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Lecture Notes in Statistical Physics compiled for undergraduate physics students.
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PBC Lecture Notes in Physics Series– Statistical Physics by Dr. Abhijit Kar Gupta 1
Statistical Physics (A set of lecture notes for undergraduate level students; for private circulation only)
Lecture Notes prepared by-
Dr. Abhijit Kar Gupta Physics Department, Panskura Banamali College
Panskura R.S., East Midnapore, W.B., India, Pin-Code: 721152 e-mail: [email protected], Ph.(res.) 033-40644589
References: 1. Perspectives of Modern Physics - Arthur Beiser (McGraw-Hill)
2. Fundamentals of Statistical and Thermal Physics - Federick Reif (McGraw-Hill Book Company, Int. Ed.) 3. Fundamentals of Statistical Mechanics
- B.B. Laud (New Age Int. Publishers) 4. Statistical Physics (Lecture notes downloaded from Internet) - Harvey Gould & Jan Tobochnik
Lecture-1
What is Statistical Physics?
Statistical Physics describes a many-particle system.
Statistical Mechanics combines statistical ideas with the laws of Mechanics.
Statistical ideas involve the concept of probability. The laws of Mechanics refer
to the Classical Mechanical (Newtonian) laws or Quantum Mechanical laws
according to the characters of the particles involved.
The laws of Classical and quantum mechanics determine the behaviour of molecules at
the microscopic level. The goal of Statistical Mechanics is to begin with the microscopic
laws of Physics that govern the behaviour of the constituents of the system and deduce
the properties of the system as a whole. Statistical Mechanics is the bridge between the
microscopic and macroscopic worlds.
Note: Comparison of Statistical Physics and Thermodynamics
Thermodynamics provides a framework for relating different macroscopic
properties of a system. Thermodynamics is concerned only with the
macroscopic variables and ignores the microscopic variables that
characterize individual molecules.
Statistical Physics and Thermodynamics both assume that the average
macroscopic properties do not change with time. We call them
Equilibrium Statistical Physics or Equilibrium Thermodynamics.
PBC Lecture Notes in Physics Series– Statistical Physics by Dr. Abhijit Kar Gupta 2
The essential steps for the statistical description of a system of particles:
1. Specification of the state of the system:
We need a method to recognize a system of particles at a particular time. This means we
have to keep track of positions and momentums of all the particles of a large system
through some way in order to understand the emerging thermodynamic and other
physical properties of the entire system.
Consider a system of gas particles. The positions and momentums of the particles may
continuously change to give a new identification of the many body system. This
„identification‟ is a „state‟ of the system in the classical sense.
Quantum mechanically the state of a system of atoms, electrons or molecules is described
by a wave function , which is a function of a set of variables.
2. Statistical Ensemble: Ensemble is a collection of a set of states (of the system of particles) under similar
external conditions.
Example Let us throw 10 dice together. At one time we may have a set such that the 1
st one shows
1 on the top face, 2nd
one 5, the 3rd
one 6 and so on. The set of all these 10 dice for one
such throwing event is what we may call a state of the system of 10 dice.
If we now repeat the performance under similar conditions, then the collection of all such
unique states form an ensemble.
3. Basic postulate of a priori probability:
When we do have a probabilistic event, we ask for the probability of occurrence of a
particular outcome. We do not have to do all the experiments to enumerate that. Instead,
we estimate this probability of occurrence of a single event through some fundamental
postulate (kind of adhoc assumption). This is the postulate of a priori probability.
Example
When we toss a coin, it is not possible to say which side of the coin will flip: „head‟ or
„tail‟. Our deterministic laws of Mechanics can not help. But our knowledge of the
physical situations in such a case leads us to expect a priori probability which is 1/2.
PBC Lecture Notes in Physics Series– Statistical Physics by Dr. Abhijit Kar Gupta 3
4. Probability Calculations:
Once the basic postulate of a priori probability is determined, we theoretically calculate
the probability (with the help of theory of probability) of the outcome of any experiment
on the many-particle system.
Lecture-2
Concept of Phase Space
Suppose a system of particles has n degrees of freedom. This means that all the particles
in the system can be described by n positional coordinates ( q ) and n linear momentum
coordinates ( p ). We can represent all these coordinates in a n2 dimensional hyperspace
and we get a point. If the system evolves with time (that is the particles in it change their
positions and momentums), the representative point moves in that space. The space such
defined is called Phase space.
Example:
If a particle moves in one dimension then we can draw two Cartesian coordinate axes
labeled by q and p . We can then specify a point ( pq, ) in this two-dimensional space. At
another time when the position ( q ) and momentum ( p ) of the particle change, the point
moves in this space. Each point in this space (Phase space, as we call it) describes the
state of the system.
Fig.1
If the particle is free (moves without constraint), it can go anywhere and can have any
amount of energy or momentum. Therefore, the representative point ( pq, ) in the phase
space can move anywhere. The phase space has infinite extent.
However, the state of the system can never be determined with absolute accuracy. The
level of accuracy depends on the experimental measurements. Suppose our measurement
yields the position and momentum coordinates in certain ranges, q to qq and p to
q
p
),( qp In this example,
the dimension of the phase space =2.
