physics of gases - princeton university

Physics of Gases

These are the statistical mechanics notes of Clay Byers andatomic spectra and structure notes of Will Coogan from Fall 2013.

Please notify Will of any errors at [email protected]

January 26, 2015

Contents

I Statistical Mechanics 1

1 Introduction 3

1.1 Course Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2 Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.3 Preliminary Mathematics and Review . . . . . . . . . . . . . . . . . 8

1.3.1 The Schrodinger Equation . . . . . . . . . . . . . . . . . . . . 8

1.3.2 Stirling’s Approximation and Distributions . . . . . . . . . . 9

1.3.3 Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . . . . 10

1.3.4 Degeneracy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2 Ensembles 15

2.1 Canonical Ensemble . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.2 Other Ensembles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3 Boltzmann, F-D, and B-E Statistics 23

3.1 Boltzmann Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3.2 Fermi-Dirac and Bose-Einstein Statistics . . . . . . . . . . . . . . . . 24

iii

iv CONTENTS

4 Monatomic and Diatomic Gases 27

4.1 Ideal Monatomic Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4.1.1 Translational Partition Function . . . . . . . . . . . . . . . . 27

4.1.2 Electronic Partition Function . . . . . . . . . . . . . . . . . . 29

4.1.3 Nuclear Partition Function . . . . . . . . . . . . . . . . . . . 29

4.2 Ideal Diatomic Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

4.2.1 1st Assumption: Born-Oppenheimer Approximation (matom me) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

4.2.2 2nd Assumption: Rigid Rotor - Harmonic Oscillator (RRHO)Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . 31

4.2.3 Vibrational Partition Function . . . . . . . . . . . . . . . . . 32

4.2.4 Rotational Partition Function: Heteronuclear . . . . . . . . . 33

4.2.5 Rotational Partition Function: Homonuclear . . . . . . . . . 34

4.2.6 Putting it all Together . . . . . . . . . . . . . . . . . . . . . . 35

II Kinetic Theory of Gases 37

5 Overview of Kinetic Theory of Gases 39

6 Classical Statistical Mechanics 43

6.1 Classical Partition Function . . . . . . . . . . . . . . . . . . . . . . . 43

6.2 Phase Space and Liouville Equation . . . . . . . . . . . . . . . . . . 46

6.3 Equipartition of Energy . . . . . . . . . . . . . . . . . . . . . . . . . 51

7 Kinetic Theory of Gases 53

CONTENTS v

7.1 Phase Space and Liouville Eq. (Take 2) . . . . . . . . . . . . . . . . 53

7.2 Reduced Distribution Functions . . . . . . . . . . . . . . . . . . . . . 55

7.3 Fluxes in Dilute Gases . . . . . . . . . . . . . . . . . . . . . . . . . . 57

7.3.1 Transport of Mass . . . . . . . . . . . . . . . . . . . . . . . . 59

7.3.2 Transport of Momentum . . . . . . . . . . . . . . . . . . . . . 59

7.3.3 Transport of Kinetic Energy . . . . . . . . . . . . . . . . . . . 59

7.4 The Boltzmann Equation . . . . . . . . . . . . . . . . . . . . . . . . 60

7.5 Consequences of the Boltzmann Equation . . . . . . . . . . . . . . . 63

7.5.1 1. Equations of Change . . . . . . . . . . . . . . . . . . . . . 63

7.5.2 2. Boltzmann H-Theorem . . . . . . . . . . . . . . . . . . . . 64

7.5.3 3. Equilibrium Solution . . . . . . . . . . . . . . . . . . . . . 65

8 Transport Processes in Dilute Gases 67

8.1 Outline of Chapman-Enskog Method . . . . . . . . . . . . . . . . . . 67

8.2 Summary of Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . 72

III Atomic Spectra and Structure 75

9 Historical Overview 77

9.1 Balmer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

9.2 Bohr . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

9.3 Sommerfeld’s ellipses . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

9.4 De Broglie’s waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

10 Atoms at Present 81

vi CONTENTS

10.1 Schrodinger Eq. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

10.2 Selection Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

10.3 Zeeman Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

IV Appendices 89

11 Appendix A 91

12 Appendix B 93

Part I

Statistical Mechanics

1

Chapter 1

Introduction

1.1 Course Overview

We develop the ability to predict macroscopic properties of a system from anunderstanding of molecular properties. The ensemble average of a mechanicalproperty is equated to the corresponding thermodynamic function. Using basicassumptions about intermolecular potential, we are able to determine transportcoefficients (viscosity, mass diffusivity, and thermal conductivity).

1.2 Collisions

Imagine a flux of j particles through some k particles, → get a change in flux (Fig.1.1)

• Probability of a collision between j and k particle is:

dPjk = nkQPjkdx (1.1)

– QPjk is cross sectional area of process P

– dx is distance

– nk is density of k-particles

Γ + dΓ = Γ− Γ · dP = (1− nkQPjkdx)Γ (1.2)

3

4 CHAPTER 1. INTRODUCTION

Figure 1.1: Flux of j-particles through k-particles.

dΓ = nkQPjkdxΓ (1.3)

dΓ

dx= nkQ

PjkΓ (1.4)

Γ(x) = Γ0e−nkQPjkx (1.5)

• If x is in units of length, nkQPjk is in units of 1

length and define a decayconstant:

λ =1

nkQPjk

≡ λmfp (1.6)

• λ is the mean free path- how long we travel before “bouncing.”

Instead of flux, say the number of j-particles lost is:

dN = NdP = nNjQ(v)dx (1.7)

• We see Q(v) (our cross section) needs to be defined → How to do it? See Fig.1.2

1.2. COLLISIONS 5

Figure 1.2: Two particles will only hit if the distance between them is less than thesum of both sphere radii.

Q = π(r1 + r2)2 (1.8)

Collision Frequency: Imagine the mean free path is equal to the distance traveledbetween collisions.

λv(∆t)

ν(∆t)=

velocity

collision frequency=

1

nQ(v)(1.9)

ν = nvQ(v) (1.10)

• ν is the frequency with which a j-particle hits a k particle.

• We can generalize our cross section as our elastic collision and all the inelasticcollisions (KE not preserved).

Q = Q(e) +∑

Q(ne) (1.11)

– Q(ne) can result from electron excitation, ionization, etc.

– Good first estimate is using Bohr radius: Q0 = πa20 = 0.88 A

2.

– Can usually go with Q ≈ 10−20 m2 as a start.

Reaction Rate: Look at our collision frequency with all the particles:

RP = njnkvQPjk(v) (1.12)


Figure 1.3: Particle with velocity u in a flow of length l.

• RP has units of 1/m3

s = densitytime .

• This is how fast we add or take away particles in the process P .

• Example reaction: A+ e− → A+ + 2e− (ionization), where A is rigid particlek and e− is particle j. If we let n0 be the density of neutrals, then:

Rion = nen0veQion(v) (1.13)

• We are removing field particles at a rate dn0dt = −Rion. We can rewrite this to

see that n0

Rion is in units of time. → Define as τP relaxation time.

– In a flow of length l, τr = lv , the resonance time shown in Fig. 1.3

– Looking at rocket for τP and τr comparison:

∗ τr > τP is equilibrium flow

∗ τP > τr is frozen flow

Elastic collisions conserve energy.Inelastic collisions transfer energy to internal modes.

1

2m1v

21 =

1

2m1V

21 +

1

2m2V

22 + U (1.14)

• U is the energy that goes to exciting particle 2

• Momentum is also conserved (m1v1 = m1V1 +m2V2)

– For max U ( dUdV1= 0), V1 = m1v1

m1+m2

– Max U∗ =(

m2m1+m2

)m1v2

12 , the larger m2

m1, the more initial KE → U .

– If m1 << m2 → U∗ =m1v2

12

1.2. COLLISIONS 7

Figure 1.4: Particle 1 collides with particle 2, transferring some of its kinetic energyto particle 2.

Non-adiabatic criterion for elastic collisions:

• Ionization/excitation happens when the incoming particles come in fastenough that electron fields can’t adjust.

• Consider the neutral atom as a harmonic oscillator with natural frequency ν0

(See Fig. 1.5).

• e− has a forcing function that applies to the electron field (has a frequency).

– Define τ ≈ av , where v is velocity and a is range of interaction.

– If 1τ << ν0, adiabatic interaction

– If the electron is oscillating much faster than the incoming particle iscoming in, it can adjust and we have no excitations.

• So aν0v 1 is the adiabatic condition

• To excite an electron, we need energy E = hν → ν0 ≈ Eh and E = ∆ε12

• So for adiabatic interactions (no excitation), a∆ενh 1, and for non-adiabatic

collisions (excitation) a∆ενh ≤ 1

• Then ∆ε at some T gives how much energy we need at that temperature.

• If ∆ε is above the first excited state, then we need to worry about more thanground state in the partition function.


Figure 1.5: An electron oscillating with natural frequency ν0.

1.3 Preliminary Mathematics and Review

Statistical mechanics studies macroscopic systems from a molecular point of view.

• Looking into equilibrium (statistical thermo) we will develop a way tocalculate and interpret thermodynamic quantities from a molecular point ofview.

• We will do this by presenting ideas in terms of quantum mechanicalproperties such as energy states, wave functions, and degeneracy.

1.3.1 The Schrodinger Equation

Know that the Schrodinger equation determines the possible energy values εjavailable to the system and these values may have an associated degeneracy Ω(εj)

We had learned F=ma, but can’t apply to very small particles because of theHeisenberg uncertainty principle (can’t know both position and momentum).

• We use quantum mechanics and probability distribution

• We can use the Hamiltonian to measure energy: HΨ = EΨ (timeindependent), where Ψ is a probability distribution.

• Then p = hk, h = h2π , k = 2π

λ → p = hλ , and h is in units of joule-seconds

1.3. PRELIMINARY MATHEMATICS AND REVIEW 9

Figure 1.6: Particle in a box: We use a wave as the most general form of wherethe particle could be, then the wave function ψn(x) = sin(nπa x) for n = 1, 2, 3, ....Note that the wave function is zero outside of the box since there is zero possibilitythe particle can be there. Now, the 1D Hamiltonian is H = − h2

2m∂2

∂x2 , so Hψ =

−−h2n2π2

2ma2 sin(nπa x), which means ε = h2n2

8ma2 .

• Energy is momentum squared divided by mass: E = p2

2m

• So with H = − h2

2m∂2

∂x2 for the 1D Schrodinger equation, then p2 = h2 ∂2

∂x2 and

p = −ih ∂∂x

• If we can write the Hamiltonian, use boundary conditions. Then we can solvefor the energy of the system. See Fig. 1.6 for an example.

1.3.2 Stirling’s Approximation and Distributions

ln(N !) = ln(N) + ln(N − 1) + ln(N − 2) + ...+ ln(1) =∑N

m=1 ln(m). The graph ofthis looks like ln(x), so integrate from 1 to N to solve:∫ N

1 ln(x)dx = N ln(N)−N + 1. But N >> 1, so drop 1 to get Stirling’sapproximation (equation 1.15).

ln(N !) ≈ N ln(N)−N (1.15)

Binomial Distribution

• With N particles, we have N ! different arrangements of those N particles.


• We can divide these into two groups of particles, such that N1 +N2 = N . Wethen have N !

N1!N2! ways to do this.

Say we have 4 coins with N1 heads and N2 tails. Table 1.1 shows the degenerateways to arrange different combinations of heads and tails.

N1 N2 N Ways to Get

0 4 4 1 = 1

1 3 4 4!3!1! = 4

2 2 4 4!2!2! = 6

3 1 4 4!1!3! = 4

4 0 4 1 = 1

Table 1.1: Degeneracies of “coin-states.”

If we have a very large N , then we can treat N1 as a continuous variable. Fig. 1.7illustrates this for a smaller number N .

Does our distribution get sharper as N → inf? Let’s look!

• f(N1) = N !N1!(N−N1)!

Stirling→ ln(f(N1)) =

N ln(N)−N − (N1 ln(N1)−N1)− ((N −N1) ln(N −N1)− (N −N1))

• How do we measure the “width” of this function? Let’s find the max value(take ∂

∂N ln(f(N1)) = 0).

