MSCPH 504

MSCPH504
MSCPH504
M. Sc. I SEMESTER
Statistical Mechanics
DEPARTMENT OF PHYSICS
SCHOOL OF SCIENCES
UTTARAKHAND OPEN UNIVERSITY
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Board of Studies
Prof. P. D. Pant Prof. S.R. Jha,
Director School of Sciences School of Sciences, I.G.N.O.U., Maidan
Uttarakhand Open University, Haldwani Garhi, New Delhi
Prof. P. S. Bisht, Prof. R. C. Shrivastva,
SSJ Campus, Kumaun University, Almora. Professor and Head, Department of Physics,
Dr. Kamal Devlal CBSH, G.B.P.U.A.&T. Pantnagar, India
Department of Physics
School of Sciences, Uttarakhand Open
University
Department of Physics (School of Sciences)
Dr. Kamal Devlal (Assistant Professor) Dr. Meenakshi Rana (Assistant Professor (AC))
Dr. Vishal Sharma (Assistant Professor) Dr. Rajesh Mathpal (Assistant Professor (AC))
Dr. Gauri Negi (Assistant Professor)
Unit writing and Editing

Editing Writing
Dr. Kamal Devlal 1. Dr. Mahipal Singh

Department of Physics Department of Physics
School of Sciences, Uttarakhand Open R.H. Govt. P.G. College, Kashipur, Uttarakhand
University 2. Dr. Yogesh Kumar
Arvino College (University of Delhi)
Delhi
3 Dr. Kishore Thapliyal
Faculty of Science,
Palacky University, 17. listopadu 12, 771 46 Olomouc,
Czech Republic
4. Dr. Ravi Shankar Kuniyal
Department of Physics, Government Post Graduate
College, Gopeshwar (Chamoli)-246401
5. Dr. Rajesh Mathpal
School of Sciences, Uttarakhand Open University
Course Title and Code : Statistical Mechanics (MSCPH504)

ISBN :
Copyright : Uttarakhand Open University
Edition : 2021
Published By : Uttarakhand Open University, Haldwani, Nainital- 263139
Printed By : MSCPH502
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MSCPH504
Statistical Mechanics
DEPARTMENT OF PHYSICS
SCHOOL OF SCIENCES
UTTARAKHAND OPEN UNIVERSITY
Phone No. 05946-261122, 261123
Toll free No. 18001804025
Fax No. 05946-264232, E. mail info@uou.ac.in
htpp://uou.ac.in
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Contents
Course 4: Statistical Mechanics Course code: MSCPH504
Credit: 3
Unit Block and Unit title Page

number number
BLOCK 1: CLASSICAL STATISTICS
1 Statistical Mechanics at a Glance 1

2 Statistical Mechanics and Thermodynamics 22
3 Maxwell-Boltzmann Statistics 42
BLOCK 2: ENSEMBLES
4 Microcanonical Ensemble 66
5 Canonical and Grand Canonical Ensemble 89
6 Grand – canonical ensemble 111

BLOCK 3: QUANTUM STATISTICS AND STATISTICS MODELS
7 Quantum Statistics 130

8 Phase Transitions 174
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UNIT 1 Statistical Mechanics at a Glance
Structure
1.1 INTRODUCTION
1.2 OBJECTIVES
1.3 BASIC CONCEPTS OF STATISTICAL MECHANICS
1.3.1 PROBABILITY
1.3.2 BASIC RULES OF PROBABILITY
1.3.3 PROBABILITY DISTRIBUTION
1.3.4 DISTINGUISHABLE AND INDISTINGUISHABLE PARTICLES
1.4 PHASE SPACE
1.5 WHAT IS AN ENSEMBLE?
1.6 DENSITY OF STATES
1.7 ERGODIC HYPOTHESIS
1.8 POSTULATE OF EQUAL A PRIORI PROBABILITY
1.9 SUMMARY
1.10 GLOSSARY
1.11 REFERENCES
1.12 SUGGESTED READINGS
1.13 TERMINAL QUESTIONS
1.14 ANSWERS
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1.1 INTRODUCTION
In our everyday life, we often come across objects in motion.Human beingswere inquisitive
to understand the motion in detail since ages. Nowadays, we categorize the branch of physics
dealing with these moving bodies as mechanics. The motion of objects, we see around us,
astronomical objects, etc., which are macroscopic in nature can be defined by the principles
of classical mechanics. One of the earliest contributions in the classical mechanics were by
Isaac Newton and Gottfried Wilhelm Leibniz, who described the motion of a body under the
action of forces. Applicability of classical mechanics is superseded by relativistic mechanics
introduced by Albert Einstein for the bodies moving near the speed of light. Similarly,
another branch of mechanics developed in the beginning of the last century for the
microscopic systems, such as subatomic particles.Therefore, all these developments widened
our horizon in understanding of the nature and motion of an object.
In the meantime, our understanding of thermodynamic propertiesdealing with the systems in
equilibrium was developed and summarized by the laws of thermodynamics. You may
already be familiar with thermodynamics and related basic key terminologies, like the
concept of system and surroundings, different types of thermal processes, laws of
thermodynamics, state function, state variables and various macroscopic properties
(macroscopic properties are the properties of matter as a whole in terms of macroscopic
variables,likedensity, volume, temperature, pressure) related through an equation of state. So
far these thermodynamic properties for the macroscopic system were not described in terms
of its microscopic constituents.Thus, statistical mechanics is that branch of modern physics
which deals with physical systems with many degrees of freedom and describes their
macroscopic properties in terms of the microscopic properties of the constituent particles.
We understood so far that our treatment of the constituent particles depends upon whether
they can individually be described by classical mechanics or quantummechanics. Depending
upon that statistical mechanics is broadly studied under two categories namely classical and
quantum statistics.The classical statistics is discussed with Maxwell-Boltzmann statistics,
whereas quantum statistics can further be categorized as Bose-Einstein and Fermi-Dirac
statistics depending upon yet another quantum feature called the spin of the particle.
Specifically, all the microscopic particles with integral (half-integral) spin governed by
quantum mechanics are studied under Bose-Einstein (Fermi-Dirac) statistics. Therefore, in
this book, we will discuss the physical properties of various systems containing a large
number of particles (atoms or molecules), comparable to Avogadro number, i.e.,
, on the basis of the properties and behavior of the microscopic constituents of
those systems. To accomplish this we will be using statistical tools and probability theory as
well asthe dynamics of microscopic particles governed by either classical or quantum
mechanics.
In this unit, we will discuss various concepts in classical statistical mechanics, like phase
space, microscopic and macroscopic variables, the concepts of ensemble, and one of the
important principles, i.e., postulate of equal a priori probability. These basic concepts are
building blocks of statistical mechanics which are helpful in understanding the forthcoming
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units and in discussing quantum statistics. In particular, we want to make a statistical analysis
of a large number of collections of identical systems and determine their most probable
behavior. In this process, we are not interested in the individual dynamics of a single particle.
In statistical mechanics, we average the properties of all the particles to studythe macroscopic
bodies they form. For instance, the temperature of a gas is found to be related to the random
motion of the gas molecules. The faster they move on average, the higher is the temperature.
1.2 OBJECTIVES
After studying this unit, you should be able to understand-
● Probability
● Probability distribution
● Distinguishable and indistinguishable particles
● Define phase space and phase points
● Ensemble
● Density of states
● Ensemble average or phase space average and time average
● Ergodic hypothesis
● Consequence of ergodic hypothesis as postulate of equal a priori probability
1.3 BASIC CONCEPTS OF STATISTICAL MECHANICS
Statistical mechanics is a branch of science which deals with the most probable behavior of
an assembly of particles. There are various examples of such systems, like gases, liquids,
solids in different forms, stellar matter, and radiation. In statistical mechanics, one tries to
apply the ideas of statistics appropriate to a system containing a large number of particles that
are individually described by the laws of mechanics.
At the end of the nineteenth century J. C. Maxwell, L. Boltzmann and J. W. Gibbs established
the fundamental pillars of statistical physics. The statistical mechanics started with the
Maxwell’s kinetic theory of gases. To be specific, the first milestone in statistical physics is J.
C. Maxwell’s statistical law for distribution of molecular velocities proposed in 1859, which
will be discussed in detail in Unit 3. This work inspired L. Boltzmannwho dedicated all his
life in developing statistical mechanics. His entropy formula was aptly inscribed on his
tombstone. Specifically, he generalized Maxwell's argument and introduced the concept of
ensembles, which will be discussed in detail in Unit 4. Although, this branch of mechanics
was christened by J. Gibbs in addition to several of his contributions, such as establishing the
equivalence of statistical physics and thermodynamics to be discussed in Unit 2. Boltzmann’s
doctoral student P. Ehrenfestalso contributed significantly in the development of statistical
mechanics, alas, both these stalwarts had same tragic demise.Einstein completedour present
day understanding of statistical mechanicsby his various theories, like the theory of
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fluctuations, diffusion, and Brownian motion. The development ofquantum theory helped in
simplifying the principles of statistical physics.
As we have already mentioned, we are going to use probability theory to quantify various
notions in statistical physics. Here, we introduce the basic conceptof probability and
probability distributionsbecause this serves as an important tool for understanding the
behavior of statistical systems.
1.3.1 Probability
Probability is a mathematical concept that deals with calculating the likelihood of a given
event’s occurrence. Thus, the probability of an event is defined as the ratio of the number of
the cases in favor to the total number of possible cases also called sample size.
… (1.1)
Example 1.1: Tossing a coin-when an unbiased coin is tossed, there are two equally probable
outcomes, heads (H) or tails (T). Thus, we can say that the probability of the coin landingH
up is , and the probability of the coin landing T up is .
Example 1.2: Throwing die- when a fair die is thrown, there are six possible outcomes: 1, 2,
3, 4, 5, and 6. The probability of each of them is .
For example, suppose we wish to obtain the probability of getting a "4" on rolling a die. The
number of ways it can happen is 1 as there is only one face with a "4" on it, and the total
number of possible outcomes is 6 only as there are six faces altogether. So the probability of
getting “4”is .
Example 1.3: Suppose there are 5 marbles (4 blue and 1 red) in a bag. The probability that a
blue marbleis picked can be calculated as:
The number of ways it can happen is 4 for fourblue marbles, and the total number of
outcomes is 5 as there are five marbles in total. So the probability is .
Self Assessment Question (SAQ) 1: A boxcontains 4 chocolates and 4 icecreams. What is

the probability of choosing 1 chocolates?
1.3.2 Basic rules of probability

There are three basic rules of probability distribution, like summation rule, multiplication
rule, and conditional probability.
• Summation rule is applicable to mutually exclusive events, i.e., happening of one
event excludes the possibility of the happening of the other.
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• The multiplication rule is applicable when the probability of occurrence of one event
does not affect the probability of occurrence of the other. This rule is also known as joint
probability, and the probability of joint occurrence of two independent events is equal to the
product of the probabilities of each of the independent events.
• The probability for an event say A to occur conditioned to the fact that another event
say B has also occurred is called the conditional probability. This is denoted as .
All the three rules discussed above can be better understood by the following examples.
Example 1.4: A card is drawn from a pack of 52 cards. The probability of getting a red ace or
a black king can be discussed as
The total number of ways in which the event can occur is N = 52.
We wish to draw a red ace. There are 2 red aces in a pack of 52 cards. Therefore, the number
of ways favorable to the first event say .
Thus, the probability of drawing a red ace .
The number of ways in which second event, i.e., drawing a black king may happen say
, as there are 2 black kings in the pack of 52 cards.
The probability of drawing a blackking .
Notice that both events are mutually exclusive. Therefore, the probability that the card drawn
is either a red ace or a black king is
Example 1.5: A bag contains 6 black marbles and 4 blue marbles. Two marbles are drawn
from the bag, without replacement. The probability that both marbles are blue may be
discussed as
We have 6 black and 4 blue marbles so there are total ten marbles in the bag.
So the probability of the first event, i.e., drawing a blue marble is .
Now there are nine marbles in the bag, so the probability of the second event, i.e., choosing a
blue marble is .
Hence, the required probability that both the marbles are blue is .
Example 1.6: Discuss the probability of drawing two kings from a deck of 52 cards.
First event (say event A) is drawing a king first, and second event (say event B) is drawing a
king second.
For the first card the chance of drawing a king is 4 out of 52 as there are 4 kings in a deck of
52 cards. So,
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After removing a king from the deck the probability of the second card drawn is less likely to
be a king as there are only 3 of the 51 cards left are kings. Therefore,
Probability of event A and event B, i.e., = probability of event A, i.e., times the
probability of event B given event A, i.e., Thus,
Hence, the chance of getting 2 kings is 1 in 221.
Example 1.7: Suppose there are 10 marbles (6 blackand 4blue) in a bag as in Example1.5.
We randomly select two marbles with replacement from the bag, i.e., once we select the first
marble we put it back in the bag and then select the second marble. The probability of getting
two blue marbles may be discussed as,
Here, drawing with replacement makes the draws independent of each other, since the color
of the first marble drawn does not change the sample size for the second draw.
The chance of drawing a blue marble on the first draw is . With the replacement of the first
marble the probability of getting a blue marble on the second draw is again . Thus, the
probability of getting two blue marbles with replacement is . Notice the

difference between the probability of two blue marbles with replacement and without
replacement.
Self Assessment Question (SAQ) 2: What is the probability of drawing a red king and then
a black 7 without replacement from a deck of 52 cards?
1.3.3 Probability distribution

In all the examples discussed in the previous subsections, you have a rule which assignsa
single real value to each outcome as probability. Notice that this number may vary for
different outcomes. Thus, this number is a variableand depends upon the outcome of a
random example, and hence, is called random variable.
For example, consider the case of tossing a coin twice in succession. The sample space in this
case is S = {HH, HT, TH, TT}. If X denotes the number of heads obtained, then X is a
random variable and for each outcome, its value is as X(HH) = 2, X (HT) = 1, X (TH) = 1, X
(TT) = 0.More than one random variable can be defined on the same sample space.Thus,
these random variables can be used to obtain probability in each case by dividing them with
sample size. This mathematical function defines the probability of the occurrence of all the
outcomesand as expected the sum of probability in all the cases is unity.
The probability distribution of a random variable can be understood by following example.
Example 1.8: Discuss the probability distribution of the number of aces when two cards are
drawn successively with replacement from a well-shuffled deck of 52 cards.
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Let the number of aces (random variable) be denoted by X which can take the values 0, 1, or
2. Since the draws are done with replacementthe two draws formindependent
experiments.Therefore,
Thus, the required probability distribution can be summarized as

0 1 2
Similarly, joint probability distributions are defined over two or more random variables.The
probability distribution over one random variable, known as marginal distribution, can be
obtained by summation of the joint probability distributions for the rest of the random
variables.
Self Assessment Question (SAQ) 3: A pair of dice is thrown thrice. Discuss the probability
distribution of the number of doublets.
1.3.4 Distinguishable and indistinguishable particles

Before discussing further, you should know one of the important differences between the
classical statistics and the quantum statistics. This difference is directly related to the nature
of particles which form the system being considered. In classical statistics, particles are
distinguishable whereas in quantum statistics particles are indistinguishable.
Consider Example 1.7 with an additional constraint, i.e., obtain probability that the blue
marble drawn in the first round is not the same as in the second round with replacement. In
principle, if you keep track of each marble after labeling them with different numbers, you
will end up with the same probability as in Example 1.5.This labeling is possible because
these marbles, being classical particles,though may have same color, shape, etc., macroscopic
properties but have uniquepositions in the bag.Further, assume the marbles in this example
are quantum particles. We know that some properties of the subatomic particles are defined
as universal constants, suchas mass, charge.Therefore, they are identical in that sense, and
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once the marble is replaced in the bag the wavefunction associated with the replaced quantum
marble overlaps with that of the remaining marbles in the bag due to spreading of the
wavefunction as time progresses.We know that one cannot distinguish overlapping
wavefunctionsso the quantum state of the composite system ofquantum marblescan be
written as superposition of all the combinations. Therefore, the probability inthis case will be
different from the classical particles.For example, in case oftwo indistinguishable quantum
particles, quantum state is obtained asthe sum (difference)of two-particle wavefunctionand
that obtained by particle exchangewhich is symmetric (antisymmetric). The symmetric
(antisymmetric) wavefunction corresponds to bosons (fermions), i.e., quantum particles with
integral (half-integral) spin. You can easily verify the significance of Pauli exclusion
principlethat two fermions cannot have the same values for all the quantum numbers. We will
return to this discussion in forthcoming units, especially Units 7-10.
Suppose you have two distinct particles which are distinguishable, like a Helium-3
atom and a Helium-4 atom. If you switch their positions then the system changes. On the
other hand, if two particles are indistinguishable, like two protons then, switching their
positions makes no physical change because you do not know whether particles switched at
all. The concepts of distinguishable and indistinguishable particles is important in statistical
mechanics. Hereafter, for discussing classical statistical mechanics the particles shall be
considered as distinguishable and later in quantum statistics you will understand the
indistinguishability of particles in detail.
Example 1.9: Let us rearrange all the letter of the word “POTATO” and compute the
probability of obtaining words starting with P.
The total number of ways the word “POTATO” can be rearrangedis .
There is only 1 possibility for the first letter to be P andthe rest of the five letters can be
chosen in 5! ways. Notice that there are two T and O in the word, so effectively there will be
only unique words.
Thus, the probability of obtaining words starting with P can be written
as .
This example shows that a repeated occurrence of letters changes the probability of a
favorable outcome.Similarly, distinguishability (indistinguishability) of particlesplays
significant role in the computation of probabilities and thus statistics of quantum and classical
particles.
1.4 PHASE SPACE

You all know, in a static system the complete position of an object or a point particle in
classical mechanics is specified by three Cartesian coordinates. This three dimensional space
is known as position space ( ) and a small volume element in position space is
defined as . If the system is dynamic then in addition to the position
coordinates, we require three components of momenta for the specification of particles in the
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system. These three mutually perpendicular momentum coordinates

constitute momentum space. A small volume element in this space is
expressed as .
In a similar way, phase space is a scheme for the specification of a system in statistical
mechanics. In this, the position of a particle is represented in terms of Cartesian coordinates
and the corresponding momentum ( ) components. Thus, the phase space is
a combined position and momentum space. A small volume element in phase space is defined
as
…(1.2)
Such a six dimensional space for a single particle or molecule is called phase space or mu-
space (µ-space). The phase space is a pure mathematical concept used to describe a single
particle. Let us divide the phase space in two dimensional energy sheet as shown in
Figure1.1. Further, if we subdivide the range of variables and into arbitrary small
discrete intervals, then the single interval is known as phase cell. The minimum size of the
phase cell in classical statistics is equal to the area of thesingle cell, i.e., .
This may be viewed as our experimental limitation in measurement.In the classical
scenario, can be chosen arbitrarily as small as possible.Further, phase cell is the volume
occupied by each phase point in the phase space. Hence the value of this elementary volume
is equal to . However, in quantum statistics,according to the Heisenberg’s uncertainty
principle the minimum size(volume) of phase cell in phase spaceis given by ,where h is
Planck’s constant.
Figure1.1: Two dimensional energy sheet of phase space representing phase cells.
Let us consider a system of molecules or particles. If each molecule is specified in phase

space by , , position and , , momentum coordinates, then for a system of N
molecules there will be 6N dimensional space, i.e., 3N position and 3N momentum
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coordinates. This space of 6N dimension is called gamma-space or Γ space. In Γ space each

point is called a representative point or phase point corresponding to a state of a system of N
particles at a particular time.The number of phase points per unit volume of the phase space
is called phase density. With the passage of time the representative point traces out a
trajectory which gives us a complete dynamics of the system.This trajectory is known as
phase space trajectory. It is also important to mention here that a phase space trajectory never
intersects itself because if it does so, then the point of intersection would denote two different
states at two different times for the same system which is not physically acceptable. In
Figure1.2, a random phase space trajectory showing the temporal evolution of systemis
depicted. Analogously, the state of the system of N particles can be represented by specifying
the state of each particle in six dimensional spaceµ -space (as discussed above). The µ -space
is represented by a cloud of N points in a six dimensional space.
Thus, the microscopic state of the complete system is specified by the phase points. These
phase points vary with time, and therefore microscopic states of a dynamic system are
differentfor each instant of time. Further, for the measurement of macroscopic variables (N,
V, P, T, E, etc.) of the whole system we need to calculate the time average of the system of
interest. Hence, each phase point represents a separate system with same macroscopic
properties, but a different microscopic state.
Figure1.2: Random trajectory in phase space.
Example 1.10: Phase space diagram of an oscillator.

Let us consider an oscillatorof mass m and spring constant k with energy
...(1.3)
We can rewrite this equation as
…(1.4)
This represents an ellipse (see Figure1.3) in the plane for constant energy . This is a
phase space diagram of an oscillator. The phase points are lying on the elliptical path with the
ellipse whose semi major axis and semi minor axis are and , respectively. Thus,
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the phase space available to the oscillator having the energy between 0 and E is the area of
the ellipse given by .
Figure1.3: Phase space diagram of an oscillator.
Self Assessment Question (SAQ) 4: Using the concept of phase space find out the region of
states accessible for a particle in an infinite square well potential defined as
Self Assessment Question (SAQ) 5: Discuss the phase space trajectory of a free particle
moving in the positive direction.
1.5 WHAT IS AN ENSEMBLE?

If we are considering a collection of particles with macroscopic properties, like energy,
volume, chemical potential,then the collection of such particles is considered as an assembly.
Further, this collection of a large number of non-interacting, independent assemblies is
known as an ensemble or statistical ensemble. The members of an ensemble are referred to as
elements or assemblies.These elements are identical in macroscopic properties, like , , ,
and differ in their microscopic properties, i.e., elements have different position and
momentum coordinates.
In other words, we can say that an ensemble is defined as a collection of a large number of
assemblies which are identical in macroscopic properties but differ in microscopic properties.
Thus, it can be viewed as numerous copies of a system or a probability distribution defining
the state of the system.
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Figure1.4: Phase space diagram of harmonic oscillator with energy between and
for infinitesimally small .
For example, consider anotherphase space trajectory of a harmonic oscillator of energy

, with infinitesimally small , represented by an ellipse in Figure1.4. Each point in
phase space between the two ellipses represent state of a harmonic oscillator with a constraint
that their energy is between and . This corresponds to our experimental
limitation in determining the energy of the system. This hypersurface, whichcorresponds to
the collection ofall the phase pointssatisfying the constraint, is phase space distribution or an
ensemble.
There are three types of ensembles, namely microcanonical, canonical and grand canonical,
which will be discussed in detail in Unit 4. These three categories are specified on the basis
of macroscopic constraints.
1.6 DENSITY OF STATES

We have already discussed that the microscopic properties of a system are represented by
phase points. The condition of an ensemble at any time can be specified by the density, which
describes the distribution of phase points in phase space. The density distribution is often
denoted by and is a function of the continuous variables and . Consequently,
the normalization condition for a closed system is , where the volume
elements are and . Specifically, the
density of distribution in phase space gives the number of states per unit volume in a given
interval of energy of the phase space. This distribution function is a function of position
coordinates and momentum coordinates. The time dependence of may be implicitly
governed by the time dependence of q and p. However, may also have an explicit time
dependence.
Hence, we can write
… (1.5)
Therefore, the number of phase points in a small volume element say is
given by
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We will further calculate an expression of density of state for a single particle of mass m with
momentum in the range p to p+dpand energy in the range E to E+dE, respectively, placed in
a phase cell of volume in phase space. The total volume of the phase space is given by
Total volume of phase space
…(1.6)
where we have used the volume of position space ,the volume of the
momentum space
Thus, we can write the total number of phase cells in the given momentum range as
…(1.7)
Using the relation between energyEand momentum p, i.e.,

,
and therefore,
Using this relation in equation (1.7), the total number of phase cells in the energy range E to
E+dE is
…(1.8)
Hence, the density of state or the total number of phase cells per unit energy range can be
obtained as
…(1.9)
1.7 ERGODIC HYPOTHESIS: Equality of ensemble average

and time average
For a large collection of particles we shall consider the average of the variables for discussing
the complete system. In the next unit, we will discuss the average values of variables in detail
by considering various other aspects of statistical mechanics, like thermodynamic probability,
concept of microstates and macrostates.
In general, the ensemble average value of a variable in an ensemble is calculated as
. … (1.10)
Here, defines the physical property for different phase points with density of phase
points which is normalized . We have already given an
example of ensemble for harmonic oscillatoras a hypersurface between two ellipses ofenergy
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and , respectively.The ensemble average takes into consideration the contribution

from each assembly in the ensemble, represented by phase points in the
hypersurface,weighed with corresponding probabilitiesin the phase space distribution.
Similarly, the time average of the variable over a complete time period is calculated by the
following equation
… (1.11)
In case of our example of harmonic oscillator, temporal evolution of a phase point on the
elliptical hypersurface traces all the phase pointsbetween energy and over a long
period of time.
Thus, time average in principle takes into consideration the contribution from all the points in
the phase space trajectory similarto the ensemble.
Ergodic hypothesis is a connection between both types of averages. It states that the ensemble
averages are equivalent to the time averages of any physical quantity to be considered.In
other words, this hypothesis says that the ensemble average of any variable is the same as the
time average of that variable for the single system.
1.8 POSTULATE OF EQUAL A PRIORI PROBABILITY

In statistical physics, we deal with systems with a very large number of particles and assume
that the system exists in an equilibrium state. All the physical properties of such a system can
be deduced by knowing the most probable or equilibrium state of the system. In such state,
the phase space distribution of the system is independent of time, i.e., the ensemble does not
evolve anymore with time.We will discuss the significance of the most probable state in
establishing the relation between statistical and thermodynamic quantities in the next unit.
The postulate of equal a priori probability is consistent with the idea of equilibrium and is
solely related to the ergodic hypothesis. It states that the probability of finding the phase point
for a given system in any region of the phase space is identical with that for any other region
of equal extension or volume. Thus, you can assign equal a priori probability to equal
volumes in phase space. In other words, for a system in equilibrium, all accessible
microstates corresponding to possible macrostates are equally probable. The concept of
microstates and macrostates in detail will be discussed in Unit 2.In case of phase space
distribution for a harmonic oscillator, as in Figure1.4, all the phase points on the ellipse are
equally likely.This priori principle is the fundamental basis of statistical mechanics. The
following examples will illustrate this postulate in detail.
Example 1.11: Suppose we pick a card at random from a well shuffled pack of 52 cards.
There is nothing that would favor one particular card over the others. Hence, because there
are 52 cards, we would expect the probability of picking the ace of spades equal to . We
could now place some constraints on the system. For instance, we could only count red cards,
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in which case the probability of picking the ace of hearts, say, would be , by the same
reasoning. In both cases, we have used the principle of equal a priori probabilities.
Example 1.12: Let us consider an example from a famous Bollywood flick “Sholay” in
which a priori probability for both head and tail on each coin toss is same based on this
principle forVeeru, while when he comes to know the secret of the coin his posteriori
probability is 1 for head and 0 for tail.
Self Assessment Question (SAQ) 6: By tossing three coins many times check the validity of
the postulate of equal a priori probability.
1.9 SUMMARY
In this unit, you have studied the basics of the statistical physics. In particular, this unit
focuses on classical statistics. You learned the concepts of phase space and phase points. This
concept also helps you to understand the density of states. In this unit, the concept of
distinguishability and indistinguishability is also discussed. We understood that the notion of
distinguishability (indistinguishability) has a profound impact on the underlying statistics.
This is because the statistics involves computation of probabilities and the rules of probability
are sensitive to distinguishability (indistinguishability) of the particles. Various other
important concepts, like probability, probability distribution, equal a priori principle, are also
discussedwith the help of numerical examples. We have briefly introduced the concept of
ensemble which will be discussed in detail in Unit 4. All the basics of statistical
thermodynamics we have discussed in this unit are very helpful in discussing and
understanding many interesting concepts of thermodynamics and statistical physics in the
forthcoming units. Many solved examples are given in the unit to make the fundamental
concepts clear. Additionally, to check your progress and understanding some self-assessment
questions are given at the end of different sections.
We will return to some of the concepts introduced in this unit, like ensembles, equal a priori
principle in Units 2-4. The rest of the units of this block (Units 2-3) are dedicated completely
to classical statistics, but many of the topics introduced here, which can be easily extended to
quantum statistics, are discussed in detail.Dedicated study of ensemble theory and different
types of ensembles will be discussed in Block 2, which contains Units 4-6. Further, Block 3
(Units 7-10) is entirely focused on quantum statistics. In the last block, which contains Units
11-12, we will discuss the application part of both classical and quantum statistics by
applying the methods of statistical mechanics to some physical situations.
1.10 GLOSSARY
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Microscopic These are defined in terms of microscopic variables, like position

properties and momentum coordinates.
Macroscopic These are defined in terms of macroscopic variables, like number
properties of particles N, volume V, temperature T, energy E.
Phase space A scheme for the specification of a system in terms of the position
coordinates and corresponding momentum ( )
components.
Phase points It represents an individual particle in the phase space.
Density of states It tells us about the distribution of phase points in phase space
Ensemble It is a collection of a large number of independent collections of
particles.
Ensemble average The value of a variable when averaged over all the states explored
within an ensemble.
Time average The value of a variable when averaged over a time period.
Ergodic hypothesis An average over time is equal to an average over all the possible
microstates.
Probability A mathematical concept that deals with calculating the likelihood
of a given event’s occurrence.
Probability A statistical distribution function that describes all the possible
distribution values and likelihoods that a random variable can take within a
given range.
1.11 REFERENCES
1. Statistical Mechanics, S. Prakash,KedarNath Ram Nath
2. Statistical Mechanics, R. K. Pathria, Butterworth-Heinemann, Oxford.
3. Fundamentals of Statistical and Thermal Physics, F. Reif, McGraw-Hill.
4. Statistical Mechanics, K. Huang, Wiley India.
5. Fundamentals of Statistical Mechanics, B. B. Laud, New Age International

1. Thermodynamics and Statistical mechanics, W.Greiner, L. Neise and H. Stocker, Springer
Science and Business Media
2. Statistical Thermodynamics and Kinetic Theory, C. E. Hecht, Freeman, New York.
3. Statistical Mechanics: an Introduction, D. H. Trevena, Elsevier.
4. Statistical Physics, L. D. Landau andE. M. Lifshitz, Elsevier.
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1.13.1 Short Answer Type
1. Explain the terms: (i) phase space and (ii) ensemble.
2. Give two examples of phase space trajectories.
3. Define phase cell.
4. What are the basic rules of probability?
5. What is ergodic hypothesis?
6. What do you understand by the density of states?
7. Write down the statement of postulates of equal a priori principle.
1.13.2 Long Answer Type

1. Discuss the probability distribution function with the help of an example.
2. Write notes on:
(i) Probability, (ii) density of states, and(iii) ensemble average and time average.
3. Discuss the phase space trajectory of a particle of mass m in a gravitational field from a
height h attime t.
4. Explain the concept of distinguishability and indistinguishability with the help of an
example.
5. Obtain an expression for the total number of phase cells per unit energy range for a single
particle in the phase space.
6. With the help of an appropriate example explain the postulate of equal a priori principle.
7. Briefly discuss the connection between ergodic hypothesis and postulates of equal a priori
principle.
1.13.3 Numerical Answer Type

1. What is the probability of drawing two queens in succession from a pack of cards?
2. Obtain the probability of drawing an ace three times in a row with replacementfrom a pack
of cards.
3.A drawer contains threered paper clips, fourgreen paper clips, and fiveblue paper clips.
After taking one paper clip another paper clip is taken from the same drawer without
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replacement. What is the probability that the first paper clip is red and the second paper clip
is blue?
4.Check the validity of the postulate of equal a priori probability by tossing two coins many
times.
1.13.4 Objective Answer Type

1. Phase space is a:
(a) momentum space (b) configuration space
(c) combinedposition and momentum space (d) four dimensional Minkowskian space
2. The statistical approach involves the concept of ………….. of distribution.
(a) sum (b) probability
(c) collection(d) particle
3. The value of the probability of an event cannot be:
(a) 1 (b) zero
(c) 1/2 (d) negative
4. What is the probability of getting a sum 9 from two throws of a dice?

