Salahaddin University-ErbilCollege of SciencePhysics Department

Course Book

Statistical Mechanics

Undergraduate Students 3rd. Year

Assistant Professor: Tariq Abdul-Hameed Abbas

M.Sc. in Cosmic Rays PhysicsPh.D. in Solid State Physics

E.mail : Mobile:

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Description:

The word “statistics” must be interpreted as applying to a technique for describing processes involving large numbers of particles whose individual interactions are known, without considering the individual behavior of each particle.

Statistical mechanics provides the framework for performing microscopic analyses. In practice, statistical mechanics regards the interaction between the particles composing a bulk sample, and predicts the equilibrium properties of the system that result from these interactions. A practical difficulty of statistical mechanics is in finding tractable mathematical approaches to describing the many interacting bodies in the system. As in the other core areas of Physics, some systems are exactly solvable, while more complex systems require approximations or numerical analysis. Statistical mechanics is one of the subjects which serve students in physics. To understand any branch in physics, students been to understand statistical mechanics first. Statistical mechanics aims at studying the macroscopic parameters of a system in equilibrium from knowledge of the microscopic properties of its constituent particles using the laws of mechanics. Statistical mechanics provides the connection between microscopic motion of individual atoms of matter and macroscopically observable properties such as temperature, pressure, entropy, free energy, heat capacity, chemical potential, viscosity, spectral, reaction rate, etc. The study of statistical mechanics is mainly classified in two categories, these are:

1- Classical statistics or Maxwell- Boltzmann statistics.

2- Quantum statistics or Bose-Einstein and Fermi- Dirac statistics.

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We will begin with the fundaments of statistical physics and brief review of thermodynamics as an introduction of some important ideas which are fundamental in our understanding of statistical mechanics.

Student’s participation in on-line discussions (or other discussions)is highly recommended. They will get credit for them on-line activities.

This is an introductory on Statistical Mechanics and Thermodynamics given to third year undergraduates.

Forms of Teaching

Different forms of teaching will be used to reach the objectives:

Main form of teaching will be used to reach the objectives is power point. Power point will be used to present the head titles, main object, definitions, illustrate examples and exercises; also we’ll use whiteboard for more illustration of difficult problems to the students.

There will be classroom discussions and the lecture will give enough background to translate, solve, analyze, and evaluate problem sets, and different issues will discuss throughout the semester. Giving homework and quizzing them to provide further challenge.

To get the best of the subjects, it’s suggested that the students attend the classes as much as possible, listen to the teacher, understand the lectures, write notes, ask questions, and try to solve problems and exercises as much as possible.

Grading

The students are required to do at least three examinations besides some homework sets which have 40 marks. There will be a final examination on 60 marks.

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Evaluation

The study evaluation will be based roughly half on monthly exams and homework, and the other half on the final exam.If the students do very poorly in either of these two categories, especially the monthly exams, they may not pass.

Required books

The books for the course are:

1- Statistical mechanics Dr. Alfred Huan2- Statistical Physics

University of Cambridge Part II Mathematical Tripos Dr David Tong3- Statistical Mechanics, 2nd edition R.K. Pathria

We based around these books. Students are encouraged to search for any other book references which contain material that may be helpful to understand the objects.

TopicsFundaments of statistical physicsStatistical mechanics science definitionStatistical basis

Standard deviation and other measures of dispersionFactorialPermutations and combinationsStirling’s approximation Thermodynamic probability distributionBinomial distributionExamples 4 weeks

Statistical distribution laws

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Basic approach in the distribution lawsPhase space Energy levels and energy statesMicrostates and macrostates Classical statistical mechanics

Maxwell-Boltzmann distributionApplications of M.B. distributionSpeed distribution lawMean, RMS, and most probable speedsDoppler broadening of spectral linesEinstein diffusion equationLimitations of M-B distributionExamples 6 weeks

Quantum statistical mechanicsBose-Einstein distribution

Applications of B.E. distributionBlack body radiation: the photon gasWien’s LawStefan-Boltzmann law Examples 4 weeks

Fermi-Dirac distributionApplications of F.D. distributionFermi level and Fermi energyFermi energy level in a metalElectrons statistical distribution: Energy distribution of electronic gasDensity of state distribution: Electrons and holes density in an

Intrinsic semiconductors

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ExamplesComparison of M-B, B-E, and F-D laws 6 weeks

Statistical mechanics and thermodynamics laws Brief review of thermodynamics

Thermal equilibrium Kinetic theory of gases Extensive parameters of thermodynamics Entropy and thermodynamics probabilityThermodynamics potentialCanonical ensembleHeat capacity or specific heat Examples 4 weeks

General problems 3 weeks Three examinations depending on time board 3 weeks

This syllabus may be subject to changes; i.e. we may take either longer or shorter time to finish the topics, if any changes happened we will be notified well in advance.

The end

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Statistical Mechanics

Fundaments of statistical physics

Statistics is the study of how to collect, organize, analyze, and interpret numerical information from data. Descriptive statistics involves methods of organizing, picturing and summarizing information from data. What is Statistical Mechanics? Science definitionStatistical Mechanics provides the connection between microscopic motion of individual atoms of matter and macroscopically observable properties such as temperature, pressure, entropy, etc. This branch of physics applies statistical principles to the mechanical behavior of large numbers of small particles (such as molecules, atoms, or subatomic particles) in order to explain the overall properties of the matter composed of such particles. ♦ Maxwell - Boltzmann statistics explain the behavior of the molecules of a gas. ♦ Bose-Einstein statistics  explains the behavior of large numbers of bosons. ♦ Fermi-Dirac statistics  explains the behavior of large numbers of particles that obey the Pauli exclusion principle (such as electrons).

We begin with the introduction of some important ideas which are fundamental in our understanding of statistical mechanics. The list of important terms is as follows:1. Microstate2. Macrostate3. Partition Function4. Derivation of Thermodynamic laws by Statistical Mechanics5. The idea of interaction(a) particle (system) - external environment(b) Particle (within a system of many particles)6. Entropy: Disorder Number7. Equipartition Theorem: the energies are distributed equally in different degrees of freedom.8. Temperature9. Distributions: Maxwell-Boltzmann, Fermi-Dirac, Bose-Einstein.

Why do we need Statistical Mechanics ?

1. Statistical Mechanics provides the microscopic basis for thermodynamics, which, otherwise, is just a phenomenological theory. 2. Microscopic basis allows calculation of a wide variety of properties not dealt with in thermodynamics.

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3. Microscopic trajectories can, in principle and in practice, be generated providing a window into the microscopic world.

Statistical distribution lawsTypes of statistics 1- CLASSICAL STATISTICS (MAXWELL – BOLTZMANN) STATISTICS.

2- QUANTUM STATISTICS (BOSE – EINSTEIN AND FERMI-DIRAC) STATISTICS.The statistical mechanics is used to determine the most probable way in which a fixed total amount of energy is distributed among the various members of an assembly of identical particles, that is, how many particles are likely to have the energy E1, how many to have energy E2, and so on. Consider three kinds of particles:1- Identical particles of any spin that are sufficiently widely separated to be distinguished. The molecules of a gas are particles of this kind and the Maxwell-Boltzman distribution law holds for them.2- Identical particles of zero or integral spin that cannot be distinguished one from another. Such particles do not obey the exclusion principle, and the Bose-Einstein distribution law holds for them. Ex. (bosons) photons or alpha particles.3- Identical particles of spin (1/2) that cannot be distinguished one from another. Such particles obey the exclusion principle, and the Fermi- Dirac distribution law holds for them. Ex. electrons or protons.

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B A SIC A P P R O A CH IN THE T H R E E ST A TIS T ICS The basic approach in three statistics is essentially the same but we get the

different results due to different assumptions. The common approach is as follows:

In any dynamic isolated system, the total number of particles (N) and the total

energy (E) has to remain constant.

N = ni = constant

dN = dni = 0 ----(1)

E = Eini = constant

dE = Eidni = 0 ----(2)

When the system is in equilibrium then it exists only in the most probable state. From the assumptions of a given kind of statistics, we calculate thermodynamical probability (P) for any given macrostate. In the most probable

state of the system (P) must be maximum. For all natural systems P is for a very large number. We, therefore, deal with lnP. For most probable state (P) is maximum. d(lnP) = 0 -------(3)

These three conditions can be incorporated into a single eq. by the method of Lagrange’s undetermined multipliers. We multiply eq. (1) with -, eq. (2) by - and add to eq. (3)

d(ln P) - dni - Ei dni = 0

d(ln P) - ( + Ei )dni = 0 ----- (4)Where and are constant quantities independent of the ni.Phase Space: a Framework for Statistics

Statistics involves the counting of states, and the state of a classical particle is completely specified by the measurement of its position and momentum. In Mathematics and Physics, phase space is a space in which all possible states of a system are represented. So, if a system consists of n particles then we need 6n co-ordinates to describe the behavior of the system in the phase space. If we know the six quantities

x,y,z,px,py,pz then we know its state.

