5. Systems in contact with a thermal...
-
Upload
truongngoc -
Category
Documents
-
view
213 -
download
0
Transcript of 5. Systems in contact with a thermal...
Phys 112 (S2009) 05 Boltzmann-Gibbs 1
5. Systems in contact with a thermal bath
So far, isolated systems (micro-canonical methods)
5.1 Constant number of particles:Kittel&Kroemer Chap. 3Boltzmann factorPartition function (canonical methods)Ideal gas (again!)
Entropy = Sackur-Tetrode formulaChemical potential
5.2 Exchange of particles: Kittel&Kroemer Chap. 5Gibbs FactorGrand Partition Function or Gibbs sum
(grand canonical methods)
5.3 Fluctuations
Phys 112 (S2009) 05 Boltzmann-Gibbs 2
Ideal Gas Thermal BathConsider 1 particle in thermal contact with a
thermal bath made of an ideal gas with a large number M of particles
We assume that the overall system is isolated and in equilibrium at temperature τ. We call U the total energy (which is constant).
We would like to compute the probability of having our particle in a state of energy ε. Because in equilibrium, all states of the overall system are equiprobable, this is equivalent to computing the number of states such that our particle has energy ε and the thermal bath E=U-ε.
Using the fact that for an ideal gas of energy E and particle number Np, the number of states is
and that
we obtain for large Np
gRes
ε
exp −ετ
⎛⎝⎜
⎞⎠⎟
U =32Npτ
g ∝ E 3/2Np
prob ε( )∝ U − ε( )32Np ∝ 1− ε
32Npτ
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟
32Np
Np→∞⎯ →⎯⎯ exp −ετ
⎛⎝⎜
⎞⎠⎟
Phys 112 (S2009) 05 Boltzmann-Gibbs
# States in configurationFundamental postulate (or H theorem): at equilibrium
Notes: • =Boltzmann factor • Z=partition function • Probability of each state not identical!
<= interaction with bath3
Boltzmann factor
R , Uo-ε S,ε
Total system =Our system S + reservoir R isolated
Configuration: 1 state of S + energy of reservoir
1× gR Uo − ε( )
Prob configuration( )∝ gR Uo − ε( )
with Z = exp −εsτ
⎛ ⎝ ⎜ ⎞
⎠ ⎟
all states s∑
⇔ Prob state ε s( ) = 1Z
exp −ε sτ
⎛⎝⎜
⎞⎠⎟=
1Z
exp −ε skBT
⎛⎝⎜
⎞⎠⎟
⇒Prob state ε1( )Prob state ε2( ) ≈ exp ∂σR
∂Uo
ε2 − ε1( )⎛⎝⎜
⎞⎠⎟= exp 1
τε2 − ε1( )⎛
⎝⎜⎞⎠⎟
⇒Prob state ε1( )Prob state ε2( ) =
gR Uo − ε1( )gR Uo − ε2( ) = exp σR Uo − ε1( ) − σR Uo − ε2( )( )
exp −ε sτ
⎛⎝⎜
⎞⎠⎟
Phys 112 (S2009) 05 Boltzmann-Gibbs
Examples2-state system
State 1 has energy 0State 2 has energy ε
4
Pr ob 2( ) =
exp −ε
τ( )
1 + exp −ε
τ( )
=1
1 + expε
τ( )
Pr ob 1( ) =1
1 + exp −ε
τ( )
ε
0
Z = 1+ exp −ετ
⎛⎝⎜
⎞⎠⎟
U = ε =ε exp − ε
τ⎛⎝⎜
⎞⎠⎟
1+ exp − ετ
⎛⎝⎜
⎞⎠⎟=
ε
exp ετ
⎛⎝⎜
⎞⎠⎟ +1
CV =dQdT
⎛⎝⎜
⎞⎠⎟V (,N )
=∂U∂T
⎛⎝⎜
⎞⎠⎟V (,N )
= kb∂U∂τ
⎛⎝⎜
⎞⎠⎟V (,N )
= kbεkbT
⎛⎝⎜
⎞⎠⎟
2 exp εkbT
⎛⎝⎜
⎞⎠⎟
exp εkbT
⎛⎝⎜
⎞⎠⎟+1
⎛
⎝⎜⎞
⎠⎟
2
CV
T
Prob(0)
Prob(ε)
1
Phys 112 (S2009) 05 Boltzmann-Gibbs
ExamplesFree particle at temperature τ
Assumption: No additional degrees of freedome.g. no spin, not a multi-atom molecule
