Transcript
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WEEK 1

Thermal & Statistical Physics

Introduction

Thermodynamics

Definition

-Thermodynamics: is the study of the macroscopic

behaviour of physical systems under the influence of

exchange of work and heat with other systems or their

environment

Thermodynamics can be defined as the science of energy.

Energy can be viewed as the ability to cause changes.

Classical thermodynamics

-thermodynamic states and properties

-energy, work, and heat

-with the laws of thermodynamics

- PV = k, a constant --- R Boyle

The 1st and 2nd laws of thermodynamics

-simultaneously in the 1850s

-W Rankine, R Clausius, and W Thomson (Lord Kelvin).

Statistical thermodynamics

-Late 19th century -- molecular interpretation of thermodynamics

-bridge between macroscopic and microscopic properties of

systems.

-The statistical approach is to derive all macroscopic properties

(T, V, P, E, S, etc.) from the properties of moving constituent

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particles and the interactions between them (including quantum

phenomena)

Note:

-The number of the elements can be very large --- impossible to

keep track of the behaviour of each element.

-A statistical property is a single measurement which gives a

physical picture of what is occurring in all the individual parts.

Chemical thermodynamics

-Is the study of the interrelation of heat with chemical reactions or

with a physical change of state within the confines of the laws of

thermodynamics.

Thermodynamic systems

STATISTICAL

MECHANICS

QUANTUM MECHANICS OF

ATOMS & MOLECULES

MARCOSCOPIC PROPERTIES:

Large number of molecules

TIME

DEPENDENT

BEHAVIOUR:

Chemical

Kinetics

EQUILIBRIUM

PROPERTIES:

Thermodynamics

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-System is the region of the universe under study.

-Surroundings - everything in the universe except the system.

-Boundary -separates the system with the remainder of the

universe.

-fixed, moveable, real, and imaginary

There are five dominant classes of systems:

1. Isolated Systems – matter and energy may not cross the

boundary.

2. Adiabatic Systems – heat must not cross the boundary.

3. Diathermic Systems - heat may cross boundary.

4. Closed Systems – matter may not cross the boundary.

5. Open Systems – heat, work, and matter may cross the

boundary

Thermodynamic parameters

Energy transfers between thermodynamic systems as the result of

a generalized force causing a generalized displacement –conjugate

variables.

SYSTEM

SURROUNDINGS

BOUNDARY

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The most common are

Pressure-volume (mechanical parameters)

Temperature-entropy (thermal parameters)

Chemical potential-particle number (material parameters)

Thermodynamic instruments

Two types: the meter and the reservoir.

A thermodynamic meter is any device which measures any

parameter of a thermodynamic system.

-The zeroth law --it is possible to measure temperature.

-An idealized thermometer--ideal gas at constant pressure.

-a barometer-constructed from a sample of an ideal gas held

at a constant temperature.

-a calorimeter is a device which is used to measure and

define the internal energy of a system.

A thermodynamic reservoir is a very large system does not alter

its state parameters when brought into contact with the test

system.

The earth's atmosphere is an example of heat reservoir.

Thermodynamic states

State - a system is at equilibrium under a given set of conditions

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The state of the system can be described by a number of intensive

and extensive variables.

The properties of the system can be described by an equation of

state which specifies the relationship between these variables.

Thermodynamic processes

-the energetic evolution of a thermodynamic system proceeding

from an initial state to a final state.

The seven most common thermodynamic processes are;

1. An isobaris process occurs at constant pressure.

2. An isochoric process, or isometric/isovolumetric process,

occurs at constant volume.

3. An isothermal process occurs at a constant temperature.

4. An adiabatic process occurs without loss or gain of heat.

5. An isentropic process (reversible adiabatic process) occurs at

constant entropy.

6. An isenthalpic process occurs at a constant enthalpy. Also

known as a throttling process.

7. A steady state process occurs without a change in the internal

energy of a system.

The laws of thermodynamics

Classical thermodynamics is based on the four laws of

thermodynamics:

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• Zeroth law of thermodynamics, stating that thermodynamic

equilibrium is an equivalence relation – about temperature

and temperature scale.

• First law of thermodynamics is about the conservation of

energy -- deals with macroscopic properties, work, energy,

enthalpy, etc

• Second law of thermodynamics is about entropy

• Third law of thermodynamics is about absolute zero

temperature --the determination of entropy.

Thermodynamic potentials

-the quantitative measure of the stored energy in the system.

The five most well known potentials are:

Internal energy U

Helmholtz free energy A + U – TS

Enthalpy H = U + PV

Gibbs free energy G = U + PV – TS

Grand potential

Potentials are used to measure energy changes in systems as they

evolve from an initial state to a final state.

Note: the term thermodynamic free energy is a measure of the

amount of mechanical (or other) work that can be extracted from

a system.

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The Kinetic Theory of Gases

Kinetic theory or kinetic theory of gases -- explain macroscopic

properties of gases, such as P, T, or V, by considering their

molecular composition and motion.

Ideal gases kinetic theory of

-modeling the gases as molecules (or atoms) in constant motion in

space

-A mathematical explanation of the behaviour of gases

-the KE depending on the temperature of the gas

The kinetc theory makes seven assumptions:

The volume occupied by the

gas molecules themselves is

negligible compared with the

volume of space between them

All the particles that

make up the gas are

identical

The distribution of

energy amoung

particles is random

There are sufficent

numbers of molecules for

the statistical average to

be meaningful.

Collisions are

all perfectly

elastic.

The molecules

travel in straight

lines between

collisions

Newtonian

mechanics can be

applied to molecule interactions

Assumptions

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The kinetic theory of gases -- deduced equations that related the

easily observable properties such as P, ρ, V and T to properties

not easily or directly observable -such as the sizes and speeds of

molecules.

Pressure

• The pressure of a gas is caused by collisions of the molecules

of the gas with the walls of the container.

• The magnitude of the pressure is related to how hard and how

often the molecules strike the wall

Absolute Temperature

• The absolute temperature of a gas is a measure of the average

kinetic energy of its' molecules

• If two different gases are at the same temperature, their

molecules have the same average kinetic energy

• If the temperature of a gas is doubled, the average kinetic

energy of its molecules is doubled

Favg

m vx

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Molecular Speed

• All the molecules ⇒average kinetic energy (and therefore an

average speed)

• the individual molecules move at various speeds, ⇒exhibit a

DISTRIBUTION of speeds

• Collisions can change individual molecular speeds but the

distribution of speeds remains the same.

• At the same temperature, lighter gases move on average

faster than heavier gases.

• The average kinetic energy, ν, is related to the root mean

square (rms) speed u

2mu2

1υ =

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Statistical mechanics

Statistical mechanics is the application of probability theory to the

field of mechanics, which is concerned with the motion of

particles or objects when subjected to a force.

-relating the microscopic properties of individual atoms and

molecules to the macroscopic or bulk properties of materials that

can be observed in everyday life

-it can be used to calculate the thermodynamic properties of bulk

materials from the spectroscopic data of individual molecules.

The fundamental postulate in statistical mechanics:

Given an isolated system in equilibrium, it is found with

equal probability in each of its accessiblemicrostates.

This postulate is necessary because it allows one to conclude that

for a system at equilibrium, the thermodynamic state (macrostate)

which could result from the largest number of microstates is also

the most probable macrostate of the system.

• Fundamental concepts: microstate, macrostates, accessible

states, ensemble, the fundamentals postulate

• Microcanonical and canonical ensembles and their

applications

• Open systems, chemical potential. Grand canonical ensemble

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• Distinguishable and indistinguishable particles

• Bose-Einstein and Fermi-Dirac statistics and their

applications

Thermal Physics

Thermal physics is the study of the statistical nature of physical

systems from an energetic perspective.

It include,

i. Thermodynamics – macroscopic theory, essentially

completed by Carnot 1824

ii. Kinetic theory – microscopic theory of molecular

distribution

i. Maxwell

ii. Boltzmann

Classical Mechanics

Quantum Mechanics

Thermodynamics

Statistical

Mechanics

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iii. Ensemble theory of statistical mechanics – all

thermodynamics informations can be derived from the

partition function – formulated by Gibbs in 1902

Macroscopic & Microscopic systems

-macroscopic when it is large enough to be visible in the ordinary

sense

-microscopic if it is roughly of atomic dimensions, or smaller

If N is the number of particles in the system

- A system is macroscopic if

1>>N

- And is microscopic if the other way around.

Microscopic descriptions

Describes the system in highest degree of detail. (e.g. positions

and velocities/momentums of all the atoms in the system).

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Macroscopic descriptions

Overall state of the system: averaging over large number of

entities (atoms or molecules). (e.g. pressure).

Instruments measure averages over a large number of atoms or

molecules.

Understanding of a particular system: to relate macroscopic

measurements to theoretical descriptions of the microscopic

entities.

The Microscopic View

Thermodynamics properties like T & P can be related to

microscopic quantities.

e.g T --in terms of a molecular velocity.

P --collisions of molecules with the walls of a container.

Levels of Description and Observation:

Mechanics: small number of particles (single molecule): single

simulation --one single molecule experiment

Stochastisch: several particles (molecules): several simulations

-- several single molecule Experiments

Statistics: huge number of particles and events: macroscopic

measurement

Wave – particle duality

-light and matter exhibit properties of both waves and particles

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Einstein: the electrons were knocked free of the metal by incident

photons, with each photon carrying an amount of energy E that

was related to the frequency, ν of the light by

E = h ν

h: Plank’s constant (6.626 x 10-34 J seconds).

de Broglie: all matter has a wave-like nature; and related

wavelength, λ and momentum, p:

p

hλ =

note: for photon; p = E / c and λ = c / ν.

where c is the speed of light in vacuum

Uncertainty principle

Heisenberg: certain specific pairs of variables cannot be measured

simultaneously with high accuracy.

e.g: within an atom - possible to measure the position (∆x), or the

momentum (∆p), of a subatomic particle (electron,..) but not

possible to measure both of them at the same time.

-because the measuring process interferes to a substantial degree

with what is being measured.

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Formulation and characteristics

At atomic dimensions

–particle is not like a hard sphere

-the smaller the dimension→the more wave-like it becomes

Using light to identify the location or motion of an electron, the

photon of light will influence the electron's motion and position.

Mathematically, the uncertainty principle looks like:

h∆x∆p ≥

Where ∆x = the uncertainty in position

∆p = the uncertainty in momentum

h= Planck's constant

A sine wave:

λλλλ - wavelength

Momentum is precisely known, p = h / λ

The position, ∆x is unknown

Adding several waves of different λ

-Localize the wave (position is certain)

-The momentum is uncertain (each different wavelength

represents different momentum)

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Application

-to caculate energy which would be required to keep a

particle within a given volume.

Example: energy to keep an electron inside an atom.

Assume: the diameter of the atom = 4 A = ∆x

∆p = h / ∆x

h = 6.626 x 10-34 J-s

∆p = h / ∆x = 6.626 x 10-34 J-s / 4 x 10-10 m

= 1. 6565 x 10-24 kgm/s

E = ½ mV2 = (mV)2/(2m) = p2/(2m)

m= 9.11 x 10-31 kg (electron mass)

1 ev = 1.6 x 10-19 J

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( )( )

ev9.4

J/ev1.6x10kg9.11x102

kgm/s1.6565x10E

1931

24

=

=−−

Quantized Energy States

Max Planck suggested that energy is transferred in “packets”

called quanta (plural).

Quantum: the smallest quantity of energy that can be emitted or

absorbed as electromagnetic energy

Bohr model (1913)

-electrons were orbiting the nucleus

-when a charge traveling in a circular path should lose energy

by emitting electromagnetic radiation

-it should end up spiraling into the nucleus (which it does not).

Classical physical laws inadequate to explain the inner

workings of the atom

-idea of quantized energy from Planck

-only orbits of certain radii, corresponding to defined

energies, are "permitted"

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-An electron orbiting in one of these "allowed" orbits:

Has a defined energy state

Will not radiate energy

Will not spiral into the nucleus

Electron energy:

RH = 2.18 x 10-18 J Rydberg constant

n: the principle quantum number, n = 1, 2, 3, …..

n --corresponds to the different allowed orbits for the electron.

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Emission of light corresponds to a transition of the atoms between

two states

Frequency of emitted light

h

EEυ&EE∆Ehυ fi

fi

−=−==

Consequence of Bohr's deduction

1. Each atom can be represented by an energy diagram

2. Distance between two energy levels gives a particular

νννν the energy of atoms is quantized

The implications --only certain photon energies are allowed

when electrons jump down from higher

levels to lower levels

→ producing the spectrum.

Energy

Ei

Ef

Light, ν

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An electron transition between quantized energy levels with

different quantum numbers n yields a photon by emission with

quantum energy Ephoton.

Ephoton = hν = E2 – E1

eVn

1

n

113.6

n

1

n

1

h

me2πhν

2

2

2

1

2

2

2

1

2

42

−=

−=

for transition from n = 2 to n = 1

n = 2

n = 1

E2

E1

n = 5

n = 4

n = 3

n = 2

The hydrogen spectrum

The energy levels

n = 1

n = 2

n = 3

n = 4

n = 5

Ionization

energy

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eV10.24

1113.6h ν =

−=

ν = 10.2 eV / h

1 eV = 1.6 x 10-19 J h = 6.626 x 10-34 J.s

ν = (10.2 x 1.6 x 10-19 J) / (6.626 x 10-34 J.s)

= 121.95 nm (UV)

Planck’s theory:

- Energy is always emitted or absorbed in whole number

multiples of hν (i.e hν, 2hν, 3hν)

-the energy levels that are allowed are ‘quantized.’

(restricted to certain quantities or values)

Quantum Harmonic Oscillator

A diatomic molecule vibrates;

- Potential energy that depends upon the square of the displacement

from equilibrium.

-But the energy levels are quantized at equally spaced values.

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The energy levels of the quantum harmonic oscillator are

π

πω

ω

2

.'

)(2

,...3,2,1,02

1

constsPlanck

frequency

nnEn

=

=

=

+=

h

h

The quantum harmonic oscillator

-the foundation for the understanding of complex modes of

vibration in larger molecules, the motion of atoms in a solid

lattice, the theory of heat capacity, etc.

