3 Equilibrium Between Two Systems - UC Berkeley Cosmology...

Phys 112 (S2012) 3 Equilibrium between 2 systems B. Sadoulet1

3 Equilibrium Between Two Systems

Chapter 2 Kittel&Kroemern+ bits and pieces e.g. from chapter 9“Microcanonical methods”

3.1 Equilibrium between 2 systems3.2 Temperature, pressure, chemical potential

Definition

3.3 Ideal gas3.4 Thermodynamics identities

The three lawsS,U,H,F,GA note about differentials

3.5 Chemical PotentialWhy is the Chemical Potential a potentialWhy is it call “Chemical”

Phys 112 (S2012) 3 Equilibrium between 2 systems B. Sadoulet

In class

2

Take 2 systems and put them in contact Put them in weak interactions

Isolated=> fixed U,V,N

Number of states

Weak interaction => (quantum) states are not modified

How many states do we have in the combined system?

U = U1 +U2V = V1 +V2Ni = Ni1 + Ni2

U1,V1, Ni1 U2 ,V2 , Ni2

States of a Combination of 2 systems

g1 U1,V1,Ni1( )g2 U2 ,V2 ,Ni2( )


In class

3



Number of states


How many states do we have in the combined system?A:

B:

C: None of those

U = U1 +U2V = V1 +V2Ni = Ni1 + Ni2

U1,V1, Ni1 U2 ,V2 , Ni2


g1 U1,V1,Ni1( )g2 U2 ,V2 ,Ni2( )

g1 U1,V1,Ni1( ) + g2 U2 ,V2 ,Ni2( )

g1 U1,V1,Ni1( ) • g2 U2 ,V2 ,Ni2( )


In class

4



Number of states


How many states do we have in the combined system?

Answer B:

U = U1 +U2V = V1 +V2Ni = Ni1 + Ni2

U1,V1, Ni1 U2 ,V2 , Ni2


g1 U1,V1,Ni1( )g2 U2 ,V2 ,Ni2( )

g1 U1,V1,Ni1( ) • g2 U2 ,V2 ,Ni2( )


Let us consider a gasConstraints: Energy U

Volume V Note: AdditiveNumber of particles of species i : Ni

Take 2 systems and put them in contact Put them in weak interactions Isolated=> fixed U,V,N

The configuration is described by U1,V1,Ni1

Weak interaction => (quantum) states are not modified multiplicity function =product of multiplicity functions

More states => Entropy increases

U = U1 +U2V = V1 +V2Ni = Ni1 + Ni2

U1,V1, Ni1 U2 ,V2 , Ni2

3.1 Thermal equilibrium between 2 Systems

g U1,V1, Ni1( ) = g1 U1,V1, Ni1( )g2 U −U1,V −V1, Ni − Ni1( )


Thermal equilibrium between 2 Systems

What is the most likely configuration (in equilibrium)?The probability of a configuration is proportional to the number of

its (quantum) states => Maximum probability is obtained for

Similarly

Since the distribution is very peaked, for all practical purpose we can say that this is the “equilibrium configuration”!

∂g∂U1

= 0, ∂g∂V1

= 0, ∂g∂Ni1

= 0

�

∂g∂U1

= ∂g1

∂U1

g2 + g1∂g2

∂U1

= ∂g1

∂U1

g2 − g1∂g2

∂U 2 U2 =U−U1

= 0 <=> ∂ logg1

∂U1

= ∂ logg2

∂U 2

but logg1 =σ1 ⇒ ∂σ1∂U1

=∂σ 2∂U2

∂σ1∂V1

=∂σ 2∂V2

∂σ1∂Ni1

=∂σ 2∂Ni2


Graphic Representation

Watch outNumber of states of the

combined system is the product of the number of states of each system

Most likely configuration= the configuration with thelargest number of states

g1 g2

U1

g1 U1( )g2 U −U1( )

g1 U1( )g2 U −U1( )

UMost likelyU1

log g1 log g2

U1

UMost likely

U1

∂ logg1 U1( )∂U1

= −∂ logg2 U −U1( )

∂U1

=∂ logg2 U2( )

∂U2 U2 =U −U1


Calculation for an ideal gas

8

http://cosmology.berkeley.edu/Classes/S2009/Phys112/Equilibrium2sysEquilbrium2sys.html


CommentsMicrocanonical methods

Compute number of states => entropy of configuration (U, V, Ni )=>T, p, µi Examples: gas system (next page) Kittel: systems of spins

Maximum probability <=> “equilibrium configuration”Strictly speaking we should be speaking of the equilibrium probability

distribution of configurationsthe system fluctuates around the configuration of maximum probability

Approximate language but does not matter because of narrowness of distribution <= Central limit theorem

At equilibrium, the entropy of an isolated system is maximum (an instance of the H theorem!)

