5. Systems in contact with a thermal bath

Transcription

1 5. Systems in contact with a thermal bath So far, isolated systems (micro-canonical methods) 5.1 Constant number of particles:kittel&kroemer Chap. 3 Boltzmann factor Partition function (canonical methods) Ideal gas (again!) Entropy = Sackur-Tetrode formula Chemical potential 5.2 Exchange of particles: Kittel&Kroemer Chap. 5 Gibbs Factor Grand Partition Function or Gibbs sum (grand canonical methods) 5.3 Fluctuations Phys 112 (S2009) 05 Boltzmann-Gibbs 1

2 In class A System in Contact with a Reservoir R, U S, The system+ reservoir are isolated and in equilibrium The system is in one state of energy How does the number of states of the reservoir evolves when increases? 2

3 In class A System in Contact with a Reservoir R, U S, The system+ reservoir are isolated and in equilibrium The system is in one state of energy How does the number of states of the reservoir evolves when increases? A: Increases B: Decreases C: It depends on the conditions 3

4 In class A System in Contact with a Reservoir R, U S, The system+ reservoir are isolated and in equilibrium The system is in one state of energy How does the number of states of the reservoir evolves when increases? B: Decreases (very fast) The total energy is U 0 = U + = Constant U = U 0 4

5 In class A System in Contact with a Reservoir R, U S, The system+ reservoir are isolated and in equilibrium How does probability for the system to be in a state of energy varies with? 5

6 In class A System in Contact with a Reservoir R, U S, The system+ reservoir are isolated and in equilibrium How does probability for the system to be in a state of energy varies with? A: Increases when increases B: Decreases when increases C: Is constant when increases 6

7 In class A System in Contact with a Reservoir R, U S, The system+ reservoir are isolated and in equilibrium How does probability for the system to be in a state of energy varies with? B: Decreases (very fast) The probability of having the system in a state of energy and the reservoir at an energy U = U 0 is proportional to the number of states g R ( U 0 ) 1 7

8 Ideal Gas Thermal Bath Consider 1 particle in thermal contact with a thermal bath made of an ideal gas with a large number M of particles We assume that the overall system is isolated and in equilibrium at temperature. We call U the total energy (which is constant). We would like to compute the probability of having our particle in a state of energy. Because in equilibrium, all states of the overall system are equiprobable, this is equivalent to computing the number of states such that our particle has energy and the thermal bath E=U-. Using the fact that for an ideal gas of energy E and particle number N p, the number of states is g Res exp g E 3/2 N p and that U = 3 2 N p we obtain for large N p prob ( ) ( U ) 3 2 N p N p 3 2 N p N p exp Phys 112 (S2009) 05 Boltzmann-Gibbs 8

9 R, U o - # States in configuration 1 g R U o Fundamental postulate (or H theorem): at equilibrium Prob configuration Notes: exp s =Boltzmann factor Z=partition function Boltzmann factor S, Probability of each state not identical! <= interaction with bath Total system =Our system S + reservoir R isolated Configuration: 1 state of S + energy of reservoir ( ) ( ) g R ( U o ) ( ) ( ) = g R ( U o 1) g R ( U o 2 ) = exp σ R ( U o 1) σ R ( U o 2 ) ( ) ( ) exp σ R ( 2 1 ) U o = exp 1 ( 2 1) Prob state 1 Prob state 2 Prob state 1 Prob state 2 Prob state s ( ) = 1 Z exp s with Z = all states s ( ) = 1 exp Z exp s k s B T Phys 112 (S2009) 05 Boltzmann-Gibbs 9

10 Examples 2-state system Z = 1+ exp exp U = = 1 + exp = C V = dq dt V (,N ) State 1 has energy 0 State 2 has energy = U T exp + 1 V (,N ) U = k b V (,N ) 1 Pr ob( 1) = 1 + exp 0 2 exp = k b k b T exp k b T ( ) k b T ( ) ( ) = exp Pr ob( 2) 1 = ( ) 1 + exp 1 + exp Prob(0) 1 C V Prob() T Phys 112 (S2009) 05 Boltzmann-Gibbs 10

11 In class A System in Contact with a Reservoir R, U S, The system+ reservoir are isolated and in equilibrium Why does the probability that S in a quantum state energy decreases exponentially with? 11

