Lecture 8 Classical vs. Statistical Thermodynamics

Ensemble averages meaning of β State functions as a function of Q meaning of γ Fluctuations

Ensemble averages

To connect with classical thermodynamics we need to know what are and

€

N = N =

NeγNe−βE i (N )

i,N

∑

Ξ=

∂ lnΞ

∂γ

€

P = P =

Pie−βE i

i

∑Q

=1

β

∂ lnQ

∂V

€

E = E =

E ie−βE i

i

∑Q

= −∂ lnQ

∂β

Number of particles

Pressure

Energy

Conservation of Energy

In canonical ensemble

Also energy of a given state is only function of N and V, Ei=Ei(N,V)

For constant N

€

E = piE i∑

€

dE = dpiE i∑ + pidE i∑

€

pi = e−βE i /Q

€

ln pi = −βE i − lnQ

€

E i = −1

β(ln pi + lnQ)

€

dE i =∂E i

∂V

⎛

⎝ ⎜

⎞

⎠ ⎟N

dV +∂E i

∂N

⎛

⎝ ⎜

⎞

⎠ ⎟V

dN

€

dE i =∂E i

∂V

⎛

⎝ ⎜

⎞

⎠ ⎟N

dV = −PidV

Conservation of Energy - 2

€

dE = dpiE i∑ + pidE i∑

€

E i = −1

β(ln pi + lnQ)

€

dE i =∂E i

∂V

⎛

⎝ ⎜

⎞

⎠ ⎟N

dV = −PidV

€

dE = −1

βdpi(ln pi + lnQ)∑ + pi

∂E i

∂V

⎛

⎝ ⎜

⎞

⎠ ⎟N

dV∑

€

dpi∑ = 0

€

d pi ln∑ pi

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟= ln pid∑ pi

€

dE = −1

βd pi ln pi∑

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟+ pi

∂E i

∂V

⎛

⎝ ⎜

⎞

⎠ ⎟N

dV∑

Heat Work

Conservation of Energy - 3

€

dE = −1

βd pi ln pi∑

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟− piPidV∑

€

dS' = −kd pi ln pi∑ ⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

€

dE =1

kβd S '

( ) − P dV

€

1

kβ= T

€

1

kT= β

Also where is the chemical potential

€

kT

= γ

Helmholz Free Energy

Statistical mechanics defined Helmholz free energy, entropy and energy

Thus

In canonical ensemble

thus

€

F = E −S

kβS = −k pi ln pi

i

∑ E = E i pi

i

∑

€

pi =e−βE i

Q→ ln pi = −βE i − lnQ

€

F =1

βpi(ln pi

i

∑ + βE i)

€

F =1

βpi(−lnQ − βE i

i

∑ + βE i) = −1

βlnQ pi

i

∑ = −1

βlnQ

€

F = −1

βlnQ = −kT lnQ

Other State Functions

We showed that

We also showed that

thus

Entropy

Pressure

€

F = −kT lnQ

€

E = −∂ lnQ

∂β

⎛

⎝ ⎜

⎞

⎠ ⎟N ,V

= −∂ lnQ

∂T

⎛

⎝ ⎜

⎞

⎠ ⎟N ,V

∂T

∂β

€

E = kT 2 ∂ lnQ

∂T

⎛

⎝ ⎜

⎞

⎠ ⎟N ,V

€

S = −F

T+

E

T= k lnQ + kT

∂ lnQ

∂T

⎛

⎝ ⎜

⎞

⎠ ⎟N ,V

€

P = −∂F

∂V

⎛

⎝ ⎜

⎞

⎠ ⎟N ,T

= kT∂ lnQ

∂V

⎛

⎝ ⎜

⎞

⎠ ⎟N ,T

Fluctuations

€

∂∂T

⎞

⎠ ⎟E Q =

∂

∂T

⎞

⎠ ⎟ E ie

−E i / kT

i

∑

€

∂E

∂T

⎛

⎝ ⎜

⎞

⎠ ⎟N ,V

Q + E ∂Q

∂T

⎛

⎝ ⎜

⎞

⎠ ⎟N ,V

= E i

∂e−E i / kT

∂T

⎛

⎝ ⎜

⎞

⎠ ⎟N ,Vi

∑

€

Q = e−E i / kT

i

∑

€

∂Q

∂T= e−E i / kT d

dTi

∑ −E i

kT

⎛

⎝ ⎜

⎞

⎠ ⎟=

1

kT 2E i

i

∑ e−E i / kT

€

E i

∂e−E i / kT

∂T

⎛

⎝ ⎜

⎞

⎠ ⎟N ,Vi

∑ =1

kT 2E i

2

i

∑ e−E i / kT

€

∂E

∂T

⎛

⎝ ⎜

⎞

⎠ ⎟N ,V

Q + E 1

kT 2E i

i

∑ e−E i / kT =1

kT 2E i

2

i

∑ e−E i / kT

€

∂E

∂T

⎛

⎝ ⎜

⎞

⎠ ⎟N ,V

+ E 1

kT 2

1

QE i

i

∑ e−E i / kT =1

kT 2

1

QE i

2

i

∑ e−E i / kT

Fluctuations -2

€

∂E

∂T

⎛

⎝ ⎜

⎞

⎠ ⎟N ,V

+ E 1

kT 2

1

QE i

i

∑ e−E i / kT =1

kT 2

1

QE i

2

i

∑ e−E i / kT

€

CV + E 1

kT 2E =

1

kT 2E 2

€

kT 2CV = E 2 − E 2 ~ N

€

E 2 − E 2

E 2~

1

N

In the thermodynamic limit, N, fluctuations in intensive thermodynamic properties are not measurable