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The position of a particle is a point having the coordinates x,y,z in ordinary three dimensional

space. This combination position and momentum space is called phase space, which is a six dimensional space. At any instant of time, suppose one of the particles have its position co-ordinates lying between x and x+dx,y and y+dy, z and z+dz and let its momentum coordinates lie between px and px+ dpx, , py and py+ dpy ,pz and pz+d pz. Thus every particle is completely

specified by a point in phase space. The volume of the cell in the phase space is

dv = dxdydzdpxdpydpz

According to the uncertainty principle

dxdpx = dydpy = dzdpz = h

Then the smallest "cell" in phase space which we can consider is constrained by the uncertainty principle to be

dvminimum = h3 =dxdpxdydpydzdpz

The minimum volume of an elementary cell in phase space in Quantum

Mechanical system is h3, where h is Planck’s constant

Total number of elementary cells in phase space = Total volume in phase space / Volume of

one elementary cell

= dxdydz dpx dpy dpz / dv

Energy levels and energy states

Consider the population cells groups in the phase space.Suppose gi is the number of cells in the phase space for i groups with energies Ei.As shown in the Fig. belowEi is the energy levels

ni is the occupation numbers

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Ei

E3 E2 E1

gi

1 2 3 4

g1=1 n1=3 (no degenerate) i=1

E

g2=4 n2=4

g3=3 n3=5

gi is the number of cells with the same energy.

The figure represents in a schematic way the concepts of energy levels, energy states, and the

degeneracy of a level. The energy levels can be thought of as a set of shelves of different

elevations, while the energy states correspond to a set of boxes on each shelf. The

degeneracy gi of level i is the number of boxes on the corresponding shelf or the number of cells.

The sum of the occupation ni over all levels equals the total number of particles,∑

ini=N

Since the particles in those states included in any level i all have the same energy Ei , the total energy of the particles in level i is Ei ni, and the total energy E of the system is ∑

iE in i=E

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Microstates and macrostates

A Microstate is defined as a state of the system where all the parameters of the constituents (particles) are specified. Many microstates exist for each state of the system specified in macroscopic variables (E, V, N,) and there are many parameters for each state. We have 2 perspectives to approach in looking at a microstate:

1. Classical MechanicsThe position (x,y,z) and momentum (Px, Py, Pz) will give 6 degrees of freedom and this is put in a phase space representation.2. Quantum MechanicsThe energy levels and the state of particles in terms of quantum numbers are used to specify the parameters of a microstate.

The macrostate is the distribution of particles in the phase state, but microstate is the way of distribution of the particles in the phase state (how the particles distributes in the phase state).Classical statistical mechanicsMaxwell – Boltzmann distribution (M-B distribution)Consider a number of N particles of energy level 1,2,3,…..,n with energy E1, E2,E3,….En. If there are gi cells with energy Ei, the number of ways in which one particle can have the energy Ei is gi . The total number of ways (Pi) that ni particles can have the energy Ei is (gi) ni. Hence the number of ways in which all N particles can be distributed among the various energies is (P1)(P2)(P3)…= (g1)n1(g2)n2(g3)n3 ………This result does not equal P, however, since we must take into account the possible permutation of the particles among the different energy levels.

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The total number of ways in which the N particles can be distributed among the possible energy levels (the most probable distribution) is: p= N !

n1 !n2 !n3!(g¿¿1)n1(g¿¿2)n2(g¿¿3)n3¿¿¿……….

OrP=N !∏i=1

N

(gini

ni! )…………(1)

No. Energy No. of particle No. of state1234...n

E1E2E3E4...En

n1n2n3n4...nn

g1g2g3g4...gn

ExampleConsider 4 cells, three distinguishable particles in one energy level. Find the probability distribution of these particles.g=4 N=3 A,B,CP =(gi)ni =(g1)n1= 43 = 64

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ABC ABC ABC ABCAB CC ABAC B. . .ExampleSuppose 4 particles can be distributed among 2 energy levels as; n1=2, g1=3, n2= 2, g2=2 . Find the number of ways of particles distribution among: a) each level, b) the two levels, c) also find the most probable distribution. Using M.B. distribution.

a) P1 = (g1)n1 = 32 = 9 P2= (g2)n2 = 22 = 4b) p=(g¿¿1)n1(g¿¿2)n2=(3¿¿2) (22 )=36¿¿¿ways. c) The most probable distribution

p= N !n1 !n2 !

(g¿¿1)n1(g¿¿2)n2 ¿¿

¿ 4 !2!2 !

(32 )(22)

¿ 4 × 3× 2× 12×12×1

(32 )(22)

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E

g2=5

g1=2 N=6

= 636 = 216 ways. Example Calculate the most probable distribution of distributing 3 objects a, b and c into two cells of two energy levels with the occupation {1, 2}. Answer:

p= N !n1 !n2 !

(g¿¿1)n1(g¿¿2)n2 ¿¿

Therefore, there are three ways (3! / 1! 2! ) (1)1(1)2 = 3 | a | b c |, | b | c a |, | c | a b |. ExampleCalculate the number of ways of distributing 20 objects into six cells of six possible energy levels with the arrangement {1, 0, 3, 5, 10, 1).

p= N !n1 !n2 !n3!

(g¿¿1)n1(g¿¿2)n2(g¿¿3)n3¿¿¿…….

(20! / 1! 3! 5! 10! 1!)(1)1(3)1…. = 931170240 ExampleGiven 6 distinguishable particles, 2 energy levels (one with a degeneracy of 2 and the other degeneracy of 5), Calculate the number of macrostates and microstates in this system.

The macrostates =7 which are:15

(6,0), (5,1),(4,2), (3,3), (2,4),(1,5) and (0,6)The microstates are: For(6,0) macrostatep= N !

n1 !(g¿¿1)n1=6 !

6 !(2)6=26=64¿ways

From(5,1) macrostatep= N !

n1 !n2 !(g¿¿1)n1(g¿¿2)n2= 6 !

5 !1!(2)5(5)1=30 ×25=9600 ¿¿ ways

And so on for the other macrostates.Maxwell - Bolt z m ann law for large number of particles

For fixed total number of particles and fixed total energy.N=∑

ini

E=∑i

ni❑i

dN = 0 and dE= 0∑

idni=0∑

i❑idni=0

For maximum distribution, the most probable distributionlnpni

=0

From equ. (1)lnp=lnN !−∑ ln ni!+∑ ni ln g i…… ………. (2)

From stirling formula16

ln ni !=ni ln ni−n i

Equation (2) becomeslnp=lnN !+∑ ni ln gi−∑ n i lnn i+∑ ni

lnp=lnN !+∑ (ni ln gi¿−ni ln ni+ni)¿

lnpni

=0+ ln gi−ni

ni−ln ni+1

¿ ln gi−1−ln ni+1=ln g i−ln ni

lnpni

=ln¿

Using Lagrange's multiple methoddlnp−dN−dE=0

∑i

( lnpni

¿d ni−d ni−❑id ni)=0¿

∑i

( lnpni

¿−−❑i)dni=0¿

lnpni

−−❑i=0 …………… ..(4)

Where and are constant quantities independent of the ni.From eqns.(3) and (4)ln

g i

ni−¿- i=0

ni=g i e−−❑i ……………………….(5)

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This law is known as Maxwell-Boltzmann law of energy distribution and ei Boltzmann factor.

This formula gives the number of particles ni that have the energy i in terms of the number of cells in phase space gi that have the energy i and the constants , .Evaluation of constants (, )We must evaluate gi, and in equ.(5). It is convenient to consider a continuous distribution of particle energies rather than the discrete set 1, 2 and 3 ,….. If n()d is the number of particles, and g()d is the number of cells in phase space whose energies lie between and +d, equ.(5) becomesn()d=g()e−¿e−¿d … … …… … ….(6 )¿ ¿

In terms of momentum, since =p2/2m eqn.(6) becomesn ( p ) dp=g ( p)e−¿ e−p2/2 mdp ¿ …………..(7)n ( p ) dp is the number of particles, and g(p)dp is equal to the number of cells in phase space in which a particle has a momentum between p and p+dp. Since each cell has the volume h3.

g ( p ) dp=∭∬dxdydz dpx dpy dpz

h3

∭dxdydz=V occupied phase volume in ordinary position space∬dpx dpy dpz=4 π p2 dp [see phase space]

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Where 4πp2 dp is the volume of a spherical shell of radius p and thickness dp in momentum space. Henceg ( p ) dp=4 πV p2 dp

h3 ………………….(8) n ( p ) dp=4 πV p2 e−¿e

−p2

2 m

h3 dp¿ …………………(9)0

❑

n ( p ) dp=N

By integrating equ.(9)N=4 π e−¿ V

h3 0

❑

p2 e−p2

2 m dp¿

N= e−¿ V

h3 ( 2 πm❑ )

3 /2

¿

Where we use the definite integral0

❑

x2 e−x2

dx=14 √ π

α 3

Or0

❑

p2 e−−p2

2 m dp=14 √ π (2m)3

β3

Hence e−α= N h3

V( β2 πm

)3 /2

n ( p ) dp=4 πN ( β2 πm

)3 /2

p2e− p2

2m dp ……………(10)19

To find β

P2=2m and dp= md√2m∈

We can write equ.(10) in the formn (∈ ) d∈=2 N β3 /2

√π√∈ e−β∈d∈……..(11)

The total energy is E=

0

∞

∈n (∈ ) d∈

¿ 2 N β3/2

√π 0

∞

∈3/2e− β∈d∈

E=32

Nβ ……………..(12)

Where we use the definite integral0

∞

x3 /2 e−x dx= 34 α2 √ π

α

If we compare eqn.(12) with the eqn. of the kinetic theory of gases E=3

2NKT

We findβ= 1

KT… ………………..(13)

Where K is Boltzmann constant and T absolute temperature.20

n(E)

E

After the evaluation of and, we can write (equ.11) the Boltzmann distribution law in its final form.n (∈ ) d∈= 2πN

(πKT )3 /2 √∈e−∈ /KT d∈ …………………(14)Boltzmann distribution of energies, this equation gives the number of particles with energies between and +d as shown in the fig.

