z(N,V,T) = X E P(E) = (N,V,< E >)e <E> kt logz(n,v,t) =log (N,V,< E >) Helmholtz free energy <E> kt E E F (N,V,T) =<E> TS = ktlogz(n,v,t)

31 Canonical ensemble N system etc. bath Ei Probability distribution ~ counting total number of arrangements (microstates) n2 n1 S = C = N i n i X i P i logp i

32 Canonical ensemble N system etc. bath Ei 0 0 n2 = 1 n1 = 1 Probability distribution = counting total number of arrangements (microstates) C = N i n i N = 2, n1 = n2 = 1 C = 2

33 Canonical ensemble N system etc. bath Ei Probability distribution = counting total number of arrangements (microstates) C = N i n i n3 = 1 n2 n1 = 1 N = 3, n1 = n2 = n3 =1 C = 3

34 Canonical ensemble C = N i n i Let s approximate this expression NX Z N logn = X=1 logx 1 dxlogx =[XlogX X] N 1 NlogN N N =e NlogN e N = N N e N C = N N i n n i i

35 Canonical ensemble logc = logn N log i n i = logn N N X = logn N N n(i) is the number of particle in the state of energy E(i) logc = logn N = logn N N X NX i P i logp i i N X i i n i logn i P i NlogP i N P i N(logP i + logn) NlogN NX i P i = P i = n i N NS

36 Canonical ensemble S is a function of P(i) At equilibrium S is maximum Maximize C (or S) under constrains: X P i =1 i X P i E i =<E> i

37 Canonical ensemble 2 constrains: 2 Lagrange multipliers: NX F = P i logp i + ( X i P i 1) + ( X i P i E i <E>) i 1 (P 1 logp 1 + P 2 logp 1 (P 1 + P 1 (P 1 E 1 + P 2 E ) = logp i = logp i E i =0

38 Canonical ensemble logp i = ( + 1) E i P i = e ( +1) e E i Z = e ( +1) P i = e Z E i Maxwell-Boltzmann distribution P(E)% E beta tune the distribution

39 Canonical ensemble Partition function X i P i =1= X i e Z E i Z = X i e E i Average energy <E>= X i P i E i = P i E ie Z E i

40 Canonical ensemble Entropy S = X i P i logp i = X i e E i Z log e Ei Z S = X i e E i Z ( E i logz) S = X e Z E i E i + X e Z E i logz = <E>+logZ S = <E>+logZ Helmholtz free energy

41 Canonical ensemble: averages Consider a quantity that depends on atomic postions and momenta: A({R i }, {P i }) In equilibrium, the average value <a> is: A({R i }, {P i })= X microstates AP i = Pmicrostates A({R E i}, {P i })e P microstates e E Ensemble average When measure quantity A in experiment or MD simulation: Z 1 0 dta({r i (t)}, {P i (t)}) 1 N t X A({R i (t)}, {P i (t)}) t Time average Under equilibrium conditions temporal and ensemble averages are equal

42 Summary Microcanonical (NVE) Canonical (NVT) Isobaric/Isothermal (NPT) Probability distribution P ({P i }, {R i })= 1 (N,V,E) P ({R i }, {P i })= E kt e Z(N,V,T) P ({R i }, {P i },V)= e E PV kt Z(N,P,T) (N,V,E) = X micro (E H({R i }, {P i })) Z(N,V,T) = X micro e E kt Z(N,V,T) = X V X e E PV kt micro Free energy (macro micro) S = klog (N,V,E) F (N,V,T) = ktlogz G(N,P,T) = ktlogz

43 Equipartition of energy: Any degree of freedom that appears squared in the Hamiltonian contributes 1/2kT of energy Equipartition of energy Average value of quantity that appears squared in the Hamiltonian < P 2 1 >= H = P R d 3N Rd 3N P P 2 1 e R d 3N Rd 3N Pe Change of variable < P 2 1 >= 3NX i=2 H kt H kt P 2 i 2m + V = P H 0 = R R d 3N Rd 3N 1 Pe H0 kt dp1 R R d 3N Rd 3N 1 Pe H0 kt dp1 e P 2 1 e P1 2 r kt kt = x2, dp 1 = dx R q dx kt x 2 kte x2 R dxx 2 e x2 R q = kt R = 1 dx kt dxe x 2 e x2 2 kt P 2 1 kt P 2 1 kt

44 Equipartition of energy 3 dimension and N particle <K>= 3 2 NkT In most cases c.m. motion is set to zero at time zero (constant of motion it remains zero) Often angular momentum is zeroed (and remains zero) Instantaneous temperature: <K>= 3N 3 kt 2 <K>= 3N 6 kt 2 T (t) = 2K(t) N eff k