• ∂f(N1)∂N

∣∣∣∣N∗

= 0 or ∂ ln(f(N1))∂N = 0→ get N∗ = N

2 as most probable state for

coins.

• If N is very large, we get a delta function.

1.3.3 Lagrange Multipliers

With f(N1, N2), then ∂f = ∂f∂N1

∂N1 + ∂f∂N2

∂N2 (general form).


Figure 1.7: The distribution of ways to arrange N for a small number N .

We also have N1 +N2 = N , or φ(N1, N2) =constant.

Then ∂φ = ∂φ∂N1

∂N1 + ∂φ∂N2

∂N2 = 0 since φ is constant.

Maximize the first equation, ∂f = 0, but now we are equal! So:

get α

∂f∂N1

+ α ∂φ∂N1

= 0∂f∂N2

+ α ∂φ∂N2

= 0(1.16)

Because∂f∂N1∂φ∂N1

=∂f∂N2∂φ∂N2

= −α, say it’s some −α constant.

Another way:

• We want max or min of f → f(x1, x2, ..., xn) =max or min

• Equations 1.17 then give constraints on our function:

constraints on our function

g1(x1, ..., xn) = c1

...

gn(x1, ..., xn) = cm

(1.17)

• Then construct f∗ such that f∗ = f − α1g1 − ...− αngn


Figure 1.8: Degenerate states as a function of nx and ny.

• Solve a system of equations now:∂f∗

∂xj= 0 j = 1, ..., n

gi = ci i = 1, ...,m(1.18)

• ∂f∗

∂xjgives a function of α’s → h(α’s), then plug into gi = ci to get α’s.

1.3.4 Degeneracy

We can have multiple microstates to make one macrostate

• Let’s say HΨ = EΨ in 2D, then a solution can be Ψ = sin(nxπa x) sin(nyπb x)

• We do some tedious math to get E = h2

8ma2 (n2x + n2

y)

• nx and ny can only be integers, so what states satisfy?

• Say√E is my radius, so make it h2

8ma2 (n2x + n2

y) = R2 as a circle (see Fig. 1.8)

• Larger circle means more possible states that can satisfy the equation/energy


– Thus the degeneracy rises with increasing values of E: Ω(E) rises forhigher E.

– At typical room temp, we get Ω(E) to be O(10N )→ one mol hasN = 6.022× 1023. Think of resulting number of microstates!

Chapter 2

Ensembles

2.1 Canonical Ensemble

How do we figure the macrostate out from what we know of microstates of asystem?microstates → macrostates

We postulate the most likely state is the average → what we measure (Gibbspostulate, very important!)

A system A with N,V, and ε.

Micro Canonical Ensemble

ANparticles

Aε = εtot

AV = Vtot

(2.1)

15

16 CHAPTER 2. ENSEMBLES

A A

A A

A A

A A

Table 2.1: A combination of a large number of systems create an ensemble.

Switching to canonical → we will have different energy states, each with their ownenergy, and a certain number of systems in this state.

1 2 3 State number

E1 E2 E3 Energy number

a1 a2 a3 Occupation number

Table 2.2: Canonical Ensemble.

• a = a1, a2, ..., aj

• Our constraints are:

–∑

j aj = A

–∑

j ajEj = εtot

• Postulate: We will say each micro state is equally probable, but the macrostates are not equally probable (think of the example from Table 1.1).

• We have distinguishable particles, and the number of ways W (a) to assemblethem is:

W (a) =A!

a1!a2!...=

A!∏j aj !

(2.2)

• This lets us translate into probabilities Pj

–ajA is the fraction of systems of the ensemble in the jth state.

2.1. CANONICAL ENSEMBLE 17

Figure 2.1: Distributions of aj .

– We want the probability of a system being in the jth state → needaverage

ajA over all distributions with equal weight to each:

Pj =ajA

=1

A

∑aW (a)aj(a)∑

aW (a)(2.3)

– We use aj(a) because it depends on the other states also.

– We want a∗ which maximizes W (a), as shown in Fig. 2.1.

Now we apply Lagrange multipliers with our two constraints (total number ofparticles is same, and total energy is the same).

∂

∂ajln[W (a)]− α

∑ak − β

∑akEk = 0 (2.4)

• α and β are the unknown constants we need.

• Equation 2.4 gives ln(a∗j )− α− 1− βEj = 0 or a∗j = e−α′e−βEj , where

α′ = α+ 1.

• If we sum both sides over j, we get eα′

= 1A

∑j e−βEj .


• Because Pj =a∗jA , we can rearrange: 1

A = eα′∑

e−βEjand a∗j = e−α

′e−βEj , so

a∗jA = e−βEj∑

j e−βEj .

• This is our probability of being in the jth state where we maximize our ways(picked a∗ that is most probable).

With our new expression, our average energy E = e−βEj∑e−βEj

And pressure ℘j =(∂E∂V

)N→ 2Ej = −℘jdVj and you can integrate.

• So average pressure ℘ =∑℘j · Pj , where Pj =

a∗jA

• ℘ =∑ − (∂E∂V )N e−βEj∑

e−βEj

Now we postulate that E = Ethermo and ℘ = ℘thermo → What we measure shouldequal our new expression from the micro world.

Doing some math now, we can see:

add these together

(∂E∂V

)N,β

= −℘+ β(E℘)− βE℘(∂℘∂β

)N,V

= E℘− (E℘)(2.5)

And find:(∂E∂V

)N,β

+ β(∂℘∂β

)N,V

= −℘

• We can see this is the same form as(∂E∂V

)T,N− T

(∂℘∂T

)N,T

= −℘

• If we flip so in terms of 1T , it gives us a positive sign:(

∂E∂V

)T,N

+ 1T

(∂℘

∂ 1T

)N,T

= −℘→ β = 1kT for some constant k.

• We find that k is constant across any system → universal → Boltzmannconstant.

2.2. OTHER ENSEMBLES 19

Let’s summarize what we have done with ensembles so far:

• Energy states from quantum mechanics

• Occupation states of those energy states by individual systems aj of anensemble

• Distributions of aj ’s: W (a)→ Maximize with a∗j

• Use Lagrange multipliers → Have an α and β → α cancels out, equation interms of β

• Solve and get β = 1kT

We will simplify our expressions by saying:

Q =∑j

e−βEj (2.6)

This is the canonical ensemble partition function, which bridges Q.M. energy statesof a macroscopic system and the thermodynamic properties of that system. If wecan get Q as a function of N,V, and T , we can calculate thermo properties in termsof quantum mechanical and molecular parameters.

Helmholtz free energy: A = −kT ln(Q) is the useful work obtainable from a closedthermodynamic system (also A = U − TS)

2.2 Other Ensembles

An ensemble is a collection of a large number of systems (A), each with the samemacroscopic properties for a particular thermodynamic system of interest.

• Ω(N,V,E): Micro canonical ensemble, an isolated system

• Q(N,V, T ): Canonical ensemble, a system in contact with a heat bath

• Ξ(V, T, µ): Grand canonical ensemble, a system in contact with a heat andparticle bath


Figure 2.2: A representation of how different types of ensembles are related.

• ∆(N,T, ℘): Isobaric-Isothermal ensemble, mostly used in chemistry, a systemwith constant ℘

We can choose any type of ensemble listed above where we hold the differentproperties (in parenthesis) constant.

• You go through the math to get expressions for thermo equations

• We can find the values they give differ, but by such a small amount that itdoesn’t matter to us

• Since it doesn’t matter, choose the ensemble that works best for us.

For µ chemical potential:∑

i µi∂Ni = 0 is equilibrium.

Table 3-1 in the McQuarrie [3] summarizes each ensemble and has fundamentalrelation associated.

2.2. OTHER ENSEMBLES 21

- Thermodynamic Aside -

From thermo:1st Law

∂E = ∂Q− ∂W (2.7)

2nd Law

∂E = T∂S − ∂W= T∂S − ℘∂Vand E = E(S, V )

so ∂E =(∂E∂S

)V∂S +

(∂E∂V

)S∂V

(2.8)

From Equations 2.7 and 2.8 you can see that ℘ =(∂E∂V

)S

and T =(∂E∂S

)V

.

What If we could come up with a function of temperature and volume? f = f(T, V )

• We get Helmholtz free energy: A(T, V ) = E − TS

• We can similarly find g(T, ℘) =⇒ entropy H(T, ℘)

- End Thermodynamic Aside -

The partition functions from each ensemble can be linked together:

for example: Q(N,V, T ) =∑E

Ω(N,V,E)e−EkT (2.9)

These ensembles are each in their own specific equilibrium → Don’t forget that it’sa mental construct!

Note: S = k ln(Ω) is engraved on Boltzmann’s grave- his greatest contribution.

Chapter 3

Boltzmann, Fermi-Dirac, andBose-Einstein Statistics

In the high temperature limit, quantum mechanics → Boltzmann statistics

3.1 Boltzmann Statistics

H ≈ Htrans +Hrot +Hvib +Helect is a good approximation for ideal gas.

We use the canonical partition function Q(N,V, T ) =∑

j e−EjkT ,

Ej = εai + εbi + ...+ εNi

• Break it up further: Q = (∑

i e− εaikT )(

∑j e−εbjkT )(

∑k e− εckkT )(...)

• If distinguishable, then we have Equation 3.1

Q(V, T ) = [q(V, T )]N and q =∑i

e−εikT (3.1)

• For indistinguishable particles, this becomes a little complicated

– We have N particles and N ! ways to arrange them, so we just sum allindices indiscriminately

23

24 CHAPTER 3. BOLTZMANN, F-D, AND B-E STATISTICS

– It is unlikely we can fill all states (many, many more states than N), sojust divide by N ! (All indices will be different, summing indiscriminatelyis ok). End up with equation 3.2 for indistinguishable particles

Q =[q(V, T )]N

N !(3.2)

Bosons: sum of protons, electrons, and neutrons is even (integral spin). Wavefunction must be symmetric under the operation of the interchange of two identicalparticles.

Fermions: sum of protons, electrons, and neutrons is odd (half integral spin). Wavefunction must be antisymmetric under the operation of the interchange of twoidentical particles.

3.2 Fermi-Dirac and Bose-Einstein Statistics

Different approach to get same results as Boltzmann (above)

Starting definitions: Ej =∑

k εknk, N =∑

k nk

• Same partition function: Q(N,V, T ) =∑

j e−EjkT , but our restriction makes us

rewrite:∑∗

nke−β

∑i εini

• Hard to evaluate, so let’s use grand canonical partition function:

Ξ(V, T, µ) =∞∑N=0

λNQ(N,V, T ) and λ = eβµ (3.3)

=⇒ Ξ =∞∑N=0

λN∑∗

nke−β

∑i εini =

∑N

∑∗

nkλ∑nke−β

∑εini (3.4)

=⇒ Ξ =∑N

∑∗

nk

∏k

(λe−βεk)nk (3.5)

Big thought step!

=⇒ Ξ =∏k

nkmax∑nk=0

(λe−βεk)nk (3.6)

3.2. FERMI-DIRAC AND BOSE-EINSTEIN STATISTICS 25

• Each set of nk can go from 0 to some max value → many combinations overN , that lets us make this simplification and get rid of our weird set notation.

- Side Note -

Let’s see why: For example, S =∑∞

N=0

∑∗nj x

n11 xn2

2 and n1 and n2 = 0, 1, and 2.

Show its equivalence:

S = (1 · 1 + 1x2 + 1x22 + x11 + x1x2 + x1x

22 + x2

11 + x21x2 + x2

1x22)

n1 + n2 = N , so all possible values of N allowed us to use different exponents up toN = 2 + 2 = 4.

