Mathematical tools and statistical physics
F.Calvayrac
*Introduction*
The purpose of statistical physics is to connect microscopic physics equations (such as the Schrödinger
equation or the Newton relation) to situations observed macroscopically, such as the laws of classical
thermodynamics, which will validate both models or calculate missing quantities or parameters from
experiments or fundamental considerations. Historically, Boltzmann was able to make great strides in
demonstrating the plausibility of the atomic hypothesis, unproven experimentally in his time.
Beyond two particles in question, the microscopic equations are not generally soluble mathematically, and
very sparingly soluble on a digital computer; situations are called "ergodic" or chaotic, which means that the
behavior of each atom or particle component system is almost unpredictable, although the equations that
describe the deterministic be in the case of classical mechanics; However, these equations are unstable, and
the slightest disturbance in the initial state of the system results in a complete divergence of solutions
( "butterfly effect") who ultimately explore randomly all the options open by conservation laws such as the
energy conservation law. The idea is then to make this weakness a strength, and calculate average macroscopic
quantities assuming microscopic quantities are random. For this, we must start with some probability reminders
that allow us to calculate the average values (or hopes), essential technique learned in this course with the usual
probability laws of physics (Boltzmann, Poisson, Gauss, Fermi Dirac, Bose-Einstein) we will successively
build expressions.
The present course (especially in terms of examples in magnetism and solid-state physics, treated in other
courses) was drawn from the following sources, plus online resources especially in English:
B.Diu D.Lederer B.Roulet Statistical Physics (Hermann Editions)
P.Atkins Physical Chemistry (Boeck Editions)
J.Rossel experimental and theoretical physics from Accurate (Editions du Griffon Neuchatel)
R.Balescu Equilibrium and nonequilibrium statistical mechanics (Wiley)
1. * probability * Reminders
The probability P is defined in the French tradition as the limit of the frequency of an event for an infinite
number of trials
Let 𝑤¯ be the number of favorable cases and 𝛺¯ the number of possible cases
𝑤¯
𝑃= ¯
𝛺
there are two possibilities:
* First case: the countable case 𝛺¯ = (𝑥! , 𝑥" , 𝑥# , , , , 𝑥$̄ )
* Second case : the uncountable case where the variable x is random, in this case as
𝑃(𝑋 ⩽ 𝑥) = 𝐹(𝑥)
&'())
𝑓(𝑥) =
&)
where 𝐹(𝑥) is the cumulative function
𝑃(𝑥 ∈ 𝐸) = 1
)∈,
𝑑 𝐹(𝑥) = 1
𝑓 (𝑥)𝑑𝑥
)∈,
In the Anglo-Saxon tradition (Bayesian) the probability of an event is the a priori probability of the event (i.e.
a value assigned before doing experiments, which can be changed when information is accumulated). This
approach is commonly used in signal processing.
1.1 * Law of Large Numbers *
As a complement to the previous definition, the law of large numbers states that the expectation of a variable
𝐸(𝑋) tends to average values, namely, the summation of the contribution of each possibility weighted by their
probabilities.
The average value of observations in the case of equiprobability reads for instance
𝑋- =
1
(𝑋 + 𝑋" + ⋯ … + 𝑋- )
𝑛 !
𝐸(𝑋) = ∫. 𝑋 𝑑𝐹(𝑋) in the continuous case
𝐸(𝑋) = ∑/ 𝑥/ 𝑓(𝑥/ )returning to the discrete case (by a Dirac comb); we find the "average" formula
example 1
Compute the expectation E (X) of a die
1 2 3 4 5 6
𝐸(𝑋) = : + + + + + @ = 3,5
6 6 6 6 6 6
example 2
Search the expectation E (X) of a roulette wheel that gives"36 times the bet" (the betting unit + 35 times this
vakye ... so the net gain is actually only 35 in the best case, and there a zero on the wheel that adds a possibility)
−1 ∗ 36 35
1
𝐸(𝑋) = :
+ @=−
37
37
37
So the more you play, the more one you are sure to lose money
The variance (standard deviation) is defined by
"
𝑉(𝑋) = 𝐸(𝑋 " ) − E𝐸(𝑋)F
1.2 Binomial Distribution * *
Let us consider n experiments giving a result "yes" or "no" with probability p.
Example
!
Consider n dice throws. We want to count the number of 1 with for example 𝑝 = the probability of having k
0
probability of success ( "yes") and (1-k) probability of failures. The order of these successes has no influence
on the result since the addition is commutative; we may for example have a "1" on the first throw OR the last
for the same amount.
The composed likelihood (with 'AND') is the product of the probabilities of each event if they are independent
(for k successes on n independent throws).
In the case of an 'OR' this probability is the sum of the probabilities of each event: it multiplies the various
possibilities by the factor
𝑛!
𝑘! (𝑛 − 𝑘)!
number of combinations = number of ways to have k successes for n trials regardless of the order.
𝐶-1 =
Finally by multiplying the elements of "and" success in the first and the second test for example and by adding
the elements of "or" (success in the first and second test OR the second and third OR the first and last ... .etc)
𝑓(𝑘, 𝑛, 𝑝) = 𝐶-1 𝑝1 (1 − 𝑝)-21
This distribution is called binomial because it reminds of the development (𝑎 + 𝑏)Exercise
Compute the expectation of this distribution 𝐸(𝑋)
-
𝐸(𝑋) = M 𝑘 𝑓(𝑘, 𝑛, 𝑝)
134
𝐸(𝑋) = ∑-134 𝑘 𝐶-1 𝑝1 (1 − 𝑝)-21
𝐸(𝑋) = ∑-134 N
1-!6! (!26)"#!
O
(-21)!1!
-!6! (!26)"#!
𝐸(𝑋) = ∑-134 N (-21)!(12!)! O
(-2!)!6 (!#%)(!26)("#!)
𝐸(𝑋) = 𝑛𝑝 ∑-13! :
(-21)!(12!)!
(-2!)!6(!) (!26)("#!#%)
@ = 𝑛𝑝 ∑-2!
134 :
(-212!)!(1)!
@
using the binomial theorem
(𝑎 + 𝑏)- = M 𝐶-1 𝑎1 𝑏 (-21)
1
thus, observing that
1=1
(-2!)
= E𝑝 + (1 − 𝑝)F
(-2!)
-2!
(𝑛 − 1)! 𝑝(1) (1 − 𝑝)(-212!)
= MP
Q
(𝑛 − 𝑘 − 1)! (𝑘)!
134
we obtain
𝐸(𝑋) = 𝑛𝑝
formula to remember for later.
Calculate now as an exercise
𝐸(𝑋 " ) = M 𝑘 " 𝑓(𝑘, 𝑛, 𝑝)
and we set 𝑚 = 𝑛 − 1𝑘 = 𝑠 + 1
and with a similar reasoning we obtain
𝐸(𝑋 " ) = 𝑛𝑝E(𝑛 − 1)𝑝 + 1F − (𝑛𝑝)"
The variance is therefore
"
𝑉(𝑋) = 𝐸(𝑋 " ) − E𝐸(𝑋)F = 𝑛𝑝E(𝑛 − 1)𝑝 + 1F − (𝑛𝑝)(𝑛𝑝) = 𝑛𝑝(1 − 𝑝)
1.3 * normal approximation*
𝑓(𝑘, 𝑛, 𝑝) = 𝐶-1 𝑝1 (1 − 𝑝)-21
is now supposed to be a real variable close to the expectation value (computed above) 𝑘 = 𝑛𝑝
the maximum is sought for 𝑓or equivalently (the logarithm being monotonous) for 𝑙𝑛(𝑓)
𝑙𝑛(𝑓) = 𝑙𝑛E𝐶-1 F + 𝑘𝑙𝑛(𝑃) + (𝑛 − 𝑘)𝑙𝑛E(1 − 𝑝)F
and since Stirling's formula (for the logarithm of the factorial) then gives approximately
𝑛!
𝐶-1 =
𝑘! (𝑛 − 𝑘)!
𝑙𝑛(𝑁!) ≈ 𝑁𝑙𝑛(𝑁) − 𝑁
𝑑𝑙𝑛(𝑓)
= 𝑙𝑛(𝑛 − 𝑘) − 𝑙𝑛(𝑘) + 𝑙𝑛(𝑝) − 𝑙𝑛(1 − 𝑝)
𝑑𝑘
if then 𝑘 = 𝑛𝑝
𝑑𝑙𝑛(𝑓)
=0
𝑑𝑘
therefore 𝑘 = 𝑛𝑝 corresponds to the maximum. Expanding 𝑙𝑛(𝑓)
to second order around this value
𝑙𝑛E𝑓(𝑘)F = 𝑙𝑛E𝑓(𝑛𝑝)F +
(𝑘 − 𝑛𝑝)" 𝑑" 𝑙𝑛E𝑓(𝑛𝑝)F
2!
𝑑𝑘 "
𝑑" 𝑙𝑛E𝑓(𝑘)F
1
1
=− −
"
𝑑𝑘
𝑘 𝑛−𝑘
if 𝑘 = 𝑛𝑝
𝑑" 𝑙𝑛E𝑓(𝑘)F
1
1
=−
=−
"
𝑑𝑘
𝑛𝑝(1 − 𝑝)
𝑉𝑎𝑟(𝑋)
this constitutes the so-called “normal” approximation (Gaussian curve).
#(!#"')(
𝑓(𝑘, 𝑛) = 𝑓(𝑛𝑝)𝑒 (!*+,(-)
Comparison of the histogram calculated with "drawers" of size 1, and obtained for the total of 100 random
numbers between 0 and 1 (experiment repeated on computer 10,000 times) and the normal approximation,
which with a variance of 10 coincides with the result of experiment. Indeed, the addition is commutative, the
result of this experiment obviously has a mean equals to 50, and is well described by the binomial distribution
(the order of the factors does not matter).
1.4 * Poisson’s distribution*
Let us assume now that 𝑓(𝑘, 𝑛) = 𝐶-1 𝑝1 (1 − 𝑝)-21
𝑝 ≪1
𝑛≫1
(Low probability event, experiment repeated many times: for example, number of horse kicks received by
soldiers in the Prussian army, first historical example of application of this law, or example of radioactive
disintegration).
So :
𝑓(𝑘, 𝑛)
𝐶-1 𝑝1 (1 − 𝑝)-21
𝑝(𝑛 − 𝑘 + 1)
= 12! 12!
=
𝑓(𝑘 − 1, 𝑛) 𝐶- 𝑝 (1 − 𝑝)-217!
𝑘 (1 − 𝑝)
8(1,-)
and if 𝑝 ≪ 1 𝑘 ≪ 𝑛 8(12!,-) =
61
This has been shown to have an expectation value of
𝐸 = 𝑛𝑝
then
𝑓(𝑘, 𝑛)
𝐸
=
𝑓(𝑘 − 1, 𝑛) 𝑘
Therefore
but
𝑓(𝑘 − 1, 𝑛)
𝐸
=
𝑓(𝑘 − 2, 𝑛) 𝑘 − 1
𝑓(0, 𝑛) = (1 − 𝑝)𝐸 1 (1 − 𝑝)𝑓(𝑘, 𝑛) =
𝑘!
secondly we assumed
𝑝≪1
𝑓(0, 𝑛) = (1 − 𝑝)- ≈ 1 − 𝑛𝑝
then
(1 − 𝑛𝑝) ≈ 𝑒 2, (1 − 𝑝)- ≈ 𝑒 2,
the two equations lead to:
𝐸 1 𝑒 2,
𝑓(𝑘, 𝑛) =
𝑘!
called Poisson’s distribution
Since
𝑒, = 1 + 𝐸 +
Normalization condition is satisfied
𝐸²
𝐸1
+. . . . + +. . . ..
