Part II: Statistical Physics Chapter 6: Boltzmann Statistics

Outline Introduction: Very brief Boltzmann Factor Average Values in a Canonical Ensemble The Equipartition Theorem

X Bai

SDSMT, Physics

Fall Semester: Oct. - Dec., 2014

X Bai Part II: Statistical Physics Outline Introduction: Very brief Boltzmann Factor Average Values in a Canonical Ensemble The Equipartition Theorem

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

X Bai Part II: Statistical Physics Outline Introduction: Very brief Boltzmann Factor Average Values in a Canonical Ensemble The Equipartition Theorem 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=U-TS), and Gibbs free energy (G=U-TS+PV) govern the processes toward equilibrium and phase transformations.

X Bai Part II: Statistical Physics Outline Introduction: Very brief Boltzmann Factor Average Values in a Canonical Ensemble The Equipartition Theorem 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.

X Bai Part II: Statistical Physics Outline Introduction: Very brief Boltzmann Factor Average Values in a Canonical Ensemble The Equipartition Theorem From an isolated system to a non-isolated 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.

Reservoir R Combined system U0 - ε

U0 = const System S

A combined (isolated) system: (1) a heat reservoir (2) a system of interest in thermal contact with the heat reservoir

X Bai Part II: Statistical Physics Outline Introduction: Very brief Boltzmann Factor Average Values in a Canonical Ensemble The Equipartition Theorem 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) . (Ω of a particular macrostate) Probability 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.

X Bai Part II: Statistical Physics Outline Introduction: Very brief Boltzmann Factor Average Values in a Canonical Ensemble The Equipartition Theorem 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: U0 = UR + US = const. Energies in the system (and therefore in the reservoir) may ﬂuctuate by a ”small” amount δ: U0 = (UR − δ) + (US + 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(Ei ) of ﬁnding the system S in a particular (microscopic) quantum state i of energy Ei ?”

X Bai Part II: Statistical Physics Outline Introduction: Very brief Boltzmann Factor Average Values in a Canonical Ensemble The Equipartition Theorem Boltzmann Factor

Now let’s figure out what this P(Ei ) is. Assume two microstates s1 and s2 in the system, corresponding to two different energy levels E(s1) and E(s2). The probability to find the system at these two states are P(s1) and P(s2).

Ωcomb.(s1, U − E(s1)) = ΩS (s1) × ΩR (U − E(s1)) (1)

∵ ΩS (s1) = 1 (system is now on a ﬁx known state - no degeneracy case.)

Ωcomb.(s1, U − E(s1)) = ΩR (U − E(s1)) = ΩR (s1) (2)

Ωcomb.(s1, U − E(s1)) P(s1) = (3) Ωcomb.(U, N, T )

And same for state s2. So, P(s ) Ω (s ) 2 = R 2 (4) P(s1) ΩR (s1)

This is the ratio. So, what is P(s1)?

X Bai Part II: Statistical Physics Outline Introduction: Very brief Boltzmann Factor Average Values in a Canonical Ensemble The Equipartition Theorem Boltzmann Factor- cont.

Since S = klnΩ, Ω = eS/k .

SR (s1)/k ΩR (U − E(s1)) = e (5)

SR (s2)/k P(s2) e [SR (s2)−SR (s1)]/k = S (s )/k = e (6) P(s1) e R 1

The diﬀerence SR (s2) − SR (s1) is the entropy change in the reservoir. It must be tiny. So, we can use the thermodynamic identity to ﬁnd the answer:

TdSR = dUR + PdVR − µdNR (7) 3 dVR ≈ A˚ ≈ 0: Volume change due to re-distribution of particles on microstates

dNR = 0, for system consisting of single atom, for example 1 1 S (s ) − S (s ) = [U (s ) − U (s )] = − [E(s ) − E(s )] (8) R 2 R 1 T R 2 R 1 T 2 1

−E(s2)/(kT ) P(s2) −[E(s2)−E(s1)]/(kT ) e = e = −E(s )/(kT ) (9) P(s1) e 1

Note: Let’s see a case in which the PdV is not negligible.

