Statistická termodynamikaStatistical thermodynamics
Peter Koš[email protected]
Lecture 1, 4.10.2015, MC260P105, WS 2016/2017If you find a mistake, kindly report it to the author :)
Course requirements
New in 2016/17: Exercises
* Within the scope of syllabus* Will be refined after first lecture* Focused on understanding, not on
memorizing formulae!* Statistical sampling of student's
knowledge by a few randomquestions and answers
* Inherently imperfect* See the web page for more
information
Oral exam
* Working out some derivations in detail* Numerical solution of relevant
problems* Practical preparation for the exam* Anything else we agree upon.
Macroscopic result
Final grade
Ultimate goal: to enjoy the subject and learn new things.
Course material
Secondary
* Wikipedia* Internet search* Beware of bugs and
incompetent content all over theinternet!
* Verify by yourself whatever youobtain from an unreliable source
* D. McQuarrie: Statistical Mechanics,University Science Books, 2000
* L. Reichl: A modern course in statisticalPhysics, Wiley, 1998
* T. Boublík: Statistická termodynamika,Academia, 1996
* Many other recognized textbooks* A nice book available online:
http://homepage.tudelft.nl/v9k6y/imsst/index.html
* Lecture notes for this course (pdf)* Lecture notes of other statmech courses,
e.g. P. Nachtigall (this course, previousyears), M. Tuckermanhttp://www.nyu.edu/classes/tuckerman/stat.mech/lectures.html
Ultimate goal: to enjoy the subject and learn new things.
Primary
Thermodynamics from first principles
ThermodynamicsMolecules
Macroscopic system
Sample in a containern=1mol, T=300K, V=1L
* Positions* Momenta* Mass* Charge, polarizability* Electronic states
* Kinetic energy* Potential energy* Interactions
* Pressure* Density* State (solid, liquid, gas)
* Entropy* Internal energy* Free energies
(Gibbs or Helmholtz)
Statistical
thermodynamics
(mechanics)
Predict thermodynamic properties
from molecular information
Thermodynamics from first principles
ThermodynamicsMolecules
Macroscopic system
Sample in a containern=1mol, T=300K, V=1L
* Positions* Momenta* Mass* Charge, polarizability* Electronic states
* Kinetic energy* Potential energy* Interactions
* Pressure* Density* State (solid, liquid, gas)
* Entropy* Internal energy* Free energies
(Gibbs or Helmholtz)
Statistical
thermodynamics
(mechanics)
Predict thermodynamic properties
from molecular information
Few parametersO(1)
ManyparametersO(1024)
Example application(current research)
Titration of a weak polyelectrolyte
0
0.2
0.4
0.6
0.8
1
-2 0 2 4
εr=801.4e-02 mol/l
Degre
e o
f io
niz
ation,
α
pH-pKA
N = 125
102050
100200
* Simple model* Molecular simulation* Dissociation equilibrium* Highly nonideal
Prediction from theory or simulation is cheap
Reduces the amount of necessary experiments
0
0.2
0.4
0.6
0.8
1
-2 0 2 4
cpol [mol/l]
εr=80
Degre
e o
f io
niz
ation,
α
pH-pKA
1.4e-016.8e-022.7e-021.4e-02
Ideal
First principles
Quantum mechanics
HΨ = EΨ
1
Schrödinger: Particle in a box
* Discrete set of allowed states* Probabilistic character
* Numerically tractable for small systems(<103 electrons)
* For bigger systems too demanding
Classical mechanics
Newton:
* Full description in terms of p, q and time* Still too demanding for a macroscopic system
~ 1024 particles
Solution: statistics!
* Macroscopic system properties represent anaverage over individual molecules
? What is the probability distribution?
Enx,ny,nz =h2
8ma2(n2x + n2y + n2z
)
1
dpdt ≡ p = F
pi,q,i , t0
1
Postulates of statistical TD
1. Ensemble average corresponds to TD average
An average value of an arbitrary mechanical property of a macroscopic system is equalto the mean value for the statistical ensemble.
An isolated system with a constant number of particles, N, volume V and energy E isequally likely to be in any of its Ω(E) available quantum states.