Noting that E= p2

2 m=1

2m v2

dE= pm

dp=mvdv

From eqn.(14) the Boltzmann distribution of particle momentums and speeds can be obtained asn ( p ) dp= √2πN

(πmKT )3/2 p2 e− p2/2mKT dp………………..(16)21

Boltzmann distribution of momentum, this is gives the number of particles having momentum between p and p+dp.n ( v )dv=√2πN m3 /2

(πKT )3/2 v2e−mv2/2 KT dv ………………(17)

Boltzmann distribution of speeds, this is gives the number of particles having speeds between v and v+dv.The Maxwell –Boltzmann distribution function (FMB)FMB which is proportional to the probability that a particle will be found in the state vhas the form, or FMB =P(V) the probability.FMB=

n (v ) dv4 π v2 dv

FMB=N ( m

2 πKT)

3 /2

e−mv2/2 KT=N ( m2πKT

)3/ 2

e−E /KT=N ( m2 πKT

)3/2

e−p2 /2 mKT………………..(18)Where 4πv2d vthe volume of a spherical shell of radius v and thickness dv in speed space.Examples

1- Find the root mean square velocity (vr.m.s) for a Maxwell- Boltzmann distribution.SolutionThe speed distribution function n (v) is denoted by dnv=n (v ) dv

n ( v )dv=F MB(4 π v2)dv From equation (18)22

n ( v )dv=4 πN ( m2πKT )

32 v2e

−mv 2

2 KT dv………… ..(20)

The average value of the square of the speed is(v¿¿2)= 1

N 0

∞

v2d nv=1N

0

∞

v2 n (v ) dv¿From equation (20)(v¿¿2)= 1

N 0

∞

v2 4 πN ( m2 πKT )

32 v2 e

−mv2

2 KT dv ¿

(v¿¿2)= 4 πNN ( m

2 πKT )32

0

∞

v4 e−mv2

2 KT dv ¿

(v¿¿2)=4 π ( m2πKT )

32 3

8 √ π ( 2 KTm

)5/2

¿ [see the useful integrals table below](v¿¿2)=3 KT

m¿

vrms=√v2=√ 3 KTm

2- Find the most probable speed (vp) for a Maxwell- Boltzmann distribution.

Solution The most probable speed in the M.B. distribution is that for which n (v) =4πv2 FMB is maximum or dnv

dv=0.

From equation (20)n ( v )=4 πN ( m

2 πKT )32 v2 e

−mv2

2KT

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dnv

dv=4πN ( m

2 πKT )32 v e

−mv2

2KT [2−v2( mKT )]=0

vp=√ 2 KTm

Useful integralsI n ( α )=

0

∞

une−α u2

du 0n 0 1 2 3 4 5In() 1

2 √ πα

12

14 √ π

α3

12α2

38 √ π

α5

1α3

3- Find the mean velocity of the molecularv. v= 1

N 0

∞

v d nv=1N

0

∞

vn (v )dv

¿ 1N

0

∞

v 4 πN ( m2 πKT )

32 v2 e

−mv 2

2 KT dv

¿4 π ( m2 πKT )

32

0

∞

v3 e−mv2

2KT dv

Using the integral for n=3 in table (1)v=√ 8 KT

πm

This is the mean or the average speed of the molecules.Note, the average energies ∈−¿¿ per particles from the fig. of energy distribution above is:

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∈−¿=E

N ¿ From eqn. (12)∈

−¿=32

1β ¿

∈−¿=3

2 KT ¿ ……………….(15) Average particles energyThe speed of a particle with the average energy of 3

2KT is

vrms=√v2=√ 3 KTm

………… ..(20) rms speed.Since ∈=1

2m v2=3

2KT . This speed is denoted vrms because is the

square root of the average of the squared particle speeds.Example:

Let vX ,vY ,vZ represent the three components of velocity of a molecule in gas. Deduce expressions for the following mean values in terms of K,T and m.

a/ ¿ vX >¿ b/ ¿ vX2 >¿

Solution :

a/ vX=¿ v X≥1/N −∞

∞

v X p(v X)d v X=−∞

∞

vX ( m2πkT )

12 e

−mV X2

2 KT d v X

=( m2 πkT )

12 −∞

∞

vX e−mV X

2

2 KT d v X=0

Since−∞

∞

vX e−mV X

2

2 KT d v X=0

b/ vX2 =¿ v X

2 >¿= −∞

∞

vX2 p(vX ¿d vX=

−∞

∞

vX2 ( m

2πkT )12 e

−mV X2

2 KT d v X

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=( m2 πkT )

12 1

2 √ (2 KT )3 πm3 = KT

m see table (1)

G r a p hi c al distri b u t ion of M-B spe e d distrib u tion

v p < < v rms

These three measures of v are not equal because the distribution is not symmetrical about its peak.

A more exact expression for the average velocity:

= =

Note that M is the molar mass and that the gas constant R is used in the expression.k is the Boltzmann constant. If you multiply k by Avogadro's number, you'll get R .

v p = =

vrms =

=

=

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Make sure that you see that v rms = , not v rms = . The latter equation reduces to v rms = , which is not the case. v rms requires the mean of the squares of the velocities. Square the velocities first, then take their mean.

ExampleCalculate the root mean square speed of methane molecules at 27°C.

vrms = √3RT/MT = 27+ 273 = 300 K, M= 16. R= 8.314 x 107 erg k-1 mol-1

vrms = √3X8.314X107X300/16 = 683.9X102 cm s -1

= 683.9 ms-1

A Maxwell-Boltzmann speed distribution changes with temperature. As discussed with the kinetic molecular theory, higher temperatures lead to higher velocities. Thus the distribution of a gas at a hotter temperature will be broader than it is at lower temperatures.

Dependence of Maxwell-Boltzmann speed distribution on Temperature

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The total area underneath the Maxwell-Boltzmann speed distributions is equal to the total number of molecules{(n/dv¿)dv }¿. If the area under the two curves is equal, then the total number of molecules in each distribution are equal. Dependence of Maxwell-Boltzmann energy distribution on Temperature

T1 < T < T2

The Maxwell-Boltzmann speed distribution also depends on the molecular mass of the gas. Heavier molecules have, on average, less kinetic energy at a given temperature than light molecules. Thus the distribution of lighter molecules like H2

is much broader and faster than the distribution of a heavier molecule like O 2 :

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Dependence of Maxwell-Boltzmann speed distribution on Molecular Mass

Example:

Let vX ,vY ,vZ represent the three components of velocity of a molecule in gas. Deduce expressions for the following mean values in terms of K,T and m.

a/ ¿ vX >¿ b/ ¿ vX2 >¿

Solution :

a/ vX=¿ v X>¿1/N −∞

∞

v X n(v X)d vX=−∞

∞

vX ( m2 π kT )

12 e

−mV X2

2 KT d v X

=( m2 π kT )

12 −∞

∞

vX e−mV X

2

2 KT d v X=0

Since−∞

∞

vX e−mV X

2

2 KT d v X=0

b/ vX2 =¿ v X

2 >¿= −∞

∞

vX2 n(vX ¿d vX=

−∞

∞

vX2 ( m

2π kT )12 e

−mV X2

2 KT d v X

=( m2π kT )

12 1

2 √ m3 π(2 KT )3 = KT

m see table (1)

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Applications of M.B distribution

The Doppler broadening of spectral lines

One of the effects which arise from the distribution of the velocities of the molecules in a hot gas at low densities is the brooding of the spectral lines which are emitted by the gas molecules. This broadening can be used as an experimental check for the validity of the Maxwell Boltzmann velocity distribution. This broadening (i.e. spread) arises from the distribution of velocities of the molecules in a gas.

