Part II: Statistical Physics


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1 Chapter 6: Boltzmann Statistics SDSMT, Physics Fall Semester: Oct.  Dec., 2014
2 1 Introduction: Very brief 2 Boltzmann Factor Isolated System and System of Interest Boltzmann Factor The Partition Function Z 3 Average Values in a Canonical Ensemble Applications 4 The Equipartition Theorem
3 Introduction In the first part of Thermal Physics, the Thermodynamics, we have learned: (1) Bulk properties of a large system, equations of state; (2) Microscopic picture of a thermal system: multiplicity, entropy, and the 2nd Law, which includes simple statistical treatment of an isolated system. (3) Thermodynamic treatment of systems interacting with each other or in contact with the heat reservoir = the maximum entropy, the minimum free energy principles, and their applications in engine and refrigerators. (4) How enthalpy (H=U+PV), Helmholtz free energy (F=UTS), and Gibbs free energy (G=UTS+PV) govern the processes toward equilibrium and phase transformations.
4 Introduction We tried to connect (2) and the rest contents by showing simple examples such as Ideal Gas, Einstein solid, and van der Waals gas/fluid = impressive connections between macroscopic and microscopic properties. In doing so, we based all arguments on a fundamental assumption: a closed (isolated) system visits every one of its microstates with equal frequency. In other words, all allowed microstates of the system are equally probable. In this course, we will develop more complicated models based on the same fundamental assumption for the study of a greater variety of physical systems.
5 From an isolated system to a nonisolated system We will introduce the most powerful tool in statistical mechanics to find the probability of finding a system in any particular microstate. To start, let s revisit the Isolated System and System of Interest. Combined system U 0 = const Reservoir R U 0  ε System S ε A combined (isolated) system: (1) a heat reservoir (2) a system of interest in thermal contact with the heat reservoir
6 Isolated System cnt. Some fundamental assumptions for an Isolated System: An isolated system in thermal equilibrium will pass through all the accessible microstates states at the same recurrence rate as it evolves over time, i.e. all accessible microstates are equally probable. Probability of a particular microstate of a microcanonical ensemble = 1 (Total number of all accessible microstates). Probability of a particular macrostate = (Ω of a particular macrostate) (Total number of all accessible microstates) The energy inside the system is conserved. A set of hypothetical systems with this probability distribution is called a mirocanonical ensemble These provides us with the basis for the study of a system we are interested in.
7 A system in thermal contact with a heat reservoir The system of interest can be any small macroscopic or microscopic object  a box of gas, a piece of solid, an atom or molecule, etc. Assuming: Interactions between the system and the reservoir are weak: with heat exchange but no affect on the microscopic structure inside the system of interest; Total energy conservation: U 0 = U R + U S = const. Energies in the system (and therefore in the reservoir) may fluctuate by a small amount δ: U 0 = (U R δ) + (U S + δ) = const. This system of interest can also be a small system: with small number of particles, small volume,... At the equilibrium between the system and reservoir, our question would be What is the probability P(E i ) of finding the system S in a particular (microscopic) quantum state i of energy E i?
8 Boltzmann Factor Now let s figure out what this P(E i ) is. Assume two microstates s 1 and s 2 in the system, corresponding to two different energy levels E(s 1) and E(s 2). The probability to find the system at these two states are P(s 1) and P(s 2). Ω comb. (s 1, U E(s 1)) = Ω S (s 1) Ω R (U E(s 1)) (1) Ω S (s 1) = 1 (system is now on a fix known state  no degeneracy case.) Ω comb. (s 1, U E(s 1)) = Ω R (U E(s 1)) = Ω R (s 1) (2) P(s 1) = Ω comb.(s 1, U E(s 1)) Ω comb. (U, N, T ) And same for state s 2. So, P(s 2) P(s = Ω R(s 2) 1) Ω R (s 1) This is the ratio. So, what is P(s 1 )? (3) (4)
9 Boltzmann Factor cont. Since S = klnω, Ω = e S/k. Ω R (U E(s 1)) = e S R (s 1 )/k P(s 2) P(s = esr (s2)/k 1) e S R (s 1 )/k = e[s R (s 2 ) S R (s 1 )]/k (5) (6) The difference S R (s 2) S R (s 1) is the entropy change in the reservoir. It must be tiny. So, we can use the thermodynamic identity to find the answer: TdS R = du R + PdV R µdn R (7) dv R Å3 0: Volume change due to redistribution of particles on microstates dn R = 0, for system consisting of single atom, for example S R (s 2) S R (s 1) = 1 T [U R(s 2) U R (s 1)] = 1 [E(s2) E(s1)] (8) T P(s 2) P(s = 1) e [E(s 2) E(s 1 )]/(kt ) = e E(s2)/(kT ) (9) e E(s 1)/(kT ) Note: Let s see a case in which the PdV is not negligible.