2!
𝑘!
:
M 𝑓 (𝑛, 𝑘) = 𝑒 , 𝑒 2, = 1
134
The Poisson distribution gives us the probability that a certain number of events occur for a given time interval.
Understandably, the probability that 0 events occur decreases rapidly with time, the probability that one event
occur is low for short time, increases close to the time interval considered, and decreases for longer time while
the probability of having more than one event increases, and so on.
Numerical approximation p = 0.02 n = 90 k = 4
graph of f (k, n) as a function of k
* 2. Distribution of micro-canonical states *
We will start by considering an isolated microscopic situation treated first in quantum mechanics, but assuming
that the atoms or molecules constituting the system are identical, but "discernible" (Boltzmannions) in the
sense that one can follow them in time and individually identify them (no overlap in between quantum wave
functions).
We therefore consider 𝑁atoms, molecules or particles whose total energy is set to 𝐸 ± 𝑑𝐸
These identical but discernable molecules are distributed on packets of quantized energy levels with
𝑛/ particles on each level of energy 𝜖/
Therefore 𝑁 = ∑/ 𝑛/ and 𝐸 = ∑/ 𝑛/ 𝜖/
We now suppose that the system is in equilibrium (does not change over time) and close to balance (only small
fluctuations are possible). We will also assume that all distributions giving the same energy 𝐸 are equally
likely.
We define the weight 𝑊 of a distribution of the molecules on the levels. The reader is encouraged to take the
example of some smaller systems (2 particles on 3 levels, 3 particles on 2 levels, etc) to be convinced of the
validity of this formula in the case of identical but distinguishable particles.
𝑁!
𝑊=
𝑛4 ! 𝑛! ! 𝑛" !. . . .
We will now seek the dominant configuration, which corresponds to the maximum of 𝑙𝑛(𝑊)
𝑙𝑛(𝑊) = 𝑙𝑛(𝑁!) − M 𝑙𝑛 𝑛/ !
after Stirling's formula for the logarithm:
𝑙𝑛(𝑁!) = 𝑁𝑙𝑛(𝑁) − 𝑁
Then
𝑙𝑛(𝑊) = 𝑁𝑙𝑛(𝑁) − M 𝑛/ 𝑙𝑛(𝑛/ )
𝑁 = M 𝑛/
/
The extremum is given by 𝑑𝑙𝑛(𝑊) = N∑/
; <= >
;-.
𝑑𝑛/ O = 0
We can not write directly that each of the partial derivative is zero because the variables are not independent
(the total is set at N). We therefore use the theory of Lagrange multipliers, which allows, by differentiating
each of the two constraints 𝑁 = ∑/ 𝑛/ and 𝐸 = ∑/ 𝑛/ 𝜖/ , by multiplying by the factors 𝛼 and 𝛽 to obtain
independent variations by adding them to the starting equation
𝑑𝑙𝑛(𝑊) = 0
so
𝑑𝑙𝑛(𝑊) = cM 𝜕 𝑙𝑛
/
∑𝜕N
then
𝑊
𝑑𝑛 e + 𝛼𝑑𝑛/ − 𝛽𝜖/ 𝑑𝑛/ = 0
𝜕𝑛/ /
?@-(?)2-. @-(-. )7A2BC.
;-.
O 𝑑𝑛/ = 0
𝑛/
– 𝑙𝑛 N O + 𝛼 − 𝛽𝜖/ = 0
𝑁
𝑛/
𝑙𝑛 N O = 𝛼 − 𝛽𝜖/
𝑁
so
𝑛/
= 𝑒 A2BC.
𝑁
and therefore since
M 𝑛/ = 𝑁
/
M 𝑁 𝑒 A2BC. = 𝑁
/
is what sets the first Lagrange multiplier
𝑒 A M 𝑒 2BC. = 1
/
A
𝑒 =
1
∑/ 𝑒 2BC.
So, the probability of finding the system in the level i is:
𝑛/
𝑒 2BC.
𝑃/ = =
𝑁 ∑/ 𝑒 2BC.
This is the most probable state of the system at equilibrium. One recognizes a Boltzmann distribution already
seen in kinetic theory of gases (Maxwell-Boltzmann distribution of velocities).
We now define the partition function (sometimes denoted by 𝑧)
𝑞 = M 𝑒 2BC.
/
We shall see by comparing some results with those of classical thermodynamics that with the Boltzmann
constant equal to the ratio of the gas constant and Avogadro's number, and the absolute temperature (in Kelvin)
𝑒 2BC.
𝑃/ =
𝑞
1
𝛽=
𝑘D 𝑇
example 1
Consider a two energy levels system with 𝜖4 = 0 𝜖! = 𝜖 and calculate the probability of finding the
system and in each of the two levels.
Solution
The partition function of this system is: 𝑞(𝜖) = 1 + 𝑒 2BC
then the probability is
𝑒 2BC.
𝑞
𝑒 2BC/
𝑃(𝜖4 ) =
1 + 𝑒 2BC
1
𝑃(𝜖4 ) =
1 + 𝑒 2BC
𝑃(𝜖/ ) =
𝑃(𝜖! ) =
𝑒 2BC
1 + 𝑒 2BC
and the sum of the probabilities is equal to 1.
example 2
Check the partition function of a quantum harmonic oscillator by setting the first level to 0 (in principle
!
𝜖/ = ℏ𝜔 N𝑖 + "O but we will take 𝜖/ = 𝑖𝜖
Solution
The partition function of a harmonic oscillator is:
:
𝑞 = M 𝑒 2/BC
/34
It may be noted that this is a geometric series of reason 𝑟 = 𝑒 2BC and first term equal to 1
Then the partition function of a harmonic oscillator is
!
𝑞 = !2E #01
example 3
Calculate the average energy 𝐸¯ of a quantum harmonic oscillator
Solution:
𝐸¯ = ∑:
/34 𝐸 . 𝑃/ (Expectation of Energy)
but 𝑃/ =
E #01.
F
𝐸 = 𝜖/
:
1
𝐸¯ = M 𝜖/ 𝑒 2BC.
𝑞
/34
Note that 𝜖/ 𝑒 2BC. = −
&E #01.
&B
! &F
&@-(F)
𝐸¯ = − F &B 𝐸¯ = − &B
!
The partition function of a harmonic oscillator was previously determined at 𝑞 = !2E #01
&@-(F)
Therefore, for N harmonic oscillators 𝐸¯ = −𝑁 &B
therefore the average energy of a system of N harmonic oscillators is
𝑁𝜖𝑒 2BC
𝐸¯ =
1 − 𝑒 2BC
At low temperatures we notice that this average energy is concentrated on the first level
2.1 * Study of an ideal gas *
Neglecting the interactions between the particles, the total energy is purely kinetic. For a particle
confined in a box of size 𝑙 by an infinite potential the Hamiltonian reads
𝐻=𝑇=
* one-dimensional case
The stationary Schrödinger equation reads
𝐻𝜓 = 𝐸𝜓;
with
𝑃"
2𝑚
𝑃"
2𝑚
𝜕
𝑃 = −𝑖ℏ
𝜕𝑥
−ℏ" 𝑑" 𝜓
𝐻𝜓 =
= 𝐸𝜓
2𝑚 𝑑𝑥 "
𝐸=
-The fixed boundary conditions are
𝜓(𝑥 = 0) = 0 and 𝜓(𝑥 = 𝑙) = 0
The solution of equation (1) is 𝜓 = 𝐴 𝑠𝑖𝑛(𝐾𝑥 + 𝜙)
The first boundary condition gives 𝜓(𝑥 = 0) = 0 ; 𝜙 = 0
The second boundary condition gives 𝜓(𝑥 = 𝑙) = 0 ; 𝐾𝑙 = 𝑛𝜋
In quantum mechanics the total presence probability is normalized by
:
1 |𝜓|" 𝑑𝑥 = 1
2:
so
𝜓 = 𝐴𝑠𝑖𝑛(𝐾𝑥 + 𝜙) = 𝐴𝑠𝑖𝑛(𝐾𝑥)
because
𝜙=0
So we re-inject this into the Schrödinger equation and we see that energy is quantized
ℏ" "
𝐸=
𝐾
2𝑚
𝑛" ℎ"
𝐸=
8𝑚𝑙 "
In three dimensions
The Hamiltonian is written as
ℏ" 𝑑"
𝑑"
𝑑"
P " + " + "Q
2𝑚 𝑑𝑥
𝑑𝑦
𝑑𝑧
The wave function can be separated in the three directions of space because the equation is linear
𝜓 = 𝜓) . 𝜓G . 𝜓H
𝐻=−
The energy of the system is
𝐸=
ℎ"
E𝑛" + 𝑛G" + 𝑛H" F
8𝑚𝑙 " )
The equation gives the energy of a perfect gas, by shifting the index a notch from 0 to the form
ℎ" "
𝐸=
𝑛
8𝑚𝑙 "
𝜖- = (𝑛" − 1)𝜖 with 𝜖! = 0
the partition function is therefore for the ideal gas
:
𝑞 = M 𝑒 2BC"
-3!
Assuming that the energy varies in a continuous manner (what is acceptable for a large energy) then
:
𝜖 𝑞 ≈ 1 𝑒 2BC" 𝑑𝑛
!
:
𝑞 = 1 𝑒 2BI-
( 2!JC
𝑑𝑛
!
:
𝑞 = 1 𝑒 2B-
(C
𝑑𝑛
4
therefore, by symmetry (even function)
:
𝜋
1 𝜋
(
1 𝑒 2A) 𝑑𝑥 = x 𝑞 = y
𝛼
2 𝛽𝜖
2:
On the other hand, with 𝜖 =
K(
LM@ (
𝑞 = 𝑙y
𝛬 = ℎy
2𝑚𝜋 𝑙
=
𝛽ℎ"
𝛬
𝛽
1
= ℎy
2𝑚𝜋
2𝜋𝑚𝑘D 𝑇
𝛬 is the thermal de Broglie wavelength. Note that this length is characteristic of the quantum side of the system:
if the system is large compared to it, the system is classic. This occurs for small values, large masses or high
temperatures.
Generalization in 3D
N
The partition function is 𝑞 = O2 from the form of the Schrödinger equation. Note for the example of
argon in standard conditions of temperature and pressure that the de Broglie wavelength is very small
compared to the average interatomic distances, so the argon atoms in these conditions can be treated by
classical mechanics. On the contrary, for helium, lighter, at low temperatures, or for an electron at room
temperature, confined in a crystal lattice, the de Broglie wavelength is greater than the characteristic
dimensions of the problem, which implies a purely quantum treatment.
Calculating the average energy of an ideal gas
Using the formulas previously seen
M 𝑛/ 𝜖/ = 𝐸
:
𝐸¯ = 𝑁 M 𝜖/ 𝑃/
/34
:
𝑁
𝐸¯ = M 𝜖/ 𝑒 2BC.
