Stat Therm 1

7/28/2019 Stat Therm 1

1/73

Supplement

Statistical Thermodynamics


2/73

Contents

1. Introduction

2. Distribution of Molecular States

3. Interacting SystemsGibbs Ensemble

4. Classical Statistical Mechanics


3/73

1. Introduction

Mechanics : Study of position, velocity, force and

energy

Classical Mechanics (Molecular Mechanics)

Molecules (or molecular segments) are treated as rigid object(point, sphere, cube,...)

Newtons law of motion

Quantum Mechanics

Molecules are composed of electrons, nuclei, ...

Schrodingers equationWave function


4/73

1. Introduction

Methodology of Thermodynamics and Statistical Mechanics

Thermodynamics

study of the relationships between macroscopicproperties

Volume, pressure, compressibility,

Statistical Mechanics (Statistical Thermodynamics)

how the various macroscopic properties arise as a consequence of the microscopic natureofthe system

Position and momenta of individual molecules (mechanical variables)

Statistical Thermodynamics (or Statistical Mechanics) is a linkbetween microscopic

properties and bulk (macroscopic) properties

Methods of QM

Methods of MM

A particular

microscopic model

can be usedr

U

P T V

Statistical

Mechanics

Pure mechanical variablesThermodynamic Variables


5/73

1.Introduction

Equilibrium Macroscopic Properties

Properties are consequence of average of individual

molecules

Properties are invariant with time Time average

Mechanical

Properties of

Individual

Molecules

position, velocity

energy, ...

Thermodynamic

Properties

temperature, pressure

internal energy, enthalpy,...

average

over molecules

average

over time

statistical

thermodynamics

2 3 4


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1. Introduction

Description of States

Macrostates: T, P, V, (fewer variables)

Microstates: position, momentum of each particles (~1023 variables)

Fundamental methodology of statistical mechanics

Probabilistic approach : statistical average

Most probable value

Is it reasonable ?

As N approaches very large number, then fluctuations are negligible

Central Limit Theorem (from statistics) Deviation ~1/N0.5


7/73


Statistical Distribution

n : number of occurrences

b : a property

b

ni

1 2 3 4 5 6

if we know distribution

we can calculate the average

value of the property b


8/73


Normalized Distribution Function

Probability Distribution Function

b

Pi

b1 b5b4b3b2 b6

i

ii

i

ii

iiiiii

bP

bn

bn

n

bnbP

1)(

)(

)()()(

i

ii

i

ii

PbFbF

Pbb

)()(

Finding probability (distribution) function is

the main task in statistical thermodynamics


9/73


Quantum theory says ,

Each molecules can have only discrete values of energies

Evidence

Black-body radiation

Planck distribution

Heat capacities

Atomic and molecular spectra

Wave-Particle duality

Energy

Levels


10/73


Configuration ....

At any instance, there may be no molecules at e0 , n1molecules at e1 , n2 molecules at e2,

{n0 , n1 , n2 } configuration

e0

e2e1

e3

e4e5

{ 3,2,2,1,0,0}


11/73


Weight ....

Each configurations can be achieved in different ways

Example1 : {3,0} configuration 1

Example2 : {2,1} configuration 3e0

e1

e0

e1 e

0

e1 e

0

e1


12/73


Calculation of Weight ....

Weight (W) : number of ways that a configuration can be achieved in

different ways

General formula for the weight of {n0 , n1 , n2 } configuration

i

in

N

nnn

NW

!

!

!...!!

!

321

Example1

{1,0,3,5,10,1} of 20 objects

W = 9.31E8

Example 2

{0,1,5,0,8,0,3,2,1} of 20 objects

W = 4.19 E10


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Principles of Equal a Priori Probability

All distributions of energy are equally probable

If E = 5 and N = 5 then

0 1 2 3 4 5

0 1 2 3 4 5

0 1 2 3 4 5

All configurations have equal probability, butpossible number of way (weight) is different.