(a) 1/8 (b) 1/6
(c) 1/9 (d) 1/12
5. What are the chances that no two boys are sitting together for a photograph if there are 5
girls and 2 boys?
(a) 2/7 (b) 5/7
(c) 4/7 (d) 1/21
6. Ensemble is a set of assemblies of a system, where each individual system may have a:
(a) different microstate but same macrostates (b) different macrostate but same microstate
(c) different macrostate and microstate (d) same macrostate and microstate
7. Ergodic hypothesis relates the equality of:
(a) phase space and mu space (b) ensemble average and time average of a physical quantity
(c) time average of a physical quantity and density of states (d) ensemble average and
density of states
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1.14 ANSWERS
1.14.1 Self Assessment Questions (SAQs):
1. There are 4 chocolatesin the box containing total number of 4 chocolates and 4 ice
creams,so the probability of getting 1 chocolatesis .
2. There are four card in the pack of cards of each value, with two red and two black.
P (red king) = 2/52. Now we have only 51 cardsand there are still two black 7s in the deck.
P (second card is a black seven) = 2/51.
Therefore, P (red king then black 7)= (2/52) (2/51)=1/663.
3. The probability of rolling a doublet is 1/6 on each throw. It follows that the probability of
no doublet on a given throw is 5/6. Therefore,
Probability of zero doublets = (5/6) (5/6) (5/6) = 125/216.
Probability of one doublet = (5/6) (5/6) (1/6)+(5/6) (1/6) (5/6) +(1/6) (5/6) (5/6) = 75/216.
Probability of two doublets = (5/6) (1/6) (1/6) + (1/6) (5/6) (1/6) +(1/6) (1/6) (5/6) = 15/216.
Probability of three doublets = (1/6) (1/6) (1/6) = 1/216.

We can verify that the sum of the probability distribution is (125+75+15+1)/216=216/216=1.
4.As we know that Hamiltonianrepresents the total energy of a system, i.e.,

.
In the question, for the given range , V=0. So the Hamiltonian or energy of
particle is
From this relation we get .Hence, the region of states accessible to the particle
is shown in dark color.
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Note that an arbitrary trajectory for the particle will be any random path in the range shown
in accessible phase space.
5. A free particle means that there is no potential(V=0) and is thus moving freely. Therefore,
the energy of free particle has the form . Now,we can draw phase space trajectory by
similar arguments as in the previous question. The trajectory is shown as
1.14.2 Terminal Questions: Numerical type questions

1. The probability of drawing two queens:
There are 4 queens in the pack of 52 cards, so the probability of drawing 1 queen = .
Now only 3 queens left and only 51 cards left in the pack. Therefore, the probability of
drawing second queen = .
Hence, the probability of drawing two queens in succession is
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P(two queens in succession) = P(first card queen) P(second card queen)

= .
2. P(Three aces in a row) = P(Card 1 is an ace) P(card 2 is an ace) P(card 3 is an ace).

We are replacing the cards, this means there will always be four aces in the deck of 52 total
cards. Hence,
P (Getting an ace) = 4/52.
P (Three aces in a row) = (4/52) (4/52) (4/52)=1/2197.
3. In question it is given that the first paper clip is not replaced. The sample space of the first
event is 12 paper clips, but the sample space of the second event is changed it is now 11
paper clips. Both the events are dependent. So,
P(red) = ( 3/12)
P (blue given red occurred) = (5/11)
P(red then blue) = P(red) P(blue given red occurred) = (3/12) (5/11) = ( 5/44).
1.14.3 Objective type questions

1. Correct option is (c), combined position and momentum space.
2. Correct option is (b), probability.
3. Correct option is (d), negative.
4. Correct option is (c), 1/9.
5. Correct option is (b), 5/7.
6. Correct option is (a), different microstate but same macrostates.
7. Correct option is (b), ensemble average and time average of a physical quantity.
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UNIT 2 Statistical Mechanics and Thermodynamics
Structure
2.1 INTRODUCTION
2.2 OBJECTIVES
2.3STATISTICAL EQUILIBRIUMS
2.4CONNECTION BETWEEN STATISTICAL AND THERMODYNAMIC QUANTITIES
2.5MICROSTATES AND MACROSTATES
2.5.1 MICROSTATE
2.5.2 MACROSTATE
2.6LIOUVILLE'S THEOREM
2.7 SUMMARY
2.8 GLOSSARY
2.9 REFERENCES
2.12 ANSWERS
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2.1 INTRODUCTION
In the first chapter, you have learnt the basics of statistical mechanics and its role in the
modern physics. You already understand that statistical mechanics deals with the most
probable behavior of the assembly of a large number of particles, usually comparable to
Avogadro number. The dynamics of individual constituent particles may be governed by
either classical or quantum mechanics depending upon their properties. You may have been
wondering why not we solve the equations of motion for all particles independently to obtain
the instantaneous state of the system and study the macroscopic properties forthis state.You
probably did not realize the complexity of the task at hand as you would have to solve atleast
1023 differential equations. This is not only impractical, we are actually not interested in the
states of the individual particles and macroscopic properties of the system only. Therefore,
we understand that many copies of the system instatistical ensemble are in different states at
any instant of time and thus provide the probability of occurrence in certain state. Thus, it not
only simplifies our lives to compute all the macroscopic properties as functions weighed with
corresponding probabilities in the phase space distribution, but has application to be
discussed in last few units of the book.
We have mentioned in the previous unit that thermodynamics deals with macroscopic
properties of the system at equilibrium. We have also mentioned that statistical mechanics is
also studied at this stationary state. Equilibrium refers to a state which does not change with
time.This state has maximum entropy and is thus most probablecondition for a closed
system.Note that this does not mean that the states of the constituent particles is not changing,
but their evolution is such that the system remains unchanged macroscopically. As we are
going to discuss the connection between statistical and thermodynamic quantities in this unit,
we should understand the idea of different types of statistical equilibriumsbefore that.
Specifically, we will be discussing thermal, mechanical, and chemical equilibriums.
Thermodynamic properties can be categorized as intensive and extensive properties. The
former one is the bulk property and does not depend on the size and nature of material, while
in contrast, the latter one is an additive property of the subsystems and increases with size.A
few examples of the intensive property are temperature, pressure, chemical potential, density;
and extensive property areenthalpy, entropy, internal energy, Gibbs energy. We will discuss
these quantities in detail in the present chapter.
We have discussed so far that the macroscopic properties of the system are concerned with
the state of the system represented by phase points and remains unchanged with change in the
state of the individual particles only. We will discuss the phase points and corresponding
properties using terminology of ensemble theory by introducing the idea of macrostate and
microstate.
2.2 OBJECTIVES
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● Concept of statistical equilibrium and its importance in statistical physics

● Establish a relation between statistical and thermodynamic quantities
● Microstates and macrostates
● Liouville's theorem and its applications
2.3 STATISTICAL EQUILIBRIUMS
We are familiar with thermodynamic equilibrium, since our graduation classes. In brief, a
system is said to be in thermodynamic equilibrium, if it satisfies all the three requirements of
equilibrium, i.e.,thermal, mechanical, and chemical equilibriums. A system is said to be in
thermal equilibrium when there is no temperature difference between the system and its
surroundings. For mechanical equilibrium there should be no unbalanced forces acting on any
part of the system or the system as a whole. While if there is no chemical reaction within the
system and there is no movement of any chemical constituent from one part of the system to
the other then it represents chemical equilibrium. In Unit 1,we have discussed the postulate of
equal a priori principle. This principle states that any arbitrary volume element dV in phase
space bounded by a surface and containing a definite number of phase points does not change
with time, i.e., equal volumes in phase space. The postulate of equal a priori principle is
consistent with the idea of equilibrium.
For a system of the ensemble to be in statistical equilibrium, two conditions should be
fulfilled. Firstly the probabilities of finding phase points in various regions of phase space
should be independent of time. Secondly the average values of the measurable properties of
the system in the ensemble should be independent of time.Therefore, for the system to be in
equilibrium, the phase space density should be independent of time. This equilibrium
condition is solely related to the Liouville's theorem. In the forthcoming sections, we will
prove the equilibrium condition with the help of Liouville's theorem.As we know that any
statistical ensemble is defined by the distribution function which characterizes it,the statistical
equilibrium is a state in which the probability distribution function of the states remains
constant over time. In a similar way, as in the case of thermodynamic equilibrium for an
ensemble to be in statistical equilibrium the system should fulfill the condition of thermal,
mechanical, and particle equilibrium.
2.4 CONNECTION BETWEEN STATISTICAL AND

THERMODYNAMIC QUANTITIES
Statistical mechanics is an experimental and empirical physics. Further, we are familiar with
thermodynamic equilibrium. In this equilibrium, there is no net macroscopic flow of matter
or energy, either within a system or between systems. Whereas, in statistical equilibrium, the
average values of all the physical quantities characterizing the state are independent of time.
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Statistical equilibrium plays the same role as thermodynamics equilibrium in

thermodynamics.
At equilibrium, in statistical mechanicsprobability is maximum, and in thermodynamics
entropy is maximum. We know thatin equilibrium the entropy of a system depends upon the
energy of the system (E), number of particles/molecules present in the system ( ), and the
external variables (say ), like pressure, volume, etc. Therefore, we can write
….(2.1)
As we have already mentioned that statistical mechanics and thermodynamics are closely
related with each other. This relation between probability and entropy is an important concept
in statistical physics. Here, we define statistical temperature , statistical pressure , and
statistical chemical potential to establish this relation. Specifically, the statistical
temperature
…(2.2)
is related to the thermodynamic temperature as , where k is the Boltzmann’s constant.

Similarly, statistical pressure
, …(2.3)
equals to the thermodynamic pressure ,and statistical chemical potential

…(2.4)
are other relevant parameters. Thermal equilibrium states same for all parts of the system,
while is constant for mechanical equilibrium at thermal equilibrium. Due to particle
diffusion allowed in particle equilibrium chemical potential is constant at thermal and
mechanical equilibrium.
Suppose we make a small insignificant reversible change in the equilibrium condition in such
a way that the new system is in equilibrium, too.This can be described as
…(2.5)
With the help of equations (2.2)-(2.4), we obtained

…(2.6)
where is generalized force.After rearrangement we can write
…(2.7)
If we assume that the number of particles are fixed and the volume is the only external
parameter, i.e., , then from equation (2.7) we get
…(2.8)
The above equation represents the change in the internal energydue to heat exchange and
change in external parameters. You should know about the first law of thermodynamics
which tells us that the change in the internal energy can be defined in terms of
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thermodynamic quantitiesas .Here, is the thermodynamic entropy.As far

as statistical mechanics is concerned, internal energy is the total energy (both kinetic and
potential energies) of all the constituentsin the system. Thus, for statistical mechanics we can
write .On comparing equation (2.8) with the relation in
thermodynamics , we have
…(2.9)
From this relation we can say that and are proportional to and , respectively. Thus, we
can write
…(2.10)
and we have
…(2.11)
It is imperative to mention here that for an ideal gas thepotential energy of the particles
vanishesas the non-interacting gas molecules are assumed to be point particles, i.e., the
diameter of the particles is negligible in comparison to mean free path,undergoing only
elastic collision with each other. Thus, internal energy is just the sum of kinetic energy of all
the gas molecules and only depends on the temperature not volume.
Comparing equations (2.8) and (2.11), we get
and …(2.12)
In a similar way, we can define other thermodynamic potentials which will be useful while
discussing different types of ensembles.
Helmholtz Free Energy (F):It definesthe work that can be extracted from a closed system at
constant volume and temperature. Mathematically,it is
…(2.13)
which on differentiation gives . After rearrangement we can write
…(2.14)
Independently, we can define
…(2.15)
to obtain
…(2.16)
This allows us to write, on comparing equations (2.14) and (2.16), as
and
…(2.17)
Enthalpy (H): It is the total of the internal energy of the system and the product of pressure
and volume. The internal energy of the system is the amount of energy required to bring the
system together. It can be described as
…(2.18)
with derivative
…(2.19)
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Further, if we write
and
…(2.20)
Thus, comparing equations (2.19) and (2.20), we have

and …(2.21)
Gibbs Free Energy (G): It is defined as the maximum reversible work that may be extracted
from a system at constant temperature and pressure. Mathematically,
and its derivative

…(2.22)
Also, we may write
and thus
…(2.23)
Hence, comparing equations(2.22) and (2.23) we have

and …(2.24)
Therefore, we have established a relation between the statistical quantities, like entropy ( ),
temperature ( ), pressure ( ), and thermodynamic quantities, like Helmholtz free energy (F),
enthalpy (H), and Gibbs free energy (G).
Self Assessment Question (SAQ) 1:Obtain the change in internal energy as in equation
(2.11) for an ideal gas.
Self Assessment Question(SAQ) 2: With the knowledge of thermodynamics discuss the
physical significance of Helmholtz free energy and Gibbs free energy.
2.5 MICROSTATES AND MACROSTATES
The aim of statistical mechanics is to serve as a bridge between the microstate and
macrostate. Let us understand these two states in detail.
2.5.1 Microstate
We have discussed in the last unit that6Ndimensional Γ space defines the position and
momentum of allN particles. Specifically, the state of such a system can be specified by each
particle's position qi and momentum pi.
The coordinates of this 6N dimensional phase space describing the state of the constituent
particles of the system are known as microstate.
2.5.2 Macrostate
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Suppose we are interested in the macroscopic variable, like energy Eofthe system,then it will
be functions of the microstate, i.e.,
. ...(2.25)
At equilibrium, the valueof E is very close to the average or mean value of E. This mean
value of the macroscopic quantities specifies the state called macrostate of the system. Thus,
the macroscopic state of the system is characterized by the macroscopic properties,like
pressure, volume, temperature. More precisely, a microscopic state is simply a point in the
phase space while a volume in the phase space represents a macroscopic state as it
corresponds to a large number of microstates.
Example 2.1: Consider 1 mol. of He gas at pressure (P) = 1 atm and temperature (T) =
298K. The given pressure and temperature defines one macrostate. Any change in one of
the macroscopic properties (say pressure or temperature) corresponds to a different
macrostate. For each macrostate, the configuration of each gas particle is the possible
microstates.
Here, one should note that a number of different microstates can have the same macrostate.
Example 2.2: Let us distribute three particles a, b, and c in two compartments. The
distribution is shown in Table 2.1.
Macrostate Microstate
Description of states Number of
Compartment 1 Compartment 2 states
(3, 0) abc 0 1
(2, 1) ab c 3
bc a
ca b
(1, 2) c ab 3
a bc
b ac
(0, 3) 0 abc 1
Table 2.1:Different possible macrostates and microstatesfor three particles distributed in two
compartments.
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From the above table, we can conclude that the system has four macrostates and eight
microstates. The macrostates (3,0) and (0,3) has 1 each, and macrostates (2,1) and (1,2) has 3
each microstates, respectively.
Example 2.3: Let us discuss the case of tossing three coins. If we toss them several times,
then each of the eight possible outcomes are microstates. One can see in Table 2.2 that the
macrostates with only two heads or tails also have three possible microstates. Note that it
confirms that there may be more than one microstates corresponding to the same macrostate.
Macrostates Microstates
First coin Second Third

coin
coin
All heads H H H
Two heads H H T
H T H
T H H
Two tails H T T
T H T
T T H
All tails T T T
Table 2.2:All possible macrostates and microstates for the tossing of three coins.
Example 2.4: Suppose there are four particles, say A, B, C, and D, and there are two cells in
which these particles are to be distributed. Let us discuss the macrostates and corresponding
microstates.
Here, we have four different particles A, B, C, and D, and there are only two cells, namely
Cell 1 and Cell 2. In Table 2.3 given below the possible arrangementsare summarized.
Description of states Number of

states
Cell 1 Cell 2
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(4, 0) ABCD 0 1
(0, 4) 0 ABCD 1
(1, 3) A BCD 4
B ACD
C ABD
D ABC
(3, 1) BCD A 4
ACD B
ABD C
ABC D
(2, 2) AB CD 6
AC BD
AD BC
BC AD
BD AC
CD AB
Table 2.3:All the macrostates and microstates of four different particles in two cells.
Therefore, we can conclude that a state of the system which is defined in terms of
microscopic variables are called microstates. Here, you should understand that microstates
are the states of the constituent particles which gives the complete specification of the
system. Whereas the states defined in terms of the macroscopic variables, like E, V, T are
called macrostates. We also conclude one of the important facts here, i.e.,there are many
possible microstates corresponding to a given macrostate.
In statistical mechanics, we also have constraints on a system. These constraints are a set of
conditions or say restrictions that must be followed by a system of interest. If we have a
system of total number of particlesNwith total energy E, then we can represent the constraints
in the form of equations known as equations of constraints as and
. Here, is the number of particles in the ith cell, and is the energy
of each particle in the ith cell. On the basis of the equations of constraints, we can categorize
all the macrostates in two types, one is accessible macrostate and the other one is inaccessible
macrostate. Let we discuss an examplebelow which clarifies these two kinds of states.
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Example 2.5: Suppose there are two particle J and K, and we wish to distribute them into two
compartments say Compartment 1 and Compartment 2. The possible arrangements are shown
in Table2.4 given below.
Description of states Number
Compartment 1 Compartment 2 of states
(2, 0) JK 0 1
(1, 1) J K 2
K J
(0, 2) 0 JK 1
Table 2.4: All the macrostates and microstates for two particles distributed in two
compartments.
If we put the constraints that no compartment should remain empty, then the macrostates (2,
0) and (0, 2) are not allowed. Only one macrostate (1, 1) will be there,which results in the
decrement of the number of macrostates and microstates. Hence, we conclude that the
macrostate which is allowed under the constraint is accessible macrostate, whereas the
macrostates which are not allowed under the constraint are inaccessible macrostates. For the
above mentioned constraints, i.e., no compartment should be empty, the macrostate (1, 1) is
accessible macrostate while (2, 0) and (0, 2) are inaccessible macrostates.
Self AssessmentQuestion (SAQ) 3: Discuss the microstates and macrostates for the
distribution of two classical (distinguishable) particles in two compartments.
SelfAssessment Question (SAQ) 4: Consider two coins and toss them many times. Discuss
the various possible macrostates and the corresponding microstates.
2.6 LIOUVILLE'S THEOREM

We have studied in the previous unit that each phase point in the phase space represents the
state of a system. The representative points in the phase space move along a path known as
phase space trajectory. This trajectory is determined from Hamilton’s equation of motion. As
a result of the motion of the phase points the density of the system in phase space, i.e., phase
space density changes with time. Liouville’s theorem is helpful in finding the rate of change
of phase density at a given point in phase space.
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Let us consider an arbitrary volume element in the hyperspace located between and
, and (see Figure 2.1). The number of phase points or systems in this
volume element are given by
…(2.26)
Here, is the phase space density, and is the volume element .
The number of phase points entering the volume element through any face will be different
from the number of phase points which are leaving the opposite face. This change in the
number of phase points per unit time is given by
…(2.27)
Further, consider two faces perpendicular to the axis, located at and .The
number of phase points entering the element in time through the face are thus
, where is velocity in the direction .
Notice that changes in position as well as momentum coordinates.Thus, at the opposite face
changes to , changes to , and changes to .Thus,
the number of phase points leaving the element through face are
, where higher-order terms are neglected.
Figure2.1:Arbitrary volume element in the hyperspace.
Hence, the net number of systems entering the volume element in the direction , i.e.,the
difference between the number of phase points entering face and leaving opposite
face ,is .Similarly, the number of phase points
entering the volume element in the direction , i.e., the difference between the number of
phase points entering face and leaving opposite face ,
is .
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Therefore, the net number of phase points entering the volume element through all its faces
comes out as
…(2.28)
which helps us conclude
. …(2.29)
This equation can also be rearranged as
.
…(2.30)
We can further use the Hamilton’s equation of motion here, i.e., the quantities
,
…(2.31)
and their derivatives
.
…(2.32)
Therefore, the first term in the bracket of equation (2.30) vanishes, and we obtain
. …(2.33)
Here, gives the rate of change in phase density at an instant of time. Equation (2.33) is
known as Liouville’s equation.
Further, we can express equation (2.33) in the form
,
…(2.34)
or
.
…(2.35)
Here, is Poisson bracket operationdefined for the
functions and .Therefore,the total time derivative of can be defined as
. …(2.36)
This equation tells us that the phase space density is an integral of motion, i.e., the total time
derivative of the phase space density vanishes along a phase space trajectory. This is
Liouville’s theorem in classical statistics.
According to Gibbs is called the principle of conservation of density in phase space,
i.e.,the density of a group of points remains constant along their phase trajectories.Further,
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considering equation (2.33) at statistical equilibrium, i.e., as is not explicitly

dependent on time,it reduces to
which will be satisfied only if . Thus, phase points are uniformly

distributed in the phase space at statistical equilibrium.
As we discussed about the statistical equilibrium in Section 2.3,herewe can conclude that is
a function of some property of the ensemble which is independent of time. This means that
the system of ensemble will be in a statistical equilibrium.
Self Assessment Question (SAQ) 5: We have discussed the Liouville’s theorem which
represents the conservation of density in the phase space by considering the time derivative of
phase density for a volume element in equation (2.27).Similarly, illustrate the conservation of
extension of phase space using the time derivative of volume element in equation (2.27).
The Liouville’s theoremplays a significant role in statistical mechanics.The theorem is valid

for both equilibrium and non-equilibrium systems. Consequently, it is of fundamental
importance for the Gibbs formulation of statistical mechanics.In other words, this theorem
states that the phase space density remains constant in phase space which suggests that the
phase space density gives the probability of finding a system at an instant of time in the phase
space.
2.7 SUMMARY
We all are too familiar with equilibrium or say thermal equilibrium, since our graduation
classes. The state of equilibrium refers to a state of a system which does not change with
time. This unit is solely based on the concept and condition of statistical equilibrium. We
have learned about the various conditions required for a system to be in the statistical
equilibrium. We also came to know about the role of entropy and probability in
thermodynamics and statistical mechanics, respectively. Further, we defined various
thermodynamic potentials and establish a relation between the statistical and thermodynamic
quantities. Here, in the present unit, we have discussed one of the most important concepts in
statistical physics, i.e., the concept of macrostates and microstates with the help of various
examples.We can also understand the postulate of equal a priori principle in terms of
microstates and macrostates, which will be discussed in detail in the next unit.Specifically,
microstates are the representative points in the phase space, and we have already discussed in
Unit 1 that these representative points are known as the phase points in the phase space.
Finally, in this unit we came to know about Liouville's theorem which suggests an equal
distribution of phase points in the given phase space at statistical equilibrium.This theorem
also tells us that if we move with a phase space point as it evolves under the Hamiltonian, the
phase space density of the neighbourhood is constant.
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2.8 GLOSSARY
Thermal equilibrium When the temperature of two or more than two systems is same
then this condition is known as thermal equilibrium.
Mechanical equilibrium Two systems are said to be in mechanical equilibrium if their

pressures are same.
Chemical equilibrium It is the state in which the chemical activities or concentrations

of the reactants and products have no net change over time
and also there is no movement of any chemical constituents
from one part of the system to the other.
Thermodynamic A system is said to be in thermodynamic equilibrium with

equilibrium another system when it achieves all three equilibriums, i.e.,
thermal, mechanical, and chemical equilibriums.
Statistical equilibrium A system is in statistical equilibrium when the probabilities of

finding the phase points in various regions of the phase space
and the average values for the measurable properties of the
system in the ensemble are independent of time.
Macrostates States which can be represented in terms of macroscopic

variable, like pressure P, volume V, number of particles N,
temperature T.
Microstates States which can be represented in terms of microscopic

variables and phase points in the phase space.
Liouville’s theorem It states that the density of particles or phase space density in
phase space is constant.
2.9 REFERENCES
1. Statistical Mechanics, S. Prakash, Kedar Nath Ram Nath
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5. Fundamentals of Statistical Mechanics, B. B. Laud, New Age International
1. Thermodynamics and Statistical mechanics, W.Greiner, L. Neise and H. Stocker, Springer

Science and Business Media
2. Statistical Thermodynamics and Kinetic Theory, C. E. Hecht, Freeman, New York.

1. Explain the term statistical equilibrium.
2. Write down the condition of statistical equilibrium.
3. Define the terms (i) thermodynamic equilibrium and (ii) macrostate and microstate.
4. State the principle of equal a priori principle in terms of microstate and macrostate.
5. Write down the statement of Liouville's theorem.
1. Differentiate between thermodynamic equilibrium and statistical equilibrium.

2. Discuss the concept of macrostates and microstates in detail. Using an example of two
dices rolled simultaneously, write the number of microstates and macrostates.
3. State and prove Liouville's theorem. Also mention the consequences of Liouville's
theorem.
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4. Consider three particles which are distributed among three energy levels. Discuss the
possible microstates and macrostates of the system.
2.11.3Numerical Answer Type

1. How many macrostates and microstates are there when a single coin is tossed three times?
2. There are two distinguishable particles to be arranged in three cells. Show the possible
arrangements, and discuss the macrostates and microstates in a tabular form.
3. A system has a single particle state with 0, 1, 2, and 3 energy units. Discuss the
macrostates and microstates for three distinguishable particles to be distributed in these
energy states such that the total energy of the system is 3 units.
4. Discuss the macrostates and microstates for tossing of four coins simultaneously.

