5.60 Spring 2007 Revised Lecture #24 page

1

STATISTICAL MECHANICS

Statistical

Mechanics

Macroscopic

Thermodynamic Properties

0 0

ln p

G H T SRT K

Δ = Δ − Δ= −

Properties of Individual

Atoms/Molecules

0

H Eψ ψ= Goal of Statistical Mechanics: to describe macroscopic, thermodynamic properties in terms of microscopic atomic & molecular properties Properties of a system can be described at two levels:

1) Macroscopic thermodynamic description e.g. p, V, n, CV, H, A, G,…..

2) Microscopic description Specify the state of each molecule! Use classical or quantum mechanics More than 1023 variables! And need to update them every 10-15 s or so!

Either classical or quantum description is impractical. Statistical mechanics describes macroscopic mechanics in statistical terms, i.e. in terms of “average” or “most probable” results. Probability of system in a state with given energy What is functional form? For independent energies εi and εj (e.g. energies of two different molecules or x and y components of kinetic energy in a single molecule, etc.) the joint probability should be the product:

Pij(εi + εj) = Pi(εi)Pj(εj)

5.60 Spring 2007 Revised Lecture #24 page

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Suggests exponential form ( )i j ji

C CCe eε ε e εε+=

We expect high-energy states to be less probable than low-energy states, and that they become more probable at high T, i.e. ratio of εi to T is what matters. ⇒ ( ) ~ εε −

ii

iC TP e

Or more conventionally

( ) εε −∝ii

ikTP e where k = R/NA = 1.38 x 10-16 erg/K is the Boltzmann constant

For two states i and j with energies εi and εj, the relative probability is

ε

ε

−

−=i

j

kTi

kTj

P eP e

To get absolute probabilities (not just relative), write

( ) ε εε − −∝ =ii i

ikT kTaP e e

Sum of probabilities for all the states

1 ε−= =∑ ∑ i kTi

i iaP e ⇒ 1

ε−=∑ i kT

i

ae

ε

ε

−

−=∑

i

i

kT

i kT

i

ePe

⇒ probability of being in state i

For a whole system or assembly of molecules, in a particular system state i (specified by indicating the state of each and every molecule) with energy Ei:

−

−=∑

i

i

E kT

i E kT

i

ePe

5.60 Spring 2007 Revised Lecture #24 page

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Partition functions The sums

ε− ≡∑ i kT

iqe Molecular partition function

and − ≡∑ iE kT

iQe Canonical partition function

measure how probabilities are partitioned among different available states. They are unitless numbers. Example: perfect atomic crystal lattice at T ≈ 0 K Set ground state energy E0 = 0 All other state energies >> kT ⇒ Q ≈ 1

( )0 0

00 10 1

− −

−− −≈ =

+=

+

E kT E kT

E kTE kT E kTe eP

ee e.

Example: mole of atoms in the gas phase at room T Could be treated quantum mechanically (particle in a box states) or classically (continuum of states of different kinetic energy). Or use “lattice” model: divide available volume into atomic size volume elements ~ 1 Å3 = 10-30 m3

If total volume ~ 1 m3, then each atom has 1030 possible locations Molecular “translational” partition function is

30, 10ε− = ≈∑ transi trans kT

iqe

For a system of N = 1024 atoms, how many microscopic states? How many ways to assign atoms to selected locations:

( )( )( ) ( ) ( ) ( )24 2410 30 1030 30 30 30 3010 10 10 10 10 10 x Ntransq= = =

Huge number! Number of distinguishable states is less if the particles are indistinguishable: Have to divide by N! = 1024!