X Bai Part II: Statistical Physics Outline Introduction: Very brief Boltzmann Factor Average Values in a Canonical Ensemble The Equipartition Theorem Partition Function for a hydrogen atom

An example with PdVR being big enough and requires a new Boltzmann Factor: When high-n states are occupied because the approximate radius of the 2 −11 electron wave function is a0n , a0 = 5 × 10 m is the Bohr radius. When keep PdV in Atomic model and Energy-level diagram for a hydrogen atom 1 dSR = T (dUR + PdVR − µdNR ), the new Boltzmann Factor becomes (at constant pressure): BoltzmannFactor = e−(E+PV )/kT . Hydrogen at ground state: 5 −30 3 −6 PV0 ≈ 10 Pa × 10 m ≈ 10 eV . When n = 10, 2 3 PV10 ≈ (10 ) PV0 ≈ 1 eV . comparing with kT at room temperature: kT ∼ 8.617 × 10−5 eV /K × 300 K ∼ 2.6 × 10−2 eV . Low temperature: kT ∼ 8.617 × 10−5 eV /K × 1 K ∼ 8.617 × 10−5 eV . X Bai Part II: Statistical Physics Outline Introduction: Very brief Boltzmann Factor Average Values in a Canonical Ensemble The Equipartition Theorem Boltzmann Factor- cont.

The Boltzmann factor is

Boltzmann factor = e−E(s)/(kT ) (10)

Rewrite Eq. (9),

−E(s2)/(kT ) P(s2) −[E(s2)−E(s1)]/(kT ) e = e = −E(s )/(kT ) (11) P(s1) e 1 P(s ) P(s ) 1 2 = 1 = = const. (12) e−E(s2)/(kT ) e−E(s1)/(kT ) Z 1 P(s) = e−E(s)/(kT ) (13) Z

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.

X Bai Part II: Statistical Physics Outline Introduction: Very brief Boltzmann Factor Average Values in a Canonical Ensemble The Equipartition Theorem Partition Function Z

1 −E(s)/(kT ) We arrived at P(s) = Z e To calculate the probability, we still need to know Z. The formula for Z can be easily obtained by the fact that

X X 1 −E(s)/(kT ) 1 X −E(s)/(kT ) 1 ≡ P(s) = e = e (14) Z Z s s s

X −E(s)/(kT ) Z = e (15) s

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.

X Bai Part II: Statistical Physics Outline Introduction: Very brief Boltzmann Factor Average Values in a Canonical Ensemble The Equipartition Theorem Boltzmann Factor- Remarks.

Several remarks about the Boltzmann factor: P(s ) 1 For the ratio 2 = e−[E(s2)−E(s1)]/(kT ), only the energy diﬀerence P(s1) E(s2) − E(s1) 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.

X Bai Part II: Statistical Physics Outline Introduction: Very brief Boltzmann Factor Average Values in a Canonical Ensemble The Equipartition Theorem 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 Ei is:

X Bai Part II: Statistical Physics Outline Introduction: Very brief Boltzmann Factor Average Values in a Canonical Ensemble The Equipartition Theorem Degenerate energy levels - cnt.

−Ei /kT P(Ei ) ∝ di e (16)

where di is the degree of degeneracy for energy level Ei . The partition function should take the form of

X −Ei /kT Z = di e (17) i

X Bai Part II: Statistical Physics Outline Introduction: Very brief Boltzmann Factor Average Values in a Canonical Ensemble The Equipartition Theorem Two ensembles

The Microcanonical ensemble and the canonical ensemble.

microcanonical ensemble canonical ensemble

For an isolated system, the multiplicity Ω For a system in thermal contact with reservoir, provides the number of accessible microstates. the partition function Z provides the # of The constraint in calculating the states: U, V, N – accessible microstates. The constraint: T, V, N const – const

For a fixed U, the mean temperature T is For a fixed T, the mean energy U is specified, specified, but T can fluctuate. but U can fluctuate. E 1 − n 1 k T Pn = - the probability of finding a system in B Pn = e - the probability of finding a Ω Z one of the accessible states system in one of these states F T,V, N k T lnZ S(U,V, N)= kB lnΩ ( )= − B - in equilibrium, S reaches a maximum - in equilibrium, F reaches a minimum

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.