VX

Observer

gas molecule

From Doppler expression

λ=λo(1−V X

c )Where

λois the emitted wave length

λ is the observed wave length

The velocity of a moving molecule toward an observer is

. V X=c (λo−λ)

λo

and

dV X=−cλo

dλ

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The negative sign only relates to the direction chosen forV X. If the probability distribution function f x (V X ) is defined as nx(vx)/N so that f x (V X ) d V Xis the probability that the x-component of the velocity is in the range VX to VX+dvX

Then, f X (V X )=( m2 πKT

)12 e

(−mV X

2

2 KT )

Note that the velocity is not three velocities but in x- direction only.

The fraction of the radiation f λ ( λ ) d λ which is received by the observer in the wavelength λ to λ+d λ is obtained by

f λ ( λ ) d λ=( m2 πKT

)12 e

(−mc2

2 KT (λ o−λ

λo)

2

) cλo

d λ

The intensity of radiation emitted in the wavelength λ to λ+d λ will be given as

I ( λ ) d λ=C f λ ( λ ) d λ

=C ( m2 πKT

)12 e

(−mc2

2 KT (λ o−λ

λo)

2

) cλo

d λ

=I (λo)e(−mc2

2KT (λo− λ

λ o)

2

)d λ

Where C is a constant and I ( λo ) is the intensity emitted per unit range of wavelength at λo.

The intensity distribution as a function of wavelength will be therefore have the form of a Gaussian curve about λo as shown

I(λo¿ I(λ ¿

31

λo λ

The Einstein diffusion equation

A relationship between the mobility μ and the diffusion coefficient D of ions in a gas will be derived as an application of the M-B statistics.

Consider a gas containing ions in electric field E.

Let n(x) ions per unit volume at a distance x from the end of the vessel.

If the charge on each ion is q then potential energy of an ion at x ,compared with that of an ion at x=0 , will be

∈¿X)= - qEX

E Field plates

X

gas

Using the Boltzmann factor for the relative probability than an ion will have a particular energy gives

n(x)n(0)

=e(∈ (0)−∈ ( X ))

KT =e−∈(X )

KT =eqEXKT

Where n(0) is the concentration of ions at X=0 and T is the temperature of the gas.

32

If the ions have a mobility μ ,there will be an ionic drift velocity μE in the direction of the field E.

The drift current per unit area per unit time is

J(drift)= n(x). μE

Also, if D is the diffusion coefficient, there will be a diffusion current across unit area at X per unit time in the direction of the concentration gradient ,which is

j(diffusion)= - D dn(x )dx

j(drift)+j(diffusion) = 0

where ,no total current flow, no total movement of ions

or

n(x) . μE = Ddn(x )dx

Substituting for n(x)

dn ( x )dx

=n ( x ) . qEKT

So μ= DqKT

Or μD

= qKT which is the Einstein diffusion equation

Some limitations of M.B law

1- It is applicable only to an isolated gas of identical molecules in equilibrium which satisfied the conditions; the gas is said to be ideal, the gas is dilute or is said to be non-degenerate.

2- The expression for the M.B count does not give correct expression for the entropy S of an ideal gas, this leads to the Gibbs paradox.

33

3- It cannot be applied to a system of indistinguishable particles.

Quantum statistical mechanics

Bose –Einstein distribution (B-E distribution)

The Bose-Einstein distribution describes the statistical behavior of integer spin particles (bosons).

Basic postulates

In B-E statistics, the conditions are:

1-The particles of the system are identical and indistinguishable.

2-Any number of particles can occupy a single cell in the phase space.

3-The size of the cell cannot be less than h3, where h is a Plank’s constant.

4-The number of phase space cells is comparable with the number of particles, i.e., the occupation index f(Ei) is ni/gi≈1.

5-B-E statistics is applicable to particles with integral spin angular momentum in units of h/2π .All particles which obey B.E statistics are known as Bosons.

The distinction between (M-B) statistics and (B-E) statistic is M-B statistic governs identical particles which can be distinguished from one another, while B-E statistic governs identical particles which cannot be distinguished.

Suppose ni indistinguishable particles (Bosons) distributed among gi cells of phase state of energy εi at i level. The first step is to determine the number of ways in which ni indistinguishable particles can distributed in gi cells.

To find gi cell we must put (gi-1) wall

To find Pi ways of distribution for ni particles on the gi cells. There are (ni+gi-1) positions and there is a particle or wall on this position.

The first particle of ni particles will occupy any of (ni+gi-1) positions.

The second particle will occupy (ni+gi-2) positions

34

The last particle can occupy any position of gi and remains (gi-1) positions for the cell walls.

ni+ gi- ni= gi

The sum of distribution ways Pi is

Pi =(n i+g i−1 ) (n i+g i−2 )−−−−−−−−gi

( gi−1 )!

Pi=(n i+g i−1 ) !

(gi−1 ) !

For indistinguishable particles

Pi=(n i+g i−1 ) !¿! (gi−1 ) !

The number of ways P in which N particles can be distributed is the product:

P=∏i=1

N

Pi

PB-E = Π (n i+g i−1 )!¿ ! ( gi−1 ) ! ------------------------ (21)

Where II denotes multiplication of terms stated above the various values of i from i=1 to N.

Example:

The energy level i contains 3 cells and 2 indistinguishable particles. Find the number of the particles distribution ways. Using B.E distribution.

Solution:

gi = 3 ni = 2 Pi =?

Pi =(n i+g i−1 ) !¿! (gi−1 ) ! =

(2+3−1 ) !2! (3−1 ) ! = 4 !

2× 2 = 6

35

This Fig. shows the distribution of 2 bosons between 3 cells.

Example:

Find the distribution way for one level contains one cell non degenerate.

Pi = (n i+g i−1 ) !¿! (gi−1 ) ! =

(¿+1−1 )!¿ !(1−1)! = ¿ !

¿ !0 ! =1

There is only one way of distribution.

To deduce B-E distribution law

Assume (n i+ gi) ≫1

Take the natural log. of both sides of equ.(21)

ln P =∑ ¿¿

From stirling’s formula and let (gi -1) ≈ gi

ln p = ∑ ¿¿ --------------22

From the maximum distribution

∂ lnpmax=0

∂ lnpmax =∑ ¿¿ = 0

Using Lagrange’s method

−β∑ ϵi ∂∋¿0 And −α∑ ∂∋¿0

36

gi

Pi 1 2 31 ••2 ••3 ••4 • •5 • •6 • •

Where β and α are Lagrange’s coefficients

By adding these values the result is:-

=∑ [ ln (n i+g i )!− ln ¿¿¿¿−α−βϵi]∂∋¿0¿

¿

ln ¿+gi¿ −α−βϵi=0

ln ¿+gi¿ =α+βϵi

¿+gi¿ =eα eβϵi

1+gi¿ =eα eβϵi

¿= gieα eβϵi−1

And β = 1KT

¿= gieα eβϵi−1

--------------------------23

This equation represents the most probable distribution of the particles among various energy levels for a system obeying B-E statistics and is therefore, known as Bose-Einstein’s distribution law for an assembly of bosons.

For photonic gas ∑ ∂∋≠ 0 because the number of photons are not constant, so from Lagrange’s method α∑ ∂∋¿0 or α =0 for photonic gas then B-E distribution for photonic gas becomes

¿= gi

eϵikT −1

--------- 24

Applications of B-E distribution

1. Black body radiation: the photon gasEverybody emits radiation which depends upon the nature and the temperature of that body.

37

The ability of a body to radiate is related to its ability to absorb radiation.

Theoretical model of a black body

Suppose a cavity of a volume (V) and contain a large number of indistinguishable photons of various frequencies (υ). The number of states g(p)dp in which a photon can have a momentum between p and p+dp is equal to twice the number of cells in phase space. This is because photons of the same frequency can have two different direction of polarization.

Hence (from eq. 8)

g ( p ) dp=8 πV p2 dph3

Where h is plank’s constant.

p=hυc andp2=

(hv)2

c2

dpdυ

=hc or dp= hdυ

c

p2dp¿ h3υ2 dυc3 and

g (υ )dυ=8 πV υ2

c3 dυ ----------------25

∑ ∂∋≠ 0 Photons may be created and destroyed (Number of photons is not constant)

From (eq. 24)

n(υ)dυ =8πVc3

υ2 dυ

ehvkT −1

----------------------- 26

Where ϵ = hυThis is shows the B-E spectral distribution.

38

λmaxλ

ϵ

The number of photons with frequencies between υ+dυ in the radiation within a cavity of volume V and at absolute temperature T.

The corresponding spectral energy per unit volume is given by:

ϵ (υ )dυ=hυn(υ)V

dυ

ϵ (υ )dυ =8 πhc3

υ3 dυ

ehvkT −1

----------27

Plank’s radiation formula

Which agree with experimental of B-E distribution of photons.