14/73

A Dominating Configuration

For large number of molecules and large number of energy

levels, there is a dominating configuration.

The weight of the dominating configuration is much more

larger than the other configurations.

Configurations

Wi

{ni}


15/73

Dominating Configuration

0 1 2 3 4 5

0 1 2 3 4 5

0 1 2 3 4 5

W = 1 (5!/5!) W = 20 (5!/3!) W= 5 (5!/4!)

Difference in Wbecomes larger when N is increased !

In molecular systems (N~1023) considering the

most dominant configuration is enough for average


16/73

How to find most dominant

configuration ?

The Boltzmann Distribution

Task : Find the dominant configuration for given N andtotal energy E

Method : Find maximum value ofW which satisfies,

i

ii

i

i

nE

nN

e 0

0

i

ii

i

i

dn

dn

e
http://www-groups.dcs.st-and.ac.uk/~history/PictDisplay/Boltzmann.html


17/73

Stirling's approximation

A useful formula when dealing with factorials of

large numbers.

NNNN ln!ln

i

i

i

i

ii

i

i

i

i

nnNN

nnnNNN

nNnnn

NW

lnln

lnln

!ln!ln!...!!

!lnln

321


18/73

Method of Undetermined

Multipliers

Maximum weight , W

Recall the method to find min, max of a function

Method of undetermined multiplier :

Constraints should be multiplied by a constant and

added to the main variation equation.

0ln

0ln

idnW

Wd


19/73


Multipliers

0ln

lnln

iii

i

i

ii

i

i

i

i

i

dndn

W

dndndndn

WWd

e

e

0ln

i

idn

We

undetermined multipliers


20/73


Multipliers

j i

jj

ii

ii

n

nn

n

NN

n

W

nnNNW

)ln(lnln

lnlnln

1ln1

lnln

N

n

N

N

NN

n

N

n

NN

iii

j

i

i

j

j

jj

i

j

j i

jjn

n

n

nnn

n

n

n

nn1ln

1ln

)ln(

N

nNn

n

W ii

i

ln)1(ln)1(lnln


21/73


Multipliers

0ln ii

N

ne ie

N

ni e

j

j jj

j

j

ee

eNenN

e

e

1

j

i

i j

i

e

e

N

n

P e

e Boltzmann Distribution

(Probability function forenergy distribution)

Normalization Condition


22/73

The Molecular Partition Function

Boltzmann Distribution

Molecular Partition Function

Degeneracies : Same energy value but different states (gj-

fold degenerate)

q

e

e

e

N

np

i

j

i

j

ii

e

e

e

j

jeqe

jlevels

jjegq

e


23/73

How to obtain the value of beta ?

Assumption :

T 0 then q 1

T infinity then q infinity

The molecular partition function gives an indication of the

average number of states that are thermally accessible to amolecule at T.

kT/1


24/73

2. Interacting Systems

Gibbs Ensemble

Solution to Schrodinger equation (Eigen-value problem)

Wave function

Allowed energy levels : En

Using the molecular partition function, we can calculate

average values of property at given QUANTUM STATE.

Quantum states are changing so rapidly that the observed

dynamic properties are actually time average over

quantum states.

i

i

i

EUm

h 22

2

8


25/73

Fluctuation with Time

time

states

Although we know most probable distribution of energies of individualmolecules at given N and E (previous sectionmolecular partitionfunction) it is almost impossible to get time average for interactingmolecules


26/73

Thermodynamic Properties

Entire set of possible quantum states

Thermodynamic internal energy

,...,...,, 111 i

,...,...,,, 321 iEEEE

i

ii tEU

1lim


27/73

Difficulties

Fluctuations are very small

Fluctuations occur too rapidly

We have to use alternative, abstract approach.

