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Statistická termodynamika Statistical thermodynamics Peter Košovan [email protected] Lecture 1, 4.10.2015, MC260P105, WS 2016/2017 If you find a mistake, kindly report it to the author :)

Statistická termodynamika Statistical thermodynamicslynette.natur.cuni.cz/~kosovan/download/2016-17/L1.pdf · Statistická termodynamika Statistical thermodynamics Peter Košovan

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Statistická termodynamikaStatistical thermodynamics

Peter Koš[email protected]

Lecture 1, 4.10.2015, MC260P105, WS 2016/2017If you find a mistake, kindly report it to the author :­)

Course requirements

New in 2016/17: Exercises

* Within the scope of syllabus* Will be refined after first lecture* Focused on understanding, not on

memorizing formulae!* Statistical sampling of student's

knowledge by a few randomquestions and answers

* Inherently imperfect* See the web page for more

information

Oral exam

* Working out some derivations in detail* Numerical solution of relevant

problems* Practical preparation for the exam* Anything else we agree upon.

Macroscopic result

Final grade

Ultimate goal: to enjoy the subject and learn new things.

Course material

Secondary

* Wikipedia* Internet search* Beware of bugs and

incompetent content all over theinternet!

* Verify by yourself whatever youobtain from an unreliable source

* D. McQuarrie: Statistical Mechanics,University Science Books, 2000

* L. Reichl: A modern course in statisticalPhysics, Wiley, 1998

* T. Boublík: Statistická termodynamika,Academia, 1996

* Many other recognized textbooks* A nice book available online:

http://homepage.tudelft.nl/v9k6y/imsst/index.html

* Lecture notes for this course (pdf)* Lecture notes of other stat­mech courses,

e.g. P. Nachtigall (this course, previousyears), M. Tuckermanhttp://www.nyu.edu/classes/tuckerman/stat.mech/lectures.html

Ultimate goal: to enjoy the subject and learn new things.

Primary

Thermodynamics from first principles

ThermodynamicsMolecules

Macroscopic system

Sample in a containern=1mol, T=300K, V=1L

* Positions* Momenta* Mass* Charge, polarizability* Electronic states

* Kinetic energy* Potential energy* Interactions

* Pressure* Density* State (solid, liquid, gas)

* Entropy* Internal energy* Free energies

(Gibbs or Helmholtz)

Statistical

thermodynamics

(mechanics)

Predict thermodynamic properties

from molecular information

Thermodynamics from first principles

ThermodynamicsMolecules

Macroscopic system

Sample in a containern=1mol, T=300K, V=1L

* Positions* Momenta* Mass* Charge, polarizability* Electronic states

* Kinetic energy* Potential energy* Interactions

* Pressure* Density* State (solid, liquid, gas)

* Entropy* Internal energy* Free energies

(Gibbs or Helmholtz)

Statistical

thermodynamics

(mechanics)

Predict thermodynamic properties

from molecular information

Few parametersO(1)

ManyparametersO(1024)

Example application(current research)

Titration of a weak polyelectrolyte

0

0.2

0.4

0.6

0.8

1

-2 0 2 4

εr=801.4e-02 mol/l

Degre

e o

f io

niz

ation,

α

pH-pKA

N = 125

102050

100200

* Simple model* Molecular simulation* Dissociation equilibrium* Highly non­ideal

Prediction from theory or simulation is cheap

Reduces the amount of necessary experiments

0

0.2

0.4

0.6

0.8

1

-2 0 2 4

cpol [mol/l]

εr=80

Degre

e o

f io

niz

ation,

α

pH-pKA

1.4e-016.8e-022.7e-021.4e-02

Ideal

First principles

Quantum mechanics

HΨ = EΨ

1

Schrödinger: Particle in a box

* Discrete set of allowed states* Probabilistic character

* Numerically tractable for small systems(<103 electrons)

* For bigger systems too demanding

Classical mechanics

Newton:

* Full description in terms of p, q and time* Still too demanding for a macroscopic system

~ 1024 particles

Solution: statistics!

* Macroscopic system properties represent anaverage over individual molecules

? What is the probability distribution?

Enx,ny,nz =h2

8ma2(n2x + n2y + n2z

)

1

dpdt ≡ p = F

pi,q,i , t0

1

Postulates of statistical TD

1. Ensemble average corresponds to TD average

An average value of an arbitrary mechanical property of a macroscopic system is equalto the mean value for the statistical ensemble.