Maxwell-Boltzmann distribution KK p 392Normalizing to N particles, the probability to find one particle per
unit volume in the velocity interval dv and solid angle dΩ is
5
Prob dVvolumeelement
, p, p + dp[ ]Momentum interval , dΩ
solid angleelement
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟=
1Z
exp −p2
2Mτ⎛⎝⎜
⎞⎠⎟
Boltzmann factordoes not vary appreciably over dp
dVp2dpdΩh3
number of spatialquantum states
in phase space element
dΩ = d cosθdφ
= Prob dVvolumeelement
, ε,ε + dε[ ]energyy interval , dΩ
solid angleelement
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟=
2M 3/2
Zexp −
ετ
⎛⎝⎜
⎞⎠⎟dV εdεdΩ
h3
= Prob dVvolumeelement
, v,v + dv[ ]velocity interval , dΩ
solid angleelement
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟=M 3
Zexp −
12Mv2
τ
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟
dVv2dvdΩh3
f (v)dvdΩ = n M
2πτ⎛⎝⎜
⎞⎠⎟
32exp −
Mv2
2τ⎛⎝⎜
⎞⎠⎟v2dvdΩ with n = N
V
Phys 112 (S2009) 05 Boltzmann-Gibbs
Another interpretation of theBoltzmann Distribution
We derived it from number of states in the Reservoir
But each excitation in the reservoir will have a typical energy
Thermal fluctuations of the order ofDifficult to get
6
≈ τ
τ
ε >> τ
Prob ∝ exp −ετ
⎛⎝⎜
⎞⎠⎟
Phys 112 (S2009) 05 Boltzmann-Gibbs
Definition
If states are degenerate in energy, each one has to be counted separately
Relation with thermodynamic functionsEnergy
Entropy
Free Energy
Chemical potential
7
Partition FunctionZ = exp −
εsτ
⎛ ⎝ ⎜ ⎞
⎠ ⎟
all states s∑
T
F =U −τσ = −τ log Z
U = ε =1Z
εs exp −εsτ
⎛ ⎝ ⎜ ⎞
⎠ ⎟
s∑ = −
∂ logZ
∂1τ
= τ2∂ logZ∂τ
σ = − pss∑ logps = −
e−ε sτ
Zs∑ log e
−ε sτ
Z
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟ = logZ +
Uτ
=∂ τ log Z( )
∂τ
⇒ σ =−∂F∂τ
U = τ 2 ∂ log Z∂τ
= −τ 2∂ F
τ⎛ ⎝
⎞ ⎠
∂τas implied by the thermodynamic identity
µ =∂F∂N τ ,V
= −τ ∂ logZ∂N
Phys 112 (S2009) 05 Boltzmann-Gibbs 8
ExamplesEnergy of a two state system
Free particle at temperature τAssumption: No additional degrees of freedom
e.g. no spin, not a multi-atom molecule
Method 1: density in phase space Energy:
Method 2: quantum state in a box: see Kittel-Krömer (chapter 3 p 73) Same result, of course!
Z = 1+ exp −ετ
⎛⎝⎜
⎞⎠⎟
U = ε =τ 2∂ logZ
∂τ=
ε exp − ετ
⎛⎝⎜
⎞⎠⎟
1+ exp − ετ
⎛⎝⎜
⎞⎠⎟=
ε
exp ετ
⎛⎝⎜
⎞⎠⎟ +1
ε =px2 + py2 + pz2
2M
Z1 =d3q d3p
h3∫ exp −ετ
⎛ ⎝ ⎜ ⎞
⎠ ⎟ =
Vh3
dpxe− px
2
2Mτ−∞∞∫
= 2πMτ
dpye−
py2
2Mτ−∞∞∫ dpze
− pz2
2Mτ−∞∞∫
⇒ Z1 = V 2πMτ
h2⎛ ⎝ ⎜ ⎞
⎠ ⎟ 32 = V Mτ
2π2⎛ ⎝ ⎜ ⎞
⎠ ⎟ 32 = VnQ
with nQ =
Mτ2π2
⎛⎝⎜
⎞⎠⎟
32 ⇒ ε = τ 2 ∂ logZ1
∂τ=
32τ
Phys 112 (S2009) 05 Boltzmann-Gibbs 9
Partition FunctionPower comes from summation methods
• Direct e.g. geometric series
• Few term approximation useful if f decreases fast
• Integral approximation
Other example: Harmonic oscillator
rii=0
∞∑ = lim
m→∞ri
i=0
m∑ = lim
m→∞
1− rm+1
1− r=
11− r
f i( ) = f 0( )i=0
∞∑ + f 1( ) + f 2( ) + ....