𝑞
/34
𝐸¯ = −
𝐸¯ = −𝑁
Differentiating the only remaining term
𝑞=
𝑉
=
𝛬#
𝑁 𝑑𝑞
𝑞 𝑑𝛽
𝑑𝑙𝑛(𝑞)
𝑑𝛽
𝑉
2
B (
ℎ# N"MQO
3
𝑙𝑛(𝑞) = 𝑙𝑛(𝑉) − E𝑙𝑛(𝛽) − 2𝑙𝑛(ℎ) + 𝑙𝑛(2𝑚𝜋)F
2
thus, with N the number of particles
3
𝐸¯ = 𝑁𝑘D 𝑇
2
-For n moles, the number of particles is
𝑁 = 𝑛𝑁R
𝑁R 𝑘S = 𝑅
We find here the heat capacity of one mole of monatomic gas corresponding to a known result in classical
thermodynamics. This is the first argument to identify the results with the thermodynamic temperature
𝑑𝐸¯
3
𝑐N = P Q = 𝑅
𝑑𝑇 N 2
-
T
for the diatomic gas we will see at the end of the course 𝑐N = " 𝑅
2.2 Pressure of the ideal gas in kinetic theory
In a volume 𝑉 of one mole of atoms of an ideal gas supposed to obey classical mechanics the number
dn of atoms having a speed 𝑉. + 𝑑𝑉. is proportional to the Boltzmann factor
𝑑𝑛 𝛼 𝑒 2BC
but
1
𝜖 = 𝑚𝑉)"
2
1 "
𝜖=
𝑝
2𝑚 )
𝑑𝑛 = 𝐶𝑒 2BC
𝑑𝑛 = 𝐶𝑒
(
%3*
2( ! 64
5
or if you prefer to express this in terms of momenta or impulsions, the normalization constant is different
% '(
𝑑𝑛 = 𝐶′𝑒
2(3!4 6
5
𝑑𝑝)
Let N be the number of atoms present in the volume considered
𝑁 = 1 𝑑𝑛
we find that by employing a formula to remember.
𝑁
𝜋
(
𝐶′ =
1 𝑒 2A) 𝑑𝑥 = x
𝛼
~2𝜋𝑚𝑘S 𝑇
Consider now a wall orthogonal to the Ox axis and count the number of atomic impacts on this wall, as well
as the momentum transferred by each shock:
𝛥𝑝
€€€€⃗) = 2𝑝)
by the parallelogram rule.
As Newton's relationship is expressed as
𝐹⃗ = 𝑚𝑎⃗
with
€⃗
𝑑𝑉
𝑑𝑡
€€€€⃗
𝑑𝑝
𝐹⃗ =
𝑑𝑡
𝑎⃗ =
𝐹⃗ ≃
𝛥𝑝⃗
𝛥𝑡
The pressure P exerted on the wall is the average force per unit area. Only particles up to the positive x will
hit the wall, to which they will therefore be transferred a force per time unit 𝛥𝑡
𝛥𝑝⃗ = −2𝑝) = 𝐹) 𝛥𝑡
Furthermore, the number of collisions is proportional to the surface considered, specifically the number of
particles 𝑑𝑧 present in a small volume of surface 𝛥𝑆 and length
𝑝)
𝑉) 𝛥𝑡 =
𝑚
𝑃 = 1−
Since 𝐹) = −
"64
𝐹)
𝐹) 𝑝)
𝑑𝑧 = 1 −
𝛥𝑆𝛥𝑡 𝑑𝑛
𝛥𝑆
𝛥𝑆 𝑚
∆V
𝑃=1
% '²4
2𝑝²)
𝑁
2
𝑒 (3!86 𝑑𝑝)
𝑚 ~2𝜋𝑚𝑘S 𝑇
It is known that
𝜋
(
1 𝑒 2A) 𝑑𝑥 = x
𝛼
and differentiating with respect to 𝛼
7:
1 𝑥 " 𝑒 2A)² 𝑑𝑥 =
4
#2
1
√𝜋𝛼 (
4
for an arbitrary volume we recover the ideal gas law :
2𝑁
With 𝛽 = 1
!
8X
#2
2
(
1
𝑁
√𝜋
√𝜋
(2𝑚𝑘S 𝑇)(
𝑃=
:
@ =2
%
𝑚~2𝜋𝑚𝑘S 𝑇 4 2𝑚𝑘S 𝑇
𝑚(2𝜋𝑚𝑘S 𝑇)( 4
𝑛
𝑃 = 𝑁𝑘S 𝑇 = 𝑅𝑇
𝑉
2.3 * Entropy in the microcanonical ensemble *
Entropy, first defined by Carnot and Clausius in the first part of the nineteenth century, is a quantity e
which predicts the direction of a spontaneous reaction, for example the direction of heat transfer between two
different temperature objects placed in contact.
Let this state function be called S ; in a heat transfer this quantity varies (see thermodynamics course). For
example, when two bodies at different temperatures are placed in contact, these temperatures are equalized
and the total entropy increases although the balance of the exchanged heat is zero; this way, the disorder has
increased
𝑑𝑄
𝑑𝑆 ⩾
𝑇
𝑑𝑆 ⩾
𝑑𝑄 𝑑𝑄
−
𝑇!
𝑇"
Boltzmann's theory allows for the connection between entropy and disorder.
At equilibrium the definition is
𝑆 = 𝑘S 𝑙𝑛𝛺
with 𝛺 the number of states accessible to the system or
𝑆 = 𝑘S 𝑙𝑛𝑊 with the weight of the most probable configuration from the previous chapter.
Another definition reads 𝑆 = ±𝑘S ∑/ 𝑃/ . 𝑙𝑛𝑃/
With 𝑃/ the probability of finding the system in state i
Here this is the negative entropy (or entropy with the - sign), assessing the amount of disorder of the system.
This amount is for example used in signal processing, or data compression software to evaluate its
effectiveness.
If the system is ordered there is only one state
𝑃! = 1 so 𝑃/ = 0 and 𝑆 = 0
!
If the system is perfectly messy all probabilities are equal ; 𝑃/ = $
𝑆 = 𝑘S 𝑙𝑛(𝛺) and we find the definition of Boltzmann.
Attempt to link these definitions together we find an explanation of classical thermodynamics, by the
results of the previous chapters.
We found for the internal energy with the energy of "zero point" constant.
𝑈 = 𝑈(0) + M 𝑛/ 𝜖/
/
∑/ 𝑛/ 𝜖/ = 𝐸
and so
𝑑𝑈 = M 𝑛/ 𝑑𝜖/ + M 𝜖/ 𝑑𝑛/
/
/
The first term ∑/ 𝑛/ 𝑑𝜖/ corresponds to a change the values to levels with a constant population (typically
by changing the size of the containment box for a perfect gas,), therefore providing energy without changing
the disorder
𝜖/ =
ℎ" "
𝑖
8𝑚𝑙 "
By identification with the first principle this amount corresponds to ordered work in the presence of
compressive forces only 𝛿𝑊 = −𝑃𝑑𝑉
The second term corresponds to changing populations at constant levels (energy change without work,
only thermal) corresponding to the quantity
M 𝜖/ 𝑑𝑛/
/
𝛿𝑄 = 𝑇𝑑𝑆
in the reversible case.
Finally we therefore have
𝑆=
but 𝛽 = 1
!
𝛿𝑄 ∑ 𝜖/ 𝑑𝑛/
=
𝑇
𝑇
8X
𝑑𝑆 = M 𝑘S 𝛽𝜖/ 𝑑𝑛/
/
From the definition by Boltzmann
Recall that as shown in the previous chapter
𝑑𝑆 = 𝑑(𝑘S 𝑙𝑛𝑊)
𝑊=
𝑁!
∏/ 𝑛/ !
since
𝑙𝑛(𝑊) = 𝑙𝑛(𝑁!) − M 𝑙𝑛 𝑛/ !
by Stirling's formula for the logarithm
𝑙𝑛(𝑁!) = 𝑁𝑙𝑛(𝑁) − 𝑁and𝑙𝑛(𝑛/ !) = 𝑛/ 𝑙𝑛(𝑛/ ) − 𝑛/
Then
𝑆 = 𝑘S 𝑙𝑛𝑊 = 𝑘S c𝑙𝑛𝑁! − M 𝑙𝑛 𝑛/ !e ≃ 𝑘S c𝑁𝑙𝑛𝑁 − 𝑁 − M 𝑛/ 𝑙𝑛𝑛/ + M 𝑛/ e
/
so since𝑁 = ∑/ 𝑛/
/
𝑆 = 𝑘S 𝑙𝑛(𝑊) = 𝑘S P𝑁𝑙𝑛(𝑁) − M 𝑛/ 𝑙𝑛(𝑛/ )Q
and since
𝜕𝑙𝑛𝑊
+ 𝛼 − 𝛽𝜖/ = 0
𝜕𝑛/
as stated earlier
𝑑𝑆 = 𝑘S 𝑑(𝑙𝑛𝑊) = 𝑘S ∑/ N
;@->
;-.
𝑑𝑛/ O = ∑/ −𝑘S (𝛼 − 𝛽𝜖/ )𝑑𝑛/ with fixed
𝑁 = M 𝑛/
𝐸 = ∑/ 𝜖/ 𝑛/
We have 𝑑𝑆 = ∑/ 𝑘S 𝛽𝜖/ 𝑑𝑛/
/
/
If we take the definition of entropy information
𝑆 = −𝑘S ∑/ 𝑃/ . 𝑙𝑛𝑃/
and
𝑛/
𝑁
𝑃/ =
we found once again the result therefore the definitions are consistent.
𝑁 = M 𝑛/
/
𝑆 = 𝑘S 𝑁𝑙𝑛𝑁 − M 𝑛/ 𝑙𝑛𝑛/
/
Furthermore we have shown that
𝑃/ =
E #01
F
𝑆 = −𝑘S 𝑁 M
/
𝑒 2BC
(−𝛽𝜖/ − 𝑙𝑛𝑞)
𝑞
and with 𝑆 = 𝑘S 𝑁𝑙𝑛𝑞 + 𝑁𝛽𝑘S ∑/ −𝜖/ 𝑃/
𝑛/
𝑁
M 𝑛/ 𝜖/ = 𝑈 − 𝑈(0) = 𝐸
𝑃/ =
𝑆 = 𝑘S 𝑁𝑙𝑛𝑞 + 𝛽𝑘S 𝐸
then
with
𝑆 = 𝑘S 𝑁𝑙𝑛𝑞 + 𝛽𝑘S E𝑈 − 𝑈(0)F
𝑈 − 𝑈(0) = −
𝑁 𝑑𝑞
: @
𝑞 𝑑𝛽
Note that the thermodynamic quantities can be calculated from the partition function; we will see that the other
usual quantities (pressure, free energy ...) may also be calculated from this same function.
Example: Take the case of the harmonic oscillator then determine the entropy
𝑃/ =
E #0 1.
F
𝑞=
1
1 − 𝑒 2B
C
1
𝜕
1
𝑆 = 𝑘S 𝑁𝑙𝑛 :
𝑙𝑛
@ − 𝛽𝑘S 𝑁
2BC
𝜕𝛽 1 − 𝑒 2BC
1−𝑒
𝑁 𝜕𝑞
𝜕𝑙𝑛𝑞
𝑈 − 𝑈(0) = − : @ = 𝑁
𝑞 𝜕𝛽
𝜕𝛽
E #01
𝑆 = −𝑘S 𝑁𝑙𝑛E1 − 𝑒 2BC F + 𝛽𝑘S 𝑁𝜖 !2E #01
BC
𝑆 = 𝑘S 𝑁 :E 01 2! − 𝑙𝑛E1 − 𝑒 2BC F@
This allows us to study the variation of entropy depending on the temperature. We see that when the latter
tends to 0, the entropy also tends to 0.