Now, basic algebra gives S = (1 + x1 + x21)(1 + x2 + x2

2)→ S =∏2k=1(1 + xk + x2

k)

- End Side Note -

Fermi-Dirac Statistics: nk = 0 or 1 (particles can’t be in same energy level)

Then nmax1 = 1 and:

ΞFD =∏k

(1 + λe−βεk) (3.7)

Bose-Einstein Statistics: nk has no restriction on the occupancy of each state, sonmaxk =∞. Then:

ΞBE =∏k

∑∞nk=0(λe−βεk)nk

• But any∑∞

j=0 xj = (1− x)−1 when x < 1, so:

ΞBE =∏k

(1− λe−βεk)−1 (3.8)

26 CHAPTER 3. BOLTZMANN, F-D, AND B-E STATISTICS

The FD and BE equations can be written together as:

ΞFDBE

=∏k

(1± λe−βεk)±1 (3.9)

So we had: N =∑

k nk = kT(∂ ln Ξ∂µ

)V,T

(from Chap. 2, Chap. 3 in McQuarrie)

= λ(∂ ln Ξ∂λ

)V,T

=∑

k

λe−βεk

1± λe−βεk.

• And comparing we see the average number of particles in the kth quantum

state is nk =λe−βεk

1± λe−βεk.

• And we get E = Nε =∑

k nkεk =∑

k

λεke−βεk

1± λe−βεk

• From Chap. 3 in the MacQuarrie ℘V = kT ln Ξ = ±kT∑

k ln[1± λe−βεk ]

In Boltzman limit, energy states number of particles, so nk → 0 since λ→ 0, sowe can write nk = λe−βεk (for small λ) and E →

∑j λεje

−βεj .

• Also, ∓V → ±kT (±λ∑

j e−βεj ) = kTλq

• We can show λq = N , so ∓V = NkT

Thus, the quantum view agrees with Boltzmann statistics!

Chapter 4

Monatomic and Diatomic Gases

4.1 Ideal Monatomic Gas

• Single molecules, room temperature, low pressures (< 1 ATM)

• No molecules in same energy state and indistinguishable: Q = qN

N ! and

q =∑e−

εjkT

• We will approximate q as separable: q = qtransqelectqnucl

4.1.1 Translational Partition Function

Going back to the particle in a box: εnxnynz = h2

8ma2 (n2x + n2

y + n2z) (for energy

states)

Then, qtrans =∑∞

nx,ny ,nz=1 e−βεnxnynz

• Writing out each piece and summing, we get:

qtrans =∑∞

nx=1 e−βh

2n2x

8ma2∑∞

ny=1 e−βh2n2

y

8ma2∑∞

nz=1 e−βh

2n2z

8ma2 =

(∑∞n=1 e

−βh2n2

8ma2

)3

We want to replace the summation with an integral, so difference between two

successive energy states must be small: ∆ = βh2(nx+1)2

8ma2 − βh2n2x

8ma2 = h2(2nx+1)kT8ma2 , which

gives very small values ∼ O(10−20).

27

28 CHAPTER 4. MONATOMIC AND DIATOMIC GASES

• Since our ∆ is small, we can say it is approximately equivalent to integrate,

so: qtrans =[∫∞

0 e−βh2n2/8ma2

dn]3

=(

2πmkTh2

)3/2a3

(using identity∫∞

0 e−ax2

= 12

√πa )

• Replace a3 with volume:

qtrans =

(2πmkT

h2

)3/2

V (4.1)

Sum of states: Worry about the sum of different energy levels, εj , not theindividual n’s that comprise our εj .

• Now another way to look at it is: qtrans =∫∞

0 ω(ε)e−βεdε, where ω(ε) is thedegeneracy.

• And ω(ε) = π4

(8ma2

h2

)3/2ε1/2∆ε+O((∆ε)2)

• As ∆ε→ 0, we get q = π4

(8ma2

h2

)3/2 ∫∞0 ε1/2e−βεdε =

(2πmkTh2

)3/2V

• εj,trans =∑

j εje−εj/kT

qtrans= kT 2 ∂ ln qtrans

∂T = 32kT (Yay!)

DeBroglie Wavelength:

Λ ≡(

h2

2πmkT

)1/2

(4.2)

• And 12mv

2 = p2

2m = εtrans

• So p2

2m = 32kT → p = (mkT )1/2, but look at Λ.

• Λ ≈ hp (units of length, Plank is units of J-s)

• We use this to see when in the Boltzmann limit Λ3

V 1, and we seeqtrans = V

Λ3 → We can say Λ a for the Boltzmann limit.

4.1. IDEAL MONATOMIC GAS 29

4.1.2 Electronic Partition Function

qelect =∑

i ω(εi)e−βεi

We can choose where we define the zero state → Say it is at ε1 = 0

qelect = ω(ε1) + ω(ε2)e−βε2 + ω(ε3)e−βε3 + . . . (4.3)

• First term is important for most gases at ordinary temps.

• Second term is added in for alkali and halogens.

4.1.3 Nuclear Partition Function

Usually just ignore (set to 1)

qnucl = ωn1 (4.4)

• You need O(1010) K temps to get next state, just set to 1 then.

Q =[qtransqelectqnucl]

N

N !

We can use this to write some of our thermodynamic equations:

• Helmholtz Free Energy: A(N,V, T ) = −kT lnQ (useful work obtainable froma closed system at constant T )

• A = −kT ln[(

2πmkTh2

)3/2 V eN

]−NkT ln(ωe1 + ωe2e

−β∆ε12)︸︷︷︸usually very small

• Pressure!: ℘ = kT(∂ lnQ∂V

)N,T

= NkTV Cool stuff!

Note: Stirling approximation gives us the “e” in the ln term in A(N,V, T ).


4.2 Ideal Diatomic Gas

In addition to translational and electronic (and nuclear) degrees of freedom,diatomic molecules also have rotational and vibrational degrees of freedom (5total).

• Using Schrodinger Eq. is extremely difficult → We will use simplification tohelp.

4.2.1 1st Assumption: Born-Oppenheimer Approximation(matom me)

Nuclei are much more massive than electrons → move slow relative to the electrons

• Now the Schrodinger eq. approximately separates into 2 simpler equations:

– Motion of electrons about the “fixed” nuclei → eigenvalues uj(r)

– Motion of the nuclei in the potential field uj(r) that is set up by theelectrons in the electronic state j (uj(r) = ground state)

First excited state is usually several eV above ground → just consider ground state

• Approximate the potential curve as a Morse Potential → u0(r)→ re ≈ 2.6A(ur is internuclear potential)

Since potential is set to u0(r)→ spherically symmetrical → split into 2 equationsusing a center of mass motion (m1 +m2) and relative motion of reduced massµ = m1m2

m1+m2

• Now we can wrote our Hamiltonian as H = Htrans +Hint with eigenvaluesε = εtrans + εint

• The partition function for a diatomic molecule is then q = qtrans · qint andfrom earlier in the chapter (monatomic gases) and using our 2 masses:

qtrans =[

2π(m1+m2)kTh2

]3/2V

4.2. IDEAL DIATOMIC GAS 31

Now we have to investigate qint to understand diatomic molecules, and we will seeit will differ for heteronuclear and homonuclear molecules.

Relative motion of the two nuclei in our potential u(r) is rotary motion andvibration

• Vibration is small, call fixed internuclear distance re (rigid dumbbell)

• Expanding u(r) about re:

u(r) = u(re) + (r − re)(∂u∂r

)r=re

+ 12(r − re)2

(∂2u∂r2

)r=re

+ . . ., simplifies:

u(r) = u(re) + 12(r − re)2

(∂2u∂r2

)r=re

+ . . .,

k = ∂2u∂r2 → u(r) = u(re) + 1

2k(r − re)2

• First derivative at re = 0 (minimum potential → ground state) and k is“force constant” (stiffness)

• This assumption gives us our second big step and is called:

4.2.2 2nd Assumption: Rigid Rotor - Harmonic Oscillator(RRHO) Approximation

Hrot,vib = Hrot +Hvib and εrot,vib = εrot + εvib and qrot,vib = qrot · qvib

• Chap. 1.3 gives rigid rotor:

εJ = h2J(J+1)2I

ωJ = 2J + 1J = 0, 1, 2, . . .

I = moment of inertia = µr2e

µ = reduced mass

• Harmonic oscillator:

εvib = hν(ν + 12)

ωn = 1n = 0, 1, 2, . . .

ν = 12π

(k

µ

)1/2

k = ∂2u∂r2


In rotation, transitions from one state to another happen by going from 1 state toadjacent, or ∆J = ±1 and energy absorbed is at frequencies of h

4π2 I ∼ microwaveregion.

In vibration, the same occurs, but only at one frequency of 12π

(kµ

)1/2∼ 1000 cm−1

(infrared).

From the spectroscopy results, we define a rotational constant B = h8π2cI

so energyof a rigid rotor becomes εJ = BJ(J + 1) (c is the speed of light).

Now we will further split the Hamiltonian by assuming elect and nuclear areseparate:H = Htrans +Hrot +Hvib +Helect +Hnucl → ε = εtrans + εrot + εvib + εelect + εnucl

and q = qtrans · qrot · qvib · qelect · qnucl.

• This is NOT exact, but stands as a useful approximation for diatomicmolecules.

• Knowing qtrans and qelect is similar to the beginning of the chapter andqnucl = 1 again. We only need to solve our vibrational and rotationalpartition functions.

We need ground states:

• Rotational: Zero energy ground state occurs at J = 0 state

• Vibrational: Zero energy ground state is bottom of internuclear potential wellof lowest electronic state: ε0 = hν

2 (εvib = hν(n+ 12)).

• Electronic: Redefine as energy of ground state as −De, and electronicpartition function is now: qelect = ωe1e

De/kT + ωe2e−ε2/kT + . . .

4.2.3 Vibrational Partition Function

Remember εvib = hν(n+ 12), ν = 1

2π

(kµ

)1/2where k is force constant and µ is

reduced mass.

qvib =∑e−βεn → qvib =

e−βhν/2

1− e−βhν(math is in McQuarrie).


• This is because we had a geometric series and can compute sum → noapproximation.

• Now, with high enough temperature, βhν 1 and we make the prior sum anintegral:

qvib = e−βhν/2∑e−βhνn

→ qvib(T ) = e−βhν/2∫∞

0 e−βhνndn = kT/hν for kT hν

• Note since we have an exact solution, we don’t need this approximation, wecan compare:

Define vibrational temperature Θvib = hν/k (values in table 6.1 in McQuarrie)

• Fraction of molecules in a state above ground state

fn>0 =∑∞

n=1

e−βhν(n+1/2)

qvib= 1− f0 = e−Θvib/T

• This number is very small for most diatomic molecules except Br2 (table 6.2in McQuarrie)

Now we can observe the rotational partition function, but need to look at 2 cases.

4.2.4 Rotational Partition Function: Heteronuclear

Partition function by summing our energy levels: qrot =∑∞

j=0(2J + 1)e−βBJ(J+1)

(β = 1kT )

• Denote Bk = Θrot (characteristic temperature of rotation). Θrot = h2/8π2Ik

• We can’t do the sum like with vibration, so if we approximate that thedistances between energy levels is small: ∆ε

kT = 2Θr(J+1)T 1 then we can say

it is an integral

• As J increases, our distances do too, but we see the terms are so small itdoesn’t matter:


qrot(T ) =∫∞

0 (2J + 1)e−ΘrJ(J+1))/TdJ

→ qrot = T/Θr = 8π2IkTh2 for Θr T

• This improves in accuracy with increaseing temperature → High Temp Limit

If we have a low temperature or a high Θr, then perform the sum directly:

qrot(T ) = 1 + 3e−2Θr/T + 5e−6Θr/T + 7e−12Θr/T ∼ within 0.1% for Θr > 0.7 T .

If we are in an intermediate range where we aren’t too small but smaller than0.7 T , then we can find an approximation using an Euler-MacLauren expansion:

qrot(T ) = TΘr1 + 1

3

(ΘrT

)+ 1

15

(ΘrT

)2+ 4

315

(ΘrT

)3+ . . . ∼ within 0.1% for Θr < T

4.2.5 Rotational Partition Function: Homonuclear

With same atoms in a molecule, we get coupling due to interlocking wave functionsand now we must pair our qrot with qnucl based on partical symmetry

Bosons: Nuclei with integral spin - Rotational levels with odd values of J getcoupled with the I(2I + 1) antisymmetric nuclear spin functions and levels witheven values of J must be coupled with the (I + 1)(2I + 1) symmetric nuclear spinfunctions:

qrot,nucl = (I+1)(2I+1)∑

J even(2J+1)e−ΘrJ(J+1)/T +I(2I+1)∑

J odd(2J+1)e−ΘrJ(J+1)/T

Fermions: Likewise, nuclei with half-integer spin have it swapped:

qrot,nucl = I(2I+1)∑

J even(2J+1)e−ΘrJ(J+1)/T +(I+1)(2I+1)∑

J odd(2J+1)e−ΘrJ(J+1)/T

• We can see that qrot,nucl 6= qrotqnucl → Have to consider nuclear contribution!

