Introduction Statistical Thermodynamics. Monday, January 6, 14

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1 Introduction Statistical Thermodynamics 1

2 Molecular Simulations Molecular dynamics: solve equations of motion Monte Carlo: importance sampling r 1 r 2 r n MD MC r 1 r 2 2 r n 2

3 3 3

4 4 4

5 Questions How can we prove that this scheme generates the desired distribution of configurations? Why make a random selection of the particle to be displaced? Why do we need to take the old configuration again? What is the desired distribution? How large should we take: delx? 5 5

6 Outline Rewrite History Atoms first! Thermodynamics last! Thermodynamics First law: conservation of energy Second law: in a closed system entropy increase and takes its maximum value at equilibrium System at constant temperature and volume Helmholtz free energy decrease and takes its minimum value at equilibrium Other ensembles: Constant pressure grand-canonical ensemble 6 6

7 Atoms first thermodynamics next 7

8 A box of particles Conservation of total energy We have given the particles an intermolecular potential Newton: equations of motion F(r) =-ru(r) m d2 r dt 2 = F(r) 8

9 Phase space Molecular: Thermodynamics: N,V,E Γ N = { r 1,r 2,,r N,p 1,p 2,,p } N { p 1,p 2,,p } N ( ) Γ N 0 point in phase space Γ N trajectory: classical mechanics ( t) { r 1,r 2,,r } N Why this one? 9

10 All trajectories with the same initial total energy should describe the same thermodynamic state { p 1,p 2,,p } N These trajectories define a probability density in phase space { r 1,r 2,,r } N 10

11 Intermezzo 1: phase rule Question: explain the phase rule? Phase rule: F=2-P+C F: degrees of freedom P: number of phases C: number of components Why the 2? 11

12 Making a gas What do we need to specify to fully define a thermodynamic system? 1. Specify the volume V 2. Specify the number of particles N 3. Give the particles: initial positions initial velocities More we cannot do: Newton takes over! System will be at constant: N,V,E (micro-canonical ensemble) 12

13 Pressure What is the force I need to apply to prevent the wall from moving? How much work I do? 13

14 Collision with a wall Elastic collisions: Does the energy change? What is the force that we need to apply on the wall? 14

15 Pressure one particle: 2 m v x # particles: ρ A v x 50% is the positive directions: 0.5 P A = F = ρ A m vx 2 Kinetic energy: UK = ½ m v 2 = ³ ₂ kb T (we define temperature) Pressure: P V = N k B T 15

16 Experiment (1) NVE1 NVE2 E1 > E2 What will the moveable wall do? 16

17 Experiment (2) NVE1 NVE2 E1 > E2 Now the wall are heavy molecules What will the moveable wall do? 17

18 Newton + atoms We have a natural formulation of the first law We have discovered pressure We have discovered another equilibrium properties related to the total energy of the system 18

19 Thermodynamics (classical) 19

20 Experiment NVE 1 NVE 2 E 1 > E 2 The wall can move and exchange energy: what determines equilibrium? 20

21 Classical Thermodynamics 1st law of Thermodynamics Energy is conserved 2nd law of Thermodynamics Heat spontaneously flows from hot to cold 21

22 Classical Thermodynamics Carnot: Entropy difference between two states: B S = S B S A = Using the first law we have: A dq rev T U = Q + W If we carry out a reversible process, we have for any point along the path du = TdS + dw If we have work by a expansion of a fluid du = TdS pdv 22

23 Hence, for the total system ds = ds L + ds H = dq Let us look at the very initial stage dq is so small that the temperatures of the two systems do not change For system H For system L 1 1 T L T H ds H = dq T H ds L = dq T L Heat goes from warm to cold: or if dq > 0 then TH > TL This gives for the entropy change: ds > 0 Hence, the entropy increases until the two temperatures are equal 23

24 Question Thermodynamics has a sense of time, but not Newton s dynamics Look at a water atoms in reverse Look at a movie in reverse When do molecules know about the arrow of time? 24

25 Thermodynamics (statistical) 25

26 Statistical Thermodynamics Basic assumption For an isolated system any microscopic configuration is equally likely Consequence... but classical All of statistical thermodynamics and equilibrium thermodynamics thermodynamics is based on laws 26

27 Ideal gas Let us again make an ideal gas Basic assumption We select: (1) N particles, (2) Volume V, (3) initial velocities + positions This fixes; V/n, U/n For an isolated system any microscopic configuration is equally likely 27

28 What is the probability to find this configuration? The system has the same kinetic energy!! Our basic assumption must be seriously wrong!... but are we doing the statistics correctly? 28

29 ... lets look at our statistics correctly What is the probability to find this configuration? Basic assumption: 1 P = total # of configurations number 1 can be put in M positions, number 2 at M positions, etc Total number of configurations: M N with the larger the volume of the gas the more configurations M = V dr 29