Statistical Mechanics

-relates the macroscopic thermodynamic properties of a system to

the ensemble behaviour of its components.

-Each thermodynamic state of the system (represented by

particular values of the state functions) is termed a macrostate.

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-For each macrostate, there can be many corresponding

microstates.

-requires a large number of particles (1 mol = 6.022 x 1023

particles)

-water = 55 mol/l, i.e. 3.3 x 1025 water molecules per liter

Thermal Concept

Temperature is the average kinetic energy per molecule of the

molecules in the substance.

Heat is the energy transferred between objects when they

change temperature, and moves from areas of high temperature to

areas of low temperature

Internal energy is the total energy related to the thermal motion

of the molecules in a substance --vibrational & translational

δW – all the levels are raised by δεi. No change in populations

δεI –No change in the energy level εi. Populations are change by

dNi

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Distribution Function (Binomial, Gaussian,..)

Probability (Mean, standard deviations,

permutation,…)

Applied

Statistics

Quantum Statistics

For system of

particles

For Fermions For Bosons

Using:

Density of state

&

Distribution function

Energy

Distribution

Follow:

The Fermi-Dirac Dist.

Appl.ication:

Electrons in metal

Follow:

The Bose-Einstein Dist.

Describe

Bose-Einstein Cond. &

Thermal radiatian

Application:

Superconductivity,

Liquid helium, ..

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Note

ωψχφ

υτσρ

πξν

µλκι

θηζε

δγβα

ΩΨΧΦ

ΥΣ

ΠΞ

Λ

Θ

∆Γ

OmegaPsiChiPhi

UpsilonTTauSigmaPRho

PioOOmicronXiNNu

MMuLambdaKKappaIIota

ThetaHEtaZZetaEEpsilon

DeltaGammaBBetaAAlpha

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Lect 2 SSF 2273 Statistical Mechanics: Introduction Statistical mechanics is a branch of physics concerned with interpreting and evaluating the properties of microscopic systems (electrons, ions, atoms, molecules,…particles) to permit the prediction of the properties of the macroscopic systems embodying these elementry ones

The Microscopic View Thermodynamics properties like T & P can be related to microscopic quantities. e.g T can be expressed in terms of a molecular velocity. P can be understood by picturing the miniature collisions of

molecules with the walls of a container. Wave – particle duality

-light and matter exhibit properties of both waves and of particles

Einstein: the electrons were knocked free of the metal by incident photons, with each photon carrying an amount of energy E that was related to the frequency, of the light by

E = h ν

h: Plank’s constant (6.626 x 10-34 J seconds).

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de Broglie: all matter has a wave-like nature; and related wavelength, λ and momentum, p:

ph

=

note: for photon; p = E / c and = c / .

where c is the speed of light in vacuum

Uncertainty principle Heisenberg: certain specific pairs of variables cannot be measured simultaneously with high accuracy. e.g: within an atom - possible to measure the position (∆x), or the momentum (∆p), of a subatomic particle (electron,..) but not possible to measure both of them at the same time. - because the measuring process interferes to a substantial degree with what is being measured. Formulation and characteristics at atomic dimensions

– particle is not like a hard sphere -because the smaller the dimension -the more wave-like it becomes

using light to identify the location or motion of an electron, the photon of light will influence the electron's motion and position.

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Mathematically, the uncertainty principle looks like:

hxp ≥

Where ∆x = the uncertainty in position ∆p = the uncertainty in momentum h= Planck's constant

A sine wave: λ - wavelength Momentum is precisely known, p = h / λ The position, ∆x is unknown Adding several waves of different λ Localize the wave (position is certain)

The momentum is uncertain (each different wavelength represents different momentum)

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Application -to caculate energy which would be required to

contain a particle within a given volume.

Example: energy to keep an electron inside an atom. Assume: the diameter of the atom = 4 A = ∆x ∆p = h / ∆x h = 6.626 x 10-34 J-s ∆p = h / ∆x = 6.626 x 10-34 J-s / 4 x 10-10 m = 1. 6565 x 10-24 kgm/s E = ½ mV2 = (mV)2/(2m) = p2/(2m) m= 9.11 x 10-31 kg (electron mass) 1 ev = 1.6 x 10-19 J

( )( )ev9.4

J/ev1.6x10kg9.11x102kgm/s1.6565x10

E1931

24

=

= −−

Quantized Energy States Max Planck suggested that energy is Quantum

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The electrons in free atoms can be found in only certain discrete energy states. These sharp energy states are associated with the orbits or shells of electrons in an atom, e.g., a hydrogen atom.

The implications of these quantized energy states is that only certain photon energies are allowed when electrons jump down from higher levels to lower levels, producing the spectrum.

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e.g Hydrogen Energy Levels

an electron transition between quantized energy levels with different quantum numbers n yields a photon by emission with quantum energy Ephoton.

Ephoton = hν = E2 – E1

n = 2

n = 1

E2

E1

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eVn1

n1

13.6n1

n1

hme2

h 22

21

22

21

2

42

−=

−=

for transition from n = 2 to n = 1

eV10.241

113.6h =

−=

ν = 10.2 eV / h 1 eV = 1.6 x 10-19 J h = 6.626 x 10-34 J.s ν = (10.2 x 1.6 x 10-19 J) / (6.626 x 10-34 J.s) = 121.95 nm (UV)

Planck’s theory: -υυυ υ

-the energy !"#$

(restricted to certain quantities or values)

Quantum Harmonic Oscillator A diatomic molecule vibrates; - potential energy that depends upon the square of the displacement from equilibrium. -But the energy levels are quantized at equally spaced values.

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The energy levels of the quantum harmonic oscillator are

The quantum harmonic oscillator

-the foundation for the understanding of complex modes of vibration in larger molecules, the motion of atoms in a solid lattice, the theory of heat capacity, etc.

quantized State & phase space From uncertainty prin. – can only identify a particle as being within a box of area ∆x∆px = h

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if position --- x, y, z momentum --- px, py, pz the particle coordinate - - in a six dimensional space ---- “phase space” uncertainty of the coordinate ∆x∆px = h ∆y∆py = h ∆z∆pz = h or ∆x∆px ∆y∆py ∆z∆pz = h3 -the particle is located within a six-dim. Quantum box or quantum state of volume h3

∆px

∆x

x

P

∆px

∆x

x

P

∆px

∆x

x

P

or

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Let say position and momentum a particle are, 0 < x < xo 0 < p < po the number of quantum state available to the particle

hpx

stateoneofareaareatotal

stateof(No. oo==

in 6-D no of quantum state = (∆x ∆y ∆z∆px ∆py ∆pz ) / h3 = (Vr Vp ) / h3 = (1/ h3) dxdydz dpxdpydpz

appro =3

33

hpdrd

statequantumno .

∆px

∆x

x

Px h

xo

po

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2

1

0 1 2 3 4 5 microstate

E

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FROM MICROSCOPIC

TO MACROSCOPIC BEHAVIOUR

1

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THE GOAL: To understand the properties of MACROSCOPIC

SYSTEMS (systems of many electrons, molecules, photons, or other constituents).

Familiar examples:

1. Air in your room (gas)

2. A glass of water (liquid)

3. A coin (solid)

4. A rubber band (polymer)

2

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Less familiar examples:

1. superconductors

2. cell membranes

3. the brain

4. the stock market

5. neutron stars

3

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Macroscopic system: the typical

questions asked.. 1. How does the pressure of a gas depend on the

temperature and the volume of its container?

2. How does a refrigerator work? How can we make it more efficient?

3. How much energy do we need to add to a kettle of water to change it to steam?

4

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Q1-Q3:

Concerned with macroscopic properties such as Pressure, Volume, Temperature and processes related to heating and work.

Relevant to thermodynamics which provides a framework for relating the macroscopic properties of a system to one another.

Thermodynamics is concerned only with macroscopic quantities and ignores the microscopic variables that characterize individual molecules.

Many of the applications of thermodynamics are to engines (internal combustion engines; steam turbines)

5

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4. Why are the properties of water different from those of steam, even though water and steam consist of the same type of molecules?

5. How and why does a liquid freeze into a particular crystalline structure?

6. Why does helium have a superfluid phase at very low temperatures? Why do some materials exhibit zero resistance to electrical current at sufficiently low temperatures?

7. In general, how do the properties of a system emerge from its constituents?

6

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Q4-Q7:

Relate to understanding the behaviour of macroscopic systems starting from the atomic nature of matter.

e.g: water consists of molecules of hydrogen and oxygen

H2O

The laws of classical mechanics (Newton’s laws) & quantum mechanics determine the behaviour of molecules at the microscopic level.

7

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8. How fast does the current in a river have to be before its flow changes from laminar to turbulent?

9. What will the weather be tomorrow?

8

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Question 8 also relates to a macroscopic system, but temperature is not relevant in this case.

Moreover, turbulent flow continually changes in time.

Question 9 concerns macroscopic phenomena that change with time. Although there has been progress in our understanding of time-dependent phenomena such as turbulent flow and hurricanes, our understanding of such phenomena is much less advanced than our understanding of time-independent systems.

9

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Goal of Statistical Mechanics :

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 a bridge between the microscopic and macroscopic worlds.

10

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For this reason we will focus our attention on systems whose macroscopic properties are independent of time and consider questions such as those in Questions 1–7.

11

The statistical approach is to derive all

macroscopic properties (T, V, P, E, S, etc.) from

the properties of moving constituent particles and

the interactions between them (including quantum

phenomena)

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Quantum (Quanta- plural)

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In physics, a quantum (plural: quanta) is the minimum amount of any physical entity involved in an interaction. Behind this, one finds the fundamental notion that aphysical property may be "quantized," referred to as "the hypothesis of quantization".[1] This means that the magnitude can take on only certain discrete values.

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A photon is a single quantum of light, and is

referred to as a "light quantum". The energy of

an electron bound to an atom is quantized,

which results in the stability of atoms, and hence

of matter in general.

As incorporated into the theory of quantum

mechanics, this is regarded by physicists as part

of the fundamental framework for understanding

and describing nature at the smallest length-

scales.

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Wave-particle duality

Every quantum object is both a particle, like a tennis ball,

and some sort of wave, like a wave in the ocean. Imagine

a tennis ball that would be in several different places at

the same time. But when you try to locate it with

measuring instruments, the quantum object is suddenly

reduced to one spot. This means that electrons, atoms,

molecules, and even photons (light particles), are small

particles and waves, both at the same time!

This is the basic property of the quantum world.

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PRACTICAL APPLICATIONS

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Quantum mechanics and CCD display

All our digital cameras are manufactured with CCD

(charge-coupled device) image sensors that work

using quantum mechanics. Light is made up of

photons, which behave as particles and waves at

the same time. When light falls upon a CCD sensor,

each photon — when it has enough energy — pulls

out one electron. These electrons are thus detected

and converted into a digital image.

This technology has enabled us to develop ultra-

sensitive cameras.

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IN LABORATORY

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21

Using electrons to observe matter

In order to overcome the limitations of optical

microscopes, physicists use quantum mechanics.

Since electrons are — like light — waves, why not

use them to light up objects? This is what lies

behind the electron microscope, which has an

amazing magnifying power.

It is one of the most popular microscopes

amongst physicists, who want to observe atoms

and molecules, and amongst biologists, who want

to observe microbes and viruses.

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http://www.youtube.com/watch?feature=player

_embedded&v=wsq7qXr9Hl0

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STATISTICAL MECHANICS Statistical mechanics is a branch of theoretical

physics (and mathematical physics) that studies, using PROBABILITY THEORY, the average behaviour of a mechanical system where the state of the system is uncertain.

Statistical mechanics is a collection of mathematical tools that are used to fill this disconnection between the laws of mechanics and the practical experience of incomplete knowledge.

IT IS A BRIGDE BETWEEN THE MICROSCOPIC AND MACROSCOPIC WORLDS.

23

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Note:

-The number of the elements can be very

large --- impossible to keep track of the

behaviour of each element.

-A statistical property is a single

measurement which gives a physical

picture of what is occurring in all the

individual parts.

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2727

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Thermodynamic potentials

-the quantitative measure of the stored energy in the system.

The five most well known potentials are:

Internal energy U

Enthalpy H = U + PV

Helmholtz free energy A + U – TS

Gibbs free energy G = U + PV – TS

Grand potential

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Potentials are used to measure energy

changes in systems as they evolve from an

initial state to a final state.

Note: the term thermodynamic free energy

is a measure of the amount of mechanical

(or other) work that can be extracted from a

system.

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WEEK 2

Statistical Mechanics: statistics for small system Probability

Mean value

-single element system

Let PS : probability the system in state S

fS : function when the system in state S

f ≅ mean value of f

Definition:

∑=S

SS fPf

Example: coin 2 states: H: head, T: tail

PH: probability of landing heads = ½

PT: probability of landing tails = ½

Let PS : probability the system in state S

fS : function when the system in state S

f ≅ mean value of f

∑=S

SS fPf

Let say f : the number of heat showing

fS = 1 for H & fS = 0 for T

The mean value for f

= PHfH +PTfT = (1/2) (1) + (1/2) (0) = ½

--the average number of heads per coin showing is ½

Example: dice

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n: no of dot upward = f

if fn = n, what is the mean value for the system?