PBC Lecture Notes in Physics Series– Statistical Physics by Dr. Abhijit Kar Gupta 4
pp respectively. Then accordingly, we divide the phase space into small cells of
width q and height p and specify the state ( pq, ) which belongs to a particular cell.
Fig.2
Area of each cell is 0hpq , a constant which has a dimension of angular momentum.
The concept of phase space for a system of particles:
If a system has „ f ‟ degrees of freedom, then the system can be described by a set of f
positional coordinates fqqq ......., 21 and f momentums fppp ........., 21 .
For a system of N free particles, Nf 3 .
Now we can construct a hypothetical phase space of f2 -dimensions: f -number of axes
for positions and f -number of axes for momentums. Therefore, the set of numbers (the
coordinates, fqqq ......., 21 , fppp ........., 21 ) can be considered as a point in the f2 -
dimensional phase space. Again, as before, the phase space is divided into cells according
to the accuracy of the measurement and we obtain the volume of a cell to be
q
p
),( pq
NOTE: The size of the cell depends upon the accuracy of measurement. More accurately w can
measure the state ( pq, ), the less will be the uncertainties q and p , the less will be
the size of a cell.
The lower limit of the size of each cell is determined by the Heisenberg Uncertainty
principle in Quantum Mechanics: 2
pq , where
2
h , h being the Planck‟s
constant.
PBC Lecture Notes in Physics Series– Statistical Physics by Dr. Abhijit Kar Gupta 5
f
ff hpppqqq 02121 ........... . Here the accuracy of measurement in any n -th
direction is nq and np such that 0hpq nn .
Phase Space for a bounded system:
One dimensional Harmonic Oscillator
The energy of the oscillator is given by
22
2
1
2kx
m
pE ,
where x is the displacement, p is the momentum and k is the spring constant.
If the total energy E of the oscillator is a constant, then it describes an ellipse in the 2-
dimensional phase space made by x and p .
In practice, the accuracy of the measurement is such that position lies between x and
xx and momentum lies between p and pp . Therefore, the energy of the
oscillator lies between E and EE . Thus the phase space is confined in the region
between two ellipses corresponding to E and EE . So, there are many different sets
of ( px, ) lying within the cells in the annular region.
Fig.3
Note that a given interval dx corresponds to a larger number of cells lying between the
two ellipses when Ax (near the maximum value of x ) than when 0x (near origin).
This indicates that we have the greater probability of finding the oscillator in the
microstates which are close A (amplitude) than close to origin. This result is supported
EE
E
dx dx
x
p
PBC Lecture Notes in Physics Series– Statistical Physics by Dr. Abhijit Kar Gupta 6
by the fact that the oscillator has small velocity near the extreme points Ax ( ); hence it
spends a longer time there than near equilibrium point ( 0x ), where it moves rapidly.
Lecture-3
Microstate: Suppose the cells in the phase space are numbered with some index n 1, 2, 3……. Then
the state of the system of particles is described by specifying a particular cell (having
some n -value) where the system is found. Each cell describes a microscopic state or
„microstate‟ of the system.
Quantum Mechanically, a quantum state is described by n , where n = 1, 2,
3…..designate possible quantum numbers.
Macrostate: The macroscopic state or „macrostate‟ of a system is defined by specifying the external
parameters of the system and any other conditions acting on the system. The external
parameters can be volume of the system or constant total energy (for an isolated system)
etc.
Corresponding to a given macrostate the system can be in any of the possible microstates.
Concept of Equilibrium: An isolated system is said to be in equilibrium when the probability of finding the system
in any one state is independent of time. All macroscopic parameters describing the
system are then also time independent.
In other words, the system is equally likely to be found in any of the states accessible to
it.
Estimation of number of Microstates Ideal Gas:
Consider an ideal monatomic gas of N molecules enclosed in a volume V . Because of
the interaction between the molecules is zero, the energy is only kinetic (Potential energy
0U ).
If the molecules are numbered as 1, 2, 3, ….. n and we associate the momentum of the
center of mass of the i -th molecule as ip , the energy of the system of noninteracting
molecules can be written as
PBC Lecture Notes in Physics Series– Statistical Physics by Dr. Abhijit Kar Gupta 7
2
12
1
N
i
ipm
E ……………………………… (1)
Now we can construct the phase space by N3 -position coordinates ( Nrrr ,......, 21 ) and
N3 -momentum coordinates ( Nppp ,......, 21 ). Each position and momentum vector has
three components in the three dimensional coordinate system.
The number of microstates )(E lying between the energies E and EE is then
equal to the number of cells in the phase space contained between E and EE . The
number of cells is proportional to the volume of the phase space.
)(E Volume of phase space.
Therefore, we now estimate the volume of phase space of the system:
Volume of phase space = 1
3....... rd 2
3 rd ……. 2
3 rd 1
3 pd 2
3 pd …… Npd 3 , ……(2)
where the integration is done over all positions and momentum coordinates for which the
total energy lies between E and EE .
Since each integral over ir extends over the volume V of the container (each molecule,
being free, can move anywhere in the closed space),
Vrd i 3 …………………………… (3)
There are N such integrals. Therefore, the contribution in the expression (2) is .NV
Now we rewrite expression (1) where we explicitly show the momentum components.
