lecture 11 statistical thermodynamics

Statistical Thermodynamics(Lecture 11)

1st semester, 2021Advanced Thermodynamics (M2794.007900)

Song, Han Ho

(*) Some materials in this lecture note are borrowed from the textbook of Ashley H. Carter.

2

è Establish the tie between macroscopic and molecular propertiesè Stochastic (probabilistic) approach à Focus on equilibrium statesè Fundamental questions

l What are the energy states available to the particles (atoms/molecules) comprising a macroscopic system?

l How are these particles distributed among these states at equilibrium to satisfy the macroscopic constraints?

IntroductionStatistical Thermodynamics

Quantum Mechanics(accessible energy states)

EN

NN

jjj

jj

=

=

å

åe

Statistical Mechanics(distribution among states)

Statistical Thermodynamics(microscopic)

Classical Thermodynamics(macroscopic)

3

è Let’s apply some elementary concepts of statistical thermodynamics to a coin-tossing experiment. (Assumption) Coins are distinguishable.

Coin-Tossing Experiment

N1: number of headsN2: number of tails (N2 = N – N1)

wk: thermodynamic probability=number of microstates in

macrostate k

smicrostate ofnumber total:or

wheremacro

1

W

=WW

= å=

N

kk

kk wwP

Statistical Thermodynamics

4

è Continue on. Macrostate (or configuration) is specified by the number of particles in each of the energy levels of the system. (= thermodynamic state in the classical theory)

Microstate is specified by the number of particles in each energy state. In general, there are more than one energy state (i.e. quantum state) for each energy level (or the energy level has degeneracy.)

First Postulate of Statistical Mechanics“In equilibrium, each microstate can occur with an equal probability.”

Coin-Tossing ExperimentStatistical Thermodynamics

5

è Continue on. The average occupation number is

Let’s generalize the coin-tossing experiment with a larger number of coins.How many ways are there to select from the N candidates N1 heads and N-N1 tails with distinguishable coins?


( ) ( ) ( ) ( ) ( )[ ]heads) ofnumber average (

21041624314161 .. 1

k

®

=´+´+´+´+´=

=W

== åå

åå

Nge

PNwN

w

wNN

kkjk

kkjk

k

kkjk

j

)!(!!

11 kkk NNN

Nw-

=


6

è Continue on. Calculate the maximum thermodynamic probability for different number of coins.(use Stirling’s formula, for large n)


6!2!2!4

4

max ==

=

w

N

70!4!4!8

8

max ==

=

w

N

300max 10

!500!500!1000

1000

»=

=

w

N

nnnn -» ln!ln

The peak always occurs at N/2, but grow rapidly and becomes much sharper as N increases. In other words, the most probable configuration is that of total randomness.


7

è Continue on. The “ordered regions” almost never occur. (w is extremely small compared to wmax)This leads to the very important conclusion!

i.e. The total number of microstates is very nearly equal to the maximum number.

For thermodynamic problem, the “outcomes” are the occupation numbers of each of n energy levels for the most probable macrostate, i.e. wmax(N1, N2, … , Nn). This most probable macrostate is the equilibrium state of the thermodynamic system.

In extending our formula of tossing coins from two levels to n levels,


maxwwk »=W å

Õ=

== n

jj

n N

NNNN

Nw

1

21 !

!!!...!

!


8

è The fundamental problem of statistical thermodynamics is to determined the most probable state or the equilibrium state, in given constraints.

à Method of Langrange multiplier

As a system proceeds toward a state of equilibrium, the entropy increases, and at equilibrium attains its maximum value. (Classical point of view)The system tends to change spontaneously from states with low thermodynamic probability to states with high probability. (Statistical point of view)

Assembly of Distinguishable Particles

energy) ofion (conservat

particles) ofion (conservat

:sconstraint under the ),...,,( Find

1

1

21max

UN

NN

NNNw

n

jjj

n

jj

n

=

=

å

å

=

=

e


9

è Boltzmann made the connection between the classical concept of entropy and the thermodynamic probability.

The entropy is an extensive property. Therefore, the combined entropy of the two subsystems is simply the sum of the entropies of each subsystem.

From the characteristic of the thermodynamic probability,

Then,

Finally,

Thermodynamic Probability and Entropy

)(wfS =

)()()(or BAtotalBAtotal wfwfwfSSS +=+=

BAtotal www =

)()( BAtotal wwfwf =

)()()( BABA wwfwfwf =+


10

è Continue on.

The only function to satisfy the above condition is the logarithm.

Thermodynamic Probability and Entropy

wkS ln=

)()()( BABA wwfwfwf =+

)10(1.38constant Boltzmann : where 1-23 -´ JKk


11

è Schrӧdinger’s equation: the governing equation for the dynamical behavior of matter on the atomic scale (just as Newton’s law for macroscopic particles)

è Time independent Schrӧdinger’s equation

è The equation has solutions only for specific values of , termed the eigenvalues of the energy.