An isolated system with a constant number of particles, N, volume V and energy E isequally likely to be in any of its Ω(E) available quantum states.

2. Equal a priori probabilities

? What is an ensemble?N, V, E N, V, E N, V, E

N, V, EN, V, EN, V, E

N, V, E N, V, E N, V, E

* Mental construction of a set of systemscharacterized by the same TD parameters

* Isolated from the universe by impermeableadiabatic walls

* Many systems in different quantum stateswhich lead to the macroscopic parameters

* Equal a priori probabilities* Thermal equilibrium

* Mental construction of a set of systemscharacterized by the same TD parameters

* Isolated from the universe by impermeableadiabatic walls

* Many systems in different quantum stateswhich lead to the macroscopic parameters

* Equal a priori probabilities* Thermal equilibrium

Various ensembles

N, V, E N, V, E N, V, E

N, V, EN, V, EN, V, E

N, V, E N, V, E N, V, E

N, V, T N, V, T N, V, T

N, V, TN, V, TN, V, T

N, V, T N, V, T N, V, T

μ, V, T μ, V, T μ, V, T

μ, V, Tμ, V, Tμ, V, T

μ, V, T μ, V, T μ, V, T

Microcanonical Canonical Grandanonical

* Systems in differentquantum states

* Impractical: cannotmeasure total energy ofa system

* Systems can exchangeenergy

* Cannot exchangeparticles

* N, V, T all measurable

* Systems can exchangeparticles

* Convenient for smallsystems in equilibriumwith a reservoir

For N →∞ they become equivalent – matter of convenience

Fluctuations can be important at small N (computer simulations)

Excursion to statistics

Probability P: discrete and continuous random variable x:

1 =∑j

Pj, 1 =∫P(x)dx

Expectation value (mean):

E[x] = µ =∑j

xjPj, E[x] = µ =∫xP(x)dx

More general – n-th moment around c:

µn =∑j

(xj − c)nPj, µn =∫(x− c)nP(x)dx

Variance – 2nd central moment:

Var[x] = σ2 =∑j

(xj−µ)2Pj, Var[x] = σ2 =∫(x−µ)2P(x)dx

• Higher moments: n = 3 skewness, n = 4 kurtosis

More on statisticsDraw N samples of a random variable (continuous case analogous)Sample mean and mean (expected) value

x = 1N

N∑j=1

xj, limN→∞

x = E[x]

Sample variance and variance

s2x =1N

N∑j=1

(xj − x)2 = x2 − x2, limN→∞

s2x(x) = Var[x]

Variance of the mean

Var(x) = Var(1N

N∑j=1

xj)= s2x

N

•With increasing N variance of the mean vanishesNumber of ways to select k elements out of N:(

Nk

)= N!k!(N− k)!

Microcanonical ensemble

? What is the distribution of states?

•Many systems, no exchange of energy or particles

• Total number of particles (spins) N• Total energy E , volume V = Nv (values unimportant)

Ising model – Lattice with spins

• n spins up (s = +1), (N− n) spins down (s = −1)• Special case – No interaction between the spins

• All states have the same energy

•Magnetization: M =∑

i si = 2n−N

Possible ways to select n spins up – statistical weight:

W(n) =(Nn

)= N!n!(N− n)!

Probability of a state with given n – binomial distribution

P(n) =W(n) /( N∑n=0

W(n))=(Nn

)(12

)N

Image source: wikipedia.org

Canonical ensemble

N, V, T N, V, T N, V, T

N, V, TN, V, TN, V, T

N, V, T N, V, T N, V, T

? What is the distribution of states?

• Ensemble composed of A systems•Can exchange heat but not particles• Total volume V = AV• Total energy E (exact value unimportant)

Quantum states of the system :State number 1 2 3 . . . mEnergy E1 ≤ E2 ≤ E3 ≤ . . . EmSystems in state j a1 a2 a3 . . . am(occupation number)

Distribution of occupation numbers(m-dimensional vector) a = aj

Possible ways to select occupation numbers:

W(a) = A!a1!a2!a3! ...am!

= A!∏iak!

1

Constraints:∑j

aj = A

∑j

ajEj = E

1

Most probable distribution

Probability of a certain occupation number:

Pj =ajA= 1

A

∑a aj(a)W(a)∑

aW(a)where aj is the average over all possible distributions

1

Mean in termsof probability:M =

∑j

MjPj

1

Knowledge of Pj allows

us to calculate any

mechanical property M

(first postulate) .