f i( ) ≈ f (x)dx0
∞
∫i=0
∞∑
Consider an harmonic oscillator ωThe probability of mode of energy sω , s integer ≥ 0 is
Prob s( ) = 1Zexp −
sωτ
⎛⎝⎜
⎞⎠⎟
Z = exp −sωτ
⎛⎝⎜
⎞⎠⎟s=0
∞
∑ =1
1− exp − ωτ
⎛⎝⎜
⎞⎠⎟⇒U = sω = τ 2
∂ log Z( )∂τ
=ω
exp ωτ
⎛⎝⎜
⎞⎠⎟ −1
Phys 112 (S2009) 05 Boltzmann-Gibbs 10
Applies to classical gases in equilibrium with no potentialStart from Boltzmann distribution Prob (1 particle to be in state of energy ε)=
with
=>
Change of variable to velocities:Probability of finding particle in v, v+dv, dΩ
Maxwell distributionParticle density distribution in velocity space Note: 1) Could get the factor in front directly as normalization
2) Note that no h factor! 3) One can easily compute Equipartition!
Maxwell Distribution K&K p. 392
1Z1exp −
ετ
⎛ ⎝ ⎜ ⎞
⎠ ⎟
Z1 =
d3q d3ph3∫ exp −
ετ
⎛ ⎝ ⎜ ⎞
⎠ ⎟ = V Mτ
2π2⎛ ⎝ ⎜ ⎞
⎠ ⎟ 32 = VnQ
For N particles, the volume density between p and p + dp = NV1nQexp −
ετ
⎛⎝⎜
⎞⎠⎟d 3ph3
=nnQexp −
ετ
⎛⎝⎜
⎞⎠⎟d 3ph3
= n 12πMτ
⎛⎝⎜
⎞⎠⎟3/2
exp −ετ
⎛⎝⎜
⎞⎠⎟d 3p
f (v)dvdΩ = n M2πτ
⎛ ⎝ ⎜ ⎞
⎠ ⎟ 32 exp −
Mv2
2τ⎛
⎝ ⎜
⎞
⎠ ⎟ v2dvdΩ
p = Mv
Mv2
2=32τ =
32kBT
Probability of finding this particle in volume dV and between p and p + dp = 1VnQ
exp −ετ
⎛⎝⎜
⎞⎠⎟
dVd 3ph3
Number of states
Phys 112 (S2009) 05 Boltzmann-Gibbs
Density of States
11
εε
D ε( )dε
ε + dε Individual States of probability ps
It is convenient to replace expressions such as ps fss∑ (fs any function of the state s)
by an integral over the energy ε : ps fss∑ ≈ ps ε( ) fs0
∞
∫ ε( )D ε( )dε
D ε( )dε is the number of states between ε and ε + dε and is called the density of states. We replace ps
sε<εs <ε+dε
∑ by ps ε( )D ε( )dε since for small enough dε, ps does not vary
Note that if you ask what is the probability p ε( )dε for the system to be betweenenergy between ε and ε + dε p ε( )dε = ps ε( )D ε( )dε i.e., p ε( ) ≠ ps ε( )For example, for Boltzmann distribution and a non relativistic ideal gas (monoatomic,spinless)
ps ε( ) = 1Zexp −
ετ
⎛⎝⎜
⎞⎠⎟ and using p = 2Mε d 3p = p2dpd cosθdϕ = p2dpdΩ
D ε( )dε =d 3x
V∫ dΩ p2dpangles∫h3
=V 4πh3
p2 dpdε
dε =V 4πh3
122M( )3/2 εdε = V
4π 2
2M
⎛⎝⎜
⎞⎠⎟3/2
εdε
Phys 112 (S2009) 05 Boltzmann-Gibbs 12
Weak interaction limit2 sub systems in weak interaction:
<=>States of subsystems are not disturbed by interaction (just their population)<=>
=> Partition functions multiply for distinguishable systems
Distinguishable systemsSuppose first that we can distinguish each of the individual
components of the system: e.g. adsorption sites on a solid, spins in a lattice
For 2 subsystems
and for N systems
Partition Function for N systems
Ψ1+2 n1,n2( ) =Ψ1 n1( )Ψ2 n2( ) with n1, n2 still meaningfull
Z = exp −ε s1
+ε s2
τ⎛
⎝ ⎜
⎞
⎠ ⎟ ⇒ Z 1+ 2( )
s2
∑s1
∑ = Z 1( )Z 2( )
Z N( ) = Z 1( )( )N
Phys 112 (S2009) 05 Boltzmann-Gibbs 13
ExamplesFree energy of a paramagnetic system
In a magnetic field B, ε = −mB if m is the magnetic moment.