Ensemble average method (proposed by Gibbs)


28/73

Alternative Procedure

Canonical Ensemble

Proposed by J. W. Gibbs (1839-1903)

Alternative procedure to obtain average

Ensemble : Infinite number of mental replica

of system of interestLarge reservoir (constant T)

All the ensemble members have the same (n, V, T)

Energy can be exchanged butparticles cannot

Number of Systems : NN


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Two Postulate

E1 E2 E3 E4 E5

time

Fist Postulate

Thelong time averageof a mechanical variable M isequal to theensemble averagein the limit N

Second Postulate (Ergodic Hypothesis)

The systems of ensemble aredistributed uniformly for (n,V,T) systemSingle isolated systemspend equal amount of time


30/73

Averaging Method

Probability of observing particular quantum state i

Ensemble average of a dynamic property

Time average and ensemble average

i

i

ii

n

nP ~

~

i

i

iPEE

i

i

in

ii PEtEU

limlim


31/73

How to find Most Probable

Distribution ?

Calculation of Probability in an Ensemble

Weight

Most probable distribution = configuration with maximum weight

Task : find the dominating configuration for given N and E Find maximum W which satisfies

W

i

in

N

nnn

N

!~!

~

!...~!~!~!

~

321

i

iit

i

i

nEE

nN

~

~~

0~

0~

i

ii

i

i

ndE

nd


32/73

Canonical Partition Function

Similar method (Section 2) can be used to get most

probable distribution

j

E

E

ii

j

i

e

e

N

nP

Q

e

e

e

N

nP

i

j

i E

j

E

E

ii

j

EjeQ



33/73

How to obtain beta ?

Another interpretation

rev

i

ii

i

i

i

ii

i

ii

i

rev

N

ii

i

ii

dqTdSdPPdPQdPPdPE

wPdVdV

V

EPdEP

ln1

)lnln(1

i

ii

i i

iiii dEPdPEPEddU )(

pdVTdSwqdU revrev

The only function that links heat (path integral) and

state property isTEMPERATURE.

kT/1


34/73

Properties from Canonical Partition

Function

Internal Energy

VNVN

qsi

Ei

VN

qsi

E

ii

i

i

QQ

QU

eEQ

eEQ

PEEU

i

i

,,

)(,

)(

ln1

1


35/73

Properties from Canonical Partition

Function

Pressure

N

i

i

iiiNi

iiNi

V

EP

wdVPdxFdE

dxFdVPw

)(

)(

dx

dViF

V

idE

i

E

Small Adiabatic expansion of system

V

QP

eV

E

QV

Q

eV

E

QeP

QP

PPP

i

E

N

i

N

i

E

N

i

i

E

i

i

i

i

ii

ln

ln1

ln

11

P

,

i


36/73

Thermodynamic Properties from


ijNVTi

i

NT

NTNV

NV

NV

N

QkT

V

Q

QkTG

QkTA

V

Q

T

QkTH

T

QQkS

T

QkTU

,,

,

,,

,

,

ln

)ln

ln

(ln

ln

)ln

ln()

ln

ln(

)

ln

ln(ln

)ln

ln(


37/73

Grand Canonical Ensemble

Ensemble approach for open system

Useful for open systems and mixtures

Walls are replaced by permeable walls

Large reservoir (constant T )

All the ensemble members have the same (V, T, i )