N
i
iziyix pppmE1
2222 ………………………………….. (4)
The sum in (4) contains N3 square terms (N-particles, 3-components each). For
E const., we may think of equation (4) as an equation for a sphere in N3 -dimensional
hyperspace. Then we shall have the radius 2/1)2()( mEER .
Note:
Here iiii dzdydxrd 3 and
iiii dpdpdppd 3 . These are the three dimensional volume elements.
PBC Lecture Notes in Physics Series– Statistical Physics by Dr. Abhijit Kar Gupta 8
Volume of a sphere in N3 -dimension is proportional to NR3 (Note: the volume of a
sphere in 3-dimension is proportional to 3r , where r is the radius).
Thus the Volume of phase space is proportional to 2/3)2( NN mEV . So now we can write
the number of microstates, 2/3)2()( NN mEVE
2/3)2()( NN mEAVE ,
where A is a constant independent of V and E . The expression tells us how the number
of accessible microstates in an ideal monatomic gas depends on the volume and energy.
Since, N is generally of the order of Avogadro‟s number (~ 2310 ) and thus very large,
)(E thus increases extremely rapidly with energy.
Lecture-4
Microcanonical Ensemble: An ensemble representing an isolated system in equilibrium is called “microcanonical”
ensemble.
If the energy of an isolated system in a state r is denoted by rE , the probability rP of
finding the system in that state is given by
rP = C (Const.) if EEEE r ,
= 0 Otherwise.
If a phase space is constructed for such a system corresponding to energy in the range
between E and EE , the system is equally likely to be found in any of the cells
(microstates). Thus the probability ( rP ) is same for any of the cells.
The constant C can be determined from the normalization condition (the system must be
found in any of the cells): 1r
rP , the sum is over all the accessible microstates in the
range between E and EE . If the system has total N -number of microstates, then we
have N
Pr
1 for any r in the energy range.
Note: Suppose A is not an isolated system and it interacts with another system B.
We can treat the combined system (A+B) to be isolated. Thus the microcanonical
ensemble can be formed for the combined system.
PBC Lecture Notes in Physics Series– Statistical Physics by Dr. Abhijit Kar Gupta 9
Canonical Ensemble:
Canonical ensemble is referred to a system when it interacts (thermally) with a heat
reservoir (called, Environment).
Probability of finding the system in any microstate r is
rE
r eCP
. ,
where C is the constant of proportionality. Now we use the Normalization condition,
1r
rP .
We get 1 rE
eC
.
r
EreC
1.
The normalized probability is
r
E
E
rr
r
e
eP
.
The parameter has the dimension of inverse of energy. In Statistical Physics one
chooseskT
1 , where k is Boltzmann constant and T is the equilibrium (absolute)
temperature of the system.
Note: (for interested students) Derivation of the Probability:
Let us consider a small system A is in thermal interaction with a heat reservoir B. Here
we consider A << B, the degrees of freedom in A is much less than that of B.
If the interaction between the two systems is weak, the energies are additive. The energy
( AE ) of A is not fixed. Whereas the total energy of the combined system (A+B) has a
constant value in the range between TE and EET .
Suppose, A is found to be in a state r of energy A
rE .
From the conservation of energy we can write:
TBA
r EEE ,
where BE is the energy of the reservoir. A
r
TB EEE .
We then have the number of accessible states to the reservoir is
PBC Lecture Notes in Physics Series– Statistical Physics by Dr. Abhijit Kar Gupta 10
)()( A
r
TB EEE
Now if the system A is in a definite state (the state is fixed), the number of states
accessible to the combined system (A+B) is also )( A
r
T EE .
In this situation, where A is fixed in a state r , the probability rP of occurrence of this is
simply proportional to the number of states accessible to the combined system (A+B).
Therefore,
CPr )( r
T EE , ………………………..(1)
where C is a constant of proportionality. We now use rE in place of A
rE for
convenience.
Now make use of the fact that A << B and thus A << (A+B), the combined system.
Hence rE << TE .
So, we can write the following Taylor expansion, considering r
TB EEE ,
...............ln
)(ln)(ln r
EE
B
T
r
T EE
EEETB
The expansion is around TB EE . The other higher order terms are neglected.
The derivative TB EE
BE
ln is evaluated at the fixed energy TB EE and thus a
constant independent of energy rE of A.
Hence we can write the above expansion as
r
T
r
T EEEE )(ln)(ln
or, rET
r
T eEEE
)()( . …………………………….(2)
Here )( TE is a constant independent of r . Combining (1) and (2) we thus have
rE
r eCP
.
EXAMPLE: A molecule in an ideal gas.
Let us think of a single molecule which is at thermal equilibrium with all the other
molecules at (absolute) temperature T and enclosed in a volumeV . This system of a
single molecule can be thought of in a canonical ensemble.
PBC Lecture Notes in Physics Series– Statistical Physics by Dr. Abhijit Kar Gupta 11
Energy of the molecule, )(2
1 222
zyx pppE .
Therefore, the probability that the molecule has momentum lying in the range xp
and xx dpp , yp and yy dpp , zp and zz dpp is
zyx
ppp
zyx dpdpdpedpdpPdp zyx )( 222
,
Considering canonical probability ( Ee ). Thus we can easily get a distribution for
velocities which are Maxwell distribution of molecular velocities.