For 1 spin Z1 = exp −−mBτ
⎛⎝⎜
⎞⎠⎟+ exp −
mBτ
⎛⎝⎜
⎞⎠⎟= 2cosh mB
τ⎛⎝⎜
⎞⎠⎟
For N spins Z=Z1N since they are distinguishable
Z = 2N coshN mBτ
⎛⎝⎜
⎞⎠⎟
⇒U = τ 2 ∂ logZ∂τ
= −NmB tanh mBτ
⎛⎝⎜
⎞⎠⎟⇒ Magnetization M = −
UVB
= nm tanh mBτ
⎛⎝⎜
⎞⎠⎟
B
M
τ
M
mBτm
Phys 112 (S2009) 05 Boltzmann-Gibbs 14
Indistinguishable SystemsIndistinguishable:in quantum mechanics, free particles are
indistinguishableWe should count permutations of states among particles only once
Therefore if we have N indistinguishable systems
This is a low occupation number approximation due to Gibbs ( no significant probability to have 2 particles in the same state: we will
see how to treat the case of degenerate quantum gases)
e.g. for 2 particles Z 1( )Z 2( ) = e−εs1τ
s1∑ e
−εs2τ
s2
∑ overcounts s1, s2 ≡ s2 , s1for s1 ≠ s2we have to divide by 2
More generally if we have N identical systems, the Nth power of Z1can be expanded as
(a1 + ...+ as + ....am )N =N !
n1!...ns !...nm !ni∑ a1
n1 ..asns ...am
nm
If we have a large number of possible states, most of the terms are of the form N !aiaj ...akN terms
corresponding to N different states which are singly occupied. If the multiple occupation ofstates is negligible we are indeed overcounting by N !
Z N( ) ≈ 1N!
Z 1( )( )N
Phys 112 (S2009) 05 Boltzmann-Gibbs 15
Ideal GasN atoms
Indistinguishable
Interpretation: nQ as quantum concentrationde Broglie wave length
λ = hMv
≈ h3Mτ
⇐ 12Mv2 ≈ 3
2τ
1
λ3≈3Mτ
h2⎛⎝⎜
⎞⎠⎟
32 ≈
Mτ
2π2⎛
⎝⎜⎞
⎠⎟
32 = nQ
ZN =1N!
Z1( )N =1N!
nQV( )N
logZN = N lognQn
⎛⎝⎜
⎞⎠⎟+ N with n = N
V= concentration
nQ =Mτ2π2
⎛⎝⎜
⎞⎠⎟
32
logN!≈ N logN − N⇒ log ZN = N log nQVN
⎛ ⎝ ⎜ ⎞
⎠ ⎟ + N
Phys 112 (S2009) 05 Boltzmann-Gibbs 16
Ideal gas lawsEnergy
equipartition of energy 1/2τ per degree of freedom
Free energy
Pressure
Entropy: Sackur Tetrode identical to our expression counting states
if extra degrees of freedom, additional terms!Well verified by experimentInvolves quantum mechanics
Chemical potential
F = −τ log Z = −τN lognQn
⎛ ⎝ ⎜
⎞ ⎠ ⎟ −τN
U = τ2∂log Z∂τ
=32Nτ =
32NkBT
σ =SkB
= −∂F∂τ
= N lognQn
⎛ ⎝ ⎜
⎞ ⎠ ⎟ +
52
⎡ ⎣ ⎢
⎤ ⎦ ⎥
S =dQT0
T∫
µ =∂F∂N
=∂ −τN log
nQVN
⎛⎝⎜
⎞⎠⎟− τN
⎛⎝⎜
⎞⎠⎟
∂N= τ log n