P1= P2 = P3 = P4 = P5 = P6 = 1/6

∑=S

SS fPf = ∑=S

S nPf =(1/6)(1)+ (1/6)(2)+(1/6)(3)

+ (1/6)(4) + (1/6)(5) + (1/6)(6) = 3 ½

if fn = (n-1)2

∑=S

SS fPf = ( )2

1∑ −=S

S nPf = (1/6)(0) + (1/6)(1)+

(1/6)(4) + (1/6)(9) + (1/6)(16) + (1/6)(25) = 6

19

Note:

f, g : functions of the states of systems

c : constant

Then ( ) gfgf +=+ and fccf =

-few elements system

Let p = the probability the criterion is satisfied

q = the probability the criterion is not satisfied

Example:

i. criterion: a flipped coin lands head-up

p = ½ and q = ½

ii. criterion: a rolled dice lands with 2 dots up

p = 1/6 and q = 5/6

p + q = 1 or q = 1- p

The probability the criterion is satisfied or not is always

one

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System: 2 identical systems (2 coins)

C1 C2

p1 + q1 = 1 and p2 + q2 = 1

for each element, the criterion is or not satisfied is,

(p1+q1)(p2+q2) = 1x1 = 12 = 1

= p1p2 + p1q2 + q1p2 + q1q2

p1p2 – both elements satisfy the criterion

p1q2 – element 1 –satisfy the criterion

element 2 - not satisfy the criterion

q1p2 - element 1 –not satisfy the criterion

element 2 - satisfy the criterion

q1q2 - both elements not satisfy the criterion

Example

system: a box with 2 air molecules inside

V1 + V2 = V and V1 = V/3

Criterion: all possible configurations

-The probability of either molecule in V1,

p1 = 1/3 & p2 = 1/3

1 2 V1 V2

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-The probability that each molecule is not in V1,

q1 = 2/3 & q2 = 2/3

-the probability for all possible configuration

(p1+q1)(p2+q2) = 1x1 = 12 = 1

= p1p2 + p1q2 + q1p2 + q1q2

-both molecules in V1: p1p2 = (1/3) (1/3) = 1/9

-mol.1 in V1 & mol.2 in V2: p1q2 = (1/3) (2/3)

= 2/9

-mol.1 in V2 & mol.2 in V1: q1p2 = (2/3) (1/3)

= 2/9

-both molecules in V2: q1q2 = (2/3) (2/3) = 4/9

Example: 2 dice D1 D2

Criterion: one dot up for D1 and D2

prob. For one dot up: p1 = p2 = 1/6

prob. For not landing with one dot up: q1 = q2 = 5/6

(p1+q1)(p2+q2) = 1x1 = 12 = 1

= p1p2 + p1q2 + q1p2 + q1q2 -one dot up for D1&D2: p1.p2 = 1/36

-one dot up for D1& not D2: p1.q2 = 5/36

-one dot up for D2& not D1: q1.p2 = 5/36

- not D1& not D2: q1.q2 = 25/36

Idntical elements: with the same probability

For 2 identical elements

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p1= p2 = p q1 = q2 = q

The possible configurations;

(p1+q1)(p2+q2) = (p + q)2

= (p2 + 2pq + q2)

p2: all satisfy the criterion

2pq: one satisfy – one does not

q2: none satisfy the criterion

For 3 identical elements

(p1+q1)(p2+q2) (p3+q3) = (p + q)3 = 1

= (p3 + 3p2q + 3pq2 + q2)

p3: all satisfy the criterion

3p2q: two satisfy – one does not

3pq2: one satisfy – two do not

q3: none satisfy the criterion

For N elements

(p1+q1)(p2+q2)…..(pN + qN) = (p + q)N but

( ) 1qp

n)!-(Nn!

N!qp n)(Nn

N

0n

N==+ −

=

…..binomial expansion

PN(n): the probability the system in state of n elements

satisfy the criterion and N-n elements do not

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nNn

N qpn)!(Nn!

N!(n)P −

−=

)!(!

!

nNn

N

− ≅ Binomial coefficient

-the number of different configurations of the

individual elements, for which n satisfy the

criterion and (N-n) do not

Example: consider 5 molecules in abox

V1 + V2 = V V1 = (1/3) V

(i) find the prob. of 2 mol. in V1 and 3 in V2

p = 1/3 & q = 2/3

N = 5, n = 2

32

53

2

3

1

2)!(52!

5!(2)P

−=

= 80 /243

(ii) number of different possible arrangements for 2

mol in V1 and 3 mol in V2

)!(!

!

nNn

N

−= 5! / (2! 3!) = 10

State of a system relative to two different criteria;

-system of N elements in a state where;

V1 V2

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n1 elements satisfy 1st criterion

n2 elements satisfy 2nd criterion

the probability PN(n1,n2)

Note: an element’s behaviour with respect to one of the

criteria does not affect its behaviour with respect to the

other ……`statistically independent criteria’

Pij: the system in state i with respect to 1st criterion and also

in state j with respect to the 2nd

Pij = Pi . Pj

Or Pijk….. = Pi . Pj. Pk, …….

Where i, j, k, ….

Example:

System: 5 mol. -find: the probability for 2 mol in V1 and 4 mol in VU

N = 5; ni = 2; nj=4

( ) ( ) nN

2

n

2

nN

1

n

155 qp!n)-(Nn!

N!.qp

!n)-(Nn!

N!.P5(4)2P2.4P −−==

V1 V2

VU

VD

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051.02

1

2

1

!1!4

!5*

3

2

3

1

!3!2

!5432

=

=

Note:

Binomial expansion – correct for any size

nNn

N qpn)!(Nn!

N!(n)P −

−=

but for large N, N! can be calculated using Stirling’s

formula

ln(m!) ≈ m ln n – m + (1/2) ln (2πm)

Probability Distribution of Discrete Random

Variables

A probability distribution is listing of all the possible

values that a random variable can take along with their

probabilities.

Example: to find out the probability distribution for the

number of heads on three tosses of a coin:

First toss.........T T T T H H H H

Second toss.....T T H H T T H H

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Third toss........T H T H T H T H

the probability distribution

No. of

heads

X

Probability

P(X)

mean or

expected

value

X. P(X)

0 1/8 0.0

1 3/8 0.375

2 3/8 0.75

3 1/8 0.375

∑X P(X) = 1.5

Mean = E(X) = ∑X.P(X)

where:

E(X) = expected value,

X = an event,

P(X) = probability of the event

Binomial Distribution:

-discrete probability distributions

-Several characteristics underlie the use of the

binomial distribution.

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Characteristics of the Binomial Distribution:

1. The experiment consists of n identical trials.

2. Each trial has only one of the two possible mutually

exclusive outcomes, success or a failure.

3. The probability of each outcome does not change

from trial to trial, and

4. The trials are independent, thus we must sample with

replacement.

Binomial Equation:

-able to identify three things:

-the number of trials

-the probability of a success on any one trial

-the number of successes desired

nNn

N qpn)!(Nn!

N!(n)P −

−=

Example: binomial distribution:

What is the probability of obtaining exactly 3 heads if a fair

coin is flipped 6 times?

Ans:

N = 6, n = 3, and p = q = .5

363 )5.01(5)!36(!3

!6)3( −−

−=P

3125.0)125.0)(125.()23)(23(

23456== o

xx

xxxx

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Binomial distributions

or

Properties of Binomial Distribution

Mean np

Variance npq

Standard deviation (npq)1/2

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Binomial Distribution: Fluctuations

Suppose the following binomial distribution;

• the probability of success: p

• the probability of failure: 1-p = q

The mean n = pN

N: the number of elements per system

n∆ : average fluctuation of n about its mean value, n

0)( =−=−=∆ nnnnn

-positive fluctuations cancel the negative one

The standard deviation

222 )()tan( nndeviatiandards −== σ

[ ] 212)()tan( nndeviatiandards −== σ

(Note: using ∑=S

SS fPf and nNn

N qpn)!(Nn!

N!(n)P −

−=

)

( )222 )( nnPnn

n

n −=−= ∑σ

( )( )∑

=

− −

−=

N

n

nNnnnqp

nNn

N

0

2

!!

!

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Solving----- Npq=2σ

or Npq=σ

relative fluctuations

NNP

q

Np

Npq

n

1∝==

σ

Example:

-system of 100 flipped coins: what are;

average number of H (head),

the standard deviation,

σ = 8

n =125 n

N=250

PN(n)

σ=2.5

n =12.5 n

N=25

PN(n)

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the relative fluctuation?

Ans: n= pN = (1/2)(100) = 50

σ = (Npq)1/2 = ((100)(.5)(.5))1/2 = 5

σ/ n = 5/50 = 10 %

- now N = 10,000 flipped coins

n= pN = (1/2)(10 000) = 5 000

σ = (Npq)1/2 = ((10 000)(.5)(.5))1/2 = 50

σ/ n = 50/5 000 = 1 %

note:

A binomial probability distribution must meet each of the

following:

1. There are a fixed number of trials

2. The trials must be independent

3. Each trial must have outcomes classified into two

categories

4. The probabilities remain constant for each trial

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Gaussian Distribution

The Gaussian distribution is useful where binomial formula

is not

The probability PN(n) ≅ P(n)

P(n) – a continuous function of n

Criteria for increase accuracy;

i) choose as smooth a function as possible

ii) choose reference point, as close to the value of n

To satisfy (i)—expand the logarithm of P(n)

To satisfy (ii) – choose reference as nmax

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interested in values of P(n) for n near nmax

Using Taylor series expansion

..)()(ln2

1)()(ln)(ln)(ln 2

2

2

+−∂

∂+−

∂+=

==nnnP

nnnnP

nnPnP

nnnn

But 0)(ln =

∂= nn

nPn max. point

2nd derivative

using ...)1(

)()1()()(=

−+

−+=

∆=

nn

nPnP

n

nP

n

nP

22

2 11)(ln

σ−=−=

∂= Npq

nPn nn

therefore 2

2)(

2

1)(ln)(ln nnnPnP −−=

σ

nmax n N

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or 2

2

2

)((

)()( σ

nn

enPnP

−−

=

to calculate P(n)

-the sum of the probabilities of all possible values of n must

be equal to 1

∑ =n

nP 1)(

∑ ∫∞

−∞=

≈∆=n n

dnnPnnP )()(

( )[ ]

1)2)(()(22

2/ === ∫∞

−∞=

−− πσσnPdnenP

n

nn

Note:

therefore

( )[ ]222/

2

1)( σ

σπ

nn

n

enP−−=∑

Because σπ2

1)( =nP

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Gaussian Distribution

Properties of Gaussian distribution

( )nnx −=

When nn =

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-3σ -2σ -1σ 0 1σ 2σ 3σ

X

f(X)

-3σ -2σ -1σ 0 1σ 2σ 3σ

X

f(X)

1σ ≅ half width at half maximum

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Probability of a measurement falling within -σ to +σ of the

mean is 0.683

Probability of a measurement falling within -2σ to +2σ of

the mean is 0.954

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Probability of a measurement falling within -3σ to +3σ of

the mean is 0.997

Normal Approximation to the Binomial

The shape of a binomial distribution depends on the values

of n and p.

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-the shape of a BD will be fairly symmetrical

-for large values of n (say, n > 20 ),

-when p is about 0.05 < p < 0.95 ).

EXAMPLE:

-flip a coin for N =20 and p = 0.5

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- binomial probability distribution → NDF

A normal density curve

-with mean np = 20(0.5) = 10

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-standard deviation (np(1-p))1/2 = 2.236

--fit with BDF

Random Walk A random process consisting of a sequence of discrete steps

of fixed length

The random walk is central to statistical physics.

-predicting how fast one gas will diffuse into another,

-how fast heat will spread in a solid,

-how big fluctuations in pressure will be in a small

container,

-and etc………..

Problem: to find the probability of landing at a given spot

after a given number of steps, or to find how far

away the girl is on average from where she

started.

…after many many steps

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The simplest random walk is a path constructed according

to the following rules:

• There is a starting point.

• The distance from one point in the path to the next is a

constant.

• The direction from one point in the path to the next is

chosen at random, and no direction is more probable

than another.

Consider each step is of length s0,

It can be either forward (right) or backward (left).

Probability of going forward (right) is p.

Probability of going backward (left) is q = 1- p

After N steps, if n are forward (right), the distance traveled

is,

S = [n – (N – n)]s0 = (2n – N) s0

The progression of the walk and it’s different outcomes are

nicely organized in Pascal’s triangle.

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This is known as a binomial distribution.

Define the probability function

fN (n) -- the probability that in a walk of N steps, end at

point n.

For the nonzero probabilities.

For a walk of no steps, f0(0) = 1.

For a walk of 1 step, f1(−1) = ½, f1(1) = ½.

For a walk of 2 steps, f2(−2) = ¼, f2(0) =2 x ¼ = ½, f2(2) = ¼.

……. f3(−3) = 1/8 , f3(−1) = 3/8 , f3(1) = 3/8, f3(3) = 1/8.

…… f4(4) = 1/16, f4(2) = ¼; f4(0) = 3/8, …….

n −5 −4 −3 −2 −1 0 1 2 3 4 5

f0(n) 1

f1(n) ½ ½

f2(n) ¼ ½ ¼

f3(n) 1/8 3/8 3/8 1/8

f4(n) 1/16 1/4 3/8 1/4 1/16

f5(n) 1/32 5/32 5/16 5/16 5/32 1/32

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Factor by (1/2) N

n −5 −4 −3 −2 −1 0 1 2 3 4 5

f0(n) 1

2f1(n) 1 1

22f2(n) 1 2 1

23f3(n) 1 3 3 1

24f4(n) 1 4 6 4 1

25f5(n) 1 5 10 10 5 1

This is Pascal’s Triangle—every entry is the sum of the

two diagonally above

These numbers are in fact the coefficients that appear in the

binomial expansion of (p + q)N

Picturing the Probability Distribution

Visualizing this probability distribution →

For 5 steps, it looks like:

-a walk of 100 steps → n steps forward and

100 − n steps backward

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-the final landing place is n − (100 − n) = 2n − 100 paces in

the forward direction.

Note that this is an even number, and goes from −100 to

+100.

The total number of 100-step walks having just n forward

steps is

(100)!/n!(100 − n)!

