“microcanonical methods” 3.1 equilibrium between 2...
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Phys 112 (F2006) 3 Equilibrium between 2 systems B. Sadoulet1
3 Equilibrium Between Two Systems
Chapter 2 Kittel&Kroemer“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 (F2006) 3 Equilibrium between 2 systems B. Sadoulet2
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
Configuration described by U1,V1,Ni1
Weak interaction => (quantum) states 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( )
Phys 112 (F2006) 3 Equilibrium between 2 systems B. Sadoulet3
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
Phys 112 (F2006) 3 Equilibrium between 2 systems B. Sadoulet4
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
Phys 112 (F2006) 3 Equilibrium between 2 systems B. Sadoulet5
3.2 Temperature,Pressure,Chemical PotentialTemperature
Definition
We will have to check that corresponds to ordinary T
PressureDefinition
Have to check that corresponds to ordinary p (page 6)In fact: Ideal gas law
Chemical potential of species iDefinition
See chapter 5
τ = kBT
�
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
Phys 112 (F2006) 3 Equilibrium between 2 systems B. Sadoulet6
Simulation
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
Phys 112 (F2006) 3 Equilibrium between 2 systems B. Sadoulet7
CommentsMicrocanonical methods
Compute number of states => entropy of configuration (U, V, Ni )=>T, p, µi
Example: gas system (next page)Kittel: systems of spins
Maximum probability <=> “equilibrium configuration”Strictly speaking we should be speaking of the equilibrium probability
distribution of configurations the system fluctuates in the set of configurations 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 dσ
dt≥ 0
Phys 112 (F2006) 3 Equilibrium between 2 systems B. Sadoulet8
3.3 Ideal GasCheck that our definitions are OK. Not in Kittel& Kroemer. cf Reif 2.5
Calculation of entropy as a function of UDone in Chapter 2 of the notes slide 13
Temperature
this is the classical result (3/2 kB T per spinless monoatomic particle)
Pressure
Law of ideal gases! Reassuring!Illustrates the power of microcanonical methods (at the price of often
difficult computations of the multiplicity of states)
1τ=∂σ∂U
=3N2U
⇒U =32
Nτ =32
NkBT
pτ=∂σ∂V
=NV
⇒ pV = Nτ = NkBT
for large N g ∝V NU3N2 σ = f N( ) + N logV + 3 / 2N logU
Phys 112 (F2006) 3 Equilibrium between 2 systems B. Sadoulet9
Is the pressure the 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
vΔtdAcosθVolume of cylinder n p( ) p2dpdΩ
density in cylinder ∫
= dϕ0
2π
∫ d cosθ 2pvcos2θ n p( ) p2dp0
∞
∫0
1
∫
Phys 112 (F2006) 3 Equilibrium between 2 systems B. Sadoulet10
Ideal Gas:Chemical Potential
We need to use full expression of entropy Notes chapter 2 slide 15
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
⎛
⎝⎜⎞
⎠⎟
Phys 112 (F2006) 3 Equilibrium between 2 systems B. Sadoulet
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
11
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!
Phys 112 (F2006) 3 Equilibrium between 2 systems B. Sadoulet12
3.4Thermodynamics 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
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=> 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
Phys 112 (F2006) 3 Equilibrium between 2 systems B. Sadoulet13
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 = TdSdQ− pdV
dW
+ µidNi = dQ + dW + µidNii∑
i∑
H =U + pV ⇒ dH = TdS +Vdp + µidNii∑
F =U − TSor τσ ⇒ dF = −SdT
σdτ − pdV + µidNi
i∑
G = F + pV ≡ µi T,P( )Nii∑ ⇒ dG = −SdT
σdτ +Vdp + µidNi
i∑
Configuration Variables
Natural variables F T ,V ,Ni( )
Natural variables U T ,V ,Ni( )
Natural variables H S, p,Ni( )
Natural variables G T , p,Ni( )
Phys 112 (F2006) 3 Equilibrium between 2 systems B. Sadoulet14
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
Phys 112 (F2006) 3 Equilibrium between 2 systems B. Sadoulet15
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
Phys 112 (F2006) 3 Equilibrium between 2 systems B. Sadoulet16
3.4 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
+ ΔΦexternal
1
2 ⇔F.dr is an exact differential
F r( ) Φ
Phys 112 (F2006) 3 Equilibrium between 2 systems B. Sadoulet17
Why is µ called Chemical Potentiak?Equilibrium with several species i
If the two systems are in equilibrium, each kind separately has to be in equilibrium
=>
Conserved quantitiesIn 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
Phys 112 (F2006) 3 Equilibrium between 2 systems B. Sadoulet18
Law of mass actionLaw of Mass Action
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∏