12 In class A System in Contact with a Reservoir R, U S, The system+ reservoir are isolated and in equilibrium Why does the probability that S in a quantum state energy decreases exponentially with? A: Because as you increase, the energy and the number of states in the reservoir increases B: Because as you increase, the energy of the reservoir decreases and the number of states in the reservoir increases C: Because as you increase, the energy and the number of states in the reservoir decreases 12

13 In class A System in Contact with a Reservoir R, U S, The system+ reservoir are isolated and in equilibrium Why does the probability that S in a quantum state energy decreases exponentially with? C: Because as you increase, the energy and the number of states in the reservoir decreases U R = U 0 ( ) = quickly increasing function of U R g R e.g., ideal gas U R 3/2 N 13

14 Examples Free particle at temperature Prob Assumption: No additional degrees of freedom volume element [ ] dv, p, p + dp, dω Momentum interval e.g. no spin, not a multi-atom molecule solid angle element = Prob dv,[, + d], dω volume element energyy interval solid angle element = Prob dv,[ v,v + dv], dω volume element velocity interval solid angle element = 1 Z exp p2 2M Boltzmann factor does not vary appreciably over dp = 2M 3/2 Z = M 3 Z exp exp dv 1 2 Mv2 dvp 2 dpdω h 3 number of spatial quantum states in phase space element ddω h 3 dvv 2 dvdω h 3 dω = d cosθdφ Maxwell-Boltzmann distribution KK p 392 Normalizing to N particles, the probability to find one particle per unit volume in the velocity interval dv and solid angle dω is 3 M 2 Mv 2 f (v)dvdω = n 2π exp 2 v2 dvdω with n = N V Phys 112 (S2009) 05 Boltzmann-Gibbs 14

15 In class Maxwell Distibution Why does the Maxwell distribution has a v 2 3 M 2 Mv 2 f (v)dvdω = n 2π exp 2 v2 dvdω with n = N V 15

16 In class Maxwell Distibution Why does the Maxwell distribution has a v 2 3 M 2 Mv 2 f (v)dvdω = n 2π exp 2 v2 dvdω with n = N V A: This has to do with the number of states in the reservoir B: The density of quantum states per unit velocity interval in the system S goes to zero as v 2 when v goes to zero C: This comes from the normalization 16

17 In class Maxwell Distribution Why does the Maxwell distribution has a v 2 B: The density of quantum states per unit velocity interval in the system S goes to zero as v 2 when v goes to zero 3 M 2 Mv 2 f (v)dvdω = n 2π exp 2 v2 dvdω with n = N V N = 1 M = 1 = 1 integrated over angles Velocity Energy / v

18 In class Maxwell Distribution Why doesn t the system S have a fixed energy? v 18

19 In class Maxwell Distribution Why doesn t the system S have a fixed energy? Which statement is the most accurate A: The system is not isolated B: The uncertainty principle prevents this C: The reservoir keeps kicking the system S D: There is constant exchange of energy between S and the reservoir, and the typical energy of these exchanges is v 19

20 In class Maxwell Distribution Why doesn t the system S have a fixed energy? The most accurate statement is D D: There is constant exchange of energy between S and the reservoir, and the typical energy of these exchanges is Example: System S is one particle in a bath of N particles 3 The average particle energy is, and in collisions the typical energy exchanged is a fraction 2 of this. More general: Energy exchange by various excitations Particle collision Electrons = Fermi Dirac bath, electron-like and hole-like excitations Photons Phonons (quantized vibrations in a solid) v 20

21 Another interpretation of the Boltzmann Distribution We derived it from number of states in the Reservoir But each excitation in the reservoir will have a typical energy Thermal fluctuations of the order of Difficult to get >> Prob exp Phys 112 (S2009) 05 Boltzmann-Gibbs 21

22 Definition Partition Function Z = exp s all states s If states are degenerate in energy, each one has to be counted separately Relation with thermodynamic functions Energy U = = 1 Z s exp s = logz s 1 = 2 logz Entropy e s s σ = p s log p s = log e s Z s Z = log Z + U = ( log Z) Free Energy F = U σ = log Z σ = F U = 2 log Z = 2 F as implied by the thermodynamic identity Chemical potential µ = F = log Z N,V N Phys 112 (S2009) 05 Boltzmann-Gibbs 22 T