1. In general the possible microstates for a given macrostate are:
(a) only 1 (b) Zero
(c) as many as possible (d) cannot say
2. The relation between the statistical entropy and thermodynamic entropy is given by
(a) (b)
(c) (d)
3. Among the following the correct expression for the entropy is
(a) (b)
(c) (d)
4. In Liouville's theorem, which of the following physical quantity remains constant along the
phase space trajectory:
(a) momentum (b) phase density
(c) energy (d) ensemble
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5. At statistical equilibrium,Liouville's theorem tells us phase points are ……… distributed in

phase space
(a) randomly (b) uniformly
(c) not uniformly (d) neither randomly nor uniformly
2.12 ANSWERS
2.12.1 Self-Assessment questions:
1. As we have already mentioned that for an ideal gas , so we can obtain
2. If we check or calculate the change in Helmholtz free energy, i.e., ,and change in Gibbs
free energy , then it provides us useful information about the system and the process
underway. For example, if changes in these quantities vanish,i.e., (at constant
temperature and volume) and (at constant temperature and pressure), then it
represents that the systemis at chemical equilibrium.The minimum value of these free
energies represent that the system is in stable condition. The change also tells us about the
direction and nature of the process taking place. If at constant temperature and
volume, the process is non-spontaneous, and corresponds to the
spontaneousprocess. Similarly, for the process is non-spontaneous (endergonic),
while is signature of the spontaneous (exergonic)process at constant temperature and
pressure. All these facts are summarized in the following table.
State Mathematical Differential equation Condition Stable

function form for change equilibrium
in free measure
energy /Type of
process
Helmholtz System is in
free energy equilibrium
(F)
Non-
spontaneous
Spontaneous
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Gibbs free System is in

energy (G) equilibrium
Non-
spontaneous
(endergonic)
Spontaneous
(exergonic)
3.Different microstates and macrostates for the distribution of two classical (distinguishable)
particles in two compartments.
Compartment 1 Compartment 2 of states
(2, 0) ab 0 1
(1, 1) a b 2
b a
(0, 2) 0 ab 1
4.Possible macrostates and the corresponding microstates for tossing of two coins.
Coin 1 Coin 2 of states
Both heads H H 1
Either head or tail H T 2
T H
Both tail T T 1
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5. Let us consider an elementary volume in the phase space, and the phase space
density can be assumed uniform throughout the phase space for this very small region. Thus,
from equation (2.27) we will have
We know in a given phase space each phase point represents a definite system, which can
neither be created nor destroyed. Therefore, , i.e., the number of phase point
remains fixed. So,
With the help of equation (2.36), the above equation reduces to
or .
This shows the conservation of extension in phase space.
2.12.2 Numerical answer type:
1. The macrostates are (three heads, two heads, one head, no head), and the corresponding
microstates are (three heads: HHH; two heads: HHT, HTH, THH; one head: HTT, THT,
TTH; no head: TTT).
2. Let the particles be R and P
Macrostates Microstates Total number

of microstates
Cell 1 Cell 2 Cell 3
Both in same RP - - 3
cell
- RP -
- - RP
Both in R P - 6
different cells
P R -
- R P
- P R
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R - P
P - R
3. Consider three particles as J, K, and L, and the energy unit is E.
Macrostates 0E 1E 2E 3E Total Number of

energy microstates
(2, 0,0, 1) JK 0 0 L 3E 3
KL 0 0 J
LJ 0 0 K
(1, 1, 1, 0) J K L 0 3E 6
J L K 0
K L J 0
K J L 0
L J K 0
L K J 0
(0, 3, 0, 0) 0 JKL 0 0 3E 1
4. There are five macrostates as (all heads, only one head, two heads, only one tail, and all
tails). The possible microstates are (HHHH, TTHT, TTTH, THTT, HTTT, HHTT, HTTH,
THHT, THTH, TTHH, HHHT, HHTH, HTHT, HTHH, THHH, TTTT).
2.12.3 Objective answer type:
1. Correct option is (c), as many as possible.
2. Correct option is (b), .
3. Correct option is (a), .
4. Correct option is (b), uniformly.
5. Correct option is (d), statistical equilibrium.
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UNIT 3 Maxwell-Boltzmann Statistics
Structure
3.1 INTRODUCTION
3.2 OBJECTIVES
3.3THERMODYNAMICAL PROBABILITY
3.3.1 MOST PROBABLE STATE
3.4CLASSICAL MAXWELL-BOLTZMANN DISTRIBUTION LAW
3.4.1 EVALUATION OF CONSTANTS AND
3.4.2 MAXWELL DISTRIBUTION LAW OF VELOCITIES
3.4.3 MAXWELL SPEED DISTRIBUTION LAW
3.5PRINCIPLE OF EQUIPARTITION OF ENERGY
3.6 CLASSICAL STATISTICS IN BRIEF
3.7 SUMMARY
3.8 GLOSSARY
3.9 REFERENCES
3.12 ANSWERS
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3.1 INTRODUCTION
We have already discussed and understand so farthe basic concepts of statistical mechanics
and probability theory. We have further elucidated the relevance of statistical equilibrium in
this discussion and stressed on the connection between statistical and thermodynamic
quantities.The principle of equal a priori probability and the ergodic hypothesis enable us to
compute statistical quantities and understand intensive and extensive thermodynamic
properties. We have also discussed that the density of a group of points remains constant
along their phase trajectorieswhich is useful in the Gibbs formulation of statistical mechanics.
Our basic tools for both classical and quantum statistics are incomplete without the
introduction of the idea of macrostates and microstates. Each phase point in the phase space
corresponds to a microstate while many such microscopic statesmay have same macroscopic
properties and thus macrostate. In the present unit, we will stress on the significance of the
number of microstates in each macrostate which is concerned with thermodynamic
probability. We will revisit the definition of the most probable state in view of the
thermodynamic probability.
We have all the tools required to understand the classical statistics governed by Maxwell-
Boltzmann distribution law. We also discuss the nature of particles this type of statistics may
be used.We further study Maxwell-Boltzmann distributions of velocity and speedin brief.
This gives us significant types of speeds namely, most probable speed, mean speed and root
mean square speed. These unique types of speeds have a prominent role in statistical
mechanics. Therefore, we will not only obtain their mathematical expressions, we will also
discuss their physical relevance with the help of corresponding examples. We conclude this
unit, as well as this block, focused on the classical statistics with the principle of equipartition
of energy over all the degrees of freedoms.
Here, it is worth stressing thatthe significance of the discussion in the present unit (and this
block) is not restricted to classical particles, but many of these basic ideas areeither directly
useful or easily extendible to quantum statistics to be discussed in the forthcoming blocks.
One more significant example of classical statistics deservesattention here could be
interaction between a quantum particle and its environment, which causes decoherence or
quantum to classical transition of quantum behavior. To perform this study the environment
is modeled as a classical system governed by Maxwell-Boltzmann statistics. There are
various physical processes relevant in different branches of physics where the topics
discussed in the present unit will be useful. For instance, obtaining Einstein coefficients and,
in turn, the idea of laser. To begin with, we start our discussion with thermodynamic
probability of macrostates.
3.2 OBJECTIVES
• Thermodynamic probability
• Most probable state
• Classical distribution law known as Maxwell-Boltzmann distribution law
• Maxwell law of distribution of velocities
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• Maxwell speed distribution law

• Principle of equipartition of energy
• Fundamental postulates of statistical mechanics
3.3 THERMODYNAMICAL PROBABILITY

In the previous unit, we have discussed the microstates and macrostates in detail. The number
of microstates corresponding to a given macrostate is called its thermodynamic probability.
Hence, the probability of a macrostate is defined as the ratio of the number of microstates in
it to the total number of possible microstates of the system.
. …(3.1)
For a given macrostate, the number of microstates is equal to the number of

possiblearrangements of various particles. Consider there are n particles to be filled in two
compartments. Let r particles are kept in Compartment 1 and the remaining (n-r) are in
Compartment 2, thenthe total number of arrangements is given by
… (3.2)
Here, the notation means we have totaln particles, and out of these we choose rparticles
to be filledin one compartment and the remaining in the second compartment.
Thus,thermodynamic probabilitygives the number of microstates in a macrostate (r, n-r).
Further, if there are c compartments, then the total number of microstates of the system is .
Therefore, the probability of macrostate (r, n-r)can be computed as
. …(3.3)
With the help of following examples we can understand the concept of thermodynamic
probability and probability.
Example 3.1: Let us distribute three distinguishable particles a, b, and c in two compartments
as shown in Table 3.1. The system has four macrostates and eight microstates.
MACROSTATE MICROSTATES
Description of state Number of states
Compartment 1 Compartment 2
(3, 0) abc 0 1
(2, 1) ab c 3
bc a
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ca b
(1, 2) c ab 3
a bc
b ac
(0, 3) 0 abc 1
Table 3.1: Distribution of three distinguishable particles a, b, and c in two compartments.
Thermodynamic probability of macrostate (3, 0) is 1 as there is only one possible microstate

corresponding to macrostate (3, 0). Similarly, thermodynamic probability of macrostates (2,
1), (1, 2), and (0, 3) are 3, 3, and 1, respectively.
Example 3.2: Calculate the thermodynamic probability for the case of three coins tossed
several times.
In this case, there are eight possible microstates. See Table3.2 below summarizing all these
possible outcomes.
First coin Second Third Number

coin of states
coin
All heads H H H 1
Two heads H H T
H T H 3
T H H
Two tails H T T
T H T 3
T T H
All tails T T T 1
Table 3.2:All possible outcomes of three coin tosses summarizing all microstates and
macrostates.
Now we can discuss the thermodynamic probability of all the macrostates.Thermodynamic

probability of both macrostates “All heads” and “All tails” is 1 as there is only one microstate
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corresponding to these macrostates.Similarly, thermodynamic probability of macrostate “Two

heads” is 3.You can verify that the thermodynamic probability of macrostate “Two tails” is 3.
In Unit 1,we have discussed the postulate of equal a priori principle and stated that it is
connected with the concept of microstates and macrostates.Further, in unit 2 we have already
discussed this principle in brief. Here, we can verify the validity of this principleusing the
following example.
Example 3.3: Let us consider three distinguishable particles, say a, b, and c, in two
compartments (see Example 3.1).
With the help of the expression for thermodynamic probability and probability, we can
calculate probabilities of different microstates as
Probability of microstate (a b c, 0) corresponding to macrostate (3,0)= .
Probability of microstate (a b, c) corresponding to macrostate (2, 1) = .
Probability of microstate (a, bc) corresponding to macrostate (1, 2) = .
Probability of microstate (0, a b c) corresponding to macrostate (0, 3) = .

Thus, we can conclude that the probability of all microstates of the system corresponding to
the possible macrostates isequal.
Self Assessment Question (SAQ) 1: Check the validity of the postulate of equal a priori
principle by considering the system of four particles a, b, c, and d in two compartments for
any two macrostates.
3.3.1 Most probable state: Equilibrium state

In statistical physics, we deal with systems with a very large number of particles and assume
that the system exists in an equilibrium state. All the physical properties of such a system can
be deduced by knowing the most probable or equilibrium state of the system. Thus, the
macrostate having maximum probabilityis termed as the most probable state.
Suppose there is a system of n distinguishable particles distributed in two compartments. Let
r particles are kept in Compartment 1, and the remaining (n-r) are in Compartment 2, then the
total number of arrangements = . Therefore, the probability of macrostate(r,n-r) is
represented as
.
…(3.4)
The probability will be maximum when . The maximum value of probability is then
given by or
…(3.5)
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We will elaborate this idea further using an example.
Example 3.4: A coin is tossed six times. What is the probability of getting (i) all heads, (ii)
three heads and three tails, (iii) four heads and two tails?Also calculate the probability of the
most probable state.
We have a coin which is tossed six times. Here, we will use equation (3.4). For there are all
heads, then . Therefore,
(i) For there are three heads and three tails, then .
Therefore,
(ii) For there are four heads and two tails, then .
Therefore,
We have already mentioned that for most probable state , and thus the result of (ii)
corresponds to the most probable state.
3.4 CLASSICAL MAXWELL-BOLTZMANN DISTRIBUTION

LAW
The distribution of classical particles is given by Maxwell-Boltzmann distribution law. In

general, we have three types of particles. The first kind of particles is identical but
distinguishable (like molecules of a gas) and the other two categories have identical but
indistinguishable particles with integer or half integer spin (shall be discussed in Block 3
under quantum statistics).
The Maxwell distribution law, we are going to discuss or derive here, tells us about the
distribution of energy among the assembly of identical particles.Let us consider a system of a
large number of particles or gas molecules (say N)with of them in different
compartments(or different energies), namely 1, 2, 3,…i, respectively.First, we will write the
basic assumptions of classical statistics as
(a) The particles or classical particles are identical and distinguishable.
(b) Each energy level or state can be occupied by any number of particles.
(c) The total number of particles in the system is constant the total number of particles
...(3.6)
or
...(3.7)
(d) The total energy of all the particles in the system is also constant
...(3.8)
or
...(3.9)
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where is the energy of the ith particles.
All these N particles are continuously moving in different directions and thus the values of
skeep changing,maintaining the values which define the state of maximum probability or
most probable state. We know that in this equilibrium state the thermodynamic
probability is maximum, which can be written in the condition
...(3.10)
where the thermodynamic probability may be written in this case as
. ...(3.11)
Taking the logarithm of both the sides, we get
. ...(3.12)
Note that the number of molecules is large which allows us to use Striling’s approximation
(i.e., ) to obtain
…(3.13)
Differentiating this equation with respect to (keep in mind that the total number of
particles N is constant) we obtain
...(3.14)
or
...(3.15)
Equations (3.7), (3.9), and (3.15) are independent of each other and they must be satisfied at
any instant. Thus, we apply the Lagrange method of undetermined multiplier, i.e., multiply
equation(3.7) by ,equation(3.9) by , and adding the resultant equations to equation(3.15),
we get
…(3.16)
where and are arbitrary constants independent of s.Similarly, all s are independent
of each other. Thus, equation (3.16) will be satisfied only when each term in the summation
is separately equal to zero, i.e.,
After rearrangement and simplification,we can rewrite this equation as

...(3.17)
This expression is known as Maxwell-Boltzmann distribution law of particles in classical
statistics.Here, the values of two constants may be further evaluated in
thenextsubsection.
This distribution law has various important applications, such as in discussing the probability
that a given assembly of molecules in equilibrium, and in understanding the equipartition
principle in detail. With the help of above distribution law, we will discuss the some
important distribution laws in statistical mechanics, i.e., the law of distribution of velocities
and speedsof molecules.
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3.4.1 Evaluation of constants and in Maxwell-Boltzmann

distribution law
For evaluating the two constants in Maxwell-Boltzmann distribution law, we assume
continuous distribution of energy and can represent equation (3.17) in the following form
…(3.18)
where and is the number of molecules with energy between and .
Now consider a system of monoatomic ideal gas of mass m defined in three Cartesian (x, y, z)
coordinates and three momentum coordinates . Then the number of molecules in
volume element with energy in the range and are
…(3.19)
The kinetic energy of these particles with velocity components is given by
Thus, we can write equation (3.19) as
…(3.20)
The total number of molecules can be obtained by integrating it over all the possible values of
position and momentum coordinates as
…(3.21)
Thus, we can write
…(3.22)
The integration over coordinate space gives us the total volume ,therefore,
…(3.23)
Thus, equation (3.23) becomes
With the help of a standard result
we get
or
…(3.24)
We know the total energy E of N molecules is
…(3.25)
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Using equations (3.19)-(3.20) in equation (3.25), we get
or
…(3.26)
As we can write this integration in the following form
…(3.27)
which allows us to use the standard result
to obtain equation (3.27) as
or
…(3.28)
With the help of equation (3.24), the expression of energy comes out as
…(3.29)
For an ideal gas we have
…(3.30)
So on comparing equation (3.29) and equation (3.30), we have
…(3.31)
Now, from equation (3.24), we get
or
or
…(3.32)
Thus, we can obtain the coefficient as

. …(3.33)
Thus, on substituting the values of constants the distribution law can be written as
. …(3.34)
3.4.1 Maxwell distribution law of velocities
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At equilibrium, the Maxwell-Boltzmann distribution law is given by equation (3.17). If we

consider the position and momentum in the range , ,
, , , , respectively. Thus, the number
of molecules in this phase space volume element is proportional to the corresponding volume
.
Therefore, the number of molecules of mass m having energy in this volume element
using equation (3.34) is given by
…(3.35)
In terms of position-velocity coordinates in the range , ,

, , , , equation (3.35) can be written
using as
...(3.36)
If we further integrate this equation with respect to the position coordinates, we will get the
number of molecules having velocity coordinates in the range , ,
and as
...(3.37)
Therefore, we can obtain the number of molecules having velocity components in the range
by integrating this equation with respect to , which becomes
thatcan be simplified as
…(3.38)
This equation gives a symmetric distribution of the number of molecules having x-component
of velocity in the range . Therefore, the probability that a molecule will have
x-component of velocity in the range can be represented as
...(3.39)
These two equations, i.e.,equations(3.38) and (3.39), representthe Maxwell distribution law
of velocities. In a similar way, we can obtain the expressions for the number of molecules
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having x and y component of velocity in the range , and ,

respectively.
It is clear from equation(3.39) that the function is symmetrically

distributed about , i.e., . This can be used to calculate the average
value of using Maxwellvelocity distribution to be zero. Further, the maximum of the
probability distribution function can be calculated using the first derivative with respect to ,
i.e., or . Thus, the maximum value of is
...(3.40)
It is clear from this equation that increases as massm increases whileT decreases.
Example 5: Calculate the velocity component for which the probability falls to times the
maximum value.
The probability distribution function is given by
The condition gives the value of maximum probability, i.e.,
We are given that
Therefore,
or
Taking logeof both the sides of this equation

.
Using the property , we obtained
Hence, this is the value of for which the probability falls to times the maximum value.
Self Assessment Question (SAQ) 2: Calculate the value of velocity component , for which
the probability falls to times the maximum value.
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3.4.3 Maxwell speed distribution law

Let us obtain the expression of the speed distribution law from the velocity distribution law
given by equation (3.37). This equation represents the Maxwell distribution law for the
number of molecules with velocity components in the interval , ,
and .For the expression of speed distribution, we change the velocity
components in terms of the speed of a molecule v and changing it over to polar coordinates,
i.e.,
and .
Using these values in equation(3.37), we get
. …(3.41)
This is the equation for the number of molecules with speeds between v and v+dv in a
direction lying within the angular ranges and ,and and . We can obtain the
number of molecules with velocity between v and v+dv (say )by integrating over the
angular part as
…(3.42)
or
. …(3.43)
The function is called the Maxwell speed distribution function, and is called the
probability distribution functiongiven by
. …(3.44)
Both the velocity and speed distributions have been experimentally verified, too.
Most probable speed ( ): It is the speed at which distribution has maximum value.
The condition for to be maximum is or This
can be simplified to , which gives or
. …(3.45)
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This is the most probable speed of molecules.
Mean or average speed ( or ):It is the mean of all the speeds at which the molecules of
the gas are moving.Thus, the mean speed is calculated
as or
which can be simplified as
. …(3.46)
Root mean square speed ( ):It is the average speed squared of the molecules of the gas.
It can be defined as which can be simplified as
or Thus, we can
compute
. …(3.47)
It is clear from equations (3.45)-(3.47) that (see Figure3.1).
Figure 3.1:Speed distribution of gas moleculesdepicting most probable , average ,

and root mean square speeds.
The variation of the speed distribution with all three types of molecular speeds, i.e., most
probable, average, and root mean square speeds,is depicted in Figure3.1. The graph shows
how the speeds of molecules are distributed for a gas. The vertical axis of the distribution
graph gives the number of molecules per unit speed, and the total area under the entire curve
is equal to the total number of the molecules in the given gas. Here, we can also see that the
most probable speed is the speed that is most likely to be found for a molecule in a gas.
Example 3.6: Calculate the most probable speed of hydrogen molecule at 270C.
Use , mass of hydrogen molecule .
We have the expression of most probable speed given by equation (3.45)
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Using mass of hydrogen molecule and

temperature . From the expression of most probable speed
Example 3.7: Calculate the root mean square value of speed of nitrogen molecules (mass of
the nitrogen molecule ) at 270C.
With the help of equation (3.47). For given temperature , we can
compute
Example 3.8: At what temperature the mean speed of hydrogen molecules will be same as
that of oxygen molecules at 350C. (Given, the molecular mass of hydrogen molecule is 2 and
of oxygen molecule is 32.)
We have the expression of mean speed given by equation (3.46).
For hydrogen molecules,
While for oxygen molecules,
Given, , , .
According to the question
, i.e., , and thus
Therefore, at the mean speed of hydrogen molecules will be same as that of

0
oxygen molecules at 35 C.
Self Assessment Question (SAQ) 3:Calculate the most probable, average, and root mean
speed of oxygen molecules at 270C. (Given, molecular mass of
hydrogen molecule is 2 and of oxygen molecule is 32.)
Self Assessment Question (SAQ) 4:Calculate the temperature at which the most probable
speed of the molecules of hydrogen gas will be double the most probable speed of oxygen
molecules at 300 K. (Given, , molecular mass of hydrogen molecule
, and molecular mass of oxygen molecule .)
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3.5 PRINCIPLE OF EQUIPARTITION OF ENERGY

This principle states that the total kinetic energy of a system in thermal equilibrium at
temperature T is equally distributed amongst the various degrees of freedom, and the average
kinetic energy associated with each degree of freedom is , where k is Boltzmann’s
constant.
With the help of the following discussion, we can understand this principle in more detail.
Consider one mole of a monoatomic gas in thermal equilibrium at temperature Tand each gas
molecule has three degrees of freedom due to its translational motion, i.e.,
...(3.48)
where is the mean square velocity of the gas molecules, and m is the mass of the molecule.
If the components of the mean square velocity of the gas molecules along the three
axesare , , and , respectively. Then the average energy of a gas molecule will be
...(3.49)
From above two equations, i.e., (3.48) and (3.49), we have
...(3.50)
However, it is clear that the molecular motion is random in nature, and there is no preferred
direction of motion. Therefore, the average kinetic energy corresponding to each degree of
freedom is same. Thus, we can write
...(3.51)
Hence, from equation(3.50) we can express
...(3.52)
This suggests that the mean kinetic energy per molecule per degree of freedom is .
Further, if we generalize this discussion for a system of N molecules of monoatomic gas then
there are 3N number of degrees of freedom, and therefore the total kinetic energy is
.For example,in case of10 Helium molecules,there will be 30 degrees of freedom.
Example 3.9: Discuss the principle of equipartition of energy for diatomic gas molecules.
The diatomic gases (like nitrogen, hydrogen, oxygen) have five degrees of freedom (three
translational and two rotational). Therefore, the total kinetic energy
is
Example 3.10: Discuss the principle of equipartition of energy for linear triatomic and non-
linear triatomic gas molecules.
The linear triatomic molecule like carbon dioxide has five degrees of freedom (three
translational and two rotational). Hence, the total kinetic energy ofa linear triatomic molecule
is .However, in case of a non-linear triatomic molecule, like water
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molecule, there are six degrees of freedom (three translational and three rotational). Hence,
the total kinetic energyof a linear triatomic molecule is .
Example 3.11: The average kinetic energy of a monoatomic gas molecule at a certain
temperature is . Calculate the temperature giventhat .
Solution: We know the average kinetic energy of a monoatomic gasis .
So, ,
Self Assessment Question (SAQ) 5: What is the total kinetic energy for a hydrogen chloride
molecule?
3.6 Classical statistics in brief

We have studiedthe first three units focused onclassical statistical mechanics. Many of the
concepts are discussedwhich are basic building blocks of classical statistical physics. Here,
we are in the position of summarizing all these important discussions as follows.
(i) Any gas under consideration is a collection of a large number of molecules in motion.
(ii) All the phase cells in a phase space are equal in size.
(iii)All the accessible microstates corresponding to all possible macrostates are equally
probable.
(iv) The equilibrium state of a gas corresponds to the macrostate of maximum probability.
(v) The total number of molecules is constant. This is in agreement with the conservation
of matter.
(vi) The total energy of the system is constant. This is in agreement with the conservation
of energy.
The conclusion of classical statistics mentioned abovewill be further helpful in constructing
the structure of quantum statistical mechanics and its applications.
3.7 SUMMARY
This unit is solely based on the concept of thermodynamic probability. In this unit, we have
learned about the concepts of thermodynamic probability and also calculated it for different
cases. Further, with the help of this concept we discussed the distribution of molecules in the
classical scenario as Maxwell-Boltzmann distribution law. This law is applicable for only
classical particles or molecules. As an application of Maxwell-Boltzmann distribution law,
we have obtained the expression for the velocity distribution law. Also, we discussed
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Maxwell speed distribution law and defined the various types of speeds namely most
probable speed, mean or average speed and root mean square speed. We have further shown
that the root mean square speed is maximum and most probable speed is minimum among all
three speeds. We also came to know about the principle of equipartition of energy which
states that for a system in thermal equilibrium, on the average, an equal amount of energy
will be associated with each independent energy state or degree of freedom associated with
the system.
As a whole we can say that we are now familiar with all the basics of statistical mechanics
after completing this unit including first two units. In the next units, we shall discuss the
concepts of ensemble and different types of ensembles in detail.
3.8 GLOSSARY
Thermodynamic probability Itis the number of microstates corresponding to a given

macrostate.
Most probable state The most probable state is a macrostate with maximum
probability.
Maxwell-Boltzmann distribution It is a probability distribution in classical statistical

law mechanics used for discussing the distribution of
speed, velocity, and energy of particles at a particular
temperature.
Equipartition principle This principle states that energy is shared equally

among all the accessible degrees of freedom of a given
state.
Degree of freedom It isdefined as the number of independent ways that

specify the orientation or dynamics of a system or
body, such as translational, rotational and vibrational.
3.9 REFERENCES
2. Statistical Mechanics, S. Prakash,KedarNath Ram Nath.
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1. Thermodynamics and Statistical Mechanics, W. Greiner, L. Neise and H. Stocker,

Springer Science and Business Media.
2. Statistical and Thermal Physics, S. Lokanathan and R. S. Gambhir, PHI Learning Private
Limited, Delhi.
4. Fundamentals of Statistical Mechanics, B. B. Laud, New Age International.
3.11.1Short Answer Type

1. What is thermodynamic probability?
2. What is most probable state?
3. Write down the expression of Maxwell-Boltzmann distribution law.
4. State the principle of equipartition of energy.
5. What is the total kinetic energy for a linear hydrogen cyanide (HCN) and
non-linear Sulfur dioxide (SO2)molecules?
6. Write down the expressions for the most probable, average, and root mean
square speeds.

1. With the help of an example, discuss the difference between probability and
thermodynamic probability.
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2. Derive the expression of Maxwell-Boltzmann distribution law for the distribution of

particles. With the help of distribution law also obtain the Maxwell velocity distribution law?
3. Obtain an expression for speed distribution of particles in classical statistical mechanics.

Also deduce the expression for the most probable, average and root mean square speeds.
4. What are the postulates of statistical mechanics?
5. State and prove the principle of equipartition principle.

1. Calculate the thermodynamic probability of various macrostates corresponding to the
distribution of two distinguishable particles into three energy levels 0, E, and 2E.
2. Consider five particles in three compartments and calculate the thermodynamic probability
for (i) 4 particles in the first compartment, one in the second, and none in the third
compartment, and (ii) All five particles in the first compartment, and none in the second and
third compartments.
3. Find out the number of ways of distributing two identical particles in three distinct energy
levels according to classical Maxwell-Boltzmann statistics.
4. Find the value of temperature at which the mean speed of hydrogen molecules is double
the mean speed of fluorine molecules at 300K.[ , molecular
mass of hydrogen molecule is 2 and of fluorine molecule is 38.]
5.If the most probable speed of nitrogen molecule is 50m/s at T K. Calculate the most
probable speed of oxygen molecule at temperature4T K. [Given, ,
mass of nitrogen molecule mass of oxygen molecule
]

1. In statistical equilibrium, the thermodynamic probability of a system is
(a) 1 (b) zero
(c) maximum (d) minimum but not 1
2. Maxwell-Boltzmann statistics is applicable for

(a) photons (b) ideal gas
(c) phonons (d) electrons

3. The value of in Maxwell-Boltzmann distribution law is
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(a) 1 (b) zero
(c) (d)
4. The most probable speed of a gas molecules of mass m at a given temperatureT K is

proportional to
(a) m (b)
(c) (d)
5. The speed of four particles are 2, 2, 3, and 3, respectively. Their root mean square speed is
(a) (b)
(c) (d)
6. The ratio between the root-mean-square speed, average speed and most probable speed
will be
(a) 1.596: 1.732: 1.414 (b) 1: 1.596: 1.414
(c) 1.732 : 1.596: 1.414 (d) 1.596: 1.732: 2.125
3.12 ANSWERS
3.12.1 Self-Assessment questions
1. The two macrostates for the distribution of four distinguishable particles a, b, c, and d in
two compartmentsmay be chosen as
Macrostate (0, 4) – One possible microstate (0, a b c d).
Macrostate (1, 3) – Four possible microstates {(a, b c d); (b, c d a); (c, d a b); (d, a b c)}.
Probability of microstate (0, a b c d) corresponding to macrostate (0, 4) = .
Probability of microstate (a, b c d) corresponding to macrostate (1, 3) = .
Probability of microstate (b, c d a) corresponding to macrostate (1, 3) = .
Probability of microstate (c, d a b) corresponding to macrostate (1, 3) = .
Probability of microstate (d, a b c) corresponding to macrostate (1, 3) = .