X Bai Part II: Statistical Physics Outline Introduction: Very brief Boltzmann Factor Average Values in a Canonical Ensemble The Equipartition Theorem In-Class Exercise

Exercise 06-01: 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.

X Bai Part II: Statistical Physics Outline Introduction: Very brief Boltzmann Factor Average Values in a Canonical Ensemble The Equipartition Theorem In-Class Exercise

Three more examples. 1. Exercise 06-02: 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 = 2.3 × 10−30 kg, its energy is therefore higher by δmc2. What was the ratio of the number of protons to the number of neutrons at T = 1011 K?

2. Exercise 06-03: 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 Self-reading, p.226 in textbook.

X Bai Part II: Statistical Physics Outline Introduction: Very brief Boltzmann Factor Average Values in a Canonical Ensemble The Equipartition Theorem 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(si ), the average value of some quantity x(si ) can be found as:

X x¯ = hx(si )i = x(si )P(si ) (18) i

1 X −βE(s ) 1 x¯ = x(s )e i , β = . (19) Z i kT i

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: P hx(si ) + y(si )i = [x(si ) + y(si )]P(si ) = P P i i x(si )P(si ) + i y(si )P(si ) = hx(si )i + hy(si )i

X Bai Part II: Statistical Physics Outline Introduction: Very brief Boltzmann Factor Average Values in a Canonical Ensemble The Equipartition Theorem Useful equations

Another useful representation for the average energy: P 1 P −βi 1 P ∂ −βi hEi = i i P(i ) = Z i i e , which is hEi = − Z i ∂β e , which is 1 ∂ P −βi hEi = − Z ∂β i e 1 ∂ ∂ hEi = − Z ∂β Z = − ∂β lnZ β=1/kT hEi = − ∂ lnZ = − ∂ lnZ. ∂ 1 − k ∂T kT (kT )2 It can be written as: 2 ∂ hEi = kT ∂T lnZ So, if we know Z = Z(T , V , N, ...), we know the average energy!

X Bai Part II: Statistical Physics Outline Introduction: Very brief Boltzmann Factor Average Values in a Canonical Ensemble The Equipartition Theorem Useful equations - degenerate energy levels

We have learned:

−Ei /kT P(Ei ) ∝ di e di : degree of degeneracy for energy level Ei

X −Ei /kT Z = di e i The alternative representation for the average energy in this case (pay attention to di ): P 1 P −βi 1 P ∂ −βi hEi = i i P(i ) = Z i i di e , which is hEi = − Z i di ∂β e , which 1 ∂ P −βi is hEi = − Z ∂β i di e 1 ∂ ∂ hEi = − Z ∂β Z = − ∂β lnZ β=1/kT hEi = − ∂ lnZ = − ∂ lnZ. ∂ 1 − k ∂T kT (kT )2 This can be written as: 2 ∂ hEi = kT ∂T lnZ

X Bai Part II: Statistical Physics Outline Introduction: Very brief Boltzmann Factor Average Values in a Canonical Ensemble The Equipartition Theorem In-Class Exercise

Exercise 06-04a: 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

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, δEi = Ei − E¯, for i = 1, 2, ..., 5. ¯ 2 (b) Computer the average of the square of the found deviations, (δEi ) . Then, computer the square root of this quantity. This quantity is called the root-mean-square (rms) deviation, or standard deviation, noted as σE .