Most probable distribution

Mean in termsof probability:M =

∑j

MjPj

1

Probability of a certain occupation number:

Pj =ajA= 1A

∑a aj(a)W(a)∑

aW(a)

The most probable distribution a∗:

Pj =1Aa∗jW(a∗)W(a∗) =

a∗jA

= ajA

1

Most probable distribution

Probability of a certain occupation number:

Pj =ajA= 1A

∑a aj(a)W(a)∑

aW(a)

The most probable distribution a∗:

Pj =1Aa∗jW(a∗)W(a∗) =

a∗jA

= ajA

Search for maximum of W(a) (equiv. lnW(a))∂

∂aj

lnW(a)− α

∑k

ak − β∑k

akEk= 0, j = 1,2, 3 ...

1

Constraints:∑j

aj = A

∑j

ajEj = AE

1

Mean in termsof probability:M =

∑j

MjPj

1

Lagrange multipliers

Most probable distribution

Probability of a certain occupation number:

Pj =ajA= 1A

∑a aj(a)W(a)∑

aW(a)

The most probable distribution a∗:

Pj =1Aa∗jW(a∗)W(a∗) =

a∗jA

= ajA

Search for maximum of W(a) (equiv. lnW(a))∂

∂aj

lnW(a)− α

∑k

ak − β∑k

akEk= 0, j = 1,2, 3 ...

Use Stirling’s approximation to lnW(a):− lna∗j − α− βEj = 0, j = 1,2, 3 ...

Finally:a∗j = e−αe−βEj, j = 1,2, 3 ...

1

Constraints:∑j

aj = A∑j

ajEj = AE

Stirling:limN→∞

lnN! = N lnN−N

1

Mean in termsof probability:M =

∑j

MjPj

1

Evaluation of α and β

Use

a∗j = e−αe−βEj and∑j

aj = A to obtain eα = 1A∑j

e−βEj,

From

Pj =a∗jA

we get Pj =e−βEj(N,V)∑je−βEj

and define Q =∑j

e−βEj(N,V)

then we can write E as

E(N,V, β) =∑j

EjPj =1Q∑j

Ej(N,V)e−βEj(N,V)

From mechanics:

dEj = −pjdV thus pj = −(∂Ej∂V

)N

hence

p =∑j

pjPj = −1Q∑j

(∂Ej∂V

)Ne−βEj(N,V)

1

Use

a∗j = e−αe−βEj and∑j

aj = A to obtain eα = 1A∑j

e−βEj,

From

Pj =a∗jA

we get Pj =e−βEj(N,V)∑je−βEj

and define Q =∑j

e−βEj(N,V)

then we can write E as

E(N,V, β) =∑j

EjPj =1Q∑j

Ej(N,V)e−βEj(N,V)

From mechanics:

dEj = −pjdV thus pj = −(∂Ej∂V

)N

hence

p =∑j

pjPj = −1Q∑j

(∂Ej∂V

)Ne−βEj(N,V)

1

Evaluation of α and β

Partition function

Partition function is the summit ofstatistical mechanics.Richard P. Feynmann: Introduction to Statistical

Mechanics, Westview Press, 1998

. . .Evaluation of α and β

Recall that

E = 1Q∑j

Ej(N,V)e−βEj(N,V) and p = −1Q∑j

(∂Ej∂V

)Ne−βEj(N,V) .

With some manipulations we can express(∂E∂V

)N,β

= −p+ βEp− βEp .

(∂p∂β

)N,V

= Ep− Ep .

Compare this with classical thermodynamics(∂E∂V

)N,T

− T(∂p∂T

)N,V

= −p =(∂E∂V

)N,T

+ 1T

(∂p∂1/T

)N,V

.

We conclude that

β = 1kT ,

where k is a constant (to be determined as Boltzmann constant kB).

