STATISTICAL MECHANICS
Mechanics
Theory of interaction between particles,
Theory of force
Classical Mechanics
Deals with classical particles
(dimension > atomic dimension
<dimension of planet )
Quantum Mechanics
Deals with quantum particles
(dimension ≤ atomic
dimension)
Statistical Mechanics
Deals with large no of particles,
paticles may be classical, may be
quantum
In classical mechanics we deal with 2/3 particles or a system having few no. of d.o.f and then
generalize the theorem for large no. of particles.
In quantum mechanics we deal with maximum 3 dimensional system.
We cannot explain the behavior of a system having large number of particles using classical or
quantum mechanics. We cannot explain specific heat of solid, behavior of thermal radiation,
thermodynamic behavior of a system by classical or quantum mechanics.
Statistical mechanics is special branch of physics which explains these behaviors using theory of
probability.
When we consider a system of large no. of particles, we cannot say the exact behavior of
microparticles, rather we can say the most probable behavior of microparticles. Stat mech is the
special branch of Physics the aim of which is to relate the macroscopic behavior in terms of
microscopic behavior.
Radioactive decay  statistical in nature
In radioactive decay we cannot say which particle decay first at which time, rather we can say av.
No. of particles decay at a given instant of time.
Concept of temperature  statistical in nature
Temperature of a system is the measure of K.E due to random motion of constituent particles
about the center of mass of the system. In a macrosystem all the micro particles are in continual
motion and undergo collision among themselves. So K.E of all the micro particles are changing
continually. In equilibrium we can measure temperature of the system. No motion, no
temperature. Concept of temperature is invalid for single particle system or system having few
no. of particles.
When we are to explain a phenomenon involving large no. of particles theory of probability is
introduced.
What is the necessity of theory of probability
If we want to analyze the motion of a macrosystem we have to know the motion of
microparticles. We have to know the coordinates and momenta of each microparticle  x, y, z,
px, py, pz  6 parameters. We have to construct equation of motion of each particle and to solve
the eq. of motion we have to know the initial conditions of position and momentum.
If we consider 1 mole of gas. There are 6.023×1023 no. of atoms. So we have to know
6×6.023×1023 no. of variables.
Very tedious job
To avoid these type of calculation theory of probability is introduced.
Statistical Mechanics predicts the motion of macrosystem without the knowledge of initial
conditions of microparticles.
Statistical Mechanics
(Theory of large no of particles)
Quantum Statistical Mechanics
Classical Statistical Mechanics
(Theory of large no of quantum particles)
(Theory of large no. of classical particles)
Bosons (indistinguishable
particles)
(i) Obey Bose-Einstein
Statistics
(ii) example: (a) Photon,
Phonon
(b) atomic neuclus having
even mass no.--C12, He4,
H2,...
Fermions (indistinguishable
particles)
(i) Obey Fermi-Dirac Statistics
Classical particles
(distinguishable particles)
(ii) Example (a) All
fundamental Particles (e, p,
n, neutrino, positron, ... )
(i) Obey Maxwell-Boltzmann
Statistics
(b) atomic nuecleus having
odd mass no.---H1, He3, Li3,....
(iii) No idea of spin
(iii) have integral spin
(iii) have 1/2 integral spin
(iv) have symmetric wave
function
(iv) have anti-symmetric
wave function
(v)Do not obey Pauli's
Exclusion principle
(v) Obey Pauli's exclusion
Principle
(ii) Example: Molecules of gas
(iv) No idea of wave function
(v) Do not obey Pauli's
Exclusion Principle
Pauli’s Exclusion Principle: No two particles can occupy one energy state.
Example:
Consider a system of 2 particles in 3 energy states.
Obtain the distribution of 2 particles in 3 energy states for (i) classical particles, (ii) Bosons, (iii)
Fermions.
Classical (distinguishable
particles)
Bosons (indistinguishable
particles)
Fermions
particles)
Particles -- A, B
Particles – A, A
Particles – A, A
0
E
2E
0
E
2E
0
E
2E
A
B
-
A
A
-
A
A
-
B
A
-
A
-
A
A
-
A
A
-
B
-
A
A
-
A
A
B
-
A
AA
-
-
-
A
B
-
B
A
AA
-
AA
AB
-
-
AB
-
AB
-
-
(indistinguishable
Possible distributions = 3
Possible distributions = 6
Possible distributions = 9
Let us define a parameter
𝛾 (𝑜𝑐𝑐𝑢𝑝𝑎𝑛𝑐𝑦 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦) =
3⁄9
𝛾𝑐𝑙𝑎𝑠𝑠𝑖𝑐𝑎𝑙 = 6⁄9 =
1
𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡ℎ𝑎𝑡 2 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑖𝑛 1 𝑒𝑛𝑒𝑟𝑔𝑦 𝑠𝑡𝑎𝑡𝑒
𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡ℎ𝑎𝑡 1 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒 𝑖𝑛 1 𝑒𝑛𝑒𝑟𝑔𝑦 𝑠𝑡𝑎𝑡𝑒
3⁄6
, 𝛾𝐵𝑜𝑠𝑜𝑛𝑠 = 3⁄6 = 1,
2
0⁄3
𝛾𝐹𝑒𝑟𝑚𝑖𝑜𝑛𝑠 = 3⁄3 = 0
𝛾𝐵𝑜𝑠𝑜𝑛𝑠 > 𝛾𝑐𝑙𝑎𝑠𝑠𝑖𝑐𝑎𝑙
Conclusion: (i) Bosons have the greater tendency to bunch together than classical particles
(ii) Fermions have the greatest tendency to remain apart
Q1.