• For most molecules at ordinary temperatures, we had Θr T , so replace the∑with an integral

• Then we can see both equations above (integral and half-integer spin)become:


qrot,nucl =(2I + 1)2T

2Θr= qrotqnucl, where qrot(T ) =

T

2Θrand qnucl = (2I + 1)2

• For this to be valid, we must have Θr/T < 0.20

Notice the similarity between the hetero- and homonuclear → a factor of 1/2,which will only work when in the High Temp Limit of Θr < 0.20 T

• This high temp limit allows us to call the factor of 2 as σ, and is oursymmetry number.

• Now, we can write:

qrot(T ) ≈ 8π2IkT

σh2≈ 1

σ

∑(2J + 1)eΘrJ(J+1)/T ,

where Θr T , σ = 1 for heteronuclear, and σ = 2 for homonuclear.

Hydrogen (H2) is a molecule where we aren’t in the high temp limit → Can’t useour equation.

• Table 6-1 in McQuarrie shows almost all molecules are at high temp (Θr T )

• Therefore, we can use the Euler-MacLauren expansion for most cases, and:

qrot(T ) = TσΘr1 + 1

3

(ΘrT

)+ 1

15

(ΘrT

)2+ 4

315

(ΘrT

)3+ . . .

• Usually only the first term is necessary.

4.2.6 Putting it all Together

We have studied each partition function and can now assemble them together:

• q(V, T ) = qtransqrotqvibqelectqnucl, giving Eq. 4.5:

q(V, T ) =

(2πmkT

h2

) 32

V︸︷︷︸qtrans

· 8π2IkT

σh2︸︷︷︸qrot

· e−βhν

2

1− e−βhν︸︷︷︸qvib

·ωe1 + ωe2e−∆ε12

kT︸︷︷︸qelec

· 1︸︷︷︸qnucl

(4.5)


• Q(N,V, T ) = qN

N !

Assumptions:

• Θr << T is the largest restriction and must hold

• Only the ground electronic state is important

• The zero of energy is the separated electronically unexcited atoms at rest intheir ground electric states

• Rigid rotor harmonic oscillator is our weakest assumption → most error

This partition function will yield results with very small errors when compared toexperimental data → sometimes < 1% error.

• Correction factors can be applied (Problems 6-23 and 6-24 in Ref. [3])

Overall, we start with the Schrodinger equation, get the εj eigenvalues, applying itto our partition functions we derived, and assume a high temperature limit.

• This is only good up to our diatomic molecule- not even looking at systems!

Note: The Hamiltonians and Energy Equations: Table 4.1

Translation Htrans = p2

2m =p2x+p2

y+p2z

2m εtrans = h2n2

8ma2 , n2 = n2x + n2

y + n2z

Rotation Hrot = 12I

[p2θ +

p2φ

sin2(θ)

]εrot = h2

2I J(J + 1), ω = 2J + 1

Vibration Hvib =p2

vib2m + 1

2kx2 εvib = hν(n+ 1

2), ω = 1

Table 4.1: Hamiltonians and Energy Equations.

Part II

Kinetic Theory of Gases

37

Chapter 5

Overview of Kinetic Theory ofGases

• Look at a classical system since we usually work in high temperature limit

• Define phase space

• Ensemble of systems → many phase points → Density in phase space (f isdistribution function)

• The equation of motion of phase space distribution is described bythe Liouville equation

– Shows the time evolution of the phase space distribution function f

– Valid for both equilibrium and non-equilibrium systems

• Next look to reduced distribution functions (not all particles interact or areimportant)

• Using Liouville equation introduce reduced distribution function f (1) and f (2)

and work into it

• We get the BBGKY Hierarchy → set of equations describingdynamics of a system of a large number of interacting particles → achain of equations f (s) depends on f (s+1)

• In dilute gases, most molecules don’t interact, so we only care about singletdistribution function f (1)

39

40 CHAPTER 5. OVERVIEW OF KINETIC THEORY OF GASES

– Once an expression for f(1)j (r,v, t) is found, you can calculate fluxes and

thus transport properties

– But getting f (1) from BBGKY is very difficult because of dependency onfollowing terms

• f (1)j can be found by deriving the Boltzmann equation, fundamental

equation for kinetic theory of gases

• Assume only 2 body interactions are important and see how number ofj-molecules in drdv change

– Account for inward and outward fluxes over dt and assume molecularchaos

• Find integrodifferential equation for f(1)j , one for each component of the

dilute gas

• Solving Boltzmann equation isn’t easy, but you can gain insight from itwithout solving!

• Enskog’s general equation of change for a property ψj shows thatmultiplying Boltzmann equation by a collisional invariant andintegrating over dvj gives transport equations

– ψj =mass gives continuity equation, ψj =momentum gives momentumequation, ψj =kinetic energy gives energy balance equations

• Boltzmann H-theorem gives dHdt ≤ 0, Boltzmann equation has

preferred direction

– This is analogous to the 2nd law of thermodynamics → we will tendtowards an equilibrium state

– While fluctuations occur (we deal with probability), we always trend toequilibrium

– Recurrence also questioned → shows it happens on order of 101019years,

so no conflict with idea of irreversibility on most physically interestingsystems

• From H-theorem, we find equilibrium gives our distribution ofvelocity as a Maxwellian distribution

• Solving Boltzmann equation is done with Chapman-Enskog methodto give successive approximations

41

• Each successive approximation of f (singlet) is added to get more refinedsolution

• Solutions of this type are in final stage of relaxation → equilibrium →hydrodynamic stage

• We find first approximation f [0] gives ideal hydrodynamic equations,p[0] = ρkT I and q[0] = 0

• Second approximation yields f [0] + f [1] giving Navier Stokes equations → getexpressions for λ, η, D12

• Third approximation with f [2] yields Burnett equations, but limited use to us

• Note: all around equilibrium is where we are useful

• The expressions for thermal conductivity λ, shear viscositycoefficient η, and binary diffusion coefficient D12. All arecomplicated integrals that end up depending onintermolecular potential u(r)

• From u(r), get deflection angle → collisional cross-sectional area → collisionintegral → aij and bij → λ, η, D12

• Often referenced from hard sphere model, but Leonard-Jones potential ispretty good

– Coeff. usually found from “one term” expansion, two is more accurate,three almost dead on.

42 CHAPTER 5. OVERVIEW OF KINETIC THEORY OF GASES

Chapter 6

Classical Statistical Mechanics

So far, we derived partition functions for linear molecules → using Schrodingerequation and molecular properties up to a dilute diatomic gas at high temperature.

• Remember for HTL: number of available molecular quantum states number of particles so states occupied likely contain only one molecule.

• In HTL, we replaced sums with integrals → gave numerically satisfactoryanswers (except vibration).

At higher temperature, average energy per molecule increases, so quantumnumbers describing motion also increase (think radius).

• (nx for translation, J for rotation) Molecules are in high quantum numberlimit, ≈ O(104) at room temperature.

• Classical behavior is obtained in limit of large quantum numbers, so can westay in context of classical mechanics? Do like Gibbs did before quantummechanics fully came about.

6.1 Classical Partition Function

We’ve used the molecular partition function q =∑e−βεj which is a sum over all

possible quantum energy states → Let’s postulate that classical partition function

43

44 CHAPTER 6. CLASSICAL STATISTICAL MECHANICS

will also be over classical energy states.

• Energy in a classical sense is a continuous function of momentum andposition (pj and qj).

• Then sum is an integral, and classical energy is Hamiltonian function: H(p, q)qclassical ≈

∫. . .∫e−βH(p,q)dpdq (conjecture)

The Hamiltonian depends on p1, p2, . . . , ps and q1, q2, . . . , qs, and s is the number ofcoordinates needed to completely specify the motion or position → s represents thedegrees of freedom of molecule.

qj are generalized coordinates, pj are conjugate momentum, choose q that isconvenient.

We had for monatomic ideal gas: qtrans =[

2πmkTh2

]3/2V (quantum partition

function)

Hamiltonian (classical) of one atom of monatomic ideal gas: Simply the KE!H = 1

2m(p2x + p2

y + p2z)

• Plugging in H to our qclass yields a sixfold integral (3 coordinates, 3momentum). Integrating gives:

qclass ≈ V[∫∞−∞ e

βp2/2mdp]3

= (2πmkT )3/2V

• This is the same as our quantum partition function without the Plank’sconstant factor.

• We can’t expect to derive a classical expression and get Plank’s constant- onthe right path

Let’s compare with rigid rotor as well to see: qrot = 8π2IkTh2 (for quantum level)

• With rigid rotor Hamiltonian: H = 12I (p2

θ +p2φ

sin2 θ)→ qclass = 8π2IkT

Harmonic oscillator: qvib = kThν (in quantum level)

6.1. CLASSICAL PARTITION FUNCTION 45

• Hamiltonian: H = p2

2µ + k2x

2 → qclass = kTν

We see that we are off by h3, h2, and h for 3, 2, and 1 degree of freedom in eachqclass.

• Also, q is dimensionless, so the factor of h will correct the units.

• Therefore, we will assume that for classical mechanics we can write:

q =∑j

e−βεj︸︷︷︸quantum

→ 1

hs

∫. . .

∫e−βH

s∏j=1

dpjdqj︸︷︷︸classical

(6.1)

This applies to a single molecule, so expanding to a system of N independentindistinguishable particles:

Q = qN

N ! , so Qclass = 1N !

∏Nj=1

[1hs

∫. . .∫e−βH

∏sNi=1 dpidqi

](rewriting an exponent)

• Hj is Hamiltonian of the jth molecule, function of pj1, . . . , pjs and qj1, . . . , qjs.

• Rewriting and relabeling indices allows us to write, with a Hamiltonian H ofthe N -body system:

Q =1

N !hsN

∫. . .

∫e−βH

sN∏i=1

dpidqi (6.2)

We conjecture this partition function is the classical partition function, and H(pq, )is the classical N -body Hamiltonian for interacting particles [(p, q) represents set ofpj , qj for entire system]

• To show what is happening with particle interactions, let us expand ourHamiltonian.

• For a monatomic gas,H(p, q) = 1

2m

∑Nj=1(Pxj + Pyj + Pzj) + U(x1, y1, z1, . . . , xN , yN , zN )

(translation)


• Our Hamiltonian now contains an internal energy that is dependent onparticles’ position in N -body system

• This potential energy describes the particle interactions

This partition function is actually the correct classical limit of Q→ Proven in 1934with rigorous method.

Plugging in our new H(p, q) into the expression for Q will give the partitionfunction for classical monatomic gas with particle interactions (translation):

Qclass =1

N !

(2πmkT

h2

)3N/2

ZN where ZN =

∫Ve−U(x1,...,zN )/kTdx1 . . . dzN (6.3)

• ZN is the classical configuration integral (focus of current research) -describes intermolecular forces.

• Without intermolecular forces, or a dilute gas, we have U → 0, ZN → V N

(Previoius Q Eq.)

It often happens that not all degrees of freedom can be treated classically (likevibration).

• We then treat the degrees we can (translation, rotation, etc.) classically andones we can’t (vibration) quantumly → then we can sayH = Hclass +Hquantum → q = qclassqquantum,qclass = 1

hs

∫e−Hclass(p,q)/kTdp1dq1 . . . dpsdqs

• Therefore, assuming a separable Hamiltonian, the entire system can bewritten as:

Q = QclassQquantum =Qquantum

N !hsN

∫e−Hclass/kTdpclassdqclass (6.4)

6.2 Phase Space and Liouville Equation

Historically, statistical mechanics was formulated by Boltzmann, Maxwell, andGibbs before quantum mechanics.