nQ
⎛
⎝⎜⎞
⎠⎟
p = −∂F∂V
=τNV
⇒ pV = Nτ = NkBT with nQ =
Mτ2π2
⎛⎝⎜
⎞⎠⎟
32
Phys 112 (S2009) 05 Boltzmann-Gibbs 17
Barometric Equation: Three different waysµ=constant
Thermodynamic equilibrium between slabs at various z
Single particle occupancy of levels at different altitude
Hydrodynamic equilibrium
Generalizes to case when the temperature is not constant
At constant temperature µtotal=C ⇔τ logn z( )nQ
= −Φ z( ) + C ⇔ n z( ) = n0 exp −mgzτ
⎛⎝⎜
⎞⎠⎟
n z( )∝ prob εK , z( )∝ exp −εK +Φ z( )
τ⎛⎝⎜
⎞⎠⎟⇒ n z( ) = n0 exp −
mgzτ
⎛⎝⎜
⎞⎠⎟
− p + dp( )dA
pdA−mgn z( )dAdz
pdA − p + dp( )dA − mgn z( )dAdz = 0⇒ 1n z( )
∂p∂z
= −mg
pV = Nτ ⇔ p z( ) = n z( )τ ⇒1
n z( )∂n∂z
= −mgτ
Phys 112 (S2009) 05 Boltzmann-Gibbs 18
Internal Degrees of Freedom (Optional)In general, we need to add internal degrees of freedom
e.g. for multi-atomic molecules Rotation ≈ 10-2 eV
Vibration ≈ 2 10-1 eV similar to harmonic oscillator
Spin in a magnetic field etc..
Rotation energy ε j = j j +1( )ε0 with multiplicity 2 j+1
Zint = 2 j+1( )j∑ exp −
j j +1( )ε0
τ⎛⎝⎜
⎞⎠⎟≈ 2 j+1( )
−1/2
∞
∫ exp −j j +1( )ε0
τ⎛⎝⎜
⎞⎠⎟dj
= τεo
exp −x( )x0
∞
∫ dx with x = j j +1( )ε0
τ dx =
2 j +1( )ε0
τdj, x0 = −
ε0
4τ
⇒ Zint ≈τεo
exp εo4τ
⎛⎝⎜
⎞⎠⎟≈τεo
for τ >>ε0 ⇒ for τ >>ε0 U =τ 2∂ logZ
∂τ= τ Crot = kB
For τ < < ε0, keep the first terms of the discrete sum Zint ≈ 1+ 3exp −2 εoτ
⎛⎝⎜
⎞⎠⎟+ ...
logZint ≈ 3exp −2 εoτ
⎛⎝⎜
⎞⎠⎟ U =
τ 2∂ logZ∂τ
≈ 6ε0 exp −2 εoτ
⎛⎝⎜
⎞⎠⎟ Crot ≈ 12kb
εoτ
⎛⎝⎜
⎞⎠⎟
2
exp −2 εoτ
⎛⎝⎜
⎞⎠⎟≈ 0
Zint = exp −nωτ
⎛⎝⎜
⎞⎠⎟n=0
∞
∑ =1
1− exp − ωτ
⎛⎝⎜
⎞⎠⎟≈
τω
for τ >>ω ⇒ Cvib = kB
-1/2 0 1 2 3 4 5 j
In most cases, the kinetic and internal degrees of freedomare independent, for one particle Z = Zkin
kinetic Zint
internal
Diatomic moleculescf. KK fig 3.9 kB
log τ
CV
kBRotation
Vibration
Slightly betterapproximationthan 0
Phys 112 (S2009) 05 Boltzmann-Gibbs 19
Exchanging Particles with an Ideal gasLet us consider again a system with an energy level ε , which can now be occupied by a variable number N of particles
We put in contact with a thermal reservoir constituting on a large number Np-N of free particles. Energy and particles can be exchanged with the reservoir.
