Part II: Statistical Physicsodessa.phy.sdsmt.edu/~bai/teaching_2014Fall_done/PHYS-343/Ther… · X...

Outline Partition Function and Free Energy Partition Function for Composite Systems Applications

Part II: Statistical PhysicsChapter 6: Boltzmann Statistics

X Bai

SDSMT, Physics

Fall, 2014

X Bai Part II: Statistical Physics


1 Partition Function and Free Energy

2 Partition Function for Composite SystemsPartition Function for system of distinguishable particlesPartition Function for system of indistinguishable particles

3 ApplicationsThe Maxwell Speed Distribution and Boltzmann FactorsMore about Ideal Gas: from Statistics Point of View



Partition Function and Free Energy

We have talked about two different systems: isolated system and a system thatis in thermal equilibrium with a heat reservoir.

isolated system with fixed energy U: The multiplicity Ω(U) (the numberof microstates corresponding to a certain macrostate) and entropyS = klnΩ play very important role.

a system that is in thermal equilibrium with a heat reservoir oftemperature T : We have learned the partition function Z(T ).

We know Z is defined as Z =∑

s e−E(s)/kT . This means, Z counts how many

microstates are accessible, weighting each one in proportion to itsprobability e−E(s)/kT (P(s) = e−E(s)/kT/Z , Z is a constant). Z(T ) is verymuch like Ω(U). Is there a quantity for a system that is in thermal equilibriumwith a heat reservoir as the entropy is for an isolated system?We have learned the Helmholtz free energy, which is defined as

F ≡ U − TS (1)(∂F

∂T

)V ,N

= −S (2)



Partition Function and Free Energy - cnt.

In Chapter 5, we discussed S ,F ,G and how they govern the process towardequilibrium.

At constant energy and volume, S tends to increase. (The Second Lawof Thermodynamics)

At constant temperature and volume, F = U − TS tends to decrease.

At constant temperature and pressure, G = U + PV − TS tends todecrease.

For a stem at a fixed temperature, when the system loses energy (in format ofheat TS) and the environment gains energy, which may gives net increase inentropy of the combined system.Let’s make a comparison:




⇒: Ω ∼ Z⇒: For an isolated system (left), S tends to increase. For a system at constanttemperature (right), F tends to decrease.⇒: Like S, F can be written as the logarithm of a statistical quantity, in thiscase Z.For a system in an environment with fixed T0: FS = US − T0SS decreases ⇔Lose some US to let SS increase.




Let’s try to relate F with Z : From Eq. (1) and use Eq. (2), we haveU = F + TSU = F − T

(∂F∂T

)V ,N

U = T 2(

FT 2 − 1

T∂F∂T

)V ,N

U = T 2 ∂∂T

(−FT

)V ,N

U = kT 2 ∂∂T

(−FkT

)V ,N

Comparing this to the mean energy we have obtained: 〈E〉 = +kT 2 ∂∂T

lnZ , wewill have lnZ = −F

kT. This gives the correlation between F and Z :

Z = exp

(−FkT

)= e−βF (6.56) in the textbook. (3)

F = −kTlnZ (6.63) in the textbook. (4)




We know, in some cases, how to calculate Z , So we can calculate F from thisequation. From F , we can calculate other quantities by using what we havelearned in Chapter 5:

Identities:F = U − TSdF = dU − SdT − TdSdU = TdS − PdV + µdNdF = −SdT − PdV + µdN

S = −(∂F

∂T

)V ,N

P = −(∂F

∂V

)T ,N

µ =

(∂F

∂N

)T ,V

(5)




Now, lets compare the Microcanonical ensemble and the canonical ensemble

microcanonical ensemble canonical ensemble

( ) Ω= ln,, BkNVUS ( ) ZTkNVTF B ln,, −=

Ω=1

nP TkE

nB

n

eZ

P−

=1

- the probability of finding a system in one of the accessible states

- the probability of finding a system in one of these states

- in equilibrium, S reaches a maximum - in equilibrium, F reaches a minimum

VUNUNV NS

TVS

TP

US

T ,,,

1⎟⎟⎠

⎞⎜⎜⎝

⎛

∂

∂−=⎟⎟

⎠

⎞⎜⎜⎝

⎛

∂

∂=⎟⎟

⎠

⎞⎜⎜⎝

⎛

∂

∂=

µ

VTNTNV NF

VFP

TFS

,,,⎟⎟⎠

⎞⎜⎜⎝

⎛

∂

∂−=⎟⎟

⎠

⎞⎜⎜⎝

⎛

∂

∂−=⎟⎟

⎠

⎞⎜⎜⎝

⎛

∂

∂−= µ



In-class exercise: Another definition of entropy

Exercise 06-05The entropy can also be defined as S = −k

∑s P(s)lnP(s), where s

represents accessible microstate in the system. P(s) is the probability ofthe system being in microstate s.

(1) For an isolated system, P(s) = 1Ω

for all microstates s. Prove in thiscase this entropy formula reduces to S = klnΩ.