2.The probability distribution function is given by
According to question
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or
Taking loge of both the sides of this equation

.
We obtained
3. Use equations (3.45), (3.46) and (3.47).Here temperature is T= 27+273=300K.
4. We have the expression of most probable speed given by equation (3.45).

For hydrogen molecules,
While for oxygen molecules,
Given, , , .
According to the question
, i.e., ,
and thus
Therefore, at the mean speed of hydrogen molecules will be double as that of

oxygen molecules at 300K.
5. HCl is a diatomic molecule having three translational and two rotational degrees of
freedom. So, the total kinetic energy of this molecule is
.
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3.12.2 Numerical answer type question:

1. Let the two particles are A and B
0E E 2E Number
of states
Total energy 0 AB - - 1
Total energy E A B - 2
B A -
Total energy 2E - AB - 3
A - B
B - A
Total energy 3E - A B 2
B A
Total energy 4E - - AB 1
As we know that thermodynamic probability is the number of microstates corresponding to a

given macrostate. Therefore,
Thermodynamic probability of macrostate “Total energy 0” = 1.
Thermodynamic probability of macrostate “Total energy E” = 2.
Thermodynamic probability of macrostate “Total energy 2E” = 3.
2. Let the five particles are a, b, c, d, and e. We can write all the microstates corresponding
to the desired macrostates.
0E E 2E Number
of states
(4,1,0) abcd e - 5
abce d -
abdc c -
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adce b -
bcde a -
(5,0,0) abcde - - 1
(i) Thermodynamic probability of macrostate (4,1,0) = 5.

(ii) Thermodynamic probability of macrostate (5,0,0) = 1.
3. Let the two particles are and .According to Maxwell-Boltzmanndistribution law a

single state can be occupied by an arbitrary number of particles. So, the distribution is given
as
Energy Energy Energy

level 1 level 2 level 3
Both in same state - -
- -
- -
Both in different states -
4.Use equation (3.46) for calculating . Here,temperature , so
and thus
.
5. From equation (3.45), for nitrogen molecules we can write
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,
or
.
Now again with the help of equation (3.45), for oxygen molecule,
Substitute the value of T from the expression of nitrogen in the expression of , we

get
3.12.3 Objective answer type questions:

1. Correct option is (c), maximum.
2. Correct option is (b), ideal gas.
3. Correct option is (d), .
4. Correct option is (b), .
6. Correct option is (c), 1.732 : 1.596: 1.414.
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UNIT 4 Microcanonical Ensemble
Structure
4.1 INTRODUCTION
4.2 OBJECTIVES
4.3 ENSEMBLE THEORY
4.3.1 MICROCANONICAL ENSEMBLE
4.3.2 CANONICAL ENSEMBLE
4.3.3 GRANDCANONICAL ENSEMBLE
4.4MICROCANONICAL ENSEMBLE
4.4.1 ENTROPY OF IDEAL GAS IN MICROCANONICAL ENSEMBLE
4.5GIBBS PARADOX
4.6 PARTITION FUNCTION AND ITS RELATION WITH THERMODYNAMIC
QUANTITIES
4.7 SUMMARY
4.8 GLOSSARY
4.9 REFERENCES
4.12 ANSWERS
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4.1 INTRODUCTION
We already understand the significance of statistical mechanics in the modern physics.

Specifically, its applicability in studying the systems with a large number of constituent
particles,where it provides us tool to study the macroscopic properties of the system in terms
of the microscopic properties of its particles. We have already learned the basic tools of
statistical physics and classical statistics in Block 1.
The statistical analysis of the systemis useful if we consider a probability distribution
describing the state of the systemcalled ensemble, which allows us to calculate the average
values of the macroscopic properties as a weighed function with the help of probability
distribution.Physically, we can consider an ensemble as multiple copies of independent
systems with same macroscopic properties but different microscopic states. There are
different types of ensembles depending uponthe macroscopic constraints. One of the most
significant ensembles describes isolated system, i.e., when the individual assemblies are
separated by a rigid, impermeable, and well insulated walls and thus the individual systems
cannot exchange energy and particles with each other. The present unit will be focused
entirely on this particular type of ensemble, i.e., microcanonical ensemble.
Ourunderstanding of the thermodynamics tells us that entropy or randomness is an extensive
property.Therefore, entropy of the systemobtained by mixing two distinct ideal gases
increases, whereas if both ideal gases are sameit is a reversible process and entropy should
not increase. Mathematically, a contradiction occurs in the latter case as it gives non-zero
change in entropy, which is termed asGibbs paradox. This is resolved by dividing the number
of microstates, used in obtaining the formula for entropy, by a factor called Gibbs factor.
Quantum mechanics providesus an explanation for this factor as the constituent particles or
gas molecules are identical.
We will further discuss partition function, which quantifies the number of microstates
accessible to the system in a given ensemble.While calculating this function, it becomes
important that whether there exist more than one state with the same energy, if so it is
referred to asdegeneracy.The partition function providesus a mathematical tool to obtain
thermodynamic quantities.Therefore, we will be obtaining partition function and its relation
to thermodynamic quantities for microcanonical ensemble. Some of the topics discussed in
this unit will be further useful in the next units and understanding the rest of the ensembles,
i.e., canonical and grand canonical ensembles. Specifically, we will obtain partition function
for canonical and grand canonical ensembles in the next two units and obtain corresponding
thermodynamic quantities. Here, we begin with a general discussion of ensemble theory.
4.2 OBJECTIVES
• Concepts of ensemble
• Microcanonical ensemble
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• Entropy of an ideal gas in microcanonical ensemble

• Gibbs paradox and Gibbs factor
• Partition function
• Relation between partition function and various thermodynamic quantities
4.3 ENSEMBLE THEORY

In statistical mechanics, we always deal with a large number of particles. For this kind of a
system, it is very difficult to study the precise microscopic details of the system. Therefore,
we need to construct the concept of ensembles in which a large number of systems with the
same macroscopic properties but different microscopic properties are considered, and the
individual system is assumed to evolve from different initial conditions. Therefore, the time
evolution of individual system will be different from the other systems. Hence,the ensemble
is a collection of copies of identical systems with the same macroscopic but different
microscopic properties, and the macroscopic variables are obtained as averages over the
system of ensemble.
In statistical mechanics, there are three types of ensembles, namely microcanonical,
canonical, and grand canonical, specified on the basis of macroscopic constraints.
4.3.1 Microcanonical ensemble

A collection of essentially independent assemblies with same energy , volume , and
number of the particles is known as microcanonical ensemble. The individual assemblies
are separated by a rigid, impermeable, and well insulated walls. The macroscopic properties,
such as , , and , are not affected by the presence of other systems. In this case, the
individual systems cannot exchange energy and particles with each other(see Figure4.1).
Here, it should be clear that independent systems essentially satisfy same macroscopic
conditions, while the individual system in the ensemble differ in microscopic conditions.
Figure4.1: Pictorial representation of microcanonical ensemble.
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4.3.2 Canonical ensemble

A collection of essentially independent assemblies with same temperature , volume , and
number of the particles is known as canonical ensemble. The individual assemblies are
separated by a rigid, impermeable, and diathermic wall. The macroscopic properties, such as
, , and , are not affected by the presence of other systems. Here, the individual systems
can exchange energy, but not particles with each other. Since energy can be exchanged
between assemblies, one could bring all of the assemblies in thermal contact with each other
(see Figure4.2).
Figure4.2: Pictorial representation of canonical ensemble.
4.3.3 Grand canonical ensemble

The grand canonical ensemble is a collection of essentially independent assemblies having
same temperature , volume , and chemical potential . The individual assemblies are
separated by a rigid, permeable, and diathermic wall. In grand canonical ensemble, the
assemblies can exchange energy as well as particles with each other (see Figure4.3).
Figure4.3:Pictorial representation of grand canonical ensemble.
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We will discuss the microcanonical ensemble in the forthcoming sections of this unit.
Whereas, the canonical and grand canonical ensembles shall be discussed in the Units 5 and
6, respectively, in more detail.
It is worth emphasizing here that the three kinds of ensembles mentioned in the previous
section are used to obtain different thermodynamics quantities.
Self Assessment Question (SAQ) 1:Discuss how microcanonical, canonical, and grand
canonical ensembles are different from each other.
4.4 MICROCANONICAL ENSEMBLE

In the previous section, we have introduced the microcanonical ensemble in brief. The
microcanonical ensemble is a system which is totally isolated from its surrounding.
Therefore, it describes an isolated system, i.e., a microcanonical ensemble is a collection of
multiple copies of the isolated system (seeFigure4.1). In microcanonical ensemble, the
Boltzmann entropy relation is significant to discuss the thermodynamics of a
system.Here, S is a thermodynamic quantity entropy and is a statistical quantity gives the
total number of microstates. This serves as a bridge between statistical mechanics and
thermodynamics, which shows that entropy is an appropriate thermodynamic function for
understanding isolated system of microcanonical ensemble.
4.4.1 Entropy of ideal gas in microcanonical ensemble

We all are familiar with ideal gases since our graduation classes. Here, we briefly discuss
what an ideal gas is. It is a gas in which the molecules interact only by elastic collisions and
there are no intermolecular attractive forces between the molecules. Ideal gas obeys general
gas equation and other gas laws. Further, the internal energy of an ideal gas is solely
temperature dependent and is independent of its volume. Therefore, the internal energy is the
only average kinetic energy of the molecules of an ideal gas.
Let us calculate the expression for the entropy of an ideal gas in microcanonical ensemble.
Suppose there are N particles or molecules of mass m each enclosed in a container of volume
V with energy . The total energy of the system is E. Therefore, the total number of
microstates is given by
…(4.1)
Here, the integral over the phase space gives the total volume, andthus
.
…(4.2)
The integration is left in momentum spaceonly, and the momenta are constrained by relation
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…(4.3)
Thisrepresents the equation of a 3N-dimensional hypersphere of radius .
We assume that the energy is not exactly constant but vary by an amount , which
satisfy . Thus, the momentum space available or accessible for the gas is a thin
shell of a 3N-dimensional hypersphere of radius and thickness . Therefore,
the surface area of this hypersphere is . (Here, we use the standard result,
the surface area of anN-dimensional hypersphere of radius Ris ).
Hence, the total number of microstates can becalculated as
or
…(4.4)
Further, using the relation , the entropy of the gas can now be written as
…(4.5)
With the help of Stirling formula we can write
Thus, the expression of entropy given by equation (4.5) now reduces as
…(4.6)
In this equation, the terms in the curly bracket are much smaller than the other terms, hence
we can neglect them, which results in
…(4.7)
This is the expression of entropy (thermodynamic) for an ideal gas in the microcanonical
ensemble.
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We have a relation between thermodynamic entropy and statistical entropy as , where

is Boltzmann’s constant, and is the statistical entropy. Therefore, the expression of
statistical entropy looks like
…(4.8)
We now establish relation with other thermodynamic quantities.

1. Internal energy:
From Unit 2, we are familiar with the statistical temperature as
We have used the fact that V and N are constant, so the remaining terms vanish.Thus, we get
or the internal energy as
…(4.9)
2. Relation between statistical temperature and thermodynamic pressure:
Using the relation introduced in Unit 2, we have
We have used the fact that E and N are constant here so the remaining terms vanish. Thus, we
get the desired relation as
or
…(4.10)
3. Chemical potential:
We have already discussed in Unit 2 that
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Therefore, the chemical potential can be defined as

…(4.11)
Example 4.1:If the available volume Vof the container of an ideal gas is doubled with no
change in the number of moleculesN,their massm, and total energyE, how will the statistical
entropy will change?
The new volume is now and the total number of molecules are constant. Therefore, the
expression of statistical entropy given by equation (4.8) becomes
This means that on increasing the volume by keeping the number of molecules constant the
statistical entropy increases.
Self Assessment Question (SAQ) 2:How will the pressure change in Example 4.1 discussed
above?
4.5 GIBBS PARADOX

We all are familiar that the mixing of two different gases is an irreversible process which
leads to an increase in entropy. The Gibbs paradox involves the contrast between the
quantities for mixing two distinct ideal gases and that for mixing of the same ideal gas. Let us
consider two different but similar ideal gases in a box of volume V with partition as shown in
Figure4.4, at the same temperature and pressure. The ideal gas in Partition 1 (Partition 2) of
volume contains number of gas molecules of mass andtotal energy
. We subsequently remove the partition and allow the gases to mix with each other.
Before mixing, the initial entropy ( of the combined system, using equation (4.7), is the
sum of the entropies of the two gases, i.e.,
.…(4.12)
After mixing, i.e., on the removal of partition, the energy, temperature, and pressure will not
change as they are at the same temperature and pressure. Thus, the volume available to both
the gases is V. Therefore, the final entropy ( ) after mixing of gases comes out as
. …(4.13)
So the change in entropy is
i.e.,
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. …(4.14)
This shows that there is a change (an increase) in entropy after mixing of the same kind of
different gases, which is what we are expecting. This change, i.e., is called the entropy of
mixing.
Figure 4.4: Pictorial representation of mixing of two ideal gases.

If we further consider that there is same ideal gas on both sides before removing the partition,
then a problem arises. On removing the partition there should be no effect on entropy as the
system remains unchanged. However, the expression of change in entropy represented by
equation (4.14) yieldsa change in entropy. Let us check this fact in detail. In the case of two
identical ideal gases, .Therefore, the final entropy is
or
…(4.15)
Here, is the total number of particles.Thus, equation (4.15) can be simplified as
…(4.16)
We can calculate , which gives us anon-zero value, i.e., there is a change in the
entropy.However, there should be no change in entropy because this is a reversible process,
i.e.,after reinserting the partition we are not able to make out if the partition was removed
before. This contradiction of mixing of identical gases is known as Gibbs paradox.
This paradox can be resolved. Gibbs suggested that while counting the number of microstates
for N molecules of anideal gas if we multiply the number of microstates by a factor known as
Gibbs factor, i.e., by , the contradiction disappears.
In this way equation (4.4) in Subsection 4.4.1, takes the following form
…(4.17)
Hence, the expression of entropy (see equation (4.5)) takes the form,
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…(4.18)
Further, on applying the Stirling’s approximation in this equation, we get an extra term
in comparison with equation (4.7). Thus,on simplification we obtain the
correct form of entropy as
…(4.17)
This equation describes the entropy of a classical ideal gas known as the Sackur-Tetrode
equation.
Hence, the entropy of two gases before mixing can be described by
…(4.18)
For two identical ideal gases, and . So equation (4.18) reduces in the
following form
Thus, the initial entropy of the combined system is
…(4.19)
This equation is similar to the equation for the entropy after mixing. So inthe case of mixing
of identical gases,there is no change in entropy after removing the partition. In a similar
way,we can also verify Sakur-Tetrode equation for the case of different types of ideal gases,
which gives the same result for the entropy of mixing.
The justification for the Gibbs factor can be provided from the quantum mechanical point of
view. Specifically, we know that the particles or atoms or molecules are considered as
identicalin quantum statistics (discussed in detail in Section 1.3.4). Consider Example 1.9 of
Unit 1 and calculate the number of words that can be obtained by rearrangement of a five
letter word. It would be , but what if all the five letters are same (say “aaaaa”) then we
divide the total number by 5!to obtain only 1 word after rearrangement, i.e., “aaaaa”.
Similarly, we should count the microstates in this case by assuming the particles (gas
molecules) are indistinguishable. Therefore, we must divide the number of microstates by ,
where N is the number of gas molecules. Thus, Gibbs paradox is resolved by taking into
consideration that gas molecules are identical.
Example 4.2:Out of the two cases given below in which case we will encounter Gibbs
paradox for mixing of gases?
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Case (a): There is an ideal gashelium (oxygen) in Partition 1 (Partition 2) of volume

containing number of gas molecules of mass and total energy before
mixing.
Case (b): There is an ideal gas helium (helium) in Partition 1 (Partition 2) of volume
contains number of gas molecules of mass and total energy before
mixing.
In case (b), we have identical gas molecules so due to the indistinguishability of gas
molecules we will encounter Gibbs paradox.
4.6 PARTITION FUNCTION AND ITS RELATION WITH

THERMODYNAMIC QUANTITIES
First of all, we define partition function, which tells us how many microstates are accessible
to the system in a given ensemble. In other words, it is a measure of volume occupied by the
system in phase space. Partition function plays a central role in statistical mechanics and is
useful in calculating various thermodynamic functions.
Consider an assembly or collection of molecules of an ideal gas in microcanonical
ensemble. The Maxwell Boltzmann distribution law can be written as,
…(4.20)
where is the energy of the ith particles, and .Therefore, the total number of
molecules in the given system is
or
…(4.21)
Thus, we introduce Z,called as partition function, as
…(4.22)
This partition function represented by equation (4.22) tells us how the gas molecules are
distributed among various energy states or the particles are partitioned among various energy
levels.In a similar way, we will obtain the two different expressions of partition function for
the other two types of ensembles, i.e., canonical ensemble in Unit 5 and grand canonical
ensemble in Unit 6.
Here, we introduce the term degeneracy, which corresponds to the situation if there are more
than one stateor level with the same energy.Taking into consideration the degeneracy of
states, the expression of partition function given by equation (4.22) modifies to
…(4.23)
where is the degeneracy factor.
In thermodynamic equilibrium, partition function which is a function of thermodynamic state
variables, like volume and temperature, describes the statistical properties of a system.
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We will further discuss that various thermodynamic variables can be expressed in terms of
the partition function or its derivatives.
1. Relation with entropy (S):
For a classical system or classical statistical mechanics the distribution of molecules can be
defined by
…(4.24)
Taking the logarithm of both sides of the equation
…(4.25)
and using Striling’s approximation, we can write
…(4.26)
Further, we know that the Maxwell Boltzmann distribution law is
…(4.27)
With the help of equations (4.25) and (4.26), we get
…(4.28)
or
…(4.29)
In equation (4.28), the last two terms represent the total energy of the molecules and the total
number of molecules, respectively, i.e.,
Therefore, equation (4.28) can be simplified as
Further, using , the partition function, we get

…(4.30)
Now we know that,the Boltzmann’s entropy relation is
…(4.31)
With the help of equation (4.30), we get
or
…(4.32)
This is the relation between partition function and entropy.
For an ideal gas, , so
…(4.33)
This equation gives the entropy of an ideal gas in microcanonical ensemble. As in
mechanics,the potential energy of a system tells us about its stability, in a similar way,
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entropy gives information about the most stable equilibrium state of the system. Also the
entropy difference determines the direction of a changein an isolated system.
2. Relation with Helmholtz free energy (F):

The Helmholtz free energy is defined as
…(4.34)
which can be obtained using equation (4.31) as
or
…(4.35)
3. Relation with pressure (P):

Pressure of a gas given by
and can be obtained in terms of the partition function

…(4.36)
4. Relation with total energy (E):

The total energy of a system of N particle is given by
…(4.37)
Here, the average energy per particle is obtained as
Further, using , it becomes
…(4.38)
For an isothermal-isochoric process, we have
or
,
and
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…(4.39)
Thus, the total energy is given by
or
…(4.40)
This is the relation between total energy and partition function.
5. Relation with enthalpy (H):
The enthalpy is defined as
and can be written in terms of the partition function as

…(4.41)
6. Relation with Gibbs potential or Gibbs free energy (G):

Gibbs potential is defined as
It can be obtained in terms of partition function as

…(4.42)
7. Relation with specific heat at constant volume ( ):
The specific heat at constant volume is defined as
and using equation (4.38), we obtain
This can be simplified to

…(4.43)
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Thus, we can conclude that if we are able to write a partition function for a given system in
microcanonical ensemble we can obtain various thermodynamic quantities of the system.
Example 4.3:Consider a system of N particles with only three possible energy levels , ,
and . Write down the expression of the partition function and average energy per particle
for this system by assuming that the particles are following Maxwell Boltzmann statistics.
For the given system of N particles with only three possible energy levels , , and . Using
equation (4.22) we can directly write the partition function as
Further, from equation (4.38), the average energymay be expressed as
Example 4.4:There are two energy levels and with degeneracy 1 and 2 for a system of
two particles which follow Maxwell Boltzmann statistics. Calculate the mean energy of the
system.
It is given that the particles are following Maxwell Boltzmann distribution law and
degeneracy of the lower energy level is 1, whereas degeneracy of the upper energy level is 2.
So, we can write partition function using equation (4.23) as
Again, using equation (4.38), we havethe mean energy of the system
Self Assessment Question (SAQ) 3: The expression of partition function for a system is
given as . Write down the expression for Helmholtz free energy.
Self Assessment Question (SAQ) 4:Obtain an expression for total energy when the partition
function of a system is given by .
4.7 SUMMARY
In this unit, we learned about the concept of ensembles and three types of ensemblesoften
used in statistical mechanics.Based on the three different types of macroscopic conditions,
these three categories of ensembles are proposed. We also discussed the microcanonical
ensemble in detail by considering the system of ideal gases in classical statistics. Yet another
important quantity in statistical mechanics, i.e., partition function, is discussed in detail along
with its relation with various thermodynamic quantities. The paradoxin deriving entropy due
to mixing of ideal gases is discussed as the well-known Gibbs paradox, which can be
resolved by dividingthe number of microstates with the Gibbs factor.Quantum mechanics
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provides an explanation of the Gibbs factor that it is due to indistinguishabilityof molecules

or particles.
This unit was further focused on the system of an ideal gas as we calculated the entropy of
ideal gases in microcanonical ensemble. We know the connection between statistical
mechanics and thermodynamics is given by the Boltzmann entropy relation.Thus, it is
important to calculate entropy in a particular type of ensemble,which helps us to calculate
other thermodynamic parameters.
In a similar way, we will discuss the remaining two kinds of ensembles, i.e. canonical and
grand canonical ensembles, in Units 5 and 6, respectively. Thus,topics discussed in this unit
will also be helpful in understanding the calculations and discussions in the forthcoming
units.
4.8 GLOSSARY
Ensemble A large number of systems with the same macroscopic

properties but different microscopic properties are
collectively called an ensemble.
Microcanonical ensemble A collection of essentially independent assemblies with

same energy E, volume V, and number of the particles N is
known as microcanonical ensemble.
Entropy A thermodynamic quantity which is a measure of the

molecular disorder or randomness of a system.
Gibbs factor For the system of indistinguishable particles or molecules

we have to multiply the expression of number of
microstates by a factor of , known as Gibbs factor.
Gibbs paradox Gibbs paradox involves the contradiction between mixing

two ideal gases of a different kind and that of mixing two
ideal gases of the same kind.
Partition function This is a function which tells us how the gas molecules are
distributed among various energy states or the particles
are partitioned among various energy levels.
4.9 REFERENCES
2. Statistical Mechanics, S. Prakash, Kedar Nath Ram Nath.
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1. Thermodynamics and Statistical Mechanics, W. Greiner, L. Neise and H. Stocker,

Springer Science and Business Media.
2. Statistical and Thermal Physics, S. Lokanathan and R. S. Gambhir, PHI Learning Private
Limited, Delhi.
4. Fundamentals of Statistical Mechanics, B. B. Laud, New Age International.

1. Define ensemble.
2. With the help of pictorial representation define microcanonical ensemble.
3. What do you understand by partition function?
4. What is the Gibbs paradox?
5. Write down the expression for entropy and Sakur-Tetrode equation after removing Gibbs
paradox.
1. Explain in brief the ensemble theory and various types of ensembles in

statistical mechanics.
2. What do you understand by ensemble? How the concept of ensemble is utilized to
obtain the priorities of a statistical system.
3. Define partition function for a microcanonical ensemble. Use it to derive
expressions for chemical potential, Helmholtz free energy, and Gibbs free energy.
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4. Obtain an expression for entropy of an ideal gas in the microcanonical

ensemble,and also establish the relation between statistical entropy and various
thermodynamic quantities.
5. Discuss the significance of partition function.Subsequently, obtain an
expression for the partition function in microcanonical ensemble and also establish its
relation with different thermodynamic variables.
6. By considering a system of an ideal gas discuss the Gibbs paradox by
deriving an expression for entropy. Obtain an expression for Sakur-Tetrode equation
after the removal of Gibbs paradox.
7. Explain Gibbs paradox. How it can be resolved by the concept of
indistinguishability of the molecules or particles.

1. Consider a system of two particles, each of which can be in one of three quantum
states of respective energies , ,and . Obtain an expression for the particle function
if the particles obey classical Maxwell Boltzmann statistics.
2. Obtain the expression of Helmholtz free energy by writing partition function for a
system of non-interacting classical particles having two energy levels with energies
and . The lower level is four fold degenerate and the upper level is doubly
degenerate.
3. Calculate the partition function of a single harmonic oscillator with energy levels
given by .
4. Consider a system of two classical particles, each of which can occupy any of the two
energy levels and . Write down the expression for the mean energy of the system
at temperature T.
5. If a non-interacting system has two energy levels ,the lower level is doubly
degenerated while that of energy is non-degenerate. Write down an expression for
partition function of this system. Also calculate the total energy and Gibbs free
energy.

1. The individual assemblies of an ensemble in which the macroscopic properties,
namely energy, volume, and the number of particles remain same are separated by
rigid
(a) impermeable and insulated wall (b) permeable and insulated wall
(c) impermeable and diathermic wall(d) permeable and diathermic wall
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2. In microcanonical ensemble, the individual systems cannot exchange

(a) particles (b) energy and particles
(c) energy (d) none of these
3. The correct form of Boltzmann entropy relation is

(a) (b)
(c) (d)
4. The entropy of the assembly of ideal gas molecules in terms of partition function Z is
given by
(a) (b)
(c) (d)
5. For removing the Gibbs paradox, one should multiply the expression of
number of microstates by a factor of
(a) (b)
(c) (d)
4.12 ANSWERS
4.12.1 Self Assessment questions:
1. We can summarize some of the important facts about the microcanonical ensemble in the
tabular form given below.
Property Microcanonical Canonical ensemble Grand canonical

ensemble ensemble
Macroscopic Energy E, volume V and Temperature T, Temperature T,

properties of number of particles N volume V and volume V and
independent number of particles chemical potential
assemblies N
Separation Rigid, impermeable and Rigid, impermeable Rigid, permeable

between insulated wall and diathermic wall and diathermic wall
independent
assemblies
Contact of Cannot interact Can interact Can interact

individual
assemblies with
each other
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Macroscopic No exchange of energy Only energy can be Both energy and

property that can and particles exchanged particles can be
be exchanged exchanged
between
individual systems
Variation in the Both remain constant Number of particles Neither number of

number of remains constant but particles nor energy
particles and not energy remains constant
energy for each
assembly
As we are restricted with microcanonical ensemble in this unit, the expression of partition
function is, in the present case. Further, with the knowledge of the partition
function for microcanonical ensemble, we can define various physical quantities in terms of
the partition function, like entropy (equation (4.32)), Helmholtz free energy (equation (4.35)),
pressure (equation (4.36)), total energy (equation (4.40)), enthalpy (equation (4.41)), Gibbs
free energy (equation (4.42)), specific heat (equation (4.43)). Similar expressions for
partition function in canonical and grand canonical ensembles will be obtained in the next
two units.
2.From equation (4.8), the pressure will remain one-half.
3. The given partition function is,

From equation (4.35) we can write
or
4. With the help of equation (4.40), we can calculate the total energy as
and thus
or
4.12.2 Numerical answer type:
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1. It is given that particle (say A and B) obey Maxwell Boltzmann statistics, i.e., any
number of particles can occupy a single energy state. The distribution is shown in the
table below.
Total energy Degeneracy
AB - - 0 1
A B - 2
B A
- AB - 1
A - B 2
B A
- A B 2
B A
- - AB 1
Now using the relation given by equation (4.23), we can obtain an expression of
partition function as
.
2.In question, we have for lower energy level, and for upper energy
level. So, with the help of equation (4.23), we can directly write the expression for
partition function as
. Thus, from equation (4.35) we obtain the expression of
Helmholtz free energy as
or
3. We have . Thus, the partition function can be written as
The summation term is a geometric series, so we can write

and
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8. Let P and Q be the two particles and the distribution is given below
Total energy Degeneracy
PQ - 0 1
P Q 2
Q P
- PQ 1
We can write down the partition function as
Thereafter, with the help of equation (4.38), we get the expression of mean energy
9. According to the question, we can write the partition function
We can use equation (4.40) for calculating the total energy as
Further, from equation (4.42), Gibbs free energy can be calculated as
4.12.3: Objective answer type:

1. Correct option is (a), impermeable and insulated walls.
2. Correct option is (b), energy and particles.
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3. Correct option is (c), .

