the maxwell-boltzmann distribution brennan...

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The Maxwell-Boltzmann Distribution

Brennan 5.4

Lecture prepared by Melanie Hill

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Maxwell-Boltzmann DistributionScottish physicist James Clerk Maxwell developed his kinetic theory of gases in 1859. Maxwell determined the distribution of velocities among the molecules of a gas. Maxwell's finding was later generalized in 1871 by a German physicist, Ludwig Boltzmann, to express the distribution of energies among the molecules.

Maxwell made four assumptions…

[1] McEnvoy, J.P. and Zarate, Oscar; “Introducing Quantum Theory;” 2004[2] http://www.hk-phy.org/contextual/heat/tep/trans01_e.html

Maxwell pictured the gas to consist of billions of molecules moving rapidly at random, colliding with each other and the walls of the container.

This was qualitatively consistent with the physical properties of gases, if we accept the notion that raising the temperature causes the molecules to move faster and collide with the walls of the container more frequently.[1]

[2]

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The diameter of the molecules are much smaller than the distance

between them

Maxwell-Boltzmann Distribution

Maxwell’s Four Assumptions[1]

xD D

The collisions between molecules

conserve energy

The molecules movebetween collisions withoutinteracting as a constantspeed in a straight line

The positions and velocitiesof the molecules are

INITIALLY AT RANDOMGreat

insight!

[1] McEnvoy, J.P. and Zarate, Oscar; “Introducing Quantum Theory;” 2004

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Maxwell-Boltzmann DistributionBy making these assumptions, Maxwell could compute the probability that a molecule

chosen at random would have a particular velocity.

This set of curves is called the Maxwell Distribution. It provides useful information about the billions and billions of molecules within a system; when the motion of an individual molecule can’t be calculated in practice. [1]

We will derive the Maxwell-Boltzmann Distribution, which will provide useful information about the energy.

[1] McEnvoy, J.P. and Zarate, Oscar; “Introducing Quantum Theory;” 2004

[1]

Raising the temperature causes the curve to skew to the right, increasing the most probable velocity.

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Maxwell-Boltzmann DistributionWhy use statistical mechanics to predict molecule behavior? Why not just

calculate the motion of the molecules exactly?

Even though we are discussing classical physics, there exists a degree of “uncertainty” with respect to the fact that the motion of every single particle at all times cannot be determined in practice in any large system.

Even if we were only dealing with one mole of gas, we would still have to determine characteristics of 6x1023 molecules!!

Maxwell’s theory was based on statistical averages to see if the macrostates, (i.e. measurable, observable) could be predicted from the microstates.

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Maxwell-Boltzmann DistributionIn Section 5.3, it was determined that the thermal equilibrium is established

when the temperatures of the subsystems are equal. So…

What is the nature of the equilibrium distribution for a system of N non-interacting gas particles?

The equilibrium configuration corresponds to the most probable (most likely) configuration of the system.

Consider the simplest case, a system of N non-interacting classical gas particles.

– Classical system:• there are no restrictions on how many particles can be put into any one state

simultaneously• the particles are distinguishable, i.e. each particle is labeled for all time

First, we’ll need to determine the number of microstates within any given configuration, i.e. the number of ways in which N objects can be arranged into n distinct groups, also called the Multiplicity Function.

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Maxwell-Boltzmann DistributionThe Multiplicity Function - determine the number of ways in which N objects

can be arranged into n containers.

Consider the first twelve letters of the alphabet:

a b c d e f g h i j k l

Arrange the letters into 3 containers without replacement. Container 1 holds 3 letters, Container 2 holds 4 letters, and Container 3 holds 5 letters.

| _ _ _ | | _ _ _ _ | | _ _ _ _ _ |

For the 1st slot, there are 12 possibilities. | c _ _ | | _ _ _ _ | | _ _ _ _ _ |

For the 2nd slot, there are 11 possibilities. | c h _ | | _ _ _ _ | | _ _ _ _ _ |

For the 3rd slot, there are 10 possibilities. | c h k | | _ _ _ _ | | _ _ _ _ _ |

etc…

There are 12! possible arrangements if the containers are ignored.