(2) For a system in thermal equilibrium with a reservoir at temperature T ,P(s) = 1

Zexp(−E(s)/kT ). Show in this case the preceding formula agrees

with what we know about entropy.



Distinguishable particles

Ztotal =∑s

exp[β(E1(s) + E2(s))] =∑s

exp[β(E1(s)]exp[β(E2(s)] (6)

If there is NO interaction between two particles,

the system state s = 1s1 , 2s2, 1s2 , 2s1E1(s) + E2(s) = E1(s1) + E2(s2),E1(s) + E2(s) = E1(s2) + E2(s1) (7)

The number of ”2-particle system” states for distinguishable particles:

Ns = Ns1 × Ns2 (8)

Ztotal =∑s1

∑s2

exp[β(E1(s1)]exp[β(E2(s2)] (9)

Ztotal = Z1Z2 (10)

which can be generalized for N particles,

Ztotal = Z1Z2 · · ·ZN (11)



Indistinguishable particles

Ψ(1,2) = Ψ(2,1): Interchanging the states of two identical particles does NOT change the state of the system

= 1

2

2

1

Ztotal =1

2Z1Z2 (12)

For system consisting of N particles, we have

Ztotal =1

N!ZN

1 (13)

where Z1 is the partition function for any one of the particles individually.But this is NOT always correct! It is only good in the case when all particlesare in different states, the low density situation. Let’s look at one simpleexample ...



The difference at the microscopic level

In dense systems, two or more identical particles may have significantprobability to occupy the same single-particle state.

S1, E(s1) S2, E(s2) S3, E(s3) S4, E(s4) S5, E(s5)

11000 01010 20000

10100 01001 02000

10010 00110 00200

10001 00101 00020

01100 00011 00002

All allowed system states

Two par3cles into 5 single par3cle states

For dis3nguishable par3cles: 2*(5+5) + 5 = 25 “system states” For iden3cal par3cles: (5+5) + 5 = 15 “system states” ≠25/2!

5 single par6cle states

2 par6cles

Each digit for each energy level/single par6cle state: 1: occupised; 0: empty

Dense systems: Quantum effects, Fermions, Bosons, and their distribution,which we will learn in the next Chapter.



Applications

We have learned several important concepts, obtained some useful tools, anddid some exercises for better understanding. Now, let’s try to apply what wehave learned ... next time.



The Maxwell Speed Distribution

(1) What is a Speed Distribution D(v)? Let’s look at a plot

Probability (v,v+dv) = (Shaded Area)/(Total Area) = Nv/N

vx

vy

vz

( )dvv

dvvv24

volume""π=

+↔

v



The Maxwell Speed Distribution- cnt.

P(v , v + dv) = D(v)dv

D(v) ∝(Prob. of a molecule

with velocity ~v

)×(

N of vectors ~vcorresponding to speed v

)(Prob. of a molecule

with velocity ~v

)∝ exp(−mv 2/2kT )(

N of vectors ~vcorresponding to speed v

)∝ 4πv 2

Putting all together, we have

D(v) = C · 4πv 2e−mv2/2kT (14)

What is the constant C - a normalization factor that makes

1 =

∫D(v)dv = 4πC

∫v 2e−mv2/2kTdv (15)

C =( m

2πkT

)3/2

(16)

D(v) =( m

2πkT

)3/2

4πv 2e−mv2/2kT (17)




D(v) =( m

2πkT

)3/2

4πv 2e−mv2/2kT

This is the Maxwell speed distribution.

The avarge speed:

v =

∫vD(v)dv (18)

v =

∫ ( m

2πkT

)3/2

4πv 3e−mv2/2kTdv (19)

v = 4π( m

2πkT

)3/2∫

v 3e−mv2/2kTdv (20)

v = 4π( m

2πkT

)3/2 1

2

∫v 2e−mv2/2kTdv 2 (21)

v = 4π( m

2πkT

)3/2 1

2(√

m/2kT )4

∫(√

m/2kTv)2e−(√

m/2kTv)2

d(√

m/2kTv)2

(22)




v = 4π( m

2πkT

)3/2 1

2(√

m/2kT )4

∫ue−udu (23)

Using formula:

∫ ∞0

xne−axdx = n!× a−(n+1)

v = 4π( m

2πkT

)3/2 1

2(√

m/2kT )4(24)

v = 2π( m

2πkT

)3/2(

2kT

m

)4/2

(25)

v = 2π

π2

( m

2πkT

)3/2(

2πkT

m

)4/2

(26)

v = 21

π

√2πkT

m(27)

v =

√8kT

πm(28)



Remarks and In-class exercise

1 mean: v =∫∞

0vD(v)dv

2 P(v > v0) =∫∞v0D(v)dv

3 P(v < v0) =∫ v0

0D(v)dv

4 P(v1 < v < v2) =∫ v2

v1D(v)dv

Exercise 06-06:Consider a mixture of Hydrogen and Helium at T=300 K. Find the speedat which the Maxwell distributions for these gases have the same value.