In this case:The total number of states accessible to the combined system includes the product of the number of states initially accessible to each of the systems. This total number of states can only increase through the exchange of energy, volume, particles

An approximate argument: the entropy of the “equilibrium configuration” is maximum by definition. Total entropy even bigger!

dσdt

≥ 0


In class Entropy of the most probable configuration

We can define the entropy of the most probable configuration as

How does this compare with the entropy of the system in equilibrium ?

10

σmax = log g U1max ,V1max ,Ni1max( )( ) = log g1 U1max ,V1max ,Ni1 max( )⎡⎣ ⎤⎦ + log g2 U −U1max ,V −V1max ,Ni − Ni1max( )⎡⎣ ⎤⎦

σ = − pi

i∑ log pi( ) = log gt( ) where gt is the total number of states


In class Entropy of the most probable configuration



Possible answersA: They are the same!B:C:

11


σ = − pi


σmax < σσmax > σ


In class



AnswerB:

Entropy of the most probable configuration

12


σ = − pi


σmax < σ

gt = g U1,V1,Ni1( )U1 ,V1 ,Ni1∑ = g1 U1,V1,Ni1( )g2 U −U1,V −V1,Ni − Ni1( )

U1 ,V1 ,Ni1∑

σ −σmax ≈ log N1( ) + log N2( ) − log N1 + N2( )⎡⎣ ⎤⎦very small compared to

σ = N lognQn

⎛⎝⎜

⎞⎠⎟+

52

⎡

⎣⎢

⎤

⎦⎥


3.2 Temperature,Pressure,Chemical PotentialTemperature

Definition

We have already checked that corresponds to ordinary T for ideal gas

PressureDefinition

Have to check that corresponds to ordinary p

Chemical potential of species iDefinition

τ = kBT U =32NkBT

�

1τ

=∂σ∂U

=> at equilibrium τ 1 = τ 2

pτ=∂σ∂V

=> at equilibrium p1 = p2

PV = Nτ = NkBT

µiτ

= −∂σ∂Ni

=> at equilibrium µi1 = µi2


Simulationhttp://cosmology.berkeley.edu/Classes/S2009/Phys112/Diffusion/GaussDiffusion.html

Wall partially transparent to particlesInitial state Final state

-5

-3

-1

1

3

5

-5 -3 -1 1 3 5-5

-3

-1

1

3

5

-5 -3 -1 1 3 5


Ideal Gas:Chemical Potential

We need to use full expression of entropy Notes chapter 2 slide 21

µ is a measure of the concentration!

µ = −τ ∂σ∂N

= −τ∂ N log 2πM

h22U3N

⎛⎝⎜

⎞⎠⎟3/2 VN

⎛

⎝⎜⎞

⎠⎟+ 52N

⎛

⎝⎜

⎞

⎠⎟

∂N= τ log n

2πMh2

2U3N

⎛⎝⎜

⎞⎠⎟3/2

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

= τ log nnQ

⎛

⎝⎜⎞

⎠⎟


Isolated => ΔU=0, ΔQ=0, ΔW=0 => Tf=Ti

Increase of entropy ?

Note Process is not a succession of equilibrium configurations (“irreversible”): T, p are not defined during transition

16

Expansion of Ideal Gas into VacuumA prototype Conceptually Important!

cf. end of chap. 6 in Kittel & Kroemer

Initial

FinalIsolated

Vi

Vf

σ = log gt = NlogV +32N logU...⇒Δσ = Nlog

VfVi

⎛ ⎝ ⎜

⎞ ⎠ ⎟ ΔS = NkB log

VfVi

⎛ ⎝ ⎜

⎞ ⎠ ⎟

Sudden!