23 Examples Energy of a two state system Z = 1+ exp U = = 2 log Z Free particle at temperature Assumption: No additional degrees of freedom e.g. no spin, not a multi-atom molecule Method 1: density in phase space Energy: = p x 2 + p 2 2 y + p z 2M d 3 q d 3 p Z 1 = h 3 exp = V h 3 dp x e p 2 x 2M Z 1 = V 2πM 2 M = V 3 = 2πM dp y e p 2 y h 2 2π 2 log Z 2π 2 = 2 1 Method 2: quantum state in a box: see Kittel-Krömer (chapter 3 p 73) Same result, of course! 3 2 = VnQ with nq = = 2M exp 1+ exp M dp z e 3 2 p z 2 2 M = exp + 1 = 3 2 Phys 112 (S2009) 05 Boltzmann-Gibbs 23

24 Partition Function Power comes from summation methods Direct e.g. geometric series Few term approximation useful if f decreases fast Integral approximation i=0 r i i=0 f ( i) f (x)dx 0 m r i m i=0 = lim = lim m 1 r m+1 1 r = 1 1 r f ( i) = f ( 0) + f ( 1) + f ( 2) +... i=0 Other example: Harmonic oscillator Consider an harmonic oscillator ω The probability of mode of energy s ω, s integer 0 is Prob( s) = 1 Z s ω exp Z = exp s ω 1 = s=0 1 exp ω log( Z ) U = s ω = 2 = ω exp ω 1 Phys 112 (S2009) 05 Boltzmann-Gibbs 24

25 Applies to classical gases in equilibrium with no potential Start from Boltzmann distribution 1 exp Prob (1 particle to be in state of energy )= Z 1 with Z 1 = d3 q d 3 p h 3 exp = V M 3 2 2π 2 = VnQ Probability of => finding this particle in volume dv and between p and p + dp = 1 exp Vn Q Change of variable to velocities: Probability of finding particle in v, v+dv, dω Maxwell distribution Maxwell Distribution K&K p. 392 For N particles, the volume density between p and p + d p = N V p = Mv Particle density distribution in velocity space Note: 1) Could get the factor in front directly as normalization 1 exp n Q d 3 p h 3 = n exp n Q d 3 p 1 = n h 3 2π M 3 f (v)dvdω = n M 2 Mv 2 exp 2π v 2 dvdω 2 3/2 dvd 3 p h 3 Number of states exp d 3 p 2) Note that no h factor! Mv 2 3) One can easily compute Equipartition! Phys 112 (S2009) 05 Boltzmann-Gibbs 25 2 = 3 2 = 3 2 k BT

26 Individual States of probability p s Density of States + d D( )d It is convenient to replace expressions such as by an integral over the energy : p s f s p s D s 0 p s f s (f s any function of the state s) s ( ) f s ( ) D( )d ( )d is the number of states between and + d and is called the density of states. We replace by p s p s s < s < +d Phys 112 (S2009) 05 Boltzmann-Gibbs ( ) D ( )d since for small enough d, p s does not vary Note that if you ask what is the probability p( )d for the system to be between energy between and + d p( )d = p s i.e., p ( ) D ( ) p s ( ) ( )d For example, for Boltzmann distribution and a non relativistic ideal gas (monoatomic,spinless) p s ( ) = 1 Z exp and using p = 2M d 3 p = p 2 dpd cosθdϕ = p 2 dpdω D( )d = V d 3 x dω p 2 dp angles h 3 = V 4π dp V 4π 1 p 2 d = ( h 3 d h 3 2 2M )3/2 d = V 4π M 3/2 d

27 In class Partition Function for 2 systems How do I get the partition function for 2 systems in weak interactions? v 27

28 In class Partition Function for 2 systems How do I get the partition function for 2 systems in weak interactions? A: the sum of partition functions B: the product of partition functions C: the sum of the logs D: the product of partition functions divided by 2! v 28

29 In class Partition Function for 2 systems How do I get the partition function for 2 systems in weak interactions? B: the product of partition functions if the system are distinguishable D: the product of partition functions divided by 2! if the systems are indistiguishable v 29