The probability of landing at 2n − 100 after a random 100-

step walk is proportional to the number of such walks that

terminate there

The probability of this occurring is,

The average distance covered after N steps is,

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and

note: pNn =

Standard deviation

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WEEK 2

1

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Example 1: COIN

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HEAD showing

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Example 2: DICE

=

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Criterion:

-something that is used as a reason for making a judgment or decision

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Example 3: 2 Identical Systems (2 COINS)

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Example 4: BOX 1

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Example 5: 2 DICE

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Identical Elements: with the same Probability

A:

B:

( p3 + 3p2 q + 3q2 p + q3 )

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C:

Binomial Expansion

Binomial Coefficient

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Example 6: BOX 2 – Consider 5 molecules in a box

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V U = VD

V 1 = 1/3 V

Let the total volume is V

or V U = 1/2 V

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17

1 COIN;

3 tosses

8 configurations= 8 microstates

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NO OF

HEADS

FIRST

THROW

SECOND

THROW

THIRD

THROW

MACRO

STATE

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NO OF

HEADS

FIRST

THROW

SECOND

THROW

THIRD

THROW

MACRO

STATE

0

1

2

3

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20

NO OF

HEADS

FIRST

THROW

SECOND

THROW

THIRD

THROW

MACRO STATE

0 T T T 1

1 T T H 2

T H T

H T T

2 H H T 3

H T H

T H H

3 H H H 4

1+3+3+1 4 MACROSTATES

= 1

= 3

= 3

= 1

= 8

MICROSTATES

TOTAL

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0(1/8)+(1)(3/8)+(2)(3/8)+(3)(1/8) = 0.0 + 0.375 + 0.75 + 0.375 = 1.5

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Binomial distribution

In probability theory and statistics,

the binomial distribution is the discrete

probability distribution of the number of

successes in a sequence of n independent

yes/no experiments, each of which yields

success with probability p.

Therewith the probability of an event is

defined by its binomial distribution.

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The binomial distribution is frequently used to

model the number of successes in a sample

of size n drawn with replacement from a

population of size N.

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24

If the sampling is carried out without

replacement, the draws are not

independent and so the resulting

distribution is a hypergeometric

distribution, not a binomial one.

However, for N much larger than n, the

binomial distribution is a good

approximation, and widely used.

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Example 8: Binomial Distribution

0.5

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N

N

N

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Example 9: system of 100 COINS

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In statistics and probability theory,

the standard deviation (SD) (represented by

the Greek letter sigma, σ) measures the

amount of variation or dispersion from the

average.

A low standard deviation indicates that the

data points tend to be very close to

the mean (also called expected value); a high

standard deviation indicates that the data

points are spread out over a large range of

values.

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A random walk is a mathematical formalization

of a path that consists of a succession of

ramdom steps.

Example of Modeled Random Walks:

i) the path traced by a molecule as it travels

in a liquid or a gas,

ii) the search path of a foraging animal,

iii) the price of a fluctuating stock

iv) the financial status of a gambler

although they may not be truly random in reality.

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The term random walk was first introduced

by Karl Pearson in 1905. Random walks have

been used in many fields:

ecology, economics, psychology, computer

science, PHYSICS, chemistry, and biology.

Random walks explain the observed

behaviours of many processes in these fields,

and thus serve as a fundamental MODEL for

the recorded STOCHASTIC ACTIVITY.

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A popular random walk model is that of a

random walk on a regular lattice, where at

each step the location jumps to another site

according to some probability distribution.

In a simple random walk, the

location can only jump to

neighbouring sites of the lattice.

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In simple symmetric random walk on

a locally finite lattice, the probabilities of

the location jumping to each one of its

immediate neighbours are the same.

The best studied example is of random

walk on the d-dimensional integer lattice

(sometimes called the hypercubic

lattice)

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One-dimensional random walk

This walk can be illustrated as follows:

A marker is placed at zero on the number line

and a fair coin is flipped. If it lands on heads,

the marker is moved one unit to the right. If it

lands on tails, the marker is moved one unit to

the left.

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After five flips, the marker could now be on

1, −1, 3, −3, 5, or −5.

Example:

With five flips, three heads and two tails,

in any order, will land on 1.

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There are :

10 ways of landing on 1 (by flipping three heads and two

tails),

10 ways of landing on −1 (by flipping three tails and two

heads),

5 ways of landing on 3 (by flipping four heads and one tail),

5 ways of landing on −3 (by flipping four tails and one

head),

1 way of landing on 5 (by flipping five heads), and

1 way of landing on −5 (by flipping five tails).

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See the figure below for an illustration of the

possible outcomes of 5 flips.

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AT

LAST!! THE END….

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WEEK 3

Internal energy -the energy stored in the motions and interactions of the

elements (e.g, atoms, molecules, electrons, etc) of a system

Degree of freedom

-ways each elements can store energy

Consider the atom of a solid

Energy of each atom = PE + KE

Vibration in 3-D

Potential energy: EP = ½ kx2 + ½ ky2 + ½ kz2

Kinetic energy: EK = ½ mvx2 + ½ mvy

2 +1/2 mvz2

= ½ (mvx)2 /m + ½ (mvy)

2/m +1/2 (mvz)2/m

=222

2

1

2

1

2

1ZYX P

mP

mP

m++

P: momentum

PE & KE in the form

E = b q2 b: constant

Each particle → 6 degrees of freedom

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N particles → 6N degrees of freedom for energy of the form

E = bq2

Equipartition theorem

the internal energy will be distributed equally among all those

degrees of freedom with energy in the form E=bq2

e.g: ideal gas

PV = mRT

In term of Boltzmann’s constant

PV = NkT

From molecular theory

2

3

1vNmPV =

2v : Average velocity

KE: 2222

2

1

2

1

2

1

2

1zyx vmvmvmvm ++=

Equip. theorem: 222

zyx vvv ==

2222

2

1

2

1

2

1

2

1xxx vmvmvmvm ++=

)2

1(3

2

1 22

xvmvm =

kTvm2

3

2

1 2 =

The average energy per degree of freedom

kTE2

1= k= 1.381 x 10-23 J/K

: Boltzmann’s constant

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KE → related to temperature

Higher KE, higher temperature

Note: when heat added, but temperature remains constant

→ new degrees of freedom, where the added energy is

stored

Changing the internal energy

- 3 ways / interactions

i. thermal interaction

-adding / removing heat from the system

ii. mechanical interactions

-doing work on the system or letting system doing

work on something else

iii. diffusive interaction

-adding or removing particles that undergo reactions

with the particles of the system

Heat Transfer

-heat transfer between A1 and A2 only

A1 A2

Q

Rigid, thermally conducting membrane

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heat in: increase particles velocity → increase temperature

heat out: lower the velocity → lower the temperature

→ changing the internal energy

Work (mechanical interactions)

Expansion – molecules strike a receding piston, loosing KE

-system does work – loosing energy

Compression – molecules strike on incoming piston – gain KE

-work done on the system – increases energy

Hot – higher velocity Cold – lower velocity

A1 A2

W W

A1 A2

W

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Particle transfer

PE: interactions of an element with its neighbours

-measured relative to a reference

-- known as zero-energy reference level

zero energy ref. level – the PE of an isolated particle

Consider: a particle trapped in a harmonic oscillator potential

well

ETH: thermal energy

A particle with energy E0 ossilates in a potential well -

ETH = E0 - µ

For N particles: ETH = E0 - µN

Chemical potential, µµµµ: the energy of the very lowest point in

the particle’s potential well.

EP

x 0

Ep = µ

EP = µ + ½ kr2

Total energy

E = EP + EK

= µ + ½ kr2 + ½ mv

2

= µ + ETH

E0

A1 A2

W

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Adding ∆N particles without adding thermal energy (constant T

and S), the internal energy of the system

( )

ST

STN

EorNE

,

,

∆∆

=∆=∆ µµ

-introduced in 1876 by the American mathematical physicist

Willard Gibbs

If to any homogeneous mass in a state of hydrostatic stress

we suppose an infinitesimal quantity of any substance to be

added, the mass remaining homogeneous and its entropy

and volume remaining unchanged, the increase of the

energy of the mass divided by the quantity of the substance

added is the potential for that substance in the mass

considered

Note: Chemical potential energy is a form of potential energy

related to the structural arrangement of atoms or molecules --the

breaking and forming of chemical bond.

The chemical potential µ of a thermodynamic system is the

amount by which the energy of the system would change if an

additional particle were introduced, with the entropy and

volume held fixed.

∆N

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Chemical potential µ is a measure of the free energy available

to do the work of moving a mole of molecules from one

location to another or through a barrier such as a cell

membrane…

Molecules tend to spontaneously move from areas of higher µ

(higher concentration) to areas of lower µ (lower

concentration), thus increasing the entropy of the universe

µA > µB: transformation of substance A into substance B, or

transport from place A to place B

µA = µB: no transformation, no transport, chemical equilibrium

µA < µB: transformation of substance B into substance A, or

transport from place B to place A.

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First law of thermodynamics

Changing the internal energy

(a) adding heat (∆Q)

(b) doing work (∆W)

(c) adding particles (∆N)

∆∆∆∆U = ∆∆∆∆Q - ∆∆∆∆W + µµµµ∆∆∆∆N --- The First Law of Therm.

Note: -∆W: work is done on the system

Note:

U – property of a system, determined by the two end

points (initial & final)

Q, W – interactions, determined by the routes taken.

Cannot be determined from the end points alone.

dU = δδδδQ - δδδδW + µµµµdN

Or dU = δδδδQ - δδδδW for non diffusive int.

To differentiate between properties and interactions;

(a) (b) (c)

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-exact differential- property functions

-inexacct diff. – non-property functions (Interactions)

Let F = f(x, y) x, y: variables

Initial state: (xi, yi)

Final state: (xf, yf)

∆F = f(xf, yf) – f (xi, yi)

F is determined by the two end points

Diff. form dyy

Fdx

x

FdF

∂+

∂=

Let a diff. eq. g(x,y)dx + h(x,y)dy -- exact or not?

EXACT – if there is a F(x, y) function which satisfy

hy

Fandg

x

F=

∂∂

=∂∂

Alternatively, use the identity

yx

F

xy

F

∂∂

∂=

∂∂

∂ 22

Note: dependent & independent variables

F = F(x, y)

x

yxF

∂ ),( y is held constant.

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TSP: WEEK 3

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Or yx

F

∂ y is held constant

Note:

The system

System characteristics

Definite boundaries in space

May be open or closed (respect to transfer of matter)

May or may not be thermally insulated

Thermodynamics state of a system

- at equilibrium, system assumed to occupy one of a set of

abstract, thermodynamics states

- each state is defined by the values of state variables

o e.g: at least two of the variables (T, V, P….)

- system state determines system properties

- two types properties

o Extensive (e.g volume, mass..)

o Intensive (e.g T, P, density…..)

Change of system state

- System traversal from initial to final state.

- Resulting change in some macroscopic system property

- Reversible or irreversible

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o Reversible – change take a path near to equilibrium

– system changes can be accurately predicted

o Irreversible – changes deviate from equilibrium –

system changes cannot be accurately predicted

Statistical Thermodynamics

- concern: behaviour of a large number of particles

- take a statistical view

- thermodynamics state – the state of system for N particles.

The states of a system

-A microstate is the state defined by specifying in detail the

location and momentum of each particle

-Each atom is an oscillator that contains equally spaced energy

levels. Atoms may be in any of these quantum states

The total number of atoms, N

The total energy the system E

-Consider 3 atoms with a total energy of 3 units.

Microstates 1, 2 & 3

En i

i

i =∑ ε

Nni

i =∑

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Microstates 4, 5, 6, 7, 8, & 9

Microstate 10

For 3 atoms with E = 3 units → 10 possible microstates

To predicts the number of microstates

(Case: 3 atoms with E = 3 units)

STEP 1

1. The 1st oscillator can be in any one of three states → 3

possible microstates.

2. The 2nd oscillator →in any one of the two states→ 2

additional microstates

3. The 3rd oscillator→only one state→ 1 additional

microstate.

The total number of microstates Ω is

6!3123 ==⋅⋅=Ω STEP 2:

-One oscillator in state 3 and the other two in state 0.

3

2

1

0

3

2

1

0

3

2

1

0

3

2

1

0

3

2

1

0

3

2

1

0

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1. The 1st oscillator → 3 possible microstates.

2. The 2nd oscillator → 2 additional microstates

3. The 3rd oscillator → 1 additional microstate.

→ Total of 6 microstates

Note: but

-reduces the configuration to 3 microstates.

3

!2

!3

2

123==

⋅⋅=Ω

STEP 3

-all three oscillators are in the same state

1. The 1st oscillator → 3 possible microstates.

2. The 2nd oscillator → 2 additional microstates

3. The 3rd oscillator → 1 aditional microstate.

→ Total of 6 microstates

Note: But

3

2

1

0

3

2

1

0

3

2

1

0

3

2

1

0

3

2

1

0

Same microstate

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14

etc

-reduces this configuration to 1 microstates

1

!3

!3

6

123==

⋅⋅=Ω

Then, in general, the number of microstates for any given

configuration is

( )( )( )( ) ( )∏==Ω

!

!

!!!!

!

321 io n

N

nnnn

N

Example:

-the number of microstates for N = 5 & E= 5

-possible configurations

-the total number of microstates is

126 = 1 + 30 + 20 + 20 +30 + 20 + 5

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Example:

4 molecules in a box

How many ways are there?

-to place 0 in L and 4 in R: 1!4!0

!4=

-to place 1 in L and 3 in R: 4!3!1

!4=

-to place 2 in L and 2 in R: 6!2!2

!4=

-to place 3 in L and 1 in R: 4!1!3

!4=

-to place 4 in L and 0 in R: 1!0!4

!4=

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16

Microstate Macrostate Degeneracy

LLLL I 1

RLLL

LRLL

LLRL

LLLR

II

4

LLRR

LRLR

RLLR

LRRL

RLRL

RRLL

III

6

LRRR

RLRR

RRLR

RRRL

IV

4

RRRR V 1

Macrostates: 5 Microstates: 16

Changing the number microstates -by changing the internal energy

-Four ways to increase the number of quantum states of

molecules

1-when a system is heated.

2-when a system expands into a vacuum of fluids mix.

3-when a solute is added to a solvent.

4-when a phase change occurs due to the input of energy

as in the first

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A system is being heated

Microstates of cold

system

Q Microstates of hot

system

A system is allowed to expand in volume

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Equilibrium For an isolated system;

When the probability of it being in the various possible

states do not vary with time.

Consider: changing the internal energy by compressing

-First the additional energy is given to particles near the

boundary

–Then shared with the rest of the system – by mutual interaction

- When all possible states are equally probable – the system is

in equilibrium

Relaxation time – the time required for a system to reach

equilibrium after being perturbed.

Quasi-static process – when system interact, the transfer of heat,

work, or particles, the process must be slower compared to the

relaxation time --- the system near equilibrium.