Thermodynamics IdentitiesZeroth Law

Two systems in equilibrium with a third one are in equilibrium with each other

First LawHeat transfer: Definition

Not an exact differential

Heat is a form of energyFundamental Thermodynamic Identity: Apply only at equilibrium (or reversible

“quasi-static” processes)

Second LawWhen an isolated system evolves from a non equilibrium configuration to

equilibrium, its entropy will increase

Third LawEntropy is zero at zero temperature ( or log of number of states occupied)=> method to compute entropy

δQ = τdσ = TdS

dU = τdσ − pdV + µidNii∑ = TdS − pdV + µidNi

i∑

�

<= dσ =1τdU +

pτdV −

µi

τdN

i

i

∑

σ 0 = log(g10g20 ) constraints removed⎯ →⎯⎯⎯⎯⎯ σ f = log g1 U1( )g2 U −U1( )⎡⎣ ⎤⎦ ≥U1

∑ log g1g2( )max ≥ σ 0

σ = − pss∑ log ps( ) only one state populated ⇒ po = 1 ⇒σ = 0

S =dQT0

T

∫ , σ =dQτ0

τ

∫

Successionof equilibria


Thermodynamics Identities (Gas)U,V, N

S, V, N (e.g., for constant volume situations)

S, P, N (e.g., for constant pressure situations)Enthalpy (KK chap. 8)

T, V, N (e.g., for constant volume situations)Helmholtz Free Energy (KK chap. 3)

T, P, N (e.g., for constant pressure situations)Gibbs Free Energy (KKchap. 9)

will be derived later

dS =1TdU +

pTdV −

µiTdNi

i∑

dU = TdS

dQ− pdV

dW + µidNi = dQ + dW + µidNi

i∑

i∑

H =U + pV ⇒ dH = TdS +Vdp + µidNii∑

F =U − TS

or τσ ⇒ dF = −SdT

σdτ − pdV + µidNi

i∑

G = F + pV ≡ µi T,P( )Ni

i∑ ⇒ dG = −SdT

σdτ +Vdp + µidNi

i∑

Configuration Variables

Natural variables F T ,V ,Ni( )

Natural variables H S, p,Ni( )

Natural variables G T , p,Ni( )

Natural variables U S,V ,Ni( )


In class QuestionHow do we measure experimentally entropy?

19


In class QuestionHow do we measure experimentally entropy?

A:

B:

C:

20

S T( ) = kbσ T( ) = dQT =0

T

∫

S T( ) = kbσ T( ) = S T = 0( ) + 1

TdU

T =0

T

∫ at constant volume

S T( ) = kbσ T( ) = CvT =0

T

∫dTT

=32N

T =0

T

∫dTT


In class AnswerHow do we measure experimentally entropy?

B:

but problem in taking not valid close to absolute zeroor for that matter Sackur Tetrod is not valid either at absolute zero

21

S T( ) = kbσ T( ) = S T = 0( ) + 1TdU

T =0

T

∫ at constant volume ≡ S T = 0( ) + dQTT =0

T

∫ ≡ S T = 0( ) + Cv T( ) dT

TT =0

T

∫

Use dU = TdS − pdV at constant volume

Cv = const. e.g., Cv = NkbT

S = Nkb lognQn

⎛⎝⎜

⎞⎠⎟+52

⎡

⎣⎢

⎤

⎦⎥ The classical approximation breaks down when n > nQ

nQ =2πMh2

2U3N

⎛⎝⎜

⎞⎠⎟3/2

= nQ =2πMh2

kbT⎛⎝⎜

⎞⎠⎟3/2

T→0⎯ →⎯⎯ 0 S T→0⎯ →⎯⎯ −∞


Exact DifferentialsExact Differential

independent of path Stokes theorem

≠ Non exact differential: dependent of path

e.g. heat transfer depends on path

AB

g x, y( ) dg =∂g∂x

dx + ∂g∂y

dy ⇒ ∂2g∂x∂y

=∂2g∂y∂x

⇔ dg = g B( )AB∫ − g A( )

dg = a x, y( )dx + b(x, y)dy with ∂a∂y

≠∂b∂x

�

dQ = TdS⇒ dQ = dU + pdV = adU + bdV

clearly ∂a∂V U

= 0 ≠ ∂b∂U V

= ∂p∂U V

e.g. for an ideal gas pV = Nτ = 23U ⇒ ∂p

∂U V

= 23V

a(x, y)dx + b(x + dx, y)dyb(x, y)dy + a(x, y + dy)dx

Difference = ∂b∂x

− ∂a∂y

⎡ ⎣ ⎢

⎤ ⎦ ⎥ dxdy

dy

dx


Differential identitiesConsequences of thermodynamic identities

non intuitive relationships that we will use often

Example: free energy (K.K. Chap. 3 p 70-71)