X Bai Part II: Statistical Physics Outline Introduction: Very brief Boltzmann Factor Average Values in a Canonical Ensemble The Equipartition Theorem In-Class Exercise

(con. from Exercise 06-04a) 2 ¯2 ¯ 2 (c) Prove in general that σE = E − (E) . (d) Check the preceding formula for the ﬁve-atom toy model.

Exercise 06-04b: (1) For any system in equilibrium with a reservoir at temperature T , prove 2 2 1 ∂2Z the average value of E is E = Z ∂β2 . q 2 (2) Deﬁne the root-mean-square (rms) deviation as σE = (Ei − E¯) ,

ﬁnd a formula for σE in terms of the heat capacity, C = ∂E¯/∂T .

X Bai Part II: Statistical Physics Outline Introduction: Very brief Boltzmann Factor Average Values in a Canonical Ensemble The Equipartition Theorem Paramagnetism

Energy

+μB

0 B -μB

Down Up State

Two-state paramagnet: Magnec dipoles in an external magnec ﬁeld (le) and energy levels of a single dipole (right)

X Bai Part II: Statistical Physics Outline Introduction: Very brief Boltzmann Factor Average Values in a Canonical Ensemble The Equipartition Theorem 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:

X −βE(i) −βµB βµB Z = e = e + e = 2cosh(βµB) (20) i The probability of the dipoles being in the ”up” and ”down” state: 1 eβµB P = e−βE↑ = (21) ↑ Z 2cosh(βµB) 1 e−βµB P = e−βE↓ = (22) ↓ Z 2cosh(βµB)

X Bai Part II: Statistical Physics Outline Introduction: Very brief Boltzmann Factor Average Values in a Canonical Ensemble The Equipartition Theorem Paramagnetsm- cnt.

The average energy: ¯ X E = Ei PEi = (−µB)P↑ + (µB)P↓ = (−µB)(P↑ − P↓) i eβµB − e−βµB E¯ = (−µB) = −µBtanh(βµB) (23) 2cosh(βµB) The total energy of a collection of N dipoles: U = −NµBtanh(βµB) (24) The mean value of a dipole’s magnetic moment along the direction of the magnetic ﬁeld B X µ¯B = µB (i)PEi = (+µ)P↑ + (−µ)P↓ i

µ¯B = µtanh(βµB) (25) The total magnetization of the sample: M = Nµtanh(βµB) (26)

X Bai Part II: Statistical Physics Outline Introduction: Very brief Boltzmann Factor Average Values in a Canonical Ensemble The Equipartition Theorem Rotation of Diatomic Molecules

12e J=3

6e J=2 Nstates=2J+1

2e J=1 0 J=0 Energy levels for the rotational states of a diatomic molecule

X Bai Part II: Statistical Physics Outline Introduction: Very brief Boltzmann Factor Average Values in a Canonical Ensemble The Equipartition Theorem Rotation of Diatomic Molecules - cnt.

1 The allowed rotational energies: E(j) = j(j + 1) ( is a constant, 1 ∝ moment of inertia ) 2 Number of degeneracy: N = 2j + 1

Assume particles occupy ”all” possible j, corresponding to high temperature:

∞ ∞ X −E(j)/kT X −j(j+1)/kT Zrot = (2j + 1)e = (2j + 1)e (27) j=0 j=0 Z ∞ Z ∞ −j(j+1)/kT −j(j+1)/kT Zrot ≈ (2j + 1)e dj = e d[j(j + 1)] 0 0 Z ∞ Z ∞ ∞ −u/kT kT −u/kT −kT −u/kT Zrot ≈ e du = e d(u/kT ) = e 0 0 0 kT Z ≈ + ( when kT ). (28) rot

X Bai Part II: Statistical Physics Outline Introduction: Very brief Boltzmann Factor Average Values in a Canonical Ensemble The Equipartition Theorem Mean rotational energy at high temperatures

is called rotational constant. It is inversely proportional to the molecule’s moment of inertia. For molecules consisting of same atoms, such as N2, O2, etc.? kT Z ≈ + (for identical atoms when kT ). (29) rot 2 Now we can calculate the average rotational energy of a molecule at high temperatures: 1 ∂Z −1 1 < E >= − = −β = = kT (30) rot Z ∂β β2 β