5.60 Spring 2007 Revised Lecture #24 page

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distinguishable particles

! indistinguishable particles

=

=

Ntrans trans

Ntrans trans

Q q

Q q N

Stirling’s approximation: lnN! ≈ NlnN – N or N! ≈ e-NNN

So

( )

( )( ) ( ) ( ) ( )

24

2424 24 2424

24 24

10301010 10 106 10 6 0.4 5.6

1024 10

1010 10 10 10

! 10

N Ntrans trans

trans N Nq qQ eN N e e

−−

= = = = = =

Still huge! So probability for any one system state is incredibly small. The probability is partitioned among a huge number of states. Example: polymer configurations including protein folding. e.g. just 4 polymer subunits with some favorable interaction energy -εint (e.g. due to H bonding) if non-covalently bonded subunits are in neighboring “lattice” sites:

Microstate i:

Energy εi: -εint 0 0 0 Degeneracy g: 1 3

In this simple example, the “configurational” molecular partition function is , int int0 0 0 03ε ε ε−

= = =+ + + +∑confi conf kT kT kTkT kT kT kT

microstatesi

q e e e e e e e

The last expression suggests writing the partition function as a sum over energy levels εi instead of individual states, if we account for their degeneracies gi:

int 03εε

ε

−= = +∑i

conf ii kTkT kT

energiesq g e e e

5.60 Spring 2007 Revised Lecture #24 page

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This can be done for the canonical partition function too, if we account for the degeneracies Ωi of system energies Ei:

− −= = Ω∑ ∑i

ii iE kT E kT

states energiesi E

Q e e

Thermodynamic properties from partition functions All the macroscopic thermodynamic properties can be derived from the microscopically based partition functions! Start with energy U = <E>, the average system energy. Use β = 1/kT.

1i

i i ii i

EU E p E E eQ

β−= = =∑ ∑

Use the following result:

, ,

i i

ii iV N V N

E EQe Eβ β

β β

− −∂ ∂= = −

∂ ∂

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∑ ∑ e

Then

( ),, ,

1 1 ln lni

E

ii V NV N V N V N

Q Q QE E e

Q Q Tβ

β β

− ∂ ∂ ∂ ∂= = − = − = −

∂ ∂ ∂ ∂

⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

∑,

T

β

( ) ( ) 2

1 1

T T kT kT

β∂ ∂= = −

∂ ∂

( )2

,

ln

V N

QU kT

T

∂=

∂

Generally U is a function of (N,V,T). This leads naturally to the Helmholtz free energy A(N,T,V). Derive A based on analogy with the Gibbs-Helmholtz relation.

A = U – TS A/T = U/T – S ( )( ) ,1

V N

ATUT

⎛ ⎞∂= ⎜ ⎟⎜ ⎟∂⎝ ⎠

5.60 Spring 2007 Revised Lecture #24 page

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( ), ,

ln ln1V N V N

QU kTβ

⎛ ⎞⎛ ⎞∂ ∂= − = − ⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

Q⇒ A = -kTlnQ

(constant of integration can be shown to be zero)

,

lnlnV N

A U QS k Q kTT T T

∂⎛ ⎞= − + = + ⎜ ⎟∂⎝ ⎠

, ,

lnT N T N

A Qp kTV V∂ ∂⎛ ⎞⎛ ⎞= − =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

, ,

lnT V T V

A QkTN N

μ ∂ ∂⎛ ⎞⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

Entropy in terms of probabilities and degeneracies

1 lniE kT

ii

E eS U A Qk kT kT Q

−

−= = +

∑

But ln iE kTiE kT e −= − so ( )ln ln

ii

E kTE kT

i

S e e Qk Q

−−= − +∑

Since 1iE kT

ii

i

eP

Q

−

=∑

∑ = , we can multiply lnQ by 1 and combine terms to get

( )ln ln lni i i

iE kT E kT E kT E kT

E kT

i i i

S e e e ee Qk Q Q Q Q

− − −− ⎛ ⎞

= − + = − ⎜ ⎟⎝ ⎠

∑ ∑ ∑i−

This is just Gibbs eq for entropy in terms of state probabilities lni i

iS k p p= − ∑

If the system is isolated, then all states have the same energy and the same probability p = 1/Ω where Ω is the number of degenerate states. Then Boltzmann expression for entropy (on his tombstone!) lnS k= Ω Can understand entropy in terms of number of different available states. This microscopic picture of entropy is at the heart of statistical mechanics.