30 Partition Function for N systems Weak interaction limit 2 sub systems in weak interaction: <=>States of subsystems are not disturbed by interaction (just their population) <=> Ψ 1+2 ( n 1,n 2 ) =Ψ 1 ( n 1 )Ψ 2 ( n 2 ) with n 1, n 2 still meaningfull => Partition functions multiply for distinguishable systems Distinguishable systems Suppose first that we can distinguish each of the individual components of the system: e.g. adsorption sites on a solid, spins in a lattice For 2 subsystems Z = exp s 1 + s2 Z ( 1 + 2) = Z ( 1)Z ( 2) s 1 s 2 and for N systems ( ) = ( Z ( 1) ) N Z N Phys 112 (S2009) 05 Boltzmann-Gibbs 30

31 Examples Free energy of a paramagnetic system In a magnetic field B, = mb if m is the magnetic moment. For 1 spin Z 1 = exp mb + exp mb = 2cosh mb For N spins Z=Z 1 N since they are distinguishable U = 2 log Z Z = 2 N cosh N mb = NmB tanh mb Magnetization M = U VB mb = nm tanh M M m B mb Phys 112 (S2009) 05 Boltzmann-Gibbs 31

32 Indistinguishable Systems Indistinguishable:in quantum mechanics, free particles are indistinguishable We should count permutations of states among particles only once s 1 s 1 s 2 e.g. for 2 particles Z ( 1)Z ( 2) = e e overcounts s 1,s 2 s 2, s 1 for s 1 s 2 we have to divide by 2 s 2 More generally if we have N identical systems, the Nth power of Z 1 can be expanded as N! (a a s +...a m ) N n = a 1 n 1..a s n s...a m m n i n 1!...n s!...n m! If we have a large number of possible states, most of the terms are of the form N!a i a j...a k N terms corresponding to N different states which are singly occupied. If the multiple occupation of states is negligible we are indeed overcounting by N! Therefore if we have N indistinguishable systems ( ) 1 ( N! Z ( 1 )) N Z N This is a low occupation number approximation due to Gibbs ( no significant probability to have 2 particles in the same state: we will see how to treat the case of degenerate quantum gases) Phys 112 (S2009) 05 Boltzmann-Gibbs 32

33 In class Why is this approximate Why is the division of the product by N! approximate? v 33

34 In class Why is this approximate Why is the division of the product by N! approximate? A: It is actually rigorous B: This is a large N approximation C: This is a low concentration approximation: no two systems in the same states D: The systems are interacting, therefore the product is approximate v 34

35 In class Why is this approximate? Why is the division of the product by N! approximate? C: This is a low concentration approximation: the probability of having two systems in the same state is negligible For 2 systems Z 1 2! exp s 1 + s2 Z ( 1+ 2) Z ( 1)Z 2 s 1 s 2 2! OK for 2 different states 1,2 is indinguishable from 2,1 exp exp has to be divided by 2! But the terms where we have the same state exp should not be divided by 2 ( ) v 35

36 N atoms Ideal Gas Indistinguishable Z N = 1 N! Z 1 ( 1 )N = ( N! n QV) N log N! N log N N log Z N = N log V n Q + N log Z N = N log n Q n + N with n = N N V = concentration M n Q = 2π 2 Interpretation: n Q as quantum concentration de Broglie wave length 3 2 λ = h Mv h 3M 1 2 Mv λ 3 3M h M 2 2π 2 = nq Phys 112 (S2009) 05 Boltzmann-Gibbs 36

37 Ideal gas laws Energy Free energy Pressure U = 2 log Z = 3 2 N = 3 2 Nk BT equipartition of energy 1/2 per degree of freedom F = log Z = N log n Q N n p = F V = N V pv = N = Nk BT with n Q = M 2π Entropy: Sackur Tetrode if extra degrees of freedom, additional terms! Well verified by experiment T S = Involves quantum mechanics 0 Chemical potential σ = S = F k B = N log n Q + 5 n 2 µ = F N = Phys 112 (S2009) 05 Boltzmann-Gibbs 37 dq T N log n V Q identical to our expression counting states N N N = log n n Q

38 In class Why use F Why not use? µ = σ N with σ = S = F k B = N log n Q n v 38

39 In class Why use F Why not use? µ = σ N Why is it different? with σ = S = N log n Q k B n Don t we get log n Q n N N = log n 3 n Q 2 A: we cannot use µ = σ N B: you can use it but with the proper independent variables being fixed C: We are missing something v 39

40 In class Why use F Why not use? Why is it different µ = σ N µ = σ N with σ = S = N log n Q k B n σ = log n Q N,V n N N = log n 3 n Q 2 σ = log n Q N,V n N N B: you can use it but with the proper independent variables being fixed µ = σ = N U,V N N log M = log n n Q = log n 3 n Q 2 2U 2π 2 3N v 40 3/2 V N = log n Q n N N 3N 2N + 5 2

