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Thermodynamics Y Y Shan

AP3290 92

Chapter 7 The Maxwell–Boltzmann statistics

In view of the examples in 4.3 of Chapter 6 where “particles” (such as dice or

tokens) are not real particles, “energy” was not assigned to individual “particles”.

However, in statistical thermodynamics, we deal with thermodynamic systems consisting

of millions of real particles with different energies. The total energy of a system is the

major factor we concern. Therefore, energies of individual particles (in different energy

states, i.e. quantum states) must be under consideration when counting microstates of

distributions for systems with certain total energies.

Consider a distribution of a system, denoted as i

a , satisfying:

(i)

Total number of particles is N ,

(ii) Total energy of the system is U .

(iii) Each particle can have energies (or energy levels) of: LL , ,,21 i

ε ε ε

(iv) The degeneracy of energy levels are: LL ,,,21 i

ggg accordingly (i.e., there are

ig states having same energy

iε ). The corresponding numbers of particles on

each energy level are: LL ,,,,321 ii

aaaaa =

Therefore a distribution ia describes :

LL

LL

LL

, , , ,

,, , ,

, , , ,

321

321

321

i

ii

i

aaaa

gggga ⇒

ε ε ε ε

∑

∑

=

=

iii

ii

aU

a N

ε :systemtheof energyTotal

:particlesof numberTotal

Eqn7-1

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AP3290 93

7.1 Counting of Microstates

(a) Microstates by Maxwell-Boltzmann counting

In a system, if particles are distinguishable and different particles can have

same quantum states (occupy same quantum states), for such a distribution i

a , the

total number of microstates can be counted and equals to:

∏=Ωi

i

a

i

MBag N

i

!! Eqn7-2

This is called Maxwell-Boltzmann counting, and such a system is called a Boltzmann

system.

(b) Microstates by Bose–Einstein counting

In a system, if particles are indistinguishable and different particles can have

same quantum states (occupy same quantum states), for such a distribution i

a , the

total number of microstate can be counted as:

∏−

−+=Ω

iii

ii

BE ga

ga

)!1(!

)!1(

This is called Bose–Einstein counting, and such a system is called a Bose system.

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AP3290 94

(c) Microstates by Fermi-Dirac counting

In a system, if particles are indistinguishable and different particles cannot have

same quantum states (can not occupy same quantum states, i.e., particles obey the Pauli

exclusion principle), for such a distribution i

a , the total number of microstate can be

counted as:

∏−

=Ωi

iii

i

FDaga

g

)!(!

!

This is called Fermi-Dirac counting, and such a system is called a Fermi system.

Example: Check Eqn7-2. For a system having 3 distinguishable “particles” (3 pieces

of dice): N=3. Now play a three-dice-throwing game, suppose that showing the following

four sides (i.e. four individual states): would have the same reward

grade 10$1 −=ε , (equivalent to energy level 1ε with degeneracy of g1=4). While

showing states would have the same reward grade 20$2 +=ε (equivalent to

energy level of 2ε , with energy degeneracy g2=2). Find the total number of possible

formations (i.e. total microstates MB

Ω ) for getting the reward of -$30, $0, +$10, +$60

(i.e. corresponding to the distributions of: 0,3=i

a , 1,2=i

a , 2,1=i

a ,

3,0=ia ) and their probabilities, accordingly.

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AP3290 95

(i) For the case of loosing -$30, i.e. showing any formations (microstates) from the

distribution 0,3=i

a , the total number of microstates: 64440,3

=×=Ω MB .

( calculated by Eqn7-2: 64

!0

2

!3

4!3

!

!03

===Ω ∏i i

a

i

MB

a

g N

i

)

(ii) For the case of winning $0, i.e. showing any formations (microstates) from the

distribution 1,2=ia , the total microstates: 96)2()44( 1

31,2

=⋅××=Ω C MB .

( calculated by Eqn7-2: 96166!1

2

!2

4!3

!!

12

=⋅===Ω ∏i

i

a

i

MBa

g N

i

)

(iii) For the case of winning $10, i.e. showing any formations (microstates) from the

distribution 2,1=i

a , the total microstates: 48)22()4( 1

3

1

3

2,1=⋅××⋅=Ω C C

MB.