The Partition Function for Ideal Gas

For a gas that has N identical molecules, the partition function is

Z =1

N!ZN

1 (29)

Z depends on the Boltzmann factors for all possible microspates:

Z =∑s

e−E(s)/kT (30)

E(s) is the total energy of a particle at state s: E(s) = Ek + Eint : transitionalenergy and internal energy (such as rotational, excitations, ...).

Z =∑s

e−E(s)/kT =∑s

e−Ek/kT e−Eint/kT (31)

Assume transitional and internal states are independent from each other,

Z =∑s

e−E(s)/kT =∑T .S.

e−Ek/kT∑I .S.

e−Eint/kT (32)

Z = ZtrZint (33)



The Partition Function for Ideal Gas - cnt.

Let’s remind what microstates the particles in a box can have.

Particle in 1-D potential well - Wavelength: λ=2L/n, n=1,2,3,… - Momentum: pn = nh/(2λ) - Energy: En=pn

2/(2m)=h2n2/8mL2

For 3-D potential well: - Energy: Enx,ny,nz= [px

2+py2+pz

2]/(2m)=h2nx2/8mLx

2 + ny2/8mLy

2 + nz

2/8mLz2

Let’s look at 1-D situation:

Z1d =∑n

e−En/kT =∑n

e−h2n2/(8mL2kT ) (34)

Since the energy levels are very close to each other, we can use integration toreplace the sum:




Z1d =

∫ ∞0

e−h2n2/8mL2kTdn =

√π

2

√8mL2kT

h2=

L

lQ(35)

where lQ is the quantum length:

lQ =h√

2πmkT(36)

The result can be extended to 3-D situation:

Etr =p2x

2m+

p2y

2m+

p2z

2m(37)

Ztr =

∫ ∞0

∫ ∞0

∫ ∞0

e−h2n2x/8mL2kT e−h2n2

y/8mL2kT e−h2n2z/8mL2kTdnxdnydnz (38)

Ztr =Lx

lQ

Ly

lQ

Lz

lQ=

V

vQ(39)

vQ =

(h√

2πmkT

)3

(40)

vQ is called the quantum volume.




vQ =(

h√2πmkT

)3

is called the quantum volume.

If we include the intrinsic states, the single-particle partition function is (UsingEq. 33 Z = ZtrZint):

Z1 =V

vQZint (41)

Therefore, the partition function for a gas of N particles is,

Z =1

N!

(V

vQZint

)N

(42)

The logarithm of it is:

lnZ = N [lnV + lnZint − lnvQ ]− lnN! (43)

For large N, using the Stirling’s approximation:

N! ≈ NNe−N√

2πN and drop√

2πN,

lnZ = N [lnV + lnZint − lnvQ − lnN + 1] (44)



The Partition Function for Ideal Gas - Applications

We have obtained the expressions for thermal properties at the beginning ofthis lecture:

F = −kTlnZ , S = −(∂F

∂T

)V ,N

, P = −(∂F

∂V

)T ,N

, µ =

(∂F

∂N

)T ,V

.

Now, we have Z = 1N!

(VvQZint

)Nor more useful, the logarithm of it:

lnZ = N [lnV + lnZint − lnvQ − lnN + 1] for ideal gas. So, we can calculate allthe thermal properties of an ideal gas.Let’s do some exercises at class:

Total energy: U = − 1Z∂Z∂β

= − ∂lnZ∂β

Heat capacity: CV = ∂U∂T

Helmholtz free energy: F = −kTlnZPressure: P

Entropy: S

Chemical potential: µ



The Partition Function for Ideal Gas - Applications

And the answers are:

Total energy: U = 32NkT + Uint

Heat capacity: CV = 32Nk + ∂Uint

∂T

Helmholtz free energy: F = −NkT [lnV + lnZint − lnvQ − lnN + 1] =−NkT [lnV − lnvQ − lnN + 1] + Fint

Pressure: P = kNTV

Entropy: S = Nk[ln(

VNvQ

)+ 5

2

]− ∂Fint

∂T

Chemical potential: µ = −kTln(

VZintNvQ

)




Several key points/steps:

(1) Partial derivative with respect to β:(∂lnZintri.∂β

)(2) Partial derivative with respect to T of

(∂lnvQ∂T

). This gives 3

2kN to make

the 52kN in Eq.(6.92) in the textbook (p.255)

(3) Compare Eq.(6.92) on p.255 with the Sackur-Tetrode Equation on p.77-78in the textbook.

Eq.(6.92): S = Nk[ln(

VN

(1vQ

))+ 5

2

]− ∂Fintrin.

∂T.

Eq.(2.49): S = Nk[ln(

VN

(4πmU3Nh2

)3/2)

+ 52

].

Please read more in the textbook, p. 254-255.



Ideal Gas - Example with intrinsic energy.

In-class Exercise 06-07: For a mole of nitrogen (N2) gas at room tem-perature and atmospheric pressure, computer U,F , S , µ. The rotationalconstant for N2 molecule is ε = 0.00025 eV . The electric ground state ofN2 is NOT degenerate.


Part II: Statistical Physicsodessa.phy.sdsmt.edu/~bai/teaching_2014Fall_done/PHYS-343/Ther… · X...

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