41 Barometric Equation: Three different ways µ=constant Thermodynamic equilibrium between slabs at various z At constant temperature µ total =C log n( z) Single particle occupancy of levels at different altitude n( z) prob( K, z) exp + Φ( z) K ( ) = n mgz 0 exp Hydrodynamic equilibrium pda mgn( z)dadz ( p + dp)da Generalizes to case when the temperature is not constant n Q = Φ( z) + C n z n z pda ( p + dp)da mgn( z)dadz = 0 1 n z pv = N p( z) = n z ( ) 1 n z ( ) ( ) = n 0 exp mgz ( ) n z = mg p z = mg Phys 112 (S2009) 05 Boltzmann-Gibbs 41

42 Internal Degrees of Freedom (Optional) In general, we need to add internal degrees of freedom In most cases, the kinetic and internal degrees of freedom are independent, for one particle e.g. for multi-atomic molecules Rotation 10-2 ev Rotation energy j = j( j + 1) 0 with multiplicity 2 j+1 Z int = ( 2 j+1) exp j( j + 1) 0 j ( 2 j+1) exp j( j + 1) 0 1/2 dj = exp( x) x 0 o dx with x = j( j + 1) 0 dx = ( 2 j + 1) 0 dj, x 0 = 0 4 o Z int exp o 4 o Vibration ev similar to harmonic oscillator n=0 1 exp ω Spin in a magnetic field etc.. for >> 0 for >> 0 U = 2 log Z = C rot = k B For < < 0, keep the first terms of the discrete sum Z int 1+ 3exp 2 o +... log Z int 3exp 2 o U = 2 log Z 6 0 exp 2 o C rot 12k o b Z int = exp n ω = 1 ω for >> ω C vib = k B Z = Z Z kin int kinetic C V internal 2 exp 2 o 0-1/ j Diatomic molecules cf. KK fig 3.9 k B Rotation Slightly better approximation than 0 Vibration k Phys 112 (S2009) 05 Boltzmann-Gibbs 42 log

43 In class Evolution of the number of states of the Reservoir R, U o -s,nio-nis) S, Total system =Our system + reservoir Configuration: 1 state s of S of energy s and # of particles N is + energy of reservoir Exchange of energy and particles Question: How does the number of states of the reservoir change when we increase the energy of the system S?(the number of particles in system S stay constant) A: the number of states decreases B: the number of states increases C: the number of states stay constant 43

44 In class Evolution of the number of states of the Reservoir R, U o -s,nio-nis) S, Total system =Our system + reservoir Configuration: 1 state s of S of energy s and # of particles N is + energy of reservoir Exchange of energy and particles Question: How does the number of states of the reservoir change when we increase the energy of the system S? (the number of particles in system S stay constant) A: the number of states decreases g( R,U 0 s ) = g R,U 0 ( )exp σ U s g( R,U 0 )exp s > 0 44

45 In class Evolution of the number of states of the Reservoir R, U o -s,nio-nis) S, Total system =Our system + reservoir Configuration: 1 state s of S of energy s and # of particles N is + energy of reservoir Exchange of energy and particles Question: How does the number of states of the reservoir change when we increase the number of particles in the system S?(the energy in system S stay constant) A: the number of states decreases B: the number of states increases C: it depends 45

46 In class Evolution of the number of states of the Reservoir R, U o -s,nio-nis) S, Total system =Our system + reservoir Configuration: 1 state s of S of energy s and # of particles N is + energy of reservoir Exchange of energy and particles Question: How does the number of states of the reservoir change when we increase the number of particles in the system S?(the energy in system S stay constant) C: it depends g( R,U 0 s, N 0 N s ) = g R,U 0 For an ideal gas µ = log True also for Bose Einstein gas ( )exp σ R U s exp σ R N N s g( R,U 0 )exp s exp µn s > 0 n n Q < 0 the number of states in reservoir decreases when N sincreases But for Fermi Dirac µ =Fermi level > 0 the number of states in reservoir increases when N s increases 46