2. Equal a priori probabilities
? What is an ensemble?N, V, E N, V, E N, V, E
N, V, EN, V, EN, V, E
N, V, E N, V, E N, V, E
* Mental construction of a set of systemscharacterized by the same TD parameters
* Isolated from the universe by impermeableadiabatic walls
* Many systems in different quantum stateswhich lead to the macroscopic parameters
* Equal a priori probabilities* Thermal equilibrium
* Mental construction of a set of systemscharacterized by the same TD parameters
* Isolated from the universe by impermeableadiabatic walls
* Many systems in different quantum stateswhich lead to the macroscopic parameters
* Equal a priori probabilities* Thermal equilibrium
Various ensembles
N, V, E N, V, E N, V, E
N, V, EN, V, EN, V, E
N, V, E N, V, E N, V, E
N, V, T N, V, T N, V, T
N, V, TN, V, TN, V, T
N, V, T N, V, T N, V, T
μ, V, T μ, V, T μ, V, T
μ, V, Tμ, V, Tμ, V, T
μ, V, T μ, V, T μ, V, T
Microcanonical Canonical Grandanonical
* Systems in differentquantum states
* Impractical: cannotmeasure total energy ofa system
* Systems can exchangeenergy
* Cannot exchangeparticles
* N, V, T all measurable
* Systems can exchangeparticles
* Convenient for smallsystems in equilibriumwith a reservoir
For N →∞ they become equivalent – matter of convenience
Fluctuations can be important at small N (computer simulations)
Excursion to statistics
Probability P: discrete and continuous random variable x:
1 =∑j
Pj, 1 =∫P(x)dx
Expectation value (mean):
E[x] = µ =∑j
xjPj, E[x] = µ =∫xP(x)dx
More general – n-th moment around c:
µn =∑j
(xj − c)nPj, µn =∫(x− c)nP(x)dx
Variance – 2nd central moment:
Var[x] = σ2 =∑j
(xj−µ)2Pj, Var[x] = σ2 =∫(x−µ)2P(x)dx
• Higher moments: n = 3 skewness, n = 4 kurtosis
More on statisticsDraw N samples of a random variable (continuous case analogous)Sample mean and mean (expected) value
x = 1N
N∑j=1
xj, limN→∞
x = E[x]
Sample variance and variance
s2x =1N
N∑j=1
(xj − x)2 = x2 − x2, limN→∞
s2x(x) = Var[x]
Variance of the mean
Var(x) = Var(1N
N∑j=1
xj)= s2x
N
•With increasing N variance of the mean vanishesNumber of ways to select k elements out of N:(
Nk
)= N!k!(N− k)!
Microcanonical ensemble
? What is the distribution of states?
•Many systems, no exchange of energy or particles
• Total number of particles (spins) N• Total energy E , volume V = Nv (values unimportant)
Ising model – Lattice with spins
• n spins up (s = +1), (N− n) spins down (s = −1)• Special case – No interaction between the spins
• All states have the same energy
•Magnetization: M =∑
i si = 2n−N
Possible ways to select n spins up – statistical weight:
W(n) =(Nn
)= N!n!(N− n)!
Probability of a state with given n – binomial distribution
P(n) =W(n) /( N∑n=0
W(n))=(Nn
)(12
)N
Image source: wikipedia.org
Canonical ensemble
N, V, T N, V, T N, V, T
N, V, TN, V, TN, V, T
N, V, T N, V, T N, V, T
? What is the distribution of states?
• Ensemble composed of A systems•Can exchange heat but not particles• Total volume V = AV• Total energy E (exact value unimportant)
Quantum states of the system :State number 1 2 3 . . . mEnergy E1 ≤ E2 ≤ E3 ≤ . . . EmSystems in state j a1 a2 a3 . . . am(occupation number)
Distribution of occupation numbers(m-dimensional vector) a = aj
Possible ways to select occupation numbers:
W(a) = A!a1!a2!a3! ...am!
= A!∏iak!
1
Constraints:∑j
aj = A
∑j
ajEj = E
1
Most probable distribution
Probability of a certain occupation number:
Pj =ajA= 1
A
∑a aj(a)W(a)∑
aW(a)where aj is the average over all possible distributions
1
Mean in termsof probability:M =
∑j
MjPj
1
Knowledge of Pj allows
us to calculate any
mechanical property M
(first postulate) .
