2. Thermodynamics. Introduction. Understanding Molecular Simulation

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1 2. Thermodynamics Introduction

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

3 How do we know our simulation is correct? Molecular Dynamics: if the force field is correct we follow the real dynamics of our system, if we simulate sufficiently long, we can compute the properties of interest Monte Carlo: what is the distribution we need to sample? how do we sample this distribution? Statistical Thermodynamics Importance Sampling

4 Outline 2. Thermodynamics 2.1. Introduction 2.2. Forces and Thermodynamics 2.3. Statistical Thermodynamics Basic Assumption Equilibrium 2.4. Ensembles Constant temperature Constant pressure Constant chemical potential

5 2. Thermodynamics 2.2 Forces and Thermodynamics

6 Atoms and Thermodynamics History: thermodynamics was first atoms came later Nicolas Léonard Sadi Carnot (wikipedia) Question: how would things have looked if atoms where first? Johannes van der Waals (wikipedia)

7 Phase space Point in phase space: Γ = { r 1,r 2,,r N,p 1,p 2,,p } N Γ N ( ) 0 Molecular dynamics: { } p,p,,p 1 2 N trajectory from classical mechanics from t=0 to t=t Γ N ( ) 0 Γ N ( ) t Γ N ( ) t { } r,r,,r 1 2 N

8 Thermodynamics? First law of thermodynamics The Gibbs phase rule Pressure Equilibrium, temperature

9 2. Thermodynamics Thermodynamics: first law of thermodynamics

10 The first law: a box of particles Our system: Isolated box with a volume V In which we put N particles the particles interact through a given intermolecular potential no external forces Newton: equations of motion F( r) = U( r) Consequence: Conservation of total energy E m d2 r dt 2 = F r ( ) NVE: micro-canonical ensemble

11 2. Thermodynamics Thermodynamics: Gibbs phase rule

12 Intermezzo 1: The Gibbs Phase Rule The Gibbs Phase Rule gives us for a thermodynamic system the number of degrees of freedom Phase rule: F=2-P+C F: degrees of freedom P: number of phases C: number of components Example: boiling water P = 2 (water and steam) C = 1 (pure water) F=2-2+1=1 Wikipedia Hence, if we fix the pressure all other thermodynamic variables are fixed (e.g., temperature and density) Question: why is there the 2?

13 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)

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

15 Discussion Question Newton s equation of motion are symmetric in time: if we look at a movie of water molecules in forward or reverse we do not see a difference! But a macroscopic system we do see the difference

16 2. Thermodynamics Thermodynamics: pressure

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

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

19 Pressure one particle: 2 m vx # particles: ρ A vx 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 kb T

20 Question How does the pressure change if we have interacting particles: because of the repulsive forces? because of the attractive forces?

21 2. Thermodynamics Thermodynamics: equilibrium

22 Experiment NVE1 NVE2 E1 > E2 What will the moveable (isolating) wall do?

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

24 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

25 Experiment NVE1 E1 > E2 NVE2 The wall can move and exchange energy: what determines equilibrium?

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

27 Classical Thermodynamics The first law: ΔU = Q + W If we carry out a reversible process: Carnot: Entropy difference between two states: ΔS = S A S B = B A dq rev T If we have work by a expansion of a fluid du = TdS + dw du = TdS pdv

28 Equilibrium H Hence, for the total system: Heat goes from warm to cold: or if dq > 0 then TH > TL This gives for the entropy change: L 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 ds H = dq T H ds L = dq T L ds = ds L + ds H = dq 1 1 T L T H ds 0 Hence, the entropy increases until the two temperatures are equal

29 2. Thermodynamics 2.3 Statistical Thermodynamics

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

31 2. Thermodynamics Statistical Thermodynamics: Basic Assumption

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

33 What is the probability to find this configuration? The system has the same energy as the previous one!! Our basis assumption states that this configuration is equally likely as any other configuration But having all atoms in the corner of our system seems to be very unlikely. and very dangerous Our basic assumption must be seriously wrong!

34 Question: How to compute the probabilities of a particular configuration? Use a lattice model to make the counting the number of possible confirmations easier Assumptions: the position of a molecule is given by the lattice site there is no limit in the number of molecules per lattice site

35 Question: what is the probability of a given configuration? Basic assumption: P = particle number 1 can be put in 1 Total # of configurations 4 M positions, number 2 at M positions, etc. For N particle the total number of configurations is: M N Hence the probability is: P = 1 M 4 Question: how does the statistics change if the particles are indistinguishable?

36 Question: What are the probabilities of these configurations?

37 and this one? Is there a real danger that all the oxygen atoms are all in one part of the room?