Fundamental Postulate

An isolated system in equilibrium is equally likely to be in

any of its accessible state

Let Ωo: the number of states accessible to the entire system

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The probability it will be found in any one of them,

o

=1

e.g

i) Calculate all possible microstates for a lattice of 3 non-

interacting particle.

ii) What is the probability that the orientation of the first

particle’s magnetic moment is + and the total magnetization is

+µ?

Ans

(i)

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Number of microstate: 8

The total magnetization of +1µ is 3 (red box) The first particle’s magnetic moment + is 2 (green box)

(ii)

3

21)1( =

+=++

stateofnumbertotal

beingofwithstatesnumberisstateofP

µµ

The spacing of states

- identify the state of a system by its internal energy, U

- several different states have the same energy –

The state is degenerate or

The energy level is degenerate

E.g three spins ½ particles in magnetic field B

microstate energy

↑ ↑ ↑ (+++) +3µB ↑ ↑ ↓ (++ -) +1µB ↑ ↓ ↑ (+- +) +1µB ↓ ↑ ↑ (-++) +1µB ↑ ↓ ↓ (+- -) -1µB ↓ ↑ ↓ (-+ -) -1µB ↓ ↓ ↑ (- -+) -1µB ↓ ↓ ↓ (- - -) -3µB

The Energy, U = +1µB state is three-times degenerate

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WEEK 3

1

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2

Show how to transform from above to below:

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The Internal energy will be distributed equally among all those degrees of freedom with energy in the form E = bq2

e.g: Ideal Gas PV = mRT

In terms of Boltzmann’s Constant PV =NkT

From molecular theory

3

By assumptions, there are no

preferred directions

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since

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11Commons is a freely licensed media file repository

Wavefunctions of a quantum harmonic oscillator

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Picture of wavefunctions (and energy levels) of quantum

harmonic oscillator, using colour scale for probability density.

Commons is a freely licensed media file repository.

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Change of System State

- System traverses from initial to final state.

-Resulting change in some macroscopic system property.

-Reversible or Irreversible.

• Reversible – change take a path near to equilibrium –

system changes can be accurately predicted.

• Irreversible – changes deviate from equilibrium – system

changes cannot be accurately predicted.

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Statistical Thermodynamics

-concern ; behaviour of a large number of

particles

- take a statistical view

-Thermodynamic state – the state of a

system for N particles

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24

Denotes the

continued product

over all values of i

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Microstates 1, 2, 3 ,4, 5, 6

Microstates 7, 8, 9

Microstate 10

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3 microstates

2 microstates1 microstate

(3+2+1) microstates

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N = 3 ; 1! ; 2! ; 3!

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Degeneracy

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In physics, two or more different

quantum states are said to be

degenerate if they are all at the same

energy level. Statistically this means that

they are all equally probable of being

filled, and in quantum mechanics it is

represented mathematically by the

Hamiltonian for the system having more

than one linearly independent eigenstate

with the same eigenvalue.

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Conversely, an energy level is said to be degenerate if it contains two or more different states. The number of different states at a particular energy level is called the level's degeneracy, and this phenomenon is generally known as a quantum degeneracy.

In quantum theory this usually pertains to electronic configurations and the electron's energy levels, where different possible occupation states for particles may be related by symmetry.

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To cause (a system) to become altered or imbalanced from a normal state.

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Example: (i) How many possible microstates are there

for a lattice of 3 non-interacting particles .

(ii) What is the probability that the orientation of the

first particle’s magnetic moment is + and the total

magnetization of +1m?

microstates magnetization

+3m

+1m

+1m

+1m

-1m

-1m

-1m

-3m

∑= 8 microstates

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ThaT’s all for now….

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1

WEEK 4 Density of state

Let ∆U = energy range of finite width

Expect: number of states ∝ ∆U

If Ω(U, ∆U): number of states between U and U+∆U

Ω(U, ∆U)= g(U) ∆U

g (U): density of state

States available to the

system

∆U

U

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Definition

The energy density of states (EDOS) function measures the

number of energy states in each unit interval of energy and in

each unit volume of the crystal

volumexEnergy

statesofnumberUg =)(

-figure

∆E = 4 -3 = 1 eV → 4 energy states

The density of states at E= (3 + 4) /2 = 3.5 eV

4

11

4)5.3(

3===

cmxeVvolumexEnergy

statesofnumberg

Example: Suppose a crystal has two discrete states (i.e. single

states) in each unit volume of crystal.

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The density-of-state function consists of two Dirac delta

functions of the form

( ) ( )21)( UUUUUg −+−= δδ

Integrating over energy gives the number of states in each unit

volume

( ) ( ) 2)( 21

00

=−+−== ∫∫∞∞

UUUUdUdUUgNv δδ

If the crystal has the size 1x4 cm3 then the total number of

states in the entire crystal must given by

∫ ==

4

0

8dVNN v

Microstates of interacting systems

2 interacting systems, A1 and A2 – heat, work & particles

Let U1: internal energy of A1

U2: internal energy of A2

A1 A2 Q

W

N

1 cm

1 cm

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4

U0 = U1 + U2 ≅ constant

The number of states

ΩO = Ω1Ω2

Let:

R for A1 = 6 degrees of freedom

R for A2 = 10 degrees of freedom

Therefore Ω1 ∝ (U1)3

Ω2 ∝ (U2)5

Assume U1 + U2 = 5 unit of energy

U1 U2 Ω1 Ω2 ΩO = Ω1Ω2

0 5 0 3125 0

1 4 1 1024 1024

2 3 8 243 1944

3 2 27 32 864

4 1 64 1 64

5 0 125 0 0

Total 3896

Since every state is equaly likely, the most probable energy

distribution

U1= 2 eu

U2 = 3 eu

Note: 1944/3896 = 0.5

-half of the time energy distribution is U1 = 2 eu

and U2= 3 eu

Now consider system with R1 = 12 and R2 = 20

Ω1 ∝ (U1)6

Ω2 ∝ (U2)10

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Assume U1 + U2 = 5 unit of energy

U1 U2 Ω1 Ω2 ΩO = Ω1Ω2

0 5 0 9.77x106 0

1 4 1 1.05x106 1.05x106

2 3 64 5.90x106 3.78x106

3 2 729 1.02x106 0.744x106

4 1 4.1x103 1 0.004x106

5 0 1.56x104 0 0

Total 5.57 x 106

For U1= 2 eu & U2 = 3 eu

-the accessible states are 68%

-the system will have this energy distribution more than

two-thirds of the time!!!

Let R1 = 120 and R2 = 200

Ω1 ∝ (U1)60

Ω2 ∝ (U2)100

U1 U2 Ω1 Ω2 ΩO = Ω1Ω2

0 5 0 7.9x1069 0

1 4 1 1.6x1060 1.6x1060

2 3 1.2x1018 5.2x1047 6.2x1065

3 2 4.2x1028 1.3x1030 5.5x1058

4 1 1.3x1028 1 1.3x1036

5 0 8.7x1041 0 0

Total 6.2 x 1065

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For U1= 2 eu & U2 = 3 eu

-the accessible states are 99.997%

-the system will have this energy distribution at any instant in

time!!!

Marcoscopic system

Number of particles ~ 1024

R1 = 6 x 1024 and R2 = 10 x 1024

( )24103

11

xUαΩ and ( )

24105

22

xUαΩ

U1 U2 Ω1 Ω2 ΩO = Ω1Ω2

0 5 0 24

1099.610 x 0

1 4 1 24

1002.610 x 24

1002.610 x 2 3

241081.110 x

241077.410 x

241058.610 x

3 2 2410861.210 x

24100.310 x

241087.510 x

4 1 24

1061.310 x 1 24

1061.310 x 5 0

241019.410 x 0 0

Total 24

1058.610 x

For U1= 2 eu & U2 = 3 eu

-the accessible states are the most probable --- i.e 24

1056.010 x

times more probable than the other distribution.

NOTE:

When 2 interacting macroscopic systems are in equilibrium, the

values of the various system variables will be such that the

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number of states available to the combined system is a

maximum.

The second law of thermodynamics As 2 interacting macroscopic systems approach

equilibrium, the changes in the system variables will be

such that the number of states available to the combined

system increases. Or, in the approach to equilibrium,

∆Ωo > 0

Note: 1st Law: - reflect inviolable fact

-work for small system

2nd Law: - based on probability

-for large system – there is some infinitesimal

probability the law be violated!!!

Heat flow and energy spreading

The fundamental science behind the second law:

Energy spontaneously disperses from being localized to

becoming spread out if it is not hindered

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Thermal contact:

-thermal energy flows from the higher occupied levels in

the warmer object into the unoccupied levels of the cooler

one until equal numbers are occupied in both bodies,

bringing them to the same temperature.

The degree of dilution of the thermal energy is

Q /T

Cold body: TCold is low; few thermal energy states are occupied,

so the amount of energy spreading can be very great.

When TCold near THot, more thermal energy states are occupied

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Entropy Entropy – (Greek for ‘turning into’)

– Is a measure of the degree of disorder of a system.

– In a reversible process entropy remains constant.

– In an irreversible process, entropy MUST increase.

Entropy can NEVER decrease in a closed system.

– In any cyclic process the entropy will either increases or

remains the same

The 2nd law of thermodynamics:

Every time energy is transformed from one state to

another, there is a loss in the amount of that form of

energy, which becomes available to perform work of some

kind. The loss in the amount of ‘available energy’ is known

as ‘entropy’

Entropy –a state variable whose change is defined for a

reversible process at T where Q is the heat absorbed

Entropy – a measure of the amount of energy which is

unavailable to do work

Entropy – a measure of the disorder of a system.

Entropy – a measure of the multiplicity of a system

Definition

Entropy just measures the spontaneous dispersal of

energy: how much energy is spread out in a process, or

how widely spread out it becomes – as a function of

temperature.

Or

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Entropy = S ≅≅≅≅ k ln ΩΩΩΩ

k: Boltzmann’s constant

Ω very big number – take lnΩ

Combine system (A0 = A1 + A2)

Ω0 = Ω1 Ω2

k lnΩ0 = k ln(Ω1 Ω2) = k lnΩ1 + k lnΩ2

S0 = S1 + S2

2nd law can be stated in terms of the entropy (alternative)

As 2 interacting macroscopic systems have reached

equilibrium, the changes in the system variables will be

such that the entropy of the combined system increases.

∆S0 > 0

When A0 = A1 + A2 have reached equilibrium, then ΩO will be

maximized

∆S0 (equilibrium) = 0

2nd law (alternative)

For any 2 interacting systems (whether in equilibrium or

not) the entropy of the combined system cannot decrease.

∆S0 > or = 0

What is entropy? How is it related to the second law?

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Entropy measures how much energy is dispersed in a particular process (at a specific temperature).

Energy spontaneously tends to flow only from being

concentrated in one place to becoming diffused or dispersed and

spread out.

e.g A hot frying pan cools down when it is taken off the kitchen

stove.

The energy concentrated inside a chemical like oil or coal

(or food) will spread out.

Cars rust, pendulums run down, people get old.

--Things like this don't happen backwards.

Like the flow of time (time arrow)

--they happen only in one sequence

The 2nd law of thermodynamics (or the law of entropy)

The universe and all of its energy systems will increase in

disorder as time moves forward.

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Disorder means the breakdown of energy into useless heat,

from which no work can be done.

Entropy is a measure of disorder

Chaos is a state of organized disorder

-entropy gives information about the evolution of an isolated

system with time → gives the direction of "time arrow"

-State which is more disordered →this state came later in time

-The second law of thermodynamics --gives the direction of

heat flow in any thermal process.

Heat naturally flows from higher T to lower T.

No natural process has as its sole result the transfer of

heat from a cooler to a warmer object.

No process can convert heat absorbed from a reservoir at

one temperature directly into work without also rejecting

heat to a cooler reservoir. That is, no heat engine is 100%

efficient.

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-Reasons -- why nature behaves in that way.

• Any process either increases the entropy of the universe -

or leaves it unchanged.

-Entropy is constant only in reversible processes

-All natural processes are irreversible.

• All natural processes tend toward increasing disorder

-energy is conserved

-but iits availability is decreased.

• Nature proceeds from

-simple to the complex

-orderly to the disorderly

-low entropy to high entropy.

• The entropy of a system is proportional to the logarithm of

the probability of that particular configuration of the

system occuringm(S ≅≅≅≅ k ln ΩΩΩΩ)

-more highly ordered the configuration of a system

-less likely it is to occur naturally

-hence the lower its entropy.

-The laws of Thermodynamics --wide range of applicatios

-Cosmology, History, Economics, Military, and to

almost everything

The 2nd law of Thermodynamics:

-the total entropy in the world is constantly increasing

-decrease in ‘available energy’.

-the unavailable energy-form works as pollution.

-the world is moving towards a dissipated state

-the pollution is constantly increasing

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14

-entropy (i.e. the ‘unavailable’ energy or pollution) tends

towards a "maximum"

-In a closed system - ultimately reached a stage where

there is no longer any difference in energy level --‘the

equilibrium state’.

-Maximum entropy -no longer ‘free energy’ is available

for work

Page 242: Sem5 thermal nota v1 dr wan

DENSITY OF STATE,

ENTROPY AND

THE SECOND LAW

WEEK 4

1

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2

In solid-state and condensed matter physics,

the density of states (DOS) of a system describes

the number of states per interval of energy at each

energy level that are available to be occupied by

electrons.

Unlike isolated systems, like atoms or molecules in gas

phase, the density distributions are not discrete like

a spectral density but continuous.

A high DOS at a specific energy level means that

there are many states available for occupation.

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3

A DOS of zero means that no states can be

occupied at that energy level.

In general a DOS is an average over the space

and time domains occupied by the system.

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Density of State

4

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5

R is the number of degrees of freedom

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eV, J, kJ

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8

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MACROSCOPIC SYSTEM

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its

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PHEWH ! !

30

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1

WEEK 5

Thermal Interaction

Axiom 1:

There exist special states of macroscopic physical

system, called equilibrium states, which can be fully

described by the internal energy, U, and set of

extensive parameters, X0, X1, ……XC.