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Maxwell-Boltzmann Distribution(cont.) The Multiplicity Function - determine the number of ways in which N objects can

be arranged into n containers

Since we care about the containers but we don’t care about the order of the letters within each container, we divide out the number of arrangements within each given container, resulting in:

720,27!5!4!3!12

=⋅⋅

There are 27,720 ways of partitioning 12 letters into the 3 containers

The Multiplicity Function:

In general, the number of distinct arrangements of N particles into n groups containing N1, N2, …, Ni, …, Nn objects becomes:

!...!...!!!

21 ni NNNNN

⋅⋅⋅⋅⋅

∏=

=⋅⋅⋅⋅⋅

= n

ii

nini

N

NNNNN

NNNNNQ

1

2121

!

!!...!...!!

!),...,,...,,(

Where Ni is the number of objects in container i

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Physically, each container corresponds to a state in which each particle can be put.• In the classical case, there are no restrictions on how many particles that can be

put into any one container or state• See Section 5.7 for the quantum case where there are restrictions for some

particles

The states are classified in terms of their energy, Ei, and degeneracy, gi

Think of the system as n containers with gi subcontainers.

So, how can we arrange Ni particles in these gisubcontainers?

Each level has gi degenerate states into which Ni particles can be arranged

There are nindependent levels

Ei

Ei+1

Ei-1

Degenerate states are different states that have the same energy level. The number of states available is known as the degeneracy of that level.

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Since we are dealing with the classical case:• there are no restrictions on how many particles can be put into any one subcontainer

or state simultaneously• the particles are distinguishable, i.e. each particle is labeled for all time

So, how many arrangements of Ni particles in these gi subcontainers are possible?

Consider an example where we have 3 particles and 2 subcontainers:

RG

B

3 particles 2 subcontainers

n containers with gi subcontainers.· · ·

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Maxwell-Boltzmann Distribution(cont.) How many arrangements of Ni particles in these gi subcontainers are possible?

RG

B

R

G B

R

GB R

G

B

R

G

B

R

G B

R

GB

RG

B

There are 8 possible arrangements, or 23

In general, the possible arrangements of Niparticles into gi subcontainers is:

iNig

Therefore, if our system has a particular state that has a particular degeneracy, there is an additional multiplicity of iN

ig for that particular state.

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Maxwell-Boltzmann Distribution(cont.) Therefore, the total Multiplicity Function for a collection of classical particles is:

Constraint 1 implies:

There are two physical constraints on our classical system:1. the total number of particles must be conserved2. the total energy of the system must be conserved

where U is the total energy for the system

∏∏ =

=

=n

i

Nin

ii

niig

N

NNNNNQ1

1

21

!

!),...,,...,,(

NNn

ii == ∑

=1φ

Constraint 2 implies:

UNEn

iii == ∑

=1ψ

The equilibrium configuration corresponds to the most probable (most likely) configuration of the system, i.e. the configuration with the greatest multiplicity! To find the greatest multiplicity, we need to maximize Q subject to constraints.

ConstantConstant

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Maxwell-Boltzmann DistributionFrom previous slide:

∑∑==

−+=n

ii

n

iii NgNNQ

11!lnln!lnln

∏∏ =

=

=n

i

Nin

ii

niig

N

NNNNNQ1

1

21

!

!),...,,...,,(

This equation will be easier to deal with if we take the logarithm of both sides:

xxxx −≈ ln!lnApplying Stirling’s approximation, for large x:

∑∑∑===

+−+−=n

ii

n

iii

n

iii NNNgNNNNQ

111lnlnlnln

0ln =∂∂

−∂∂

+∂∂

jjj NNQ

Nψβφα

In order to maximize, we need to make use of Lagrange multipliers and Constraints 1 and 2:

Almost exact because we’re dealing with verylarge x, 1023 particles!