2

!1

4!3

!!

21

=⋅⋅===Ω ∏i

i

a

i

MBa

g N

i

)

(iv) For the case of winning $60, i.e. showing any formations (microstates) from the

distribution 3,0=i

a , the total microstates: 82223,0

=××=Ω MB .


2

!0

4!3

!!

30

===Ω ∏i

i

a

i

MBa

g N

i

)

So: the probability of $30-lost: 64/(64+96+48+8)=29.6%, winning-nothing:

96/216=44.4%, $10-win:22.2%, and $60-win: 3.7%.

Conclusion: Don’t gamble.

7.2 Maxwell-Boltzmann statistics

7.2.1 Maxwell-Boltzmann distribution

For a Boltzmann system of N particles i

a , the total number of microstates is:

∑∑∏ ===Ωi

iii

ii

i

a

i aU a N a

g N

i

ε , , !

!

Under what circumstance, Ω will reach to its maximum?

According to the principle of equal a priori probabilities, if the number of

microstates of a distribution becomes maximum, this distribution is the most probable

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AP3290 96

distribution. It has the highest probability to be observed. This most probable

distribution is called the Maxwell-Boltzmann distribution.

Thus, our goal is to find out what i

a will be ( ?)=i

a to make the total number

of microstatesΩ

become maximum.

Brief derivation of Maxwell-Boltzmann distribution:

From Eqn.7-2, and using Stirling's formula: )1for( ),1(ln!ln >>−= x x x x , we

can obtain:

∑ ∑+−=Ωi i

iiii gaaa N N lnlnlnln Eqn7-5

When Ω gets to maximum, Ωln will also become maximum. The

mathematical condition to get the maximum of Ωln is: 0ln

=Ω

ida

d .

From Eqn.7-5, let 0ln

=Ω

ida

d , by use of the Lagrange method,

the solutioni

a (i.e. the number of particles occupying energy leveli

ε ) is obtained (skip

the details): N f a ii )(ε =

Z

eg f

i

ii

βε

ε −

=)( Eqn.7-6

is the probability of a particle occupying the energy leveli

ε , and is called the Boltzmann

distribution function. The function ∑ −=

ii

ieg Z βε

is called the partition function. In

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AP3290 97

terms of energy-level-occupation, “i” denotes the ith energy level. The factor β is called

Boltzmann factor and can be proved to equal tokT

1, and k is called Boltzmann constant.

For each energy level iε , there are ig individual states (quantum states), in termsof individual-state-occupation, The probability of an individual quantum state “j” to be

occupied is:

Z

e

g

f P

j

j

j

j

βε ε ε

−

==)(

)(

7.2.2 The partition function Z

In statistical physics, the partition function

∑ −=

ii

ieg Z βε

Eqn.7-7

is an important quantity that encodes the statistical properties of a system in

thermodynamic equilibrium. Most of the thermodynamic variables of the system, such as

the internal energy, free energy, entropy, and pressure, can be expressed in terms of the

partition function or its derivatives. The partition function can be related to

thermodynamic properties because it has a very important statistical meaning.

1)( Since === ∑∑ −

Z

Z

Z

eg f

i

i

i

i

i βε

ε

The partition function thus plays the role of a normalizing constant, ensuring that the

probabilities add up to one. This is the reason for calling Z the "partition function": it

encodes how the probabilities are partitioned among the states to be occupied at different

energy levels ( iε ), based on their individual energies. The letter Z stands for the German

word Zustandssumme, “sum over states”.

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AP3290 98

7.3 Application of Boltzmann statistics--I

We have been emphasizing the importance of the partition function Z, which

encodes the statistical thermodynamic properties and is the key to calculating all

macroscopic thermodynamic properties of the system.

When we know , we can calculate all macroscopic properties of the system -

the internal energy, free energy, entropy, and pressure.

