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CHAPTER 9
Statistical Physics
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9.0
9.1
9.2
9.3
9.4
9.5
9.6
9.7
Intro
Historical Overview
Maxwell Velocity Distribution
Equipartition Theorem
Maxwell Speed Distribution
Classical and Quantum Statistics
Fermi-Dirac Statistics
Bose-Einstein Statistics
Ludwig Boltzmann, who spent much of his life studying statistical
mechanics, died in 1906 by his own hand. Paul Ehrenfest, carrying on his
work, died similarly in 1933. Now it is our turn to study statistical
mechanics. Perhaps it will be wise to approach the subject cautiously.
- David L. Goldstein (States of Matter, Mineola, New York: Dover, 1985)
9.0 : What are we dealing with?
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Physics : object, state, equation of motion, …
~infinitely many objects in some cases (e.g., gas)
Practically it is neither possible to track the motions of the
constituent particles nor interesting to know
Physical properties (states) defined in average sense. It is what
we measure or see.
Need to introduce concepts of statistics and probability
Eventually, we want to learn quantum statistics (developed in
20C)
The fundamental assumption
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Probability? Statistics?
Rolling n dices. p ~ (# of possible events)/6n
The fundamental assumption : each (accessible) ‘event’ has the
same probability
Probability ⋉ # of events
Ideal gas system as an example - event
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Consider a gas molecule. How is an ‘event’ defined for the molecule?
Event = microstate
Six parameters for a gas molecule — the position (x, y, z) and the
velocity (vx, vy, vz)
These parameters define six-dimensional phase space
Small unit volume (ℏ/2𝑚) in the phase space defines an event.
v
microstate
(event)
∆𝑝∆𝑥 = 𝑚∆𝑣∆𝑥 =
ℏ
2
ℏ
 Minimum volume ∆𝑣∆𝑥 = 2𝑚
x
Do all the events have the same probability?  no!
Ideal gas system – isolated system
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microstates for a system have the same probability if the system is
isolated  no influence from outside
Isolated  energy is constant  states with the same energy are
accessible
All the accessible states have the same probability  just count the
# of states!
v
accessible states
microstate (event)
x
• Does a molecule in a container has a fixed energy?  No
• What went wrong?  molecule interacts with the container, more ‘accessible’
states  not an isolated system.
• How are we going to deal with the situation?
Making an isolated system
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(Container + molecule) is an isolated system  we can use the
fundamental assumption for this system
However, we are only interested in the molecule, not the container.
For example, what is the probability for the molecule to have (x,v)?
 count # of accessible states for the system for given (x,v) of the
molecule
# of accessible states for the system = # of accessible states for the
container
This is an isolated system and we know how to deal with it
Yet, this remains to be what we want to know 
• How do we find # of states for the container?
Entropy & probability
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Entropy S defined as 𝑑𝑆 = 𝛿𝑄𝑟𝑒𝑣 /𝑇 (in relation to the 2nd law of
thermodynamics)
In statistical mechanics, 𝑆 = 𝑘𝐵 𝑙𝑛Ω or Ω = 𝑒 𝑆/𝑘𝐵 where Ω is the number
of microstates (for an isolated system)
1
𝑇
=
𝜕𝑆
𝜕𝐸
or Ω~𝑒 𝐸/𝑘𝐵𝑇
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From thermodynamics,
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Ω𝑡𝑜𝑡 = Ω𝑏𝑜𝑥 Ω𝑝𝑡𝑙 = Ω𝑏𝑜𝑥 ( the state for the molecule is fixed)
Combining above three gives 𝑝~Ω𝑏𝑜𝑥 ~𝑒 𝐸𝑏𝑜𝑥 /𝑘𝐵𝑇 ~𝑒 (𝐸𝑡𝑜𝑡 −𝐸)/𝑘𝐵𝑇 ~𝑒 −𝐸/𝑘𝐵𝑇
Probability for a system (molecule) in thermal equilibrium with a
reservoir (box) at a state is proportional to 𝑒 −𝐸/𝑘𝐵𝑇  Boltzmann factor
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𝑝𝑖 =
𝑒 −𝐸𝑖 /𝑘𝐵 𝑇
𝑗𝑒
−𝐸𝑗 /𝑘𝐵 𝑇
Container has energy Etot-E
This has energy E
9.1 : Historical Overview
Benjamin Thompson (Count Rumford)
 Put forward the idea of heat as merely the motion of individual
particles in a substance
James Prescott Joule
 Demonstrated the mechanical equivalent of heat
James Clark Maxwell
 Brought the mathematical theories of probability and statistics to
bear on the physical thermodynamics problems
 Showed that distributions of an ideal gas can be used to derive the
observed macroscopic phenomena
 His electromagnetic theory succeeded to the statistical view of
thermodynamics
Historical Overview
Einstein
 Published a theory of Brownian motion, a theory that supported
the view that atoms are real
Bohr
 Developed atomic and quantum theory
9.2: Maxwell Distribution of Molecular Speed
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Ideal gas in a box.