47 Exchanging Particles with an Ideal gas Let us consider again a system with an energy level, which can now be occupied by a variable number N of particles We put in contact with a thermal reservoir constituting on a large number N p -N of free particles. Energy and particles can be exchanged with the reservoir. We consider a single state with N particles and an energy N We can compute the number of states of the overall system as a function of Ν and. M 2( U N) 2π 2 3(N g = g Res 1 = e 5/2 p N) N p N V N = e 5/2 1+ N p N 5/2 3/2 N p N N p N M 2π 2 2U 3N p N p V Direct Counting <= or Seckur Tetrode 3/2 N p N 1 N U 3/2 N p N = g Res ( ) N exp µ N p e n N p 5/2 Q n exp N log n Q N U = 3 n exp exp N ( µ ) lim 1+ a N exp a N 2 N p b N exp( N logb) Phys 112 (S2009) 05 Boltzmann-Gibbs 47 N ( ) N

48 R, U o - Phys 112 (S2009) 05 Boltzmann-Gibbs Gibbs factor System in contact with thermal bath (reservoir) S, Exchange of particles # States in configuration 1 g R ( U o s, N oi N is ) Prob( configuration) g R ( U o s, N oi N is ) At equilibrium Pr ob( state 1, N i 1 ) g Pr ob( state 2, N i 2 ) = R ( U o 1, N oi N i1 ) σ g R ( U o 2, N oi N i 2 ) = exp R σ ( 1 2 ) R ( N i1 N i 2 ) U i N i Prob state s with, N is Notes: ( ) = 1 ( s, N is ) µ i N is Z exp i with Z = all states s,n is Absolute activity λ i = exp µ i Z = grand partition function s, N is exp ( ) µ i N is i 48 Total system =Our system + reservoir Configuration: 1 state s of S of energy s and # of particles N is + energy of reservoir = 1 s, N is Z exp k B T ( ) µ i N is i

49 Definition Grand Partition Function Z = all states s and occupation N is exp i ( ) µ i N is s, N is Note: (s,ν) depends on N i both through through increase of N, and the interactions In weak interaction limit (ideal gas) and only 1 species, ( ) = N s s s, N s Z = exp µ s energy of each level Prob N s particles in state s µ i N i Note If the N is =N i are constant, Z = exp i Z ( N i ) the probabilities probabilities are the same as with Z s,n s This formalism is useful only if the N is change. ( ) N s ( ) = 1 Z exp ( s µ ) N s Phys 112 (S2009) 05 Boltzmann-Gibbs 49

50 Examples few state systems Kittel-Kroemer Gas adsorption (heme in myoglobin, hemoglobin, walls) Assume that each potential adsorption site can adsorb at most one gas molecule. What is the fraction of sites which are occupied? State 1= 0 molecule => energy 0 State 2= 1 molecule => energy Fraction occupied = f = Z = 1+ exp µ ( ) ( ) = 1 exp( µ ) +1 exp µ 1 + exp µ = same form as Fermi- Dirac Phys 112 (S2009) 05 Boltzmann-Gibbs 50

51 In class How is µ determined? A: The density of O 2 in the blood B: The vapor pressure in the atmosphere C: Not enough information to say v 51

52 In class How is µ determined? A: The density of O 2 in the blood <= dissolved gas is in equilibrium with absorption sites If ideal (applies equally well to gas and dissolved gas and any low concentration species) µ = log n p = log pv = N n = N gas n Q n Q V = P Langmuir adsorption isotherm f = p n Q exp ( ) + p v 52

53 In class Comparison with 2 level system Adsorption Fraction occupied = f = exp µ 1 + exp µ 2 level system: no µ! ( ) ( ) = 1 exp( µ ) +1 = same form as Fermi- Dirac exp Pr ob( 2) = 1 + exp ( ) ( ) = 1 ( ) 1 + exp 53

54 In class Adsorption dependences Adsorption Fraction occupied = f = exp µ 1 + exp µ Adsorption dependence on µ? if µ increases (higher concentration or vapor pressure) A: Fraction of occupied sites increases B: Fraction of occupied sites decreases C: Fraction is constant ( ) ( ) = 1 exp( µ ) +1 = same form as Fermi- Dirac 54

55 In class Adsorption dependences Adsorption Fraction occupied = f = Adsorption dependence on µ? if µ increases (higher concentration or vapor pressure) A: Fraction of occupied sites increases Denominator decreases -> e.g., oxygen mask ( ) ( ) = 1 exp( µ ) +1 exp µ 1 + exp µ = same form as Fermi- Dirac 55