Maxwell identities (K.K. Chap. 3 p. 71): Advanced!Consider e.g.,

∂F∂T V ,Ni

= −S⇔ ∂F∂τ V ,Ni

= −σ ∂F∂V τ ,Ni

= − p ∂F∂Ni τ ,V

= µi

�

U = TS +F = −T ∂F∂T

+F = −T 2∂ FT⎛ ⎝ ⎜

⎞ ⎠ ⎟

∂T= −τ 2

∂ Fτ⎛ ⎝ ⎜

⎞ ⎠ ⎟

∂τ

F(T,V, N), S(T,V, N), p(T,V, N)∂2F∂T∂V

=∂2F∂V∂T

⇒∂S∂V

=∂p∂T


Why is the Chemical Potential a Potential?

PotentialRecall: let us consider a force field . It derives from a potential if

independent of path (Stokes’ theorem) = potential energy difference between point 1 and 2

Raising the potential energy of a systemLet us consider an isolated system at zero potential energy

Let us then raise it at uniform potential energy per particleThe entropy is not changed by uniform potential (number of states not changed)

F = −

∇Φ⇔ Φ2 − Φ1 = −

F ⋅dr

1

2

∫

Uo µo =∂Uo∂N σ ,V

Uo →U = Uo + NΔΦ

µtotal =

∂U∂N σ ,V

= µointernal + ΔΦ

external

1

2 ⇔F.dr is an exact differential

Fr( ) Φ


Conservation of Chemical PotentialEquilibrium with several species i

If the two systems are in equilibrium, each kind separately has to be in equilibrium

=>

Conserved quantities (K&K chapter 9)In a reaction between species, the number of disappearing particles or

molecules is related to the number of produced particles or molecules

The probability distribution at equilibrium will be sharply peaked around the configuration of maximum total entropy :

orwith the constraints

µi 1( ) = µi 2( ) ∀i21µi 1( ) µi 2( )

ν1A1 +ν2A2 ↔π 3A3 +π 4A4⇔ νi Ai

i∑ ↔ 0 with ν3 = −π3 , ν4 = −π 4

δσ =∂σ∂NA1

δNA1 +∂σ∂NA2

δNA2 +∂σ∂NA3

δNA3 +∂σ∂NA4

δNA4 = 0

µ1δNA1 + µ2δNA2 + µ3δNA3 + µ4δNA4 = 0

δNA1ν1

=δNA2ν2

=δNA3ν3

=δNA4ν4

δNA1ν1

=δNA2ν2

=δNA3ν3

=δNA4ν4

⇒ ν iµii∑ = 0

or ν iµiinitial∑ = π iµi

Final∑

Conservation of chemical potential


Law of mass actionLaw of Mass Action (K&K chapter 9)

Consider the reaction

or

=>

or

νiAii∑ ↔ 0

ν iµi

i∑ = 0 with µi = τ log ni

nQi

⎛

⎝⎜⎞

⎠⎟ninQi

⎛

⎝⎜⎞

⎠⎟i∏

νi

= 1

niνi

i∏ = K τ( ) with K τ( ) = nQi( )

i∏ νi

niνi

initial i∏

njπ j

final j∏

= K τ( ) with K τ( ) =nQi( )νi

initial i∏

nQj( )π j

final j∏


Is the pressure the force per unit area? A last task: Show that the pressure we compute is indeed the

average force per unit area cf. Kittel and Kroemer Chapter 14 p. 391

Describe the particles by their individual density in momentum space (ideal gas)

If the particles have specular reflection by the wall, the momentum transfer for a particle arriving at angle θ is

Integration on angles gives

that we would like to compare with the energy density

θ

�

non relativistic⇒ pv = 2ε ⇒ pressure P = 23u (energy density)

u = 32NVτ ⇒ P = N

Vτ = same pressure as thermodynamic definition = τ∂σ

∂V U,Nultra relativistic ⇒ pv = ε ⇒ P =

13u

2 pcosθ

�

23× 2π pv n p( )p 2dp

0

∞∫

!

�

u = ε n p( )d 3 p0

∞∫ = 4π ε n p( )p 2dp0

∞∫

density in d 3p = n p( ) p2dpdΩ

vΔt

dA

P =ForcedA

=d ΔpΔtdA

=1

dAΔt2pcosθ

Momentum transfer vΔtdAcosθ

Volume of cylinder n p( ) p2dpdΩ

density in cylinder ∫

= dϕ0

2π

∫ d cosθ 2pvcos2θ n p( ) p2dp0

∞

∫0

1

∫

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