10 Partition Function for a hydrogen atom An example with PdV R being big enough and requires a new Boltzmann Factor: When highn states are occupied because the approximate radius of the electron wave function is a 0n 2, a 0 = m is the Bohr radius. When keep PdV in ds R = 1 T (du R + PdV R µdn R ), the new Boltzmann Factor becomes (at constant pressure): BoltzmannFactor = e (E+PV )/kt. Atomic model and Energy level diagram for a hydrogen atom Hydrogen at ground state: PV Pa m ev. When n = 10, PV 10 (10 2 ) 3 PV 0 1 ev. comparing with kt at room temperature: kt ev /K 300 K ev. Low temperature: kt ev /K 1 K ev.
11 Boltzmann Factor cont. The Boltzmann factor is Rewrite Eq. (9), Boltzmann factor = e E(s)/(kT ) (10) P(s 2) P(s = 1) e [E(s 2) E(s 1 )]/(kt ) = e E(s2)/(kT ) e E(s 1)/(kT ) (11) P(s 2) e = P(s 1) E(s 2)/(kT ) e = 1 = const. E(s 1)/(kT ) Z (12) P(s) = 1 Z e E(s)/(kT ) (13) We arrived at the most useful formula in all of statistical mechanics. Please memorize it. It is also called the Boltzmann distribution, or the canonical distribution.
12 Partition Function Z We arrived at P(s) = 1 Z e E(s)/(kT ) To calculate the probability, we still need to know Z. The formula for Z can be easily obtained by the fact that 1 s P(s) = s 1 Z e E(s)/(kT ) = 1 Z e E(s)/(kT ) (14) s Z = s e E(s)/(kT ) (15) Z is just the sum of all Boltzmann Factors. Several remarks: 1 Z is a constant  independent of particular state s. But it depends on temperature. 2 Assuming the energy of ground state is zero, the Boltzmann Factor of ground state is 1. Boltzmann Factors for excited states are less than 1. 3 At very low temperature, Z 1. 4 At high temperatures, Z can be very big.
13 Boltzmann Factor Remarks. Several remarks about the Boltzmann factor: 1 For the ratio P(s 2) P(s 1 ) = e [E(s 2) E(s 1 )]/(kt ), only the energy difference E(s 2) E(s 1) makes contributions. 2 We do not have to know anything about the reservoir except that it maintains a constant temperature T. 3 We made the transition from the fundamental assumption for an isolated system to The system of interest which is in thermal equilibrium with the thermal reservoir : The system visits each microstate with a frequency proportional to the Boltzmann factor. 4 An ensemble of identical systems all of which are in contact with the same heat reservoir and distributed over states in accordance with the Boltzmann distribution is called a canonical ensemble.
14 Degenerate energy levels If SEVERAL quantum states of the system (different sets of quantum numbers) correspond to the SAME energy level, this level is called degenerate. The probability to find the system in one of these degenerate states is the same for all the degenerate states. Thus, the total probability to find THE SYSTEM in a state with energy E i is:
15 Degenerate energy levels  cnt. P(E i ) d i e E i /kt where d i is the degree of degeneracy for energy level E i. (16) The partition function should take the form of Z = i d i e E i /kt (17)
16 Two ensembles The Microcanonical ensemble and the canonical ensemble. microcanonical ensemble canonical ensemble For an isolated system, the multiplicity Ω provides the number of accessible microstates. The constraint in calculating the states: U, V, N const P = 1 n Ω For a fixed U, the mean temperature T is specified, but T can fluctuate.  the probability of finding a system in one of the accessible states For a system in thermal contact with reservoir, the partition function Z provides the # of accessible microstates. The constraint: T, V, N const For a fixed T, the mean energy U is specified, but U can fluctuate. Pn = 1 e Z S ( U V, N) = k lnω F( T, V, N) = kb T lnz, B  in equilibrium, S reaches a maximum  in equilibrium, F reaches a minimum En kb T  the probability of finding a system in one of these states For the canonical ensemble, the role of Z is similar to that of the multiplicity Ω for the microcanonical ensemble. F(T,V,N)=f(Z) gives the fundamental relation between statistical mechanics and thermodynamics for given values of T, V, and N, just as S (U,V,N) = S(Ω) gives the fundamental relation between statistical mechanics and thermodynamics for given values of U, V, and N. F: is the Helmholtz Free Energy! We will learn more about the partition function and free energy in later sections.