Energy and particles can be exchanged

Number of Systems : NN


38/73


Similar approach as Canonical Ensemble

We cannot use second postulate because systems are not isolated

After equilibrium is reached, we place walls around ensemble and treat

each members the same method used in canonical ensemble

Afterequilibrium

T,V, T,V,N1

T,V,N2

T,V,N3

T,V,N4

T,V,N5

Each members are(T,V,N) systems Apply canonical ensemble methods for each member


39/73


Weight and Constraint

W

Njj

Nj

j

Nn

Nn

,

,

)!(

!)(

Nj

jt

Nj

jjt

Nj

j

NNnN

NVENnE

Nn

,

,

,

)(

),()(

)(

Number of ensemble members

Number of molecules after

fixed wall has been placed

Method of undetermined multiplierwith, ,g

NVNE

j eeeNn

j g

),(*

)(N

Nj

NVNE

NVNE

jj

jee

eeNnNnNP

j

j

,

),(

),(* )()()(

g

g

N


40/73


Determination of Undetermined Multipliers

Nj

jj VNENPEU,

),()(

Nj

jj

Nj

jj VNdENPNdPVNEdU,,

),()()(),(

Nj

j

j

Nj

jj dVV

VNENPNdPNPNdU

,,

),()()(ln)(ln

1g

Nj

NVNE ee j

,

),( g

dNpdVTdSdU Comparing two equation gives,

kT

g

kT

1

Nj

kTNkTVNE ee j

,

//),(

Grand Canonical Partition Function


41/73

4. Classical Statistical Mechanics

The formalism of statistical mechanics relies very much at the

microscopic states.

Number of states , sum over states

convenient for the framework of quantum mechanics

What about Classical Sates ?

Classical states

We know position and velocityof all particles in the system

Comparison between Quantum Mechanics and Classical Mechanics

EH QM Problem Finding probability and discrete energy states

maF CM Problem Finding position and momentum of individual molecules


42/73

Newtons Law of Motion

Three formulations for Newtons second law of motion

Newtonian formulation

Lagrangian formulation

Hamiltonian formulation

ii

i

ii

t

mt

Fp

pr

ijj

iji

zyx

zyx

ppp

rrr

1

),,(

),,(

FF

pp

rr

),...,,(2

),(

energy)alPE(potentienergy)KE(kinetic),(

21 N

i i

iNN

NN

Um

H

H

rrrp

pr

pr

i

i

i

i

H

H

rp

p

r


43/73

Classical Statistical Mechanics

Instead of taking replica of systems, use abstract phase spacecomposed of momentum space and position space (total 6N-space)

Np

Nr

1t

2t

1

2

NN rrrrpppp ,...,,,,,...,,, 311321

Phase space


44/73


Classical State : defines a cell in the space (small volume ofmomentum and positions)

Ensemble Average

simplicityforState"Classical" 33 rpdddrdrdrdqdqdq zyxzyx

dEdEU Nn )()(lim)(1lim 0 P

dN )(P Fraction of Ensemble members in this range(to +d)

dkTH

dkTHdN

)/exp(...

)/exp()(P

Using similar technique used for

Boltzmann distribution


45/73



dkTH )/exp(...TPhase Integral

dkTHc )/exp(...QCanonical Partition Function

dkTH

kTEi

i

T

)/exp(...

)/exp(

limcMatch between Quantum

and Classical Mechanics

NFhN!

1c

For rigorous derivation see Hill, Chap.6 (Statistical Thermodynamics)


46/73


Canonical Partition Function in Classical Mechanics

dkThN NF )/exp(...!1

HQ


47/73

Example ) Translational Motion for

Ideal Gas

N

i i

i

N

i i

iNN

NN

m

pH

Um

H

H

3 2

21

2

),...,,(2

),(

energy)alPE(potentienergy)KE(kinetic),(

rrrp

pr

pr

N

N

NV

N

i

N

i

NN

i

i

N

Vh

mkT

N

drdrdrdp

m

p

hN

drdrdpdpm

p

hNQ

2/3

2

0

321

3

3

11

2

3

2

!

1

)

2

exp(

!

1

......)2

exp(...!

1

No potential energy, 3 dimensional

space.

We will get ideal gas law

pV=nRT


48/73

Semi-Classical Partition Function

The energy of a molecule is distributed in different modes

Vibration, Rotation (Internal : depends only on T)

Translation (External : depends on T and V)

Assumption 1 : Partition Function (thus energy distribution) can be separated

into two parts (internal + center of mass motion)

),(),,(

)exp()exp()exp(

int

intint

TNQTVNQQ

kT

E

kT

E

kT

EEQ

CM

i

CM

ii

CM

i


49/73


Internal parts are density independent and most of the

components have the same value with ideal gases.