30 What is the probability to find the 9 molecules exactly at these 9 positions? V V N 30

31 What is the probability to find the 9 molecules exactly at these 9 positions? V V N 31

32 What is the probability to find the 9 molecules exactly at these 9 positions? V V N 32

33 What is the probability to find the 9 molecules exactly at these 9 positions? V V N 33

34 Question Is it safe to be in this room? 34

35 What is the probability to find this configuration? exactly equal as to any other configuration!!!!!! This is reflecting the microscopic reversibility of Newton s equations of motion. A microscopic system has no sense of the direction of time Are we asking the right question? 35

36 Are we asking the right question? These are microscopic properties; no irreversibility Thermodynamic is about macroscopic properties: Measure densities: what is the probability that we have all our N gas particle in the upper half? N P(empty) x x 0.5 x

37 Summary On a microscopic level all configurations are equally likely On a macroscopic level; as the number of particles is extremely large, the probability that we have a fluctuation from the average value is extremely low Let us quantify these statements 37

38 Basic assumption E 1 > E 2 NVE 1 NVE 2 Let us look at one of our examples; let us assume that the total system is isolate but heat can flow between 1 and 2. All micro states will be equally likely!... but the number of micro states that give an particular energy distribution (E1,E-E1) not so, we observe the most likely one... 38

39 In a macroscopic system we will observe the most likely one P(E 1,E 2 )= N 1 (E 1 ) N 2 (E E 1 ) E 1 =E E 1 =0 N 1(E 1 ) N 2 (E E 1 ) The summation only depends on the total energy: P(E 1,E 2 )=C N 1 (E 1 ) N 2 (E E 1 ) ln P(E 1,E 2 )=ln C + ln N 1 (E 1 )+ln N 2 (E E 1 ) We need to find the maximum d ln P(E 1,E 2 ) de 1 = d ln N (E 1,E 2 ) de 1 = 0 d [ln N 1 (E 1 )+ln N 2 (E E 1 )] de 1 = 0 39

40 We need to find the maximum d [ln N 1 (E 1 )+ln N 2 (E E 1 )] = 0 de 1 d ln N 1 (E 1 ) de 1 = d ln N 2(E E 1 ) de 1 As the total energy is constant E 2 = E E 1 de 1 = d(e E 1 )= de 2 Which gives as equilibrium condition: d ln N 1 (E 1 ) de 1 = d ln N 2(E 2 ) de 2 40

41 Let us define a property (almost S, but not quite) : Equilibrium if: or d lnn 1 ( E 1 ) S * = lnn( E) ( ) = d lnn E 2 2 de 1 de 2 * S 1 = S * 2 E 1 N1 E,V 1 2 N2,V 2 And for the total system: S * = S 1 * + S 2 * For a system at constant energy, volume and number of particles the S * increases until it has reached its maximum value at equilibrium What is this magic property S *? 41

42 Defined a property S * (that is almost S): S * (E 1, E E 1 ) = lnℵ(e 1,E E 1 ) = lnℵ 1 (E 1 ) + lnℵ 2 (E E 1 ) = S * 1 (E 1 ) + S * 2 (E E 1 ) Why is maximizing S * the same as maximizing N? The logarithm is a monotonically increasing function. Why else is the logarithm a convenient function? Makes S* additive! Leads to extensivity. Why is S * not quite entropy? Units! The logarithm is just a unitless quantity. 42

43 Thermal Equilibrium (Review) NVE 1 NVE 2 E 1 > E 2 Isolated system that allows heat flow between 1 and 2. ℵ(E 1,E E 1 ) = ℵ 1 (E 1 )iℵ 2 (E E 1 ) Number of micro states that give an particular energy distribution (E 1,E-E 1 ) is maximized with respect to E 1. 43

44 For a partitioning of E between 1 and 2, the number of accessible states is maximized when: * S 1 = S * 2 E 1 N1 E,V 1 2 N2,V 2 What do these partial derivatives relate to? Thermal equilibrium --> Temperature! Temperature de = TdS-pdV + T = E S V,N i M i=1 or µ i dn i 1 T = S E V,N i 44

45 Summary Statistical Mechanics: basic assumption: all microstates are equally likely Applied to NVE Definition of Entropy: S = kb ln Ω Equilibrium: equal temperatures 45

46 Question How large is Ω for a glass of water? For macroscopic systems, super-astronomically large. For instance, for a glass of water at room temperature Macroscopic deviations from the second law of thermodynamics are not forbidden, but they are extremely unlikely. 46

47 Systems at Constant Temperature (different ensembles) 47

48 The 2 nd law Entropy of an isolated system can only increase; until equilibrium were it takes its maximum value Most systems are at constant temperature and volume or pressure? What is the formulation for these systems? 48