The derivatives of the 1st and 2nd constraints are zerobecause the constraints are equal to constants

All we are doing is adding and subtractingconstants multiplied by zero!

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0ln =∂∂

−∂∂

+∂∂

jjj NNQ

Nψβφα

Substituting in ln Q and Constraints 1 and 2:

011ln1ln =−++

⋅+⋅− j

jjjj EN

NNg βα

Taking the derivative, noting that N is constant and the only terms that are nonzero are when i = j:

0lnlnln111 11

=

∂∂

−

∂∂

+

+−+−

∂∂ ∑∑∑ ∑∑

=== ==

n

iii

j

n

ii

j

n

i

n

ii

n

iiiii

j

NEN

NN

NNNgNNNNN

βα

∑∑∑===

+−+−=n

ii

n

iii

n

iii NNNgNNNNQ

111lnlnlnln

0lnln =−+− jjj ENg βα

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To find β, we need to determine the total number of particles, N, in the system and the total energy, E, of the system.

If the states are closely spaced in energy, they form a quasi-continuum and the total number of particles, N, is given by:

Where D(E) is the Density of States Function found in Section 5.1

∫∞

=0

)()( dEEDEfN

0lnln =−+− jjj ENg βα

jj

j EgN

βα −=ln

jE

j

jj e

gN

Ef βα−==)(

What are α and β?

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Using the standard integral:

∫∞

+− +Γ

=0

1

)1(n

axn

andxex

23

232)(

πhEmED =E

j

j egN

Ef βα−==)(

∫∞

=0

)()( dEEDEfN

∫∞

−=0

23

232 dEeEemN Eβα

πh

2323

23

23

2

βπα

Γ

= emNh

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Maxwell-Boltzmann DistributionThe total energy, E, of the system can be found by:

Using the standard integral again:

∫∞

+− +Γ

=0

1

)1(n

axn

andxex

23

232)(

πhEmED =E

j

j egN

Ef βα−==)(

∫∞

=0

)()( dEEDEEfE

∫∞

−=0

23

23

232 dEeEemE Eβα

πh

2523

23

25

2

βπα

Γ

= emEh

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Maxwell-Boltzmann DistributionThe average energy per particle is given as:

Where kB is the Boltzmann constant.

Equating the average energy per particle with the ratio of our equations for E and N:

TkNE

B23

=

NETk

em

em

NE

B ==

Γ

Γ

=23

23

2

25

2

2323

23

2523

23

βπ

βπ

α

α

h

h

KeVkB

5106175.8 −×=

TkB23

23

25

23

25 =

Γ

⋅

Γ

β

β

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TkB23

23

25

23

25 =

Γ

⋅

Γ

β

β

TkB23

2323

23

=

Γ⋅

Γ⋅

β

TkB23

23

=β

TkB

1=β

( ) ( )zzz Γ=+Γ 1

Use the relationship[1]:

[1] http://mathworld.wolfram.com/GammaFunction.html eq. 37

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Maxwell-Boltzmann DistributionNow we have β but we still need α. α will be derived in Section 5.7.

TkB

µα =

Where µ is the chemical potential. In Section 5.8, we will find out that the chemical potential, µ, is exactly equal to the Fermi Energy, EF — Leaving us with an α of:

(Eq. 5.7.26)

TkE

B

F=α

From previous slides:

TkB

1=βjE

j

jj e

gN

Ef βα−==)(

TkEE

j

jj

B

jF

egN

Ef−

==)(

Substituting α and β into f(Ej):

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Reversing terms in the numerator of the exponent:

This distribution gives the number of particles in the jth state, where the jth state has a degeneracy, gj.

If we want to find the probability of finding the particle in the jth state, we need to normalize. We’ll start by dividing the number of particles in the jth state, Nj, by the total number of particles, N.

Using Constraint 1 and f(E):

∑∑=

−

=

==n

j

TkE

j

n

jj

Bj

egeNN11

α

TkEE

j

jj

B

jF

egN

Ef−

==)(

( )TkEE

j

jj

B

Fj

egN

Ef−−

==)(

It’s intuitive! For a given temperature, the chance of higher energy states being occupied decreases exponentially.