Quantum States and Energy Levels

( ) 082

22 =-÷÷

ø

öççè

æ+Ñ yfepy

hm

Time independent Schrӧdinger’s Equation

( ) fe +=+= mp /2/1energy potentialenergy kinetic 2

particle) a finding ofdensity y probabilit :|| (function wave 2yy =zyx yyyy =

zyx eeee ++=

zyx ffff ++=

)( fe -


12

è Apply Schrӧdinger’s equation to the translational kinetic energy states of a dilute monatomic ( ) particle in the absence of external force fields (“particle in a box” à on the boundary)

è Substituting into Schrӧdinger’s equation and separating variables,

with the boundary conditions:

Quantum States and Energy LevelsStatistical Thermodynamics

only ,0 treef ==0=y

x

y

z ab c

)()()( zyx trtrtr yyyy =

)()()( zyx trtrtrtr eeeee ++==

)(8)()(

12

2

2

2

xhm

dxxd

x trtr

tr

epyy ÷÷

ø

öççè

æ-=

0)()0( == atrtr yy

13

è The solution to the equation,

è Applying B.C.

è Similarly for y and z:

è Total translational energy:


)(8)()(

12

2

2

2

xhm

dxxd

x trtr

tr

epyy ÷÷

ø

öççè

æ-=

0, 0)0( =Þ= Btry

÷÷ø

öççè

æ+= B

hxmxAx tr

tr 2

2 )(8sin)( epy

0)(8sin 0)( 2

2

=÷÷ø

öççè

æÞ=

hxmaAa tr

trepy

3,...) 2, ,1( 8

)( )(82

22

2

2

=÷÷ø

öççè

æ÷÷ø

öççè

æ=Þ= x

xtrx

tr nan

mhxn

hxma epep

÷÷ø

öççè

æ÷÷ø

öççè

æ= 2

22

8)(

bn

mhy y

tre ÷÷ø

öççè

æ÷÷ø

öççè

æ= 2

22

8)(

cn

mhz z

tre

úúû

ù

êêë

é÷÷ø

öççè

æ+÷÷ø

öççè

æ+÷÷ø

öççè

æ÷÷ø

öççè

æ=++= 2

2

2

2

2

22

8)()()(

cn

bn

an

mhzyx zyx

trtrtr eeee

14

è The total solution:

è Let

è For , there are numerous combinations ofà numerous quantum states with the same energyà the energy levels are degenerateà degeneracy = number of quantum states with the same energy =


úû

ùêë

é÷øö

çèæ

úû

ùêë

é÷÷ø

öççè

æúû

ùêë

é÷øö

çèæ==

cznA

byn

AaxnAzyx z

zy

yx

xtrtrtrpppyyyy sinsinsin)()()(

1,,n energy whe slationalpoint tran-zero finite a is therebox) ain particle (no ,, allfor 0 0,, ofany If

=Þ

=Þ=

zyx

zyx

nnnzyxnnn y

3/1Vcba ===

( )2223/2

2

8 zyxtr nnnmVh

++÷÷ø

öççè

æ=e

consttr =e zyx nnn ,,

trg

15


6 =Þ trg

3or 2 ,1,, =zyx nnnnx ny nz

1 2 31 3 22 1 32 3 13 2 13 1 2

)(or 14 tr222 fixednnn zyx ==++ e

3/2

2222

3/2

2

tr 814)(

8

mVhnnn

mVh

zyx =++=e

1,, =zyx nnn 1 8

3)(8

3/2

2222

3/2

2

tr =Þ=++=Þ trzyx gmVhnnn

mVhe

For zero-point energy,

16

è In quantum theory, the energy levels are discrete, but in classical physics, the energy levels are continuous. Why is the discrepancy?

Let’s consider translational energy mode.At room temperature, the mean kinetic energy of a helium gas atom is,

Then, the nominal quantum number nj for one-liter volume of helium gas is,

This implies that there are so many energy levels below this small energy value (6.2x10-21J), and thus the energy levels are very closely spaced and may be treated as an energy continuum.

Density of Quantum StatesStatistical Thermodynamics

( ) 92/1

2

3/22

3/2

2

107.28 8

´»÷÷ø

öççè

æ ´=®÷÷

ø

öççè

æ=

hmVnn

mVh tr

jjtree

JkTtr2123 102.6293)1038.1(

23

23 -- ´=´´´==e

17

è Under the conditions that the quantum numbers are large and the energy levels are very close together, we can regard the n’s and the ɛ’s as continuous functions.Let’s evaluate the density of states g(ɛ), or degeneracy.

The possible combination (nx, ny, nz) in a given ɛtr,correspondsto points on a sphere of radius R in (nx, ny, nz) space.Thus,


2tr2

3/22222 8 R

hmVnnnn zyx º=++= e

eee

eeeee

eeee

dd

)(dN

)(N.T.O.Hdd

)(dN)(N

)(N)d(Nd)ε( g

»

-úûù

êëé ++=

-+=

18

è Continue on.Define N(ɛ), as the number of states contained within the octant of the sphere of radius R.

Then,

To complete our discussion, we need to consider additional energy states regarding spin!


23tr

23

23 8

634

81 /

/

hmVR)(N eppe ÷øö

çèæ=×=

eepeeeee dm

hVd

d)(dNd)(g // 21

tr23

324 ==

fermions half-onespin for 2bosons zerospin for 1

24)( 2/1tr

2/33

==

=

s

s

s dmhVdg

gg

eepgee

lecture 11 statistical thermodynamics

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