49 fixed volume but can exchange energy Constant T and V 1 We have our box 1 and a bath Total system is isolated and the volume is constant First law Second law ds 0 Box 1: constant volume and temperature du = dq pdv = 0 1 st law: du 1 + du b = 0 or du 1 = du b The bath is so large that the heat flow does not influence the temperature of the bath + the process is reversible 2 nd law: ds 1 + ds b = ds 1 + du b T TdS 1 du

50 fixed volume but can exchange energy Constant T and V 1 Total system is isolated and the volume is constant Box 1: constant volume and temperature 2 nd law: TdS 1 du 1 0 Let us define the Helmholtz free energy: A For box 1 we can write da 1 0 d(u 1 TS 1 ) 0 Hence, for a system at constant temperature and volume the Helmholtz free energy decreases and takes its minimum value at equilibrium A U TS 50

51 Canonical ensemble 1/k B T Consider a small system that can exchange heat with a big reservoir Hence, the probability to find E i : 51 Boltzmann distribution 51

52 Thermodynamics What is the average energy of the system? Compare: 52 52

53 Thermodynamics First law of thermodynamics Helmholtz Free energy: 53 53

54 What is the average energy of the system? Compare: Hence: 54 54

55 Atoms? We have assumed that we can count states Quantum Mechanics: energy discreet What to do for classical model such as an ideal gas, hard spheres, Lennard-Jones? Energy is continue: potential energy kinetic energy Particle in a box: ε n = ( nh)2 8mL 2 What are the energy levels for Argon in a 1-dimensional box of 1 cm? 55

56 What are the energy levels for Argon in a 1-dimensional box of 1 cm? Argon: m=40 g/mol= kg h= J s ε n = ( nh)2 8mL 2 ε n = n 2 ( J ) Kinetic energy of Ar at room temperature J Many levels are occupied: only at very low temperatures or very small volumes one can see quantum effect! q translational = ( nh) 2 3D: q translational = 2π mk T B h 2 q translational = n=1 ( nh) 2 e 8mL 2 k B T e 8mL 2 k T B dn q translational = 2π mk T B 0 h V V = Λ 3 De Broglie wavelength 1 2 L 56

57 Partition function; q = e E n = e E n dn n=1 Hamiltonian: H = U kin +U pot = Z 1,V,T = C e p 2 i 2m k B T dp 3N p i 2 2m i +U pot e ( r N ) U p ( r) k T B dr 3N One ideal gas molecule: Up(r)=0 q translational = IG Z 1,V,T 2π mk T B h 2 = C e 2m k T B dp 3 2 V = V Λ 3 p 2 dr = CV ( 2π mk B T ) 3 2 IG Z 1,V,T = CV ( 2π mk B T ) 3 2 = V Λ 3 57

58 N gas molecules: Z N,V,T = 1 h 3N Z N,V,T = 1 h 3N N! e e p 2 2m k B T dp N p 2 2m k B T dp N e U( r) k T B dr N e U( r) k T B dr N WRONG! Particles are indistinguishable Configurational part of the partition function: Q N,V,T = 1 Λ 3N N! e U( r) k T B dr N 58

59 Question For an ideal gas, calculate: the partition function the pressure the energy the chemical potential 59

60 ideal gas molecules: IG Q N,V,T = V N Λ 3N N! All thermodynamics follows from the partition function! Free energy: F IG IG = k B T lnq N,V,T Pressure: Energy: F IG = F 0 + k B TN ln ρ p = F V T,N E = F T 1 T E = 3 2 Nk BT V,N = k B TN 1 V ( ) = k B TN ln Λ 3 ln V N = 3k B N ln Λ 1 T V,N Λ = h 2 2π mk B T

61 Chemical potential: µ i = F N i T,V,N j βf = N ln Λ 3 + N ln βµ = ln Λ 3 + ln ρ + 1 N V βµ IG = βµ 0 + ln ρ 61 61

62 Summary: Canonical ensemble (N,V,T) Partition function: Probability to find a particular configuration Free energy 62 62

63 Summary: micro-canonical ensemble (N,V,E) Partition function: Probability to find a particular configuration Free energy 63 63

64 Other Ensemble 64

65 Other ensembles? COURSE: In the thermodynamic limit the thermodynamic properties are MD and MC different independent of the ensemble: so buy a bigger computer ensembles However, it is most of the times much better to think and to carefully select an appropriate ensemble. For this it is important to know how to simulate in the various ensembles. But for doing this wee need to know the Statistical Thermodynamics of the various ensembles

66 Example (1): vapour-liquid equilibrium mixture T L L+V V Measure the composition of the coexisting vapour and liquid phases if we start with a homogeneous liquid of two different compositions: composition 66 How to mimic this with the N,V,T ensemble? What is a better ensemble? 66

