Part II: Statistical Physics - Department of...

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

Part II: Statistical PhysicsChapter 6: Boltzmann Statistics

X Bai

SDSMT, Physics

Fall Semester: Oct. - Dec., 2014

X Bai Part II: Statistical Physics


1 Introduction: Very brief

2 Boltzmann FactorIsolated System and System of InterestBoltzmann FactorThe Partition Function Z

3 Average Values in a Canonical EnsembleApplications

4 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 2ndLaw, which includes simple statistical treatment of an isolated system.

(3) Thermodynamic treatment of systems interacting with each other or incontact with the heat reservoir =⇒ the maximum entropy, the minimum freeenergy principles, and their applications in engine and refrigerators.

(4) How enthalpy (H=U+PV), Helmholtz free energy (F=U-TS), and Gibbsfree energy (G=U-TS+PV) govern the processes toward equilibrium and phasetransformations.



Introduction

We tried to connect (2) and the rest contents by showing simple examples suchas Ideal Gas, Einstein solid, and van der Waals gas/fluid =⇒ impressiveconnections 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. Inother words, all allowed microstates of the system are equally probable.

In this course, we will develop more complicated models based on the samefundamental assumption for the study of a greater variety of physical systems.



From an isolated system to a non-isolated system

We will introduce the most powerful tool in statistical mechanics to find theprobability of finding a system in any particular microstate. To start, let’srevisit the Isolated System and System of Interest.

Combined system U0 = const

Reservoir R

U0 - ε

System S ε

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



Isolated System- cnt.

Some fundamental assumptions for an Isolated System:

An isolated system in thermal equilibrium will pass through all theaccessible microstates states at the same recurrence rate as it evolvesover 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 systemswith this probability distribution is called a mirocanonical ensemble

These provides us with the basis for the study of a system we are interested in.



A system in thermal contact with a heat reservoir

The system of interest can be any small macroscopic or microscopic object - abox of gas, a piece of solid, an atom or molecule, etc. Assuming:

Interactions between the system and the reservoir are weak: with heatexchange but no affect on the microscopic structure inside the system ofinterest;

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 numberof 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?”



Boltzmann Factor

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

P(s1) =Ωcomb.(s1,U − E(s1))

Ωcomb.(U,N,T )(3)

And same for state s2. So,

P(s2)

P(s1)=

ΩR(s2)

ΩR(s1)(4)

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



Boltzmann Factor- cont.

Since S = klnΩ, Ω = eS/k .

ΩR(U − E(s1)) = eSR (s1)/k (5)

P(s2)

P(s1)=

eSR (s2)/k

eSR (s1)/k= e [SR (s2)−SR (s1)]/k (6)

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

TdSR = dUR + PdVR − µdNR (7)

dVR ≈ A3 ≈ 0: Volume change due to re-distribution of particles on microstates

dNR = 0, for system consisting of single atom, for example

SR(s2)− SR(s1) =1

T[UR(s2)− UR(s1)] = − 1

T[E(s2)− E(s1)] (8)

P(s2)

P(s1)= e−[E(s2)−E(s1)]/(kT ) =

e−E(s2)/(kT )

e−E(s1)/(kT )(9)

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



Partition Function for a hydrogen atom

An example with PdVR being big enough and requires a new BoltzmannFactor: When high-n states are occupied because the approximate radius of theelectron wave function is a0n

2, a0 = 5× 10−11 m is the Bohr radius.

When keep PdV indSR = 1

T(dUR + PdVR − µdNR),

the new Boltzmann Factor becomes(at constant pressure):BoltzmannFactor = e−(E+PV )/kT .

Hydrogen at ground state:PV0 ≈ 105Pa× 10−30m3 ≈ 10−6 eV .When n = 10,PV10 ≈ (102)3PV0 ≈ 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 .

Atomic model and Energy-‐level diagram for a hydrogen atom



Boltzmann Factor- cont.