Axiom 2:

For all system in equilibrium there exists a function of

the extensive parameters, called the entropy, S. If

there are no internal constraints on the system, the

extensive parameters can take the values that

maximise S over the possible states with internal

constraints.

The functional relation between S and the extensive

parameters

S = S (U, X0, X1,…..XC)

Note: X0 can be V; X1 can be N,….

2 systems interacting thermally

NVU

S

,

∂ no mechanical interactions

no exchange of particle

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TSP-WEEK 5

2

A0 = A1 + A2

When A1 and A2 are in equilibrium;

The entropy S0 is at it maximum value

0

,1

0 =

NVU

S

If S0 = f (U1)

Using S0 = S1 + S2 and dU1 = -dU2

NVNVNV

U

S

U

S

U

S

,1

2

,1

1

,1

0

∂+

∂=

0

,2

2

,1

1

,1

0 =

∂−

∂=

NVNVNVU

S

U

S

U

S

NVNV

U

S

U

S

,2

2

,1

1

∂=

∂ (*)

For 2 systems in thermal equilibrium,

the porperty NV

U

S

,

∂ is the same— Temperature, T

A1 A2 A0

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TSP-WEEK 5

3

Temperature is defined as TU

S

NV

1

,

From (*) 21

11

TT=

Note: when 2 systems are in thermal equilibrium, their

temperatures are equal.

Therefore, for 2 systems of the same temperature in thermal

contact,

NVNV

U

S

U

S

TT,2

2

,1

1

21

110

∂−

∂=−=

( )

NVNVU

S

U

SS

,1

0

,1

21

∂=

+∂=

Entropy combined system maximum at equilibrium

Let A1 in thermal equilibrium with A3

A2 in thermal equilibrium with A3

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4

T1 = T3 and T2 = T3

Therefore T1 = T2

Zeroth Law

If two systems are each in thermal equilibrium with a

third system, then they are in thermal equilibrium with

each other.

Note:

- temperature is the indicator of thermal equilibrium

- all parts of a system must be in thermal equilibrium if

the system is to have a definable single temperature.

Entropy – a measure of the number of accessible states Ω

T

1 - measures the Ω varies with the internal energy, U

( )

NV

NV

UkUU

S

T,

,

)(ln1

Ω∂

∂=

∂=

But 2)()(

R

UconstU =Ω

Therefore

( )

∂=Ω

∂= 2

,)ln()(ln

1R

NVUconstk

UUk

UT

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TSP-WEEK 5

5

( )

=+

∂=

U

kRUconst

U

kR 1

2ln)ln(

2

U

Rk

T

2

1

1= or kRTU

2

1=

U is measured to the zero-energy reference level, Nµ

Uthermal = Utotal - Nµ

Utotal = Uthermal + Nµ

= NkRT µ+2

1

if each particle – v degrees of freedom

R = vN

)2

(

2

µ

µ

+=

+=

kTv

N

NNkTv

Utotal

many common process, µ does not change or very little

)(

2TNk

vU ∆≈∆

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6

Heat Flow

A1 and A2 -- not yet in thermal equilibrium

T

UU

U

SS

NV

∆=∆

∂=∆

,

02

2

1

1210 >

∆+

∆=∆+∆=∆

T

U

T

USSS

1st Law ∆U1 = - ∆U2

011

21

10 >

−∆=∆

TTUS

If T2> T1, then ∆U1>0

If T1> T2, then ∆U1<0

Note:

if 2 interacting systems are not yet in thermal

equilibrium, then the 2nd law demands that the energy

must flow from the hotter system to the cooler one,

and not vice versa.

A1 A2 A0

T1 T2

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7

Phase Transition (add heat, T unchanged)

NVU

S

T ,

1

∂= at higher T, the

NVU

S

,

∂is smaller

From S = k ln Ω(U)

Where Ω(U) = (const)UR/2

S = ½ Rk lnU

TU

Rk

U

S

NV

1)

1(

2,

==

Undergoing phase transition—

Add heat, temperature remains constant

TU

Rk 1)

1(

2= --- why??

TU

S

NV

1

,

=

S

U (∝ T)

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8

2 reasons;

i) as heat is added, R increases, R/U constant

ii) the added heat – releases particles from potential

wells ( the zero-energy reference is higher)

Uthermal = Utotal - Nµ

from Ω(U) = (const)UR/2 , Ω(U) = (const)(U-Nµ)R/2

S = ½ Rk ln (U- Nµ)

TNU

Rk

U

S 1

)(2=

−=

µ

2 phase region U

S

∂ constant

boiling

melting

S

U

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9

Heat Reservoirs (add heat, T unchanged)

Heat reservoirs – sufficiently large systems that their

temperatures do not change when a small amount of heat is

added to or removed from them.

∆Q – small amount of heat.

Taylor series expansion

....)(2

1)()()( 2

,

2

2

,

+∆

∂+∆

∂+=∆+ Q

UQ

U

SUSQUS

NVNV

ignoring (∆Q)2 and higher

)()()(

,

QU

SUSQUS

NV

δδ

∂+=+

T

QdS

QT

USQUS

δ

δδ

=

=−+ )(

1)()(

----- for any system, large or small

T

QdS

δ=

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10

Thermal interaction with reservoir

Let 1U : average Internal Energy for A1

S0: entropy of the combined system (A1+A2)

Ωo: accessible states for the combined system

From fundamental postulate,

P (U1) = (const) Ωo (U1)

And S0 = k ln Ωo or k

S

e0

0 =Ω

k

US

econstUP

)(

1

10

)()( =

Entropy of the combined system

S0 (U1= 1U +∆U1)

Using Taylor series expansion

...)(2

1)()()( 2

12

1

0

2

1

1

0

1101110

11

+∆

∂+∆

∂+==∆+= U

U

SU

U

SUUSUUUS

UU

at equilibrium, S0 is maximum

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11

0

11

0 =

UU

S

And

∂+

∂=+

∂=

1

2

1

1

1

21

1

2

2

1

0

2

)(U

S

U

S

USS

UU

S

∂=

∂−

∂=

2112

2

1

1

1

11

TTUU

S

U

S

U

T2, reservoir temperature, -- constant

0

1

21

=

TU

And since TRk

U2

=

2

11111 22

1

U

Rk

U

Rk

UTU−=

∂=

Therefore 2

1

2

1

0

2

21

U

Rk

U

S

U

−=

And,

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TSP-WEEK 5

12

2

12

1

1101110 )(4

)()( UU

RkUUSUUUS ∆−==∆+=

Put this into the probability equation

212

1

10

21

210

)(4

)(

))(4/()(

1

)(

)()(

UU

R

k

US

k

UURkUS

eeconst

econstUPi

∆−

=

=

22

1 2/)()(

σUeconst

∆−=

With

1

2U

R=σ

The probability the system has energy in the range U1 and

U1+dU1 is propotional to the size of the range dU1

1

2/)(

11

221)()( dUeconstdUUP

U σ∆−=

1)(

1

11 =∫U

dUUP

∫∫∆−∆− =

1

1

1

221

11

2/)()()(

U

Ua

U

UdUeconstdUeconst

σ

With 22

1

σ=a

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13

aconstdUeconst

U

Ua π)()(

1

1

1 =∫∆−

Therfore 22

1)(

1)(

πσπ

π

==

=

aconst

aconst

For system in equilibrium with reservoir, the probability

that the system has energy in the range U and U+dU is

dUedUUP

U 22 2/)(

22

1)( σ

πσ

∆−=

With UR

2=σ and )( UUU −=∆

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TSP-WEEK 5

14

e.g: 2 systems & let ev1=ε for degree of freedom

R = 8 and R = 800

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15

Heat Capacity (The Storage of Thermal Energy)

-a measure of how much heat energy must be added

in order to raise its temperature by one standard unit.

General Definition

• If CT = Total Amount of Heat Energy required to raise

the temperature of some System by 1 oC, therefore Q

= CT ∆T ; where CT is heat capacity (SI: J/ oC)

SPECIFIC HEAT CAPACITY

If c = Amount of Heat Energy per kilogram that is

required to raise the temperature of one kilogram of the

substance 1 oC, therefore q = m c ∆T ; where c is

specific heat capacity (SI: J/kg.oC)

MOLAR HEAT CAPACITY

If c = Amount of Heat Energy per mole that is required

to raise the temperature of 6.022x10 23 molecules of the

substance 1 oC, therefore Q = n c ∆T; where c molar heat

capacity (SI: J/mole/ oC)

VOLUMETRIC HEAT CAPACITY of a SUBSTANCE

If cv = Amount of Heat Energy per unit volume that is

required to raise the temperature of one cubic meter of

the substance 1 oC, therefore Q = V cv ∆T, where cv is

volumetric heat capacity (SI: J/m3.oC )

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16

Definition

Xconstantatcapacityheat:

X

XT

QC

∂=

Two types of heat capacity

Heat capacity at constant pressure,

P

PT

QC

∆=

Heat capacity at constant volume

V

VT

QC

∆=

∆Q

∆Q

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Third law of thermodynamics (Nernst's theorem)

Order & disoder

two different gasses separated by a partition will mix when

the partition is removed, increasing system disorder.

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Entropy (S) is the thermodynamic state function that

describe the amount of disorder, -- a large value for entropy

means high degree of disorder

Entropy changes

An increase in disorder results in an increase in entropy.

- S increases when solid→liquid, liquid → gas

- S decreases when gas dissolves in a solvent

- S increase as temperature increase

Adding heat to the system,

-increases the number of the accessible states,

-increases disorder,

-increases the entropy.

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Removing heat from the system,

-decreases the number of accessibles states

-decreases the disorder

-decreases the entropy

remove the heat until,

The number of accessibles state Ω = 1

From S = k ln Ω = k ln (1) = 0

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third law of thermodynamics the entropy of any pure, perfect crystalline element or

compound at absolute zero (0 K) is equal to zero.

-all molecular motion stops at 0K

-represent perfect order

Absolute zero

Definition: The lowest possible temperature allowed by the

laws of thermodynamics.

At this temperature, molecules would possess the absolute

minimum KE allowed by quantum mechanics.

The temperature is equivalent to -273.15°C or 0K (kelvin).

At absolute zero, the entropy of any system vanished.

Or

The temperature at which all possible heat has been

removed from an object.

Absolute zero cannot be reached experimentally, although

it can be closely approached.

Cornell and Wieman cooled a small sample of atoms down

to only a few billionths (0.000,000,001) of a degree above

Absolute Zero!

Note: entropy T

QdS

δ=

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21

δQ = mC dT T

mCdTdS =

∫ ∫∫ ===∆T

dTmC

T

mCdTdSS

if C : constant

heating a sample from absolute zero to finite temperature T,

increases the entropy by,

∫∫∫ ===∆TTT

T

dTmC

T

mCdTdSS

000

= ST – S0 = ST – 0 = ST

Example:

A sample with specific heat c = 5T1/2 J/kg.K is heated to

300 K. Determine the entropy of the sample at 300 K

KkgJT

dTT

T

cdT

T

Q

dSSSSS

TT

T

KK

./1735

300

0

2/1

00

0

0300300

====

=∆=−=

∫∫∫

δ

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Example:

The reaction A + B → C is carried out at 300 K. How much

heat is released (per kg) of C if cA = 5T1/2 J/kg.K, cB = 8T1/3

J/kg.K and cC = 15 T1/3 J/kg.K.

Ans: ∆Q = T∆S and ∆S = SC – (SA + SB)

SA (300K) = 173 J/kg.K

SB (300K) =

SC (300K) =

∆S =

∆Q =

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WEEK 5

THERMAL INTERACTION

1

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2

AXIOM or Postulate

is a proposition that is not proved or demonstrated but

considered to be either self-evident, or subject to necessary

decision.

Therefore its truth is taken for granted and serves as a

starting point for deducing and inferring other truths theory

(dependent).

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AXIOM 1 /Postulate 1

AXIOM 2/Postulate 2

For all system in equilibrium there exists a function of the extensive parameters, called the entropy, S. If there are no internal constraints on the system, the extensive parameters can take the values that maximize S over the possible states with internal constraints.

3

There exist special states of macroscopic physical

system, called equilibrium states, which can be fully

described by the internal energy, U, and a set of

extensive parameters, X0, X1, X2,….Xc

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4

Macroscopic property

(physical system)Equilibrium State (ES)

fully described by

microscopic properties

(extensive thermodynamic

properties) U , V , N , S , …..

Intensive thermodynamic

properties

Extensive thermodynamic

properties

Independent of the size of the

thermodynamic system : P, T, r

Dependent : V , S, U, H, N, …..

= S maximised

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U1 U2

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Note: when 2 systems are in equilibrium, their temperatures are equal.

Therefore, for 2 systems of the same temperature in thermal contact,

Entropy of combined system is maximum at equilibrium

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T1 = T3T2 = T3

Therefore T1 = T2

T3

T2

A2

A3

T2

T3

T1

A1

A3

T3

Let A1 in thermal equilibrium with A3 A2 in thermal equilibrium with A3

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Zeroth Law:

If two systems are each in thermal

equilibrium with a third system, then they

are in thermal equilibrium with each other.

Note:

i. Temperature is the indicator of thermal equilibrium

ii. All parts of a system must be in thermal

equilibrium if the system is to have a definable

single temperature.

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Entropy - a measure of the number of accessible states Ω

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Note: if 2 interacting systems are not yet in thermal equilibrium, then the

2nd law demands that the energy must flow from the hotter system to the

cooler one, and not vice versa.

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Add Heat, temperature remains constant

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Ū1

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probability

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where cv is volumetric heat capacity

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cv

cp

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Order & Disorder

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Entropy changes

An Increase in disorder results in an increase in

entropy.

- S increases when solid – liquid, liquid – gas

- S decreases when gas dissolves in a solvent

- S increases as temperature increases

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J/K

Example:

A sample with specific heat c = 5T ½ J/kg/K is heated to 300 K from

0 K. Determine the entropy of the sample at 300 K.

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161J/kg.K

301 J/kg.K

301 – (173 + 161)= - 32.74 J/kg.K

-32.74 x 300 = - 9.8 kJ/kg

(heat is given out by the system)

J/K

kJ

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THAT’S ALL FOR NOW!!