Lecture-5
Calculation of Physical quantities in Canonical Ensemble:
Probability
r
E
E
rr
r
e
eP
= Z
e rE
,
where
r
EreZ is called Partition function.
kT
1 , dimension is inverse of energy.
k = Boltzmann constant
T = Absolute temp.
Average Energy:
r
E
r
r
E
r
r
rr
r
e
Ee
EPE
,
where the sums are over all accessible microstates r of the system.
To simplify, we use a trick.
r
r
EEe r
=
r
E
r
re
=
r
Ere
=
Z
PBC Lecture Notes in Physics Series– Statistical Physics by Dr. Abhijit Kar Gupta 12
Hence, we obtain
E =
Z
Z
1 =
Zln.
Dispersion of Energy:
2222222 EEEEEEEEE
Here
r
E
r
r
E
r
r
e
Ee
E
2
2 .
We use the trick again to simplify:
r
E
r
r
E
r
r
E rrr eEeEe
2
2
Z
E12
r
Ere
2
= 2
21
Z
Z
Therefore, we can write
222
EEE2
21
Z
Z
2
2
1
Z
Z=
2
2 lnln
1
ZZ
Z
Z.
We can also write the above in terms of average energy:
.ln2
EZE
When we use the relationkT
1 , we get
2E =
E =
T
T
E = VCkT 2 .
PBC Lecture Notes in Physics Series– Statistical Physics by Dr. Abhijit Kar Gupta 13
Therefore, the energy fluctuation in the canonical ensemble is proportional to the specific
heat.
Example #1 A system with two energy levels is in thermal equilibrium with a heat reservoir at
temperature 600 K. The energy difference between the levels is 0.1 eV. Find (i) the
probability that the system is in the higher energy level and (ii) the temperature at which
this probability will equal 0.25.
The system is in canonical ensemble.
(i) 1
1
1
EE
E
ee
eP
…………….(1)
6001038.1
106.116
13
kT
EE ………….(2)
Putting (2) in (1) we get 126.0P
(ii) 1
125.0
/
kTEe, Put
TkT
E
16
13
1038.1
106.1 and we obtain 1056T K.
Lecture-6
The Equipartition of Energy:
According to Classical Mechanics, for a system having f -generalized coordinates and
f -generalized momenta, the Energy can be written as
)......,,.....,( 2121 ff pppqqqEE
Now let us assume that the energy obeys the following:
(i) Energy is additive.
This means the total energy E can be written as the sum of energies for all f -modes.
fiE .............21 ,
where ,..., 21 are the energies corresponding to the 1st mode, 2
nd mode and so on. The
above expression can be written as
EE i ,
where the i -th mode is separated out and the rest is included in E .
PBC Lecture Notes in Physics Series– Statistical Physics by Dr. Abhijit Kar Gupta 14
(ii) Energy is quadratic in momentum.
We can write 2)( iii pp , is a constant.
If the system is in equilibrium at the absolute temperature T , the energy should be
distributed according to the canonical ensemble.
Now the mean value of the energy i for the i -th mode can be written as
ff
E
ffi
E
i
dpdpdpdqdqdqe
dpdpdpdqdqdqe
.........
........
2121
2121
ff
E
ffi
E
dpdpdpdqdqdqe
dpdpdpdqdqdqe
i
i
.........
........
2121
)(
2121
)(
fiif
E
i
fiif
E
ii
dpdpdpdpdqdqdqedpe
dpdpdpdpdqdqdqedpe
i
i
.............
...........
11121
11121
i
ii
dpe
dpe
i
i
i
i
dpe
dpe
i
i
Suppose, idpeX i
X
X
i
= Xln
Now we evaluate the integral:
PBC Lecture Notes in Physics Series– Statistical Physics by Dr. Abhijit Kar Gupta 15
idpeX i
=
i
pdpe i
2=
0
2
2 i
pdpe i
We substitute 2
ipy .
iidppdy 2
= idpy
2/1
2
= idpy 2/12/1
2
idp = dyy 2/1
2
1
X =
0
2
2 i
pdpe i
= 2 dyye y 2/1
02
1
= dyye y )12/1(
0
1
= 2/11
=
1 =
.
Therefore, we can write
lnlnln
i
=
ln
2
1 =
2
1.
This is Equipartition theorem of Classical Statistical Mechanics.
Theorem: The mean value of each independent quadratic term (corresponding to each mode) in the
energy of a canonical system is kT2
1.
Example: Harmonic Oscillator
Energy 22
2
1
2kx
m
pE .
The energy term has two square terms (two modes) in it. Therefore, the mean energy for
the oscillator is E = kT2
12 = kT .
This result can be derived directly:
kTi2
1
PBC Lecture Notes in Physics Series– Statistical Physics by Dr. Abhijit Kar Gupta 16
E =
dpdxe
Edpdxe
E
E
= .kT
Application: A molecule in an ideal gas.
Energy, 222
2
1zyx ppp
m
The single molecule has three modes (the degrees of freedom).
According to equipartition theorem, the average energy for each mode is kT2
1.
.2
3
2
1.3 kTkT
One mole of gas contains N (Avogadro number ~ 2410 ) number of molecules.
Therefore, the mean energy per mole of gas is
NkTkTNNE2
3)
2
3( .
