Statistical Physics
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Topics

Introduction

The Boltzmann Distribution

The Maxwell Distribution

Summary
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Introduction
We believe we now have the basic laws that, in
principle, can be used to predict the detailed
behavior of an arbitrarily large assembly of
atoms and molecules
But even a tiny piece of matter consists of
millions of atoms
In practice, the complexity of the calculation is
far beyond the capability of any conceivable
computer and we need a different approach
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Introduction
Happily, for most applications we are not
interested in the precise behavior of each
atom, but only the collective behavior of the
assembly, which can be described with only a
few variables, such as temperature, pressure
and volume
Statistical physics is the study of the collective
behavior of large assemblies of atoms and
molecules using probabilistic reasoning
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The Boltzmann Distribution
The Austrian physicist
Boltzmann asked the following
question: in an assembly of
atoms, what is the probability
that an atom has total energy
between E and E+dE?
His answer: Pr( E )  f B ( E )
where
fB (E )  Ae
 E / kT
Ludwig Boltzmann
1844 - 1906 5
The Boltzmann Distribution
fB (E )  Ae
 E / kT
is called the
Boltzmann distribution,
e-E/kT is the Boltzmann
factor and
k = 8.617 x 10-5 eV/K
is the Boltzmann constant
The Boltzmann distribution
applies to identical, but
distinguishable particles
Ludwig Boltzmann
1844 - 1906 6
The Boltzmann Distribution
The number of particles with energy E is given
by
n ( E )  g ( E ) f B ( E )  A g ( E )e
 E / kT
where g(E) is the statistical weight, i.e., the
number of states with energy E.
However, in classical physics the energy is
continuous so we must replace g(E) by g(E)dE,
which is the number of states with energy
between E and E + dE. g(E) is then referred to
as the density of states.
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The Boltzmann Distribution
Example 8-1
The Law of Atmospheres
Classically, the total energy of a gas molecule
of mass m, near the Earth’s surface, is
where z is the vertical
2
p
E 
 m gz distance above the ground
2m
z
Wanted: the
So we can write
fraction
of
2
 p / 2 m kT  m gz / kT
particles
f B  Ae
e
between
z and z+dz 8
The Boltzmann Distribution
Example 8-1
The Law of Atmospheres
A basic rule of probability is:
sum, or integrate, over quantities whose values
are either unknown or not of interest.
z
We are interested only in z. After
integrating the Boltzmann distribution with
respect to p we get
 m gz / kT
fB  A 'e
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The Boltzmann Distribution
Example 8-1
The Law of Atmospheres
The fraction of molecules between z and z + dz
is
f B ( z ) dz 
z
mg
e
 m gz / kT
dz
kT
At T = 300K, the ratio of the fraction
at z = 1000 m to that at z = 0 m is
just fB(1000) / fB(0) = 0.893
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The Boltzmann Distribution
Example 8-2
H Atoms in First Excited State
At temperature T, the atoms of a gas will
occupy different energy levels.
For hydrogen, the energy difference E2 - E1
between the 1st excited state and the ground
state is 10.2 eV. What is the ratio of
the number of atoms in the 1st excited state to
the number in the ground state at T= 5800 K
(the temperature of the Sun’s “surface”)?
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The Boltzmann Distribution
Example 8-2
1. Number of atoms in state E
n ( E )  g ( E ) f B ( E )  A g ( E )e
2.
 E / kT
Ratio of number of atoms in E2 and E1
n 2 / n1  g 2 f B ( E 2 ) / g 1 f B ( E 1 ) 
g2
e
 ( E 2  E1 ) / kT
g1
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The Boltzmann Distribution
Example 8-2
3. Ratio of statistical weights. The degeneracy
for each orbital quantum number l is given by
2l+1. For the ground state of hydrogen, l = 0,
which gives 1. For the 1st excited state l = 0
and l = 1, which gives 4. But for each of
these states the electron has 2 spin states.
So we have g1 = 2 and g2 = 8. So
g2
g1

8
2
4
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The Boltzmann Distribution
Example 8-2
4. For T = 5800 K (kT ≈ 0.5 and
DE = E2-E1 = 10.2 eV) we have
n2
g 2  ( E 2  E1 ) / kT

e
n1
g1
 4e
 1 0 .2 / 0 .5
 10
8
Even at the Sun’s surface there are relatively
few atoms in the 1st excited state. This is
because the energy gap DE >> kT
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The Maxwell Distribution
An important application of the Boltzmann
distribution is the distribution of molecular
speeds v in a gas of N molecules:
 m 
n ( v ) dv  4  N 

 2  kT 
3/2
2
v e
2
 m v / kT
dv
This distribution, in fact, was derived by
James Clerk Maxwell before Boltzmann’s work.
But it is an important (and famous) special case.
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The Maxwell Distribution
Most probable
Average
RMS
Different summaries
of the molecular speed
computed from the
Maxwell
distribution
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The Maxwell Distribution
Example
The average speed of a nitrogen molecule
at T = 300 K is given by
v 
8kT
m
With k = 1.39 x 10-23 J/K and
m = 4.68 x 10-26 kg one gets
<v> = 475 m/s = 1700 km/h
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Summary


Statistical physics is the study of the
collective behavior of large assemblies of
particles
Ludwig Boltzmann derived the following
energy distribution for identical, but
distinguishable, particles
 E / kT
fB (E )  Ae

The Maxwell distribution of molecular speeds
is a famous application of Boltzmann’s general
formula
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