38 Are we asking the right question? Thermodynamic is about macroscopic properties: These are averages over many configurations 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

39 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

40 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 now quantify this

41 Question If all configurations are equally likely what will be then the energy we will observe in the two boxes?

42 2. Thermodynamics 2.2 Statistical Thermodynamics: Equilibrium

43 Discussion: equilibrium (1) 1 We have a closed and isolated system 2 Basic Assumption: every configuration is equally likely Question: How do we know the system is in equilibrium? How can we determine the macroscopic properties (e.g. density of the two systems)

44 Discussion: equilibrium (2) We have a closed and isolated system, but heat can flow between system 1 and 2 NVE1 NVE2 Basic Assumption: every configuration is equally likely: Questions: How do we know the system is in equilibrium? how does this tell us what will what will be the macroscopic properties (e.g., temperature) of the two systems?

45 Solution: 1 2 NVE1 NVE2 All micro states are equally likely!... but the number of micro states that give an particular density or energy distribution over the 2 systems are not...

46 Macroscopically we will observe the most likely one P( E 1,E ) The probability to find E1 in 2 volume 1 and E2 in volume 2 N ( 1 E ) The number of configurations 1 that result in an energy E1 in volume 1. NVE1 NVE2 E 2 = E E 1 The total energy E is constant P( E 1,E ) 2 = E 1 =E E 1 =0 N ( 1 E 1 )N ( 2 E E ) 1 N ( 1 E 1 )N 2 E E 1 ( ) = CN ( 1 E 1 )N ( 2 E E ) 1 Experimentally we will observe the most likeley configuration; which is given by the maximum We need to find: dp ( E 1,E ) 2 = 0 de 1

47 Finding: dp ( E 1,E ) 2 = 0 de 1 with: P( E 1,E ) 2 = CN ( 1 E 1 )N ( 2 E E ) 1 Is equivalent in finding: or: with E2=E-E1 ( ( )) dln P E 1,E 2 = 0 de 1 ( ( )) ( ( )) dln N 1 E 1 + dln N E E 2 1 = 0 de 1 de 1 ( ( )) ( ( )) dln N 1 E 1 dln N E E 2 1 = 0 de 1 de 2 The solution of this equation gives the energies in volume 1 and 2 that are most likely, i.e., the largest number of configurations have these energies

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

49 Defined a property S * (that is almost S): ( ) = ln N( E 1,E E ) 1 S * E 1,E E 1 Question 1: Why is maximising S * the same as maximising N? Answer: The logarithm is a monotonically increasing function. Question 2: Why is the logarithm a convenient function? Answer: makes S* additive! Leads to extensivity. Question 3: Why is S * not quite entropy? Answer: Units! The logarithm is just a unitless quantity. S = k B S * = k B ln( N( E) ) ( )

50 For a partitioning of E between 1 and 2, the number of configuration is maximized when: S 1 * E 1 = S * 2 E N 1,V 2 1 N 2,V 2 What do these partial derivatives relate to? Temperature de = TdS pdv + i=m i=1 µ i dn i T = E S N i,v or 1 T = S E N i,v Thermal equilibrium equal temperature of system 1 and 2! S = k B S * = k B ln( N( E) )

51 Question: How large is N(E) for a glass of water? How to estimate N(E) Number of molecules of the order Make a grid of 10 6 cells (100x100x100) Summary: N( E) ( 10 6 ) For macroscopic systems N(E) is super-astronomically large Macroscopic deviations from the second law of thermodynamics are not forbidden, but they are extremely unlikely.

52 2. Thermodynamics 2.4 Ensembles

53 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? When do we have equilibrium for such a system?

54 Other ensembles? In the thermodynamic limit the thermodynamic properties are independent of the ensemble: so buy a bigger computer 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.

55 Example (1): vapour-liquid equilibrium mixture P L T=constant L+V V composition vapour(v)-liquid (L) Phase diagram of a binary mixture at (fixed) temperature T Question: if we have a 50%-50% mixture how many degrees of freedom if we have V-L coexistence? Experiment: determine the composition of the coexisting vapour and liquid phases if we start with a homogeneous liquid of two different compositions but the same temperature: How to mimic this with the N,V,T ensemble? What is a better ensemble?

56 Example (2): adsorption in porous media Question: What are the equilibrium conditions water adsorbed in clay layers Experiments: 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?