(i) The statistical internal energy U

In order to demonstrate the usefulness of the partition function, let us calculate the

thermodynamic value of the internal energy.

ln)(

,

β β ε ε ε ε

ε β

βε

βε βε

∂

∂−=

∂

∂−====

−=∂

∂

=

∑∑∑

∑∑−

−−

Z N

Z

Z

N eg

Z

N Nf aU

eg

Z

eg Z

iii

iii

iii

iii

ii

i

ii

Therefore, the statistical expression of internal energy is:

ln

β ∂

∂−=

Z N U Eqn7-3a

This means that when the partition function Z is known, the internal energy U can be

calculated. Therefore, the specific heatV

C can be calculated by:

V

V T

U C

∂

∂=

Example: Einstein’s specific heat of solids

Since an atom in a solid, having three degrees of freedom, can vibrate

independently in three dimensions, it should really be considered to be equivalent to three

separate 1D quantum harmonic oscillators (QHO refer to AP3251). Therefore, Einstein's

model pretty much said that a solid of n atoms comprised 3n number of 1D quantum

harmonic oscillators, and assuming that all the 3n oscillators have the same oscillating

frequency of ωo. For each 1D QHO:

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AP3290 99

...,3,2,1,0 ,)2

1( +∞=+= iii ω ε h , it is not degenerated, i.e., different states

have different energies, or energy levels).

The partition function for QMO (where the degenercy 1=i

g ) can be obtained as:

( )0

0

000

1

2

1

0

2

1

0

)2

1(

ω β

ω β

ω β ω β ω β

βε

h

h

hhh

−

−∞

=

−−∞

=

+−−

−==== ∑∑∑

e

eeeeeg Z

i

ii

i

i

ii

Here we applied the formula: )1(for,1

1

0

<−

=∑∞

=

x x

xn

n , and here 10 <= − ω β he x .

Now useEqn7-3-a, replacing N → 3n, the statistical internal energy U of a system having

the total number of quantum harmonic oscillators of N=3n is:

−−−

∂

∂−=

∂

∂−=

−)1ln(

2

13

ln0

0

ω β ω β β β

hh en

Z N U

1

3

2

3

0

00

−+=∴

ω β ω

h

hh

e

nnU

(here kT / 1= β , which will be proved later). Thus, the specific heatV

C can be

calculated:

2

20

1

30

0

−

=

∂

∂=

kT

kT

V

V

e

e

kT nk

T

U C

ω

ω

ω

h

h

h

Eqn7-3b

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AP3290 100

(ii) To calculated the statistical pressure P

Since the statistical internal energy: ∑=i

iiaU ε . Its differential form:

∑∑ +=+=i

ii

i

iia d adadU dU dU ii

ε ε ε

Compare it with the 1st law of thermodynamics:

PdV dQdW dQdU −=+=

Since the heat absorbing (or releasing) will excite the particle to different energy levels,

resulting in the change of particle numbers ida at energy level iε . Whereas the

quantized energies of the particles iε depends on the volume (See AP3251: particle in a

potential box:

++=2

3

2

2

2

2

2

1

222

2 L

n

L

n

L

n

m

z y xnnn z y x

hπ ε , 321 L L LV = ), so that the change of

energy level id ε is related to the change of the volume. We get:

∑

∑

=−

=

i

ii

i

ii

d aPdV

dadQ

ε

ε

,

Eq7-3c

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AP3290 101

ln1 )(1

)(1

)(

Z d N dZ Z

N egd Z

N

ed g Z

N d

Z

eg N

d Nf d aPdV

ii

ii

ii

i

iii

iii

i

i

i

β β β

β ε

ε ε ε

βε

βε

βε

−=−=−=

−==

==−

∑

∑∑

∑∑

−

−

−

The pressure can be expressed and calculated statistically by:

V

Z N P

∂

∂=

ln

β Eq7-3d

(iii) Determination of Boltzmann factor β

From the first law: PdV dQdU −= ,

PdV dU dQ += , Using equations 7-3a, and 7-3d

)(lnln1ln

ln

Z d N

N Z

N d dV V

Z N Z N d dQ

β β β

β β β +

∂

∂−=

∂

∂+

∂

∂−=

,lnln1

+∂

∂

−⋅=∴ Z N Z

N d dQ β β β Eq.7-3e

dS T dQ ⋅=

It can be concluded that the Boltzmann factor is reverse proportional to temperature

,1

T ∝ β which can be written as:

constantthecallediswhere

,1

Boltzmannk kT

≡ β

Eq7-3f

From Eq.7-3e, entropy can be statistically expressed as:

∂

∂−=

β β

Z Z Nk S

lnln

Eq.7-3g

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AP3290 102

7.4 Application of Boltzmann statistics—II---mono-atomic ideal gases

(i) Semi-classical approximation

For classical gases (mono-atomic ideal gases), their molecules can be regardedclassical particles. The energy levels are becoming so condensed that they turn out to be

continuous energy band.