The molar specific heat at constant volume is
RNKT
EC
V
V2
3
2
3
, '' R is known as gas constant.
The fluctuation (dispersion) in energy can be calculated as
2E = VCkT 2 = NkkT
2
32 = 22
2
3TNk where we consider the ideal gas at equilibrium
is in canonical ensemble and we use the result derived earlier.
The relative fluctuation is
2/1
2/1
222/1
2
3
2
2
3
2
3
NNkT
TNk
E
E
The above shows that the relative fluctuation (in energy) becomes less and less important
as the system size increases. This is negligible when 1N .
PBC Lecture Notes in Physics Series– Statistical Physics by Dr. Abhijit Kar Gupta 17
Lecture-7
Maxwell-Boltzmann (MB) Statistics:
MB statistics is applicable to Classical particles which are identical and distinguishable.
Consider a system of N -particles (molecules in general) which can be found in any of its
microstates having energy E .
Now the individual particles may have some discrete energy states 1 , 2 , 3 ….and so
on so that there are 1n particles of energy 1 , 2n particles of energy 2 ….up to kn
particles of energy k .
Therefore, we can write
ki nnnn ..........21 = N ………………………..(1) [ Conservation of particles]
iin = kknnn ...........2211 = E ……………….(2) [ Conservation of Energy]
Let us assume that ig be the number of microstates each of which has the same
energy i . Then the number of ways for a single particle to occupy any of the states is ig
(degeneracy).
The number of ways for two particles to have the same energy i is then ig ig = 2
ig .
The number of ways for in particles to have the same energy i = in
ig .
So the number of ways for 1n particles to have energy 1 , 2n particles to have energy 2
and so on is
1P = kn
k
nnggg ..........21
21 =
k
i
n
iig
1
.
NOTE: Suppose we have 2 distinguishable particles A and B and 3 Boxes.
Therefore, we have 2n and 3g .
Let us see how we do distribute them:
AB
AB
AB
A B
B A
A B
B A
A B
B A
Total number of ways = 9 = 23 = ng .
PBC Lecture Notes in Physics Series– Statistical Physics by Dr. Abhijit Kar Gupta 18
To calculate the all possible ways of how N particles are distributed among k -energy
states we need to consider the possible ways of selecting the group of particles.
The total number of ways the first group of 1n -particles can be selected from the total N
particles is )!(!
!
111 nNn
NCn
N
.
Next 2n -particles can be selected from the rest of )( 1nN particles in
2
1 )(
n
nNC
=
)!(!
)!(
212
1
nnNn
nN
ways.
In this way we can select 1n , 2n , ….. kn particles out of total N particles in the
following ways:
2P = 1n
N C1
221
3
21
2
1 ).....()()(...........
k
k
n
nnnN
n
nnN
n
nNCCC
= !)!......(
)!.......(............
)!(!
)!(
)!(!
!
1121
221
212
1
11
kk
k
nnnnN
nnnN
nnNn
nN
nNn
N
= !!!......!
!
121 kk nnnn
N
=
k
i
in
N
1
!
!.
Therefore, the total number of ways P of the distribution is
P = 21 PP
=
k
i
in
N
1
!
!
k
i
n
iig
1
. This is termed as Thermodynamic Probability.
Taking logarithm on both sides, we get
i
n
i
i
iignNP ln!ln!lnln
= i i
iii gnnN ln!ln!ln …………………………..(3)
For large number of particles we can consider 1N and also each 1in .
So, we use Sterling’s formula:
nnnn ln!ln , for 1n .
Thus equation (3) becomes
i
ii
i
i
i
ii gnnnnNNNP lnlnlnln
PBC Lecture Notes in Physics Series– Statistical Physics by Dr. Abhijit Kar Gupta 19
= i
ii
i
ii gnnnNN lnlnln ……………………………..(4)
[Since Nni
i ]
Now if we have the distribution P to be the most probable one maxP , then
0lnlnlnln max i
i
ii
i
ii
i
i ngnnnnP
[Here N and ig ‟s are constants.]
Now we have,
i
i
i nn
n 1
ln .
0ln i
i
i
ii nnn [ Since i
inN const ]
So, we can write
0lnln i
iii
i
i ngnn ………………………………………(5)
We have other two variational equations (from eqn. (1) and (2) )
0..........21 k
i
i nnnn ……………………………….(6)
0........2211 kki
i
i nnnn ………………………..(7)
We combine three variational equations (5), (6) and (7) by multiplying (6) by , (7) by
and add them up with (5):
0lnln i
i
iii ngn ………………………………...(8)
[Note: This method is called the method of Lagrange’s undetermined multipliers]
In the above equation (8), we can choose and such that all the in ‟s become
independent. In such a case the quantities in the brackets must be zero for each i .
Hence,
0lnln iii gn
ieegn ii
…………………………………………………..(9)
We write
This is Maxwell-Boltzmann distribution; „ if ‟ is termed as MB-function.
ieeg
nf
i
ii
PBC Lecture Notes in Physics Series– Statistical Physics by Dr. Abhijit Kar Gupta 20
Further we can rewrite the distribution considering the following:
ieegNn
i
i
i
i
iegeNi
i
i
iieg
Ne
Hence the normalized MB-distribution becomes:
i
i
ii
i
eg
Nen
.
