Statistical Thermodynamics CHEN 689 Fall 2015 Perla B. Balbuena 240 JEB [email protected] 5-3375 1

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Statistical ThermodynamicsCHEN 689Fall 2015

Perla B. Balbuena240 JEB

[email protected]

mailto:[email protected]

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Goals of Statistical Thermodynamics

• Based on a microscopic description, predict macroscopic thermodynamic properties

• Example: what is pressure?– Molecular simulations– Probabilistic description

• Objective: Determine probability distributions and

average values of properties considering all possible states of molecules consistent with a set of constraints (an ensemble of states)

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Connection between microscopic state and macroscopic state

• Example: NVE microcanonical ensemble – how to describe an “ideal gas” of

identical indistinguishable particles; – And a real condensable gas??

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Quantum mechanical description of microstates

2223/2

2

8 zyxlll lllmV

hzyx

Cubic box side L = V1/3

h = Planck’s constantm = mass of the particlelx, ly, lz: quantum numbers (0,1, 2,…)

Consider a macroscopic system

• N particles, volume V, and given certain forces among the particles

• Schrödinger equation:

• For an ideal gas:

• Monoatomic gas:

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jjN jEH ˆ J = 1, 2, 3,…

N

iij VNE

1

),(

Ignore electronic states and focus on translational energies

Cubic container of side L1/3

2223/2

2

8 zyxlll lllmV

hzyx

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energy states and energy levelsEnergy state lx ly lz

A 2 1 1

B 1 2 1

C 1 1 2

2223/2

2

8 zyxlll lllmV

hzyx

compute :e

each energy state gives the same energy level;this jth energy level has a degeneracy (wj) of 3

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distribution of non-interacting identical

molecules (indistinguishable) in energy states

we define a set of occupation numbers: n( n1, n2, n3…) each number is associated with the ith molecular state

jj

iji nE

energy of ith microstate

(all molecular energy states) W degeneracy is the number

of microstates that havethis energy level

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Statistical Mechanics Postulates

• All microstates with the same E, V, N are equally probable

• The long time average of any mechanical property in a real macroscopic system is equal to the average value of that property over all the microscopic states of the system (ergodic hypothesis)– each state weighted with its probability of

occurrence, provided that all the microstates reproduce the thermodynamic state and environment of the actual system

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Boltzmann energy distribution

• how to assign probabilities to states of different energies?

• Assume a system at N, V, in contact with a very large heat bath at constant T

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pA(En)

pB(Em)

since A and B are totally independent from each other,

pA(En) pB(Em) is the probability of finding the complete system with Aand B in the specified energy sub-states

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how the energy of the composite system would change if we change En without changing Em?

as a first approximation we assume that the energy levels areclosely spaced ( continuous function)

The Boltzmann distribution of energyConsider two macroscopic subsystems (no need to be identical) inside an infinite thermostatic bath (i.e., their temperatures are perfectly controlled and fixed). If the energy in A fluctuates, this has no effect on B, and vice-versa

The Boltzmann distribution of energy

Probability that subsystem A is in a microstate sA with energy En

A np E

This is NOT the probability of finding subsystem A with energy En

Several microstates of subsystem A may have the same energy

These that have the same energy are equally probable (postulate 1)


Probability that subsystem A is in an energy level of energy En

A n A nE p E

is the degeneracy of energy level En in subsystem A

A nE


Probability that subsystem B is in a microstate sB with energy Em

B mp E

This is NOT the probability of finding subsystem B with energy Em.

Several microstates of subsystem B have the same energy.

These that have the same energy are equally probable (postulate 1)


Probability that subsystem B is in an energy level of energy Em

B m B mE p E

is the degeneracy of energy level Em in subsystem B

B mE


These probabilities in subsystems A and B are independent of each other. Remember our initial assumption: if the energy in A fluctuates, this has no effect on B, and vice-versa


John enters a draw with 8% chance of winning.

Mary enters another, independent draw with 5% chance of winning.

What is the probability that John AND Mary win?


John enters a draw with 8% chance of winning.

Mary enters another, independent draw with 5% chance of winning.

What is the probability that John AND Mary win?

& 0.08 0.05 0.004 0.4%p J M


The probability of finding the composite system in a particular microstate sAB is only a function of the total energy of the composite system, EAB (because of postulate 1).

Suppose:

AB n mE E E

AB n m A n B mp E E p E p E


Other states with same energy of the composite system, EAB, will have the same probability (again because of postulate 1)

For example:

AB n m A n B m

A n B m

p E E p E p E

p E p E


Let us now examine another issue: what is the effect of changing the value of En without changing Em?

m m

AB n m AB n m n m

n n m nE E

AB n m

n m

p E E dp E E E E

E d E E E

dp E E

d E E


Using:

m

m

AB n m A nB E

n nE

p E E dp Ep

E dE

AB n m A n B mp E E p E p E

We also have that:


Combining the results of the two previous slides:

m

AB n m A nB E

n m n

dp E E dp Ep

d E E dE

An analogous development for the effect of Em gives:

n

AB n m B mA E

n m m

dp E E dp Ep

d E E dE


Then:

m n

A n B mB E A E

n m

dp E dp Ep p

dE dE

1 1

n m

A n B m

n mA E B E

dp E dp E

p dE p dE

ln lnA n B m

n m

d p E d p E

dE dE


This equation deserves careful examination:

ln lnA n B m

n m

d p E d p E

dE dE

Its left hand side is independent of subsystem BIts right hand side is independent of subsystem A

Then, each side of the equation should be independent of both subsystems and depend on a characteristic that they share


The two subsystems share their contact with the thermal reservoir that keeps their temperature constant and equal

ln lnA n B m

n m

d p E d p E

dE dE

: a function of temperature

The minus sign is introduced for convenience


Integrating these two differential equations:

nEA n Ap E C e

, : integration constantsA BC C

mEB m Bp E C e


But the result of adding the probabilities of each subsystem should be equal to 1

states n of states n of states n ofsystem A system A system A

1n nE EA n A Ap E C e C e

states n ofsystem A

1

n

AE

Ce

Canonical partition function


But the result of adding the probabilities of each subsystem should be equal to 1

states m of states m of states m ofsystem B system B system B

1m mE EB m B Bp E C e C e

states m ofsystem B

1

m

BE

Ce

Canonical partition function


The canonical partition function of a system is:

,

states i

, , iE V NQ V N e The probability of a particular microstate i with energy Ea is:

, ,

,

states i

, ,i

E V N E V N

iE V N

e ep E

Q V Ne

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Statistical Thermodynamics CHEN 689 Fall 2015 Perla B. Balbuena 240 JEB [email protected] 5-3375 1