4. Correct option is (d), .
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UNIT 5: Canonical Ensemble
Structure:
5.1 Introduction:
5.2 Objectives
5.3 Canonical ensemble
5.3.1 Application
5.4 Energy fluctuation in the canonical ensemble
5.5 Linear harmonic oscillator
5.6 Summary
5.7 Glossary
5.8 Terminal questions
5.8.1 Multiple Choice Questions
5.9 References
5.10 Suggested readings
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5.1 Introduction:
The given unit discuss an important ensemble of statistical mechanics, canonical ensemble.
Among all ensembles, canonical ensemble is considered as more appropriate. It plays a
central role in statistical mechanics. Reasonably it should come first and rest other ensemble
(micro-canonical and grand-canonical) should be followed as a special case. In the case of
canonical ensemble, system is considered in thermal equilibrium with a heat bath (taken as a
closed system).
The partition function is used as an important parameter to solve various thermodynamic

quantities for canonical ensemble. Also, this helps to explain energy fluctuations as well. The
difficulties and limitations raised by the Boltzmann’s statistical mechanics was removed by
the Gibb’s theory which was based on the ensemble concept. The Boltzmann’s method is
applicable only for the systems having small number density almost no interactions between
particles while Gibb’s ensemble theory shows its validity for both the classical and quantum
systems.
5.2 Objectives:
Students will understand:
1. Concept of canonical ensemble.

2. Role of partition function.
3. The way to compute the various thermodynamic variables with the help of partition
function.
4. Energy fluctuations calculations.
5. The case of Linear Harmonic Oscillator.
5.3 CANONICAL ENSEMBLE: -

It is a collection of essentially independent system having same temperature T. Exchange of
energy of ensemble is explained by temperature.
It is similar to partial open system. Ensemble are separated by rigid impermeable but
conducting walls.
Here we make walls conducting so energy is going to exchange. The physical quantity which
defines the exchange of energy at equilibrium is temperature. The natural variable are [N, V,
T].
The energy of this system is variable and can take up values anywhere b/w to .
Let have a subsystem S1 in S2 which is a heat reservoir from which we can draw as much heat
as we want without changing anything E2>>E1.
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S2
E2
E1
S1
Assuming system is closed then E1 + E2 = Constant
System as a whole is micro-canonical ensemble
E1 + E2 = E = Constant .... (1)
Since the reservoir is much layer them the system any value of E1 will be a small fraction
of E.
….(2)
Let the number of states available to the reservoir be (E2). The larger the number of
states available to the reservoir, the larger the probability of the reservoir assuming that
particular energy value E2.
The phase density of the subsystem is proportional to the number of

microstates available to both reservoirs
Ρ(p1 , q1) (E2) …(3)
(E – E1)
But the entropy is defined as,
S2 (E – E1) = K ln (E – E1)
As E1<< E, hence the above expression is expanded
S2 (E – E1) S2 (E) –
S2 (E) – E1/T
K ln (E – E1) = S2 (E) – E1/T
K ln (E – E1) =
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using eqnn (3)
ln ρ(p1, q1)
ρ(p1, q1) exp. * exp.
If ‘H’ be the Hamiltonian of the system then,
ρ(p1, q1) exp.
where ‘i' is dummy indices
ρ(p,q)
i.e. ρ(p, q) phase density is dependent on exponential term and hence ρ is not a constant. As
tempertaure increases, the phase density decreases i.e. all the states in canonical ensemble do
not have equal probability.
or ρ(p, q)= A
The ensemble average <F> of a given physical quantity F(p, q) which may be different for
system in different microstates is
This gets reduced to the form:
dω = d3Np d3Nq (volume element of phase density)
Here, denominator specifies the normalization constant or partition function which should be
a dimensionless quantity.
To get rid of the dimension of ‘dω’ we will use the relation
N! is multiplied as the particles are indistinguishable. This is done for correct Boltzmann
counting and is known as Gibb’s paradox.
Hence,
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Now, the denominator is dimensionless and the expression for partition function is
Where, the integration goes over the whole space. Once the partition function is known, it
tells us how the states are divided and this makes averages normalised.
5.3.1 Application:-
I. An Ideal Gas:-
Consider a system of N-identical particles which are non-interacting. The particles confined
to a space of volume V and in equilibrium at temperature T.
…(1)
Partition function in this case is
Since potential energy depends only on q so
d3qk = VN
To find d3p:-
Consider d3x = 4 x2 dx for spherical coordinates
Replacing d3x by d3p
But 4 is soild angle
Let we get
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___________(a)
To make QN dimensionless in more obvious way we introduce De-Broglie wavelength.
Now thermal wavelength
But its correct definition is
In above expression, we have all the three variables of canonical ensemble they are (N, V, T)
In 1-D;
and for distinguishable particle
Note:- 1. If partition function of electron & proton together but non-interacting is to find
than potential energy of e-& potential energy of p.
2. Steps to write partition function:
(a) Write integral over dω
(b) Make it dimensionless
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(c) To make dω dimensionless is multiplied
(d) Check whether system have distinguishable or indistinguishable

particles.
(i) Helmholtz Free Energy: -
Using equ. (a), we write the expression for partition function
A (N, V, T) = - KT ln QN (V, T)
but from equ. (a),
A(N, V, T) = -kT ln
= kT ln N! – NKT ln
Using Stirling formula,
= kT (N ln N – N) – NKT ln NKT ln
= NKT (ln N – 1) – NKT ln V + NKT ln h3 NKT ln
= NKT
or A (V, N, T) = NKT … (b)
The above expression is helpful to derive rest of thermodynamic parameters.
(ii) Chemical Potential: -
It can be defined as: -
Using equ. (b)
A (V, N, T) = NKT
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= NKT ln
Let us suppose
Or
= kT
(iii) Pressure:-
from equ. (a),
Now,
(iv) Entropy:-
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from equ. (a),
Let
or
or … (c)
It is also known as Sackur – Tetrode equation for the entropy.
(v) Internal Energy:-
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It is a function of temperature alone independent on volume & pressure.
(Average Energy per system in the Ensemble)
but
The Hamiltonian for a canonical ensemble is not conserved (as there is exchange of
energy)
Multiplying and dividing by to make it dimensionless.
<H> = (internal energy)
But
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but
or A= TS
Now substituting the value of ‘A’ from equ. (b) and ‘S’ from equ. (c), we get
5.4 Energy Fluctuation in the Canonical Ensemble:

The average of energy is not a good option because positive or negative energies
will cancel out each other and total energy will be zero. So to get rid of –ve sign and
this problem we take square of energies and find out the average mean square
energies.
It is defined as the root mean square deviation & Hamiltonian from its average
value <H>
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…(A)
We’ve already proved that
……(1)
By definition
…(2)
Combining equ. (1) & (2) as defined by (A)
<H2> + <H>2
Consider,
{Using equ. (2)}
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…(3)
Now using (1) & (3) in equ. (A)
using equ. (1) here,
= kT2 CV
or
Here, temperature is an intensive quantity. Energy & CV are extensive quantity. If E

N then CV N,
Here we’re dealing with macroscopic system which has number of molecules of as
out 1024. The energy fluctuation is of the order 10-12i.e. they are very tiny we can
say that canonical and thermodynamic ensemble are approximately equivalent.
For a perfect gas:-
Here, is the fudge factor. If this factor doesn’t come then result is wrong.
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5.5 Linear Harmonic Oscillator:-

Classical expression for Hamiltonian is
where,
k = spring constant
m = mass of the particle
q = A cos (ωot + )
A = Amplitude of Vibration
ωo = Angular frequency of Vibration
Since,
Hamiltonian = QN = = Free Energy A(T, V, N) we use above Hamiltonian

as,
The partition function for 1-Oscillator is
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Solving integration, we have
where,
For N-independent Oscillator
&other thermodynamic variables,
The internal energy, is an example of the general equipartition theorem in which

each coordinate or monentum appear as a quadratic term in Hamiltonian such as(
) contributes to the average energy in case classical:
In 3-Dim space, just replace N by 3N in the above expressions.
# For Quantum – Mechanical situation, energy sign values of a 1-Dim harmonic

oscillator is given by,
for single-Oscillator partition function
For N-one dimensional Oscillators.
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QN = (Q1)N from which the thermodynamic variables follows as

,
QN = (Q1)N =
For the Helmholtz free energy of the system,
A = NKT ln [2 sin h ( )]
P=0
& C = Cp = Cv =
C = NK
For , & the exponantials can be expanded, thus we find classical

equipartition result C = NK.
5.6Summary:
In the given unit you have studied aboutcanonical ensemble. Out of all the ensembles,
canonical ensemble is considered as the most appropriate one. Ideally this ensemble
considered as first and rest other ensemble should be followed as a special case. In a given
unit, the theory of canonical ensemble is discussed in which the system is considered in
thermal equilibrium with a heat bath (taken as a closed system).Here, you have also learnt
about the partition function which is used as an important parameter to solve various
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thermodynamic quantities for canonical ensemble. An important application of canonical

ensemble is highlighted.You have learnt how tocalculatethe energy fluctuations in canonical
ensemble. Also, linear harmonic oscillator case discussed as an important example.To check
your progress, multiple choice questions and terminal questions are given in the unit.
5.7 Glossary:
Ensemble: It is consisting of a large number of virtual copies of a system, considered all at
once, each of which represents a possible state that the real system might be in.
Also, a statistical ensemble is a probability distribution for the state of the system.
Density of states: It is the number of microstates or the number of independent quantum

states of an N particle system per unit energy range. In other words, the density of
states, denoted by g(E), indicates how densely packed quantum states in a particular
system. Macroscopically, we can define: the density of states can be treated as a
continuous function of the internal energy of the system.
Microstate of a system: The microstate of a system at any time is given by specifying the
maximum possible information about the system, e.g. the position and velocity of
each molecule. It is a particular quantum state of a system.
Canonical Ensemble: It is a statistical representative of a system in equilibrium (thermal

contact) with a heat reservoir (bath) at some fixed temperature. It has a fixed N, V
and T but variable E. The system can exchange energy with the heat bath, so that
the states of the system will differ in total energy. This is a closed system.
5.8 Terminal Questions:

Q 1. Find the average energy of an ideal classical gas in a canonical ensemble at temperature
T.
Q 2. A system in contact witha heat reservoir at temperature T has two accessible energy
states with energies 0 and 0.2 eV. If the probability of the system being in the higher
energy state is 0.2, find the temperature of the heat reservoir.
Q3.Show that <( E)3>=T4 + 2T3Cv at constant volume for a system of canonical
ensemble.
Q 4. How does the probability of a microstate of a system in canonical ensemble vary with
the energy of state? Show the variation with the help of a graph.
Q 5. Find out the equation of state of an ideal classical gas in canonical ensemble.
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Q 6. Show that the mean square fluctuations in the energy of a system in a canonical
ensemble is proportional to the heat capacity of the system.
Q 7. Using the method of canonical ensemble. calculate the partition function, average
energy and specific heat of a system consisting of N noninteracting quantum
harmonic oscillators and show that these expressions do reduce to the corresponding
classical results in the appropriate classical limit.
Q 8. Consider non-interacting particles subjected to a harmonic potential. Calculate the

canonical partition function
(a) for a single particle
(b) for two distinguishable particles
(c) for two spinless fermions
(d) for two spin-zero bosons
(e) for two spin-1/2 fermions.
Compare the internal energies and entropies in these various cases. Study the limit T → 0, T
→ ∞, and ђ = 0 and interpret the results physically.
Q9.Give an account of Gibb’s canonical ensemble.
Q10.Prove that energy fluctuations in canonical ensemble are related to the specific heat.
5.8.1 Some Multiple Choice Questions for Practice:
Q1. A collection of independent ensembles having the same temperature T, volume V and
chemical potential μ is known as
(a) microcanonical ensemble (b) macrocanonical ensemble
(c) canonical ensemble (d) grand canonical ensemble
Q2. A canonical ensemble provides a model for
(a) an equilibrium system with fixed volume and number of molecules and which exchanges
energy with the outside world
(b) an equilibrium isolated system with fixed volume, number of molecules and energy
(c) an equilibrium system with fixed volume and which can exchange energy and matter with
the surroundings
(d) a system at constant pressure
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Q3. In a Maxwell-Boltzmann system with two states of energies ϵ and 2ϵ, respectively and a
degeneracy of 2 for each state, the partition function is
(a) 2e- 2ϵ/kT (b) 2e- 3ϵ/kT
(c) e-ϵ/kT +e- 2ϵ/kT (d) 2(e-ϵ/kT +e- 2ϵ/kT )
Q4. An ensemble of systems is in thermal equilibrium with a reservoir for which kT

=0.025eV. State A has an energy that is 0·1 eV above that of state B. If it is assumed the
systems obey Maxwell-Boltzmann statistics and that the degeneracies of the two states are
the same, then the ratio of the number of systems in state A to the number in state B is
(a) e+0.25 (b) 1 (c) e -0.25 (d) e- 4
Q5. Consider a system consisting of two distinguishable particles and having two energy
states 0 and ϵ. The partition function of the system is given by
(a) 1+ (b) (c) (1+ )2 (d)
Q6. The classical statistics is valid under the following condition
(a) nλT3 = 1 (b) nλT3 = 0 (c) nλT3<<1 (d) nλT3>>1
Q7. The partition function of a single one-dimensional harmonic oscillator is
(a) (b)
(c) (d)
Q8. The number of microstates Ω(E) for an ideal gas of N monoatomic molecules is related
to energy as
(a) Ω(E) α E3N/2 (b) Ω(E) α EN/2 (c) Ω(E) α EN (d) Ω α E
Q9. The partition function (Z) of a system of particles is related to average energy <E>
(a)<E> = (b) <E> = - (c) <E> = (d) <E> = -
Q10. N distinguishable particles are distributed among three states having energies 0, kBT
and 2kBT respectively. If the total equilibrium energy of the system is 151.23 kBT, the
number of particles of the system is about
(a) 152 (b) 264 (c) 356 (d) 635
Q11. Which of the following statements is true?

(a) In a micro-canonical ensemble the total no. Of particles N and the energy E are
constants while in a canonical ensemble N and temperature are constants.
(b) In a micro-canonical ensemble the total no. Of particles N is constant but the
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energy E is variable while in canonical ensemble N and T are constants.

(c) In a macro-canonical ensemble N and E are constants while in a canonical
ensemble N and T both vary.
(d) In a micro-canonical ensemble N and E are constants while in a canonical
ensemble N is a constant but T varies.
Q12. An isolated system consists of two non-interacting spin (1/2) particles a & b fixed in
space& kept in magnetic field b. Find out the total no. of microstates allowed to the
system
(a) 2 (b) 4 (c) 3 (d) 0
Q13. What is the total energy of the system when one particles is in spin down state.
(a) 0 (b) 2μB (c) -2μB (d) 4μB
Q14. The partition function of two independent (non-interacting) system i and j is given by
(a) Zij = Zi + Zj (b) Zij = Zi – Zj(c) Zij = Zi x Zj (d) Zij = Zi / Zj
Q15. A gas of N non-interacting particles is in thermal equilibrium at temperature T. Each

particle can be in any of the possible non-degenerate states of energy 0, 2ε and 4ε.
The average energy per particle of the gas, when βε<<1, is
(a) 2ε (b)3ε (c) 2ε/3 (d) ε
Q16. Which of the following are the state functions of a grand canonical ensemble?
(a) E,V,N (b) T,V,μ (c) T,V,N (d) E,N,T
Q17. The relative fluctuation in energy of a system in a canonical ensemble is proportional to
(a) 0 (b) N (c) N1/2 (d) N-1/2
Q18. In a canonical ensemble at equilibrium, F is
(a) 0 (b) constant (c) maximum (d) minimum
Q19. In a monatomic gas, the first excited state is only 1.5 eV above the groundstate,
whiletheother excited states are much higher up. The ground state is doubly-
degenerate, while the first excited state has a four-fold degeneracy. If now, the gas is
heated to a temperature of 7000 K, the fraction of atoms in the excited state will be
approximately
(a) 0.07 (b) 0.14 (c) 0.42 (d) 0.3
Q20. A system consists of N weakly interacting subsystems, each with two internal quantum
states with energies 0 and ϵ. The internal energy for this system at absolute
temperature T is equal to
(a) Nϵe-ϵ/kT (b) 3/2 NKT (c) (d)
Q21. Consider an ensemble of quantum particles each of which can be in one of two states of
energy E1 and E2. This system is in equilibrium at temperature T= 300 K. Let N1 and N2
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denote the average number of particles in the two states. If the ratio N2/N1 is 1/e, the
frequency of the radiation corresponding to transition between the two states is approximately
(a) 62X109 Hz (b) 62X1011 Hz (c) 62X1013 Hz(d) 62X1015 Hz
Q22. Two non-interacting particles are distributed in three distinct states. Let Pc be the
probability for both of them in the same state in case particles are distinguishable and Pb the
probability for them to be in the same state in case they are indistinguishable bosons. The
ratio Pc/Pb is
(a) 3/2 (b) 1 (c) 2/3 (d) 1/3
Q23. An isolated system has N non-interacting particles. If each particle can exist in three
states, the entropy of the system according to Boltzmann’s prescription is given by
(a) NkB ln 2 (b) NkB ln 3 (c) kBln(3N) (d) 3kBln(N)
Q24. The classical statistics is valid under the following condition
(b) nλT3 = 1 (b) nλT3 = 0 (c) nλT3<<1 (d) nλT3>>1
Q25. The number of microstates Ω(E) for an ideal gas of N monoatomic molecules is related
to energy as
(b) Ω(E) α E3N/2 (b) Ω(E) α EN/2 (c) Ω(E) α EN (d) Ω α E
Q26. An ensemble is said to be in statistical equilibrium if the phase point density
(a) is zero (b) varies linearly with time
(c) varies inversely with time (d) is time independent
5.9 References:
1. Statistical Mechanics by R. K. Pathria
2. Statistical Mechanics by K. Huang
3. Statistical Mechanics by L. D. Landau and E. M. Lifshitz
4. Fundamentals of Statistical and Thermal Physics by E. Reif
5. Thermodynamic, Kinetic Theory and Statistical Thermodynamics by F. W. Sear and
G. L. Salinger
6. Statistical Mechanics by GeetaSanon
7. Statistical Mechanics by R. H. Fowler
8. Introductory Statistical Mechanics: R. Bowley and M. Sanchez (Oxford Univ.
Press)
9. Statistical Physics: F. Mandl (Wiley)
10. Problems and Solutions on Thermodynamics and Statistical Mechanics: Lim Yung-
Kou
(Sarat Book House)
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11. Elementary Statistical Physics by C. Kittel

12. Statistical Mechanics by R. Kubo
5.10 Suggested Readings:

1. Statistical Physics: Berkeley Physics Course, F. Reif, (McGraw-Hill)
2. An Introduction to Statistical Physics: W.G.V. Rosser (Wiley)
3. An Introduction to Thermal Physics: D. Schroeder (Pearson)
4. Concepts in Thermal Physics: Blundell and Blundell (Oxford Univ. press)
5. Statistical and Thermal Physics: Loknathan and Gambhir (PHI)
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UNIT 6 GRAND – CANONICAL ENSEMBLE
Structure
6.1 Introduction
6.2 Objectives
6.3 Grand – canonical ensemble
6.4 Definition of G.C. partition function
6.5 Deriving the ideal gas equation for G.C.E. for a perfect gas
6.6 Mean Energy for a G.C.E.:-
6.7 Particles Fluctuations in Grand Canonical Ensembles
6.8 Energy Fluctuations in Grand Canonical Ensemble:
6.9 Comparison & Ensembles:
6.10 Summary
6.11Glossary
6.12 Terminal questions
6.12.1 Multiple Choice Questions
6.13 References
6.14 Suggested readings
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6.1 Introduction:
The given unit discuss another important ensemble of statistical mechanics, grand canonical
ensemble which allows the subsystem to exchange both energy and number of particles. It
plays a central role in statistical mechanics where we need an ensemble having neither energy
nor number of particles are fixed. In the case of canonical ensemble, system is considered in
thermal equilibrium with a heat bath (taken as a closed system) while grand canonical
ensemble is taken in contact with both heat and particle bath (open system).
The partition function is used as an important parameter to solve various thermodynamic

quantities for grand canonical ensemble as well. Here, both N and E may take any value from
zero and infinity. To know the actual picture,we have explained energy and particle
fluctuations. In the grand canonical systems, the distribution function and the phase space
both will have dependence on the number of particles in the subsystem so the number of
particles plays very crucial role in grand canonical ensemble.
6.2 Objectives:
Students will understand:
6. About grand canonical.

7. Role of G.C. partition function.
8. Ideal gas equation for G.C.E. for a perfect gas
9. Mean Energy for a G.C.E
10. Energy and particle fluctuations in Grand Canonical Ensembles.
11. Comparison of all ensembles.
6.3 GRAND – CANONICAL ENSEMBLE:

It is a large collection of copies of subsystems. For the macroscopic system, the number of
copies has to be 1024 equivalent to N. The characteristics of grand canonical ensemble are:
(i) The walls are rigid specified by volume.

(ii) They are conducting specified by temperature.
(iii) They are permeable specified by chemical potential.
E constant, T = constant
N constant, μ = constant
V = constant
Therefore, [V, T, μ] are the controlling variables here.
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Let us conrider a subsystem S1 under study. It is immersed in S2 where S2 consists

of heat reservoirs and particle reservoirs. Both together form a system S specified
by (N, V, T). S1& S2 are kept at same temperature and since T is a defined
quantitytherefore both S1& S2 are canonical ensembles.
S2
T
S1
S1 has H(p, q, N1)
S2 has H(p2, Q2, N2) are respectively Hamiltonians.
The restrictions here are:-
N1 + N2 = N S1<< S2
V1 + V2 = V N2>> N1, V2>> V1
These are very strong inequalities which are resulted of reservoir.
and are the phase densities of subsystem S1& S2 respectively.
Probability that there are N1particles in volume V1 with coordinates

p1, q1, n1
Since this subsystem is immersed in S2 the is dependent in the

partition function of S1& S2 but how are they related?
They can’t be directly multiplied. Since we’re not interested in the path of S2 but the
result does not depend upon path. So, we integrate it over the whole space of path.
dp2 dq2
QN is the normalisation factor which we obtained for normalised ρ.
To remove this proportional sign we’ve to do the normalisation of that we divide

it by QN.
dp2 dq2
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Since we know that partition function is a dimensionless quantity which can be

written as
…(a)
N >> N1& V >> V1, expand using Taylor’s expansion
A (N2, V2, T) = A (N – N1, V – V1, T)
= A (N, V, T) - N1
= A(N, V, T) – N1μ + V1P
Now equ. (a) becomes,
Now,
Similarly,
Where is the quantity which is defining that the ensemble is Grand

CanonicalFugacity.
We define z= as the fugacity which tells us that there’s an exchange of matter as

well as energy.
Fugacity is the signature of G.C.E.
This is the phase density of the system.
According to normalisation condition the phase density integrated over entire space
has to be 1.
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This quantity is independent of p, q, N it can be written outside the

summation & integral sign.
6.4 Definition of G.C. partition function:
Our G.C.E partition function is the fugacity multiplied by old canonical partition
function.
Now,
Now, we’ve the following results
ST = kT ln = Canonical
A = -kT ln QN = Micro-canonical
PV = kT ln = Grand Canonical
“A partition function helps us to lead the ideal gas equation”.
e.g., For a perfect gas
ln = N {dimensionless}
Using (A),
PV = NKT Ideal gas equation.
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6.5 Deriving the ideal gas equation for G.C.E. for a perfect gas:-
The value of QN =
for perfect gas,
The G.C. partition function,
By definition,
Hence,
…(1)
Wa’ve found earlier
…(2)
on equating (1) & (2), we’ve
…(3)
# If our G.C.E. is in equilibrium then there must be some mean of N.
By definition,
………(4)
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Diff. w.r.t. z
------(5)
on comparing (4) & (5) we obtain,
…(6)
on using (1) in (6) we obtain
-------(7)
on substituting (7) equ.in (3) we obtain
Where z is independent of while QN is independent of z
PV = NKT
This is equation of state for a perfect gas G.C.E.
6.6 Mean Energy for a G.C.E.:-

To prove <H> =
PROOF:- The mean H is defined by,
We’ve lnΞ=ln
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Diff. w.r.t.
On comparing eqnn (8) & (9) we obtain
<H> =
Now, using this relation we can find out the value of mean energy for perfect gas.
It is the energy for on ideal gas
6.7 Particles Fluctuations in Grand CanonicalEnsemble

We have proved that the energy fluctuations are very small for canonical ensembles.
We’ll prove the same thing for Grand Canonical. Let N(bar) is the total phase
density integrated over entire phase volume.
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when we say, then
− β PV
∞
∫e e − βH
N = ∑ NZ N
dpdq
N =0 N!h 3 N
− βH
∞
= ∑ NZ N ∫e dpdq / N !h 3 N
N =0 e βPV
∞
QN
= ∑ NZ N
N =0 e βPV -------------(10)
Using (A), we’ve
exp
Substitute this in equ. (10)
We define,
Substitute (12) in (11)
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…(13)
We’ve expressed average number of particle in terms of Grand Canonical partition

function.
Now finding the value for <N2>
On Diff. w.r.t.Z, equ. (12) we get
When we Substitute (15) in (14), we obtain
…(16)
Equ. (16) – [(13)]2
…(17)
we ought to simplify equ. (17). For this diff. (13) w.r.t.Z again,
Multiply by z,
…(18)
When we Substitute (18) in (17) we obtain
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<N2> - <N>2 = Z <N> …(19)
= fugacity
Since we know that exponentials and logarithmic are of dimensionless quantity, this
implies that is dimensionless
Z = It is the signature of temperature & exchange of energy.
Equ.(19) in terms of will be
…(20)
So far, we’re doing statistical mechanics. Now we want to go to thermodynamics. In

thermodynamics we replace <N> by its thermodynamic limit N.
…(21)
<N> ~ N Here
We’ve the relation,
Nd = VdP – SdT
( = = Specific volume)
Diff. adore eqn nw.r.t.
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[V = N]
…(22)
Put (22) in (21)
But which is referred to as isothermal compressibility.
Isothermal compressibility is the change in volume w.r.t. pressure keeping

temperature and N constant.
Hence,
The fluctuations of particles for G.C.E. is
…(23)
This again shows that particle fluctuation will vary with times. This result
is similar to the result obtained for C.E.
EQUIVALENCE CRITERIA “All the ensembles are equivalent becauseall have very tiny
energy fluctuations”.
This is equivalence criteria for ensembles.
In equ.(23), χT is an intensive parameter.
V is an Extensive parameter.
Determining particle fluctuation for perfect gas:-
For a perfect gas,
PV = NKT
V=
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To obtain χT differentiate above equation,
for perfect gas
So the fluctuation = …(24)
So for a perfect gas, fluctuations are equal to ordinarily. But there may be
exceptions too like the one which is met in the case of phase transitions.
In these conditions compressibility of a given system can become extremely large as

is evident by an almost flattening of isothermal.
The point in P— V Curve where phase of a substance changes from phase 1 to 2 is

the critical point. Here is very large. So near the critical point, the fluctuations
will be the order of .
Phase – 2
Critical point
P
Phase – 1
V
Far away from the critical points, fluctuations are ordinary. The large fluctuations
near critical point account for the existence of phenomenon like ‘Critical
opalescence’.
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6.8 Energy Fluctuations in Grand Canonical Ensemble:

The mean value of energy is given by,
The denominator term as normalisation constant = 1
…(25)
For G.C.E., the normalised value of
Z N e − βH
ρ ( p, q, N ) = /Ξ
N !h 3N
Substitute ρ and Ξinequ. 25,
On differentiating equ. 26 w.r.t.
using (28) in (27) we obtain
…(29)
Now finding <H2>
On differentiating equ. (28) again w.r.t. we get
This is nothing but N1 of <H2>. Hence we obtain
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…(30)
To find energy fluctuation subtract (30) – (29)2
<H2> - <H>2 = …(31)
This quantity has to be related to a Thermodynamics quantity and since for

canonical ensemble was directly related to CV. Here too we expect the same
relation,
Differentiate (29) w.r.t.
…(31)
On comparing (30) & (31), we’ve
<H2> - <H>2 = …(32)
using the result for canonical ensemble,
Here too, we’ve energy fluctuations,
So, the energy fluctuations are same for the Canonical & Grand Canonical but only
when we’re away from the critical point.
6.9 Comparison of Ensembles:
S.No. MicroCanonical Canonical Ensemble Grand Canonical

Ensemble Ensemble
1. Walls are rigid, Walls are rigid, Rigid, permeable &

impermeable & wall impermeable & diathermal diathermal walls
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insulated.
2. Constant parameter are E, Constant parameter are T, Constant parameter are

V, N V, N T, V, μ
3. No exchange of energy & Only energy can exchange Both energy & particle
particle with surroundings. with surrounding not can exchange with
particle. surroundings.
4. ρ (E )=constant ρ= A exp(-E/kT) ρ= exp(Ω+Nμ-E/kT)
5. Thermodynamic state Thermodynamic state Thermodynamic state

function is S (E, V, N) function is A (T, V, N) or function is ϕ(T, V, μ) or
F PV
6. Partition function is QN (T, V, N) or Ω Ξ(T, V, μ )
Ω(E, V, N)
7. S = k ln Ω A = - kT ln QN(or Ω) PV= kT ln Ξ
6.10Summary:
In the given unit you have discussedanother important ensemble of statistical mechanics, i.e.,
grand canonical ensemble which allows the subsystem to exchange both energy and number
of particles. Here, you focussed on the ensemble where neither energy nor number of
particles are fixed. You have also discussed and analysedin the case of grand canonical
ensemble the system is taken in contact with both heat and particle bath (open system). Here,
in this unit you have discussed the partition function which is used to solve various
thermodynamic quantities for grand canonical ensemble. Here, you have analysed both N and
E may take any value from zero and infinity. You have also learnt how to computeenergy and
particle fluctuations to know the actual picture.Here, in this unit, you have discussed the
grand canonical system, the distribution function and the phase space which have dependence
on the number of particles in the subsystem so the number of particles plays very crucial role
in grand canonical ensemble. At the end, you discussed the comparison of an ensembles.
Finally, to check your progress, multiple choice questions and terminal questions are given in
the unit.
6.11 Glossary:
Grand Canonical Ensemble: System separated by a rigid, impermeable and diathermal
walls.
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Partition Function: Indicate that how the gas molecules of an assembly are distributed
among the different energy levels.
Fugacity: It refers to the activity or the ability of a gas to expand.It is more convenient to
workwith the function which is called fugacity.
Chemical potential (μ): It is the change in the energy of the system as a result of the change
in its number ofparticles, when every other thermodynamical variable that describes
the state of thesystem, such as entropy, volume, etc., is kept constant.It has the units
of energy/molecule, or otherwise stated, for a single species of particlesin a G.C.E. -
<μ<0 for bosons and - <μ< for fermions.
6.12 Terminal Questions:

Q1. Consider a grand-canonical system in which a state with energy 2kT and 21 particles has
a probability of 0.004, while a state with energy 3kT and 20 particles has a
probability of 0.002. Find the temperature of the system if it has a chemical
potential value -1.15 X 10-21 J. Also find out the value of grand partition function of
the system.
Q 2. Find the grand partition function of 1 mole of argon gas at 300 K and 1 atm pressure if it
has a chemical potential value -0.212.
Q 3. Derive an expression for grand canonical partition function and hence grand canonical
distribution.
Q 4. Show that the relative fluctuation in the particle number in a system in grand canonical
ensemble is inversely proportional to the square root of mean particle number in the
system, N.
Q 5.Consider a grand-canonical system in which a state with energy 2kT and 21 particles has
a probability of 0.004, while a state with energy 3kT and 20 particles has a
probability of 0.002. Find the temperature of the system if it has a chemical
potential value -1.15 X 10-21 J. Also find out the value of grand partition function of
the system.
Q 6. Find the grand partition function of 1 mole of argon gas at 300 K and 1 atm pressure if it
has a chemical potential value -0.212.
Q 7.Derive an expression for grand canonical partition function and hence grand canonical
distribution.
Q8.Show that the relative fluctuation in the particle number in a system in grand
canonicalensemble is inversely proportional to the square root of mean particle
number in the system, N.
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Q9.Deduce the equation of state and the entropy of an ideal classical gas in a grand canonical
ensemble. Show that the results are the same as obtained in a canonical ensemble.
What do you infer from this?
Q10.Obtain the expression for the chemical potential μ (T, P) for an ideal gas of non-
relativistic particles in a grand canonical ensemble.
6.12.1 Some Multiple Choice Questions for Practice:
Q1. Which of the following statements is true?