The Boltzmann factor is

Boltzmann factor = e−E(s)/(kT ) (10)

Rewrite Eq. (9),

P(s2)

P(s1)= e−[E(s2)−E(s1)]/(kT ) =

e−E(s2)/(kT )

e−E(s1)/(kT )(11)

P(s2)

e−E(s2)/(kT )=

P(s1)

e−E(s1)/(kT )=

1

Z= const. (12)

P(s) =1

Ze−E(s)/(kT ) (13)

We arrived at the most useful formula in all of statistical mechanics. Pleasememorize it.It is also called the Boltzmann distribution, or the canonical distribution.



Partition Function Z

We arrived at P(s) = 1Ze−E(s)/(kT )

To calculate the probability, we still need to know Z . The formula for Z can beeasily obtained by the fact that

1 ≡∑s

P(s) =∑s

1

Ze−E(s)/(kT ) =

1

Z

∑s

e−E(s)/(kT ) (14)

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 ontemperature.

2 Assuming the energy of ground state is zero, the Boltzmann Factor ofground 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.



Boltzmann Factor- Remarks.

Several remarks about the Boltzmann factor:

1 For the ratio P(s2)P(s1)

= e−[E(s2)−E(s1)]/(kT ), only the energy difference

E(s2)− E(s1) makes contributions.

2 We do not have to know anything about the reservoir except that itmaintains a constant temperature T.

3 We made the transition from ”the fundamental assumption for anisolated system” to ”The system of interest which is in thermalequilibrium with the thermal reservoir”:The system visits each microstate with a frequency proportional tothe Boltzmann factor.

4 An ensemble of identical systems all of which are in contact with thesame heat reservoir and distributed over states in accordance with theBoltzmann distribution is called a canonical ensemble.



Degenerate energy levels

If SEVERAL quantum states of the system (different sets of quantumnumbers) 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 samefor all the degenerate states. Thus, the total probability to find THE SYSTEMin a state with energy Ei is:



Degenerate energy levels - cnt.

P(Ei ) ∝ die−Ei/kT (16)

where di is the degree of degeneracy for energy level Ei .

The partition function should take the form of

Z =∑i

die−Ei/kT (17)



Two ensembles

The Microcanonical ensemble and the canonical ensemble.

microcanonical ensemble canonical ensemble

( ) Ω= ln,, BkNVUS ( ) ZTkNVTF B ln,, −=

Ω=1

nP TkE

nB

n

eZ

P−

=1- the probability of finding a system in

one of the accessible states - 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!

For an isolated system, the multiplicity Ω provides the number of accessible microstates. The constraint in calculating the states: U, V, N – const

For a fixed U, the mean temperature T is specified, but T can fluctuate.

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.

- in equilibrium, S reaches a maximum - in equilibrium, F reaches a minimum

We will learn more about the partition function and free energy in later sections.



In-Class Exercise

Exercise 06-01: Prove that the probability of finding an atom in any par-ticular energy level is P(E) = (1/Z)e−F/(kT ), where F = E − TS and theentropy of a level is k times the logarithm of the number of degeneratestates for that level.



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 thesame particle, called the nucleon. Since the neutrons mass is higher thanthe 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 atT = 1011 K?

2. Exercise 06-03: Use Boltzmann factors to derive the exponential formulafor the density of an isothermal atmosphere. Assume the temperature Tdoes not vary with z:

3. Thermal excitation of atoms Self-reading, p.226 in textbook.



Average Values

What is the average value? If the systems in an ensemble are distributed overtheir accessible states in accordance with the distribution P(si ), the averagevalue of some quantity x(si ) can be found as:

x = 〈x(si )〉 =∑i

x(si )P(si ) (18)

x =1

Z

∑i

x(si )e−βE(si ), 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(si ) + y(si )〉 =