Most probable distribution
Mean in termsof probability:M =
∑j
MjPj
1
Probability of a certain occupation number:
Pj =ajA= 1A
∑a aj(a)W(a)∑
aW(a)
The most probable distribution a∗:
Pj =1Aa∗jW(a∗)W(a∗) =
a∗jA
= ajA
1
Most probable distribution
Probability of a certain occupation number:
Pj =ajA= 1A
∑a aj(a)W(a)∑
aW(a)
The most probable distribution a∗:
Pj =1Aa∗jW(a∗)W(a∗) =
a∗jA
= ajA
Search for maximum of W(a) (equiv. lnW(a))∂
∂aj
lnW(a)− α
∑k
ak − β∑k
akEk= 0, j = 1,2, 3 ...
1
Constraints:∑j
aj = A
∑j
ajEj = AE
1
Mean in termsof probability:M =
∑j
MjPj
1
Lagrange multipliers
Most probable distribution
Probability of a certain occupation number:
Pj =ajA= 1A
∑a aj(a)W(a)∑
aW(a)
The most probable distribution a∗:
Pj =1Aa∗jW(a∗)W(a∗) =
a∗jA
= ajA
Search for maximum of W(a) (equiv. lnW(a))∂
∂aj
lnW(a)− α
∑k
ak − β∑k
akEk= 0, j = 1,2, 3 ...
Use Stirling’s approximation to lnW(a):− lna∗j − α− βEj = 0, j = 1,2, 3 ...
Finally:a∗j = e−αe−βEj, j = 1,2, 3 ...
1
Constraints:∑j
aj = A∑j
ajEj = AE
Stirling:limN→∞
lnN! = N lnN−N
1
Mean in termsof probability:M =
∑j
MjPj
1
Evaluation of α and β
Use
a∗j = e−αe−βEj and∑j
aj = A to obtain eα = 1A∑j
e−βEj,
From
Pj =a∗jA
we get Pj =e−βEj(N,V)∑je−βEj
and define Q =∑j
e−βEj(N,V)
then we can write E as
E(N,V, β) =∑j
EjPj =1Q∑j
Ej(N,V)e−βEj(N,V)
From mechanics:
dEj = −pjdV thus pj = −(∂Ej∂V
)N
hence
p =∑j
pjPj = −1Q∑j
(∂Ej∂V
)Ne−βEj(N,V)
1
Use
a∗j = e−αe−βEj and∑j
aj = A to obtain eα = 1A∑j
e−βEj,
From
Pj =a∗jA
we get Pj =e−βEj(N,V)∑je−βEj
and define Q =∑j
e−βEj(N,V)
then we can write E as
E(N,V, β) =∑j
EjPj =1Q∑j
Ej(N,V)e−βEj(N,V)
From mechanics:
dEj = −pjdV thus pj = −(∂Ej∂V
)N
hence
p =∑j
pjPj = −1Q∑j
(∂Ej∂V
)Ne−βEj(N,V)
1
Evaluation of α and β
Partition function
Partition function is the summit ofstatistical mechanics.Richard P. Feynmann: Introduction to Statistical
Mechanics, Westview Press, 1998
. . .Evaluation of α and β
Recall that
E = 1Q∑j
Ej(N,V)e−βEj(N,V) and p = −1Q∑j
(∂Ej∂V
)Ne−βEj(N,V) .
With some manipulations we can express(∂E∂V
)N,β
= −p+ βEp− βEp .
(∂p∂β
)N,V
= Ep− Ep .
Compare this with classical thermodynamics(∂E∂V
)N,T
− T(∂p∂T
)N,V
= −p =(∂E∂V
)N,T
+ 1T
(∂p∂1/T
)N,V
.
We conclude that
β = 1kT ,
where k is a constant (to be determined as Boltzmann constant kB).