(a) In a micro-canonical ensemble the total no. Of particles N and the energy E are
constants while in a canonical ensemble N and temperature are constants.
(b) In a micro-canonical ensemble the total no. Of particles N is constant but the
energy E is variable while in canonical ensemble N and T are constants.
(c) In a macro-canonical ensemble N and E are constants while in a canonical
ensemble N and T both vary.
(d) In a micro-canonical ensemble N and E are constants while in a canonical
ensemble N is a constant but T varies.
Q2. Which of the following are the state functions of a grand canonical ensemble?
(a) E,V,N (b) T,V,μ (c) T,V,N (d) E,N,T
Q3. The chemical potential μ of a system in grand canonical ensemble is
(a) positive (b) fixed (c) variable (d) 0
Q4. The probability of finding subsystem with n atoms of a perfect gas in grand canonical
ensemble is ( is mean number of atoms present)
(a) (b) (c) (d)
Q5. Two examples of G.C.E. are
(a) an ideal gas (b) a photon gas (c) radioactive decay (d) a paramagnetic solid
Q6. Entropy (s) is related to the Grand partition function by the relation
(a) (b) (c) (d)
Q6. The probability of a microstate in G.C.E. is proportional to
(a) (b) (c) 1/Ω (d)
Q7. In grand-canonical ensemble the expression for probability distribution is
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(a) (b) (c) (d)
Q8. The grand partition function of a system whose canonical partition function is ZN is
(a) (b) (c) (d)
6.13 References:
13. Statistical Mechanics by R. K. Pathria
14. Statistical Mechanics by K. Huang
15. Statistical Mechanics by L. D. Landau and E. M. Lifshitz
16. Fundamentals of Statistical and Thermal Physics by E. Reif
17. Thermodynamic, Kinetic Theory and Statistical Thermodynamics by F. W. Sear and
G. L. Salinger
18. Statistical Mechanics by GeetaSanon
19. Statistical Mechanics by R. H. Fowler
20. Introductory Statistical Mechanics: R. Bowley and M. Sanchez (Oxford Univ.
Press)
21. Statistical Physics: F. Mandl (Wiley)
22. Problems and Solutions on Thermodynamics and Statistical Mechanic: Lim Yung-
Kou (Sarat Book House)
23. Elementary Statistical Physics by C. Kittel
24. Statistical Mechanics by R. Kubo
6.14 Suggested Readings:

1. Statistical Physics: Berkeley Physics Course, F. Reif, (McGraw-Hill)
2. An Introduction to Statistical Physics: W.G.V. Rosser (Wiley)
3. An Introduction to Thermal Physics: D. Schroeder (Pearson)
4. Concepts in Thermal Physics: Blundell and Blundell (Oxford Univ. press)
5. Statistical and Thermal Physics: Loknathan and Gambhir (PHI)
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UNIT 7: QUANTUM STATISTICS
STRUCTURE
7.1 Introduction
7.2 Objectives
7.3. Construction of Symmetric and Antisymmetric Wave functions:
7.4. Average Value and Quantum Statistics
7.5. Fermi – Dirac Statistics
7.6. Bose – Einstein Statistics
7.7. Maxwell Boltzmann Distribution Law of Velocities
7.7.1. Evaluation of constants A and
7.7.2. Number of molecules having velocity within c and c + dc
7.7.3. Experimental proof of Maxwell’s distribution law from the finite breadths if spectral
lines
7.8. A comparison of the three statistics
7.9. Black Body Radiation and the Planck Radiation Law
7.10. Ideal Bose Einstein Gas
7.11. Energy and Pressure of the Gas
7.12. Gas Degeneracy
7.12.1. Degeneracy for molecular hydrogen
7.12.2. Degeneracy for helium
7.12.3. Bose Einstein Condensation
7.13. Ideal Fermi Dirac Gas
7.14. Energy and Pressure of the Gas
7.14.1. Case of slight Degeneracy
7.14.2. Case of Strong Degeneracy
7.14.3. Expression of E and P in terms of Fermi Energy ef
7.15. Thermodynamic Function of Degenerate Fermi – Dirac Gas
7.15.1. Thermal Capacity
7.15.2. Entropy
7.15.3. Helmholtz Free Energy
7.16. Compressibility of Fermi Gas
7.17. Electron Gas
7.18. Summary
7.19. Glossary
7.20. Terminal Questions
7.21. References
7.22. Suggested Readings
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7.1 INTRODUCTION
In Maxwell-Boltzmann statistics or Bose – Einstein statistics, there is no restriction on the
particles to be present in any energy state. But in case of Fermi-Dirac statistics, applicable to
particles like electrons and obeying Pauli Exclusion Principle (no two electrons in an atom
have same energy state), only one particle can occupy a single energy state. In Maxwell –
Boltzmann statistics Particles are distinguishable and only particles are taken into
consideration. In Fermi – Dirac statistics Particles are indistinguishable and quantum states
are taken into consideration. In Bose – Einstein statistics Particles are indistinguishable and
quantum states are taken into consideration. The most important application of Bose-statistics
to electromagnetic radiation in thermal equilibrium, called black body radiation. The
deviations from perfect gas behaviour exhibited by Bose Einstein gas is termed as ‘gas
degeneracy. As the temperature is lowered, beginning at T = T0 the molecules fall rapidly
into the ground state. There is a sort of condensation into this state. This phenomenon is
known as Bose Einstein Condensation. The temperature T0 at which the Bose Einstein
condensation begins depends upon the density of the gas.
7.2 OBJECTIVES
After studying this unit, you should be able to-
• understand Symmetric and AntisymmetricWavefunctions,

• understand and use Fermi – Dirac Statistics
• understand and use Bose – Einstein Statistics
• understand Ideal Bose Einstein Gas
• understand Gas Degeneracy
• understand Ideal Fermi Dirac Gas
• understand Electron Gas
7.3. CONSTRUCTION OF SYMMETRIC AND

ANTISYMMETRIC WAVEFUNCTIONS:
We know that the number of ways in which the indices can be interchanged will give the
number of solutions of the Schrödinger equation of the system. For a two particle system {let
the wave functions be [1, s1: 2,s2]there will be 2! Solutions, similarly for n particle system
there will be n! Solutions. A linear combination of these solutions will be solution of
Schrödinger equation.
As an example, we can consider two and three particle functions with 2 and 6solutions
respectively. The Schrödinger wave equation for two –particle system is
…(1)
The two degenerate solutions are and (2,1). Now the symmetric wave function will
be given by
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…(2)
And ant symmetric function by
…(3)
Where (1,2) and (2,1) are unnormalized wave functions.
Similarly, in case of three – particle system, there are six ways of exchanging the indices of
particles,
And the symmetric combinations are
…(4)
And antisymmetric combinations are
…(5)
Thus, antisymmetric wave function can be built by adding all the functions with even number
of interchanges and subtracting the sum of all those with an odd number of interchanges.
The two particles will be distinguishable from each other if the sum of the probabilities of the
individual wave functions in two states is equal to the probability derived by the symmetrised
wave function i.e., equations (2) and (3).
Thus, |
=| …(6)
Where Re is the real part of {2
This is possible if the overlap of wave functions and is zero or 2 Re

{ =0. In the way when the coordinates (space and spin) of two particles are not
the same between exchange degenerate functions, the interference term 2Re
becomes zero and particle coordinates do not overlap.
7.4. AVERAGE VALUE AND QUANTUM STATISTICS:

The average value of a dynamical quantity (a) in the same is given by
…(1)
Where is the operator corresponding to the dynamical quantity? Remembering that the
denominator integral is unity i.e.,
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…(2)
For example, the average value of the x component of momentum is given by
..(3)
In case of free particle
…(4)
The symmetric and antisymmetriceigen functions give the following results
…(5)
And
It is clear from equation (5) that remains unchanged due to exchange of coordinates
of particles because exchange of particles leaves both and unaltered. In case of
antisymmetric solution, the exchange of coordinates changes the sign of and and
hence the net result is that remains unaltered. We, therefore, conclude that the
exchange of particles leaves the average value unaffected. Therefore; from the quantum
mechanical point of view, the similar particles leaves the average value unaffected.
Therefore; from the quantum mechanical point of view, the similar particles cannot be
distinguished.
7.5. FERMI – DIRAC STATISTICS:

In Maxwell-Boltzmann statistics or Bose – Einstein statistics, there is no restriction on the
particles to be present in any energy state. But in case of Fermi-Dirac statistics, applicable to
particles like electrons and obeying Pauli Exclusion Principle (no two electrons in an atom
have same energy state), only one particle can occupy a single energy state. The distribution
of four particles (a,b,c and d) among two cells x and y, each having 4 energy states. Such that
there are three particle in cell x while one particle in cell y is shown in fig 4.
In this case there will be 4 4 = 16 possible distribution.
We consider a general case. This statistics is applied to indistinguishable particles having half
integral spin. Though the particles are indistinguishable, the restriction imposed is that only
one particle will be occupied by a single cell. The situation of distribution is as follow:
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Energy level e1, e2, …,ei,………..ek
Degeneracy g1, g2, …….,gi,…..gk
Occupation no. n1, n2,……,ni,….nk
So in case of Fermi – Dirac statistics, we have the problem of assigning ni indistinguishable

particles to gi distinguishable levels under the restriction that only one particle will be
occupied by a single level; obviously, gi must be greater than or equal to ni, because there
must be atleast one elementary wave function available for every element in the group.
Thus in Fermi – Dirac statistics, the conditions are:
(1) The particles are indistinguishable from each other i.e., there is no restriction
between ways in which ni particles are chosen.
(2) Each sublevel or cell may contain 0 or one particle. Obviously gi must be greater
than or equal to ni.
(3) The sum of energies of all the particles in the different quantum groups, taken
together, constitution the total energy of the system.
Now the distribution of ni particles among the gi states can be done in the following way:
We easily find that the first particle can be put in any one of the ith level in gi ways.
According to Pauli exclusion principle no more particles can be assigned to that filled state.
Thus we are left with (gi – 1) states in (gi – 1) ways, and soon. Thus the number of ways in
which ni, particles can be assigned to gi states is
gi(gi – 1)(gi – 2)…………(gi – ni + 1),
= …(1)
The permutations among identical particles do not give distribution, and hence such
permutations must be excluded from equation (1), which can be done on dividing it by ni !.
Thus we have the required number as
…(2)
The total number of eigen states for whole system is given by
…(3)
The probability of the specific state being proportional to G will be
…(4)
To obtain the condition of maximum probability, we proceed as follows:
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Talking log of equation (4), we have
…(5)
Using stirling approximation, equation (5) reduces to
Log (i) = + constant …(6)
Differentiating equation (6) with respect to ni, we get
= - log(
=
…(7)
The condition of maximum probability gives
…(8)
Introducing the auxiliary conditions,
…(9)
=
…(10)
And applying the Lagrange method of undetermined multipliers i.e., multiplying equations
(9) by and adding the resulting expression to equation (8), we have
…(11)
Since s can be treated as arbitrary
Log
Or
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Or ni =
…(12)
This is the most probable distribution according to Fermi – Dirac statistics.
7.6. BOSE – EINSTEIN STATISTICS:

In Maxwell-Boltzmann statistics, the particles are distinguishable from each other. Thus if the
two particles interchange their positions or energy state, a new state is generated. But in case
of Bose – Einstein statistics, the particles are indistinguishable. So the interchange of two
particles between two energy states will not produce any new state.
We consider a general case. Suppose the system contains n independent identical particles.
Let these particles be divided into quantum groups or levels such that there are n1,n2,ni…
number of particle in groups whose approximate constant energies are
respectively. n1,n2,ni are the occupation numbers of levels i.e, the numbers of levels i.e, the
numbers of levels i.e., the number of particles that are in that level. Again let there be gi
single particle states in the ith level, and we speak of this as the degeneracy or the statistical
weight of the ith level. This situation is as follows:
Energy level
Degeneracies
Occupation no.
In case of Bose – Einstein statistics, we have the problem of assigning n indistinguishable

particles to gi distinguishable levels when there is no restriction as to the number of particles
that can occupy one level. Gi is also termed as density of states function which gives number
of one particle states per unit energy range.
Thus in Bose – Einstein statistics, the conditions are:
(1) The particles are indistinguishable from each other so that there is no distinction
between the different ways in which ni particles can be chosen.
(2) Each cell or sublevel of iTh quantum state contains 0,1,2…,ni identical particles.
(3) The sum of energies of all particles in the different quantum groups, taken
together, constitutes the total energy of the system.
For this distribution, let us imagine a box divided into gi sections and the particles to be
distributed among these sections. The choice that which of the compartments will have the
sequence, can be made in gi ways. Once this has been done, the remaining (gi – 1)
compartments and ni particles i.e., total particles (ni + gi – 1) can be arranged in any order i.e.,
number of ways of doing this will be equal to (ni + gi – 1)!. Thus the total number of ways
realizing the distribution will be
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(ni + gi – 1)!
…(1)
The particles are indistinguishable and therefore rearrangement of particles will not give rise
to any distinguishable arrangement. There are ni !permutations which correspond to the same
configuration, hence term indicated by (1) should be divided by ni !.
Secondly, the distributions which can be derived from one another by mere permutation of
the cells among themselves, does no produce different states, the term (1) should also be
divided by gi!.
We thus obtain the required number of ways as
Or
There will be similar expressions for various other quantum states. Therefore, the total
number of ways in which n1 particles can be assigned to the level with the energy
and so on is given by the product of such expressions as given below:
G=
= …(2)
According to the postulate of a priori probability of eigen states, we have the probability (i)
of the system for occurring with the specified distribution to the total number of eigen states,
i.e.,
(i) = …(3)
So to obtain the condition of maximum probability, we proceed as follows:
Taking log of equation (3), we have
…(4)
Using the Stirling’s approximation, equation (4) becomes
…(5)
Here we have neglected 1 in comparison to ni and gi as they are very large numbers.
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=- …(6)
The condition of maximum probability gives
…(7)
The auxiliary condition to be satisfied are
…(8)
= . …(9)
Applying the Lagrange method of undetermined multipliers i.e., multiplying equation (8) by
and equation (9) by , and adding the resulting expressions to equation (7), we get
…(10)
As the variations are independent of each other,
or 1+
-1
This represents the most probable distribution of the elements among various energy levels
for a system obeying Bose – Einstein statistics.
7.7. MAXWELL BOLTZMANN DISTRIBUTION LAW OF

VELCOTITIES:
The molecules of a gas do not all move with the same speed due to frequent collisions and so
their velocities vary. The manner in which the molecules of a gas are distributed over the
possible velocities from zero to very high values was first worked out by Maxwell
distribution law. In deriving this law certain assumptions are made which are as follows:
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(1) The gas in an enclosure maintained at a constant temperature assumes a steady

state in which it is in thermal equilibrium with its surroundings. In the steady state the
density of the gas remains uniform on an average throughout the gas.
(2) The velocities along the three coordinate axes are independent if each other, i.e,
the probability of a molecule along any axis will depend on the velocity along that axis
only and not on the other.
(3) The probability of the velocity of a molecule lying between certain limits is a
function of velocity and of the limits considered.
Let u,v,w be the component velocities of a molecule along X, Y, and Z axes, respectively as
shown in fig 7. The number of molecules per c.c. having the velocities lying between u and
u+du can be denoted by nudu. Obviously, nu must be some function of u say n f(u) where n is
the number of molecules per e.e. and f(u) is a function of u to be determined. The probability
that any molecule selected at random will have velocities lying between u and u + du is f(u)
du. Hence the probability that molecule may have its velocity simultaneously between u and
(u + du), v and (v + du), and w and (w + dw) is
F(u)f(v)f(w)du dv dw.
All the molecules whose velocity components lie in the range u and (u + du),v and (v + dv),
w and (w + dw) will be contained in the element of volume du dv dw. The chance that a
single velocity of value c ends in this volume elements du dv dw is given by the assumption
no. (3) as convenience we write the function F(c) as (c2)
f(u) f(v) f(w) du dv dw = (c2) du dv dw
Or f(u) f(v) f(w) = (c2) = (u2 + v2 + w2).

…(2)
Where c2 = u2 + v2 + w2.
The equation (2) determines the nature of distribution law. Differentiating equation (2), we
get
D [f(u) f(v) f(w)= d[(c2)]=0.
D[(c2)] = 0, because for a particular value of c, (c2) is constant and its differentiate will be
zero.
Differentiating , we get
F(u)f(v)f(w) du + f(u) f(v)f(w)dw+f(u)f(v)f(w)dw = 0
Dividing by f(u) f(v) f(w), we get
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…(3)
Again c2 = u2 + v2 + w2 on differentiation gives
0 = 2u dx + 2u dv + 2w dw
Or 0=
…(4)
Where is an arbitrary constant.
Adding equation
du + .
…(5)
According to assumption no. (2), the velocity components are independent of each other,
hence equation (5) can only be satisfied when each the term is separately equal to zero, i.e.
…(6)
Now integrating equation 6(a), we have
Where log A is constant of integration, thus
Log {
Or f(u) = Ae-ku2/2=Ac-u2/u2 where

..(7)
Similarly, writing for other two components
…(8)
= …(9)
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From equation (7), (8) and (9), we have
F(u), f(v), f(w) =

…(10)
Thus the probability that a molecule has the velocity between u and v + du is given by
F(u) du = Acu2/a2 du
…(11)
7.7.1. EVALUATION OF CONSTANTS A AND : The only problem remains

to evaluate the values of the constants A and interms of known quantities.
Let n be the number of molecules per c.c. of the gas with all possible velocities from - to +
. Then the number of molecules per c.c. with velocity components between u and v + du. v
and u + dv and, w and w + dw is n f(u)f(v)f(w) du dv dw. Hence
Or
…(12)
The value of the definite integral
Or A=
In order to consider the value of , we calculate the pressure extorted by the gas on the walls
of the enclosure. Let us consider the case of a molecule moving along X axis with velocity u.
the molecule will collide with the surface of the enclosure which is perpendicular to X axis
and will be reflected back with velocity – u. the change in the momentum of the molecule
will be 2mu. The components v and w do not contribute any thing to the pressure for this
surface. We know that the pressure is equal to the change in momentum suffered by the
molecules striking per unit area of the wall per second if nu be the number of molecules per
unit volume having velocity u, then the number of impacts on an area t will be nu u A t.
Now the pressure P on the wall, average through the time interval t is given by
P A t= .
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Or P = 2m
…(13)
The number of molecules per c.c. having velocity components lying between u and v + du is
given in accordance with equation (11) by
Nu = nf(u) = n
=n
…(14)
Substituting the value of nu from equation (14) in equation (13), we have
P = 2m
Replacing the sign of summation by integration, we have
P = 2m du
= 2mn.
= 2mn.
Because du=
We also know that P = nkT
nkT =
Or
…(15)
Again A=
….(16)
Now, the number of molecules dn having velocity components lying between u and v + du, v
and v + dv, w + dw is given by
Dn = n(fu)duf(v)dvf(w)dw
=nA3
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=n
…(17)
This is known as Mawell’s distribution law of velocities.
7.7.2. NUMBER OF MOLECULES HAVING VELOCITY WITHIN C

AND C + DC.
Let us consider that the molecules having velocity within c and c + dc lie between two
concentric spheres of radius c and c + dc with a common centre O as shown in fig 8.
The volume of the shaded portion will be
On neglecting small terms.
This volume is the same as du dv dw in equation (17). Hence
dn = n
…(18)
This is Maxwell – Boltxmann distribution law for molecular velocities. This can also be put
in the following form
…(18)
This is Maxwell – Boltzmann distribution law for molecular velocities. This can also be put
in the following form.
dnc = n
dnc = n
…(19)
7.7.3. EXPERIMENTAL PROOF OF MAWELL’S DISTRIBUTION LAW

FROM THE FINITE BREADTHS IF SPECTRAL LINES:Theoretically s spectral
line should have zero breadth and uniform intensity but it is observed that when the spectral
line is viewed with the help of high resolving power instruments, it has large number of lines
with varying intensity. This phenomenon is direct consequence of the Maxwell’s law of
distribution of velocities and can be explained as follows:
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We know that when atoms or molecules come from high energy levels to low energy levels,
they emit spectral lines. If the atoms or molecules were at rest, the frequency emitted by them
would have a single frequency. But as the molecules are moving with different velocities,
they emit different frequency. When a source emitting radiation is moving away from the
observer with components velocity v0, its wavelength, as observed by the observer, is given
by
Where the wavelength of radiation corresponding to definite velocity v0 and c is is the

velocity of light.
Now the spectral range corresponding to the molecules having velocities in the range dvo, will
be
d
…(20)
If I is a vector representing the intensity of the spectral line, then multiplying equation (20)
by I, we have
Id …
(21)
The number of molecules having velocities between v0 and v0 + dv0 is given by
If y represented the amount of radiation emitted by a single molecules, then the intensity of
spectral line will be
.
…(22)
From equations (21) and (22), we have
. I
Or =
[because v0 = ( c from equation =
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Or I=
If then constant representing the intensity
Corresponding to molecules of velocity vO
Therefore
I = IO exp. ….(23)
If I is plotted as a function of a graph of the form shown I figure 9 is obtained. It is

observed from the graph that the intensity is decreasing on both the sides of the maximum.
The distance on the wavelength scale in either side where I = Io is known as half width and
is denoted by
When I =
Now from equation (23) we get
exp.
Or .
Solving we have
B2 =
Or b=
Or
….(24)
Where A is atomic weight
Equation (24) shows that the half width is inversely proportional to the atomic weight i.e, the
lighter elements will have greater width. Experimentally it is observed that hydrogen lines are
very much broad while spectral lines due to heavy elements like Hg and Cd are quite sharp.
This verifies the Maxwell’s law of distribution of velocities.
7.8. A COMPARISON OF THE THREE STATISTICS

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A comparison of the three statistics:
Maxwell – Fermi – Dirac Bose – Einstein

Boltzmann
1. Particles are Particles are Particles are

distinguishable indistinguishable indistinguishable
and only particles and quantum states and quantum states
are taken into
are taken into are taken into
consideration.
consideration. consideration.
2. There is no
restriction on the Only one particles No restriction on the
number of may be in a given number of particles
particles in a quantum state. in a given quantum
given state. state.
3. Applicable to Applicable to
ideal gas Applicable to photons and
molecules. electrons and symmetrical
4. Volume in six elementary particles. particles.
dimensional space Volume in phase
is not known. space is known, Volume in phase
( space is know, (h3).
5. Internal energy of
ideal gas The energy at
Even at absolute
molecules at absolute zero is
absolute zero is zero, the energy is
not zero. taken to be zero.
taken as zero.
6. – At high At high
temperatures, Fermi temperatures, Bose-
7. The most distribution Einstein distribution
probable approaches
approaches Maxwell
distribution is
– Boltzmann Maxwell-Boltzmann
given by
distribution. The distribution. The
most probable most probable
distribution is given distribution is given
by by
7.9. BLACK BODY RADIATION AND THE PLANCK

RADIATION LAW:
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The most important application of Bose-statistics to electromagnetic radiation in thermal

equilibrium, called black body radiation. In quantum theory, radiant energy occurs in energy
packets or photons or light quanta of energy hv/c where v is the frequency and c is the
velocity. Photons have zero – rest mass and a spin quantum number of 1: like all particles
with spin 1, they obey Bose – Einstein statistics.
Consider a black body radiation chamber of volume V containing radiation in equilibrium

with the walls at temperature T. Let Uv dv represent the energy density of radiation of
frequency lying between frequencies v and v + dv. The problem is to find out Uv as a
function of T. This was first solved by Max Planck through the hypothesis of linear harmonic
oscillator possessing discrete energy values.
In the momentum space, particles within a small volume are indistinguishable. It

therefore, represents an eigen state. At any instant, all particles, having their momenta
between p and p + dp will lie within a cell of volume 4 Therefore the total number of
eigen states is given by
G(p) dp = …(1)
For a photon
P=
dp = …(2)
Substituting this value of dp in equation (1), we have
G(v)dv = 4 …(3)
There is a duplication of states for the two independent directions of polarization. Therefore
G(v) dv = 8 …(4)
Equation (4) represents the total number of eigen states lying in the frequency range v and v
+ dv. Introducing the result in Bose – Einstein distribution law, we get
Dn= . …(5)
Dn = 8 .
Or …(6)
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The left hand side of equation (6) represents the number of photons per unit volume.
Multiplying by hv, the energy of photon, it gives energy density Uvdv in the specified
frequency range substituting , we have
Uvdv= . …(7)
We know that Moreover, if we put , equation (7) is just Planck radiation

formula, , means, the conditions = 0 should be dropped.
Every process of emission, in nature, results in the creation of photons and similarly, every
absorption process results in the absorption of photons, which may be converted into other
forms of energy. Under these conditions, the restriction is no longer applicable. The
term then, does not involve in undermined multiplier and equation (7) reduces to
Uv dv = …(8)
Which is Planck’s law,
When hv<<kT, the term (ehv/kT-1) = hv/kT and hence equation (8) can be written as
. Uvdv = Rayeligh Zean’s law, …(9)
For hv>>kT, equation (8) becomes
Uv dv = Wien’s Law …(10)
The total energy density is
on substituting =x
= bT4 where b = …(11)
Equation (11) is Stefan – Boltzmann law.
7.10. IDEAL BOSE EINSTEIN GAS
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In Bose – Einstein distribution, we consider a system of identical (indistinguishable),

independent, no interacting particles of integral spin (bosons) that have symmetrical wave
function.
7.11. ENERGY AND PRESSURE OF THE GAS:

Consider a prefect Bose – Einstein gas of n bosons. Let these particles be distribution among
states such that there are n1,n2,n3 no of particles In quantum state whose approximate constant
energies are e1,e2,e3 respectively.
For a perfect Bose – Einstein gas, consisting of material particles, the formula for the most
probable distribution is
Ni = …(1)
Where gi is measure of degeneracy. Since the gas is ideal the interaction between the particles
is assumed to be negligible so that the energy may be regarded as entirely translational in
character. The results thus obtained will be applicable to a monatomic gas. Equation (1) can
be written as
Ni = . …(2)
Where; for convenience, we have put D = e .
We know that for particles in a box of ‘normal size’, the translational energy levels are
closely spaced so that we can integrate over phase space instead of summing over particle
states. The number of particle states g(p)dp lying between momentum p and p + dp is
determined from
Giving …(3)
Where g1 (=2s +1) is the spin degeneracy factor (arising due to the spin s1, of the particle).
Equation (2) can now be written as
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Since , we can write number of particles lying between energy range and
from equation (4), i.e.
Where has been substituted
Let us put x=
Dx= .
So that we express dn(e) =gi
From thermodynamically properties of diatomic molecules, we note that translational

partition function is
Z2 =
It follow then dn =
On integration total number of particles is given by
N=
= …(6)
And energy E=
= kT
= . kT …(7)
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We shall now evaluate the integrals of equation (6) and (7). For a Bose Einstein system
should be positive otherwise the value of
Would become negative at sufficiently low energies which is physically unacceptable. When
a is positive D > 1. The integral of equation (6) is
And similarly, integral of equation (7) is
…(9)
Using relation (8), we get from equation (6),
…(10)
And from equations (9) and (7),
E= …(11)
Substituting value of giZI from equation (10) in equation (11), we get
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= …(12)
If we take only first term in equation (10) then
D=
Which, when substituted in equation (12), gives
E=
For one gram, molecule of the gas, since nk = R, we write energy expression as
E=
= …(13)
Where n has been substituted for bracketed terms.
7.12. GAS DEGENERACY

The pressure of a gas can be calculated from the relation.
P= …(14)
Which means, in order to calculate pressure, we must first set up a relation in energy E and
volume V.
Let us consider a particle of mass m enclosed in a container of volume abc. The wave
function of the particle must satisfy the Schrödinger equation
. …(15)
Where V* (x,y,z) represents the potential energy of the particle which for single particle in
the box should be zero. Therefore
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- . …(16)
To solve this equation let us write in terms of variable.
, …(17)
So that this equation (16) assumes the form
- …(18)
Which means
…(19)
We consider first equation (19), i.e.,
The general solution of which can be written as
…(20)
In which the constant A, B and C are to be determined by the application of boundary

conditions.
We know that the probability of finding the particle at any point (x,y,z) is equal to
| at the point. Therefore which is a function of x – coordinate only, would
determine the probability of the particle being found somewhere along the X- axis. Since the
walls (x = 0 and x = a) is equal to zero.
Thus X(x) = 0 at
Applying X(x) = 0 at x = 0, to equation (20), we get
0 = A sin C,
Giving sin C = 0 or C = 0.
Now applying X (x) = 0 at x = a to the same equation we find
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0 = A sin (Ba + C)
Giving
Ba + C = rx where rx, is an integer.
Or B= since C is zero.
Therefore equation (20) finally assumes the form
X(x) = A sin …(21s)
Similarly, for the remaining two equation of relation (19), we can find
Y(y) = A sin …(21b)
Z(z) = A sin …(21c)
Substituting (21a), (21b) and (21c) in equation of relation (19), we find
Ex= ,
Ey= ,
Ez= ,
Giving E = Ex + Ey + Ez
, .
If container is cubic, a = b = c = l,
So that E=
Further for cube of side l, volume is
V = l3
So that l2 = V2/3
And taking =r2. …(22)
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Thus we have established a relation between energy and volume. Now from equation (14),
we obtain the pressure of ideal Bose gas, thus
P=-
=-
Which is in agreement with classical result, Using equation (13), pressure of an ideal Bose
gas is
P=
= …(23)
We find from equation (13) and (23) that there are deviations in energy and pressure from the
values for ideal gas behaviour. An additional factor n occurs which can be held responsible
for these deviations. The deviations from perfect gas behaviour exhibited by Bose Einstein
gas is termed as ‘gas degeneracy’. The gas degeneracy is obviously a function of 1/D. Also
…(24)
This shows that for particles of small mass at low temperature and small volume or high
pressure, the gas degeneracy will be more marked. The deviations due to gas degeneracy are
rather small as compared to those due to Van der Waal’s forces and that is why it is not
possible to observe this effect under normal conditions.
7.12.1. DEGENERACY FOR MOLECULAR HYDROGEN:Molecular

hydrogen obeys Bose – Einstein statistics. The boiling temperature of hydrogen at
atmosphere pressure is 20.38K which is the lowest temperature at which hydrogen can exist
as a gas. For molecular hydrogen.
N = 6.02 1023
M=2 1.67 10-24 gram. (Mass of proton)
V = 1400 e.e. (Molar volume)
K = 1.38 10-16erg/K
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And h = 6.62 10-37 erg. Sec.
Spin statistical weight
Gx = (2s + 1)
= 1 for paraform
= 3 for orthoform
Taking gx = 1,
We find that
= 0.84 10-2.
So that pressure
P=
= .
We note that it differs slightly from PV = RT, the perfect gas equation.
Further, , we conclude that gas degeneracy in the case of molecular
hydrogen is small and impossible to be observed.
7.12.2. DEGENERACY FOR HELIUM: Helium also obeys Bose – Einstein

statistics. For helium molecular weight is two times that of hydrogen, it can exist in gaseous
states at much lower temperatures ( the boiling point being 4.2K at atmosphere pressure) and
the molar volume is 345e.e. we find
Which is appreciable as compared to the case of molecular hydrogen and that is why there is
some possibility of observing gas degeneracy in the gas of helium.
7.12.3. BOSE EINSTEIN CONDENSATION

We have
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We note that at temperature nearing zero i.e, T = 0,1/D will have a large value which means
the degeneracy will be more marked. It means the gas will deviate highly form its perfect gas
behaviour. The reason is as follows:
While arriving at the Bose – Einstein distribution, we have assumed that, because of the
closeness of the energy levels, we could always replace the discrete distribution by a
continuous distribution giving.
I.e. we have changed summation into integration. As long as the change in occupation
number, ni from state to state is small compared to the number of particles in the state, this is
a valid procedure. If, however, the temperature in an ideal Bose gas is lowered to zero, the
particles will begin to crowd into a few levels and the above condition will be violated. This
means that when we are working at low temperatures, we must be careful in replacing the
summation by integration. Now, from equation (5) the number of particles lying between
energy range and is given by
Where g(
We note that for ground state while actually it should have been unity
because there is one state at Therefore above distribution fails to give the number of
states at e = 0, while state, called the ground state, is very important at low temperatures
because a large number of particles occupy it at such temperatures. We further note that if e
=0, the above distribution holds good ass g (e) = 0. Therefore above distribution can still be
applied for all states except ground state which should be treated separately. For a single
state, we write
For ground state e1 = e0 = 0 and gi = 1. Therefore, number of particles in ground state.
For ground state and gi = 1. Therefore, number of particles in ground state
Therefore, for the total number of particle states we write
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From equation (6) Value x is e/kT,
And Zt = V.
Further n = n0 + gizt F3/2 (a),

…(26)
Where F2/3(
…(27)
When
So that F2/3(0) =
= 2.612
Let T = T0 when or D = 1 so that that from equation (10) we get
N = gi (Zt) T– To
= gs(Zt)t=to F3/2 (0).

…(29)
Therefore gs(Zt)t-to F.=3/12
Putting the value gs from equation (26) in equation (26), we find
N = n0 + n
= no + n
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= no + n
Or n = n – n0 + n
From equation (27), we find that with D > 1. F3/2(a) will have less value than when D = 1,
i.e.
F3/2/ (a) <F3/2(0)
When
This means that n given by equation (29) acquires maximum value when a = 0. Thus for a = 0
the maximum number of particles occupying states ground state is given by
N=n … (30)
And the rest of the particles given by
N0 = n – n = n [I – T/T0)3/2]. … (31)
When T = T0
Must condense into ground state.
Thus from equation (31) we note that, as the temperature is lowered, beginning at T = T0 the
molecules fall rapidly into the ground state. There is a sort of condensation into this state.
This phenomenon is known as Bose Einstein Condensation. The temperature T0 at which the
Bose Einstein condensation begins depends upon the density of the gas. If we consider liquid
helium to be a gas, we would obtain a value of about 3.14K for T0.
It is found experimentally that liquid helium does undergo a rather unusual transition at
2.19K below this temperature, the liquid helium displays the properties of a super fluid. It is
generally, agreed that this transition in liquid helium is associated with a Bose – Einstein
conservation.
For temperatures above T0 – must decrease in order to keep n < (n0 + n)< n. But if
becomes significantly different from zero, then no becomes very small. We, therefore, assume
for temperatures above T0 that n0 = 0. Therefore for T >T0
N0 = 0.
And n = n = giF3/2(a).
From equation (29) when n = n1 we find that
N=n
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Or F3/2(1) =F3/2(0) . …
(32)
Which when substituted in the expression for n given the total number of particle states equal
to
N = giZi 3.162 …
(33)
7.13. IDEAL FERMI DIRAC GAS

In Fermi – Dirac distribution, we consider a system of identical, independent, non –
intersecting particles sharing a common volume and obeying ant symmetrical statistics so that
the spin is half/integral and then according to the Pauli principle, the total wave function is
ant symmetrical on interchange of any two particles.
7.14. ENERGY AND PRESSURE OF THE GAS:

Participants of Fermi – Dirac assembly are the particles with half integral spin like electrons,
protons and neutrons. The general expression for the most probable distribution in energy for
the Fermi – Dirac gas is,
Ni =
Which, on putting D = , if of form
Ni = …
(1)
Since in the denominator factor + 1 occurs, need not be restricted to positive value only but
may assume negative value as well unlike Bose Einstein gas.
The number of one particle states lying between momentum p and p + dp is determined from
Gp = gs
Giving gpdp = gs …
(2)
Where gs = (2s + 1) is the spin degeneracy factor (Arising due to the spin s1 of desertions).
Since and dp = the number of states in the energy range between

e and (e + de) will be, on using equation (2),
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G(e)de = gs = gs
…(3)
Where g (e) is termed as density of states function.
From equation (1) and (3) we get the number of particles in the energy range between e and
(e + de), as
Dn(e) = gs
…(4)
Where we have substituted
. Let us put x = e/kT
And dx = de/kT
Equation (4) then becomes
Dn = gs
= gs.
From the thermodynamic properties of diametric molecules, we note that translational

partition function is
It follows then
Therefore total number of particles
… (5)
And energy E= .
= kT. … (6)
We shall evaluate the integral in equation (5) and (6) for both values of , i.e, when a is
positive and again when it is negative. When s is positive, D is greater than one and the
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condition so obtained is referred to as slight or weak degeneracy. Condition corresponding to

a negative i.e, D less than one is referred to as strong degeneracy.
7.14.1. CASE OF SLIGHT DEGENERACY:

For this case, integral in equation (5) can be expressed as
So that total number of particles is given by
Further the integral in equation (6) can be solved as follows:
So that the total energy is given by
… (8)
Putting the value of gsZt from equation (7) into equation (8), we get
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Which assumes the form … (9)
After putting D = from equation (7) which has been approximated only upon the first
term of the expansion.
The pressure of the ideal Fermi gas can now be obtained by the relation
…(10)
In which nk = R, for one note of the ideal Fermi gas.
We infer from equations (9) and (10), that ideal Fermi gas deviates from perfect gas
behaviour and this deviation, as we know, is called degeneracy, obviously, degeneracy is a
function of or e-x. Smaller is the value of D or greater the value of more marked will be
the degeneracy.
7.14.2. CASE OF STRONG DEGENERACY:
When a is large and negative e-a>>1 or D is much less than one. This increases the value of
and hence the degeneracy will become more prominent. Further
From equation (7),
= … (11)
Which shows that a gas will be highly degenerate at low temperature and high density ( we
shall discuss this case of strong degeneracy at two temperature ranges. Firstly when T = 0 i.e,
at absolute zero and secondly when temperature is above absolute zero but degeneracy is still
considerably high i.e., D is still less than unity.
(i) At T = 0: From equation (11) we note that when T = 0, D = 0, so that equation (5)
assumes the form
N= .
Since D = 0, we can replace the upper limit by 1/D. Therefore
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N= = … (12)
Which gives
With
We find that . … (13)
Which is a measure of degeneracy of the ideal Fermi gas at T = 0.
From equation (6) we can obtain the energy of the Fermi gas ar absolute zero. Putting E as E0
and D = 0 in equation (6) we get
E0 = kT.
Where E0 is termed as zero point energy of a highly degenerate Fermi gas. Replacing again
the upper limit by 1/D and solving the interval, we arrive at.
Which, on using equation (13), becomes
… (14)
And the corresponding zero point pressure will be
From equation (14) and (15) we find that a highly degenerate Fermi Dirac gas would have a
residual zero point energy and pressure – the so called zero point pressure – even at the
absolute zero of the temperature, quite unlike a Bose Eisntein gas where all the particles are
condensed to the ground state with e = 0 at T = 0.
7.14.3. EXPRESSION OF E AND P IN TERMS OF FERMI ENERGY EF:
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From equation (3), the total number of energy states lying between 0 and specified value ef
can be obtained as
Gf = gs .
Or Gf = …(15a)
Further, in Fermi Dirac distribution, not more than one particle is to be occupied by a given
cell which is also obvious from
Ni = .
Which gives ni = gi since at T = 0, D = 0.
Therefore taking gf = ni, we write
N=
Or ef= …(16)
Where the quantity ef is called the Fermi energy and it represents the energy of the highest
level filled at T = 0K for the given assembly.
From equation (13) and (16), we find that
…(17)
And from equation (14) and (16), zero point energy is
…(19)
If we define the Fermi temperature as
Equation (17) becomes
From which we conclude that gas is degenerate when T <<Tf.
(ii) At T above absolute zero but D << 1: From eq. (4), the number of particles lying in the
energy range between c and (c +de)is
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We have seen in chapter 6 art 6, 10, that
But from equation (16).
Ef =
Giving
So that dn (e) =3n
= … (20)
Which gives the total number of particles, as
N (e) = ….(22)
To evaluate the integrals of equations (21) and (22) we should solve the integral of the type
I=
Where is a simple function of e such that if e = 0. Such integrals can be

expanded using the method of Taylor’s series expansion i.e,
… (23)
Where etc denote the first, third … differentiate of the function we shall confinite
ourselves only to the first two terms of the expansion (23).
Observing equation (32), we write
So that and
Therefore we write equation (20) as
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Or
Giving
=1– +… … (24)
Taking into consideration only the first two terms of the expansion. We can write
Neglecting higher order terms.
Thus
We make here a crude approximation by putting in the second term on right side of
above equation
(25)
Now putting equation (25) in equation (24), we get
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… (26)
Neglecting the higher order terms.
Since equation (22) involves the integral of the same type, we write, using the expansion
(23), as
E=
Where
So that and
Therefore, E=
Putting the value of from equation (26) and from eq. (25), we get
Leaving from the very beginning, the higher order terms.
Therefore E= … (27)
This is the approximate energy of a highly degenerate gas. The corresponding pressure will
be
= … (28)
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7.15. THERMODYNAMIC FUNCTION OF DEGENERATE

FERMI – DIRAC GAS:
Energy and pressure expressions have already been obtained [equations (27) and (28)].
7.15.1. THERMAL CAPACITY, CV:At such low temperatures, the gas will have a
thermal capacity given by
CV =
=nei
= …
(1)
We have
CV =
From equation (2), we infer that heat capacity per particle would be small for large deviation
( and small particle mass m. If the particles are electrons then equation (2), gives electronic
specific heat. We note that it is proportional to the absolute temperature and therefore, at
ordinary temperatures, the contribution to the specific heat of metals due to electrons would
be negligible as compared to the contribution due to the atoms since atomic specific heat is
proportional to T3 while at very low temperatures electronic specific heat will be significant.
7.15.2.Entropy:Entropy S. can be obtained from
Using equation (2), If we use equation (1),
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…(3)
7.15.3. HELMHOLTZ FREE ENERGY:
neif
Which is an expression for Helmholtz free energy.
7.16. COMPRESSIBILITY OF FERMI GAS

Here we shall derive an expression for compressibility of Fermi gas at absolute zero. The
energy at absolute zero is given by
E0 =
Now P0 =
= (where h = h/2
So,
Or
Now compressibility is given by
It has been observed that the compressibility of alkali metals is close to the compressibility of
an electron gas.
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7.17. ELECTRON GAS:-

A metal can be considered to be composed of a system of fixed positive nuclei and a number
of mobile electrons referred to as electron gas.
To study the properties of an electron gas at low temperatures in the region T – 0 we shall
revise the earlier discussion. For electrons s = so that gz = 2s + 1 = 2, and therefore from
equation
10 (=n/V)=
…(1)
EO = nei
…(2)
Further, from equations (15a) and (16) of, we get
Gf = ni
Which means that in the limit T – 0 each one of the states is occupied fully upto the energy
level ef whereas all the states above this energy level are empty.
The degeneracy factor of an electron gas for electrons m = 9.1 10-28gm, and g = 2, we get
Taking a typical metal of atomic weight 100 and density 10 so that volume of gm. Atom be
10e.e. and the number of electrons, assuming one free electrons per atom, is 6.02 .
Then,
Which means degeneracy is sufficiently high. It shows clearly that for electron gas, the
classical statistics is not valid and can be applied only at temperatures of the order of 105K
(because only then D will approach unity). Therefore at low and other ordinary working
temperatures, it is necessary to use Fermi – Dirac statics to study the electron gas in the
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metals. At low temperatures electrons contribution to the specific heat of metals is given by
equation
But we have
So that CV= .
Using the above value of 1/D, we find that
CV =
Putting nk =R gas constant
= 1.978 cal deg-1 mol-1
= 2 cal deg-1 mol-1
We get electronic specific heat
CV =
= 1.5 cal/gm. Atom.
Pressure of the electrons gas can be obtained from equation
P0 =
= using g = 2.
For a metal of atomic weight 100 and density
P0 – 105 atoms
Which means at normal temperature, the pressure of the gas is sufficiently high.
7.18. SUMMARY
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In this you unit you have learnt about construction of symmetric and antisymmetric wave
functions. In this you unit you have learnt about different types of distribution method such as
Maxwell-Boltzmann, Bose – Einstein and Fermi-Dirac statistics. You also learnt about
different type of particle such as Boson and Fermions. You also learnt about Ideal Bose
Einstein Gas and Ideal Fermi Dirac Gas. . You also learnt about Gas Degeneracy as an
example for molecular hydrogen and for helium. You also learnt about Compressibility of
Fermi Gas and electron gas.
7.19. GLOSSARY
Degeneracy - different states of equal energy
Condensation - change of the physical state of matter from the gas phase into the liquid
phase
7.20. TERMINAL QUESTIONS

1. What do you understand by a symmetric and anti-symmetricwave functions?
2. What is gas degeneracy explain it for He atom?
3. Explain the following-
(i)Bose Einstein statistics (ii) Electron gas
4. Compare all the three statistics Bose Einstein, Fermi Dirac and Maxwell Boltzmann. Show
that at low temperature all the statistics give same result.
5. What do you understand by Fermi Dirac statistics? Find an expression for it?
6. What do you understand by Bose Einstein statistics? Find an expression for it?
7. What do you understand by electrons gas? Show that at normal temperature, the pressure
of the gas is sufficiently high.
7.21. REFERENCES
1. Statistical mechanics by R.K. Pathria
2. A textbook of Statistical mechanics by Suresh Chandra and Mohit Kumar Sharma
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UNIT 8 PHASE TRANSITIONS

Structure
8.1 Introduction
8.2 Objectives
8.3 Phase Transition
8.3.1 Classifications of Phase Transitions
8.4 Theory of Yang and Lee
8.4.1 Theorem 1
8.4.2 Theorem 2 (First order Phase Transitions)
8.5 Second Order Phase Transitions
8.6 Ising Model: Phase Transitions of Second Kind
8.7 Ising Model: One Dimensional
8.8Weiss Theory of Ferro-magnetism
8.9 Landau’s Theory
8.10Viril Equation of States
8.11 Summary
8.12 Glossary
8.13 Terminal Questions
8.14 Answers
8.15 References
8.16Suggested Readings
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8.1 INTRODUCTION
In the previous unit, you have studied the quantum statistics of identical
particles, symmetric and anti-symmetric wave functions, degeneracy, Bose-Einstein
condensation etc. The phase transition is very important term in statistical mechanics. In this
unit, you will study about phase transitions, various types of phase transitions, Ising model,
Landau’s theory, Weiss theory of ferro-magnetism, Virial equation of states etc.In the unit,
you will also study the significance of phase transitions.
8.2 OBJECTIVES
After studying this unit, you should be able to-
• understand phase transitions

• understand Ising model
• know about ferro-magnetism
8.3PHASE TRANSTION
In this section, let us know about phase transition. The term phase transition
(or phase change) is most commonly used to illustrate transitions between solid, liquid, and
gaseousstates of matter, as well as plasma in rare cases. A phase of a thermodynamic system
and the states of matter have uniform physical properties. During a phase transition of a given
medium, certain properties of the medium change, often discontinuously, as a result of the
change of external conditions, such as temperature, pressure, or others. For example, a liquid
may become gas upon heating to the boiling point, resulting in an abrupt change in volume.
The measurement of the external conditions at which the transformation occurs is termed the
phase transition. Phase transitions commonly occur in nature and are used today in many
technologies.
Phase transitions occur when the thermodynamic free energy of a system is non-
analytic for some choice of thermodynamic variables (cf. phases). This condition generally
stems from the interactions of a large number of particles in a system, and does not appear in
systems that are too small. It is important to note that phase transitions can occur and are
defined for non-thermodynamic systems, where temperature is not a parameter. Examples
include: quantum phase transitions, dynamic phase transitions, and topological (structural)
phase transitions. In these types of systems other parameters take the place of temperature.
For instance, connection probability replaces temperature for percolating networks.
At the phase transition point (for instance, boiling point) the two phases of a
substance, liquid and vapor, have identical free energies and therefore are equally likely to
exist. Below the boiling point, the liquid is the more stable state of the two, whereas above
the gaseous form is preferred.
It is sometimes possible to change the state of a system diabatically (as opposed to

adiabatically) in such a way that it can be brought past a phase transition point without
undergoing a phase transition. The resulting state is metastable, i.e., less stable than the phase
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to which the transition would have occurred, but not unstable either. This occurs in
superheating, supercooling, and supersaturation, for example.
Liquid
Solid A Gas
Figure 1: Phase diagram
Simple substances are capable of existing in phases of three types- solid, liquid and
gas. The three lines, in a phase diagram (Figure 1) separating these phases are called phase
equilibrium lines. The common point A, where three lines meet, is called ‘triple point’ at this
unique temperature and pressure all three phases can coexist in equilibrium with each other.
Point C is the critical point at which liquid gas equilibrium line ends. The change in volume
ΔV between liquid and gas has then approached zero; beyond C there is no further phase
transition since there exists only one ‘fluid phase’
The following figure 2 shows the curve of constant temperature T for an equation of
state p = p (v, t) describing the fluid state of a substance. In the shaded region mixture of two
phases can coexist along the horizontal line. If at the given temperature T, the pressure is
sufficiently low so that p < p1, the curve gives in, correspondingly, a unique value of v. There
exists then a well defined single phase. Here the slope of the curve ≤ 0 as necessary for the
stability condition. Also | | is relatively small so that compressibility of this phase is large
as in a case of a gaseous phase.
If at the given temperature T the pressure is sufficiently high that p > p2, then there
exists again a single phase with a unique value of v. The stability condition ≤ 0 is again
satisfied but | | is relatively large so that compressibility of this phase is relatively small as
would be the case for a liquid phase.
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p2
p1
O v1 v2
Figure 2: Fluid state representation
For intermediate pressure range p1< p < p2. At the given temperature T, there are
three values of v for each pressure p. In addition, the stability condition ≤ 0 is violated in
the region v1< v < v2 where curve has positive slope. This is the region where mixture of two
phases coexist along the horizontal line.
8.3.1 Classifications of Phase Transitions

Let us discuss the classifications of phase transitions. There are two classifications of
phase transitions. Now we shall discuss these phase transitions one by one.
Ehrenfest classification
Paul Ehrenfest classified phase transitions based on the behavior of the

thermodynamic free energy as a function of other thermodynamic variables. Under this
scheme, phase transitions were labeled by the lowest derivative of the free energy that is
discontinuous at the transition. First-order phase transitions exhibit a discontinuity in the first
derivative of the free energy with respect to some thermodynamic variable. The various
solid/liquid/gas transitions are classified as first-order transitions because they involve a
discontinuous change in density, which is the (inverse of the) first derivative of the free
energy with respect to pressure. Second-order phase transitions are continuous in the first
derivative (the order parameter, which is the first derivative of the free energy with respect to
the external field, is continuous across the transition) but exhibit discontinuity in a second
derivative of the free energy. These include the ferromagnetic phase transition in materials
such as iron, where the magnetization, which is the first derivative of the free energy with
respect to the applied magnetic field strength, increases continuously from zero as the
temperature is lowered below the Curie temperature. The magnetic susceptibility, the second
derivative of the free energy with the field, changes discontinuously. Under the Ehrenfest
classification scheme, there could in principle be third, fourth, and higher-order phase
transitions.
Though useful, Ehrenfest's classification has been found to be an incomplete method

of classifying phase transitions, for it does not take into account the case where a derivative
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of free energy diverges (which is only possible in the thermodynamic limit). For instance, in
the ferromagnetic transition, the heat capacity diverges to infinity. The same phenomenon is
also seen in superconducting phase transition.
Modern classifications
In the modern classification scheme, phase transitions are divided into two broad
categories, named similarly to the Ehrenfest classes:
First-order phase transitions are those that involve a latent heat. During such a
transition, a system either absorbs or releases a fixed (and typically large) amount of energy
per volume. During this process, the temperature of the system will stay constant as heat is
added: the system is in a "mixed-phase regime" in which some parts of the system have
completed the transition and others have not.[4][5] Familiar examples are the melting of ice or
the boiling of water (the water does not instantly turn into vapor, but forms a turbulent
mixture of liquid water and vapor bubbles). Imry and Wortis showed that quenched disorder
can broaden a first-order transition. That is, the transformation is completed over a finite
range of temperatures, but phenomena like supercooling and superheating survive and
hysteresis is observed on thermal cycling.
Second-order phase transitions are also called "continuous phase transitions". They
are characterized by a divergent susceptibility, an infinite correlation length, and a power law
decay of correlations near criticality. Examples of second-order phase transitions are the
ferromagnetic transition, superconducting transition (for a Type-I superconductor the phase
transition is second-order at zero external field and for a Type-II superconductor the phase
transition is second-order for both normal-state—mixed-state and mixed-state—
superconducting-state transitions) and the superfluid transition. In contrast to viscosity,
thermal expansion and heat capacity of amorphous materials show a relatively sudden change
at the glass transition temperature which enables accurate detection using differential
scanning calorimetry measurements. Lev Landau gave a phenomenologicaltheory of second-
order phase transitions.
Apart from isolated, simple phase transitions, there exist transition lines as well as
multicritical points, when varying external parameters like the magnetic field or composition.
Several transitions are known as infinite-order phase transitions. They are continuous
but break no symmetries. The most famous example is the Kosterlitz–Thouless transition in
the two-dimensional XY model. Many quantum phase transitions, e.g., in two-dimensional
electron gases, belong to this class.
The liquid–glass transition is observed in many polymers and other liquids that can be
supercooled far below the melting point of the crystalline phase. This is atypical in several
respects. It is not a transition between thermodynamic ground states: it is widely believed that
the true ground state is always crystalline. Glass is a quenched disorder state, and its entropy,
density, and so on, depend on the thermal history. Therefore, the glass transition is primarily
a dynamic phenomenon: on cooling a liquid, internal degrees of freedom successively fall out
of equilibrium. Some theoretical methods predict an underlying phase transition in the
hypothetical limit of infinitely long relaxation times. No direct experimental evidence
supports the existence of these transitions.
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8.4 Theory of Yang and Lee