We consider a single state with N particles and an energy NεWe can compute the number of states of the overall system as a
function of Ν and ε .
gRes
N
exp −ε − µ( )N
τ⎛⎝⎜
⎞⎠⎟
g = gRes ×1 = e5 /2
M2π2
2 U − Nε( )3(Np − N )
⎛
⎝⎜⎞
⎠⎟
3/2
Np − NV
⎛⎝⎜
⎞⎠⎟
⎡
⎣
⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥
Np −N
<=Direct Counting
or Seckur Tetrode
= e5 /2 1+ NNp − N
⎛
⎝⎜⎞
⎠⎟
5 /2⎡
⎣⎢⎢
⎤
⎦⎥⎥
Np −NM2π2
2U3Np
⎛
⎝⎜⎞
⎠⎟
3/2
Np
V
⎡
⎣
⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥
Np −N
1− NεU
⎛⎝⎜
⎞⎠⎟3/2⎡
⎣⎢⎢
⎤
⎦⎥⎥
Np −N
=
Np→∞
U =32Npτ
⎯ →⎯⎯⎯ e5 /2nQn
⎡⎣⎢
⎤⎦⎥
Np
exp −N lognQn
⎛⎝⎜
⎞⎠⎟exp −
Nετ
⎛⎝⎜
⎞⎠⎟∝ exp −
N ε − µ( )τ
⎛⎝⎜
⎞⎠⎟⇐
lim 1+ aN
⎛⎝⎜
⎞⎠⎟N
N→∞⎯ →⎯⎯ exp a( )
bN ≡ exp −N logb( )
Phys 112 (S2009) 05 Boltzmann-Gibbs
S,ε
20
Gibbs factorSystem in contact with thermal bath (reservoir)
Exchange of particles
# States in configuration
At equilibrium
Notes: • Z = grand partition function
Total system =Our system + reservoirConfiguration: 1 state s of Sof energy εs and # of particles Nis+ energy of reservoir
R , Uo-ε
1× gR Uo − ε s ,Noi − Nis( ) Prob configuration( )∝ gR Uo − ε s ,Noi − Nis( )
⇒
Pr ob state ε1
, Ni 1
( )
Pr ob state ε2
, Ni 2
( )=
gR
Uo− ε
1, N
oi− N
i1( )
gR
Uo− ε
2, N
oi− N
i 2( )
= exp −∂σ
R
∂Uε 1 − ε 2( ) −
∂σR
∂Ni
Ni1 − Ni 2( )
i
∑⎛ ⎝ ⎜ ⎞
⎠ ⎟
Absolute activity λi = exp µi
τ⎛ ⎝
⎞ ⎠
⇔ Prob state s with ε,Nis( ) = 1Z
exp −ε s,Nis( ) − µiNis
i∑
τ
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟ =
1Z
exp −ε s,Nis( ) − µiNis
i∑
kBT
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟
with Z = exp −ε s,Nis( ) − µiNis
i∑
τ
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟all states s,Nis
∑
Phys 112 (S2009) 05 Boltzmann-Gibbs 21
Grand Partition FunctionDefinition
Note: ε(s,Ν) depends on Ni both through through increase of N, and the interactions
In weak interaction limit (ideal gas) and only 1 species ,
Note If the Nis =Ni are constant,
the probabilities probabilities are the same as with Z
This formalism is useful only if the Nis change.
Z = expµiNi
i∑
τ
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟ Z Ni( )
Z = expµiNis −
i∑ ε s,Nis( )
τ
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟ all states s
and occupation Nis
∑
ε s,Ns( ) = Ns ε senergy of each level
⇒ Z = exp −ε s − µ( )Ns
τ⎛⎝⎜
⎞⎠⎟s,Ns
∑
Prob Ns particles in state s( ) = 1Z
exp −ε s − µ( )Ns
τ⎛⎝⎜
⎞⎠⎟
Phys 112 (S2009) 05 Boltzmann-Gibbs 22
Examples few state systems Kittel-Kroemer 140-145
Gas adsorption (heme in myoglobin, hemoglobin, walls)Assume that each potential adsorption site can adsorb at most one gas molecule. What is the fraction