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Solving for eα:

Substituting eα into f(Ej):

∑∑=

−

=

==n

j

TkE

j

n

jj

Bj

egeNN11

α

∑=

−=n

j

TkE

jB

j

eg

Ne

1

α

TkE

jjjB

j

eegNEf−

== α)(

∑=

−

−

=n

j

TkE

j

TkE

jj

Bj

Bj

eg

eNgN

1

Normalizing:

∑=

−

−

=n

j

TkE

j

TkE

jj

Bj

Bj

eg

egNN

1

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This normalized distribution is the probability, Pj, of finding the particle in the jth state with energy Ej:

Since j is just a dummy index, the probability of finding of a particle having an energy Eris:

∑=

−

−

=n

j

TkE

j

TkE

jj

Bj

Bj

eg

egNN

1

∑=

−

−

==n

j

TkE

j

TkE

jjj

Bj

Bj

eg

egNN

P

1

∑=

−

−

=n

j

TkE

j

TkE

rr

Bj

Br

eg

egP

1

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Maxwell-Boltzmann DistributionIf the degeneracy factor is 1, i.e. no repeatable arrangements, the probability becomes:

If we want to find out the mean value of a physical observable, y, we can make use of the following equation derived in Chapter 1:

For example, the mean energy of a system can be determined by:

∑∑

=

rr

rrr

P

yPy

∑=

−

−

=n

j

TkE

TkE

rB

j

Br

e

eP

1

∑

∑−

−

=

r

TkE

r

TkE

r

Br

Br

e

eEE

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∑

∑−

−

=

r

TkE

r

TkE

r

Br

Br

e

eEE


The classical partition function:

∑−

=r

TkE

Br

eZ

This denominator occurs very often—every time a mean value is calculated in the Maxwell-Boltzmann Distribution. Since it occurs so often, it is given a special name, the classical Partition Function, Z.

We can use the classical partition function to easily calculate the mean value of the energy.

First, we need to note the following relationships:

∑∑ −−

∂∂

−=r

E

r

Er

rr eeE ββ

β

∑ −=r

EreZ β

TkB

1=β; ⇒

Zeer

E

r

E rr

βββββ

∂∂

−=

∂∂

−=∂∂

− ∑∑ −−

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Z

ZE β∂

∂−

=


Substituting into the mean energy equation:

ZeEr

Er

r

ββ

∂∂

−=∑ −

∑ −=r

EreZ β

ZZ

Eβ∂∂

−=1

∑

∑−

−

=

r

TkE

r

TkE

r

Br

Br

e

eEE

( )ZE lnβ∂∂

−=

The mean energy equation

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Maxwell-Boltzmann DistributionSummarizing:

∑ −=r

EreZ β

∑

∑−

−

=

r

TkE

r

TkE

r

Br

Br

e

eEE

( )ZE lnβ∂∂

−=

∑=

−

−

=n

j

TkE

j

TkE

rr

Bj

Br

eg

egP

1

TkB

1=β

( )TkEE

j

jj

B

Fj

egN

Ef−−

==)( TkE

B

F=α

( )TkEE

B

F

eEf−−

=)(

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Maxwell-Boltzmann DistributionSummarizing:

( )ZE lnβ∂∂

−=

• For a given temperature, the chance of higher energy states being occupied decreases exponentially

• The area under the curve, i.e. the total number of molecules in the system, does NOT change

• The most probable energy value is at the peak of the curve, when dP/dE = 0

• The average energy value is greater than the most probable energy value

• If the temperature of the system is increased, the most probable energy and the average energy also increase because the distribution skews to the right BUT the area under the curve remains the same

http://www.bustertests.co.uk/studies/kinetics-collision-theory-maxwell-boltzmann-distribution.php

∑=

−

−

=n

j

TkE

j

TkE

rr

Bj

Br

eg

egP

1

( )TkEE

B

F

eEf−−

=)(

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