In this section, we shall discuss theory of Yang and Lee. This theory is a
mathematical description of phase transition which becomes possible because we can
characterize a phase transition as the manifestation of a certain singularity or discontinuity in
the equation of state-
…..(1)
where , v the specific volume – a parameter independent of the total volume, V. Z is

the grand partition function, z is the fugacity defined by-
z =
…..(2)
It can be made clear that both P and v are analytic functions of z in a region of the
complex z-plane that includes the real positive axis. For any finite value of V, there are no
real positive roots of the equation Z (z, V) = 0. This means that regarded as a function of
complex variable z the zeros of the function Z(z, V) are distributed in the complex z-plane
but are never on the positive real axis. But as V increases, number of zeros increases and their
positions may move about in the complex z-plane. In the limit as V→ ∞ some of the roots
may converge towards the positive real axis. In this limit, the equation of state is given by-
…..(3)
We may now hope to find possible singularities in the equation of state that
can be identified as phase transitions. Suppose that in the complex z-plane there is a region R
which contains a segment of the real positive z-axis and which is free of zeros of Z(z,v) for
all V. It is reasonable to expect that as V→∞, the stability condition holds. In that
case region R represents a single phase. If there are many overlapping regions R, each region
may be expected to correspond to a phase of the system. Therefore, to study phase transition,
we study the behavior of equation of state as z goes from one R region to another. In this
reference, there are two theorems- theorem 1 and theorem 2. Let us discuss these two
theorems.
8.4.1 Theorem 1
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exists for all z> 0. This limit is independent of the shape of the
volume V and is a continuous, non-decreasing function of z.
It is assumed that as V →∞ the surface area of V increases no faster that V2/3.
8.4.2 Theorem 2 (First order Phase Transitions)
This theorem tells about first order phase transition. Let us discuss theorem 2.
Let us consider a region R in the complex z- plane that contains a segment of the
positive real axis and contains no root of the equation Z(z, v) = 0 for any V, then for all z in
the region R the quantity V-1 log Z(z, v) converges uniformly to a limit as V→∞. This limit is
an analytic function of z for all z in region R.
Now let us discuss possible behaviors of equation of state consistent with above
theorems. Suppose region R includes complete positive z-axis as shown in figure 3. Then the
system is always in a single phase and if a phase of the system be defined as the collection of
thermodynamic states corresponding to values of zlying in the single region R then P and v in
O Z-Axis
Zeros of Z (z, v)
Figure 3: Region R that is free of zeros of Z (z, v)
the equation of state can be written as-
β P (z) = …..(4)
On the other hand, if a zero of Z (z, V) approaches the point z0 on the real positive z-axis as
V→∞ then there will be two regions R1 and R2 in which theorem 2 holds separately [as
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shown in figure (4) ]. At z= z0, P(z) must be continuous though its derivative dP(z)/dz may be
discontinuous. An example of such a behavior is shown in figure(5), where for z < z0 system
possesses one phase while for z > z0, the other phase.
R1 R2
Z-Axis
Z= Z0
Zeros of Z (z, v)
Figure 4: Two regions R1 and R2each free of zones of Z (z, v)
P(z) 1/v(z)
Discontinuity
z z
z0 z0
P(v)
b a
vb va
Figure 5: Equation of state of a system with two phases connected by a first order
transition
At z = z0, 1/v(z) is discontinuous. Thus we obtain first order phase transition between two
phases (first order derivative is discontinuous at z = z0). This is shown infigure 5 exhibiting a
discontinuity at z = z0. If dP(z)/dz is continuous at z = z0 but second derivative d2P(z)/dz2 is
discontinuous then we have second order phase transition. In the next section, we shall study
second order phase transitions in detail.
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8.5 Second Order Phase Transitions

Now in this section, let us study second order phase transition. If in the above example,
dP(z)/dz is continuous at z = z0 but second derivative d2P(z)/dz2 is discontinuous then we
have second order phase transition. The transition of liquid helium I into helium II for the
common isotope of He4 is a well known example of second order phase transition. The other
example of second order phase transition is the transition from non-ferromagnetic state to a
ferromagnetic state. When a magnetic material changes its state from a ferromagnetic
material to a paramagnetic material, the symmetrical arrangement of the elementary magnetic
moments undergoes a discontinuity jump and the symmetry changes. Therefore, the second
order phase transitions are usually associated with the abrupt changes in various properties
characterizing the symmetry of the body. Thus, we can conclude that phase transition of
second is continuous in the sense that the state of the body changes continuously but
discontinuous in the sense that symmetry of the body changes discontinuously.
8.6 Ising Model: Phase Transitions of second kind

In this section, we shall study Ising Model and phase transitions of second kind. In phase
transition of second kind, the state of the body changes continuously. Let us consider a
ferromagnetic substance like iron or nickel. In the absence of any external magnetic field,
some of the spins of the atoms become spontaneously polarized in the same direction, below
the curie temperature Tc. This generates a macroscopic magnetic field. The spontaneous
magnetization, so generated, vanishes if temperature is greater than Tc because then thermal
energy makes some of the aligned spins to flip over. In this way, the spinsget oriented at
random and no net magnetic field is produced. As the curie temperature is approached from
both sides the specific heat of the metal approaches infinity. The transition from non-
ferromagnetic state to the ferromagneticstate called the phase transition of second kind is
associated with some kind of change in the symmetry of the lattice; for example, in
ferromagnetism the symmetry of the spins is involved. In Ising model, the system considered
is an array of N fixed points called lattice sites that form an n-dimensional periodic lattice (n
= 1, 2, 3). Associated with each lattice site is a spin variable si (i=1,2,…….N) which is
anumber that is either +1 or -1. There are no other variables. If si = +1, the ith state is said to
have spin up and si = -1, it is said to have spin down. A given set of {si} specifies a
configuration of the whole system whose energy is defined to be –
EI{si} = - …..(5)
where the subscript I stands for Ising and the symbol < i, j> denotes a nearest-neighbor pair
of spins. There is no distinction between < i, j> and <j, i>. εij is the interaction energy and μH
is the interaction energy associated with an external magnetic field H. For spontaneous
magnetization, H = 0. μij and H are given constants. We apply the model to the case is
isotopic interaction so that all εij have the same value ε. We write for energy-
EI{si} = -
Or EI{si} = - …..(6)
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The case ε> 0 corresponds to ferromagnetism and the case ε< 0 to anti-ferromagnetism. In the
former case neighbouring spins tend to be parallel while in the latter case they tend to be
antiparallel. In equation (6), the sum over <i, j> contains γN/2 terms where γ is the number of
nearest neighbours of any given site (coordination number of the lattice). Thus in the Ising
model equation (6), geometry of the lattice enters through γ and interaction energy εij.
We consider only the case of ε > 0. The partition function can be written as-
Z= …..(7)
where each Si ranges independently over the values ± 1. Hence, there are 2N terms in the
summation. Thermodynamic functions e.g. internal energy, heat capacity can be calculated
through Z but it is enormously not easy to calculate the partition function. There are so many
approximate methods for this.
8.7 Ising Model: One Dimensional

One-dimensional Ising model is a chain of N spins, each spin interacting only with its two
nearest neighbours. Ignoring external magnetic field H, we can write for the energy of the
configuration specified by as-
EI = - ε
…..(8)
Applying periodic boundary condition,
Si+1 = Si
The partition function is-
Z=
=
…..(9)
where, we have used exp (css, ) = ec (ss, = 1) = ( cosh c + ss, sinh c)
e-c (ss, = -1)
which holds because ss, can only be + 1 or – 1. The expansion of products in equation (9)
gives a sum of terms, each of which is a product of the form-
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(cos βε) N-s ( sinh βε)s ( Si Si+1 ……….. Sj Sj+1)

…..(10)
Figure 6: N Ising spins arranged in a ring
Graphically these terms can be displayed by thick (corresponding to factor SS’ sinh βε ) and
thin (corresponding to factor cosh βε) links forming the ring. It can be argued that non-zero
terms are the first term ( cosh βε)N and the last term (sinh βε)N so that-
Z = 2N
= 2N ( cosh βε)N , for N >>1.

…..(11)
because cosh βε >> for βε = ≠ ∞ for T ≠0.
Therefore, Helmholtz free energy for the system is written as-
F = -kT = - N k T
…..(12)
We can write the energy of the system as-
E = = - N ε tanh
…..(13)
Obviously, in this case, there is no transition temperature. Therefore, one dimensional Ising
model cannot be ferromagnetic.
8.8 Weiss Theory of Ferro-magnetism

To explain the phenomenon of ferromagnetism, Weiss proposed a hypothetical
concept of ferromagnetic domains. He postulated that the neighboring atoms of the
ferromagnetic materials, due to certain mutual exchange interactions, form several numbers
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of very small regions, called domains. Weiss theory of ferromagnetism is also called domain
theory of ferromagnetism. It has following points:
• The domains which are aligned approximately along the direction of the applied
magnetic field grow in size at the cost of unfavorably oriented domains, that is, those align
opposite to the field direction get reduced. In other words, the domain boundaries move so as
to expand the favorable domains.
• Also domains rotate and orient themselves in the direction of the external magnetic
field.
In the presence of the weak external field, the magnetization in the material occurs mostly by
the process of domain growing, but in the strong magnetic field the material is magnetized
mostly by the process of domain alignment. When the magnetic field is removed, the domain
boundaries do not recover their original positions and thus, the material is not completely
demagnetized, but some residual magnetism remains in it.
When the magnetizing field is weak; the magnetic polarization of the sample changes as a
result of motion of domain walls. In this process, domains whose axes are parallel or at a
small angle with the magnetic field grow at the expenses of those which are unfavorably
oriented [ Figure 7 (b)]. This growth is reversible so long as the magnetic field stays very
small; if we turn the magnetic field off, the magnetization will return to zero. This part of
magnetization curve is marked in figure 7. In this region of field strength, magnetization is
not a continuous process but takes place in a series of infinite small discrete steps because of
the irregularities in crystal structure which might be due to strains, dislocation, impurities,
dirt and imperfections. This can be shown by inserting the sample in a coil connected to an
amplifier and loudspeaker, If the magnetic field surrounding the speaker is slowly increased,
individual clicks are heard in a speaker which represent small discrete flux-increments. This
is known as Barkhausen effect.
H =0 H1 H2 H3
H2> H1 H3> H2
(a) (b) (c)

(d)
Figure 7: Magnetic domain
The step like nature of the magnetization curve, it viewed on a macroscopic scale, would
appear as shown in figure 8.
B c
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Figure 8: Step-like nature of magnetization curve
Now, let us explain the curve as shown in figure 8. It is obvious from the
curve that near the knee of the curve, we have moved all domains walls and magnetized each
crystal of the sample in its best direction [figure 7 (c)]. The use of a very powerful magnetic
field then causes domain rotation, during which the magnetic polarization assumes a direction
parallel to the applied magnetic field as in figure 7 (d) and this accounts for the curve c in
figure 8. In this way, the motion of the domain walls accounts for portion of the
magnetization curve below the knee of the curve while domain rotation accounts for the
portion above the knee. If the applied magnetic field is removed, there is little change in the
domain structure so that the magnetization remains quite high until reverse magnetic fields
are applied and this accounts for the hysteresis.
Weiss introduced the idea of domains. Weiss assumed that in ferromagnetic

substances, there must exist a molecular field produced at any point by all the neighboring
molecules which is proportional to the magnetization vector M, i.e.
Molecular field M
=αM
In this way, the actual magnetic field acting upon a dipole is the sum of the
applied field H and that arising from the presence of neighbouring dipoles i.e. effective
magnetizing field may be expressed as-
Hi = H + α M
…..(14)
assuming that the contributions of the neighbouring dipoles to the effective field is
proportional to the magnetic polarization. This form of dependence is not self evident but
was chosen by Weiss because it leads to the desired form. The factor of proportionalityα is
known as the molecular field coefficient.
When the molecular field coefficient α is positive, the possibility of

spontaneous magnetization of these domains arises even in the absence of external magnetic
field. The value of such a spontaneous magnetization due to internal molecular field may be
found by putting the external field H to zero in the above relation. This reduction can also
determine the condition for spontaneous magnetization.
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Since the domains are assumed to obey the general theory of paramagnetism,
therefore, we can write-
M/MS = coth x- 1/x …..(15)
Where x = m Bi /kT = μ0 m Hi/ kT = μ0 m ( H + α M)/ kT

…..(16)
MS = nm represents the saturation value of I, n = total number of molecules or atoms per unit
volume of the substance, m = magnetic moment of the atom, k = Boltzmann’s constant and
rest symbols have their usual meaning.
Since the dipole aligning force Hi = H + α M depends partially upon the

degree of alignment , any increase in magnetic polarization will result in an increase in the
aligning force, which in turn, will result in an increase in the magnetic polarization etc. Thus,
magnetization once induced, could occur without the application of an external field. Let us
now examine the conditions under which such spontaneous magnetization could take place.
When the applied field is zero i.e. H = 0 the equation (16) reduces to-
x = μ0 m α M /kT = α (μ0 M MS)/ (n k T)

…..(17)
Since MS = n m, therefore equation (17) may be expressed as-
M/MS = (n k T ) x / (α μ0MS2)
…..(18)
Equations (15) and (18) constitute a pair of equation which may be solved simultaneously for
two variables M/MS and x.
Figure (9) represents the curves corresponding to two equations (15) and (18)
; equation (15) gives the Langevin’s curve 1 while equation (18) gives a straight line 2
passing through the origin whose slope is equal to (n k T)/ (α μ0 MS2) and increases with T.
These two curves intersect at origin O and at another point A so that solutions of these
equations are-
M/MS = 0 or AP
M/MS A
(1)
(2)
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Figure 9: Curves corresponding to equations (15) and (18)
ButM/MS = 0 cannot be a true solution, since there is spontaneous magnetization by

supposition. Hence, the only correct solution is M/MS = AP i.e. non-zero point of
intersection of two curves. We can conclude that as the slope of the straight line curve (2)
increases with T, if this slope coincide or becomes greater than that tangent at the origin to
the Langevin curve, then there will not exist a non- zero point of intersection of two curves;
hence the spontaneous magnetization will not occur. On the other hand, we know that the
slope of the tangent at the origin to the Langevin curve is 1/3, since M / MS = x/3, when x is
small. From these considerations, the condition of stable spontaneous (nkT/ αμ0MS2) < 1/3
magnetization may be expressed as T < (αμ0MS2/ 3nk).
But αμ0MS2/ 3nk = Θ, the Curie point ; hence T < Θ. Hence, below Curie
point Θ a state of magnetization is possible without an applied field. The degree of
spontaneous magnetization depends on the temperature approaching the saturation value as
the temperature approaches absolute zero, since the non-zero point intersection will take
place at-
M/MS =1 for x = μ0 m Hi /k T → ∞ when T = 0.
From the relation Θ = αμ0MS2/ 3n k, we may note that the absolute value of
Θ is greater if the magnetic moment of domains (MS) and molecular field coefficient α have
large values. If Θ is greater, greater will be the transition temperature. Incidentally, the Curie
temperature for ferromagnetic elements Fe, Co and Ni are 7700C, 11310 C and 3580 C
respectively. Above the Curie point Θ, the spontaneous magnetization no longer occurs and
ferromagnetic properties disappear and the substance becomes paramagnetic. At
temperature, not too near the transition point the Curie-Weiss law is obeyed.
There are some shortcomings of Weiss theory. This theory could not explain
why and how internal fields between the molecules of ferromagnetic materials possess such
large values and why the linear relationship expressed by Curie-Weiss law breaks near the
Curie point.
8.9 Landau’s Theory

The Landau theory of phase transitions is based on the idea that the free energy can
be expanded as a power series in the order parameter m. ... The parameters that are input into
the form are also used to plot the temperature dependence of the order parameter, the free
energy, the entropy, and the specific heat.
At a first-order phase transition, an order parameter like the magnetization is

discontinuous. At a critical point, the magnetization is continuous – as the parameters are
tuned closer to the critical point, it gets smaller, becoming zero at the critical point. However,
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experiments on the liquid-gas phase transition and on three-dimensional magnets (and exact
computations like Onsager and Yang’s for the two-dimensional Ising model) both point that
even though the magnetization is continuous, its derivative is not. In mathematical language
the magnetization is a continuous function, but not analytic. For example, at h → 0+ in the
Ising magnet in 3d, the magnetization vanishes as T → Tc from below as-
M ∝ (Tc − T)0.315 T < Tc, 3d Ising …..(19)
where all evidence suggests that the exponent is not even a rational number. In the two-
dimensional Ising model, the exact computations give
M ∝ (Tc − T)1/8 T < Tc, 2d Ising …..(20)
Even though the exponent is rational, the function is decidedly not analytic.
This was (and remains) very strange compared to most of physics. The partition function of
any finite system is a continuous function of all the parameters. Thus if any non-analyticity
occurs, it must be a property of taking an infinite number of degrees of freedom. We usually
take this limit out of necessity – it’s not possible to follow 1023 (or for that matter even 100)
particles individually, even with a computer. Even Monte Carlo simulations can at best do
thousands of particles. Now we’re saying that at a critical point, the limit we so desperately
need to take is suspect. Since dimensionalanalysis arguments rely on analyticity, these are
also suspect. Of course, at the end of the day all formulas are dimensionally consistent. What
happens though is that at and near critical points, a hidden parameter is necessary for
describing the physics. This 1 parameter arises from the short-distance physics – even if we
are interested in describing long-distance physics, critical physics necessarily involves all
length scales! To understanding how that happens requires considerable effort – this is why
Wilson won a Nobel Prize, and why many others provided essential ingredients. The first
major step toward theoretical understanding came from Landau, and his approach is still
called today Landau theory, or Landau-Ginzburg theory. Sometimes it is also called
GinzburgLandau theory, because the two wrote a paper applying these ideas to
superconductivity.
Landau theory is an effective theory for what happens at and near the critical
point. The experimental fact that very different systems can have quantitatively identical
critical behavior suggests that one does not need to worry about every single detail of the
system to understand this behavior. We gave an explicit example of how if we ignored many
details of the liquid-gas system, we could obtain a lattice gas that was identical to the Ising
model. This provides a suggestion as to why the universality occurs; Landau theory is the
first serious attempt to derive a theory that will describe the critical behavior quantitatively.
Landau theory only describes the universal behavior of a system; by construction, it cannot
for example give non-universal numbers like the value of Tc for a given system. But one of
the miracles of critical behavior is that it can give precise results for the universal behavior. It
is important to emphasize that Landau’s original (genius) idea for an effective theory was and
remains completely correct. It’s just that the naive computations do not give the right
answers. To be precise, in the next section, I will describe how the effective theory can arise
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from taking a specific approximation called mean-field theory. This approximation breaks
down in low dimensions, for reasons explained by Ginzburg. But one of the beautiful aspects
of Landau theory is that it makes deriving the consequences of mean-field theory really easy.
The whole point is that the effective theory is independent of the details, so one can just guess
what it is based on the symmetries and degrees of freedom of the system. Landau theory is an
effective theory of the order parameter. To be precise about it, one first decides what the
appropriate order parameter is to describe the phase transition. In one phase, the order
parameter is non-vanishing, in another it vanishes. In a ferromagnetic spin system, this very
naturally is the magnetization . In an antiferromagnetic systems, there are a variety of
possibilities, such as the staggered magnetization, which describes a transition away from
N´eel order. Another possibility is that no local order parameters change values at a phase
transitions. One example of such is known as “topological order”, where only non-local order
parameters characterize the transition. One of Landau’s insights was an easy way to see how
the non-analyticity arises. The basic assumption of Landau theory is that at a fixed value of
the order parameter, the free energy as a function of the order parameter is analytic, both in
the parameters such as J and T, and in the order parameter itself.
8.10Viril Equation of States

Because the perfect gas law is an imperfect description of a real gas, we can
combine the perfect gas law and the compressibility factors of real gases to develop an
equation to describe the isotherms of a real gas. This Equation is known as the Virial
Equation of state, which expresses the deviation from ideality in terms of a power series in
the density.
…..(21 )
where B is the second virial coefficient, Cis called the third virial coefficient, etc.
The second Virial coefficient represents the initial departure from ideal-gas behavior
and describes the contribution of the pair-wise potential to the pressure of the gas. The third
virial coefficient depends on interactions between three molecules. The jth virial coefficient
can be calculated in terms of the interaction of j molecules in a volume V. The second and
third Virial coefficients give most of the deviation from ideal up to 100 atm. It is
important to note
that value of the virial coefficients are temperature dependent.
Because = …..(22)
We can rearrange the viril equation to-
…..(23)
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Or = R T Z
…..(24)
Where Z =
…..(25)
The virial Equation of state is a model that attempts to describe the properties of a
real gas. If it were a perfect model, the viril equation would give results identical to those of
the perfect gas law as the pressure of a gas sample approached zero. For the viril equation to
collapse to the perfect gas law, all of the viril coefficients would need to have a value of zero
at the same temperature. This is an unlikely occurrence but because the second term in the
viril equation B = is the largest termin the equation ( , we can focus
on the temperature at which B is zero. This temperature is called the Boyle temperature (TB)
and it is the temperature at which the repulsive forces between the gas molecules exactly
balance the attractive forces between the gas molecules. The equation (23) can be written as-
…..(26)
In which case, Z =
…..(27)
Then,
…..(28)
Example 1: Calculate the pressure exerted by 0.275 moles of N2 gas in a 0.500 L flask at 273
K using the first two terms of the series of viril equation. The value of B for N2 at 273 K is –
10.5 cm3/mole.
Solution: Given, V = 0.275 L = 5 × 10-4 m3
B = - 10.5 cm3/mole = - 1.05 × 10-5 m3/mole
= 1.82 × 10-3 m3/ mole
Now applying equation
P × 1.82 × 10-3 m3/ mole = 8.3145 Joule/mole K × 273 K
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Or P = 1240 k Pa
For a perfect gas, the pressure would be 1250 kPa.
Self Assessment Question (SAQ) 1:Calculate the pressure exerted by 0.35 moles of CO2 gas
in a 0.5 L flask at 273 K. The value of B for CO2 at 273 K is -149.7 cm3/mole.
8.11 Summary
In this unit, you have studied about phase transition and its classification.
You have studied that during a phase transition of a given medium, certain properties of the
medium change, often discontinuously, as a result of the change of external conditions, such
as temperature, pressure, or others. You have discussed Ehrenfest and Modern classifications
in detail. In the unit, you have also discussed the theory of Yang and Lee, Ising model etc. To
explain the phenomenon of ferromagnetism, you have studied Weiss theory of ferro-
magnetism in the unit. Weiss postulated that the neighboring atoms of the ferromagnetic
materials, due to certain mutual exchange interactions, form several numbers of very small
regions, called domains. You have also discussed and analyzed the Landau’s theorem and
Virial equation of states. The Landau theory of phase transitions is based on the idea that the
free energy can be expanded as a power series in the order parameter m. ... The parameters
that are input into the form are also used to plot the temperature dependence of the order
parameter, the free energy, the entropy, and the specific heat.To present the clear
understanding and to make the concepts of the unit clear, some solved examples are given in
the unit. To check your progress, self assessment questions (SAQs) and terminal questions
are given in the unit.
8.12 Glossary
Generate - create, make, cause
Macroscopic- Visible to naked eye without use of any instrument.
Spontaneous- natural, impulsive
Magnetization- Difference between the ratio of the magnetic induction to the permeability
and
the magnetic intensity. It represents departure from randomness of magnetic
domains.
Align- line up, ally
8.13 Terminal Questions

1. What is meant by phase transition? Explain first order and second order phase transitions.
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2. What are the phase transitions of first and second kind? Discuss Ising model for phase
transitions of second kind.
3. Give a brief account of one-dimensional Ising model.
4. What are phase transitions? How Yang and Lee theory may be used to explain phase
transition?
5. Explain Landau theory of phase transitions.
6. Write notes on-
(a) Virial equation of states (b) Weiss theory of ferro-magnetism
7. Discuss Weiss’s modification to explain the phenomenon of spontaneous magnetization in

ferromagnetic substances.
8. Establish Virial equation of states. What are the advantages of this equation of states?
8.15 Answers
Self Assessment Questions (SAQs):
1. Given, V = 0.5 L = 5 × 10-4 m3 , B = - 149.7 cm3/mole = - 1.497 × 10-4 m3/mole, n = 0.35

moles
= 1.42 × 10-3 m3/ mole
Now applying equation
P × 1.42 × 10-3 m3/ mole = 8.3145 Joule/mole K × 273 K
Or P = 1420 k Pa
For a perfect gas, the pressure would be 1420 kPa.
8.16 References
1. Statistical Mechanics; S.L. Gupta, V. Kumar, Pragati Prakashan, Meerut
2. Statistical Mechanics; Satya Prakash, Kedar Nath Ram Nath, Delhi
3. Statistical Mechanics; R K Patharia, Academic Press
8.17 Suggested Readings
1. Elementary Principles in Statistical Mechanics; Josiah Willard Gibbs, Charles Scribner’s
Sons
2. Course of Theoretical Physics; Lev Landau, Addison-Wesley, Pergamon Press
3. Introduction to Modern Statistical Mechanics; David Chandler, Oxford University Press
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MSCPH504

Department of Physics (School of Sciences)

Unit writing and Editing

Dr. Kamal Devlal 1. Dr. Mahipal Singh

Course Title and Code : Statistical Mechanics (MSCPH504)

Unit Block and Unit title Page

1 Statistical Mechanics at a Glance 1

6 Grand – canonical ensemble 111

7 Quantum Statistics 130

UNIT 1 Statistical Mechanics at a Glance

1.3 BASIC CONCEPTS OF STATISTICAL MECHANICS

Self Assessment Question (SAQ) 1: A boxcontains 4 chocolates and 4 icecreams. What is

1.3.2 Basic rules of probability

Hence, the chance of getting 2 kings is 1 in 221.

probability of getting two blue marbles with replacement is . Notice the

1.3.3 Probability distribution

Thus, the required probability distribution can be summarized as

1.3.4 Distinguishable and indistinguishable particles

1.4 PHASE SPACE

system. These three mutually perpendicular momentum coordinates

Let us consider a system of molecules or particles. If each molecule is specified in phase

coordinates. This space of 6N dimension is called gamma-space or Γ space. In Γ space each

Figure1.2: Random trajectory in phase space.

Example 1.10: Phase space diagram of an oscillator.

Figure1.3: Phase space diagram of an oscillator.

1.5 WHAT IS AN ENSEMBLE?

For example, consider anotherphase space trajectory of a harmonic oscillator of energy

1.6 DENSITY OF STATES

Using the relation between energyEand momentum p, i.e.,

1.7 ERGODIC HYPOTHESIS: Equality of ensemble average

and , respectively.The ensemble average takes into consideration the contribution

1.8 POSTULATE OF EQUAL A PRIORI PROBABILITY

Microscopic These are defined in terms of microscopic variables, like position

2. Statistical Mechanics, R. K. Pathria, Butterworth-Heinemann, Oxford.

3. Fundamentals of Statistical and Thermal Physics, F. Reif, McGraw-Hill.

4. Statistical Mechanics, K. Huang, Wiley India.

5. Fundamentals of Statistical Mechanics, B. B. Laud, New Age International

1.12 SUGGESTED READINGS

2. Statistical Thermodynamics and Kinetic Theory, C. E. Hecht, Freeman, New York.

3. Statistical Mechanics: an Introduction, D. H. Trevena, Elsevier.

4. Statistical Physics, L. D. Landau andE. M. Lifshitz, Elsevier.

1.13 TERMINAL QUESTIONS

1.13.2 Long Answer Type

1.13.3 Numerical Answer Type

1.13.4 Objective Answer Type

4. What is the probability of getting a sum 9 from two throws of a dice?

Probability of zero doublets = (5/6) (5/6) (5/6) = 125/216.

Probability of three doublets = (1/6) (1/6) (1/6) = 1/216.

4.As we know that Hamiltonianrepresents the total energy of a system, i.e.,

similar arguments as in the previous question. The trajectory is shown as

1.14.2 Terminal Questions: Numerical type questions

Hence, the probability of drawing two queens in succession is

P(two queens in succession) = P(first card queen) P(second card queen)

2. P(Three aces in a row) = P(Card 1 is an ace) P(card 2 is an ace) P(card 3 is an ace).

1.14.3 Objective type questions

2. Correct option is (b), probability.

3. Correct option is (d), negative.

4. Correct option is (c), 1/9.

5. Correct option is (b), 5/7.