Lecture-8
The Maxwell-Boltzmann distribution in continuous form can be written as:
deegdn )()( [Note eqn. (9) in Lecture 7]…………………(1)
We have to evaluate the constants and .
For the molecules in a gas we can easily take the continuous distribution as there are
usually a large number of molecules.
Equation (1) is the number of molecules whose energies lie between and d .
In terms of molecular momentum we can write
,)()( 2/2
dpeepgdppn mp
as we have .2
2
m
p
Now, dppg )( is the number of cells in the phase space within the range of momentum
p and dpp .
dppg )( = 3h
dpdpdxdydzdp zyx, where 3h is the volume of each cell.
Here Vdxdydz = Volume occupied by the gas in ordinary position space.
dppdpdpdp zyx
24 = Volume of the spherical shell of radius p and thickness dp
in momentum space.
Hence, 3
24)(
h
dpVpdppg
PBC Lecture Notes in Physics Series– Statistical Physics by Dr. Abhijit Kar Gupta 21
dppn )( = 3
2/2 2
4
h
dpeeVp mp
.
Next we note that
0
)( Ndppn Total number of molecules.
0
2/2
3
24dpep
h
VeN mp
=
2/3
3
2
m
h
Ve
Hence,
2/33
2
mV
Nhe
.
Therefore we have the momentum distribution,
dpepm
Ndppn mp 2/2
2/32
24)(
.
Since we have mp 22 and
m
mddp
2 , the corresponding energy distribution can
be written as
de
Ndn
2/32)( .
To find the other constant we employ the last formula above and compute the total
energy E of the system of molecules.
Total energy, E =
0
)( dn
=
deN
0
2/32/32
=
N
2
3.
According to Equipartion theorem, E = NkT2
3.
Therefore, we have kT
1.
Thus the MB-distribution law now looks like:
de
kT
Ndn kT/
2/3
2)( .
This is the number of molecules with energies between and d in a sample of ideal
gas that consists a total of N molecules and whose absolute energy is T .
Home Work: Find out the Maxwell-velocity distribution from the above formula.
PBC Lecture Notes in Physics Series– Statistical Physics by Dr. Abhijit Kar Gupta 22
Lecture-9
Quantum Statistics: Applicable to quantum particles which are identical but indistinguishable.
Bose-Einstein (BE) Statistics:
Let us assume ig to be the number of states that have same energy i .
ig is the degeneracy.
We determine the number of ways to which in indistinguishable particles can be
distributed in ig cells. Here the selection of in particles from total N particles does not
count as the particles are indistinguishable.
Now, if in particles are lined up in ig cells then the group of particles are separated by
( 1ig ) partitions. Therefore, the total number of ways of arranging in particles in ig
cells can be found out by permuting ( 1 ii gn ) objects among themselves.
Total number of permutations among ( 1 ii gn ) objects is equal to ( 1 ii gn )!
But out of these, in ! permutations of the in indistinguishable particles among themselves
and ( 1ig )! permutations among themselves are irrelevant.
Therefore, the possible distinguishably different arrangements of in indistinguishable
particles among ig cells are
)!1(!
)!1(
ii
ii
gn
gn.
Hence, the „thermodynamic probability‟ P of the entire N particles is
P =
i ii
ii
gn
gn
)!1(!
)!1(.
Example:
4g , 10n
3 partitions. Particles are shown to be grouped in 2+3+4+1 ways.
PBC Lecture Notes in Physics Series– Statistical Physics by Dr. Abhijit Kar Gupta 23
Taking natural logarithm on both sides we get
i
iiii gngnP )!1ln(!ln)!1ln(ln
)!1ln(!ln)!ln( iiii gngn
[ Since, ( ii gn )>>1 ]
We use Sterling‟s formula to have
i
iiiiiiiiii gnnngngngnP )!1ln(ln)()ln()(ln .
Conditions for the most probable distribution is maxPP . For that small changes in in
any of the in ‟s do not affect the value of P .
Hence we can write
i
iiii
i
iiiiii
ii
ii nnnnn
nnngnngn
gnP 0ln1
)ln()(
1)(0ln max
[Note: ig ‟s do not change]
The result is :
0ln)ln( i
i
iii nngn …………………(1)
Now we incorporate the conditions for the conservation of particles and conservation of
energy:
0i
in …………………………………..(2)
0 i
i
i n ………………………….……..(3)
The above three equations (1), (2) and (3) can be combined to form one by the method of
Lagrange‟s method of undetermined multipliers. Let us multiply (2) by and (3) by
and then add them up with (1):
0ln)ln( i
i
iiii nngn
Since the in ‟s are independent, the quantity in the brackets must vanish for each value
of i . Hence,
0ln
i
i
ii
n
gn
ieen
gn
i
ii
1
iee
gn i
i This is Bose-Einstein (BE) distribution law.
PBC Lecture Notes in Physics Series– Statistical Physics by Dr. Abhijit Kar Gupta 24
Considering kT
1 we write:
1
1/
kT
i
ii
ieeg
nf
, BE occupation number.
Particles which follow BE statistics are called „Bosons‟. For example, Photons, Phonons,
Mesons etc.
Photon Statistics:
For photons total number is not conserved: i
in const.