3. * * Canonical Ensemble
The system previously called microcanonical ensemble (whose energy is fixed up to a small variation) does
not allow us to easily study the effects of temperature on the thermodynamic functions. This ensemble is in
classical thermodynamics, "the universe." In most common cases, it is better to consider a much smaller
system, included in the above, assuming the total energy is not fixed but the temperature is, up to a small
fluctuation. This constitutes a thermostatic system, included in a large insulated thermostat, whose total energy
is fixed, and large enough to absorb the system temperature fluctuations. This set, in statistical physics, is
called canonical ensemble.
Gibbs, by a genius masterstroke, has considered an imaginary set of replicas of an interacting system (hence
the name "ensemble") which can exchange energy (here the energy of each replica is not fixed but that of all
replicas is, forming a large system, this time considered a micro-canonical system).
Let us consider an ensemble of replicas forming a micro-canonical system. The weight of a configuration is
given by the following equation, with Ñ the number of replicas and 𝑛/ the number of times each replica
occurs
𝜔=
Ñ!
-/ !-% !-( !....!
as seen in the chapter on the microcanonical ensemble.
The probability of finding the replica in the state is the same :
𝑛/
𝑒 2B,.
𝑃/ =
=
𝑄
Ñ
with 𝐸/ the energy of the replica. The canonical partition function is obtained by summing over the replicas
and not on the possible states of an individual particle.
𝑄 = M 𝑒 2B,.
Assume equipartition of energy between the replicas. In doing so note that the average internal energy of a
replica is
[
𝐸=Ñ
with
𝑈 = 𝑈(0) + ∑ 𝑃/ 𝐸/
then
𝑈 = 𝑈(0) + ∑ 𝐸/
E #09.
\
This result is quite similar to the one obtained in the previous chapter.
Similarly assume for now
𝜔 = 𝑊Ñ
for the entropy of a replica or 𝑆 = 𝑘S 𝑙𝑛𝑊
𝑆 = 𝑘S
𝑙𝑛𝜔
Ñ
in the previous chapter we established the expression of entropy;
𝑆 = 𝑘S 𝛽E𝑈 − 𝑈(0)F + Ñ𝑘S 𝑙𝑛𝑞
We can use the formulas for energy and entropy in the canonical ensemble forgetting any mention of the
number of replicas
𝑈 − 𝑈(0)
𝑆=
+ 𝑘S 𝑙𝑛𝑄
𝑇
𝑄 = 𝑞Ñ
For N particles the energy of replica being 𝐸/ , we get by summing over levels:
𝐸/ = 𝜖/ (1) + 𝜖/ (2)+. . . . +𝜖/ (𝑁)
so
𝑄 = M 𝑒 2B IC. (!)7C. (")7⋯.7C. (?)J
/
by summing over the replicas
𝑄 = M 𝑒 2BC(!) . 𝑒 2BC(") . . . . . 𝑒 2BC(?)
/
But as each particle visit all states (ergodicity) can swap the sum of the replicas and the product on the levels
(to be convinced, take a sample of 2 levels and 3 replicates for example, as we will do in the chapter on quantum
statistics)
𝑄 = E∑/ 𝑒 2BC. F. E∑/ 𝑒 2BC. F. . E∑/ 𝑒 2BC. F
𝑄 = 𝑞?
With identical particles but obeying classical mechanics we will rigorously show in the chapter on quantum
statistics that
𝑄=
F:
?!
Example: take the perfect gas with 𝑁particles
N
𝑞 = O2
is the partition function of a particle
𝑞?
𝑁!
is the partition function for 𝑁 indistinguishable particles
𝑄=
the entropy is therefore
𝑆=
[2[(4)
X
+ 𝑘S 𝑙𝑛𝑄
with Stirling's formula
3
1
𝑆 = 𝑘S 𝑇𝑁. + 𝑁𝑘S 𝑙𝑛𝑞 − 𝑘S (𝑁𝑙𝑛𝑁 − 𝑁)
2
𝑇
#
N
𝑆 = " 𝑛𝑅 + 𝑛𝑅 :𝑙𝑛 O2 − (𝑙𝑛(𝑛𝑁R ) + 1)@
if we set
2
3
= 𝑙𝑛𝑒 (
2
;
𝑒 (.N
𝑆 = 𝑛𝑅𝑙𝑛 #
𝛬 𝑛𝑁R
For a perfect gas, therefore constituting the formula of Sackur-Tetrode giving the absolute entropy of an ideal
gas.
;
𝑒 ( . 𝑘S 𝑇
𝑆 = 𝑛𝑅𝑙𝑛
𝑃𝛬#
𝑃𝑉 = 𝑛𝑅𝑇
If we move from a volume 𝑉! to another volume 𝑉" at constant T, we have:
N
𝛥𝑆 = 𝑛𝑅𝑙𝑛 N(
%
We can also resolve this issue using the first law of thermodynamics.
𝑑𝑈 = 𝑑𝑄 + 𝑑𝑊 = 𝑛𝐶N 𝑑𝑇
because the temperature T is constant 𝑑𝑈 = 0
𝑇𝑑𝑄 = −𝑑𝑊
𝑑𝑄 = 𝑃𝑑𝑉
𝑑𝑄 = 𝑛𝑅𝑇
But
&N
N
𝑑𝑄 = 𝑇𝑑𝑆
and therefore 𝑑𝑆 = 𝑛𝑅
&N
N
𝛥𝑆 = 𝑛𝑅𝑙𝑛
𝑉"
𝑉"
We are reassured on the consistency of statistical physics and classical thermodynamics. Note that without the
term in the partition function of indistinguishable particles we would not have had this result
4. * semi-classical Approach *
In the phase space a particle is defined by two generalized coordinates (𝑞/ , 𝑞˙^ )(Positions and
velocities), and in Hamiltonian by its positions and generalized impulsions. For particles in the usual threedimensional space, the only certainty is that if the system is isolated (translationally invariant in time) the total
energy is fixed. It can also, case by case, display conserved amounts related to other invariants as rotational
invariance which will reveal the conservation of total angular momentum, for example.
Example: Harmonic oscillator in 1D
𝐻=
𝑃" 1
+ 𝑘𝑥² = 𝐸
2𝑚 2
From the above equation it follows that the phase plane is an ellipse defined by
𝑋" 𝑌"
+
=1
𝑎" 𝑏 "
𝑎" =
1
2𝑚𝐸
𝑏" =
1
2𝐸
6𝑁
In the case of a multi-dimensional system or a plurality of particles, the phase space is a Cartesian product of
these ellipses, forming a kind of doughnut or torus in the dimensional space. If there is any term of coupling
between the equations of motion of each particle, leading to chaotic trajectories, these ellipses will fill up , and
each region of the phase space respecting the conservation of energy will be covered by the system.
Thus any area dS will be covered after a finite time; if one wants to calculate the temporal average of a physical
quantity using
1 X
𝑏 = 1 𝑏 E𝑞/ (𝑡), 𝑝/ (𝑡)F𝑑𝑡
𝑇 4
we can transform this time average spatial average over the phase space according to the weights F of the
area, and integrating over the phase space. That's the whole idea of statistical physics, as presented in the
introduction.
1 𝐹 (𝑞/ , 𝑝/ ). 𝑏(𝑞/ , 𝑝/ )𝑑𝑞/ 𝑑𝑝/ (𝑞/ , 𝑝/ )
However, the inequality of Heisenberg says that uncertainty about the positions and conjugate pulses is such
that
𝛥𝑞/ . 𝛥𝑝/ ⩾ ℎ
In the so-called semi-classical approach it is assumed that the uncertainty is envenly divided, and we define
the cells of the space volume of phase (in three-dimensional space). Boltzmann had an intuition of this result
without knowledge of quantum mechanics; by postulating
𝑑𝜔 = ℎ#?
𝑆 = 𝑘S 𝑙𝑛𝛺
with the volume accessible to the system of the phase space by means of the energy
conservation. Thus, in the micro-canonical ensemble we will define the volume accessible to the system.
𝛺=
𝑑𝑉
ℎ#?
Example of an ideal gas in quantum mechanics
We have already shown that
𝑛!" 𝑛"" 𝑛#"
𝑘" = 𝜋" P " + " + " Q
𝑙!
𝑙"
𝑙#
2𝑚
𝐸
ℏ"
𝑘" =
Calculate the number of triplets providing a given energy. The number of triplets is approached by 𝑛! , 𝑛" , 𝑛# ⩾
0
Comuting 1/8 the volume of the corresponding ellipsoid in the phase space
with
1 4𝜋𝑘 # 𝑙! . 𝑙" . 𝑙#
𝑛1 = .
.
8 3
𝜋#
𝑉 = 𝑙! . 𝑙" . 𝑙#
𝑛1 =
𝑘" =
N
0Q(
"M
ℏ(
𝑘#
𝐸
%
2𝑚 (
𝑘 = : " 𝐸@
ℏ
The number of states on the surface of the ellipsoid is therefore for a variation 𝑑𝐸
&-
=
&,
&-
&,
.
&1 &,
N
"M !
N
2
%
= 0Q( . 3𝑘 " . x ℏ( . " 𝐸 2(
&,
&-
&- &1
= `Q( ℏ2 . (2𝑚)( . √𝐸
Hence the number of states accessible to the system is
𝑑𝑛 =
N
`Q( ℏ2
2
. (2𝑚)( . √𝐸𝑑𝐸
In the semi-classical approach one is interested in the number of cells between E and E + dE
𝑑𝑛 = 𝑉
&64 .&6< .&6=
K2
If one has a perfect gas, the particles are free and independent, so we have a sphere in the phase space.
𝑑𝑝) . 𝑑𝑝G . 𝑑𝑝H = 𝑑𝑝# = 𝑝" 𝑠𝑖𝑛(𝜃)𝑑𝑝𝑑𝜃𝑑𝜙 = 4𝜋𝑝" 𝑑𝑝
because we rotational invariance and we can therefore integrate over both polar angles
therefore, the number of states at the surface of the sphere in the phase space is
𝑉
𝑑𝑛 = # . 4𝜋𝑝" 𝑑𝑝
ℎ
6(
then since 𝐸 = "M
𝑝 = √2𝑚𝐸
1 %
𝑑𝑝 = √2𝑚. 𝐸 2( 𝑑𝐸
2
So:
N
%
!
𝑑𝑛 = K2 . 4𝜋. 2𝑚𝐸. √2𝑚. " 𝐸 2( 𝑑𝐸
with
N
2
𝑑𝑛 = K2 . (2𝑚)( . √𝐸𝑑𝐸
ℎ = 2𝜋ℏ
we find the same expression than in the quantum approach, however, it must be multiplied by 2 for spin
𝑑𝑛 =
2
𝑉
( . √𝐸𝑑𝐸
(2𝑚)
.
4𝜋 " ℏ#
5. * The grand canonical ensemble *
In the grand canonical ensemble, let µ design the chemical potential Consider that volume and
temperature are fixed. The number of particles is now variable: this ensemble can be seen as the inclusion of
an opened test tube within a closed and thermostatically controlled system (canonical ensemble), itself included
in an energy preserved universe (micro-canonical ensemble).