17 InClass Exercise Exercise 0601: Prove that the probability of finding an atom in any particular energy level is P(E) = (1/Z)e F /(kt ), where F = E TS and the entropy of a level is k times the logarithm of the number of degenerate states for that level.
18 InClass Exercise Three more examples. 1. Exercise 0602: At very high temperature (as in the very early universe), the proton and the neutron can be thought of as two different states of the same particle, called the nucleon. Since the neutrons mass is higher than the protons by δm = kg, its energy is therefore higher by δmc 2. What was the ratio of the number of protons to the number of neutrons at T = 1011 K? 2. Exercise 0603: Use Boltzmann factors to derive the exponential formula for the density of an isothermal atmosphere. Assume the temperature T does not vary with z: 3. Thermal excitation of atoms Selfreading, p.226 in textbook.
19 Average Values What is the average value? If the systems in an ensemble are distributed over their accessible states in accordance with the distribution P(s i ), the average value of some quantity x(s i ) can be found as: x = x(s i ) = i x(s i )P(s i ) (18) x = 1 Z i x(s i )e βe(s i ), β = 1 kt. (19) 1 The average value depends on the entire distribution, not just the peak, or the width of the distribution. 2 The average value is an additive quantity: x(s i ) + y(s i ) = i [x(s i) + y(s i )]P(s i ) = i x(s i)p(s i ) + i y(s i)p(s i ) = x(s i ) + y(s i )
20 Useful equations Another useful representation for the average energy: E = i ɛ ip(ɛ i ) = 1 Z i ɛ ie βɛ i, which is E = 1 Z E = 1 Z β i e βɛ i E = 1 Z = lnz Z β β E β=1/kt = lnz = lnz. kt 1 k 2 T It can be written as: E = kt 2 T lnz (kt ) i So, if we know Z = Z(T, V, N,...), we know the average energy! β e βɛ i, which is
21 Useful equations  degenerate energy levels We have learned: P(E i ) d i e E i /kt Z = i d i e E i /kt d i : degree of degeneracy for energy level E i The alternative representation for the average energy in this case (pay attention to d i ): E = i ɛ ip(ɛ i ) = 1 Z i ɛ id i e βɛ i, which is E = 1 Z i d i β e βɛ i, which is E = 1 Z β i d ie βɛ i E = 1 Z = lnz Z β β E β=1/kt = lnz = lnz. kt 1 k (kt ) 2 T This can be written as: E = kt 2 T lnz
22 InClass Exercise Exercise 0604a: On fluctuations The most common measure of the fluctuations of a set of numbers away from the average is the standard deviation. E 7eV 4eV 0 A system with 5 particles on 3 energy levels (a) For the system shown above, computer the deviation of the energy from the average energy, δe i = E i Ē, for i = 1, 2,..., 5. (b) Computer the average of the square of the found deviations, (δe i ) 2. Then, computer the square root of this quantity. This quantity is called the rootmeansquare (rms) deviation, or standard deviation, noted as σ E.
23 InClass Exercise (con. from Exercise 0604a) (c) Prove in general that σe 2 = E 2 (Ē) 2. (d) Check the preceding formula for the fiveatom toy model. Exercise 0604b: (1) For any system in equilibrium with a reservoir at temperature T, prove the average value of E 2 is E 2 = 1 Z 2 Z β 2. (2) Define the rootmeansquare (rms) deviation as σ E = (Ei Ē)2, find a formula for σ E in terms of the heat capacity, C = Ē/ T.