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WEEK 6

Mechanical Interaction (Work)

The system exerts force F against the piston of area A, and the piston moves distance ds, The work done by the system,

δW = F. ds = (P/A) ds = P (A ds) = PdV

Note: δW is inexact, can be changed into exact diff. dV

δW. (1/P) ….. exact

And δQ .(1/T) ….. exact , dS ,entropy 1st Law (exact differential form)

dU = TdS – PdV + µdN (*) Where U = U (T,S,P)

µ = µ(P,V, N)

T = T (P,µ, N) ……..etc..

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From (*) NVS

UT

,

∂=

NSV

UP

,

∂=

VSN

U

.

∂=µ

change in entropy, dN

TdV

T

PdU

TdS

µ−+=

1 form (*)

if S = S(U, V, N)

dNN

SdV

V

SdU

U

SdS

VSNSNV ,,,

∂+

∂+

∂=

and NVU

S

T ,

1

∂= ,………..thermal interaction

NUV

S

T

P

,

∂= , ……..mechanical (work) int.

VUN

S

T ,

∂=−

µ ……….diffusive (particles) int.

mechanical interaction,

NUV

S

T

P

,

∂=

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NUNU V

kTV

kTP,,

lnln

Ω∆=

Ω∂=

or

1

212 lnlnlnln

Ω

Ω=Ω−Ω=Ω∆=

kT

VP

or kTVP

e/

1

2 ∆=Ω

Ω

1st Law: ∆U =∆Q - p∆V + µ∆N

but ∆U = ∆N = 0, then ∆Q = P∆V

kTQ

e/

1

2 ∆=Ω

Ω

Expansion When heated – materials tend to expand

Gases: increase the force & frequency of collisions with the container wall, -- pushing outward

Solid & liquid: osscillate with greater amplitudes, increase the intermolecular spacing --- expand !!!

Coefficient of volume expansion, β ---- a measure of the relative increase in volume per unit increase in temperature.

PT

V

V

∂≡

Note: generally β = β(T, P) , but most expansions are at atmospheric pressure, --- const. pressure.

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For small change in temperature, V = V (T)

TVTT

VV

P

∆=∆

∂=∆ β

α, coefficient of linear expansion a measure of relative increase in length (X)

α = α (T, P)

TXTT

XX

P

∆=∆

∂=∆ α

relation : α and β

V′= V + ∆V = V ( 1 + β ∆T ) (*)

But V′= X′Y′Z′=X(1+α∆T) Y(1+α∆T) Z(1+α∆T)

= XYZ (1+α∆T)3

= V(1+α∆T)3 = V(1 + 3α∆T +… ) (**)

for small ∆T, … ∆T2 ≅ 0

therefore , from (*) & (**), β = 3α

Isothermal compressibility (symbol: κ ) -a measure of relative change in volume per unit increase in pressure (temperature: constant)

TP

V

V

∂−≡

1κ κ = κ (T, P) and V = V( P, T )

PVPP

VV

T

∆−=∆

∂=∆ κ

bulk modulus :TV

PV

∂−=

κ

1

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The Diffusive interaction

Chemical Potential (symbol: µ) (in many cases, Free Energy is used instead of chemical potential)

The chemical potential : the change in the energy of the sytem when an additional constituent particle is introduced, with the entropy and volume held fixed.

VSN

U

,

∂=µ

• It express how eager system is for particles.

• In equilibrium, it is equal in two systems placed in diffusive contact.

• Particles move form a region of high chemical potential to a region of low chemical potential.

• It can be found by differentiating themodynamic potentials with respect to N.

∆N

∆U is realesed when ∆N energyless particles are added

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∆N : a known number of particles are added into the system

The gain in thermal energy is ∆U

µ ∆N = - ∆U

note: - ∆U : inside the potential well

N

U

∆−=µ

µ and temperature

+

+

+ −

+ −

+ −

− +

− +

− +

− +

− +

+

+

+

+

+

+ −

+

cold

hot

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µ and pressure at higher P, particles are closer , increase the strength of

interactions :attractive (reduce µ ) or repulsive (increase µ)

µ and particles concentration will increase or decrease --- depending on the nature of the interaction. the chemical potential increases as

• the internal energy, U, of the phase increases,

• the entropy, S, of the phase decreases at a given temperature, T.

• the volume, V increases for a given pressure, P. components that possess HIGHER U are destabilized relative to those with LOWER U components with LOWER S are destabilized relative to those with HIGHER S. Equilibrium Conditions

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Let A1 and A2 : interacting – thermally,mechanically and diffusively

1st Law NVPSTU ∆+∆−∆=∆ µ

or NT

VT

PU

TS ∆−∆+∆=∆

µ1

law of conservation

12

12

12

NN

VV

UU

∆−=∆

∆−=∆

∆−=∆

the change in entropy due to redistribution of energy, volume, or number of particles,

∆S0 = ∆S1 + ∆S2

1

2

2

1

11

2

2

1

11

21

0

11N

TTV

T

P

T

PU

TTS ∆

−−∆

−+∆

−=∆

µµ..(*)

at equilibrium, entropy is maximum

1

2

2

1

11

2

2

1

11

21

0

110 N

TTV

T

P

T

PU

TTS ∆

−−∆

−+∆

−==∆

µµ

therefore

011

21

=−TT

02

2

1

1 =−T

P

T

P

02

2

1

1 =−TT

µµ

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and

T1 = T2 P1 = P2 µ1 = µ2 For 2 systems interacting thermally, mechanically and diffusively are in equilibrium. Approach to equilibrium (not yet in equilibrium)

∆S0 > 0 Let A1 and A2 : interacting – thermally,mechanically and diffusively

1st Law: 111111 NVPQU ∆+∆−∆=∆ µ

using (*)

( )

1

2

2

1

11

2

2

1

1

11111

21

0

11

NTT

VT

P

T

P

NVPQTT

S

−−∆

+∆+∆−∆

−=∆

µµ

µ

( ) ( ) 01111

121

2

121

2

1

21

0 ⟩∆−−∆−+∆

−=∆ N

TVPP

TQ

TTS µµ

therefore

011

1

21

⟩∆

− Q

TT , ( ) 01

121

2

⟩∆− VPPT

( ) 01

121

2

⟩∆− NT

µµ

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from the three equations;

1. if T1> T2, ∆Q1 < 0, and if T2 >T1, ∆Q1>0 interacting thermally: heat flow from hotter towards the cooler, NOT vice versa

2. if P1> P2, ∆V1 > 0, and if P2 >P1, ∆V1 < 0 int. Mechanically: volume is gained by the system having higher pressure at the expense of the other, and NOT vice versa.

3. if µ1> µ2, ∆N1 < 0, and if µ2 >µ1, ∆N1>0 int. diffusively: particles flow from the system with

higher µ toward the one with lower µ, and NOT vice versa.

µ and Ω Let A1 and A2 : interacting – thermally,mechanically and diffusively

dNPdVTdSdU µ+−=

and

VSN

U

,

∂=µ

1st Law

dNT

dVT

PdU

TdS

µ−+=

1

if U and V constants

VUN

S

T ,

∂=−

µ or

VUN

ST

,

∂=− µ

but S = k ln Ω

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11

VUNkT

,

ln

Ω∂=− µ ………….($)

Ω, accessible states – increasing function of number of particles.

µ neg. or dS/dN pos. – attract particles

µ pos. or dS/dN neg. – release particles

let ∆N – energyless particles added to a system at constant volume. From ($)

N

kT∆−=Ω∆

µln

but 1

212 lnlnlnln

Ω

Ω=Ω−Ω=Ω∆

)/(

1

2 kTNe

∆−=Ω

Ω µ

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12

Ideal Gas

a hypothetical gas with molecules of negligible size that exert no intermolecular forces their energies are entirely kinetic for 1 molecule, mass m and momentum p

222

2

2

1

2

1

2

1

2ZYX p

mp

mp

mm

p++==ε

each molecule : 3 degrees of freedom for N molecules : 3N degrees of freedom number of quantum states available (6 dim.)

31h

pppzyx zyx

particles

∆∆∆∆∆∆=Ω

= (const) ∆x∆y∆z∆px∆py∆pz for N particles

iziyixiii

N

i

N

i

iparticlesN dpdpdpdzdydxh

∏ ∏= =

=Ω=Ω

1 13

1

limit: volume V and let U = total energy of the gas

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13

∫∏=

N

i

iziyix

N

particlesN dpdpdpVh 1

3

1……(*)

since mUppppP Nzxzyx 2.... 22

2

2

1

2

1

2

1 =+++++

-integral in (*) is equivalent to over the surface of a 3N dimension sphere of radius (2mU)1/2

-surface area of 3N-dim. Sphere ∝ (radius)(3N-1) or

∝ (radius)3N

NN

particlesN mUV3

)2(∝Ω

2/32/3)2)(( NNN

particlesN UVmconst=Ω

2/3)( NN

particlesN UVconst=Ω

UNkVNkconstkS gasidealgasideal ln2

3ln)(ln ++=Ω=

using

TU

S

V

1=

∂ and T

P

V

S

U

=

V:constant

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14

TU

Nk

U

S

V

12

3

00 =++=

NkTU

2

3=

U:constant

T

P

V

Nk

V

S

U

=++=

∂00

NkTPV = ………..ideal gas law …….. gas model Real gas

- degrees of freedom may be larger than 3

NkTU2

υ=

where υ is degrees of freedom (we will use R for gas constant)

- mutual interactions and sizes cannot be ignored le v : molar volume R = NAk : gas constant Pv = RT

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Compress gas, will become liquid, and cannot compress anymore b = molar volume of liquid phase – limit on the molar volume

v → v-b mutual interaction – reduce the velocity of molecules hitting the wall or pressure sensor real pressure is higher than the measured one

P →( P + mutual attraction)

Mutual attraction ∝ 1/(v2 ) Incorporated the two effect into the gas law

( ) RTbv

v

aP =−

+

2

van der Waals equation of state Liquid No good model for liquid yet Can use van der Waals equation with modification

-liquid phase, the volume ≅ b -pressure term

)(22

bfv

aP

v

aP +

+→

+

where f(b) : due to other interactions (liquid)

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since v ≅ b : constant for liquid

.)()(22

constbfb

abf

v

a≅+≈+

the modified van der Waals eq, (P + const.) b = RT Solids Models for solids – have more than one components One property of one component at a time Ex. The heat capacity of the lattice alone Solid – a lattice of atomic masses coupled by spring

Potential energy, εP = (1/2) kx2 When one atom vibrating – send wave down the lattice Quantum energy of the wave: called phonon Phonon travel throughout the solid’s volume – phonon gas Metal – conduction band: electrons in the conduction band are mutually shared by all atoms – known as electron gas

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Example of application of the models 1. heat capacity

let N: constant dQ = dU + PdV

υ: degree of freedom

U = (1/2) υNkT + Nµ

dQ = (1/2) υNk dT + N dµ+ P dV

change in N dµ is small

dQ = (1/2) υ R dT + P dV

per mol: dq = (1/2) υ R dT + P dv …. (*)

R = NA k -- gas constant Molar heat capacity at constant volume

RT

qc

V

V2

υ=

∂=

since v= v(T,P) --- molar volume

dPP

vdT

T

vdv

TP

∂+

∂=

from (*), at constant P ------ dP = 0

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dTT

vPRdTdq

P

∂+=

2

υ

PP

PT

vPR

T

qc

∂+=

∂=

2

υ

or

P

VPT

vPcc

∂+=

molar heat capacity at constant pressure or

P

VPT

vPcc

∂=−

for gases, change in volume at constant pressure is large For solid & liquid, change in volume at constant pressure is smaller -----cp & cv nearly the same. Example

1. Calculate cp-cv for an ideal gas Ans: Pv = RT or v = RT/P

P

VPT

vPcc

∂=−

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P

R

T

v

P

=

RP

RP

T

vPcc

P

vp ==

∂=−

2. Calculate cp-cv for va der Waals gas

( ) RTbvv

aP =−

+

2

a and b are constant Ans:

( ) RdTdvv

aPbvdv

v

adP =

++−

23

2

−+

−−=

12

1

)(

2v

b

Pv

a

dPbvRTPdv

−+

∂=−

12

12 v

b

Pv

a

R

T

vPcc

p

vp

note: for P >> a/v2 and v >> b

Rcc vp =− ……ideal gas

for P << a / v2 and v ≅ b

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0

2

≅=−

pva

Rcc vp

………liquid

3. Find an expression for the isothermal compressibility of an ideal gas

Ans:

TP

v

v

∂−=

1κ compressibility

Pv = RT Pdv = RdT – vdP

vP

vP

T

−=

∂ or PP

v

v T

11=

∂−=κ

4. 4. Find an expression for the isothermal compressibility of a van der Walls gas

Ans: ( ) RTbv

v

aP =−

+

2

−+

−−=

12

1

)(1

2 v

b

Pv

aPv

dPbvRTdv

v

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21

−+

=

∂−=

12

1

111

2v

b

Pv

a

vb

PP

v

v T

κ

Assigment 1 Suppose the equation of state for some system was

beTPaV =

−3

12

where a and b are constants

i. write this in differential form, expressing dV in terms of dT and dP

ii. Express the isothermal compressibility of this system in terms of (T, V, P)

iii. Express the coefficient of volume expansion for this system in terms of (T, V, P)

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MECHANICAL & DIFFUSIVEINTERACTIONS

WEEK 6

1

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2

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3

from (*)

U,N U,V

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Comparing the above 2 equations:

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g

But most expansions are at atmospheric pressure , - constant pressure.

Coefficient of volume expansion, - a measure of the relative increase in

volume per unit increase in temperature.

When heated – materials tend to expand

Gases: - increase the force & frequency of collisions with the container wall,

-- pushing outward.

Solid & liquid: - oscillate with greater amplitudes, increase the intermolecular

spacing – expand!!

Expansion

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• Particles move from a region of high chemical potential to a region of

low chemical potential.

• It express how eager a system is for particles.

•In equilibrium, it is equal in two systems placed in diffusive contact.

•It can be found by differentiating thermodynamics potentials with

respect to N.