1
Alternative evaluation of β
Recall that
E = 1Q
∑j
Ej(N,V)e−βEj(N,V)
Consider an ideal gas: kinetic energy per particle, ε = 32kBT:
ε =∑
j εjNj∑jNj
and εj =p2j,x2m +
p2j,y2m +
p2j,z2m
Combinig the above equations, we get (equivalently for x, y, z)
ε
3 =∑
jp2j,x2me−β
∑j p2j,x
2m∑j e−β
∑j p2j,x
2m
≈
∞∫0p2xe−β
∑j p2j,x
2m dpx
2m∞∫0e−β
∑j p2j,x
2m dpxthen use
∞∫0x2e−ax2dx =
√π
16a3
∞∫0e−ax2dx =
√π4a
to obtain
ε = 32β and finally β = 1
kBT
1
Introducing entropy
Total derivative of E:
dE =∑j
EjdPj+∑j
PjdEj =∑j
EjdPj+∑j
Pj(∂Ej∂V
)NdV
1
Absorbed heat
Reversible work
Introducing entropy
Total derivative of E:
dE =∑j
EjdPj+∑j
PjdEj =∑j
EjdPj+∑j
Pj(∂Ej∂V
)NdV
Rearrange the equation for Pj:
Ej =1β
(− lnPj − lnQ
)⇒
∑j
EjdPj =1βd(−∑j
Pj lnPj)
because∑j
dPj = 0
From thermodynamics:
dqrev = TdS ⇒ dS = d(−kB
∑j
Pj lnPj)
Finally:
S = −kB∑j
Pj lnPj+ C with C = 0 from third law of TMD
1
Absorbed heat
Reversible work
Thermodynamic quantities from Q(N,V,T)
Internal energy
U = E = 1Q∑j
Eje−βEj(N,V) = kBT2(∂ lnQ∂T
)N,V
= kBT(∂ lnQ∂ lnT
)N,V
.
Pressure:
p = p = −1Q∑j
(∂Ej∂V
)Ne−βEj(N,V) = kBT
(∂ lnQ∂V
)N,T
.
Entropy:
S = −kB∑j
Pj lnPj = kB lnQ+ kBT(∂ lnQ∂T
)N,V
.
Enthalpy:
H = U+ pV = kBT[(
∂ lnQ∂ lnT
)N,V
+(∂ lnQ∂ lnV
)N,T
]
1
From previous slide:
U = kBT2(∂ lnQ∂T
)N,V
= kBT(∂ lnQ∂ lnT
)N,V
, p = kBT(∂ lnQ∂V
)N,T
,
S = kB lnQ+ kBT(∂ lnQ∂T
)N,V
, H = kBT[(
∂ lnQ∂ lnT
)N,V
+(∂ lnQ∂ lnV
)N,T
].
We continue with Gibbs free energy:
G = H− TS = −kBT[lnQ+
(∂ lnQ∂ lnV
)N,T
]and Helmholtz free energy:
A = E− TS = −kBT lnQ
1
More thermodynamic quantities . . .
From previous slide:
U = kBT2(∂ lnQ∂T
)N,V
= kBT(∂ lnQ∂ lnT
)N,V
, p = kBT(∂ lnQ∂V
)N,T
,
S = kB lnQ+ kBT(∂ lnQ∂T
)N,V
, H = kBT[(
∂ lnQ∂ lnT
)N,V
+(∂ lnQ∂ lnV
)N,T
].
We continue with Gibbs free energy:
G = H− TS = −kBT[lnQ+
(∂ lnQ∂ lnV
)N,T
]and Helmholtz free energy:
A = E− TS = −kBT lnQ
1
More thermodynamic quantities . . .
A is a characteristicfunction of canonical
ensemble.
More thermodynamic quantities . . .
From previous slide:
U = kBT2(∂ lnQ∂T
)N,V
= kBT(∂ lnQ∂ lnT
)N,V
, p = kBT(∂ lnQ∂V
)N,T
,
S = kB lnQ+ kBT(∂ lnQ∂T
)N,V
, H = kBT[(
∂ lnQ∂ lnT
)N,V
+(∂ lnQ∂ lnV
)N,T
].
We continue with Gibbs free energy:
G = H− TS = −kBT[lnQ+
(∂ lnQ∂ lnV
)N,T
]and Helmholtz free energy:
A = E− TS = −kBT lnQ
Replace sum over states j by sum overeigenvalues E with degeneracy Ω(N,V,E):
Q =∑j
e−βEj =∑E
Ω(N,V,E)e−βE
1
Rest of the course:how to obtain partition functionfrom microscopic parameters.
With Q, we have nowreached the summit.
Before that:partition functions of some othercommon ensembles.