The changes in calculations using Boltzmann statistics will be:

1. The energy degeneracy: pd h

V g i

3

3byreplaced ⋅ , the number states in the

volume element in momentum space, pd 3 , V is the volume of the system.

2. The Boltzmann distribution function:

Z

pd e

h

V f

Z

eg f

i

i

i

3

3)( )(

βε βε

ε ε −−

=⇒= Eq.7-4a

3. The partition function will be changed to:

∫∑ −

−=⇒= pd e

h

V Z eg Z kT

ii

i 3

3

ε

βε Eq.7-4b

4. The number of particles on leveli

ε becomes the number of particles in the energy

range: ε ε ε d +→

Z

pd e

h

V N a

Z

eg Nf a

i

i

ii

3

3)( )(

βε βε

ε ε −−

=⇒== Eq.7-4c

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AP3290 103

m

p p p

m

pvvvmmv

z y x

z y x2

2

or),(2

1

2

1 where

22222222

++==++== ε ε

, and pd 3

is the volume element in momentum space.

systemcoordinatesphericalinsin

systemcoordinateCartesianin23

3

dpd d p pd

dpdpdp pd z y x

φ θ θ =

=

(ii) The partition function of ideal gases (mono-atomic particles)

From Eq.7-4b, we obtain the partition function of ideal gas in Cartesian coordinates:

∫∫∫∫++

−−

== z y xmkT

p p p

kT dpdpdpeh

V pd e

h

V Z

z y x

23

3

3

222ε

π =∫+∞

∞−

− dxe x2

use

Thus, the partition function of the ideal gas can be calculated:

V dpdpdpe

h

V Z z y x

mkT

p p p z y x

2

3

23

222

−++

−

== ∫∫∫ αβ Eq.7-4-d

kT h

m 1,

2constantthewhere

2

3

2 =

= β

π α

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AP3290 104

(iii) “State equation of ideal gas” in its statistical form , the determination of

Boltzmann constant k, and the calculation of the internal energy of ideal gas

Using Eq7-3-d, the gas pressure P can be statistically calculated by:

V

Z N

P ∂

∂

=

ln

β

[ ]V

NkT

dV

V d NkT V

V NkT P ==

∂

∂=∴

)(ln)ln( α

NkT PV = Eq.7-4-f

This is the statistical expression of the state equation of ideal gas.

Compare Eq.7-4f with the experimental state equation: nRT PV = ,

where n is mol number, i.e. nR Nk = . So the Boltzmann constant can be obtained:

123

23

11

00

10381.1 / 10023.6

314.8 −−−−

×=×

==== JK mol particles

mol JK

N

R

nN

nR

N

nRk

For n=1mol gas, N=N 0 is the Avogadro’s number. R is the Gas constant.

(iv) The internal energy of ideal gas

The internal energy of ideal gas can be calculated statistically by Eqn7-3-a , Eq.7-4-d

V dpdpdpeh

V Z

z y x

mkT

p p p z y x

2

3

2

3

222

−++

−

==

∫∫∫ αβ

kT N N

V N Z

N U 2

3

2

3)ln(ln

2

3

ln⋅==

+−

∂

∂−=

∂

∂−=

β α β

β β

Example: Prove that, for mono-atomic ideal gas, 667.1==V

P

C

C γ

nR Nk kT N dT

d

dT

dU C

V

V 2

3

2

3

2

3==

⋅=

=

From classical thermodynamics (see chapter 2): nRC C V P =−

Therefore ,2

5nRnRC C

V P =+=

667.13

5===

V

P

C

C γ

ap3290_chapter_7_i_2009

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