What is the expected # of particles with the speed between 𝑣 and
𝑣 + 𝑑𝑣?
Molecules do not interact with each other. We can focus on one
particular molecule
Six parameters (position (x, y, z) and velocity (vx, vy, vz)) for the
molecule  a state is a small box in the six-dimensional phase
space
Position can be neglected because the energy depends only on the
velocities (remember 𝑒 −𝐸/𝑘𝐵𝑇 ?)
We can focus only on this
and ignore the rest
Velocity Distribution
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Define the velocity distribution function 𝑓(𝑣) ( because it is the
velocity that defines a state)
𝑓(𝑣)𝑑 3 𝑣 = probability of finding the particle with the velocity
within 𝑑 3 𝑣 at 𝑣. (𝑑 3 𝑣 = 𝑑𝑣𝑥 𝑑𝑣𝑦 𝑑𝑣𝑧 )
2
2
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𝑓(𝑣)𝑑 3 𝑣 ∝ 𝑒 −𝑚𝑣 /2𝑘𝐵𝑇  𝑓 𝑣 𝑑 3 𝑣 = 𝐶𝑒 −𝛽𝑚𝑣 /2 where 𝛽 = 𝑘𝐵 𝑇
All we need to is to find C from 𝑓 𝑣 𝑑3 𝑣 = 1
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Because v2 = vx2 + vy2 + vz2,
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Rewrite this as the product of three factors.
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 𝐶 = 𝐶′3
Maxwell Velocity Distribution
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g(vx) dvx is the probability that the x component of a gas
molecule’s velocity lies between vx and vx + dvx.
 if we integrate g(vx) dvx over all of vx, it equals to 1
then
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&
The velocity distribution function 𝑓 𝑣 is
9.3: Maxwell Speed Distribution
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It is useful to turn the velocity distribution into a speed
distribution as we often do not care about the direction.
F(v) dv = the probability of finding a particle with speed between
v and v + dv.
The volume of the spherical shell is 4πv2 dr.
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Maxwell speed distribution:
F ( v) dv = 4p C exp (- 12 b mv 2 ) v 2 dv
It is only valid in the classical limit.
Maxwell Speed Distribution
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The most probable speed v*, the mean speed
mean-square speed vrms are all different.
, and the root-
Maxwell Speed Distribution
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Most probable speed (at the peak of the speed distribution):
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Mean speed (average of all speeds):
2
32
é
ù
æ
ö
æ bm ö
1
1
4
ú
v = 4p C ê
=
8
p
=
ç
÷ ç
÷
2
1
è 2p ø è b m ø
2p
êë 2 ( 2 b mv) úû
kT
m
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Root-mean-square speed (associated with the mean kinetic energy):
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Standard deviation of the molecular speeds:
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σv in proportion to
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Kinetic energy :
9.4: Equi-partition Theorem
Equipartition Theorem:
 ‘In equilibrium, a mean energy of ½ kT per molecule is associated
with each independent quadratic term in the molecule’s energy.’
Examples:
1. U = ½ kx2 + p2/2m  kT
2. U = (px2 + py2 + pz2)/2m  3kT/2
Equipartition Theorem – Ideal gas
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In a monatomic ideal gas, each molecule has
K = 12 mv 2 = 12 m ( vx2 + vy2 + vz2 )
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There are three degrees of freedom.
Mean kinetic energy is 3( 12 kT ) = 23 kT
In a gas of N helium molecules, the total internal energy is
U = NE = 23 NkT
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The heat capacity at constant volume is
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For the heat capacity for 1 mole,
CV = (¶U ¶T )V = 23 Nk
cV = 23 N A k = 23 R =12.5J K
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The ideal gas constant R = 8.31 J/K.