6.2. PHASE SPACE AND LIOUVILLE EQUATION 47

• It was based in classical mechanics, which is still a most useful limit → usedfor many systems today

Consider: System with N interacting particles, each with s degrees of freedom tocompletely describe its position.

• To describe all molecules, need l = sN coordinates: q1, q2, . . . , ql give spatialorientation of N -body system.

• Also have conjugate momentum pj , with l = sN momenta, for 2l totalcoordinates of system.

• The 2l coordinates, along with the equations of motion, completely describethe system and determine the future and past course of the system.

Let’s construct a conceptual Euclidean space of 2l = 2sN dimensions, one axis foreach p and q.

• This phase space describes the state of an N -body system under classicalconsideration.

• A single phase point at time t completely specifies the state, and as thesystem evolves through time, the dynamics of the system is described by themotion of the phase point through phase space.

The trajectory of the phase point is given by Hamilton’s equations of motion:

qJ =∂H

∂pj, pJ = −∂H

∂qjand j = 1, 2, . . . , l = sN (6.5)

Now consider an ensemble of systems in phase space, for simplicity say amicrocanonical ensemble.

• Each ensemble has the same fixed N,V,E, but their own p, q.

• We will have a large number A of these isolated systems with fixedmacroscopic properties.


Each system in the ensemble has its own phase point in the same phase space todescribe it.

• The entire ensemble appears as a “cloud” of phase points, and as timeevolves, each point will trace its own independent trajectory (since they areall isolated)

Postulate of equal a priori probabilities state that for a microcanonical ensemble,the density is uniform over the “surface” of constant energy in phase space (valueof energy of isolated system)

• Therefore, as long as (p, q) for each system satisfies N,V,E, then that state isequally probable as others → every classical state is equally probable (likequantum states)

Define a number density for “cloud” as f(p, q, t) (distribution function) so # ofsystems in ensemble have phase points in dpdq about the point p, q at time t isf(p, q, t)dpdq →

∫. . .∫f(p, q)dpdq = A

• Then ensemble average of any function φ(p, q) isφ = 1

A

∫. . .∫φ(p, q)f(p, q)dpdq

• Gibbs Postulate equates this ensemble average to the correspondingthermodynamic function

The equations of motion determine each phase point trajectory, so they alsodetermine density f(p, q, t).

• If initial condition is known, then we have f(p, q, t)→ controlled by laws ofmechanics, not arbitrary!

• The time dependence of f is given by the Liouville Eq., now we derive (likefluids).

Considering a small volume element (δp1 . . . δplδq1 . . . δql) about the point(p1, . . . , pl, q1, . . . , ql) and the number of phase points within the volumeδN = f(p1, . . . , pl, q1, . . . , ql, t)δp1 . . . δplδq1 . . . δql (see Fig. 6.1)

6.2. PHASE SPACE AND LIOUVILLE EQUATION 49

Figure 6.1: Elemental volume of phase space.

• This number will change as points move → consider points through facesperpendicular to q1-axis.

• Using linear expansions of f and q1 to linear terms in δq1, we can get a netflow of phase points in the q1 direction:

Net Flow = −(∂f∂q1q1 + f ∂q1∂q1

)δp1 . . . δplδq1 . . . δql

• This can be found in all p and q directions (equal status to all coordinates inphase space) and we will get the change through all faces, which must equalthe change of δN with time:d(δN)dt = −

∑lj=1

[f(∂qj∂qj

+∂pj∂pj

)+(∂f∂qjqj + ∂f

∂pjpj

)]δp1 . . . δplδq1 . . . δql

We can simplify using our Hamiltonian Eq. of motion and dividing by the “volumeelement” to get the rate of change of density itself around the pointp1, . . . , pl, q1, . . . , ql (∂f∂t ). This describes the evolution of phase point density:

∂f

∂t= −

l∑j=1

(∂f

∂qjqj +

∂f

∂pjpj

)(6.6)

• We have ∂f∂t to indicate we fixed attention on a given stationary point in

phase space.

• This can also be written in a more conventional form using Eqs. for qj and pj :

∂f

∂t= −

l∑j=1

(∂H

∂pj

∂f

∂qj− ∂H

∂qj

∂f

∂pj

)(6.7)


or in cartesian coordinates:

∂f

∂t+

N∑j=1

pjmj· ∇rjf +

N∑j=1

Fj · ∇pjf = 0 (6.8)

This is our Liouville Eq., the most fundamental equation of classical statisticalmechanics.

• In cartesians, ∇rj is gradient with respect to spatial variables in f , ∇pj isgradient with respect to momentum variables in f , and Fj is the total forceon the jth particle.

Since f depends on p, q, and t, our Liouville Eq. looks like a total derivative, andrearranging gives df

dt = 0, so density in the neighborhood of any moving phase pointis constant along its trajectory.

• Like density of incompressible gas: and given an initial condition:f(p, q, t) = f(p0, q0, t0)

• As time progresses, an elemental volume will distort, but the phase pointswill stay within, no change in density.

• Gibbs called this the “conservation of extension in phase space”:δpδq = δp0δq0 at all t.

A corollary to this says two different phase space coordinates describing the samesystem have, for q1, . . . , q3n, p1, . . . , p3n and Q1, . . . , Q3n, P1, . . . , P3n:dq1 . . . dq3ndp1 . . . dp3n = dQ1 . . . dQ3ndP1 . . . dP3n.

• For example, a single particle in 3 dimensions can be equally described by(x, y, z) or (r, θ, φ):dpxdpydpzdxdydz = dprdpθdpφdrdθdφ

• This only holds true because we use conjugate momenta → velocitieswouldn’t allow this.

6.3. EQUIPARTITION OF ENERGY 51

6.3 Equipartition of Energy

In classical statistical mechanics → high temp limit:∑→∫

and only needHamiltonian, not eigenvalues of quantum problem

• Consider average energy of a molecule in a system of independent molecules:

ε =∫∫

He−βHdp1...dqs∫∫e−βHdp1...dqs

.

• Multiplying by N gives total energy of system, differentiating by T givesconstant volume heat capacity: CV = dε

dT .

• If Hamiltonian is of the formH(p1, . . . , qs) =

∑mj=1 ajp

2j +

∑nj=1 bjq

1j +H(pm+1, . . . , ps, qn+1, . . . , qs) and

aj , bj are constants, then each quadratic term contributes kT/2 to energy andk/2 to CV .

• This equipartition of energy is a consequence of quadratic terms, not due toclassical statistical mechanics.

Monatomic ideal gas: H = 12m(p2

x + p2y + p2

z)→ each atom gives 3kT/2 to energyand 3k/2 to CV (like Chap. 4).

Rigid rotor: H = 12I (p2

θ +p2φ

sin2 θ)→ still gives kT to energy becuase θ not φ in

“constant” terms on p2φ.

Note: only holds in classical limit: ∆εkT 1→Not true for vibrational degree of

freedom.

• Chap. 4 had CV = 52Nk + Nk(Θv/T )2eΘv/T

(eΘv/T−1)2 for RRHO approximation. 52Nk

from translation and rotation.

• From harmonic oscillator Hamiltonian, expect to give Nk to CV , but itwasn’t true → deviations!

• Then the high ΘV in some gases causes a deviation from CV = 72Nk total.

The Liouville Eq. serves as a great starting point for most approaches.

Chapter 7

Kinetic Theory of Gases

We just went over phase space and distribution functions in phase space.

• Also derived Liouville Eq. (Eq. of motion that phase space distributionfunction must satisfy).

• We didn’t look at it too much because we were looking only at equilibriumstatistical mechanics

We will review these concepts again and introduce reduced distribution functionsand derive BBGKY Hierarchy.

• Then we get to derive physical, yet approximate, equations for thedistribution function of gases.

• This will be the Boltzmann Eq. - the central Eq. for the rigorous kinetictheory of gases.

7.1 Phase Space and Liouville Eq. (Take 2)

Consider a system of N point particles - classical dynamical state of 3Nmomentum and 3N position (p and q).

53

54 CHAPTER 7. KINETIC THEORY OF GASES

• We construct a 6N -dimensional space with coordinatesq1, q2, . . . , p1, . . . , p3N → one point completely specifies the microscopicdynamical state of our N -particle system (one point consists of all particlesand their p’s and q’s)

• Through time, the phase point moves through phase space in a mannerdictated by Eq. of motion for the system.

In reality, we don’t know (or care) about the 6N coordinates of a macroscopicsystem- we just know a few macroscopic mechanical properties of the system, suchas energy, volume, velocity, etc.

• There are a great number of points in phase space compatible with the fewvariables we know about the system → the set of all those phase pointsconstitutes an ensemble of systems.

• Number of systems in ensemble →∞, so is very dense: Define density ofphase points (distribution function) as the fraction of phase points containedin the volume dq1dq2 . . . dp3N as fN (dq1dq2 . . . dp3N , t).

• Simplify with fN (p, q, t) and volume element (dp, dq) to write with ease

The density (distribution function) fN (p, q, t) is normalized such that∫fN (p, q, t)dpdq = 1.

Each phase point moves in time according to Eqs. of motion of the system → fNobeys same Eqs. of motion.

Number of phase points within some arbitrary volume V isn = N

∫V fN (p, q, t)dpdq where N is number of phase points in phase space.

• Rate of change of phase points in V is: ∂n∂t = N

∫V∂fN (p,q,t)

∂t dpdq, and sincephase points aren’t created or destroyed, the rate of change of n must be therate of phase points flowing through surface of V .

• Rate of flow of phase points: NfNu, where u = (q1, . . . , p1, . . . , p3N ) -“velocity” through phase space.

• Integrate flow over surface: ∂n∂t = −N

∫S fnu · dS (negative for outward flow).

7.2. REDUCED DISTRIBUTION FUNCTIONS 55

• Use Gauss theorem, equate our two ∂n∂t Eqs., then conservation of phase

points:∂fN∂t

+∇ · (fNu) = 0 (7.1)

∇ · (fNu) =∑3N

j=1∂∂qj

(fN qj) +∑3N

j=1∂∂pj

(fN pj) =∑3Nj=1

[∂fN∂qj

(qj) + ∂fN∂pj

(pj)]

+∑3N

j=1

[∂qj∂qj

+∂pj∂pj

]fn

• Plugging in Hamilton’s Eqs. of motion (pj = −∂H∂qj

and qj = ∂H∂pj

) gives second

summation = 0 and we have:

∂fN∂t

+

3N∑j=1

(∂H

∂pj

∂fN∂qj− ∂H

∂qj

∂fN∂pj

)= 0 (7.2)

which can be written using a Poisson bracket:

∂fN∂t

+ H, fN = 0 (7.3)

This is our Liouville Eq., the most fundamental Eq. in statistical mechanics.

In cartesian coordinates: ∂fN∂t +

∑Nj=1

pjmj· ∇rjfN +

∑Nj=1 Fj · ∇pjfN = 0 (Fj is

total force on jth particle, gradient with respect to spatial or momentum variables).

• Often write Liouville Eq. as i∂fN∂t = LfN with Liouville operator

L = −i(∑N

j=1pjmj· ∇rj +

∑Nj=1 Fj · ∇pj

).

• The Liouville operator is defined in such a way to allow us to bring LiouvilleEq. into form of Schrodinger Eq.

7.2 Reduced Distribution Functions

Once we have a distribution function fN (p, q, t), we can compute the ensembleaverage of any dynamical variable A(p, q, t).


• Our equation for ensemble average is 〈A(t)〉 =∫A(p, q, t)fN (p, q, t)dpdq.

• It turns out that dynamical variables of interest are functions of either the p’sand q’s of just a few particles or they can be written as a sum over suchfunctions - overkill here! We just need a few terms.

• An example is intermolecular potential, depending on only pair-wisepotentials and can be summed:〈U〉 =

∑i,j

∫. . .∫u(ri, rj)fN (r1, . . . ,pN , t)dr1 . . .pN for paired coordinates

ri and rj

Let’s define reduced distribution function f(n)N (r1, . . . , rn,p1,pn, t) as:

f(n)N (r1, . . . , rn,p1,pn, t) = N !