Let us treat the volume 𝑉 of the thermostated system in the canonical ensemble, consisting of the grand
canonical system studied, the volume thereof 𝑉! being assumed to be small. Let us design the volume of the
rest of the system 𝑉"
𝑉! + 𝑉" = 𝑉
The same applies to the number of particles
𝑁 = 𝑁! + 𝑁"
Let us note 𝑃! the probability of finding the grand canonical system in the considered state. This
probability is proportional to the number of states accessible to the system, by studying the micro-canonical
ensemble in which it is included. So
>
𝑃! ∝ 𝑒 !8
reversing the Boltzmann formula for the entropy of the micro-canonical system. Let us consider 𝑉 constant. A
development around the reference value of the canonical system gives 𝑆 = 𝑆(𝐸, 𝑁)
𝑆 = 𝑆4 − 𝐸! N
;a
O
;,% (?,N)
− 𝑁! N
;a
O
;?% (,,N)
Let us calculate both partial derivatives separately.
From the first principle with 𝑉 𝑁 constant
𝑑𝐸 = 𝑑𝑄 + 𝑑𝑊 = 𝑑𝑄 = 𝑇𝑑𝑆
therefore in these conditions.
𝜕𝑆
1
=
:
@
𝜕𝐸! (?,N) 𝑇
By definition of the chemical potential, considering work done by other means than pressure forces
𝑑𝐸 = 0 = 𝑑𝑄 + 𝑑𝑊bVKEc = 𝑇𝑑𝑆 + 𝑑𝑊bVKEc = 𝑇𝑑𝑆 − 𝜇𝑑𝑁
and so.
𝜇
𝑑𝑆 = 𝑑𝑁!
𝑇
𝜕𝑆
𝜇
=
:
@
𝜕𝑁! (,,N) 𝑇
Finally
𝑆 = 𝑆4 −
,%
X
+𝜇
?%
X
%
𝑃! ∝ 𝑒 !8
9
:
da/ 2 % 7e % f
6
6
We introduce the grand partition function 𝛯 to calculate the proportionality constant by normalizing
the probabilities:
𝛯 = ∑𝑒
with
𝑃=
E
2g
9#?:
h
!8 6
9 #?:.
#@ .
A
!8 6
i
M 𝑃/ = 1
/
7:
𝛯 = M 𝑒 Be? .
?34
M
𝑒 2B,
,,?8/)E&
Be?
𝛯 = ∑7:
. 𝑄?
?34 𝑒
with 𝑄? the canonical partition function
or with 𝑧 = 𝑒 Be the fugacity of the molecules
?
𝛯 = ∑7:
?34 𝑧 . 𝑄?
For an ideal gas 𝑧 = 1
In the chapter on entropy it was shown that:
𝑆=
[2[(4)
X
+ 𝑘S 𝑙𝑛𝑄
If we calculate the Gibbs free energy by the relationship
𝐹 = 𝑈 − 𝑇𝑆
we get
𝐹 = −𝑘S 𝑇𝑙𝑛𝑄
or we have the classical result
𝑑𝐹 = 𝑑𝑈 − 𝑇𝑑𝑆
or by the first principle
𝑑𝐹 = 𝑑𝑊 + 𝑑𝑄 − 𝑇𝑑𝑆
𝑑𝑄 = 𝑇𝑑𝑆
𝑑𝐹 = 𝑑𝑊 = −𝑃𝑑𝑉
So, it is and again possible to calculate an additional variable (pressure) from the partition function :
;'
𝑃 = − N;NO
X3jb-kVl-V
One can deduce the enthalpy 𝐻 and free enthalpy 𝐺 from the following equations:
𝐻 = 𝑈 + 𝑃𝑉 and 𝐺 = 𝐹 + 𝑃𝑉
If now we introduce the grand potential defined from the grand partition function
𝛺 = −𝑘S 𝑇. 𝑙𝑛𝛯
Let us calculate the average number of particles in the grand canonical ensemble
𝑁 = ∑ 𝑁 . 𝑃(𝑁) = ∑7:
?34
! !
?E #0(9#?:)
i
;
2B(,2e?)
𝑁 = B . i ∑7:
?34 ;e 𝑒
!
;
!
𝑁 = B . ;e 𝛯. i
;
𝑁 = 𝑘S 𝑇. ;e 𝑙𝑛𝛯
𝑁=−
𝜕𝛺
𝜕𝜇
and therefore, we obtain the average number of particles from the grand potential. We will see that this
formula is useful in the study of the adsorption of molecules of a catalyst, but also in the study of quantum
statistics of the next chapter.
6. * Quantum Statistics *
6.1 General
In quantum mechanics, one typically studies the probability of presence of a particle and between 𝑥
and 𝑥 + 𝑑𝑥. The pprobability density (only quantity to have a physical meaning) is given in representation 𝑥
by 𝜌(𝑥)𝑑𝑥
This is obtained from the complex wave vector 𝛹 by 𝜌 = |𝛹(𝑥)|² = 𝛹 * (𝑥)𝛹(𝑥)
More formally, one introduces an abstract wave vector .|𝛹> (independent of the representation) which can for
example describe several particles simultaneously.
For example, for two particles 1 and 2 the wave function allows to calculate the probability of finding the
particle 1 in €€€⃗
𝑟! and the same time the particle 2 in €€€€⃗
𝑟" ; it is not necessarily a product of individual wave
functions: there may be correlation or anti-correlation of the particles, which are not necessarily independent,
let alone if there is interaction between them, as the Coulomb interaction between electrons:
|𝛹> = |𝛹(𝑟€€€⃗,
𝑟" ≠ |𝛹! (𝑟€€€⃗)>.|𝛹
€€€⃗)>
! €€€⃗)>
!
" (𝑟
"
It is therefore understandable that this quantum phenomenon will change the way of counting the weight of
each configuration in statistical physics.
These two particles can be either:
-different (e.g. an electron and a proton, ...) ; then Boltzmann statistics can be used as seen in previous chapters
-identical (e.g. two electrons)
In this case, there are two possibilities, either the particles are:
-discernable, such as numbered lottery balls, or two clearly separated electrons (at sufficient distance so that
their wave functions do not overlap in practice)
-indiscernible; such as two electrons close enough at the initial time that their wave functions overlap, and
therefore one cannot know which is which.
We introduce the permutation operator 𝑃!" of two electrons, due to which the presence probability
density does not change under permutation if the particles are indiscernible.
𝑃!" 𝜌(𝑟€€€⃗,
𝑟" = 𝜌(𝑟€€€⃗,
𝑟"
! €€€⃗)
! €€€⃗)
For the wave function if we apply twice the operator we must find the initial situation
𝑃!" 𝑃"! |𝛹> = |𝛹>
so there are two possibilities
𝑃!" |𝛹> = −|𝛹>
𝑃!" |𝛹>= + |𝛹>
A "-" defines fermions and "+" bosons
spin-statistics theorem
The spin (intrinsic rotation) of elementary particles is quantified: integer or half-integer ℏ
An integer spin can correspond to a compound particle of bosons and fermions.
For example, as mediators of fundamental interactions, in the "standard model" of particle physics:
-Spin "0": Higgs boson, recently discovered experimentally
-Spin "1": the mediators of the weak nuclear interaction (W and Z bosons) and strong interaction (gluons)
-Spin "2": graviton (mediator of gravity, not yet demonstrated experimentally).
And as particles, we have:
Spin "1/2":
-The baryons (protons and neutrons)
-The leptons (electrons) ; as well as photons (mediators of electromagnetism)
Note again that we can associate particles that will in turn compound their spins; thus, combining fermions
(typically two electrons in the case of superconductivity, or a nucleus having a number of hadrons associated
with an even number of outer electrons as in the case of rubidium) we can obtain a boson.
6.2 Partition function and quantum factor
Consider a wave function built from two individual wave functions:
|𝛹> =
!
√"
(|𝛹! (𝑟€€€⃗)
€€€⃗)>
− |𝛹! (𝑟€€€⃗)
€€€⃗)>)
! > .|𝛹" (𝑟
"
" > .|𝛹" (𝑟
!
Warning: this wave function, without necessarily describing exactly the reality, is not a product of individual
wave functions which would be an even worse approximation! For details, refer to the chapters on the HartreeFock method and its extensions in the course of physics / quantum chemistry and microscopic modeling of
matter.
Let P12 be the permutation function
So we have: 𝑃!" |𝛹> = −|𝛹>
If we consider two levels (2 electrons)
if |𝛹! > = |𝛹" > then |𝛹> = 0
which is unacceptable given the requirement of standardization of presence probability in the entire space.
This is the Pauli principle, which states that two electrons cannot occupy the same quantum
configuration (if you consider spin of +1/2 and -1/2).
One can establish the expression of the weight W of a conformation taking more account of the possible energy
degeneracy 𝑔/ of a level (ie when levels can be classified by ' quantum number' 𝑛/ , but naming levels with the
same energy under the same label). In the case of hydrogen for instance, to a first approximation, level are
degenerate 2𝑛/" times.
-In the case of Boltzmannions (distinguishable particles)
the expression of the weight W is
∏/ 𝑔/-. 𝑁!
𝑊=
∏/ 𝑛/ !
-In the case of fermions, the expression of the weight W is
𝑊=˜
/
-In the case of bosons W is:
𝑊=˜
/
𝑔/ !
𝑛/ ! (𝑔/ − 𝑛/ )!
(𝑛/ + 𝑔/ − 1)!
𝑛/ ! (𝑔/ − 𝑛/ )!
Consider the example of two particles spread over two quantum energy configurations and not degenerate. For
Boltzmannions, all possibilities are allowed, and there are two ways, the particles being discernible, to obtain
total energy 𝜖 and one way to get the energy 𝜖𝜖 and 0.
In the case of fermions, only one possibility is allowed (total energy 𝜖 ) if it does not want that two particles
end up on the same level. For bosons, all possibilities are allowed, but the particles are indistinguishable so
there is only one way to get energy 𝜖. Let us summarize the values of the partition function 𝑄 in these three
cases:
two Boltzmannions
𝑊=
𝑁!
𝜋. 𝑛! !
𝑄 = 1 + 2𝑒 "#$ + 𝑒 "%#$
two Fermions
two Bosons
𝑄 = 1 + 𝑒 "#$ + 𝑒 "%#$
𝑄 = 𝑒 "#$
𝑄 ≠ -1 + 𝑒 "#$ .
%
%
here 𝑄 ≠ -1 + 𝑒 "#$ .
%
𝑄 = -1 + 𝑒 "#$ . = 𝑞%
only in this case the
factorization is possible
general formula
For the Boltzmannions
𝑄=M
-B
𝑁!
. 𝑒 2B ∑B -B CB
𝜋A . 𝑛A !
as already seen and recognized over the example
𝑄 = ™∑A 𝑒 2BCB š
?
For fermions or bosons the canonical partition function is difficult to express as a product, but using the grand
canonical partition function we find
:
𝛯 = M 𝑧 ? 𝑄?
?34
:
𝛯 = M 𝑧 ? M 𝑒 2B ∑B -B CB
?34
-B
𝑁 = M 𝑛A
A
by swapping sum and product
:
𝛯 = M M 𝜋A E𝑧𝑒 2BCB F
-B
?34 -B
:
𝛯 = 𝜋A ME𝑧𝑒 2BCB F
-B
?34
In the case of bosons
The product goes from 0 to infinity since all quantum conformations are allowed.