56 In class Adsorption dependences Adsorption Fraction occupied= f = 1 exp µ ( ) + 1 = p n Q exp ( ) + p Adsorption dependence on? if increases (keeping pressure constant) A: Fraction of occupied sites increases B: Fraction of occupied sites decreases C: It depends on the sign of µ 56

57 In class Adsorption dependences Adsorption Fraction occupied= f = 1 exp µ ( ) + 1 = p n Q exp ( ) + p Adsorption dependence on? if increases B: Fraction of occupied sites decreases µ = log n Q n > 0 < 0 =-binding energy exp( ) < 1 1 p p Fraction occupied= f = exp( µ = ) + 1 n exp ( ) + p n + p Q Q (the larger the binding energy, the bigger is the fraction of occupied site) f p n Q + p n n Q from above At large, this is small! Most of the sites are emptied by the thermal fluctuation 57

58 Ionized impurities in a semiconductor cf Kittel and Kroemer p370 Donor= tends to have 1 more electron (e.g. P in Si) State 1= ionized D + : no electron => energy= 0 State 2=neutral 1 extra electron spin up => energy = d State 3= neutral 1 extra electron spin down => energy = d 1 Fraction ionized f = 1+ 2 exp d µ 2 spin states when donor is neutral d = energy of electron on donor site Acceptor= tends to have 1 fewer electron (e.g. Al in Si) State 1= ionized A - : one electron completing the bond=> energy= a State 2= 1 neutral: missing a bond electron spin up => energy =0 Z =1+ 2exp µ d K&K use ionization energy I=- d State 3= 1 neutral: missing a bond electron spin down => energy =0 exp µ a 1 Fraction ionized f = exp µ = a exp µ a 2 spin states when acceptor is neutral a = energy of electron on acceptor site µ determined by electron concentration in semiconductor = Fermi level Phys 112 (S2009) 05 Boltzmann-Gibbs 58

59 Grand Partition Function Relation to thermodynamic functions Mean number of particles Entropy σ = N i = N is p s,nis = s N is s N is p s,nis log( p s,nis ) = log Z s N is Z exp N is i µ i N is s, N is ( ) ( ) µ i N is i + = log Z = log Z µ i,v V,µ i Energy U = = 2 log Z + µ i N i = 2 i + i µ i µ i log Z Free Energy Grand Potential Ω defined as F = U σ = log Z + µ i N i G = F + pv i Ω F µ i N i i = log Z Phys 112 (S2009) 05 Boltzmann-Gibbs 59

60 In class Note Why ( ) while = log Z? σ = log Z V,µ i Ω F µ i N i i 60

61 In class Note Why ( ) while = log Z? σ = log Z V,µ i Ω F µ i N i i A: Just an accident B: F = Ω + µ i N i and σ = F i = Ω since µ i N i i C: This is a consequence of the Thermodynamic identity = 0 61

62 In class Note Why ( ) while = log Z? σ = log Z V,µ i Ω F µ i N i C: This is a consequence of the Thermodynamic identity df = σd pdv + µ i dn i σ = Ω V,µi B: is terribly wrong µ i N i i V,µ i 0 i You are mixing independent variables σ = F V,Ni i Ω = F µ i dn i dω = σd pdv N i dµ i i = Ω V,µi First expression is not useful because N i fluctuates i 62

63 Microscopic exchanges Fluctuations In a system in equilibrium, exchanges still go on at the microscopic level! They are just balanced! Macroscopic quantities= sum of microscopic quantities N x mean e = s U = = s p s s s Z But with a finite system, relative fluctuations on such sum is of the order 1/ N e.g., fluctuation on total energy Variance not entropy! σ 2 E = ( E E ) 2 = E 2 E 2 2 e s = s Z e s s s s Z Computation with partition function; By substitution we can see that σ 2 E = E 2 E 2 = 2 U 2 2 log Z = Similarly when there is exchange of particles N = log Z log Z and σ 2 N = N 2 N 2 µ = = N µ µ µ,v Phys 112 (S2009) 05 Boltzmann-Gibbs 63 s 2 U = E = 2 log Z