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

58 2. Thermodynamics Ensembles: constant temperature

59 Canonical ensemble: classical thermodynamics 1 Our entire system is isolated (NVE), but our subsystems (box 1 and bath) can exchange energy First law du = TdS pdv fixed volume but can exchange energy Second law Box 1: constant volume and temperature 1 st law: du = 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 = ds 1 + ds b 0 or ds 0 ds 1 + ds b = ds 1 + du b T = ds 1 du 1 T 0 Giving: TdS 1 du 1 0 we have a criteria that only depends on box 1

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

61 Canonical ensemble: statistical mechanics E 1 Hence, the probability to find E 1 : Consider a small system that can exchange energy with a big reservoir =1/kBT lnω( E 1,E E ) 1 = lnω E ( ) lnω lnω E 1,E E 1 E ( ) ( ) lnω E E +! 1 If the reservoir is very big we can ignore the higher order terms: = E 1 k B T ( ) = Ω ( E,E E ) 1 1 ( ) P E 1 i Ω E i,e E i ( ) Ω( E) ( ) Ω( E) = Ω E,E E 1 1 Ω E i,e E i i ( ) ( ) = C Ω E,E E 1 1 Ω E ( ) exp E 1 P E 1 k B T β=1/kbt exp βe 1

62 Thermodynamics What is the average energy of the system? E ( ) E i P E i = i i i E i exp βe i Classical thermodynamics: F T 1 T F T 1 T = T 2 = F + TS = U ( ) ( ) exp βe i i ( ) = ln exp βe i β F T T = T 2 F T T βf = lnq NVT F T = lnq NVT β df = SdT pdv F T N i,v = S The link between statistical and classical thermo

63 From states to 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?

64 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 q translational = n=1 e ( nh) 2 8mL 2 k B T Many levels are occupied: only at very low temperatures or very small volumes one can see quantum effect! q translational = ( nh) 2 8mL e 2 k T B dn q translational = 2πmk T B 0 h L 3D: q translational = 2πmk B T h L 3 = V Λ 3 de Broglie wavelength

65 Partition function: q = E n = e k T B dn e k B T n=1 0 E n we assume that the potential energy does not depend on the velocity Hamiltonian: H = U kin + U pot = N p i 2 ( ) + U i=i 2m pot r N one atom: Z 1,V,T = C e H k B T dp 3 dr 3 p 2 = C e 2mk T B dp 3 e U pot ( r) k B T dr 3 one ideal gas atom: Up(r)=0 Z 1,V,T ideal gas = C e p2 2mk B T dp 3 1dr 3 p 2 ( ) 3 2 = CV e 2mk T B dp 3 = CV 2πmk B T Compare: q translational = 2πmk B T h V = V Λ 3 if we define C=1/h 3 ideal Z gas 1,V,T = CV ( 2πmk B T ) 3 2 = V Λ 3

66 N gas molecules: wrong: particles are indistinguishable Z N,V,T = 1 h 3N e H k B T dp 3N dr 3N if we swap the position of two particles we do not have a new configuration! Z N,V,T = 1 h 3N N! e H k B T dp 3N dr 3N Configurational part of the partition function: Q N,V,T = 1 Λ 3N N! e U( r) k T B dr 3N

67 Probability to find a particular configuration Partition function: 1 Q N,V,T = Λ 3N N! e U( r) k T B dr 3N Probability to find configuration R N P( R N ) = 1 1 Q N,V,T Λ 3N N! δ ( R N r N )e U( r N ) k T B dr 3N e U( R N ) k B T As expected we get the Boltzmann factor

68 Intermezzo: Stirling s approximation In general N is a large number: ln( N! ) = lnn + ln( N 1) + ln( N 2) +!+ ln1 ( ) N = ln( i) ln N! lnx dx n=1 1 N = xlnx x 1 N = NlnN N + 1 NlnN Hence, for large N we will use: ln( N! ) NlnN

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

70 Ideal gas molecules: All thermodynamics follows from the partition function! Free energy: This is the (absolute) reference state of the free energy: F 0, which only depends on temperature Pressure: Thermo: Energy: Thermo: U = Giving: ideal Q gas N,V,T = 1 Λ 3N N! F ideal gas = k B T lnλ 3 k B T ln V N N! p = F V F T 1 T = F 0 + k B TNln N V U = 3 2 Nk B T T,N e 0 dr 3N = V N Λ 3N N! = F 0 + k B TNlnρ p = k B TN V U = 3k B N lnλ 1 T with h 2 lnλ = ln 2πmk B T For N! we can use Stirling s approximation 1 2 Ideal gas law = C ln 1 T

71 Chemical potential: Thermo µ = F N T,V For an ideal gas we have: βf ideal gas = NlnΛ 3 + Nln N V βµ ideal gas = βµ 0 + lnρ