𝛯 = 𝜋A .
1
1 − 𝑧𝑒 2BCB
(Recognize the sum of a geometric progression.)
The average number of particles is, as seen in the chapter on the grand canonical ensemble
𝜕
𝑁¯ = 𝑧 𝑙𝑛(𝛯)
𝜕𝑧
and therefore
𝜕
1
𝑙𝑛 :𝜋A .
@
𝜕𝑧
1 − 𝑧𝑒 2BCB
𝜕
𝑁¯ = −𝑧 M 𝑙𝑛 E1 − 𝑧𝑒 2BCB F
𝜕𝑧
𝑁¯ = 𝑧
-B
from where
𝑧𝑒 2BCB
𝑁¯ = M P
Q
1 − 𝑧𝑒 2BCB
A
In the case of fermions
𝑛A is valid only 0 or 1 given the Pauli exclusion principle, therefore, the partition function can be written as:
𝛯 = 𝜋A . E1 + 𝑧𝑒 2BCB F
Therefore, the average number of particles is:
𝜕
𝑁¯ = 𝑧 𝑙𝑛 N𝜋A . E1 + 𝑧𝑒 2BCB FO
𝜕𝑧
from where
𝑁¯ = 𝑧
𝜕
M 𝑙𝑛 E1 + 𝑧𝑒 2BCB F
𝜕𝑧
-B
𝑧𝑒 2BCB
𝑁¯ = M P
Q
1 + 𝑧𝑒 2BCB
A
The average number of particles on a level is given by:
:
1
𝑛¯A = M 𝑧 ? M 𝑛A 𝑒 2B-B CB
𝛯
?34
-B
𝑛¯A = −
Factoring the quantum factor we get:
1 𝜕
𝑙𝑛𝛯
𝛽 𝜕𝜖A
𝑧𝑒 2BCB
1 + 𝑧𝑒 2BCB
𝑧𝑒 2BCB
𝑛¯A =
1 − 𝑧𝑒 2BCB
1
𝑓(𝜖A ) = !
𝑒BCB ± 1
H
1
𝑓(𝜖A ) = B(C 2e)
𝑒 B
±1
𝑛¯A =
with a "+" in the case of fermions and "-" in the case of bosons
-In the case of bosons one talks about the Bose-Einstein distribution (1924)
-In the case of fermions, the factor is called the Fermi-Dirac distribution (1926)
If 1 ≪ 𝑒 B(CB 2e) (large temperatures for instance)
The quantum factor is approximately
𝑓(𝜖A ) ≈ 𝑒 Be 𝑒 2BCB
therefore, we get the Boltzmann distribution in both cases.
In the case of Fermi-Dirac distribution the quantum factor can also be written in a form with the so-called
"Fermi energy" 𝜖' , identified as the chemical potential, and corresponding to the energy of the last electron
"stacked" at zero temperature .
1
𝑓(𝜖A ) =
B(C
1 + 𝑒 B 2CC )
1
𝛽=
𝑘S 𝑇
•
Representation of the variation factor of the quantum Fermi-Dirac function for different temperatures
(or values) for a Fermi energy of 1 in arbitrary units. We see that at low temperatures it goes quickly
from a factor of 1 to a 0 factor around the Fermi energy, and that at high temperatures it approaches a
Boltzmann factor (exponential decay)
In the case where the particles are bosons the quantum factor is written in the form:
1
𝑓(𝜖A ) = B(C 2e)
𝑒 B
−1
For fermions, so if 𝑇 → 0
𝛽→∞
𝜖A < 𝜖' : 𝑓(𝜖A ) → 1
𝜖A > 𝜖' : 𝑓(𝜖A ) → 0
the fugacity was defined by
𝑧 = 𝑒 Be
and then if we find the Maxwell-Boltzmann distribution for
𝑧≪1
1
≫1
𝑧
𝑓(𝜖A ) ≈ 𝑧𝑒 2BCB
The average number of particles is
𝑁¯ = M 𝑛¯A = 𝑧 M 𝑒 2BCB = 𝑧𝑞
A
A
For the ideal gas, the microcanonical partition function is 𝑞 with de Broglie wavelength as already seen.
𝑉
𝑞= #
𝛬
%
𝛽 (
𝛬 = ℎ:
@
2𝑚𝜋
so :
𝑁¯ =
with a volume density of particles 𝑛
𝑧=
The condition gives for 𝑧 ≪ 1
𝑧𝑉
𝛬#
𝑁¯𝛬#
= 𝑛𝛬#
𝑉
𝑛𝛬# ≪ 1
The degeneration temperature T is defined b 𝑛𝛬# = 1
Whence
2
𝛽 (
𝑛ℎ :
@ =1
2𝑚𝜋
(
2𝑚𝜋𝑘S 𝑇 = (𝑛ℎ# )2
#
(
ℎ" 𝑛2
𝑇=
2𝑚𝜋𝑘S
Example
Consider the case of helium whose lattice parameter is 𝑎 = 4𝐴˚. It is found that the degeneration
temperature 𝑇 ≈ 3𝐾 and the thermal wavelength 𝛬 ≈ 5𝐴˚
The grand partition function is related to the partition function 𝑄? in the canonical ensemble.
7:
𝛯 = M 𝑧 ? . 𝑄?
?34
For 𝑁 ≈ 𝑁¯
𝑙𝑛(𝛯) = 𝑙𝑛(𝑄? ) + 𝑁𝑙𝑛(𝑧)
and
𝑙𝑛(𝑄? ) = 𝑙𝑛(𝛯) − 𝑁𝛽𝜇 = ± M 𝑙𝑛 E1 ± 𝑧𝑒 2BCB F − 𝑁𝛽𝜇
A
with (+) in the case of fermions and (-) in the case of bosons
If 𝑧 ≪ 1
𝑙𝑛(𝑄? ) = ± ME±𝑧𝑒 2BCB F − 𝑁𝛽𝜇 = ME𝑧𝑒 2BCB F − 𝑁𝛽𝜇
A
A
developing the logarithm of the first order
Now
𝑞 = M 𝑒 2BCB
A
𝑙𝑛(𝑄? ) = 𝑧𝑞 − 𝑁𝛽𝜇
secondly
𝑁 = M 𝑛¯A = M 𝑧 𝑒 2BCB = 𝑧𝑞
A
A
𝑙𝑛(𝑄? ) = 𝑁 − 𝑁𝛽𝜇
𝑧 = 𝑒 Be → 𝛽𝜇 = 𝑙𝑛(𝑧)
and therefore as
𝑙𝑛(𝑄? ) = 𝑁 − 𝑁𝑙𝑛(𝑧)
𝑁
𝑧=
𝑞
𝑙𝑛(𝑄? ) = 𝑁 − 𝑁𝑙𝑛𝑁 + 𝑁𝑙𝑛𝑞
hence and therefore as announced in the chapter on the canonical ensemble 𝑄? decreases because the particles
are indistinguishable
𝑙𝑛(𝑄? ) = 𝑁 − 𝑁𝑙𝑛(𝑁) + 𝑁𝑙𝑛(𝑞) = 𝑙𝑛(𝑁!) + 𝑙𝑛(𝑞? )
𝑞?
𝑄? =
𝑁!
Further developing the logarithm of the grand partition function:
:
𝑙𝑛(𝛯) = ± M 𝑙𝑛 E1 ± 𝑧𝑒
A
with (+) for fermions and (-) for fermions.
In the case of the ideal gas
2BCB
F = M(±1)@7!
@3!
𝑧@
M 𝑒 2B@CB
𝑙
A
M 𝑒 2B@CB =
A
We get for the average number of particles
𝑁¯ = 𝑧
𝑉
2
(
𝑙 𝛬#
𝑙
𝜕𝑙𝑛(𝛯)
𝜕𝑧
For the ideal gas the number density is obtained
:
𝑁¯
1
𝑧@
𝜌 = = # M(±1)@7! ;
𝑉 𝛬
𝑙(
@3!
Stopping at the order 𝑙 = 1 these developments for 𝑙𝑛𝛯
𝜌=
and so
𝑧
𝛬#
𝑙𝑛(𝛯) = 𝑧𝑞 =
𝑧𝑉
𝛬#
𝑧 = 𝜌𝛬#
𝑙𝑛(𝛯) = 𝜌𝑉
and by setting the pressure in the grand canonical ensemble.
𝑃𝑉
𝑙𝑛(𝛯) =
𝑘S 𝑇
then finally
oN
= 𝜌𝑉
1 X
8
𝑃 = 𝜌𝑘S 𝑇
𝑇
𝑁𝑅𝑇
𝑃 = 𝑁R 𝑁𝑘S =
𝑉
𝑉
with 𝑁 the number of moles
So at order 1, we find the relationship of perfect gases.𝑃𝑉 = 𝑁𝑅𝑇
To order 𝑙 = 2
1
𝑧"
P𝑧
±
;Q 𝑉
𝛬#
2(
𝑧𝑉
𝑃𝑉
𝛬# 𝑃
𝜌 𝑉 = 𝑧𝑞 = # =
𝑧=
𝛬
𝑘S 𝑇
𝑘S 𝑇
𝜌=
𝑃𝑉
1 𝛬# 𝑃
𝛬0 𝑃"
𝜌" 𝛬# 𝑃𝑉
𝜌𝛬#
= #
±
=1± ;
;¡ = 𝜌 ±
;
𝑘S 𝑇 𝛬 𝑘S 𝑇 𝑘 " 𝑇 " 2(
2( 𝜌𝑘S 𝑇
2(
S
(+) For fermions and (-) for bosons.
If the particles are quantum, there is a pressure correction in relation to the usual law of classical ideal
gas, even in the case of a quantum ideal gas. We find repulsive fermions and attractive bosons.
For a degenerate system one must take into account the density of states:𝑔(𝜖)
The number of total particles is therefore
:
𝑁 = 1 𝑔 (𝜖)𝑓(𝜖)𝑑𝜖
4
For a non-relativistic ideal gas, we have seen that the density of states is given by the spin s of the particle and.
2
(2𝑠 + 1)
(
𝑔(𝜖) =
.
2𝜋𝑉
𝜖(2𝑚)
√
ℎ#
𝜖=
𝑝"
2𝑚
Generalization of expression 𝑓(𝜖)
!
𝑓(𝜖) = E 0(1#?)7D
b = 0: Maxwell Boltzmann
b = 1: Fermi – Dirac
b = -1: Bose-Einstein
For electrons (fermions) at zero temperature (where the chemical potential is identified with the Fermi energy)
one thus finds
CC
2
2
𝑣
8𝜋
𝑁 = 1 N2 # √𝜖2𝜋(2𝑚)( O 𝑑𝜖 = # 𝑉(2𝑚𝜖' )(
ℎ
3ℎ
4
then the density of the system. It is found that the Fermi energy is a function of the electron density; the socalled theory of "density functional theory", widely used today in physics and quantum chemistry, proceeds
from a generalization of such remarks in the case of real fermion gas.
(
(
1 3𝑁ℎ# 2
𝜖' =
P
Q = 𝑐𝜌2
2𝑚 8𝜋𝑉
𝑁
𝜌=
𝑉
Returning to the case of bosons the total number of particles can be divided into:
𝑁 = 𝑁4 + 𝑁′
𝑁4 : Number of particles in the ground state clearly where a discrepancy occurs in the numerator of the
expression (Bose-Einstein condensate).