Equipartition Theorem - Rigid Rotator
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For diatomic gases, consider the rigid rotator model.
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The molecule rotates about either the x or y axis.
2
2
1
1
I
w
and
I
w
The corresponding rotational energies are 2 x x
2 y y
There are five degrees of freedom (three translational and two
rotational).
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Molar Heat Capacity
Kvib = 12 kz 2 + 12 mvz2  two more from here
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The heat capacities of diatomic gases are temperature dependent,
indicating that the different degrees of freedom are “turned on” at
different temperatures.
Example of H2
Why does it turn on?
Boltzmann factor summary
Isolated system
(Heat) Reservoir
at T
Reservoir and system A
are in thermal contact.
A
The system that
we are interested in
Reservoir is so large compared to system A
that its temperature does not change.
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The system A can be at states with different energies
The probability for the system at a state with energy EA is proportional
to exp(-bEA)
9.5: Classical vs. Quantum Statistics
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So far, we have classical distribution (statistics), which is valid
when T is high and density is low (QM effect is weak)
When T is low (kBT < DE), the effect of the discrete energy levels
shows up (remember the UV side of the Blackbody radiation?).
For example, it is the reason why heat capacity CV  0 when T
 0. (rotation and vibration contributions in the previous page)
There is more! When density is very high or T is low, there is a
good chance that more than 1 particle occupy the same quantum
(micro) state.
The distribution is strongly dependent on the the character of the
particles (Fermion or Boson). It drastically changes the number
of states and , as a result, the classical theory has to be
modified.
Classical Distributions
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Boltzmann showed that the statistical factor exp(−βE) is a
characteristic of any classical system.
 this is regardless of how quantities other than molecular
speeds may affect the energy of a given state
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(Maxwell)-Boltzmann factor for classical system:
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The energy distribution for classical system:
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n(E) dE = the number of particles with energies between E + dE
g(E) = the density of states, is the number of states available per
unit energy range
FMB tells the relative probability that an energy state is occupied at
a given temperature
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Counting : distinguishable & indistinguishable
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Indistinguishability of identical particles changes the way we
count the number of states.
It is what makes quantum statistics different from classical
statistics (of course, when it matters).
Example : The possible configurations for distinguishable and
indistinguishable particles in either of two energy states:
State 1
State 2
AB
State 1 State 2
XX
A
B
B
A
X
X
XX
AB
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The probability of each is one-fourth (0.25) or one-third (0.33)
Fermion vs Boson
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That is not all! Fermions are subject to the Pauli exclusion principle and
two Fermions cannot occupy the same state.
Bosons
Fermions
State 1 State 2
State 1 State 2
XX
X
X
X
X
‘state’ is for a single
particle
XX
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3 events are possible for Bosons but only 1 event is possible for
Fermions. Each event has equal probability (fundamental assumption)
Fermions:
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Particles with half-spins. Two or more Fermions are not allowed to occupy
the same state (Pauli exclusion principle).
Bosons:
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Particles with zero or integer spins that do not obey the Pauli principle.
Multiple particles are allowed to occupied the same state
‘an event’ refers to a state for the whole system
Quantum Distributions
Consider a system with two identical Bosons with 3 possible states
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event1 event2
event3
event4
event5
event6
2e
e
0
Etotal
0
P
Ce0
e
Ce-be
2e
2e
3e
4e
Ce-2be
Ce-2be
Ce-3be
Ce-4be
What is the expected
# of particles n(E)
at each state?
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Bose-Einstein distribution:
Density of state or DOS
1
where FBE =
(BBE normalization constant)
BBE exp ( b E ) -1
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For Fermions, Fermi-Dirac distribution :
1
F
=
where
FD
BFD exp ( b E ) +1
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Both distributions reduce to the classical Maxwell-Boltzmann distribution
when exp(βE) is much greater than 1.
Quantum Distributions
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The normalization constants for the distributions depend on the
physical system being considered.
Because Bosons do not obey the Pauli exclusion principle, more
bosons can fill lower energy states.
Three graphs coincide at high energies – the classical limit.
 Maxwell-Boltzmann statistics may be used in the classical limit.
9.6: Fermi-Dirac Statistics
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Electrons determine the properties of solids
Electrons are Fermions, and their density is very high.
