stat phys lecture notes for undergrad students

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.

mailto:[email protected]


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.


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.


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.


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


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


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.


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.


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


)()( 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.


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


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 .


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 .


(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:


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 =

.


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


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 .


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 .


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 .


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


= 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


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


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.


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.


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.


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, ….. .


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:


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


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”.


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 ,


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:


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


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


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


=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

stat phys lecture notes for undergrad students

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