(N−n)!

∫. . .∫f

(n)N (r1, . . . ,pN , t)drn+1 . . . drNpn+1dpN

• In this, we’ve reduced the number of variables needed! Also, drop subscriptand write as f (n)(rn,pn, t).

We usually only need f (1) and f (2), so let’s derive an equation for both.

• Writing the force in the Liouville Eq. Fj as the sum of forces due to othermolecules in the system

∑i Fij and an external force Xj , then applying some

math and integration, we get:

∂f (n)

∂t+

n∑j=1

pjmj· ∇rjf

(n) +n∑j=1

Xj · ∇pjf(n) +

n∑i,j=1

Fi,j · ∇pjf(n)

+n∑j=1

∫∫Fj,n+1 · ∇pjf

(n+1) drn+1dpn+1 = 0 (7.4)

This is the Bogoliubov, Born, Green, Kirkwood, Yvon or BBGKY Hierarchy.

• This hasn’t been truncated by superposition yet, but we can deriveapproximate Eqs.

• We’ve done all this independent of density → now let’s specialize to dilutegases.

7.3. FLUXES IN DILUTE GASES 57

7.3 Fluxes in Dilute Gases

In dilute gases, most molecules aren’t interacting with any others, just travelingbetween collisions.

• This means macroscopic properties of the gas depend upon only the singlet

distribution function f(1)j (r,p, t).

• This is for species j with position r and momentum p→ only looking atdilute gases, so let’s drop superscript.

• Also write our Eqs. with velocity in space rather than momentum

f(1)j → fj(r,vj , t).

• We shall normalize fj such that the integral of this distribution function overall velocities is the number density of j particles at position r at time t:ρj(r, t) =

∫fj(r,vj , t)dvj .

• Furthermore, if N is the total number of j molecules in the system, thenNj =

∫∫fj(r,vj , t)drdvj (integral of ρj over dr position).

Now we go on to define a number of important average velocities.

The linear velocity vj is the velocity of a molecule of species j with respect tofixed coordinate system.

• Average velocity is then given by: vj(r, t) = 1ρj

∫vjf(r,vj , t)dvj .

• This represents the macroscopic flow of species j.

The mass average velocity is defined by: v0(r, t) =∑j mjρjvj∑j mjρj

.

• Note that the denominator is the mass density ρm(r, t) =∑

jmjρj .

• This velocity v0(r, t) is often called the flow velocity or stream velocity.

The peculiar velocity Vj is the velocity of a molecule relative to the flow velocity.


Figure 7.1: Cylinder containing all molecules of species j with velocity Vj , whichcross the surface dS during the time interval dt.

• This is defined as Vj = vj − v0.

The average of the peculiar velocity is the diffusion velocity:Vj = 1

ρj

∫(vj − v0)fj(r,vj , t)dvj .

• It is easy to show that∑

j ρjmjVj = 0 (all diffusion sums to zero → differentfrom the average).

Our ultimate goal is to get transport coefficients of a fluid.

• Let’s designate the molecular properties of mass, momentum, and kineticenergy as a collective ψj .

• Now, we will derive fluxes of these properties for the species j: imagine thesurface dS shown in Fig. 7.1.

The surface dS moves with velocity v0, unit normal n.

• All the molecules that have velocity Vj = vj − v0 and that cross dS in timeinterval (t, t+ dt) must have been in a cylinder of length |Vj |dt and base dS.

• The volume of this cylinder will then be (n ·Vj)dSdt.

7.3. FLUXES IN DILUTE GASES 59

• The number of molecules per unit volume is fjdvj , so the number that crossdS in dt with relative velocity Vj is (fjdvj)(n ·Vj)dSdt.

• We then have a flux of fj(n ·Vj)dvj , and if each molecule carries property ψjthen we can say the flux of the property is ψjfj(n ·Vj)dvj , and the total fluxacross the surface is:

total flux =∫ψjfj(n ·Vj)dvj = n ·

∫ψjfjVjdvj = n ·ψj .

The vector ψj =∫ψjfjVjdvj is the flux vector associated with property ψj , and

the component of this vector in any direction is the transport of the property ψj inthat direction.

7.3.1 Transport of Mass

ψj = mj → ψj =∫mjfjVjdvj = ρjmjVj ≡ jj

7.3.2 Transport of Momentum

ψj = mjVjx → ψj = mj

∫VjxfjVjdvj = ρjmjVjxVj

• This is the flux of the x-component of momentum relative to v0: a pressure!

Example pressure components

(pj)xx = ρjmjVjxVjx

(pj)xy = ρjmjVjxVjy(7.5)

More general partial pressure tensor of jth species

pj = ρjmjVjVj (7.6)

7.3.3 Transport of Kinetic Energy

ψj = 12mjV

2j → ψj =

mj2

∫v2jVjfjdvj = 1

2ρjmjV 2j Vj = qj

• qj is the heat flux vector of the jth species.


We now see that once we have an expression for fj(r,v, t), we can calculate all thefluxes and hence get all the transport properties of a dilute gas.

• We now need fj or at least an equation to give us fj as its solution.

• All we have for now is the BBGKY hierarchy with f (1), but it also containsf (2), and as mentioned no one has found a way to uncouple the equation.

• We next derive an equation for fj , the Boltzmann Eq.: The fundamental Eq.of the rigorous kinetic theory of gases.

7.4 The Boltzmann Equation

Using a simple physical argument, we will derive the Boltzmann Eq. for fj (firstorder f (1), drop superscript).

• Imagine a volume dr dvj at point (r,v) in phase space for a dliute gas (only2-body interactions are important).

• The number of j molecules in the element dr dvj is given by fjdr dvj .

• If we had no collisions, all particles at the starting CV will move with theequations of motion to a second point at t+ dt, this second point being(r + vjdt,vj + 1

mjXjdt).

• Since all points in dr dvj stay within over t+ dt, we havef(r,vj , t) = f(r + vjdt,vj + 1

mjXjdt, t+ dt).

That assumed no collisions, but that isn’t true! Some molecules leave or enter thestream due to collisions throughout the movement.

• Denote j molecules lost from the position range (r, r + vjdt) and velocityrange (vj ,vj + dvj) because of collisions with i molecules over time interval

(t, T + dt) as Γ(−)ji dr dvj dt.

• We similarly denote j molecules joining the stream from collisions with i

molecules as Γ(+)ji dr dvj dt.

7.4. THE BOLTZMANN EQUATION 61

Figure 7.2: Collisions of molecules of type i with one molecule of type j. The distanceA is essentially the intermolecular distance at which the potential begins to “takehold.”

• Putting the collision terms inoto our “flow equation” of particles over dt givesus:

fj(r + vjdt,vj + 1mj

Xjdt, t+ dt)dr dvj =

fj(r,vj , t)drvj +∑

i(Γ(+)ji − Γ

(−)ji )dr dvj dt

Expanding the left side gives:

∂fj∂t + vj · ∇rfj +

Xj

mj· ∇vjfj =

∑i(Γ

(+)ji − Γ

(−)ji )

• LHS is change in fj due to collisionless motion, RHS is change in fj due tocollisions.

• Notice this is very similar to the Liouville Eq. for f(1)j .

We now must find explicit expressions for the collision terms.

• Consider a fixed j particle with incoming i particle with relative velocity gijwith impact parameter in range of db around b.


• A is the range of intermolecular potential → when collisions would start tohappen.

• Then any i molecule in the cylinder will collide with the fixed j moleculeduring the time dt.

The probable number of i molecules in the cylindrical shell is 2πfi(r,vi, t)gijb db dtwith gij = |gij |.

• The total collisions that would occur with the fixed j molecule is then2πdt

∫∫fi(r,vi, t)gijb db dvi.

• But since j isn’t fixed, we have a probability of them being at the r pointwith vi → fj(r,vj , t).

• We can put terms together to write Γ(−)ji = 2π

∫∫fjfigijb db dvi.

Note that we assurred that the mean number of i molecules about the fixed j isgiven by fjfi.

• We are assuming they are uncorrelated → known asmolecular chaos assumption (not strictly true).

• This is the “weakest” point in derivation, but very important to our work.

We can do the same process for the reversed case → particles introduced to streamfrom collisions.

• Rewriting the equation by utilizing primes for quantities before collisionswhich go into b, vi, vj after collision (eg. b′, v′j , etc.) and invoking Liouvilletheorem (dp dq − dp′ dq′), we get:

Γ(+)ji = 2π

∫∫f ′if′jgijb db dvi (f ′ means its velocity is primed).

We can now plug Γ(+) and Γ(−) back into our earlier equation to get theBoltzmann Eq.:

7.5. CONSEQUENCES OF THE BOLTZMANN EQUATION 63

∂fj∂t

+ vj · ∇rfj +1

mjXj · ∇vjfj = 2π

∑i

∫∫f ′if ′j − fifjgijb db dvi (7.7)

This is an integrodifferential Eq. for f(1)j , and have one for each component j of the

gas.

• Eqs. of motion (thus intermolecular potential) enter the equation through f ′iand f ′j because the postcollisional velocities (v′i and v′j) depend onprecollisional velocities (vi and vj) through the Eqs. of motion governing thecollision.

• Also, RHS after integration yields variables of r, vj , and t, which is what ison LHS → good!

This was the simple physical derivation → standard way of going about it.

• We find though that it is very difficult to solve, which we will do later, but wecan still find important information from this equation as is!

7.5 Consequences of the Boltzmann Equation

We will find three primary consequences from the Boltzmann Eq. without solvingit.

7.5.1 1. Equations of Change

First, we will look at the relation to the equations of change (continuum mechanicsEqs.).

Multiplying the Boltzmann Eq. by ψi and integrating over dvi for fi will yield anequation where we split the LHS → looks like 3 different averaging terms for ψi.

• By doing the math to average, we can plug the 3 relations back into our new


Eq. to get:

∂(ρiψi)

∂t+∇r · ρiψivi − ρi

∂ψi∂t

+ vi · ∇rψi +Xi

mi· ∇viψi

= 2π∑j

∫∫∫ψi(f

′if′j − fifj)gijb db dvi dvj (7.8)

• This is Enskog’s general Eq. of change for the property ψi.

We find that for ψi = mass, momentum, or energy (collisional invariants) thenRHS = 0.

• What we can then do is plug in ψi = mi (or mivi or 12miv

2i ) and sum over i’s

to get:

– ψi = mi → ∂ρi∂t +∇r · (ρivi) = 0→ ∂ρ

∂t +∇r · (ρv0) = 0, Continuity Eq.for the entire system!

– ψi = mivi → sum over i→ ρmDv0Dt =

∑j ρjXj −∇p

– ψi = 12miv

2i → sum over i→ ρm

DEDt = −p : ∇u−∇ · q, where : denotes

the standard tensorial manipulation A : B =∑

i

∑j AijBji

7.5.2 2. Boltzmann H-Theorem

Next, we will discuss the Boltzmann H-Theorem by considering a single componentsystem.

• The Boltzmann H-function is defined as H(t) =∫∫

f(r,v, t) ln[f(r,v, t)]dr dv.

• Taking derivatives, equating to Boltzmann Eq. and other math will give us:

dHdt = 2π

4

∫∫∫ln(ff1

f ′f ′1

)f ′f ′1 − ff1gb db dr dv1

• Integrand is of the form of −(x− y) ln(xy

), which for all x and y is either

zero or negative.

• Therefore, we get the result that dHdt ≤ 0

7.5. CONSEQUENCES OF THE BOLTZMANN EQUATION 65

H is bounded, so it must approach a limit as t→∞, which we find gives us dHdt = 0,

so we have equilibrium or steady state with f ′f ′1 = ff1 or ln f ′+ ln f ′1 = ln f + ln f1.

• This tells us that f is a summational invariant, but the only summationalinvariants of bimolecular collisions of spherically symmetric molecules aremass, momentum, and kinetic energy.

