“microcanonical methods” equilibrium between 2 systems...

Phys 112 (S2014) 3 Equilibrium between 2 systems B. Sadoulet��1

3 Equilibrium Between Two Systems

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

Equilibrium between 2 systems Temperature, pressure, chemical potential

Definition

Ideal gas Thermodynamics identities

The three laws S,U,H,F,G A note about differentials

Chemical Potential Why is the Chemical Potential a potential Why is it call “Chemical”


Let us consider a gas Constraints: Energy U

Volume V Note: Additive Number of particles of species i : Ni

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

!Theorem: at equilibrium !!!Formal definition of temperature, pressure, chemical potential

=> thermodynamic identity

U1,V1, Ni1 U2 ,V2 , Ni2

Thermal equilibrium between 2 Systems

∂σ 1

∂U1

= ∂σ 2

∂U2

≡ 1τ

∂σ 1

∂V1= ∂σ 2

∂V2≡ pτ

∂σ 1

∂Ni1

= ∂σ 2

∂Ni2

≡ − µτ


Let us consider a gas Constraints: Energy U

Volume V Note: Additive Number 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 interactions => (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

Thermal equilibrium between 2 Systems 2

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


Thermal equilibrium between 2 Systems3

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 out Number of states of the

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

Most likely configuration = the configuration with the largest number of states

g1 g2

U1

g1 U1( )g2 U −U1( )

g1 U1( )g2 U −U1( )

UMost likely U1

log g1 log g2

U1

UMost likely

U1

∂ logg1 U1( )∂U1

= −∂ logg2 U −U1( )

∂U1

=∂ logg2 U2( )

∂U2 U2 =U −U1

Phys 112 (S2014) 3 Equilibrium between 2 systems B. Sadoulet

Calculation for an ideal gas

��6

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 Kittel: systems of spins

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

distribution of configurations the system fluctuates around the configuration of maximum probability

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


CommentsAt 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


Temperature,Pressure,Chemical PotentialTemperature

Definition !

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

Pressure Definition !

Have to check that corresponds to ordinary p

Chemical potential of species i Definition !

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 particles Initial 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

⎛

⎝⎜⎞

⎠⎟


What happens? !

��12

Expansion of Ideal Gas into VacuumA prototype Conceptually Important!

cf. end of chap. 6 in Kittel & Kroemer

Initial

FinalIsolated

Vi

Vf

Sudden!


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 !

��13

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 Law Heat transfer: Definition

Not an exact differential

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

“quasi-static” processes)

!Second Law

When an isolated system evolves from a non equilibrium configuration to equilibrium, its entropy will increase

!Third Law

Entropy 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

∫

Succession of 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( )


Exact DifferentialsExact Differential !

! independent of path Stokes theorem

!!!!

AB

g x, y( ) dg =∂g∂x

dx +∂g∂y

dy ⇒∂2g∂x∂y

=∂2g∂y∂x

⇔ dg = g B( )AB∫ − g A( )

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


Exact Differentials 2≠ Non exact differential: dependent of path

!!

e.g. heat transfer depends on pathdg = 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


Differential identitiesConsequences of thermodynamic identities

non intuitive relationships that we will use often

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

∂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


Differential identities 2Maxwell identities (K.K. Chap. 3 p. 71): Advanced!

Consider e.g., 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?

Potential Recall: 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 !

!!

!F = −

!∇Φ⇔ Φ2 − Φ1 = −

!F ⋅d!r

1

2

∫

1

2 ⇔!F.d!r is an exact differential

F!r( ) Φ


Why is the Chemical Potential a Potential?

Raising the potential energy of a system Let us consider an isolated system at zero potential energy !!Let us then raise it at uniform potential energy per particle

The entropy is not changed by uniform potential (number of states not changed)

Uo µo =∂Uo∂N σ ,V

Uo →U = Uo + NΔΦ

µtotal! =

∂U∂N σ ,V

= µointernal! + ΔΦ

external!


Conservation of Chemical PotentialEquilibrium with several species i

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

=> !

!Conservation of the Chemical (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

!

µi 1( ) = µi 2( ) ∀i21µi 1( ) µi 2( )

ν1A1 +ν2A2 ↔π 3A3 +π 4A4⇔ νi Ai

i∑ ↔ 0 with ν3 = −π3 , ν4 = −π 4

Theorem:At equilibriumν iµi

i∑ = 0

or ν iµiinitial∑ = π iµi

Final∑

Conservation of chemical potential


Conservation of Chemical Potential!

Conserved quantities (K&K chapter 9) From ! The probability distribution at equilibrium will be sharply peaked around the

configuration of maximum total entropy : !!or with the constraints

ν1A1 +ν2A2 ↔π 3A3 +π 4A4

or ν iAii∑ ↔ 0 with ν3 = −π 3 , 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

ninQi

⎛

⎝⎜⎞

⎠⎟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

Physics is simple: force from particles on the wall!

θ

density in d 3p = n p( ) p2dpdΩ

vΔt

dA


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

∫

“microcanonical methods” equilibrium between 2 systems...

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