1

Alternative evaluation of β

Recall that

E = 1Q

∑j

Ej(N,V)e−βEj(N,V)

Consider an ideal gas: kinetic energy per particle, ε = 32kBT:

ε =∑

j εjNj∑jNj

and εj =p2j,x2m +

p2j,y2m +

p2j,z2m

Combinig the above equations, we get (equivalently for x, y, z)

ε

3 =∑

jp2j,x2me−β

∑j p2j,x

2m∑j e−β

∑j p2j,x

2m

∞∫0p2xe−β

∑j p2j,x

2m dpx

2m∞∫0e−β

∑j p2j,x

2m dpxthen use

∞∫0x2e−ax2dx =

√π

16a3

∞∫0e−ax2dx =

√π4a

to obtain

ε = 32β and finally β = 1

kBT

1

Introducing entropy

Total derivative of E:

dE =∑j

EjdPj+∑j

PjdEj =∑j

EjdPj+∑j

Pj(∂Ej∂V

)NdV

1

Introducing entropy

Total derivative of E:

dE =∑j

EjdPj+∑j

PjdEj =∑j

EjdPj+∑j

Pj(∂Ej∂V

)NdV

1

Absorbed heat

Reversible work

Introducing entropy

Total derivative of E:

dE =∑j

EjdPj+∑j

PjdEj =∑j

EjdPj+∑j

Pj(∂Ej∂V

)NdV

Rearrange the equation for Pj:

Ej =1β

(− lnPj − lnQ

)⇒

∑j

EjdPj =1βd(−∑j

Pj lnPj)

because∑j

dPj = 0

From thermodynamics:

dqrev = TdS ⇒ dS = d(−kB

∑j

Pj lnPj)

Finally:

S = −kB∑j

Pj lnPj+ C with C = 0 from third law of TMD

1

Absorbed heat

Reversible work

Thermodynamic quantities from Q(N,V,T)

Internal energy

U = E = 1Q∑j

Eje−βEj(N,V) = kBT2(∂ lnQ∂T

)N,V

= kBT(∂ lnQ∂ lnT

)N,V

.

Pressure:

p = p = −1Q∑j

(∂Ej∂V

)Ne−βEj(N,V) = kBT

(∂ lnQ∂V

)N,T

.

Entropy:

S = −kB∑j

Pj lnPj = kB lnQ+ kBT(∂ lnQ∂T

)N,V

.

Enthalpy:

H = U+ pV = kBT[(

∂ lnQ∂ lnT

)N,V

+(∂ lnQ∂ lnV

)N,T

]

1

From previous slide:

U = kBT2(∂ lnQ∂T

)N,V

= kBT(∂ lnQ∂ lnT

)N,V

, p = kBT(∂ lnQ∂V

)N,T

,

S = kB lnQ+ kBT(∂ lnQ∂T

)N,V

, H = kBT[(

∂ lnQ∂ lnT

)N,V

+(∂ lnQ∂ lnV

)N,T

].

We continue with Gibbs free energy:

G = H− TS = −kBT[lnQ+

(∂ lnQ∂ lnV

)N,T

]and Helmholtz free energy:

A = E− TS = −kBT lnQ

1

More thermodynamic quantities . . .

From previous slide:

U = kBT2(∂ lnQ∂T

)N,V

= kBT(∂ lnQ∂ lnT

)N,V

, p = kBT(∂ lnQ∂V

)N,T

,

S = kB lnQ+ kBT(∂ lnQ∂T

)N,V

, H = kBT[(

∂ lnQ∂ lnT

)N,V

+(∂ lnQ∂ lnV

)N,T

].

We continue with Gibbs free energy:

G = H− TS = −kBT[lnQ+

(∂ lnQ∂ lnV

)N,T

]and Helmholtz free energy:

A = E− TS = −kBT lnQ

1

More thermodynamic quantities . . .

A is a characteristicfunction of canonical

ensemble.

More thermodynamic quantities . . .

From previous slide:

U = kBT2(∂ lnQ∂T

)N,V

= kBT(∂ lnQ∂ lnT

)N,V

, p = kBT(∂ lnQ∂V

)N,T

,

S = kB lnQ+ kBT(∂ lnQ∂T

)N,V

, H = kBT[(

∂ lnQ∂ lnT

)N,V

+(∂ lnQ∂ lnV

)N,T

].

We continue with Gibbs free energy:

G = H− TS = −kBT[lnQ+

(∂ lnQ∂ lnV

)N,T

]and Helmholtz free energy:

A = E− TS = −kBT lnQ

Replace sum over states j by sum overeigenvalues E with degeneracy Ω(N,V,E):

Q =∑j

e−βEj =∑E

Ω(N,V,E)e−βE

1

Rest of the course:how to obtain partition functionfrom microscopic parameters.

With Q, we have nowreached the summit.

Before that:partition functions of some othercommon ensembles.