2-Wien’s Law:

If we express the frequency (ν) in eq. (27) in terms of wave length (λ), this equation becomes

ϵ ( λ ) dλ=8 πh λ−3

ehc

λkT−1

dλ ----------------------------- 28

whereν= cλ

This eq. shows that when the temperature increased the λ max . for the maximum energy decreased as shown in the fig. below.

39

Set dϵ (λ)dλ

=0

for λ= λ max

hcλ max KT

=4.965by approximate method

λmax T = Constant -------------------------------------- 29

This is Wien’s displacement law.

It expresses the fact that the peak in black body spectrum shifts to shorter wave lengths as temperature is increased.

3-Stefan-Boltzmann law:

From eq. 27 the total energy density ϵ within the total energy density ϵ within the cavity is the integral of the density over all frequencies

ϵ=0

∞

ϵ ( ν ) d ( ν )=8 πhC3

0

∞ ν3d ν

ehνKT −1

Suppose hνKT

=a variable x so dx= hKT

dν

Multiply the above equation by (hKThKT

¿¿4 , so

ϵ=8 π K4 T 4

h3 c3 0

∞ X3

ex−1dx=constant T 4

When T ≈0 the integral has the value3 ! π4

90= π4

15 .

The emissivity (e) of black body is the energy emitted in sec. from a surface area

and is equal to (cϵ4 )

e=2π5 K4 T 4

15 h3 c2 ∨e=σ T 4 ---------------- 30

40

Stefan – Boltzmann Law. The energy radiated by a black body per second per unit area is proportional to T4 . σ is Stefan Constant =5.67x10-8 W/m2K4

Useful integrals: 0

∞ Xn

ex−1=Γ (n+1)ξ(n+1)

When n = 3, ξ (4 )= π4

90∧Γ (4 )=3 !

We can use Stefan-Boltzmann law to estimate the temperature of the earth from first principles. The sun is a ball of glowing gas of radius Rs=7x105 km and surface temperature Ts=5770 k. Its luminosity is

Ls=4πRs2 σTs

4

According to the Stefan-Boltzmann. The earth is a globe of radius Re=6000 km located an average distance r=1.5x108 km from the sun. The earth intercepts an amount of energy

Pe=Ls(πRe2/r2)/4π

Per second from the sun's radioactive: i.e., the power output of the sun reduced by the ratio of the solid angle subtended by the earth at the sun to the total solid angle 4π.The earth absorbs this energy, and then re-radiates it at longer wavelengths. The luminosity of the earth is

Le=4πRe2σTe

4,

According to the Stefan-Boltzmann law, where Te is the average temperature of the earth's surface. Here, we are ignoring any surface temperature variations between polar and equatorial regions, or between day and night. In steady-state, the luminosity of the earth must balance the radioactive power input from the sun, so equating Le and Pe we arrive at

Te=(Rs/2r)1/2Ts

Remarkably, the ratio of the earth's surface temperature to that of the sun depends only on the earth-sun distance and the solar radius. The above expression yields Te=279 k or 6 oC. This is slightly on the cold side, by a few degrees, because of the greenhouse action of the earth's atmosphere, which was neglected in our

41

calculation. Nevertheless, it is quite encouraging that such a crude calculation comes so close to the correct answer.

Fermi-Dirac Distribution (carriers concentration)

The Fermi-Dirac distribution applies to fermions, particles with half-integer spin which must obey the Pauli exclusion principle.

Basic postulates

Fermi- Dirac statistics is apply to indistinguishable particles which are generated by the Pauli exclusion principle which states that no two particles can occupy the same quantum state, so the number of allowed level cannot be more than the cells number (g i) in that level (gi ≥ ni )

The F-D distribution law is the same as B-E distribution law except that now each cell (that is, Quantum state) can be occupied by at most one particle.

If there are (gi) cells having the same energy ϵi and if there are a number of indistinguishable particles (ni) a, b, c, there are (gi ) position (cells) for (a) leaving (gi -1) position for( b) and (gi-2) position for (c ) and g i – (ni-1) positions for the last particle. (gi-ni) positions not occupied.

The sum of the distribution ways is

gi ( gi−1 ) (gi−2 )−−−−−−−−−−gi−(¿−1 )= gi !( gi−¿) !

Because the particles are indistinguishable

Pi= gi!¿ ! ( gi−¿ )!

The number of ways P in which N particles can be distributed is

P=∏i=1

N

Pi

P=∏ gi!¿ ! (gi−¿ )!

−−−−−−−−−−30

42

Example:

Suppose the energy level i contain 3 cells and two indistinguishable particles. Find the number of the distribution ways, by F-D distribution.

PF-D ¿gi!

¿ ! (gi−¿ )!= 3!

2! (3−2 )!=3 x 2 x 1

2 x1 x1=3

• •• •

• •Example:

Suppose in the above example there are 3 particles and 2 cells

PF-D ¿gi!

¿ ! (gi−¿ )!= 2!

3 ! (2−3 ) !=? impossible for F-D distribution

To find the Fermi-Dirac distribution law

Take the natural logarithm of the both sides of eq. 30

ln P=¿∑ [ lngi !−lnni !−ln ( gi−¿ ) !]¿

, then from Stirling's formula

ln P=¿∑ [ gilngi−nilnni−(gi−¿) ln (gi−¿ )]¿

For maximum probability, small changes dni in any of the individual ni’s must not alter.

dln P max=∑ ¿¿

Taking into account the concentration of particles and energy using Lagrange’s coefficients and adding

−α∑ dni=0∧−β∑ ϵi dni=0

to equation 3, we get

∑ ¿¿

43

ln gi−¿¿ −α−βϵi=0

gi¿ −1=eα eβ ϵi

¿= gieα eβϵi+1

Substituting β= 1KT

¿= gi

eα eϵi

KT +1−−−−−−−−−32

This equation represent the most probable distribution of the particles among various energy levels for a system obeying Fermi- Dirac statistics and is therefore known as Fermi- Dirac distribution law.

α usually negative and can be defined in terms of Fermi energy ϵf.ϵf = ± αKT

So eq. 32 becomes

¿= gi

e(ϵi−ϵf )

KT +1−−−−−−−−−33

This equation is very important in the free electron theory of metals.

The function ( ¿gi

¿=f ¿) is the occupation probability of any cell at the energy ϵi.

This Function is called Fermi-Dirac distribution or probability function and gives the probability that a quantum state at energy ϵ will be occupied by an electron. Or f(ϵ) is the ratio of filled to total quantum states at energy ϵ.

The most probable distribution function is

f ( ϵ )= 11+¿¿

44

F(ϵ)

1

ϵ0

Where ϵf is Fermi energy. It is the maximum level that the electron occupied at absolute zero temperature, below this level all states filled and above which all states empty at T=0 K.

To understand the meaning of the distribution function and Fermi level

At T approaches 0K T ≈0

When ϵ<ϵf

e(ϵ−ϵf )

KT =e−∞=0

f (ϵ) =1 Or %100

When ϵ > ϵf f (ϵ) = 0

The Fermi-Dirac probability function versus ϵ at T=0K is shown below

At T > 0 K

If ϵ = ϵf

f ( ϵ )= 11+e0 =

11+1

=12ϵf

The probability of a state occupied at ϵ = ϵf = 1/2

for all temperatures above 0K

The F-D distribution is plotted for several temperatures, assuming the Fermi energy is independent of temperature is shown below

45

T = 0

T

F(ϵ)

1

1/2

0 ϵf

For T > 0 K there is no zero probability above ϵf and some energy states below ϵf

will be empty

Or that means some electrons have jumped to high energy levels with increasing thermal energy. It should be noted that the Fermi-Dirac distribution function f(ϵ) does not give the number of particles with energy ϵi, but the number of particles

occupying the state i which has an energy ϵi . In order to find the number of

particles ni with energy ϵi we need to multiply the f(ϵ) by number of states gi

having the energy ϵi

ni = gi f(ϵ)

Example:

Calculate the probability that an energy level 3kT above Fermi energy is occupied by an electron. Let T = 300 K.

f ( ϵ )= 1

1+e(ϵ−ϵf )

KT ϵ - ϵf = 3KT

f ( ϵ )= 1

1+e3 KTKT

=0.0474=%4.74 less thanunity

The probability for an energy level above ϵf being occupied increase with T and

below ϵf being empty increase with T.

46

T1

ϵ

T = 0 K

T2>T1

T2

F(ϵ) [1-f(ϵ)]

1

1/2

0 ϵf

Note that the probability of a state a distance dϵ above ϵf being occupied is the

same as the probability of a state a distance dϵ below ϵf being empty.

The function f(ϵ) is symmetric with the function [1- f(ϵ)] about the Fermi energy. As shown in the Figure below

Example:

Assume that the F-D energy level is 6.25 eV and that the electrons follow Fermi distribution. Calculate T at which there is a one percent probability that a state 0.3 eV below the Fermi energy level will not contain an electron or empty.