• Therefore, ln f must be a linear combination of these three quantitites: m,mv, 1

2mv2

7.5.3 3. Equilibrium Solution

By performing some math it is possible to solve for the 3 unknown parameters ofour quantities of mass, momentum, and K.E. and we find:

fequilib = fMaxwell(r,v) = ρ( m

2πkT

)3/2e−

mV 2

2kT (7.9)

• The equilibrium solution to the Boltzmann Eq. is the classical expression forthe Maxwellian distribution of velocities!

In classical mechanics, the equations don’t have a preferred direction in time, butthe H-theorem has shown that the Boltzmann Eq. does have a preferred direction∂H∂t ≤ 0.

• Recall fluxes Γ(−) and Γ(+) → they are probabilities of a collision, so we mustsay ∂H

∂t ≤ 0 on average.

• Possible for it to increase, but very small on average, so while ∂H∂t > 0 at

some point, it gets lost in data.

Recurrance (reversibility) is also an issue, saying a closed mechanical system willreturn to original state (closely to)

• Boltzmann calculated that this recurrance time can be on the order of 101010

years! → ∂H∂t ≤ 0 is ok


In dilute gases, Boltzmann Eq. reigns supreme, but we must have chaos (nomulti-body dependence) for it to hold true.

Chapter 8

Transport Processes in DiluteGases

We derived the Boltzmann Eq. and found some consequences of it without solvingfor f .

• We will now solve for it using the Chapman-Enskog method (long andcomplex, we will summarize).

8.1 Outline of Chapman-Enskog Method

The types of solutions obtained by this method are called normal solutions, inwhich the spacial and time dependence of f (1)(r,v, t) appear implicitly throughlocal density, flow velocity, and temperature.

• These solutions describe the final state of relaxation of a dilute gas in itsequilibrium state.

• This final stage is called the hydrodynamic stage.

Chapman-Enskog expands f (1) in a series of the form (and dropping thesuperscript) (single particle): f = 1

ξ f[0] + f [1] + ξf [2] + . . .

67

68 CHAPTER 8. TRANSPORT PROCESSES IN DILUTE GASES

• ξ is an ordering parameter that is set to unity later on, [0], [1], . . . are mathorders, not number of particles.

• We can plug this into the Boltzmann Eq. to get a set of equations for all thef [j].

• We solve each one successively and uniquely → f [0] is easy to solve and givesus:

f [0] = fMaxwell = ρ(r, t)

(m

2πkT (r, t)

)3/2

e−m[v−v0(r,t)]2

2kT (r,t) (8.1)

The f [j] are constructed in such a way that the Eqs. of change from the BoltzmannEq. (continuum mechanics Eqs.) keep the same form for increasing mathematicalorders of f (f [0], f [1], . . .).

• However, the values of the parameters that appear in them, namely heat fluxvector q and pressure tensor p do vary (depends on order of f).

• Using our expressions for p = m∫

VVf dv and q = m2

∫V 2Vf dv and

plugging in f [0], we find that p[0] = ρkT I and q[0] = 0→ Plugging these intothe equations of change, we find that we get the ideal hydrodynamic Eqs.

• These equations specify ρ, v0, and T , therefore also specify f [0] through theabove Eq. (Maxwellian distribution).

To solve for f [1], which is much more complicated, let f [1] = φf [0] for some functionφ.

• We then have to solve for φ by plugging φf [0] into the Boltzmann Eq. → getan Eq.:

2π∫∫

f [0]f[0]′

1 (φ′ + φ′1 − φ− φ1)gb db dv1 ≡ −ρ2I(φ)

= ∂f [0]

∂t + v · ∇rf[0] + 1

mX · ∇vf[0]

• I(φ) represents some linear integral operator on φ, and we find the existancecondition for φ is just the ideal hydrodynamic equations.

We can further rewrite ∂f [0]

∂t since it depends on ρ,v0, and T :

8.1. OUTLINE OF CHAPMAN-ENSKOG METHOD 69

∂f [0]

∂t = ∂f [0]

∂ρ∂ρ∂t +∇v0f

[0] · ∂v0∂t + ∂f [0]

∂T∂T∂t

• We find the time derivatives of ρ,v0, and T through the equations of change,thus getting:

ρ2I(φ) = −f [0][(W 2 − 52)V · ∇ lnT + b : ∇v0]

• Where W is reduced velocity, W =(m

2kT

)1/2V and b = 2(WW − 1

3W2I)

• This ρ2I(φ) equation suggests we look for a solution φ of the form:

φ = −1ρ

(2kTm

)1/2A(W )W · ∇ lnT − 1

ρB(W )(WW − 13W

2I) : ∇v0

In this equation of φ, A(W ) and B(W ) are unknown scalar functions of W , andsince I(φ) is a linear integral operator, we get two separate integral equations: onefor A(W ) and one for B(W ).

• Furthermore, φ can be used to calculate the next approximation of p and qby using their Eqs.:

q[1] = −23k2Tmρ∇T

∫(Exp. of A(W ) and f [0])dv and

p[1] = −2kT5ρ (Exp. of B(W ) and f [0])dv

• Thus we can solve for q[1] and p[1] if we have solved the integral equations forA(W ) and B(W ).

• Remember, since we build f = 1ξ f

[0] + f [1] + ξf [2] + . . ., then we build

p = p[0] + p[1] + . . . and q = q[0] + q[1] + . . ..

• Comparing q[0] + q[1] to q with Fourier’s heat transfer law (q = −λ∇T ), wefind:

Thermal contuctivity: λ = 23k2Tmρ

∫A(W )W 2(W 2 − 5

2)f [0]dv

• And similarly with p[0] + p[1] to the Newtonian pressure tensor(p = p+ (2

3η − κ)I− 2η sym(∇u)):

Shear viscosity coefficient:η = kT

5ρ

∫f [0]B(W )(WW − 1

3W2I) : (WW − 1

3W2I)dv


We found that f [0] gives us the ideal hydrodynamic equations, but now f [0] + f [1]

gives the Navier-Stokes equations! (f [0] + f [1] + f [2] yields Burnett Eqs., but it istedious to calculate, and its applications haven’t been too successful yet)→ Need tostay near equilibrium with our equations.

• We’ve seen that our two expansion to f (1) ∼ f [0] + f [1] yields good resultsnear equilibrium.

Now, we must solve for A(W ) and B(W ) to get our transport coefficients λ and η.

• This is a very complex process that has been worked out by expanding theexpressions using Sonine polynomials (doing 2 terms):

λ = 75k2T8m

[1a11

+a2

12a11

a11a22−a212

+ . . .]

η = 5kT2

[1b11

+b212b11

b11b22−b212+ . . .

]• The aij and bij appearing here are complicated integrals over the dynamics of

bimolecular collisions and hence contain the information about theintermolecular potential, uij(r).

• These can be reduced to linear combinations of a set of collision integrals

Ω(l,s)ij for two components i, j.

a11 = 4Ω(2,2) b11 = 4Ω(2,2)

a12 = 7Ω(2,2) − 2Ω(2,3) b12 = 7Ω(2,2) − 2Ω(2,3)

a22 = 774 Ω(2,2) − 7Ω(2,3) + Ω(2,4) b22 = 301

12 Ω(2,2) − 7Ω(2,3) + Ω(2,4)

Table 8.1: Results from Chapman and Cowling.

• And we define the collisions between molecules of type i and type j as theintegral:

Ω(l,s)ij =

(2πkTµij

)1/2 ∫∞0

∫∞0 e−γ

2ijγ2s+3

ij (1− cosl χ)b db dγij

• µij is reduced mass for components i, j, χ = χ(b, gij) is the deflection angle,

gij is the relative velocity, γij =( µij

2kT

)1/2gij is a reduced relative velocity.

8.1. OUTLINE OF CHAPMAN-ENSKOG METHOD 71

These integrals for Ω(l,s) must be evaluated numerically.

These last 4 sets of equations then give λ and η in terms of the intermolecularpotential.

• We see the intermolecular potential by rewriting the collision integral Ω(l,s)ij as:

Ω(l,s)ij =

(kT

2πµij

)1/2 ∫∞0 e−γ

2ijγ2s+3

ij Q(l)(γij) dγij

with cross sectional area: Q(l)(γij) = 2π∫∞

0 (1− cosl χ)b db

and deflection angle: χ(b, γij) = π − 2b∫∞rm

dr/r2[1− b2

r2−uij(r)

kTγ2ij

]1/2

• γ2ij = µg2

2kT , µ is reduced mass, rm is distance of closest approach of collidingmolecules.

- Big Picture -

If we know the intermolecular potential uij(r), then we can calculate the deflectionangle χ, then cross section Q, the collisional integral Ω, and use that for the aijand bij terms to finally solve for our transport coefficients λ and η, which we foundthe form of from the Chapman-Enskog expansion of the Boltzmann Eq. (aroundequilibrium).

- End Big Picture -

As we found λ and η, we can also find diffusion D12 and thermal diffusion DT1

(much more complex though).

D12 = 316ρµ12

kT

Ω(1,1)12

• Note, all these equations and results have been for the first two termsf [0] + f [1] → usually good enough.


8.2 Summary of Formulas

To take a simple first look, we can calculate values using arigid sphere of diameter σ.

• This is the most simplistic model for the intermolecular potential uij(r)(nothing, then instant collision).

• We then solve through for Q and Ω with our rigid sphere model to gethard/rigid sphere reference values for transport coefficients of thermalconductivity, shear viscosity, and binary diffusion:

λ = 75k2T8ma11

= 2532CvN

(πkTm

)1/2 1πσ2

η = 5kT2b11

= 516

(πmkT )1/2

πσ2

D = 38mρ

kTΩ(1,1) = 3

8

(πkTm

)1/2 1πσ2ρ

For most intermolecular potentials, we can write the collision integrals above in areduced form.

• We use the rigid sphere model as a reference, and write collision integrals in anormalized form by utilizing reduced parameters.

• If the intermolecular potential can be written in the form of u(r) = εf(rσ

),

where ε is a characteristic temperature (energy), and σ is a characteristicdistance (hard sphere diameter), then:

r∗ = rσ , b∗ = b

σ , u∗ = uε , T ∗ = kT

ε , g∗2 = 12µg2

ε

These reduced quantities represent (physically) the deviation of any particularmolecular model from the hard sphere model.

Therefore, using our reduced cross sectional area → reduced collision integral, wefind generalizations of the previous 3 expressions, and when Ω(l,s)∗=1 they reduce tothe rigid sphere results:

λ =25

32

CvN

(πkT

m

)1/2 1

πσ2Ω(2,2)∗

8.2. SUMMARY OF FORMULAS 73

η =5

16

(πmkT )1/2

πσ2Ω(2,2)∗

D =3

8

(πkT

m

)1/2 1

ρπσ2Ω(1,1)∗

These results utilized the single-term expansion of A(W ) and B(W ) rather thantwo terms since one term expansions turn out to be satisfactory for most transportcoefficients.

• The thermal diffusion ratio is an exception, which requires us to cary outmore terms of the expansion.

• This means thermal diffusion is more sensitive to the form of theintermolecular potential.

• Using experimental data, it is possible to use the thermal diffusion ratio todetermine the best fit uij(r).

Many different models for intermolecular potential have been proposed, but we stillfind deviations as no single one has provided a best fit.

• Some examples: square well, Lennard-Jones, rigid spheres, Kihara potential.

• All of these are obtained from experimental data.


Figure 8.1: The first order approximations provide a good agreement with experi-mental results.

Part III

Atomic Spectra and Structure

75

Chapter 9

Historical Overview

9.1 Balmer

Balmer found the wavelengths of visible light emitted by hydrogen (in E(n) toE(2) transitions) follow the following pattern:

λ =n2

1

n21 − 4

G, where n1 = 3, 4, 5 and G is a constant.

It is also written in terms of the wave number ν (spatial frequency= 1λ = k

2π ):

ν = R

(1

22− 1

n21

), where R is the Rydberg number.

When n2 = 2 is replaced with 1, 3, 4, or 5, you get the Lyman, Paschen, Bracket,and Pfund series respectively, where the Lyman series is in the far UV and the restare infrared.