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DU is released when DN energyless

particles are added

DN : a known number of particles

are added into the system.

The gain in thermal energy is DU.

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> < 0

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Six-dimensional space

Six-dimensional space is any space that has six

dimensions, that is, six degrees of freedom, and that

needs six pieces of data, or coordinates, to specify a

location in this space. There are an infinite number of

these, but those of most interest are simpler ones that

model some aspect of the environment. Of particular

interest is six-dimensional Euclidean in which 6-

polytopes and the 5-sphere are constructed.

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Thermal

interaction

Mechanical

interaction

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29

Let

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34

but

hence

so

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van

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Assignment (individual) Exercise

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Week 8

1

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2

ENSEMBLE : Synonyms

1. totality, entirety, aggregate

ahn-sahm-buh l,

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THERE ARE 3 TYPES OF ENSEMBLES:

MICROCANONICAL

CANONICAL

GRAND CANONICAL

3

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4

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5

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6

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U constant;

N, V, E constant;

Isolated

U constant;

T not constant;

In contact with

Heat reservoir

T , m constant;

U, N not constant;

In contact with Heat and

Particle reservoir

SUMMARY

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8

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9

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10

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11

Consider TWO cases:

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12

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Ui : ground stateUj: excited state& Pi: probability in the ground state i

Pj: probability in the excited state j

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In solid-state physics, the electronic

band structure (or simply band

structure) of a solid describes those

ranges of energy that an electron within

the solid may have (called energy

bands, allowed bands, or simply bands),

and ranges of energy that it may not

have (called band gaps or forbidden

bands).

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22

Band theory derives these bands and band gaps

by examining the allowed quantum

mechanical wave functions for an electron in a

large, periodic lattice of atoms or molecules.

Band theory has been successfully used to explain

many physical properties of solids, such as

electrical resistivity and optical absorption, and

forms the foundation of the understanding of

all solid-state devices (transistors, solar cells, etc.).

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- Normally a system can have many excited

states and many ground states.

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24

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25

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-how many particles can be in any quantum state? Total N is fixed

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27

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28

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29

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Energy and Quadratic Forms

A simple model of a crystal--masses m on springs with coefficients k.

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34

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35

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separation

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37

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38

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39

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40

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41

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42

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43

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A LITTLE CAPTION OF

TANGKUBAN PARAHU, INDONESIA (2009)

44

THE END

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Week 8

1

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2

Statistical mechanics

a branch of theoretical

physics and chemistry (and mathematical physics)

that studies, using probability theory, the average

behaviour of a mechanical system where the state

of the system is uncertain.

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3

A common use of statistical mechanics is in

explaining the thermodynamic behaviour of large

systems.

Microscopic mechanical laws do not contain concepts

such as temperature, heat, or entropy,

however, statistical mechanics shows how these

concepts arise from the natural uncertainty that arises

about the state of a system when that system is

prepared in practice.

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4

The benefit of using statistical mechanics

is that it provides exact methods to

connect thermodynamic quantities (such

as heat capacity) to microscopic

behaviour,

whereas in classical thermodynamics the

only available option would be to just

measure and tabulate such quantities for

various materials.

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5

Statistical mechanics also makes it possible

to extend the laws of thermodynamics to cases

which are not considered in classical

thermodynamics,

for example microscopic systems and other

mechanical systems with few degrees of freedom.

This branch of statistical mechanics which treats

and extends classical thermodynamics is known

as statistical thermodynamics or equilibrium

statistical mechanics.

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ENSEMBLE : Synonyms

1. totality, entirety, aggregate

ahn-sahm-buh l,

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THERE ARE 3 TYPES OF ENSEMBLES:

MICROCANONICAL

CANONICAL

GRAND CANONICAL

7

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8

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9

NVE ensemble)

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In statistical mechanics, a microcanonical ensemble is the statistical ensemble that is used to represent the possible states of a mechanical system which has an exactly specified total energy.

The system is assumed to be isolated in the sense that the system cannot exchange energy or particles with its environment, so that (by conservation of energy) the energy of the system remains exactly known as time goes on.

The system's energy, composition, volume, and shape are kept the same in all possible states of the system.

10

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The macroscopic variables of the microcanonical ensemble are quantities such as

(i) the total number of particles in the system (symbol: N),

(ii) the system's volume (symbol: V) each which influence the nature of the system's internal states,

(iii) as well as the total energy in the system (symbol: E).

This ensemble is therefore sometimes called the NVE ensemble, as each of these three quantities is a constant of the ensemble.

11

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In simple terms, the microcanonical ensemble is defined by assigning an equal probability to every microstate whose energy falls within a range centered at E. All other microstates are given a probability of zero. Since the probabilities must add up to 1, the probability P is the inverse of the number of microstates W within the range of energy,

P= 1/W The range of energy is then reduced in width until it

is infinitesimally narrow, still centered at E. In the limit of this process, the microcanonical ensemble is obtained.

12

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13

2. CANONICAL ENSEMBLE:

The system is in thermal equilibrium with a heat reservoir at temperature T.

Hence the temperature of the system is constant; but the energy of the system is not constant.

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In statistical mechanics, a canonical ensemble is the statistical ensemble that is used to represent the possible states of a mechanical system which is in thermal equilibrium with a heat bath.

The system is said to be closed in the sense that the system can exchange energy with a heat bath, so that various possible states of the system can differ in total energy. The system's composition, volume, and shape are kept the same in all possible states of the system.

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The thermodynamic variable of the canonical ensemble is the absolute temperature (symbol: T).

The ensemble is also dependent on mechanical variables such as

(i) the number of particles in the system (symbol: N)

(ii) the system's volume (symbol: V), each which influence the nature of the system's internal states.

This ensemble is therefore sometimes called the NVT ensemble, as each of these three quantities is a constant of the ensemble.

15

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In simple terms, the canonical ensemble assigns a probability P to each distinct microstate given by the following exponential:

where E is the total energy of the microstate, and k is Boltzmann's constant.

16

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The number A is the free energy (specifically, the Helmholtz free energy) and is a constant for the ensemble.

However, the probabilities and A will vary if different N, V, T are selected.

The free energy A serves two roles: to provide a normalization factor for the probability distribution (the probabilities, over the complete set of microstates, must add up to one); and, many important ensemble averages can be directly calculated from the function A(N, V, T).

17

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3. GRAND CANONICAL ENSEMBLE:

The system is in contact with both a heat reservoir

and a particle reservoir.

(i) the U and N of the system are not

constant.

(ii) the T and the m are constant.

(the m is the energy required to add a particle to the

system)

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In simple terms, the grand canonical ensemble assigns a

probability P to each distinct microstate given by the

following exponential:

where N is the number of particles in the microstate

and E is the total energy of the microstate.

k is Boltzmann's constant.

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In statistical mechanics, a grand canonical ensemble is the statistical ensemble that is used to represent the possible states of a mechanical system of particles that is being maintained in thermodynamic equilibrium (thermal and chemical) with a reservoir.

The system is said to be open in the sense that the system can exchange energy and particles with a reservoir, so that various possible states of the system can differ in both their total energy and total number of particles. The system's volume, shape, and other external coordinates are kept the same in all possible states of the system.

20

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The thermodynamic variables of the grand canonical ensemble are chemical potential (symbol: µ) and absolute temperature (symbol: T).

The ensemble is also dependent on mechanical variables such as volume (symbol: V) which influence the nature of the system's internal states.

This ensemble is therefore sometimes called the µVT ensemble, as each of these three quantities are constants of the ensemble.

21

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U constant;

N, V, E constant;

Isolated system

U not constant;

T constant;

In contact with

Heat reservoir

Closed system

T , m constant;

U, N not constant;

In contact with Heat and

Particle reservoir

Open system

SUMMARY

µVT ensembleNVT ensembleNVE ensemble

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24

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Consider TWO cases:

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Ui : ground stateUj: excited state& Pi: probability in the ground state i

Pj: probability in the excited state j

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33

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34

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35

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36

In solid-state physics, the electronic

band structure (or simply band

structure) of a solid describes those

ranges of energy that an electron within

the solid may have (called energy

bands, allowed bands, or simply bands),

and ranges of energy that it may not

have (called band gaps or forbidden

bands).

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37

Band theory derives these bands and band gaps

by examining the allowed quantum

mechanical wave functions for an electron in a

large, periodic lattice of atoms or molecules.

Band theory has been successfully used to explain

many physical properties of solids, such as

electrical resistivity and optical absorption, and

forms the foundation of the understanding of

all solid-state devices (transistors, solar cells, etc.).

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38

- Normally a system can have many excited

states and many ground states.

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39

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40

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41

-how many particles can be in any quantum state? Total N is fixed

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42

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43

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44

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45

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46

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Energy and Quadratic Forms

A simple model of a crystal--masses m on springs with coefficients k.

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48

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49

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separation

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52

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53

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55

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56

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57

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58

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A LITTLE CAPTION OF

TANGKUBAN PARAHU, INDONESIA (2009)

59

THE END

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Week 10

1

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2

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3

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4

S

S

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5

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6

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= =

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8

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9

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10

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11

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14

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15

Let’s say :

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Total no of accessible states?

16

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Total no of accessible states?

17

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18

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19

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20

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21

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TO WRITE MAXWELL-BOLTZMANN DISTRIBUTION FUNCTION:

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23

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THE END….

44

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WEEK 11

1

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2

Partition functions describe the statisticalproperties of a system in thermodynamic equilibrium. It is a function of temperatureand other parameters, such as the volumeenclosing a gas.

Most of the aggregate thermodynamicvariables of the system, such as the total energy, free energy, entropy, and pressure, can be expressed in terms of the partition function or its derivatives.

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3

There are actually several different types of partition functions, each corresponding to different types of statistical ensemble (or, equivalently, different types of free energy.)

The canonical partition function applies to a canonical ensemble, in which the system is allowed to exchange heat with the environment at fixed temperature, volume, and number of particles.

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4

The grand canonical partition functionapplies to a grand canonical ensemble, in which the system can exchange both heat and particles with the environment, at fixed temperature, volume, and chemical potential.

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5

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6

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This is the reason for calling Z the "partition function": it encodes how the probabilities are partitioned among the different microstates, based on their individual energies.

The letter Z stands for the German word Zustandssumme, "sum over states". This notation also implies another important meaning of the partition function of a system: it counts the (weighted) number of states a system can occupy. Hence if all states are equally probable (equal energies) the partition function is the total number of possible states. Often this is the practical importance of Z.

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8

From

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9

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10

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11

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25

The End

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WEEK 12

1

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2

occupied by

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3

1.FERMI-DIRAC DISTRIBUTION

2.BOSE-EINSTEIN DISTRIBUTION

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4

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5

THE FERMI-DIRAC DISTRIBUTION

Assumptions:

1. The particles are identical and indistinguishable

2. They obey the Pauli exclusion principle.

This means that no quantum state can accept more than

one particle, taking spin into account. Such particles have

half-integer spin and are given the generic name

FERMION.

Examples of fermions: electrons, positrons, protons,

neutrons, and muons.

An important application of Fermi-Dirac statistics is to the

behaviour of free electrons in metals and semi-conductors.

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6

The Pauli exclusion principle is

the quantum mechanical principle that no

two identical fermions (particles with half-

integer spin) may occupy the same quantum

state simultaneously.

A more rigorous statement is that the

total wave function for two identical fermions

is anti-symmetric with respect to exchange of

the particles. The principle was formulated by

Austrian physicist Wolfgang Pauli in 1925.

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7

Integer spin particles, bosons, are not subject

to the Pauli exclusion principle: any number of

identical bosons can occupy the same

quantum state, as with, for instance, photons

produced by a laser and Bose–Einstein

condensate.

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8

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9

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14

BOSE-EINSTEIN DISTRIBUTION

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15

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16

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For the continuous energy spectrum,

Bose-Einstein

distribution

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56

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57

PHEW…!!

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BOSE EINSTEIN CONDENSATION

WEEK 13

1

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2

Here we shall be concerned with a gas of non-interacting

particles (atoms or molecules) of comparatively large mass

such that quantum effects only become important at very low

temperatures. The particles are assumed to comprise an

ideal Bose-Einstein gas. The discussion is relevant to ,

which undergoes a remarkable phase transition known as

Bose-Einstein condensation.

This phenomenon is intimately related to the

superfluidity of liquid helium at low temperatures.

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3

As noted earlier, bosons are particles of integral spin that obey Bose-Einstein statistics. There is no limit to the number of bosons that can occupy any single particle state.

Consider an ideal boson gas consisting of N

bosons in a container of volume V held at

absolute temperature T. The Bose-Einstein

continuum distribution is

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4

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5

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6

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29

THE END

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WEEK 14

BLACKBODY RADIATION

1

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2

Statistical thermodynamics is applicable to

(i)radiant energy (ii) material particles.

Familiar observation:

a hot body loses heat by radiation. The energy loss is

attributable to the emission of electromagnetic waves

from the body.

The distribution of energy flux over the wavelength spectrum does

not depend on the nature of the body but does depend on its

temperature.

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3

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4

Now: concern with the thermodynamic properties of electromagnetic radiation

in thermal equilibrium.

The radiation can be regarded as a photon gas. Consider an enclosure or

cavity of volume V at a constant temperature T. The walls of the cavity are

thermally insulated and perfectly reflecting. Since the system is isolated, it has

a fixed energy U.

However, the photons emitted by one energy level may be absorbed at

another, so the number of photons is not constant. This means that the

restriction does not apply.

Photons are bosons of spin 1 and hence obey Bose-Einstein statistics.

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5

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6

What happens to this radiation?

• The radiation is absorbed in the walls of the cavity.

• This causes a heating of the cavity walls.

• Atoms in the walls of the cavity will vibrate at frequencies

characteristic of the temperature of the walls.

• These atoms then re-radiate the energy at this new

characteristic frequency.

• The emitted “thermal” radiation characterizes the

equilibrium temperature of the black-body.

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EXAMPLE:

Assume that the radiation from the Sun can be regarded as blackbody

radiation. The radiant energy per wavelength interval has a maximum at

480 nm. Estimate the temperature of the sun.

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AT LAST!

THANK YOU

FOR YOUR

PATIENCE

GUYS…


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