64 Fermi Dirac Examples Fermions= particles with half integer spin Pauli exclusion principle: each state contains at most one particle For a state of energy Z = 1 + exp µ N = log Z µ,v σ N 2 = N µ = exp µ = 1+ exp µ exp µ exp µ + 1 = 1 exp µ + 1 = Prob( 1 particle) = N 1 N 2 ( ) N Less fluctuation than a Poisson distribution Phys 112 (S2009) 05 Boltzmann-Gibbs 64

65 Bose Einstein Bosons= particles with integer spin For a state of energy [ Z = exp µ ]N 1 = N 1 exp µ N = log Z exp µ 1 = µ,v 1 exp µ = exp µ 1 σ N 2 = N µ = exp µ exp µ 1 = N 1 + N 2 ( ) N More fluctuation than a Poisson distribution Phys 112 (S2009) 05 Boltzmann-Gibbs 65

66 In class Note Does a system in thermal contact with a reservoir goes to a state of minimum energy? 66

67 In class Note Does a system in thermal contact with a reservoir goes to a state of minimum energy? A: Of course, this is a law of physics! B: Yes, because this maximizes the number of states in the reservoir C: No, because of the constant exchange of energy. D: No, because of the Boltzmann factor 67

68 In class Note Does a system in thermal contact with a reservoir goes to a state of minimum energy? C or D C: No, because of the constant exchange of energy. The system is constantly kicked D: No, because of the Boltzmann factor The Boltzmann factor does express this fact A: Of course, this is a law of physics! is wrong because the system not only can emit energy (and go to minimum energy) but also can receive energy from the reservoir. B: Yes, because this maximizes the number of states in the reservoir The state of S of minimum energy is the most probable but other states are possible with smaller probabilities 68

69 Minimization of the Free Energy Consider a system in interaction with reservoir imposing temperature Constant exchange of energy (+ in some case volume, particles) => Fluctuations, probability distribution Contrary to a system which is able to loose energy to vacuum, system does not evolve to state of minimum energy (constantly kicked)! Contrary to an isolated system, does not evolves to a configuration of maximum entropy! (The entropy of the combination of the reservoir and the system is maximized) It evolves to a configuration of minimum Free Energy Balance between tendency to loose energy and to maximize entropy. More rigorously We cannot define the temperature of the system out of equilibrium. So we introduce F L ( U S ) U S R σ S U S G L ( ) Landau Free Energy (constant volume) ( U S,V s ) U S R σ S ( U S,V S ) + p R V S Landau Free Enthalpy (constant pressure) They are minimized for the most probable configuration( equilibrium)! Can be generalized to any variable! (Will be useful for phase transitions) Phys 112 (S2009) 05 Boltzmann-Gibbs 69

70 Landau Free Energy Let us consider a system S exchanging energy with a large reservoir R. (We are no more considering a single state for system S). Constant volume σ tot ( U S ) = σ R U U S F L σ R = σ R U ( ) + σ S ( U S ) ( U ) σ R U U S + σ S ( U S ) ( ) 1 ( U S ) R U S + σ S = σ R ( U ) 1 F L ( U S ) with R ( U S ) U S R σ S ( U S ) Landau Free Energy R, U-U s Most probable configuration maximizes σ tot i.e. minimizes F L Note that this implies F L F L ( U S ) = 0 σ S ( U S ) = 1 U s U S R e.g. for a perfect gas 3/2 MU S 3π σ S ( U S ) 2 N = N log + 5 (Sackur Tetrode with N S = 2U s 2 V 3N ) S,U S ( U S ) U S U S = 3 2 N R Phys 112 (S2009) 05 Boltzmann-Gibbs 70

71 Exchange of Particles If we exchange particles, what is minimized is Ω L ( U S, N S ) U S R σ S ( U S, N S ) µ R N S Landau Grand Potential in a way similar to the previous slide σ tot ( U S, N S ) = σ R U U S, N N S σ R = σ R U ( ) + σ S ( U S, N S ) ( U ) σ R U U σ R S N N + σ U S S (, N S S ) ( ) 1 ( U S, N S ) ( ) 1 U S + µ R N S + σ S R R ( ) with = σ R U Ω L U S, N S R Most probable configuration maximizes σ tot i.e. minimizes Ω L Note that this implies Ω L ( U S, N S ) = 0 σ S ( U S, N S ) = 1 Ω L ( U S, N S ) = 0 σ S ( U S, N S ) = µ R U s U S R N s N S R ( U S, N S ) Phys 112 (S2009) 05 Boltzmann-Gibbs 71