of sites which are occupied?State 1= 0 molecule => energy 0State 2= 1 molecule => energy ε
If ideal (applies equally well to gas and dissolved gas and any low concentration species)
Langmuir adsorption isotherm
Z = 1+ exp µ − ε
τ⎛⎝⎜
⎞⎠⎟
Fraction occupied = f =exp µ − ε
τ( )1 + exp µ − ε
τ( )=
1
exp ε − µτ( ) +1
= same form as Fermi- Dirac
f = p
τ nQ expετ( ) + p
µgas = τ log nnQ
⎛
⎝ ⎜ ⎞
⎠ ⎟ = τ log p
τnQ
⎛
⎝ ⎜ ⎞
⎠ ⎟ ⇐ pV = Nτ n =
NV
=Pτ
Phys 112 (S2009) 05 Boltzmann-Gibbs 23
Ionized impurities in a semiconductor cf Kittel and Kroemer p370
Donor= tends to have 1 more electron (e.g. P in Si)State 1= ionized D+ : no electron => energy= 0State 2=neutral 1 extra electron spin up => energy =εd
State 3= neutral 1 extra electron spin down => energy =εd
Acceptor= tends to have 1 fewer electron (e.g. Al in Si)State 1= ionized A- : one electron completing the bond=> energy= εa
State 2= 1 neutral: missing a bond electron spin up => energy =0 State 3= 1 neutral: missing a bond electron spin down => energy =0
µ determined by electron concentration in semiconductor = Fermi level
Z=1+ 2exp −εd −µτ
⎛⎝⎜
⎞⎠⎟
Fraction ionized f = 1
1+ 22 spin stateswhen donoris neutral
exp − εd − µτ
⎛⎝⎜
⎞⎠⎟
εd = energy of electron on donor site K&K use ionization energy I=-ε d
Fraction ionized f =exp − εa − µ
τ⎛⎝⎜
⎞⎠⎟
exp − εa − µτ
⎛⎝⎜
⎞⎠⎟ + 2
2 spin stateswhen acceptor
is neutral
=1
1+ 2exp − µ − εaτ
⎛⎝⎜
⎞⎠⎟
εa = energy of electron on acceptor site
Phys 112 (S2009) 05 Boltzmann-Gibbs 24
Grand Partition FunctionRelation to thermodynamic functions
• Mean number of particles
• Entropy
• Energy
• Free Energy
• Grand Potential Ω defined as
σ = − ps,NisNis∑
s∑ log ps,Nis( ) = logZ−
µi
τNis
i∑ +
ετ
=∂ τ logZ( )
∂τ V ,µi
U = ε = τ 2 ∂ logZ
∂τ+ µi Ni
i∑ = τ 2 ∂
∂τ+ τµi
∂∂µii
∑⎛⎝⎜
⎞⎠⎟logZ
F =U − τσ = −τ log Z+ µi Ni
i∑ G = F + pV
Ω ≡ F − µi Ni
i∑ = −τ logZ
Ni = Nis ps,NisNis∑
s∑ =
Nis
Zexp
µiNis −i∑ ε s,Nis( )
τ
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟Nis
∑s∑ = τ ∂ logZ
∂µi τ ,V
Phys 112 (S2009) 05 Boltzmann-Gibbs 25
FluctuationsMicroscopic exchanges
In a system in equilibrium, exchanges still go on at the microscopic level! They are just balanced!
Macroscopic quantities= sum of microscopic quantities ≈ N x mean
But with a finite system, relative fluctuations on such sum is of the order 1/√N
e.g., fluctuation on total energy
Computation with partition function; By substitution we can see that
Similarly when there is exchange of particles
U = ε = ε s pss∑ = ε s
e−ε sτ
Zs∑
σE2 = E − E( )2 = E2 − E 2 = εs
2 e−εsτ
Zs∑ − εs
e−εsτ
Zs∑⎛
⎝
⎜ ⎜ ⎜
⎞
⎠
⎟ ⎟ ⎟
2U = E =
τ 2∂ log Z∂τ
σE2 = E2 − E 2 =
τ2∂U∂τ
=τ 2∂τ
2∂log Z∂τ
∂τ
Variancenot entropy!