For molecules consisting of same atoms, such as N2, O2, degree of degeneracy reduces by a factor of 2. The partition function is: kT 1 Zrot ≈ + 2 = 2β (for identical atoms when kT ). 1 ∂Z −1 1 < E >= − = −2β = = kT (31) rot Z ∂β 2β2 β Four things to address:

X Bai Part II: Statistical Physics Outline Introduction: Very brief Boltzmann Factor Average Values in a Canonical Ensemble The Equipartition Theorem 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. 2 ∂ We can also use hEi = kT ∂T lnZ to calculate the mean rotational energy: ∂ kT < E >≈ kT 2 ln rot ∂T 1 k < E >≈ kT 2 rot kT

< Erot >≈ kT , ( when kT ). (32)

X Bai Part II: Statistical Physics Outline Introduction: Very brief Boltzmann Factor Average Values in a Canonical Ensemble The Equipartition Theorem System with more degrees of freedom

A non-spherical nuclear: Two types of motions Mean ﬁeld → single particle states (n, l, s; Zsingl.). Rotation of the nucleus →: collective rotational states (J, Zrot ). What is the partition function?

−Ei /kT X −Ei /kT P(Esingl.) ∝ di e → Zsingl. = di e i

−Ej /kT X −Ej /kT P(Ecol.) ∝ dj e → Zcol. = dj e j

1 −Ei /kT 1 −Ej /kT P(Esingl., Ecol.) = P(Esingl.) ·P(Ecol.) = di e dj e Zsingl. Zcol.

1 −(Ei +Ej )/kT P(Esingl., Ecol.) = di dj e Zsingl.Zcol.

Therefore, the partition function of the system is: Z = Zsingl. · Zcol. Example: Problem 6.48, p.255 in the textbook.

X Bai Part II: Statistical Physics Outline Introduction: Very brief Boltzmann Factor Average Values in a Canonical Ensemble The Equipartition Theorem The Equipartition Theorem

The Equipartition theorem (Chapter 1): At temperature T, the average energy 1 of any quadratic degree of freedom is 2 kT . - This can be proved based on 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 non-quadratic temperature-dependent forms of energy, the total thermal energy in the system is 1 U = Nf kT (33) thermal 2 I also made the following remarks: It only applies to systems in which the energy is in the form of quadratic degree of freedom: E(q) = cq2. 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.

X Bai Part II: Statistical Physics Outline Introduction: Very brief Boltzmann Factor Average Values in a Canonical Ensemble The Equipartition Theorem The Equipartition Theorem

Here we give a proof using Boltzmann factors. Consider a system with single degree of freedom.

X X 2 Z = exp(−βE(q)) = exp(−βcq ) (34) q q

1 X 2 Z = exp(−βcq )δq (35) δq q 1 Z +∞ Z = exp(−βcq2)dq (36) δq −∞ 1 Z +∞ Z = √ exp[−(pβcq)2]d(pβcq) δq βc −∞ 1 Z +∞ Z = √ exp(−u2)du (37) δq βc −∞ p 1 r π π/c Z = = Cβ−1/2, C = (38) δq βc δq

X Bai Part II: Statistical Physics Outline Introduction: Very brief Boltzmann Factor Average Values in a Canonical Ensemble The Equipartition Theorem The Equipartition Theorem

1 ∂ Now, we can use hEi = − Z ∂β Z

1 ∂ E¯ = − Cβ−1/2 (39) Cβ−1/2 ∂β 1 −1 E¯ = − β−3/2 (40) β−1/2 2 1 1 E¯ = β−1 = kT (41) 2 2 (42)

This is the Equipartition Theorem for 1-DoF system.

Next time, we will continue our discussion about the partition function, its relation with free energy, and what it looks like for composite systems.

X Bai Part II: Statistical Physics