Physical Chemistry
Lecture 5
Probability and Boltzmann’s
Distribution
Probability
Likelihood of an event
Described by a number between 0 and 1


0 – No chance of the event ever happening
1 – Absolute certainty that the event happens
Traditional “events”

Coin toss
 Two equally likely events
 PH = PT = ½

Die toss
 Six equally likely events
 P1 = P2 = P3 = P4 = P5 = P6 = 1/6
Probabilities are only “exact” in the limit of an infinite number
of trials
Multiple-elementary-event
probability
Events must be uncorrelated


The likelihood of the first event bears no relation to the
likelihood of the second event
Poverall = PAPB
Sequence of tosses of two dice



Toss two die simultaneously, one red and one blue
P{1,2} = P1P2 = (1/6)(1/6) = 1/36
Events are distinguishable
 One event is on the red die, the other on the blue
Does not apply to correlated events

Second event requires a certain result of the first to
determine its probability (correlation)
Probability of sequences
Multiple-event strings


Consider more than one event as the unit -- sequence
Probability of the sequence depends on the events in
the sequence
Toss of two unloaded dice


Probability of any elementary sequence is 1/36
All elementary sequences equally likely
Sequences may have properties in common



{1,2} and {2,1} both sum to 3
Probability of throwing a sequence adding to 3 with two
dice is P = P{1,2} + P{2,1} = n3 Pelementary = 2 (1/36) = 1/18
Probability of throwing a sequence adding to 2 with two
dice is P = P{1,1} = n2 Pelementary = 1 (1/36) = 1/36
Probability of sum on throws of
two dice
0.18
0.16
Probability of Sequence
0.14
0.12
0.1
0.08
0.06
0.04
0.02
0
2
3
4
5
6
7
Sum of Two Dice
8
9
10
11
12
Sequences and permutations
Sequence of events






Probability of a single sequence
is a product of the probabilities
of the events in the sequence
If order of occurrence in the
sequence is not important, one
must consider all related
sequences equivalent
Each different sequence is a
permutation
Number of permutations of a
sequence of N objects is N!
Like determining the number of
ways to put N objects in exactly
N boxes
Number of permutations of the
numbers on two dice that add
up to 7 is 6 by counting.
Psequence
61
52
43
34
25
16
= P1 P2 P3 
Sequences that add to 7 for throwing two dice.
Permutations and configurations
Number of permutations of
the order of M objects in a
sequence is M!
Number of permutations of
N objects out of M in a
sequence is smaller than the
number of permutations of
the whole set by (M-N)!
Configuration is an
unordered arrangement


Several permutations of a
sequence may correspond
to the same configuration
N! permutations correspond
to the same configuration
n perm, M
n perm, N
nconfig
= M!
=
M!
( M − N )!
=
M!
N !( M − N )!
Probability of configuration
Probability of a
configuration found by


Identifying the
probability of the most
elementary sequence
Multiplying by the
number of permutations
of the sequence in the
configuration
P( N ; M ) =
=
M!
Pelementary
N !( M − N )!
M!
P N (1 − P ) M − N
N !( M − N )!
Dominant configuration
 ∂nconfig

 ∂N

 = 0

1
0.9
0.8
Probability/Probability 50%
Dominant configuration has
the largest number of
permutations of any
configuration
Likelihood of a configuration
(% heads for two dice
shown) depends on the
number of events
Dominant configuration is
totally dominant for most
cases involving macroscopic
numbers of particles
0.7
N =10
0.6
0.5
0.4
0.3
0.2
N = 100
0.1
0
0
10
20
30
40
50
% Heads
60
70
80
90
100
Configurations and distributions
Flipping a coin produces a
limited number of outcomes


Number of permutations for
a sequence of N flips
Like putting the flips into
two different boxes
n perm, 2 (nH ) =
N!
=
nH !nT !
N!
nH !( N − nH )!
Consider putting particles
into many boxes numbered
1,2,3,…




Each configuration has a
number of particles in each
box
Often called a distribution
Number of permutations for
a particular distribution is
easily generalized from two
boxes (coin flips)
Dominant configuration has
1 particle in each box
n perm, N ({ni }) =
N!
n1!n2 !n3!
Constrained and unconstrained
maxima
What is the shortest
distance between two
cities?

Draw a straight line and
measure the length
What is the shortest
distance between two
cities, subject to the
requirement one must stay
on roads?


May not be a straight line
Example: distance from
Newark to Baltimore
Always encounter
constraints in solving
problems
Lagrange’s method to find
maximum subject to a
constraint
Finding Boltzmann’s distribution
Two constraints


Number of particles is
constant
Total energy is constant
Create a function that
represents the number
of permutations AND
the constraints


α and β are Lagrange
multipliers
Solve for maximum of
this function to find the
constrained maximum of
the distribution function
∑n
i
= N
i
∑nε
i i
= E
i


N!

G ({ni }) = ln
n
!
n
!
n
!

 1 2 3 




+ α  ∑ ni − N  + β  ∑ niε i − E 
 i

 i

 ∂G 

 = 0
∂
n
 i n j
Boltzmann’s distribution
Solving the derivative
equations is straightforward
Result is the dominant
distribution subject to a
constant number of particles
and a constant total energy
Depends on the
undetermined multiplier, β,
and the total number of
particles, N
For distributions of large
numbers of particles, this
distribution is the most
dominant
β can be shown to be related
to temperature: β = 1/kT
ni
=
N
− βε i
e
− βε j
∑e
j
Summary
Statistics of random processes predicts the
dominance of Boltzmann’s distribution
To arrive at Boltzmann’s distribution, must
constrain the system to


Specific number of particles
Specific total energy
Dependence of temperature through the
Lagrange multiplier
Applicable when the system contains a very
large number of particles