[1−f (ϵ ) ]=1− 1

1+e(ϵ−ϵf )

KTϵ=ϵf– 0.30 = 6.25-0.3 = 5.95 eV

0.01 =1- 1

1+e(5.95−6.25)

KT KT = 0.0653 eV T = 756 K

When ϵ- ϵf>> KT we can neglect the unity

f(ϵ) = e−(ϵ−ϵf )

KT

This is Maxwell –Boltzmann approximation to Fermi-Dirac function. And this is valid when

e(ϵ−ϵf )

KT ≫1

47

Occupied probability empty probability

F(ϵ) [1-f(ϵ)]

F(ϵ)

1

1/2

0 ϵf

M-B approximation

F-D probability

The F-D probability function and M-B approximation is shown in the fig.

Applications of F-D probability

Density of state distribution:

Electrons and holes density in an intrinsic semiconductors

The number of levels N (∈ )d∈for the electrons between ∈∧∈+d∈ is

N (ϵ ) d ( ϵ )=A ϵ12 dϵ

Where A=constant=8√2 π me∗¿32

h3 ¿

me∗¿❑¿ : The effective mass of electron

The parameter that relates the acceleration of an electron in the conduction band of crystal for an external force: a parameter that takes into account the effect of internal force in the crystal.

Now take the original point ϵ = 0 at valence band ϵ = ϵv = 0 and at the conduction band ϵ = ϵc = ϵg

Where ϵg is the band gap energy

See the figure below

48

0

1- For conduction band :

N (ϵ ) dϵ=A(ϵ−ϵg )12 dϵ

The free electron density or the thermal equilibrium electrons in the conduction band (n) is

n=ϵg

∞

N (ϵ ) f (ϵ )dϵ

n=ϵg

∞

A (ϵ−ϵg)12 1

1+e(ϵ−ϵf )

KT

dϵ

n=Nc e−( ϵg−ϵf

KT )Approximately

The Fermi-energy is in the forbidden energy band gap.

For electron in the conduction band ϵ > ϵc .

49

ϵ +ϵ

ϵgϵf

ϵcϵ=ϵc =ϵg

ϵ=ϵg

N(ϵ)

ϵ=ϵv =0

ϵf= 1ϵf =0

-ϵ

Variation of density of states in conduction and valence band

ϵv

If (ϵc-ϵf)>>KT. Then (ϵ- ϵf) >>KT, so Fermi probability function reduce to the Boltzmann approximation.

Nc=2( 2 π me∗KTh2 )

3/2

=constantand called the effective density of state function in

conduction band

2-for valence band:

N (ϵ ) dϵ=A(0−ϵ )1/2dϵFor the holes in the valence band the function f(ϵ) is symmetrical with the function [1-f(ϵ)].

The holes density or thermal equilibrium concentration of the holes in the valence band(p) is

p=−∞

0

N (ϵ ) [1−f (ϵ ) ]dϵ

p=−∞

0

A(0−ϵ )1/2 [1− 1

1+e( ϵ−ϵf )

kT

]dϵ

p=N ve−( ϵf

KT )approximatly

N v=2(2 π mp¿ KTh2 )

3 /2

=constant and called the effective density of states in the

valence band

50

N(ϵ)

ϵϵfdϵ

f(ϵ)

dϵ ϵf ϵ ϵfdϵ ϵϵ

N(ϵ)f(ϵ)

For intrinsic semiconductor

n = p = ni

¿2=np

¿=n=p=√np

¿=¿

Example:

Fermi level position in intrinsic semiconductors

p=N ve−( ϵf

KT ) andn=Nc e−( ϵg−ϵf

KT )

n = p, ∴ N v e−( ϵf

KT ) = Nc e−(ϵg−ϵf

KT )

lnN v - ( ϵfKT

) = lnNc -¿)

ϵf = ϵg2

+ KT2

KT ln( NcNv

)

N v=2(2 π mp¿ KTh2 )

3 /2

And Nc=2( 2π me∗KTh2 )

3/2

ϵf = ϵg2

+ 34

KT ln( mp¿

me¿ )

where mp¿=me¿ϵf = ϵg2

Also ϵf = ϵg2 when T = 0

As a result n=p=(NvNc)1 /2 e−( ϵg

2KT ) n=p ≈ Nc e−( ϵg

2 KT )

51

Variation of density of state with ϵ Fermi Dirac probability with ϵ Electron Density with ϵ

Comparison of M-B, B-E, and F-D laws

M-B B-E F-DClassical particles Bosons electronsdistinguishable Indistinguishable IndistinguishableAny spin Spin=0, 1, 2, ……… Spin=1

2, 32

, 52

……

No high limit for the number of particles for each case

No high limit for the number of particles for each case

One particles for each case

f (∈ )=ni

gi= 1

eα e∈ i

Tk

f (∈ )=ni

gi= 1

eα e∈ i

Tk−1f (∈ )=

ni

gi= 1

eα e∈ i

Tk +1

In these formulas ni is the number of particles whose energy is ∈i and gi is the

number of states that have the same energy ∈i.The quantity f (∈ )=ni

gi is called

occupation index of a state of energy ∈i , is therefore the average number of particles in each of the states of the energy. The ratio between the occupation indices of two energy levels ∈i and ∈ j is called Boltzmann factor

f (∈i)f (∈ j)

=e(∈ j−∈i)

kT

All the three distributions can be represented by a single equation

g i

ni=[e

α+∈i

Tk +x ¿

If x=0 M-B x=-1 B-E x=1 F-D

Extensive parameters of thermodynamics

In the energy representation ,the fundamental equation of a pure fluid is given by the relation :

U=U(S,V,N) energy representation

S=S(U,V,N) entropy representation

dU =¿

52

This equation describes a thermodynamic process.

From a comparison with the usual expression for the conservation of energy

∆ u=∆ Q+∆W mechanical0+∆ W chemical

du=TdS−PdV+μdN

We have the following definitions of the intensive parameters or thermodynamic fields:

Temperature T=¿

Pressure P=−¿

Chemical potential μ=¿

The functions

T=T(S,V,N) , P=P(S,V,N) and μ =μ (S,V,N)

are the equations of state in the energy representation. The knowledge of a single equation of state is not enough to obtain a fundamental equation. However, two equations of state are already enough ,since

T=T(S,V,N) , P=P(S,V,N) and μ =μ (S,V,N)

are homogeneous functions of zero degree of their variables.

In the entropy representation

S=S(U,V,N)

dS=¿

We have the equations of state in the entropy representation

¿

¿

¿

Entropy and thermodynamics probability (the relation)

53

In an isolated system, the entropy (s) will increase when in irreversible processes ,and it will be maximum at equilibrium state .The thermodynamic probability (P) will also be ,maximum at equilibrium state . So there will be a relation between (S) and (P)

S=f (P)

In the case of two thermodynamic systems A and B when in contact with each other the total entropy and the thermodynamic probability of that two systems are

P=PA+PB

ST=SA+SB

f ( PA PB )=f ( PA )+ f ¿

The only function which satisfy this relation is logarithmic

S=K ln P

Where K is constant

But the first law of thermodynamic for isolated system is

du=Tds+PdV

1T=¿

This equation shows how the macro state system for the temperature inter in the statistical mechanics.

From M-B distribution

d ln P=∑ lngi

nid ni

Also

lng i

ni=β Ei−ln α

54

A → B

UA,NA,VA ← UB,NB,VB

d ln P=∑ β Ei d n i−ln α∑ dn i

The result :

d ln P=βdu

β=d ln Pdu

But

S=K ln P

β= 1K

¿

To know the constant K ,suppose N molecules of an ideal gas in a volume V and the same molecules was made to occupy a bigger volume VO , so the relative probability to find one

molecule in the smaller volume instead of the bigger volume is (VV O

¿ ,and this probability for

two molecules is ( VV O

)2

and for N molecules is ( VV O

)N

,or

PPo

=( VV O )

N

ln P−ln PO=N ( lnV −ln V O )

S−SO=K N ¿

¿

But from Maxwell relation in thermodynamics

¿

PT

=K NV

But the ideal gas equation is

PT

=K NV

K=K =Boltzman constant

55

Or the constant K is Boltzman constant k

β= 1KT

And

ni=NZ

g ie−Ei

KT From Boltzman equation

S=K ln P

Z=∑ gi e−Ei

KT

Z is called partition function.

−Ei

KT is called Boltzman coefficient which is an important function decreases with decreasing T

and increasing Ei .

We can define variable thermodynamic coordinate in terms of Z .