24 Paramagnetism Energy +μb B 0  μb Down Up State Two state paramagnet: Magne:c dipoles in an external magne:c field (leb) and energy levels of a single dipole (right)
25 Paramagnetsm cnt. The system has two microstates with energy µb and +µb, where µ is the magnetic moment of the dipoles. Now we can see how easy it is to get the probabilities and mean values: The Partition Function: Z = i e βe(i) = e βµb + e βµb = 2cosh(βµB) (20) The probability of the dipoles being in the up and down state: P = 1 Z e βe = P = 1 Z e βe = e βµb 2cosh(βµB) e βµb 2cosh(βµB) (21) (22)
26 Paramagnetsm cnt. The average energy: Ē = i E i P Ei = ( µb)p + (µb)p = ( µb)(p P ) Ē = ( µb) eβµb e βµb 2cosh(βµB) The total energy of a collection of N dipoles: = µbtanh(βµb) (23) U = NµBtanh(βµB) (24) The mean value of a dipole s magnetic moment along the direction of the magnetic field B µ B = i µ B (i)p Ei = (+µ)p + ( µ)p µ B = µtanh(βµb) (25) The total magnetization of the sample: M = Nµtanh(βµB) (26)
27 Rotation of Diatomic Molecules E 12e J=3 6e J=2 N states =2J+1 2e 0 J=0 J=1 Energy levels for the rotational states of a diatomic molecule
28 Rotation of Diatomic Molecules  cnt. 1 The allowed rotational energies: E(j) = j(j + 1)ɛ (ɛ is a constant, 1 moment of inertia ) 2 Number of degeneracy: N = 2j + 1 Assume particles occupy all possible j, corresponding to high temperature: Z rot = (2j + 1)e E(j)/kT = (2j + 1)e j(j+1)ɛ/kt (27) Z rot Z rot j=0 0 0 Z rot + kt ɛ j=0 (2j + 1)e j(j+1)ɛ/kt dj = e uɛ/kt du = kt ɛ 0 0 e j(j+1)ɛ/kt d[j(j + 1)] e uɛ/kt d(uɛ/kt ) = kt e uɛ/kt ɛ 0 ( when kt ɛ). (28)
29 Mean rotational energy at high temperatures ɛ is called rotational constant. It is inversely proportional to the molecule s moment of inertia. For molecules consisting of same atoms, such as N 2, O 2, etc.? Z rot + kt (for identical atoms when kt ɛ). (29) 2ɛ Now we can calculate the average rotational energy of a molecule at high temperatures: < E rot >= 1 Z Z β 1 = ɛβ ɛβ = 1 = kt (30) 2 β For molecules consisting of same atoms, such as N 2, O 2, degree of degeneracy reduces by a factor of 2. The partition function is: Z rot + kt = 1 (for identical atoms when kt ɛ). 2ɛ 2ɛβ Four things to address: < E rot >= 1 Z Z β 1 = 2ɛβ 2ɛβ = 1 = kt (31) 2 β
30 Mean rotational energy at high temperatures  cnt. At high T, the average rotational energy (thus the heat capacity) is the same for identical particle system. At lower T, things are more complicated when multiple particle may fight to occupy single particle states. Strict treatment of Eq. (27) will be needed. Quantum effects will appear. We will learn how to handle this in Chapter 7. We can also use E = kt 2 lnz to calculate the mean rotational T energy: < E rot > kt 2 T ln kt ɛ < E rot > kt 2 1 k kt ɛ ɛ < E rot > kt, ( when kt ɛ). (32)
31 System with more degrees of freedom A nonspherical nuclear: Two types of motions Mean field single particle states (n, l, s; Z singl. ). Rotation of the nucleus : collective rotational states (J, Z rot). What is the partition function? P(E singl. ) d i e E i /kt Z singl. = i d i e E i /kt P(E col. ) d j e E j /kt Z col. = j d j e E j /kt P(E singl., E col. ) = P(E singl. ) P(E col. ) = 1 Z singl. d i e E i /kt 1 Z col. d j e E j /kt P(E singl., E col. ) = 1 Z singl. Z col. d i d j e (E i +E j )/kt Therefore, the partition function of the system is: Example: Problem 6.48, p.255 in the textbook. Z = Z singl. Z col.
32 The Equipartition Theorem The Equipartition theorem (Chapter 1): At temperature T, the average energy of any quadratic degree of freedom is 1 kt.  This can be proved based on 2 principles in Statistical Physics which we will learn later in this semester. For a system with N particles, each with f DoF, and there is NO other nonquadratic temperaturedependent forms of energy, the total thermal energy in the system is I also made the following remarks: U thermal = Nf 1 kt (33) 2 It only applies to systems in which the energy is in the form of quadratic degree of freedom: E(q) = cq 2. It is about thermal energy of the system  those changes with temperature, not the total energy. Degree of freedom: different systems require specific analysis: vibration, rotation,... The NDoF of a system may also vary as temperature changes.
33 The Equipartition Theorem Here we give a proof using Boltzmann factors. Consider a system with single degree of freedom. Z = q exp( βe(q)) = q exp( βcq 2 ) (34) Z = 1 δq Z = 1 δq Z = exp( βcq 2 )δq (35) q + 1 δq βc 1 Z = δq βc Z = 1 δq + exp( βcq 2 )dq (36) + exp[ ( βcq) 2 ]d( βcq) exp( u 2 )du (37) π βc = Cβ 1/2, C = π/c δq (38)
34 The Equipartition Theorem Now, we can use E = 1 Z β Z Ē = 1 Cβ 1/2 β Cβ 1/2 (39) This is the Equipartition Theorem for 1DoF system. Ē = 1 β 1/2 1 2 β 3/2 (40) Ē = 1 2 β 1 = 1 kt (41) 2 (42) Next time, we will continue our discussion about the partition function, its relation with free energy, and what it looks like for composite systems.
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