Classical Statistical Mechanics
in the Canonical Ensemble:
Application to the Classical
Ideal Gas
Canonical Ensemble in Classical
Statistical Mechanics
• As we’ve seen, classical phase space for a
system with f degrees of freedom is f generalized
coordinates & f generalized momenta (qi,pi).
• The classical mechanics problem is done in
the Hamiltonian formulation with a
Hamiltonian energy function H(q,p).
• There may also be a few constants of motion
(conserved quantities):
energy, particle number, volume, ...
The Canonical Distribution in
Classical Statistical Mechanics
The Partition Function has the form:
Z ≡ ∫∫∫d3r1d3r2…d3rN d3p1d3p2…d3pN e(-E/kT)
A 6N Dimensional Integral!
• This assumes that we’ve already solved the classical
mechanics problem for each particle in the system
so that we know the total energy E for the N particles
as a function of all positions ri & momenta pi.
E  E(r1,r2,r3,…rN,p1,p2,p3,…pN)
CLASSICAL
Statistical Mechanics:
• Let A ≡ any measurable, macroscopic
quantity. The thermodynamic average of
A ≡ <A>. This is what is measured. Use
probability theory to calculate <A> :
P(E) ≡
[E/(k
T)]
e
/Z
B
<A>≡ ∫∫∫(A)d3r1d3r2…d3rN d3p1d3p2…d3pNP(E)
Another 6N Dimensional Integral!
Relationship of Z to Macroscopic Parameters
Summary of the Canonical Ensemble
(Derivations are in the book! Results are general & apply
whether it’s a classical or a quantum system!)
• Mean Energy: Ē  E = -∂(lnZ)/∂β
<(ΔE)2> = [∂2(lnZ)/∂β2]
β = 1/(kBT), kB = Boltzmann’s constant.
• Entropy:
S = kBβĒ + kBlnZ
An important, frequently used result!
Summary of the Canonical Ensemble:
•Helmholtz Free Energy:
F = E – TS = – (kBT)lnZ
Note that this gives: Z = exp[-F/(kBT)]
dF = S dT – PdV, so
S = – (∂F/∂T)V, P = – (∂F/∂V)T
• Gibbs Free Energy:
G = F + PV = PV – kBT lnZ.
•Enthalpy:
H = E + PV = PV – ∂(lnZ)/∂β
Summary of the Canonical Ensemble:
• Mean Energy:
Ē = – (lnZ)/
= - (1/Z) Z/
• Mean Squared Energy:
2
2
E = rprEr / rpr = (1/Z) 2Z/ 2.
• nth Moment:
n
n
n
n
n
E = rprEr / rpr = (-1) (1/Z) Z/
• Mean Square Deviation:
(ΔE)2 = E2 - (Ē)2 = 2lnZ/ 2 = Ē/ .
• Constant Volume Heat Capacity
CV = ( Ē/ T)V = ( Ē/
)(d /dT) = -
The Classical Ideal Gas
• So, in Classical Statistical Mechanics, the
Canonical Probability Distribution is:
P(E) =
-E/(kT)
[e
]/Z
Z ≡ ∫∫∫d3r1d3r2…d3rN d3p1d3p2…d3pN e(-E/kT)
• This is the tool we will use in what follows.
• As we.ve seen, from the partition function
Z all thermodynamic properties can be
calculated: pressure, energy, entropy….
• Consider an Ideal Gas from the point of
view of microscopic physics. It is the
simplest macroscopic system.
• Therefore, its useful use it to introduce the
use of the
Canonical Ensemble in Classical
Statistical Mechanics.
• The ideal gas Equation of State is
PV= nRT
n is the number of moles of gas.
• We’ll do Classical Statistical Mechanics,
but very briefly, lets consider the simple
Quantum Mechanics of an ideal gas & the
take the classical limit.
• From the microscopic perspective, an ideal
gas is a system of N non interacting
Particles of mass m confined in a volume
V = abc. (a, b, c are the box’s sides)
• Since there is no interaction, each molecule
can be considered a “Particle in a Box” as
in elementary quantum mechanics.
• Since there is no interaction, each
molecule can be considered a “Particle
in a Box” as in elementary quantum
mechanics.
• The energy levels for such a system
have the form:
where nx, ny, nx = integers
• The energy levels for each molecule in the
Ideal Gas have the form:
(1)
l
(nx, ny, nx = integers)
• The Ideal Gas molecules are non
interacting, so the gas Partition Function
has the simple form:
Z = (q)N (2)
where q  One Particle Partition Function
• Ideal Gas Partition Function:
Z = (q)N (2)
q  One Particle Partition Function
• Using (2) in the
Canonical Ensemble
formalism gives the
expressions on the
right for: mean energy
E, equation of state P
& entropy S:
• The Partition Function for the 1- dimensional
particle in a “box” under the assumption that the
energy levels are so closely spaced that the sum
becomes an integral over phase space can be written:
q   e Ei / kBT
i


n2 h2 
  exp  
dn 
2
 8ma kBT 
0
2 ma 2 kBT
h
(3)
• For the 3 – dimensional particle in a “box”, th
3 dimensions are independent so that the
Partition Function can be written as the product
of 3 terms like equation (3). That is:
 2mkbT  2 2 2
 2mkbT 
q  q x q y qz  
 abc  
V
h
h




3
3
(4)
 2mkbT  2 2 2
 2mkbT 
q  q x q y qz  
 abc  
V
h
h




3
3
• Now, using the
Canonical Ensemble
expressions from
before:
• Mean Energy & the
• Equation
To obtain, for one
mole of gas:
of State
can be obtained (per
mole):
3RT
E
2
RT
P
V
• The Entropy can also be obtained:
3/ 2


E   5
 V  4 m U

S ( N ,V ,U )  N k B ln  
  
2

  N  3h N   2 

(5)
• As first discussed by Gibbs, Entropy in
Eq. (5) is NOT CORRECT!
• Specifically, its dependence on particle
number N is wrong!
 “Gibbs’ Paradox”
in the first part of Ch. 7!