By differentiate Z in terms T at constant V we get

¿

¿ 1K T 2 ∑ gi Ei e

−Ei

KT ¿ ZNK T 2∑ Ei ni=

ZuNK T 2

U=NK T 2( ∂ ln Z∂ T )

V

where U is the internal energy , but

ni

g i=N

Ze

−Ei

KT

K ln P=S

And

ln P=∑ ni lngi

ni

56

S=K∑ ni ln(g i

ni¿)¿

¿ K ln ZN ∑ d ni+¿ 1

T ∑ E ini ¿

S=KN ln ZN

+ UT

From the first law of thermodynamic

TdS=dU+PdV −μdN

P=−¿

¿−[ ∂∂V

(U−TS)]T ,N

The quantity (U−TS) is called Helmholtz free energy function (F)

F=U −TS

dF=dU−TdS−SdT

From the first law of thermodynamics

dF=−PdV −SdT +μdN

From this equation we get

P=−( ∂ F∂ V

)T , N

S=−( ∂ F∂T

)V , N

μ=−( ∂ F∂ N

)T , V

There fore

S=S (T ,V , N )

P=P (T ,V , N )

μ=μ (T ,V , N )

57

These equation are equations of state in the representation of Helmholtz. From these equations and taking the crossed derivatives ,we have three Maxwell relations

( ∂ S∂ V

)T , N

=( ∂ P∂ T

)V , N

−( ∂ S∂ N

)T ,V

=( ∂ U∂ T

)V ,N

−( ∂ P∂ N )

T ,V=( ∂ U

∂ V )T , N

From Helmholtz function (free energy)

F=U −TS

S=NK ln Z+ UT

F=U−NKT ln Z−U

F=−NKT ln Z

pressure=−( ∂ F∂V

)T , N

P=NKT ( ∂ ln Z∂ V )

T , N

And also

μ=−( ∂ F∂ N )

T ,V=−KT ln Z

ln Z=−μKT

Z=e− μKT

ni

g i=N e

(μ−E i)KT

58

Where the relative ni

N the average of fractional number of the particles in the level (i) also (

ni /N

gi¿ is the relative fraction for the particles for one energy state at any level.

Thermodynamic potential

Instead of working with energy representation or entropy representation ,it is convenient to work with independent variables of easier experimental access ,such as T,P

T=T (U , S )

We have the following thermodynamic potentials

1-Helmholtz free energy

F (T ,V , N )=U−TS

Where the variable S has been replaced by the temperature T

2-Enthalphy

H (S ,P , N )=U+PV

Where the volume has been replaced by the pressure P

3-Gibbs free energy

G (T , P , N )=U−TS+PV

4-Grand potential

Q (T ,V ,M )=U−TS−μN

Canonical ensembleThe canonical ensemble in statistical mechanics is a statistical ensemble representing a probability distribution of microscopic states of the system. For a system taking only discrete values of energy, the probability distribution is characterized by the probability pi of finding the system in a particular microscopic state i with energy level Ei, conditioned on the prior knowledge that the total energy of the system and reservoir combined remains constant. This is given by the Boltzmann distribution,

59

where

is the normalizing constant explained below (A is the Helmholtz free energy function). The Boltzmann distribution describes a system that can exchange energy with a heat bath (or alternatively with a large number of similar systems) so that its temperature remains constant. Equivalently, it is the distribution which has maximum entropy for a given average energy .

It is also referred to as the NVT ensemble: the number of particles (N), the volume (V), of each system in the ensemble are constant, and the ensemble has a well-defined temperature (T), given by the temperature of the heat bath with which it would be in equilibrium.

The quantity k is the Boltzmann constant, which relates the units of temperature to units of energy. It may be suppressed by expressing the absolute temperature using thermodynamic beta,

.

The quantities A and Z are constants for a particular ensemble, which ensure that Σpi is normalized to 1. Z is therefore given by

.

This is called the partition function of the canonical ensemble. Specifying this dependence of Z on the energies Ei conveys the same mathematical information as specifying the form of pi above.

The canonical ensemble (and its partition function) is widely used as a tool to calculate thermodynamic quantities of a system under a fixed temperature. This article derives some basic elements of the canonical ensemble. Other related thermodynamic formulas are given in the partition function article. When viewed in a more general setting, the canonical ensemble is known as the Gibbs measure, where, because it has the Markov property of statistical independence, it occurs in many settings outside of the field of physics.

60

Deriving the Boltzmann factor from ensemble theory

Let be the energy of the microstate and suppose there are members of the ensemble residing in this state. Further we assume the total number of members in the ensemble, , and the total energy of all systems of the ensemble, , are fixed, i.e.,

Since systems in the ensemble are indistinguishable, for each set , the number of ways of shuffling systems is equal to

So for a given , there are rearrangements that specify the same state of the ensemble.

The most probable distribution is the one that maximizes . The probability for any other distribution to occur is extremely small in the limit . To determine this distribution, one should maximize with respect to the 's, under two constraints specified above. This can be done by using two Lagrange multipliers and . (The assumption that would be invoked in such calculation, which allows one to apply Stirling's approximation.) The result is

.

This distribution is called the canonical distribution. To determine and , it is useful to introduce the partition function as a sum over microscopic states

Comparing with thermodynamic formulae, it can be shown that , is related to the absolute temperature as, . Moreover the expression

61

is identified as the Helmholtz free energy F. A derivation is given here. Consequently, from the partition function we can obtain the average thermodynamic quantities for the ensemble. For example, the average energy among members of the ensemble is

.

This relation can be used to determine . is determined from

.

Heat capacity or specific heat

Suppose that a body absorbs an amount of heat and its temperature consequently rises by . The usual definition of the heat capacity, or specific heat, of the body is

If the body consists of moles of some substance then the molar specific heat (i.e., the specific heat of one mole of this substance ) is defined

In writing the above expressions, we have tacitly assumed that the specific heat of a body is independent of its temperature. In general, this is not true. We can overcome this problem by only allowing the body in question to absorb a very small amount of heat, so that its temperature only rises slightly, and its specific heat remains approximately constant. In the limit as the amount of absorbed heat becomes infinitesimal, we obtain

In classical thermodynamics, it is usual to define two specific heats. Firstly, the molar specific heat at constant volume, denoted

62

and, secondly, the molar specific heat at constant pressure, denoted

Consider the molar specific heat at constant volume of an ideal gas. Since , no work is done by the gas, and the first law of thermodynamics reduces to

It follows that

Now, for an ideal gas the internal energy is volume independent. Thus, the above expression implies that the specific heat at constant volume is also volume independent. Since is a function only of , we can write

The previous two expressions can be combined to give

for an ideal gas.

Let us now consider the molar specific heat at constant pressure of an ideal gas. In general, if the pressure is kept constant then the volume changes, and so the gas does work on its environment. According to the first law of thermodynamics,

63

The equation of state of an ideal gas tells us that if the volume changes by , the temperature changes by , and the pressure remains constant, then

The previous two equations can be combined to give

Now, by definition

so we obtain

for an ideal gas. This is a very famous result. Note that at constant volume all of the heat absorbed by the gas goes into increasing its internal energy, and, hence, its temperature, whereas at constant pressure some of the absorbed heat is used to do work on the environment as the volume increases. This means that, in the latter case, less heat is available to increase the temperature of the gas. Thus, we expect the specific heat at constant pressure to exceed that at constant volume, as indicated by the above formula.

The ratio of the two specific heats is conventionally denoted . We have

for an ideal gas. In fact, is very easy to measure because the speed of sound in an ideal gas is written

64

where is the density.

Statistically calculation of specific heats Now that we know the relationship between the specific heats at constant volume and constant pressure for an ideal gas, it would be nice if we could calculate either one of these quantities from first principles. Classical thermodynamics cannot help us here. However, it is quite easy to calculate the specific heat at constant volume using our knowledge of statistical physics. Recall, that the variation of the number of accessible states of an ideal gas with energy and volume is written

For the specific case of a monatomic ideal gas, we worked out a more exact expression for

where is a constant independent of the energy and volume. It follows that

The temperature is given by

so

Since, , and , we can rewrite the above expression as

65

where is the ideal gas constant. The above formula tells us exactly how the internal energy of a monatomic ideal gas depends on its temperature.

The molar specific heat at constant volume of a monatomic ideal gas is clearly

This has the numerical value

Furthermore, we have

and

Example:

A system consist of (N) particles with two states of kinetic energy ± E .prove that T can be define as

1T= K

2Eln( NE−U

NE−U )Solution:

Z=∑ gi e−Ei

KT =e−EKT +e

EKT

66

ni=NZ

e−Ei

KT

U=∑ ni Ei=n1 E1+n2 E2

¿ NZ

(Ee ¿¿−EKT

−EeE

KT )¿

U=EN Ee−EKT −Ee

EKT

Ee−EKT +Ee

EKT

( EN−U ) Ee−EKT =(U+EN ) Ee

EKT

( NE−UNE−U )=e

2 EKT

ln (NE−UNE−U )= 2 E

KT

1T= K

2Eln( NE−U

NE−U )Example:

If the value of the possible energy for a system of particles are 0,E,2E,……,NE

a/prove that the partition function Z when gi =1 is

Z= 1

1−e−EKT

b/calculate the mean energy value for these particles.