N =
τ∂ logZ
∂µ τ ,V
and σ N2 = N 2 − N 2 =
τ∂τ∂ logZ
∂µ∂µ
=τ∂ N∂µ
Phys 112 (S2009) 05 Boltzmann-Gibbs
ExamplesFermi Dirac
Fermions= particles with half integer spinPauli exclusion principle: each state contains at most one particle
26
For a state of energy ε
Z = 1+ exp −ε − µτ
⎛⎝⎜
⎞⎠⎟
N =τ∂ logZ
∂µ τ ,V
=exp − ε − µ
τ⎛⎝⎜
⎞⎠⎟
1+ exp − ε − µτ
⎛⎝⎜
⎞⎠⎟=
1
exp ε − µτ
⎛⎝⎜
⎞⎠⎟ +1
= Prob 1 particle( )
σ N2 =
τ∂ N∂µ
=exp ε − µ
τ⎛⎝⎜
⎞⎠⎟
exp ε − µτ
⎛⎝⎜
⎞⎠⎟ +1
⎡⎣⎢
⎤⎦⎥
2 = N 1− N( ) ≤ N Less fluctuation than a Poisson distribution
Phys 112 (S2009) 05 Boltzmann-Gibbs
ExamplesBose Einstein
Bosons= particles with integer spin
27
For a state of energy ε
Z = exp −ε − µ[ ]N
τ⎛⎝⎜
⎞⎠⎟N
∑ =1
1− exp − ε − µτ
⎛⎝⎜
⎞⎠⎟
N =τ∂ logZ
∂µ τ ,V
=exp − ε − µ
τ⎛⎝⎜
⎞⎠⎟
1− exp − ε − µτ
⎛⎝⎜
⎞⎠⎟=
1
exp ε − µτ
⎛⎝⎜
⎞⎠⎟ −1
σ N2 =
τ∂ N∂µ
=exp ε − µ
τ⎛⎝⎜
⎞⎠⎟
exp ε − µτ
⎛⎝⎜
⎞⎠⎟ −1
⎡⎣⎢
⎤⎦⎥
2 = N 1+ N( ) ≤ N More fluctuation than a Poisson distribution
Phys 112 (S2009) 05 Boltzmann-Gibbs 28
Minimization of the Free Energy Consider a system in interaction with reservoir imposing
temperatureConstant exchange of energy (+ in some case volume, particles)=> Fluctuations, probability distributionContrary to a system which is able to loose energy to vacuum, system does
not evolve to state of minimum energy (constantly kicked)!Contrary to an isolated system, does not evolves to a configuration of
maximum entropy! (The entropy of the combination of the reservoir and the system is maximized)
It evolves to a configuration of minimum Free Energy“Balance between tendency to loose energy and to maximize entropy.”
More rigorouslyWe cannot define the temperature of the system out of equilibrium. So we
introduce
They are minimized for the most probable configuration(≈equilibrium)! Can be generalized to any variable! (Will be useful for phase transitions)
FL US( ) ≡US − τ Rσ S US( ) Landau Free Energy (constant volume)GL US ,Vs( ) ≡US − τ Rσ S US ,VS( ) + pRVS Landau Free Enthalpy (constant pressure)
Phys 112 (S2009) 05 Boltzmann-Gibbs 29
Landau Free EnergyLet us consider a system S exchanging energy with a large reservoir R.(We are no more considering a single state for system S).
S,USR , U-Us
FL
US
Constant volumeσ tot US( ) = σ R U −US( ) +σ S US( )
σ R U( ) − ∂σ R
∂UUS +σ S US( )
= σ R U( ) − 1τ RUS +σ S US( )
= σ R U( ) − 1τ R
FL US( ) with
FL US( ) ≡US − τ Rσ S US( ) ≡ Landau Free EnergyMost probable configuration maximizes σ tot i.e. minimizes FL US( )Note that this implies
∂FL US( )∂Us
= 0⇔∂σ S US( )∂US
=1τ R
e.g. for a perfect gas
σ S US( ) = N log
MUS
3π2N⎛⎝⎜
⎞⎠⎟3/2
NV
⎛⎝⎜
⎞⎠⎟
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥
+52
⎧
⎨⎪⎪
⎩⎪⎪
⎫
⎬⎪⎪
⎭⎪⎪
(Sackur Tetrode with τS =2Us
3N)
⇒US =32Nτ R
Phys 112 (S2009) 05 Boltzmann-Gibbs
Exchange of ParticlesIf we exchange particles, what is minimized is
in a way similar to the previous slide
30
ΩL US ,NS( ) ≡US − τ Rσ S US ,NS( ) −µRNS Landau Grand Potential
σ tot US ,NS( ) = σ R U −US ,N − NS( ) +σ S US ,NS( ) σ R U( ) − ∂σ R
∂UUS −
∂σ R
∂NNS +σ S US ,NS( )
= σ R U( ) − 1τ RUS +
µR
τ RNS +σ S US ,NS( )
= σ R U( ) − 1τ R
ΩL US ,NS( ) with
Most probable configuration maximizes σ tot i.e. minimizes ΩL US ,NS( )Note that this implies
∂ΩL US ,NS( )
∂Us
= 0⇔∂σ S US ,NS( )
∂US
=1τ R
∂ΩL US ,NS( )∂Ns
= 0⇔∂σ S US ,NS( )
∂NS
= −µR
τ R