Hence, we put 0 in the BE distribution formula and the distribution becomes
1
1
ief i
For continuous distribution:
de
dfkT 1
1)(
/ , where we put h for photons and we get
1
1)(
/
kThef
Application: Planck Radiation Formula.
Lecture-10
The Planck Radiation Formula: Planck Considered that the electromagnetic waves emitted from a black body does have
energy which is quantized in units of h , where is the frequency of the wave and h is
Planck‟s constant.
The black body is assumed to be a cavity whose walls are constantly emitting and
absorbing radiation. It has been observed that each wave originates in an atomic
oscillator, the energy of which is written as nhn , where n 0, 1, 2, ….. .
PBC Lecture Notes in Physics Series– Statistical Physics by Dr. Abhijit Kar Gupta 25
The average energy of an oscillator is
1/
kThe
h
, considering Photon statistics.
To derive a formula for the Spectral energy density it is assumed that the cavity has some
volume V and it contains a large number of indistinguishable photons of various
frequencies.
To count the photons we have to estimate the number of cells in the phase space
(constructed by positions and momentum of all photons).
If each cell has the infinitesimal volume 3h then the number of cells in the phase space
where the momentum lies between p and dpp is
3)(
h
dpdpdxdydzdpdppg
zyx .
Here,
Vdxdydz = volume of the cavity
dppdpdpdp zyx
24 = volume of the spherical cell of radius p and thickness
dp in the momentum space.
Therefore,
3
24)(
h
dpVpdppg
.
Since the momentum of a photon is c
hp
, we have
3
232
c
dhdpp
.
dc
Vdg 2
3
4)( .
Note: Every substance emits electromagnetic radiation, the character of which
depends upon the nature and temperature of the substance.
An ideal solid body which absorbs all radiation incident upon it, regardless
of frequency is called a black body.
Model of a Black body:
PBC Lecture Notes in Physics Series– Statistical Physics by Dr. Abhijit Kar Gupta 26
This is the number of cells in the frequency between and d .
The spectral energy density du )( is the energy per unit volume between and d
V
dgdu
)(2)( ,
where the factor 2 comes from the fact that each cell can be occupied by two photons of
two different directions of polarizations.
1
8)(
/
2
3
kThe
hd
cdu
1
8)(
/
3
3
kThe
d
c
hdu
. This is Planck Radiation formula.
Wien’s Displacement Law:
In the Planck‟s radiation formula above we put kT
h (dimensionless quantity).
We arrive at
1
18)(
3
3
ed
h
kT
h
kT
c
hdu
=
1
8 34
33
4
e
dT
hc
k
The plot of the above equation looks like
max
44
33
8 T
u
k
hc
PBC Lecture Notes in Physics Series– Statistical Physics by Dr. Abhijit Kar Gupta 27
The maximum in the above curve occurs at 3max which means we have
kT
h max Constant. This is Wien’s displacement law.
It quantitatively expresses the empirical fact that the peak in the black-body spectrum
shifts to progressively shorter wavelengths (higher frequencies) as the temperature is
increased.
Stefan-Boltzmann Law:
The total energy density within the cavity over all frequencies can be obtained as
0
34
0
33
4
1
8)(
e
dT
hc
kduu .
The definite integral is just some constant.
Hence we can write
4)( TTu ,
where is a universal constant. This is Stefan-Boltzmann law which states that the total
energy density is proportional to the fourth power of the absolute temperature of the
cavity walls.
Raleigh-Jeans Formula:
For high enough temperature ( T large) one can approximate
kT
he kTh 1/ .
Therefore, Planck Radiation formula can be approximated as
.
8
/
8)(
3
23
3 c
kTd
kTh
d
c
hdu
This is Rayleigh-Jeans formula which was originally obtained by considering the
continuous distribution of energy due to Classical Physics.
According to Rayleigh-Jeans formula:
2)( u which means 0)( u as 0 and )(u as .
But experimentally one finds that 0)( u as .
This is called Rayleigh-Jeans catastrophe or “ultraviolet catastrophe”.
PBC Lecture Notes in Physics Series– Statistical Physics by Dr. Abhijit Kar Gupta 28
Lecture-11
Fermi-Dirac (FD) Statistics:
Particles are identical, indistinguishable and are governed by Pauli‟s Exclusion principle.
No two particles can occupy same energy state. That is each cell can be occupied by
maximum one particle.
Assume that there are ig cells having the same energy i and in particles are to occupy
the ig cells. This means in cells are filled and )( ii ng cells are vacant.
(We assume ii ng .)
Now total ig cells (some are filled and some are vacant) cane be arranged in !ig
different ways among themselves. But the !in permutations of the filled cells and
)!( ii ng permutations of the vacant cells are irrelevant as the particles (and cells) are
indistinguishable.
Therefore, the numbers of distinct arrangements are
)!(!
!
iii
i
ngn
g
.
The „thermodynamic probability‟ P of the entire distribution of particles is the product
.)!(!
!
i iii
i
ngn
gP
Taking natural logarithm on both sides,
i
iiii ngngP )!lnln(!ln!lnln .
Using Stirling‟s formula:
i
iiiiiiiiiiii ngngngnnngggP )()ln()(lnlnln
For the distribution to represent maximum probability maxPP , small variation in in the
particle numbers must not alter P .