Solution:

a/ Z=∑i=0

i=n

gi e−EKT

¿1+e−EKT +e

−2 EKT +e

−3 EKT +…+e

−nEKT

If e−EKT =x

67

Z=1+X+X 2+ X3+…+Xn

¿ 11−X

= 1

1−e−EKT

Because (x)e−EKT is less than one when E¿ KT

b/

ln Z=−ln (1−X )

∂ ln Z∂ T

= EKT

1

eE

KT

U=NK T 2( ∂ ln Z∂ T )

V= NE

eE

KT −1

E=UN

= E

eE

KT −1

Example:

A system of N distinguishable particles distributed on two energy levels E1=0 and

E2=E in equilibrium thermodynamic at temperature T . calculate

a/partition function Z .

b/the distribution function between the two levels .

c/ the internal energy for the system .

d/the energy of the system.

Solution:

a/

Z=∑ e−Ei

KT =1+¿e−EKT ¿

b/

68

ni

N= 1

1+e−EKT

n1

N= 1

1+e−EKT

n2

N= 1

1+eE

KT

n1

n2=

1+eE

KT

1+e−EKT

c/ U=∑ ni Ei=n2 E= NE

1+eE

KT

d/ F=−NKT ln Z

S=−¿

Connection of PM.B with thermodynamics

PM .B=N !∏i

g in !

ni !

ln P=N ln N−N+∑ ¿¿

ln P=N ln Z+ EKT

S=K lnW

S=NK ln Z+ ET

Free energy F=E−TS F:free energy

F=E−TNK ln Z−E

F=−TNK ln Z

Pressure ; P=−( ∂ F∂ V )

T p:pressure

69

P=NKT ( ∂ ln Z∂ V )

T

E=N E

E=−N ∂ ln Z∂ β E:total energy

OR

E=−N T2 ∂ ln Z∂ T

H=NK T2[ ∂∂ T

(log Z)]+RT

H: enthalpy

Boltzmann's Entropy Relation

According to thermodynamics, the function entropy S of a system is related with temperature T by the relation.

1T= ∂ S

∂ E

According to the approach of statistical mechanics ,using equations we write

1T=Kβ=K ∂

∂ Elog W

By equating

∂ S∂ E

=K ∂∂ E

logW

Integrating, we getS=k log W

A liter . Boltzmann assumed that entropy S of a physical system in a definite state is a function of the maximum probability W of that state. Thus

S=F (W )

Where S is entropy and W is the thermodynamical probability of the state.

70

Let us consider two completely independent systems A and B having entropies S1 and S2 respectively. Since entropy is an extensive ( i . e additive) quantity, the entropy of the two systems together will be

S=S1+S2

If W 1 , is the thermodynamic probability of system A and W2 that of system B, then the probability W will be the product of their probabilities, i.e.

W =W 1∗W 2

Thus S=f (w )= f (W 1∗W 2)

S1=f (W 1 )

S2=f (W 2 )

Dividing, then we get

f (W 1∗W 2 )=f (W 1)+ f (W 2 )

Differentiating partially w.r. t. W 1 and W 2 ,we get

W 2´f (¿W 1∗W 2)= f́ (W 1)¿

W 1´f (¿W 1∗W 2)= f́ (W 2)¿

Then we get

f́ (W 1)

f́ (W 2)=

W 1

W 2

W 1 f́ (W 1 )=W 2´f (¿W 2)=…=K ¿

K being any constant

´f (¿W 1)=

KW 1

¿

´f (¿W 2)=

KW 2

¿

And so on

71

By integrating ,we get

f (W 1 )=K log(W 1)+C 1

f (W 2 )=K log (W 2)+C 2

And in general

f (W )=log W +C

Where C is a constant of integration

Thus

S=K log (W )+C

The constant of integration C is evaluated by putting the condition that at absolute zero ( T=O K ) the entropy S=0 and W=1.

C=0

Then

S=K logW

This is the required relation between entropy and thermodynamic probability and is called the Boltzmann's entropy relation. Here k is universal constant known as Boltzmann's constant. The constant of integration is treated as zero because we are interested in changes in entropy and not in its absolute values. Thus, the Boltzmann s entropy relation states that the entropy of a system is proportional to the logarithm of thermodynamic probability of that system.The law that the entropy of the universe always tends to increase, now means that W always tends to increase, which in turn means simply that the universe always changes towards statistically more probable states.

Partition Function And Its Relation With Thermodynamics Quantities:(a) Partition function (z)Let us consider an assembly of ideal gas molecules obeying classical statistics i.e. Maxwell Boltzmann distribution law.

Using this distribution law ,let ni molecules occupy ith state with energy between Ei and Ei+d Ei and degeneracy gi then

ni=g i e−α e

−E i

KT

¿ gi A e−β Ei

72

Where

A=e−α ,β= 1KT

So the total number of molecules in the system

N=∑i

ni

¿ A∑i

gi e−β Ei

NA

=∑i

gi e−β Ei=Z

Where Z is called Boltzmann's partition function or simply the partition function.The term

∑i

g ie−βE i

represents the sum of all the gi e−β Ei terms for energy state of a given molecule.

Consequently the quantity Z indicates that how the gas molecules of a system are distributed or partitioned among the various energy levels and hence is called a partition function.

partition function, Z=∑i

gi e−β Ei

taking ∑i

g i=1 we may write Z=∑i

e−β E i

(b) Correlation with Thermodynamic Quantities( i ) With Entropy, SAccording to Boltzmann's entropy relation between entropy and the maximum thermodynamic probability W is given by S = k log W

For a classical .system, the total number of molecules are

N=∑i

ni

We have W =N !∏i

gini

ni!

Taking logarithm we have

log W =log N !∑i

¿¿

Using Sterling approximation

73

log n!=n log n−n

We get

log W =N log N−N+¿∑i

¿¿¿

According to Maxwell-Boltzmann distribution law, for the most probable configuration, we have

ni=g i e−α e−β Ei

¿ gi A e−β Ei

Substituting for ni

log W =N log N−N+¿∑i

¿¿¿

¿ N log N−N+¿∑i

ni log gi−∑i

ni log g i−∑i

ni log A+∑i

n i β E i+∑i

ni ¿

∑i

ni =N, the total energy of the gas molecules of the system

∑i

ni E i = E, the total energy of the gas molecules of the system.

We get

log W =N log N−N−¿∑i

ni log A+ βE+N ¿

¿ N log N−¿∑i

n i log A+βE ¿

¿ N log N−¿ N log A+βE ¿

¿ N log NA

+βE

¿ N log Z+βE

Because NA

=Z ,the partition function

Substituting, we get

S=K logW

¿ KN log Z+KβE

74

¿ KN log Z+K 1KT

E

¿ KN log Z+ ET

But for an ideal gas

E=32

NKT

Substituting, we get

S=K Nlog Z+ 32

NK

This equation gives the entropy of the system of ideal gas molecules

(ii) with Helmholtz free energy, F

Helmholtz free energy F is given by the relation

F=E−TS

¿ E−T ¿

¿ E−KTN log Z+E

¿−KTN log Z

(III)with total energy, E

Average energy of a system of N particles is given by

E= EN

=∑

ini Ei

∑i

ni

¿∑

ig i EA e−β Ei

∑i

gi A e−β Ei

75

¿∑

ig i E e− β Ei

Z

Since Z=∑i

gi e−Ei

KT

For isothermal isochoric transformation

( ∂ Z∂ T

)V=∑

i

gi∗E i

K T2 e−Ei

KT

¿ 1K T 2 ∑

igi∗e

−Ei

KT ∗EiK T2( ∂ Z∂ T

)V=∑

ig i Ei e

−E i

KT

¿ Z E

E=K T 2

Z ( ∂ Z∂ T )

V

Multiplying by N, we get N E= NK T 2

Z ( ∂Z∂ T )

V

Thus ,the total energy ,E Is given by

E=N E= NK T 2

Z ( ∂ Z∂ T )

V

E=K T 2( ∂∂ T

log Z )V

(iv) with enthalpy, H

Enthalpy is given by

H=E+PV =E+RT For ideal gas PV=RT

¿ NK T 2( ∂∂T

log Z)V+RT

(v)with Gibbs potential, G

Gibbs potential is given by

76

G=H−TS¿ NK T 2( ∂∂T

log Z)V+RT−T (KN log Z+ E

T)

¿ NK T 2( ∂∂T

log Z)V+RT−TKN log Z−E

¿ NK T 2( ∂∂T

log Z)V+RT −TKN log Z−NK T2[ ∂

∂T( log Z )]

¿ RT−NKT log Z

(vi) with pressure , P, of the gas

Pressure is given by

P=−( ∂ F∂ V )

V

¿ NKT [ ∂∂ V

(log Z)]N

(vii)with specific heat at constant volume, CV

CV=( ∂ E∂ T

)V

¿ ∂∂T [NK T 2{ ∂

∂T( log Z )}

V ]

¿ NK [2T ∂∂ x

(log Z )+T 2 ∂2

∂ T 2 ( log Z)]V

77