For solids and polymeric molecules, this assumption is not

valid any more.

),0,(),,( intint TNQTNQ


50/73


Assumption 2 : for T>50 K , classical approximation can be

used for translational motion

N

N

i

NNiziyix

N

N

i

iziyix

CM

drdrdrkTUZ

mkT

h

ZN

drkTUdpmkT

ppp

hNQ

rrrUm

pppH

321

2/12

3

33

222

3

321

222

.. .)/(.. .

2

!

)/(.. .)2

exp(.. .!

1

),...,,(2

ZQN

Q N3

int

!

1

Configurational Integral


51/73


52/73

Another, Different Treatment

St ti ti l Th d i


53/73

Statistical Thermodynamics:

the basics

Nature is quantum-mechanical

Consequence:

Systems have discrete quantum states.

For finite closed systems, the number ofstates is finite (but usually very large)

Hypothesis: In a closed system, every

state is equally likely to be observed. Consequence: ALL of equilibrium

Statistical Mechanics and

Thermodynamics

E h i di id l

, but there are not


54/73

Basic assumptionEach individual

microstate is

equally probable

,

many microstates that

give these extreme

results

If the number of

particles is large (>10)

these functions are

sharply peaked

Does the basis assumption lead to something


55/73


that is consistent with classical

thermodynamics?

1E

Systems 1 and 2 are weakly coupled

such that they can exchange energy.

What will beE1?

1 1 1 1 2 1,E E E E E EW W W BA: each configuration is equally probable; but the number of

states that give an energyE1 is not know.

2 1E E E

E E E E E EW W W


56/73

1 1 1 1 2 1,E E E E E EW W W

1 1 1 1 2 1ln , ln lnE E E E E EW W W

1 1

1 1

1 ,

ln ,0

N V

E E E

E

W

lnW1E

1 E1

N

1,V

1

lnW2EE

1 E1

N

2,V

2

0

1 1 2 2

1 1 2 1

1 2, ,

ln ln

N V N V

E E E

E E

W W

Energy is conserved!dE1=-dE2

This can be seen as an

equilibrium condition

,

ln

N V

E

E

W

1 2

Entropy and number of


57/73

Conjecture:

Almost right.

Good features:

Extensivity

Third law of thermodynamics comes for free

Bad feature:

It assumes that entropy is dimensionless but (forunfortunate, historical reasons, it is not)


configurations

lnS W


58/73

We have to live with the past, therefore

With kB= 1.380662 10-23 J/K

In thermodynamics, the absolute (Kelvin)

temperature scale was defined such that

But we found (defined):

lnBS k E W

,

1

N V

S

E T

,

ln

N V

E

E

W

dE TdS-pdV idNi

i1

n


59/73

And this gives the statistical definition of temperature:

In short:

Entropy and temperature are both related to

the fact that we can COUNT states.

,

ln1 BN V

EkT E

W

Basic assumption:1. leads to an equilibrium condition: equal temperatures

2. leads to a maximum of entropy

3. leads to the third law of thermodynamics

N b f fi i


60/73

How large is W?

For macroscopic systems, super-astronomically large.

For instance, for a glass of water at room temperature:

Macroscopic deviations from the second law of

thermodynamics are not forbidden, but they are

extremely unlikely.

Number of configurations

252 1010 W

C i l bl


61/73

Canonical ensemble

Consider a small system that can exchange heat with a big reservoir

iE iE E

lnln lni iE E E E

E

WW W

ln i i

B

E E EE k T

W W

1/kBT

Hence, the probability to findEi:

exp

exp

i i B

i

j j Bj j

E E E k TP EE E E k T

W W

expi i BP E E k T


Example: ideal gas


62/73

Example: ideal gas

31

, , d exp!