9.2 Bohr

Bohr said electrons orbit the nucleus at quantized angular momenta:

mvr = nh

77

78 CHAPTER 9. HISTORICAL OVERVIEW

Figure 9.1: H-atom spectrum.

9.3. SOMMERFELD’S ELLIPSES 79

Then, he solved for the Rydberg number classically using a force balance betweenthe centripetal acceleration and the electric force for the quantized radii resultingfrom the quantized angular momenta. His result matched the experimentalRydberg number.

9.3 Sommerfeld’s ellipses

There was no reason to assume electrons orbited in circles, rather than the moregeneralized ellipses, so Sommerfeld used action integrals (

∫p dq) and found the

principal quantum number n corresponded to the major axis of the ellipse (from∫pr dr) and a new quantum number k corresponding to the minor axis (from∫pφ dφ). k was poorly indexed, starting at 1, so now we use l = k − 1. l is the

azimuthal quantum number and accounted for small variations in λ for transitionsof equivalent n values.

9.4 De Broglie’s waves

De Broglie hypothesized that the kinetic energy of the electrons was equivalent tothe energy of a wave:

E = hν

The quantized orbits were a result of all waves destroying each other except forstanding waves. The orbits must then exist where a quantized number ofwavelengths are equal to the circumference:

2πr = nλ→ mv = hλ

80 CHAPTER 9. HISTORICAL OVERVIEW

Chapter 10

Atoms at Present

10.1 Schrodinger Eq.

Electrons don’t actually orbit the nucleus, there is a “probability cloud” of wherean electron might be. For example, the probability of finding an electron at a givenradius is calculated from

∫r2ΨΨ∗ dr using Ψ from the Schrodinger equation:

ihΨt = HΨ

For a single, non-relativistic particle: ihΨt =

[− h

2

2µ∇2 + V (r, t)

]Ψ(r, t)

Time independent: (E − V )Ψ(r) = − h2

2µ∇2Ψ(r)

Using the Schrodinger equation, we determine new definitions for the quantumnumbers (see Table 10.1).

A nodal surface is a surface at which there is 0 probability of finding an electron.These are at fixed radii from the nucleus unless they go through the midpoint; thenthey are conic sections or planes. The values of n and l can be better understood byseeing how the value of

∫r2ΨΨ∗ dr varies as a function of r (Figs. 10.1 and 10.2).

Understand why the most probable value and the average value are not the same.

81

82 CHAPTER 10. ATOMS AT PRESENT

QuantumNumber

Name Interpretation Selection RulesPossibleValues

n principalnumber of nodal

surfaces +1∆n = any ≥ 0

l azimuthalnumber of nodalsurfaces through

midpoint

∆l = ±1 unlessthere is an E-field(then 0 possible)

0, . . . , n− 1

m magneticcomponent of l

alongquantization axis

∆m = ±1 if∆J = 0,

∆m = 0,±1 if∆J = ±1

−l, . . . , l

s electron spin up or down spin ∆s = 0 ±12

L∑l ∆L = ±1 integer

S∑s ∆S = 0

half integeror integer

J L + S ∆J = 0,±1 ≥ 0

Table 10.1: Quantum numbers and interpretation.

Figure 10.1: There are n− 1 surfaces with no probability of containing an electron.There are l of these surfaces going through the midpoint (not fixed radius).

10.2. SELECTION RULES 83

Figure 10.2: There are n− 1 surfaces with no probability of containing an electron.There are l of these surfaces going through the midpoint (not fixed radius).

10.2 Selection Rules

Selection rules are outlined in Table 10.1. They are illustrated by Grotriandiagrams, examples of which are in Figs. 10.3 and 10.4. The frequency of a photonis correlated to the change in energy when an electron moves from one state toanother (∆E = hν). The energy of each level is along the left side, as the the valueof n. The columns correspond to l = 0, 1, 2 . . ..

Important: Note that for hydrogen the different l states have nearly identicalenergy for an equivalent n value. In general, the larger the atomic number, thegreater the variation in energy for a given n value (compare with helium forinstance). This concept can be used to show that n = 2 for the outermost lithiumelectron. The entire second row of the period table can be compared side by side inGrotrian diagrams normalized by dividing all energies by (Z − p)2, where Z is theatomic number and p is the effective charge that the outermost electron sees.When you do this, the outermost electron of lithium is sitting at about n = 1.5, soit could belong to n = 1 or n = 2. By comparing with subsequent elements, it isclear that it is an n = 2 state.

Back to the Grotrian diagrams. There are letters corresponding to the l values.These are S for sharp (l = 0), P for principal (l = 1), D for diffuse (l = 2), and F


Figure 10.3: Grotrian diagram of hydrogen.

10.2. SELECTION RULES 85

Figure 10.4: Grotrian diagram of helium.


for fundamental (l = 3). There is also G for “comes after F in the alphabet.” Tothe top left of these letters in Fig. 10.4 is the tuplet value, which is to say howmany values of J or possible, but this is a lie. Look at the S column oforthohelium. There is only one way to add the vector 1 to 0. There are, however, 3different ways of adding S to L that satisfy the requirements of J for all subsequentl states (see next paragraph). They have slightly different energy levels and resultin three slightly different wavelengths in the transition from P to S. Note that forother transitions, such as F to D, as many as 6 bands are possible for triplets. Thenumber to the lower right hand corner of the l letter is the magnitude of J:

tupletlJ

The above paragraph takes some time to understand. First, understand whatvalues you can get for J. L is fixed, and J must be an integral addition to S.Example: L = 2 and S = 5

2 . So J must have half integer values. The biggest it canbe is 9

2 , then 72 ,

52 ,

32 ,

12 . Again, this is technically a sextuplet, but we only have 5

terms. This is because 2− 52 is the smallest J can be, but since J cannot be

negative, we omit that last value.

What’s with the dotted line in Fig. 10.4? In Grotrian diagrams, we are onlylooking at the outer electron. All other electrons are assumed to be in groundstate, which is generally true until the outer electron is ionized. The dotted linedenotes the difference between the two electrons of helium having the same oropposite spin (for ortho and parahelium respectively). This is why there is non = 1 state for the outer electron in orthohelium. That would put two electrons inthe exact same state, violating the Pauli exclusion principle.

Metastability: This is where an electron is NOT in the ground state, but cannotmove to the ground state without first receiving energy. You can see that whenn = 2 and l = 0, there is either no ground state (ortho), or an electron would haveto get energy before it can emit (para).

Know at least the following ionization potentials:

H: 13.6 eV, He: 24.5 eV, Li (first): 5.4 eV, Li (second): 76 eV

Also understand that selection rules are frequently broken, especially for heavierelements. The rules simply describe the most probable transitions predicted by theSchrodinger equation.

10.3. ZEEMAN EFFECT 87

10.3 Zeeman Effect

In the presence of a magnetic field, spectral lines split into several bands, and theirseparation from their 0-field value is proportionate to the strength of the magneticfield. This effect is useful for measuring the intensity of magnetic fields in sun spotsand elsewhere. The magnetic field exerts a torque on the magnetic dipole resultingfrom the electron’s “orbit.”

The perturbation in energy for a given electron in the direction of the quantizationaxis is:

H1 = −(µl + µs) ·Bext, where µs = − e

mS and µl = − e

2mL.

µ is on the order of 10−23, soH1

His usually small for a reasonable B.

Part IV

Appendices

89

Chapter 11

Appendix A: Chinese NapkinFormulas

ν = nQ(v)v (11.1)

λ =1

nQ(v)(11.2)

radius of H = 0.529 A (11.3)

λroom = 10−6 m (11.4)

PSTP = 105 Pa (11.5)

1 eV = 11, 600 K (11.6)

R = 1.1× 107 m−1 (11.7)

91

92 CHAPTER 11. APPENDIX A

c =

√8kT

πm(11.8)

Pj =ajA

=1

A

∑aW (a)aj(a)∑

aW (a)(11.9)

k = 1.3806× 10−23 J/K (11.10)

ΞFDBE

=∏k

(1± λe−βεk)±1 (11.11)

Q(N,V, T ) =qN

N !(11.12)

Npalm = 1021 atoms of air (11.13)

q =∑j

e−βεj = qtransqint (11.14)

qtrans =

[2πmkT

h2

]3/2

V (11.15)

qvib =kT

hν=

T

Θvib(11.16)

qrot =T

σΘrot, σ = 2 when homonuclear (11.17)

∂fj∂t

+ vj · ∇rfj +Xj

mj· ∇vfj = change in fj due to collisions (11.18)

Chapter 12

Appendix B: Principle Topics inPlain English

Have a spiel put together for each of the following:

BBGKY BBGKY offers a complete solution to the Liouville Eq. in terms ofreduced distribution functions.

Boltzmann Equation The Boltzmann Eq. is the most fundamental Eq. of therigorous kinetic theory of gases. It describes the statistical behavior of athermodynamic system not at thermodynamic equilibrium. It can be written in theform of ∂f

∂t = ∂f∂t external force

+ ∂f∂t diffusion

+ ∂f∂t collisions

. For the collision term,Boltzmann applied the molecular chaos assumption, which assumes 2-body onlycollisions between particles that are uncorrelated prior to collision. This assumptionis valid since we are only dealing with dilute gases (singlet distribution function).The solution to the Boltzmann Eq. is found via the Chapman-Enscog method.

Chapman-Enskog Method This method of solving the Boltzmann Eq.expands f (1) such that f = 1

ξ f[0] + f [1] + ξf [2] + . . ., where the first term describes a

Maxwellian distribution, the combined first two terms describe the Navier StokesEqs., and all three terms form the Burnett Eqs..

93

94 CHAPTER 12. APPENDIX B

• Assume a solution of the form f [1] = φf [0].

• Use Eqs. of change to solve for φ in terms of A(W ) and B(W ) (functions ofreduced velocity).

• Use Fourier’s Law to get λ(A(W )) (thermal conductivity).

• Use Newtonian Pressure tensor to get η(B(W )) (coefficient of shear viscosity).

• Expand∫A(W ) and

∫B(W ) using Sonine polynomials in terms of aij and

bij .

• End up needing just intermolecular potential for χ (deflection angle) for Q(collisional cross section area) for Ω (collision integral) to get λ, η, and D(diffusion coefficient). So you get gas properties from first principles!

Ensemble A collection of a large number of systems (A), each with the samemacroscopic properties for a particular thermodynamic system of interest.

• Ω(N,V,E) - microcanonical ensemble, isolated system

• Q(N,V, T ) - canonical ensemble, system in contact with heat bath

• Ξ(V, T, µ) - grand canonical ensemble, system in contact with heat andparticle bath

• ∆(N,T, ℘) - isobaric-isothermal ensemble, used in chemistry (constant ℘)

Liouville Equation The Liouville Eq. is the most fundamental Eq. of statisticalmechanics. It is completely equivalent to the 6N Hamilton Eqs. of motion of theN -body system. The equation says that the density in the neighborhood of anyselected moving phase point is a constant along the trajectory of that phase point,and the cloud of phase points behaves as an incompressible fluid (dfdt = 0). This isproven by the fact that the evolution of f obeys the N -dimensional continuity Eq..

Partition Function Describes the statistical properties of a system atequilibrium. It serves as a bridge between the quantum mechanical energy states ofa macroscopic system and the thermodynamic properties of that system.

95

Phase Space If you take an N -particle system, you can define the generalizedcoordinates and conjugate momenta of every particle as a single point in phasespace. Further, you can use an ensemble of such points representing the samemacroscopic mechanical properties, and the density of these phase points(assuming you take the limit of them) is a distribution function f . Since frepresents all phase points with the same macroscopic mechanical property, thedistribution of phase points will be invariant with respect to time. So df

dt = 0 and

we can solve for ∂f∂t with respect to p and q to find the Liouville Eq.

96 CHAPTER 12. APPENDIX B

Bibliography

[1] E. Y. Choueiri, Chapter 1: Collisional Processes, Princeton University (2012).

[2] G. Herzberg, Atomic Spectra and Structure, Dover Publications (1937).

[3] D. A. McQuarrie, Statistical Mechanics, University Science Books (2000).

97

physics of gases - princeton university

Documents