72 Summary: Canonical ensemble (N,V,T) Partition function: Q N,V,T = 1 Λ 3N N! e U( r) k T B dr 3N Probability to find a particular configuration: P( R N ) e U( R N ) k B T Ensemble average: A N,V,T = 1 Λ 3N N! A( r)e U( r) k T B dr 3N = A r Q N,V,T ( )e βu ( r ) dr 3N e βu ( r ) dr 3N Free energy: βf = lnq NVT

73 Summary: micro-canonical ensemble (N,V,E) Partition function: Q N,V,E = 1 h 3N N! δ ( E H( p 3N,r 3N ))dp 3N dr 3N Probability to find a particular configuration P( P N,R N ) 1 Free energy S = k B lnq NVE

74 2. Thermodynamics Ensembles: constant pressure

75 Constant: T and p 1 We have our system (1) and a bath (b) Total system is isolated and the volume is constant fixed N but can exchange energy + volume First law Second law Box 1: constant pressure and temperature 1 st law: du 1 + du b = 0 or du 1 = du b dv 1 + dv b = 0 or dv 1 = dv 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: ds 1 + ds b = ds 1 + du b T + p T dv b 0 TdS 1 du 1 pdv 1 0 du = dq pdv = 0 ds 0

76 1 fixed N but can exchange energy + volume Total system is isolated and the volume is constant 2 nd law: Box 1: constant pressure and temperature TdS 1 du 1 pdv 1 0 d( U 1 TS 1 + pv ) 1 0 Let us define the Gibbs free energy: G For box 1 we can write 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

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

78 lnω( V V 1,E E ) 1 = lnω( V,E) lnω E V,N E 1 lnω V E,N V 1 +! lnω( V V 1,E E ) 1 = lnω( V,E) 1 k B T E p 1 k B T V +! 1 ( ) ln Ω V V,E E 1 1 Ω( V,E) = E 1 k B T pv 1 k B T Hence, the probability to find E i,v i : ( ) = Ω ( V V,E E ) 1 1 P V 1,E 1 i,j ( ) Ω V V i,e E j ( ) Ω( V,E) ( ) = Ω V V,E E 1 1 Ω V V i,e E j i,j ( ) Ω V,E = Ce ( ) 1 k B T E 1 +pv 1 P( V 1,E ) 1 e β ( E 1 +pv 1 )

79 Partition function: ( ) = e Δ N,p,T i,j ( ) 1 k B T E i +pv j Ensemble average: V = ( ) V j e 1 k B T E i +pv j i,j e 1 k B ( T E i +pv j ) i,j = k B T dg = SdT + Vdp ( ) p lnδ N,p,T µ i dn i N,T Thermodynamics V = G p N,T Hence: G = k B T lnδ( N,p,T )

80 Summary In the classical limit, the partition function becomes Q( N,p,T ) = 1 Λ 3N N! dv e βpv dr N e βu ( r N ) The probability to find a particular configuration: P( r N,V ) e ( ) β pv+u rn The link to thermodynamics: G = k B T lnδ( N,p,T )

81 2. Thermodynamics Ensembles: constant chemical potential

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

83 Constant: T and μ 1 Total system is isolated and the volume is constant exchange energy First law du = TdS pdv + µdn = 0 and particles 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 1 = dn b 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 = ds 1 + ds b = ds 1 + du b T µ dn b T 0

84 ds = ds 1 + ds b = ds 1 + du b T µ dn b 0 T We can express the changes of the bath in terms of properties of the system ds 1 + du 1 T + µ dn 1 T 0 d TS 1 U 1 + µn 1 d U 1 TS 1 µn 1 For the Gibbs free energy we can write: ( ) 0 ( ) 0 Giving: G U TS + pv G = µn 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

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

86 lnω( N N 1,E E ) 1 = lnω( N,E) lnω E E 1 lnω N N +! 1 lnω( N N 1,E E ) 1 = lnω( N,E) E 1 k B T + µn 1 k B T +! ( ) ln Ω N N,E E 1 1 Ω( N,E) = 1 k B T E µn 1 1 Hence, the probability to find E 1,N 1 : ( ) ( ) = Ω ( N N,E E ) 1 1 P N 1,E 1 i,j ( ) Ω N N i,e E j ( ) Ω( N,E) ( ) = Ω N N,E E 1 1 Ω N N i,e E j i,j ( ) Ω N,E = Ce ( ) 1 k B T E 1 µn 1 P( N,E) β( E µn) Ce

87 In the classical limit, the partition function becomes Q ( µ,v,t ) = N=0 e βµn Λ 3N N! dr N βu( ) rn e The probability to find a particular configuration: P( N,r N ) e β U( rn ) µn