N N

NQ N V T U r

N r

3 3

1d 1

! !

NN

N N

V

N N

rFree energy:

3

3 3

ln !

ln ln ln ln

N

N

VF N

NN N N N

V

Pressure:

T

F NP

V V

Energy:

3 3

2B

F NE Nk T

Thermo recall (3)

Helmholtz Free energy:

T

F PV

d d dF S T p V

1F T F E

T

Energy:

Pressure


63/73

Ensembles

Micro-canonical ensemble:E,V,N

Canonical ensemble: T,V,N

Constant pressure ensemble: T,P,N

Grand-canonical ensemble: T,V,

Each individual

, but there are not


64/73

Basic assumptionEach individual

microstate is

equally probable

many microstates that

give these extreme

results

If the number of

particles is large (>10)

these functions are

sharply peaked



65/73

p g

that is consistent with classical

thermodynamics?

1E

Systems 1 and 2 are weakly coupled

such that they can exchange energy.

What will beE1?

1 1 1 1 2 1,E E E E E EW W W BA: each configuration is equally probable; but the number of

states that give an energyE1 is not know.

2 1E E E

E E E E E EW W W


66/73

1 1 1 1 2 1,E E E E E EW W W

1 1 1 1 2 1ln , ln lnE E E E E EW W W

1 1

1 1

1 ,

ln ,0

N V

E E E

E W

lnW1E

1 E1

N

1,V

1

lnW2EE

1 E1

N

2,V

2

0

1 1 2 2

1 1 2 1

1 2, ,

ln ln

N V N V

E E E

E E

W W

Energy is conserved!dE1=-dE2

This can be seen as an

equilibrium condition

,

ln

N V

E

E

W

1 2



67/73

Conjecture:

Almost right.

Good features:

Extensivity

Third law of thermodynamics comes for free

Bad feature:

It assumes that entropy is dimensionless but (forunfortunate, historical reasons, it is not)

py

configurations

lnS W


68/73

We have to live with the past, therefore

With kB= 1.380662 10-23 J/K

In thermodynamics, the absolute (Kelvin)

temperature scale was defined such that

But we found (defined):

lnBS k E W

,

1

N V

S

E T

,

ln

N V

E

E

W

dE TdS-pdV idNi

i1

n


69/73

And this gives the statistical definition of temperature:

In short:

Entropy and temperature are both related to

the fact that we can COUNT states.

,

ln1 BN V

EkT E W

Basic assumption:

1. leads to an equilibrium condition: equal temperatures

2. leads to a maximum of entropy

3. leads to the third law of thermodynamics

N b f fi ti


70/73

How large is W?

For macroscopic systems, super-astronomically large.

For instance, for a glass of water at room temperature:

Macroscopic deviations from the second law of

thermodynamics are not forbidden, but they are

extremely unlikely.

Number of configurations

252 1010 W

C i l bl


71/73

Canonical ensemble

Consider a small system that can exchange heat with a big reservoir

iE iE E

lnln lni iE E E E

E

WW W

ln i i

B

E E EE k T

W W

1/kBT

Hence, the probability to findEi:

exp

exp

i i B

i

j j Bj j

E E E k TP EE E E k T

W W

expi i BP E E k T


Example: ideal gas


72/73

p g

31

, , d exp!

N N

NQ N V T U r

N r

3 3

1d 1

! !

NN

N N

V

N N

rFree energy:

3

3 3

ln !

ln ln ln ln

N

N

VF N

NN N N N

V

Pressure:

T

F NP

V V

Energy:

3 3

2B

F NE Nk T

Thermo recall (3)

Helmholtz Free energy:

T

FP

V

d d dF S T p V

1F T F E

T

Energy:

Pressure


73/73

Ensembles

Micro-canonical ensemble:E,V,N

Canonical ensemble: T,V,N

Constant pressure ensemble: T,P,N

Grand-canonical ensemble: T,V,

Stat Therm 1

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