𝑁p = 1
:
4
p
:
𝑁 =1
4
𝑔(𝜖)
. 𝑑𝜖
𝑒B(C2e) − 1
%
2
2𝜋𝑉𝜖 ( (2𝑚)(
¡ 𝑑𝜖
ℎ# (𝑒B(C2e) − 1)
in the case of non-relativistic ideal gas
𝜖=
𝑑𝜖 =
𝑥
𝛽
𝑑𝑥
= 𝑘S 𝑇𝑑𝑥
𝛽
%
2
:
2
2𝜋𝑉
𝑥 ( (𝑘S 𝑇)(
𝑁′ = # (2𝑚)( 1
¡ 𝑑𝑥
ℎ
𝑒 ()2Be) − 1
4
2
𝑁′ =
%
"N "QM18 X ( :
)(
N K( O ∫4 PE (4#0?)2!Q 𝑑𝑥
√Q
We define a new function 𝐹#⁄" (𝛼)
𝐹2 =
(
2
√𝜋
%
:
1
𝑥(
¡ 𝑑𝑥
𝑒 ()7A) − 1
4
𝛼 = −𝛽𝜇
So (result obtained by numerical computation)
: %
2
𝐹2 = P NM 𝑒 2(r7!)A O 1 𝑥 ( 𝑒 2(r7!)) 𝑑𝑥Q ⩽ 𝐹#⁄" (0) = 2,612
(
√𝜋
4
2
From where 𝑁′ ⩽
"MQ1 X (
𝑉 N K( 8 O . 2,612
2
N is proportional to 𝑇 (
Below the critical temperature 𝑇j 𝑁4 will begin to rise to ensure that
𝑁 = 𝑁4 + 𝑁′
2
𝑁′
2𝑚𝜋𝑘S 𝑇j (
= 2,612 :
@
𝑉
ℎ"
Case of helium
if 𝜌 = 0,15 𝑔⁄𝑐𝑚# 𝑇j = 3,24𝐾
In practice, measurement gives 𝑇j = 2,19𝐾 but this approximate theory is still satisfactory (hypothesis of the
ideal gas)
Calculation of internal energy U
:
𝑈 = 1 𝜖 𝑓(𝜖)𝑔(𝜖)𝑑𝜖
4
2
:
2𝜋𝑉 (2𝑚)( √𝜖𝜖
𝑈=1
𝑑𝜖
ℎ# 𝑒B(C2e) − 1
4
𝑥
𝛽
𝑑𝑥
𝑑𝜖 =
𝛽
𝜖=
2
3𝑘S 𝑇𝑉(2𝑚𝜋𝑘S 𝑇)(
𝑈=
¡ 𝐹T⁄" (𝛼)
2ℎ#
with
𝐹; (𝛼) =
(
but
2
:
4
3√𝜋
𝑥(
1
𝑒 (A7)) − 1
4
¡ 𝑑𝑥
𝐹; (𝛼) ⩽ 𝐹; (0)
(
Moreover
(
𝐹; (0) ≈ 1,341
(
so
2
3𝑘S 𝑇𝑉(2𝑚𝜋𝑘S 𝑇)d(f
𝑈⩽
¡ 1,341
2ℎ#
Calculating the heat capacity
𝐶N = :
T #
𝜕𝑈Ml)
@
𝜕𝑇 N
N
2
The maximum energy is given by 𝐶N = " . " 𝑘S 𝑇 K2 (2𝜋𝑚𝑘S 𝑇)( . 1,341
Now
2
2𝜋𝑚𝑘S (
𝑁
𝑇s :
@ 2,612 =
"
ℎ
𝑉
2
#2
2
15 1,341
15 1,341
𝑇 (
(
𝐶N =
.
𝑘 𝑁𝑇 ( 𝑇s =
.
𝑅𝑛 : @
4 2,612 S
4 2,612
𝑇s
2
(
2
𝑇 (
𝐶N = 1,93𝑅𝑛 : @
𝑇s
We thus obtain a heat capacity which goes to 0 when the temperature goes to 0, without there being the same
type of discontinuity at the critical temperature than in a so-called "first order" phase transition (such as melting
water) where the heat capacity has a discontinuity corresponding to the latent heat of fusion.
* 7. Review on the example of the ideal gas and real gas *
7.1 Born-Oppenheimer approximation of a molecule forming a perfect gas
Consider a single molecule free to move in a given volume (ideal gas), formed of atoms, themselves formed
of nuclei and electrons. In order of increasing accuracy in the approximation:
-The Electrons are at least 1,836 times lighter than the nuclei so we can consider that nuclei are immobile
when studying molecules.
-We can consider that the electrons follow the nuclei.
-In view of the value of the de Broglie wavelength at normal temperatures, electrons are obviously rather
quantum and the positions of the nuclei rather governed by classical mechanics.
-The wave function can be factored as 𝛷 = 𝛷E . 𝛷with 𝛷E wavefunction of electrons and 𝛷- wavefunction of the nuclei
Strictly speaking it would be better to solve a full Schrödinger equation, but this requirement is necessary only
in very special cases (chemical reactions by femtosecond laser directed at the atomic scale, for example). For
the positions of the nuclei, it is already sufficiently demanding to solve the electronic Schrödinger equations
(parameterized by the positions of the fixed nuclei)𝐻E 𝛷E = 𝜖rE 𝛷E
Finally in this approximation we can write an approximate way the total energy of the molecule in the form
𝜖r = 𝜖rX + 𝜖rt + 𝜖rN + 𝜖rE
𝑅 → 𝑅𝑜𝑡𝑎𝑡𝑖𝑜𝑛 𝑇 → 𝑇𝑟𝑎𝑛𝑠𝑙𝑎𝑡𝑖𝑜𝑛 𝑉 → 𝑉𝑖𝑏𝑟𝑎𝑡𝑖𝑜𝑛 𝑒 → 𝐸𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑖𝑐
The partition function is thus written:
N
2BCD
𝑞 = ∑:
= 𝑞X . 𝑞t . 𝑞N . 𝑞, with the translational partition function 𝑞X = O2 of the ideal gas as previously
r34 𝑒
established, and the rotation function
𝑞t = M(2𝑗 + 1) 𝑒 2BKjSr(r7!)
r
taking into account the degeneracy of the quantum rotator.
We use the moment of inertia for a diatomic molecule. This partition function is not analytic, one must use a
numerical approximation, for example by summing over the first levels (3-4).
In the case of a diatomic molecule 𝐵 =
ℏ
`Qju
𝐼
:
(
𝑞t ≈ 1 (2𝑗 + 1) 𝑒 2Rr 𝑑𝑗 ≈
4
2
1
ℎ𝑐𝛽𝐵
%
! 1X (
Q (
O
v Kj
RSs
We can generalize the expression by 𝑞t = N O N
For a polyatomic molecule whose moments of inertia are 𝐴, 𝐵, 𝐶 and 𝜎 the number of symmetry operations (3
for ammonia, for example).
The vibrational partition function is written as 𝑞N = 𝑞N (1)𝑞N (2)..
But we can also generalize the expression with N the number of atoms. 𝑞N 𝑞N ≈ (𝑞N )#?20
We saw that the vibrational partition function starts as low temperature. At high temperatures:
1
1
𝐸N = :𝜈 + @ ℏ𝜔 = :𝜈 + @ ℎ𝑐𝜈
~
2
2
2!
~
𝜈 is in 𝑐𝑚
𝑞N =
1
≈1
1 − 𝑒 2BKjw~
1
𝑞N ≈
𝛽ℎ𝑐𝜈
~
Similarly, for low temperatures, the electronic partition function is the Fermi-Dirac one, with supposed
degeneration of the two lower levels, separated by a given energy 𝜖; this situation is most common because at
sufficiently high temperature to energize the following levels, we often already have dissociation of the
molecule (average energy greater than the bonding energy).
𝑞, = 𝑔, E1 + 𝑒 2BC F
Heat capacity
𝐶N = :
and
𝜕𝑈
𝜕𝜖
@ = −𝑘S 𝛽"
𝜕𝑉 N
𝜕𝛽
𝜖=
−1 𝜕𝑞
: @
𝑞 𝜕𝛽 N
Gradually, as the temperature increases, it "wakes up" in turn the various terms of the partition function to pass
the dissociation threshold for a few thousand K. Note that the heat capacity of a gas such as air cannot be
considered constant if we approach these temperatures, for example in internal combustion engines.
7.2 Real Gases
We talk about real gas when there are interactions between the molecules constituting it (collisions ...). Real
gases approach the ideal gas when going to low densities because interactions between molecules are weaker;
however, without collisions, one can notice that all the hypotheses of statistical physics are false, because there
would not be evolution towards equilibrium without exchange of momentum between molecules or atoms, or
allowing ergodicity to calculate average.
From the macroscopic point of view a deviation from the ideal gas law reads as
with corrections
oN3
tX
S
s
S
s
3
3
3
3
= 1 + N + N ( +. . . . N + N ( +. . ..
First order this equation can be rewritten
with
𝑃(𝑉M − 𝑏) = 𝑅𝑇
𝑏=
𝐵
𝑉M
Second order
this development should coincide with that of the equation of van der Waals
𝑎
P𝑃 + " Q (𝑉M − 𝑏) = 𝑅𝑇
𝑉M
at high temperature 𝑃𝑉M = 𝑅𝑇
dimensional representation of the curves P, V, T governed by the dimensionless equation of Van der Waals
(expressed in terms of volumes, critical pressures and temperatures).
At high temperatures we found curves of near perfect gas below the critical point, however, with a coexistence
between liquid and vapor given the unstable nature of the curves. This coexistence can be calculated by the
"Maxwell construction."
7.3 Partition function of real gas
As we have already seen that the partition function is written in three dimensions (ignoring the degrees of
freedom of rotation, vibration, electronic motion) with 𝑁 the number of atoms or molecules and 𝑄 =
y
?!O2:
𝑍 = 1 … 1E𝑒 2BN: 𝑑𝑟€€€⃗𝑑𝑟
€€€€⃗F
! €€€⃗
" … . . 𝑑𝑟
?
𝑉? the potential energy of interaction of the atoms.
In the case of the ideal gas this integral is reduced to the volume.
𝑉? = 0
𝑍 = 𝑉?𝛬
The only contribution to the Hamiltonian of the ideal gas is the kinetic energy term that gives the term which
we have already seen.
Note that 𝑉? depends on all the coordinates of the atoms simultaneously.
For example, in the case of metals, one has quasi-free valence electrons which are all involved in the
electrical and thermal conductivity and ensure the chemical bond. Note that we see these electrons form a
plasma on the surface of the metal. This plasma is opaque to certain electromagnetic frequencies, so that we
see metallic luster appear if the metal surface is not oxidized.
For insulators (ionic systems, covalent, rare gases ...), we can decompose 𝑉? in several potential 2
body , 3 body , etc., modeling the bond between atoms or molecules (e.g., considering the relative orientation
of the two molecules such as water)
1
1
𝑉? = M 𝑉r E𝑟€⃗,
€⃗,
€€€⃗F
^ €𝑟z⃗F + M 𝑉/r1 E𝑟
^ €𝑟z⃗, 𝑟
1 + ⋯.