Hence,
i
i
iii nngnP )ln(ln0ln max …………………………(1)
We also consider the conservation of particles and of energy (as before),
0i
in …………………….(2)
0i
ii n ………………..…(3).
We combine the above three equations by multiplying (2) by , (3) by and then
adding them up with (1):
0)ln(ln i
i
iiii nngn ,
PBC Lecture Notes in Physics Series– Statistical Physics by Dr. Abhijit Kar Gupta 29
where and are Lagrange‟s undetermined multipliers.
Since the in ‟s are independent, the quantity in the brackets must vanish for each value
of i . Therefore,
0ln
i
i
ii
n
ng
ieen
ng
i
ii
1 ieen
g
i
i .
1
iee
gn i
i , where
kT
1 .
This is Fermi-Dirac (FD) distribution law.
FD distribution in continuous form:
1
1
1
1)(
/
kTeeeef
, )(f is called occupation index.
Fermi Energy: To evaluate the constant we examine the condition of „electron gas‟ at low
temperatures. For small T , the occupation index )(f is 1 at 0 and )(f =1 up to a
certain energy F , called Fermi energy. For F , )(f drops to zero rapidly.
To incorporate the above we set kT
F .
The occupation index is now
1
1
)(
)()(
/)(
kTFeg
nf
.
At T o0 K
1)( f when F
= 0 when F .
As the temperature increases, the occupation index changes from 1 to 0 more and more
gradually.
But at any T we have 2
1)( f when F .
The occupation index for FD distribution at KT o0 and at a higher temperature is
drawn below:
PBC Lecture Notes in Physics Series– Statistical Physics by Dr. Abhijit Kar Gupta 30
Lecture-12
Comparison among three distributions:
kT
i
ii
ieeg
nf
/ MB
1
1/
kTiiee
f
BE
1
1/
kTiiee
f
FD
Occupation index if signifies the average number of particles in each of the states of that
energy i . This does not depend upon how the energy levels of a system of particles are
distributed. Hence, this provides a convenient way of comparing the essential features of
the three distribution laws.
For MB distribution: The ratio of two occupation indices for two energy levels i and j does not depend
upon the parameter .
.)(
)( //)( kTEkT
j
i eef
fji
This is the relative degree of occupancy for two energy states. This formula is
particularly useful for BE and FD distributions as we often ask this question for quantum
states. But we can arrive at the above formula there only when BE and FD distributions
behave like MB distribution under certain conditions.
BE distribution:
KT o0
KT o0
)(f
F
2
1)( f
1
0
PBC Lecture Notes in Physics Series– Statistical Physics by Dr. Abhijit Kar Gupta 31
When kTi , the BE distribution approaches MB distribution.
1
1/
kTiiee
f
kTiee
/
1
= kTiee
/ .
FD distribution:
1
1/)(
kTiiFe
f
for Fi .
At low temperatures ( )0T virtually all the energy states are filled with the occupation
index dropping rapidly towards zero near Fermi energy ( F ).
At high temperatures ( T large) the occupation index is sufficiently small at all
energies.
Then we can write
kTFeef/)(
)( , the FD distribution becomes similar to the MB.
Calculation of Fermi Energy:
1
1
)(
)()(
/)(
kTFeg
nf
Suppose a system (a metal sample for example) contains N free electrons.
)(f
0T
0T
0T
1T
2T
)(f
210 TT
PBC Lecture Notes in Physics Series– Statistical Physics by Dr. Abhijit Kar Gupta 32
For the calculation of Fermi energy ( F ) we consider that all energy states are filled up
starting from 0 up to F . That means 1)( f , F 0 .
So,
0
)( Ndg .
To calculate dg )( we start evaluating dppg )( .
3)(h
dpdpdxdydzdpdppg
zyx , integrated over the phase space.
We know
Vdxdydz (Volume of the system) and
dppdpdpdp zyx
24 (Volume of the momentum space contained in the
Spherical cell between p and dpp )
3
2 .42)(
h
Vdppdppg
=
3
2 .8
h
Vdpp.
The factor 2 comes from the fact that each electron has two possible spin states (+1/2
and 2/1 ).
As we know mp 22 , we get dmdpp 2/132 )2( .
dg )(
dh
Vm 2/1
3
2/328.
Now,
N
dh
Vm F
0
2/1
3
2/328 =
2/3
3
2/3
3
216F
h
Vm
.
Hence,
3/223/22
8
3
28
3
2
m
h
V
N
m
hF , where
V
N is density of free electrons.
Example:
In copper atom the electron configuration is such that we can assume each atom is
contributing one free electron to the electron gas.
Therefore, the electron density can be written as
Volume
atomsofNumber
= molemass
volumemassmoleatoms
/
)/()/(
= )(
)()(
copperofmassAtomic
atomscopperofDensityNumberAvogadro
PBC Lecture Notes in Physics Series– Statistical Physics by Dr. Abhijit Kar Gupta 33
=3
323
105.63
1094.810023.6
= 8.5 2810 electrons/m 3
3/2
28
31
234
8
105.83
1011.92
)1063.6(
F = 1.13 1810 Joule = 7.04 eV
Fermi Temperature, kT FF / ~ 10 5 Ko .
We have used:
341063.6 h Joule-sec
m = 9.11 3110 kg, electron mass
k = 1.381 2310 Joule/ Ko