Name the statistics obey by each of the following particles
(i)neutrino, (ii) Alpha particle, (iii) muon (μ meson )(s=1/2), (iv) Photon (S=1) (v) neutron, (vi) 𝜋meson(pion)(S=0), (vii) electron, (viii) ideal gas molecules, (ix) proton, (x) He atom, (xi) positron,
(xii) He3, (xiii) C12, (xiii) O16
Q2. Which statement is correct?
(a)
(b)
(c)
(d)
Bosons and Fermions are indistinguishable particles
Bosons and Fermions obey Pauli’s Exclusion Principle
Bosons and Fermions are distinguishable particles
Bosons and Fermions have the tendency to bunch together
Q3. Which statement is wrong?
(a)
(b)
(c)
(d)
MB statistics deals with distinguishable particles
FD statistics deals with indistinguishable particles
BE statistics deals with indistinguishable particles
Bosons are subjected to Pauli’s exclusion Principle
Q4. The number of possible arrangement of 2 fermions in 3 cells is
(a) 9,
(b) 6,
(c) 3,
(d) 1
Q5. The no. of possible arrangement of 2 fermions in 2 cells is
(a) 4,
(b) 3,
(c) 2,
(d) 1
Q6. 𝑯𝒆𝟑 and muon are
(a) Fermions, (b) Bosons,
(c) Fermions and Bosons respectively, (d) classical particles
Q7.
Two particles are to be distributed in 2 nondegenerate energy states. Find the number
of distributions according to MB, BE and FD statistics. Show the distributions diagrammatically.
Q8.
In a system of 2 particles, each particle can be in any of 3 possible quantum states. Find
the ratio of the probability that the two particles occupy the same state to the probability that
the two particles occupy different states for MB, BE and FD statistics.
We cannot explain Specific heat of solid, behavior of thermal radiation, thermodynamic behavior
of a system by classical or quantum mechanics. Statistical Mechanics can explain these behavior
using theory of probability.
Macrosystem, microparticle, macroproperties, microproperties, macrostate,
microstate, Thermodynamic probability
Macroscopic System – System having large no. of particles
Microparticles – The constituent particles of a macroscopic system
Macrosystem
Metal
Blackbody chamber
Solid, liquid, gas
Micropartcles
electron
photon
Atoms, molecules, ions
Macroproperties – The properties which describe a system as a whole not the individual
particles. Example: Pressure, Volume, Temperature, no. of particles, energy, etc. These are
also called bulk properties or thermodynamic properties.
Microproperties – The properties which describe the individual particles. Example: position,
orientation, momentum, etc.
Macrostate – The state characterized by macroproperties
Micro state – The state characterized by microproperties at a single instant.
To understand macroproperties, microproperties, macrostate, microstate let us consider an
example.
Let us consider a chamber consisting of monatomic gas molecules. If it is left to itself for a
sufficiently longtime, the pressure and temperature will stabilize over the entire volume of the
chamber. The state characterized by pressure, volume, temperature is called macrostate. If the
pressure, volume, temperature will remain unchanged with time the macrostate is called
equilibrium macrostate. In equilibrium macrostate, the constituent particles are in continual
motion and undergo collision among themselves and with the wall of the container. If the
position, orientation are changed, we will get different microstates. The state characterized by
position, orientation, momentum are called microstates. So, we can conclude that any
equilibrium macrostate is associated with a large no. of microstates.
In equilibrium macrostate, particles are distributed among different microstates. Whatever the
distributions, there are some restrictions imposed on the distributions of particles in different
microstates. These restrictions are called constraints. There are two constraints.
(i)
Conservation of number of particles:
If n1no. of particles in state 1
n2no. of particles in state 2 and so on
Then, whatever the distribution
n1 + n2 +… = N (total no.of particles)(constant)
i.e, total no. of particles remain constant
(ii)
Conservation of energy:
If n1no. of particles in state 1
e1 energy of state 1
n2no. of particles in state 2
e2 energy of state 2 and so on
Then, n1e1 + n2e2 + …. = E (total energy)(constant)
i.e, total energy is constant
Whatever the distributions of particles in different microstates, two constraints must be
followed.