 Fermi-Dirac distribution is extremely important.
EF is called the Fermi energy.
When E = EF, FFD = ½.
In the limit as T → 0,
FFD =
1
exp éëb ( E - EF )ùû +1
At T = 0, fermions occupy the lowest energy levels available to
them.
Near T = 0, there is little chance that thermal agitation will kick a
fermion to an energy greater than EF.
Fermi-Dirac Statistics
T=0
T>0
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As the temperature increases from T = 0, the Fermi-Dirac factor “smears out”
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Fermi temperature, defined as TF ≡ EF / k
T = TF
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T >> TF
When T >> TF, FFD approaches a simple decaying exponential or MaxwellBoltzmann.
Failure of Classical Theory for Electrons
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With the assumption that electrons are like ideal gas in solid,
Paul Drude (1900) could explain various physical properties
solids (such as electrical conductivity).
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Failure of the Drude picture :
Predicted
measured
Heat capacity
(electron)
3R/2
~0.02R at RT
Heat conductivity
~T-1/2
~T-1
What is the problem?
Electron in solid as free particle in a box
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We would like to calculate the density of states
The allowed energies for electrons are (see example 6.10)
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Rewrite this as E = r2E1
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The parameter r is the “radius” of
a sphere in phase space.
The “volume” is πr 3.
The exact number of states up
to radius r is N = 2 1 4 p r 3 .
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r
( )( 8)( 3
spin
)
Density of states for free electron in 3D
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Rewrite as a function of E:
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At T = 0, the Fermi energy is the energy of the highest occupied
level.
Total of electrons
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Solve for EF:
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The density of states with respect to energy in terms of EF:
Quantum Theory of Electrical Conduction
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At T = 0,
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The mean electronic energy:
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Internal energy of the system:
3
U = NE = NEF
5
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Only those electrons within about kT of EF will be able to absorb thermal
energy and jump to a higher state. Therefore the fraction of electrons
capable of participating in this thermal process is on the order of kT / EF.
Electron distribution in metal
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In general,
where α is a constant > 1
kBTF = EF
TF = 80,000 K for Cu
uF = 1.6x106 m/s
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The exact number of electrons depends on temperature.
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Heat capacity is
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Molar heat capacity is
9.7: Bose-Einstein Statistics for blackbody
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Not as important as Fermi-Dirac
Blackbody Radiation can be derived using BE statistics.
We consider radiation in metallic box (as we considered earlier in
blackbody radiation)
Then, we calculate the density of states as we did in FD distribution
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The density of states g(E) is
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The Bose-Einstein factor:
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Use the Bose-Einstein distribution because photons are bosons
with spin 1.
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Bose-Einstein Statistics for blackbody
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Convert from a number distribution to an energy density
distribution u(E).
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Multiply by a factor E/L3
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For all photons in the range E to E + dE
En ( E ) 8p 3 1
u(E) =
= 3 3 E E/kT
3
L
hc
e -1
8p E 3 dE
u ( E ) dE = 3 3 E/kT
h c e -1
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Using E = hc/λ and |dE| = (hc/λ2) dλ
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In the SI system, multiplying by c/4 is required.
Liquid Helium
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Has the lowest boiling point of any element (4.2 K at 1 atmosphere
pressure) and has no solid phase at normal pressure
The density of liquid helium as a function of temperature:
Superfluid transition
The specific heat of liquid helium as a function of temperature:
The temperature at about 2.17 K is referred to as the critical
temperature (Tc), transition temperature, or lambda point.
 What happens at 2.17 K is a transition from the normal phase to the
superfluid phase.
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Superfluid & transition
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Superfluid liquid helium is referred to as a Bose-Einstein
condensation.
o
not subject to the Pauli exclusion principle
o
all particles are in the same quantum state
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By using BE statistics with BBE = 1, we get
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Rearrange this,
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The result is T ≥ 3.06 K.
The value 3.06 K is an estimate of Tc.
Bose-Einstein Condensation in Gases
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Coulomb interaction among gas particles makes it difficult to
obtain high densities for Bose-Einstein condensation.
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In 1995, laser cooling and magnetic trap were used to cool gas of
87Rb atoms to about 20 nK. What remained was an extremely
cold, dense cloud with Tc = 170 nK.