∑i [x(si ) + y(si )]P(si ) =∑

i x(si )P(si ) +∑

i y(si )P(si ) = 〈x(si )〉+ 〈y(si )〉



Useful equations

Another useful representation for the average energy:〈E〉 =

∑i εiP(εi ) = 1

Z

∑i εie

−βεi , which is 〈E〉 = − 1Z

∑i∂∂β

e−βεi , which is

〈E〉 = − 1Z∂∂β

∑i e−βεi

〈E〉 = − 1Z∂∂β

Z = − ∂∂β

lnZ

〈E〉 β=1/kT= − ∂

∂ 1kT

lnZ = − ∂

− k(kT )2 ∂T

lnZ .

It can be written as:〈E〉 = kT 2 ∂

∂TlnZ

So, if we know Z = Z(T ,V ,N, ...), we know the average energy!



Useful equations - degenerate energy levels

We have learned:

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

Z =∑i

die−Ei/kT

The alternative representation for the average energy in this case (payattention to di ):〈E〉 =

∑i εiP(εi ) = 1

Z

∑i εidie

−βεi , which is 〈E〉 = − 1Z

∑i di

∂∂β

e−βεi , which

is 〈E〉 = − 1Z∂∂β

∑i die

−βεi

〈E〉 = − 1Z∂∂β

Z = − ∂∂β

lnZ

〈E〉 β=1/kT= − ∂

∂ 1kT

lnZ = − ∂

− k(kT )2 ∂T

lnZ .

This can be written as:

〈E〉 = kT 2 ∂∂T

lnZ



In-Class Exercise

Exercise 06-04a: On fluctuationsThe most common measure of the fluctuations of a set of numbers awayfrom the average is the standard deviation.

A system with 5 particles on 3 energy levels

E

7eV

4eV

0

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



In-Class Exercise

(con. from Exercise 06-04a)(c) Prove in general that σ2

E = E 2 − (E)2.(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

the average value of E 2 is⟨E 2⟩

= 1Z∂2Z∂β2 .

(2) Define the root-mean-square (rms) deviation as σE =√⟨

(Ei − E)2⟩,

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



Paramagnetism

B

Energy

State Up Down

+μB

0

-‐μB

Two-‐state paramagnet: Magne:c dipoles in an external magne:c field (leB) and energy levels of a single dipole (right)



Paramagnetsm- cnt.

The system has two microstates with energy −µB and +µB, where µ is themagnetic moment of the dipoles. Now we can see how easy it is to get theprobabilities and mean values:The Partition Function:

Z =∑i

e−βE(i) = e−βµB + eβµB = 2cosh(βµB) (20)

The probability of the dipoles being in the ”up” and ”down” state:

P↑ =1

Ze−βE↑ =

eβµB

2cosh(βµB)(21)

P↓ =1

Ze−βE↓ =

e−βµB

2cosh(βµB)(22)



Paramagnetsm- cnt.

The average energy:

E =∑i

EiPEi = (−µB)P↑ + (µB)P↓ = (−µB)(P↑ − P↓)

E = (−µB)eβµB − e−βµB

2cosh(βµB)= −µBtanh(βµB) (23)

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

µB =∑i

µB(i)PEi = (+µ)P↑ + (−µ)P↓

µB = µtanh(βµB) (25)

The total magnetization of the sample:

M = Nµtanh(βµB) (26)



Rotation of Diatomic Molecules

Energy levels for the rotational states of a diatomic molecule

E

12e

6e

2e 0

Nstates=2J+1

J=1

J=2

J=0

J=3



Rotation of Diatomic Molecules - cnt.