Accessible microstates The microstates which are permitted by the constraints
Non-accessible microstates The microstates which are not permitted by the constraints
All the accessible microstates in a given macrostate are equally probable.
Thermodynamic probability No of accessible microstates in a given macrostate large no.
𝑾=
𝑵!
𝒏𝟏 ! 𝒏𝟐 ! 𝒏𝟑 !…
subject to the condition that 𝑛1 + 𝑛2 + ⋯ = 𝑁
Relation between thermodynamic probability and entropy:
Entropy  measure of disorder.
For an irreversible process entropy is increasing. In equilibrium S becomes maximum. In
equilibrium W is max  corresponds to most probable distribution
Botzmann assumed that
𝑺 = 𝒌 𝒍𝒏𝑾 , k  Botzmann constant = 1.38×10-23 J/K
Using this relation many thermodynamic properties can be explained
Problem 1: 3 distinguishable particles each of which can be in one of the E, 2E, 3E, 4E energy
state of total energy 6E. Find all possible number of distributions of all particles in the energy
states. Find no of microstate in each case.
Answer
Method 1
Let n1, n2, n3, n4 be no. of particles in four energy states E, 2E, 3E, 4E respectively
According to the problem
n1+n2+n3+n4 = 3
(i)
 total no. of particles
n1×E + n2×2E + n3×3E + n4× 4E = 6E
(ii)
 total energy
E
2E
3E
4E
Total energy=6E
n1
n2
n3
n4
Total no of particles = 3
Obtain n1, n2, n3, n4 so that (i) and (ii) will be satisfied.
The possible distributions are
(1) n1 = 0, n2 = 3, n3 = 0, n4 = 0 (0,3,0,0)
(2) n1 = 1, n2 = 1, n3 = 1, n4 = 0 (1,1,1,0)
(3) n1 = 2, n2 = 0, n3 = 0, n4 = 1 (2,0,0,1)
No. of microstates in each case:
𝑊0,3,0,0 =
3!
=1
0! 3! 0! 0!
𝑊1,1,1,0 =
3!
=6
1! 1! 1! 0!
𝑊2,0,0,1 =
3!
=3
2! 0! 0! 1!
Total no. of microstates = 10
(1,1,1,0) is the most probable distribution because W  max for this distribution
Problem: 2(a) 5 identifiable particles are distributed in 3 nondegenerate levels with energies 0,
E & 2E. Determine the most probable distribution for a total energy 3E.
(b) In this problem what is the most probable distribution for a total energy 4E.
Answer:
(a) Let n1, n2, n3 be no. of particles in four energy states 0, E, 2E respectively
According to the problem
n1+n2+n3 = 5
(i)
 total no. of particles
n1×0 +n2×E + n3×2E = 3E
(ii)
 total energy
0
E
2E
Total energy = 3E
n1
n2
n3
Total no of particles = 5
Obtain n1, n2, n3 so that (i) and (ii) will be satisfied.
The possible distributions are
(1) n1 = 2, n2 = 3, n3 =0
(2) n1 = 3, n2 = 1, n3 = 1
No. of microstates in each distribution:
𝑊2,3,0 =
𝑊3,1,1 =
5!
= 10
2! 3! 0!
5!
3!1!1!
= 20
 Wmax
So, (3,1, 1) is the most probable distribution
(b) Hint:
𝑊2,2,1 =
5!
= 30
2! 2! 1!
Problem 3. Two particles are to be distributed in 3 levels having energies 0, E, 2E, so that the
total energy is 2E. Determine the no. of ways in which the particles can be distributed if they are
(i) classical, (ii) identical fermions, (iii) identical Bosons.
What type of particles has the minimum entropy?
Method 2
Classical (a,b)
0
E
2E
N1
0
1 (a)
1(b)
N2
2 (a,b)
0
0
N3
0
1 (b)
1(a)
0
E
2E
N1
0
1 (a)
N2
2 (a,a)
0
N3
0
1 (a)
0
E
2E
N1
1 (a)
N2
0
N3
1 (a)
Total no of Total
particles
energy
2
2E
2
2E
2
2E
2
2E
Distribution No.
of
(Macrostate) Microstates
(0,2,0)
(1,0,1)
(1,0,1)
1
1
1
Total no of Total
particle
energy
2
2E
2
2E
2
2E
distribution
No.
of
Microstates
(0,2,0)
(1,0,1)
1
1
Total no of Total
particle
energy
2
2E
2
2E
distribution
No.
of
Microstates
(1,0,1)
1
Bosons (a,a)
Fermions(a,a)
SminWmin
W  Wmin for fermions
So, fermions has the min entropy.
Problem 4. A system of 4 particles has energy levels with energies 0, 1, 2, 3 units. Total energy
of the system is 3 units. List the accessible microstates if the particles are distinguishable.