67 Example (2): swelling of clays Deep in the earth clay layers can swell upon adsorption of water: How to mimic this in the N,V,T ensemble? What is a better ensemble to use? 67 67

68 Ensembles Micro-canonical ensemble: E,V,N Canonical ensemble: T,V,N Constant pressure ensemble: T,P,N Grand-canonical ensemble: T,V,µ 68 68

69 Constant pressure 69

70 Experimental Example P T Phase equilibria x1 x2 G L+G L x How do we measure vapor-liquid equilibria for a mixture? How to mimic this experiment in NVT conditions? Better solution: NPT ensemble 70

71 fixed N but can exchange energy + volume 1 We have our box 1 and a bath Total system is isolated and the volume is constant First law du = dq pdv = 0 Second law ds 0 Box 1: constant pressure and temperature 1 st law: du 1 + du b = 0 or du 1 = du b The bath is very large and the small changes do not change P or T; in addition the process is reversible 2 nd law: dv 1 + dv b = 0 or dv 1 = dv b ds 1 + ds b = ds 1 + du b + p T T dv b 0 TdS 1 du 1 pdv

72 fixed N but can exchange energy + volume 1 Total system is isolated and the volume is constant Box 1: constant pressure and temperature Let us define the Gibbs free energy: G For box 1 we can write 2 nd law: TdS 1 du 1 pdv 1 0 d(u 1 TS 1 + pv 1 ) 0 G U TS + pv dg 1 0 Hence, for a system at constant temperature and pressure the Gibbs free energy decreases and takes its minimum value at equilibrium 72

73 N,P,T ensemble Consider a small system that can exchange volume and energy with a big reservoir The terms in the expansion follow from the connection with Thermodynamics: S = k B lnω We have: S U V,N i ds = 1 T du + p T dv = 1 T and µ i T dn i S V E,N i = p T 73

74 lnω( V V i, E E ) i = lnω V, E ( ) lnω E lnω( V V i, E E ) i = lnω( V,E) E i k B T V,N p k B T V i E i lnω V E,N V i +! Hence, the probability to find E i,v i : 74

75 Partition function: Δ( N, P,T ) = exp i, j Ensemble average: V j exp E i i, j k V = B T pv j k B T Δ N, p,t ( ) E i k B T pv j k B T = k B T ln Δ p T,N Thermodynamics Hence: G k B T dg = SdT + Vdp + V = G p T,N µ i dn i = ln Δ N, p,t ) 75

76 Summary In the classical limit, the partition function becomes The probability to find a particular configuration: 76 76

77 grand-canonical ensemble 77

78 Grand-canonical ensemble Classical A small system that can exchange heat and particles with a large bath Statistical Taylor expansion of a small reservoir 78

79 exchange energy and particles 1 Constant T and μ Total system is isolated and the volume is constant First law du = TdS pdv + µdn = 0 Second law ds 0 Box 1: constant chemical potential and temperature 1 st law: du 1 + du b = 0 or du 1 = du b dn 1 + dn b = 0 or dn b = dn 1 The bath is very large and the small changes do not change μ or T; in addition the process is reversible 2 nd law: ds 1 + ds b = ds T b du b µ b T b dn b 0 79

80 ds 1 + ds b = ds T b du b µ b T b dn b 0 We can express the changes of the bath in terms of properties of the system ds 1 1 T du + µ 1 T dn 0 d ( TS U + µn ) G U TS + pv G = µn ( ) 0 d U TS µn For the Gibbs free energy we can write: Giving: d ( pv ) 0 or or pv = U TS µn d ( pv ) 0 Hence, for a system at constant temperature and chemical potential pv increases and takes its maximum value at equilibrium 80

81 lnω( E E i, N N j,) = lnω E, N µ,v,t ensemble Consider a small system that can exchange particles and energy with a big reservoir ( ) lnω E V,N The terms in the expansion follow from the connection with Thermodynamics: ds = 1 T du + p T dv µ T dn E i lnω N E,V S = k B lnω N j +! S U V,N = 1 T S N E,V = µ T 81 81

82 lnω( E E i, N N j,) = lnω E, N ( ) lnω E V,N E i lnω N E,V N j +! lnω( E E i, N N ) j = lnω( E, N ) E i k B T + µ k B T N j ( ) ln Ω E E, N N i j Ω E, N ( ) = E i k B T + µ k B T N j Hence, the probability to find E i,n j : P( E i, N ) j = k,l ( ) Ω E E i, N N j ( ) Ω E E k, N N l exp E i k B T + µn i k B T 82

83 μ,v,t ensemble (2) In the classical limit, the partition function becomes The probability to find a particular configuration: 83 83

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