1 The allowed rotational energies: E(j) = j(j + 1)ε (ε is a constant,∝ 1

moment of inertia)

2 Number of degeneracy: N = 2j + 1

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

Zrot =∞∑j=0

(2j + 1)e−E(j)/kT =∞∑j=0

(2j + 1)e−j(j+1)ε/kT (27)

Zrot ≈∫ ∞

0

(2j + 1)e−j(j+1)ε/kTdj =

∫ ∞0

e−j(j+1)ε/kTd [j(j + 1)]

Zrot ≈∫ ∞

0

e−uε/kTdu =kT

ε

∫ ∞0

e−uε/kTd(uε/kT ) =−kTε

e−uε/kT

∣∣∣∣∞0

Zrot ≈ +kT

ε( when kT ε). (28)



Mean rotational energy at high temperatures

ε is called rotational constant. It is inversely proportional to the molecule’smoment of inertia.For molecules consisting of same atoms, such as N2, O2, etc.?

Zrot ≈ +kT

2ε(for identical atoms when kT ε). (29)

Now we can calculate the average rotational energy of a molecule at hightemperatures:

< Erot >= − 1

Z

∂Z

∂β= −εβ −1

εβ2=

1

β= kT (30)

For molecules consisting of same atoms, such as N2, O2, degree of degeneracyreduces by a factor of 2. The partition function is:Zrot ≈ + kT

2ε= 1

2εβ(for identical atoms when kT ε).

< Erot >= − 1

Z

∂Z

∂β= −2εβ

−1

2εβ2=

1

β= kT (31)

Four things to address:



Mean rotational energy at high temperatures - cnt.

At high T , the average rotational energy (thus the heat capacity) is thesame 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) willbe needed. Quantum effects will appear. We will learn how to handle thisin Chapter 7.

We can also use 〈E〉 = kT 2 ∂∂T

lnZ to calculate the mean rotationalenergy:

< Erot >≈ kT 2 ∂

∂TlnkT

ε

< Erot >≈ kT 2 1kTε

k

ε

< Erot >≈ kT , ( when kT ε). (32)



System with more degrees of freedom

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

P(Esingl.) ∝ die−Ei/kT → Zsingl. =

∑i

die−Ei/kT

P(Ecol.) ∝ dje−Ej/kT → Zcol. =

∑j

dje−Ej/kT

P(Esingl.,Ecol.) = P(Esingl.) · P(Ecol.) =1

Zsingl.die−Ei/kT 1

Zcol.dje−Ej/kT

P(Esingl.,Ecol.) =1

Zsingl.Zcol.didje

−(Ei+Ej )/kT

Therefore, the partition function of the system is: Z = Zsingl. · Zcol.

Example: Problem 6.48, p.255 in the textbook.



The Equipartition Theorem

The Equipartition theorem (Chapter 1): At temperature T, the average energyof any quadratic degree of freedom is 1

2kT . - 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 othernon-quadratic temperature-dependent forms of energy, the total thermal energyin the system is

Uthermal = Nf1

2kT (33)

I also made the following remarks:

It only applies to systems in which the energy is in the form of quadraticdegree of freedom: E(q) = cq2.

It is about ”thermal energy” of the system - those changes withtemperature, not the total energy.

Degree of freedom: different systems require specific analysis: vibration,rotation, .... The NDoF of a system may also vary as temperaturechanges.




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

Z =∑q

exp(−βE(q)) =∑q

exp(−βcq2) (34)

Z =1

δq

∑q

exp(−βcq2)δq (35)

Z =1

δq

∫ +∞

−∞exp(−βcq2)dq (36)

Z =1

δq√βc

∫ +∞

−∞exp[−(

√βcq)2]d(

√βcq)

Z =1

δq√βc

∫ +∞

−∞exp(−u2)du (37)

Z =1

δq

√π

βc= Cβ−1/2, C =

√π/c

δq(38)




Now, we can use 〈E〉 = − 1Z∂∂β

Z

E = − 1

Cβ−1/2

∂

∂βCβ−1/2 (39)

E = − 1

β−1/2

−1

2β−3/2 (40)

E =1

2β−1 =

1

2kT (41)

(42)

This is the Equipartition Theorem for 1-DoF system.

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


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