2
8
/,r
excluding 𝑖 = 𝑗
/,r,1
Most of the time however we limit in two bodies modeling potential, e.g. electrostatic ion in the case of systems
with exponential repulsive term of the "Born-Mayer" form to prevent a total contraction of a purely attractive
potential.
In the case of rare gas such as Ne; Ar, Kr (neither too light as helium to require a quantum treatment nor too
heavy to require a treatment of relativistic electrons).
𝑉? can be written as: (divided by 2 to avoid double counting)
1
𝑉? = M 𝑉/r E𝑟€⃗,
€⃗^ − 𝑟€z⃗®F = M 𝑉/r E®𝑟€⃗^ − 𝑟€z⃗®F
^ €𝑟z⃗F = M 𝑉/r E®𝑟
2
/{r
/{r
/|r
he potential Lennard-Jones is often chosen; it has a good descriptive power and presents analytical
regularities, for example, unlike so-called potential of hard spheres. Note that this potential has a long-range
attractive part corresponding to a model of the dipole-dipole interaction between neutral atoms and a shortrange repulsive term, the sum of both to provide a balance between distance atoms and describe phenomena
such as liquefaction for sufficiently low temperature gases.
𝑟 2!"
𝑟 20
𝑉/r (𝑟) = 4𝜖 ¯N O
−N O °
𝜎
𝜎
the radial distribution function 𝑔(𝑟⃗) is then introduced which will allow us to compute the canonical ensemble
in the number of pairs of atoms at the distance 𝑟⃗ from one another.
𝑔(𝑟⃗) =
with the number density.
𝜌=
1
M 𝛿 E𝑟⃗ + €𝑟⃗^ − 𝑟€z⃗F
𝑁−1
𝑁
= 1 M 𝛿 (𝑟€⃗^ − 𝑟⃗)𝑑𝑟⃗
𝑉
The average is calculated as in previous chapters from a Boltzmann factor; kinetic energy contributions
compensate the denominator and nominator and we get:
𝑉
1
1. . . 1 M 𝛿 E𝑟⃗ + €𝑟⃗^ − 𝑟€z⃗F 𝑒 2BN: 𝑑𝑟€€€⃗𝑑𝑟
€€€€⃗
! €€€⃗.
" . . . . 𝑑𝑟
?
𝑁(𝑁 − 1) 𝑍
Note that all terms of the sum are identical to the first, and thereforeType equation here.
𝑉
𝑔(𝑟⃗) = 1. . . 1 𝛿(𝑟⃗ + 𝑟€€€⃗! − €€€⃗)
𝑟" 𝑒 2BN: 𝑑𝑟€€€⃗𝑑𝑟
€€€€⃗
! €€€⃗.
" . . . . 𝑑𝑟
?
𝑍
Noting that the potential in the case of rare gas depends only on the mutual distance between two atoms, the
two first integrals can be simplified and obtaining by shifting the origin then posing |𝑟€€€⃗" − €€€⃗|
𝑟! = 𝑟⃗
𝑔(𝑟⃗) =
𝑔(𝑟⃗) =
𝑉"
1. . . 1 𝑒 2BN: 𝑑𝑟€€€⃗𝑑𝑟
€€€€⃗
# €€€⃗.
` . . . . 𝑑𝑟
?
𝑍
;@-(\)
Recall that the gas pressure is expressed in the canonical ensemble by 𝑃 = 𝑘S 𝑇 N
;N
For a perfect gas partition function include the macroscopic gas law.
𝑉?
𝑄=
𝑁! 𝛬#?
𝑘S 𝑇𝑁
𝑃=
𝑉
𝑃𝑉 = 𝑛𝑅𝑇
For a real gas, the partition function is
𝑍
𝑍𝑉 ?
=
𝑁! 𝛬#? 𝑁! 𝑉? 𝛬#?
𝑁 𝜕𝑙𝑛 𝑍⁄𝑉 ?
𝑃 = 𝑘S 𝑇 P +
Q
𝑉
𝜕𝑉
𝑄=
but
𝑍 = 1. . . 1E𝑒 2BN: 𝑑𝑟€€€⃗𝑑𝑟
€€€€⃗F
! €€€⃗.
" . . . . 𝑑𝑟
?
Note that changing the variable to explicitly integrate the volume, i.e. by asking
c⃗
𝑟⃗′ = N %⁄2 we obtain
%
𝑍 = 𝑉 ? 1. . . 1 N𝑒 2B(
∑.FD N.D IN %⁄2 ~c⃗p. 2c⃗pD ~J
𝑑𝑟⃗′! 𝑑𝑟⃗′" . . . . . 𝑑𝑟⃗′? O
By posing 𝑟′ = ®𝑟⃗′/ − 𝑟⃗′r ®
%
𝑍 = 𝑉 ? 1. . . 1 N𝑒 2B(
and differentiating with respect to 𝑉 we obtain
∑.FD N.D IN %⁄2 cpJ
𝑑𝑟⃗′! 𝑑𝑟⃗′" . . . . . 𝑑𝑟⃗′? O
O
X
𝜕𝑉/r E𝑉 !⁄# 𝑟′F 𝑟′ 2B% ∑.FD N.D IN %⁄2 cpJ
𝜕 𝑍⁄𝑉 ?
𝛽
= 1. . . 1 − M
𝑒 (
𝑑𝑟⃗′! 𝑑𝑟⃗′" . . . . . 𝑑𝑟⃗′?
𝜕𝑉
6
𝜕𝑟′
𝑉
/|r
by returning to dimensioned variables and 𝑟 = ®𝑟€⃗^ − €𝑟z⃗®
𝜕𝑉/r (𝑟) 𝑟 2B% ∑.FD N.D (c)
𝜕 𝑍 ⁄𝑉 ?
1
𝛽
= ? 1. . . 1 − M
𝑒 (
𝑑𝑟€€€⃗𝑑𝑟
€€€€⃗
! €€€⃗.
" . . . . 𝑑𝑟
?
𝜕𝑉
𝑉
6
𝜕𝑟 𝑉
/|r
and
𝑔(𝑟⃗) =
𝑉
1
1. . . 1 M 𝛿 E𝑟⃗ + €𝑟⃗^ − 𝑟€z⃗F 𝑒 2BN: 𝑑𝑟€€€⃗𝑑𝑟
€€€€⃗
! €€€⃗.
" . . . . 𝑑𝑟
?
𝑁(𝑁 − 1) 𝑍
In the first equation we get𝑟⃗
𝑑𝑉/r (𝑟)
𝜕 𝑍⁄𝑉 ? −𝑍𝛽𝑁(𝑁 − 1)
=
1
𝑟
𝑔(𝑟)𝑑𝑟⃗
𝜕𝑉
6𝑉?7"
𝑑𝑟
therefore as
?
𝑃 = 𝑘S 𝑇 N N +
;@-y ⁄N :
;N
Oby deriving the logarithm
𝑑𝑉/r (𝑟)
𝑁 𝛽𝑁(𝑁 − 1)
𝑃 = 𝑘S 𝑇 P −
1𝑟
𝑔(𝑟)𝑑𝑟⃗Q
"
𝑉
6𝑉
𝑑𝑟
assuming 𝑁 ≈ 𝑁 − 1 ; 𝜌 =
?
N
𝑃 = 𝑘S 𝑇 P𝜌 −
𝑑𝑉/r (𝑟)
𝜌"
1𝑟
𝑔(𝑟)𝑑𝑟⃗Q
6𝑘S 𝑇
𝑑𝑟
Let us recall the expression of pairs density function
𝜌𝑔(𝑟⃗) =
!
?2!
∑ 𝛿 E𝑟⃗ + 𝑟€⃗^ − 𝑟€z⃗Fso𝑔(𝑟⃗) =
N
?(?2!)
∑ 𝛿 E𝑟⃗ + €𝑟⃗^ − 𝑟€z⃗F
If the gas is not perfect but still dilute, only pair interactions will occur
𝑔(𝑟⃗) =
with
𝑉
1. . . 1 𝛿(𝑟⃗ + 𝑟€€€⃗! − €€€⃗)
𝑟" 𝑒 2BN: 𝑑𝑟€€€⃗𝑑𝑟
€€€€⃗
! €€€⃗.
" . . . . 𝑑𝑟
?
𝑍
1
1
𝑉? = M 𝑉/r E𝑟€⃗,
€⃗^ − €𝑟z⃗®F = M 𝑉/r E®𝑟€⃗^ − €𝑟z⃗®F = M 𝑉/r (𝑟)
^ €𝑟z⃗F = M 𝑉/r E®𝑟
2
2
/{r
/{r
/|r
/|r
and finally remains only
𝑔(𝑟) = 𝑒 2BN.D(c) assuming that the probability of finding a pair to the distance 𝑟 is given by a Boltzmann
factor. The normalization of pairs density function is clearly correct.
Using this formula, repeating the expression of the pressure, followed by an integration by parts
𝑃 = 𝑘S 𝑇 P𝜌 +
If we find the case of the ideal gas 𝑉/r = 0
𝜌"
1E1 − 𝑒 2BN.D (c) F 𝑑𝑟⃗Q
2
Example of the gas hard spheres: for 𝑉/r → ∞ 𝑟 < 𝜎
for 𝑉!" = 0 𝑟 > 𝜎
𝑃 = 𝑘S 𝑇 P𝜌 +
so 𝜌 =
𝜌"
𝜌" 4 #
1 𝑑 𝑟⃗Q = 𝑘S 𝑇 P𝜌 +
𝜋𝜎 Q
2
2 3
?
N
𝑃 = 𝑘S 𝑇
by writing
𝑁
𝑁2 #
𝜋𝜎 @
:1 +
𝑉
𝑉3
1
?"
1+N#
developing at first order
or
𝜋𝜎 #
≈1−
𝑁2 #
𝜋𝜎
𝑉3
2
𝑃 :𝑉 − 𝑁 𝜋𝜎 # @ = 𝑁𝑘S 𝑇
3
𝑃(𝑉 − 𝑏) = 𝑛𝑅𝑇
with 𝑏𝑁 the half volume of hard spheres.
So we recover the equation of Van der Waals simplified in this case, and we made the link between microscopic
and macroscopic vision. We could find the rest of the equation taking into account an attractive potential.
USTH
Master NANO - Mathematical methods
1. A walker performs a one dimensional random walk, made of steps of length l at
timesteps τ
(a) Show that after N steps the probability of having done (N + S)/2 steps to the
left and (N − S)/2 steps to the right is
P(S) =
N!
N−S
N
( N+S
2 )!( 2 )!2
(b) Suppose that N is large in comparison to 1 and S and using Stirling’s formula
deduce that
S2
P(S)
)=−
ln(
P0
2N
with P0 constant
(c) Deduce that the probability of being at abscissa x = Sl is
1 1 1 − x2
) 2 e 4Dt
P(x) = (
2 πDt
with ct = Nl and D = 12 cl the diffusion constant
(d) Check that P verifies the diffusion equation (Fick’s law)
∂P
∂2 P
=D 2
∂t
∂x
2. Consider a set of identical, slightly coupled harmonic oscillators with Hamiltonian
H=
− k εT
Each oscillator has a probability e
b
p2 1 2
+ kx
2m 2
of having energy ε.
By introducing β = kb1T , and derivating with respect to this parameter compute averages of x2 and p2 along
R R
< F >=
x
−
p F(x, p)e
R R
ε(x,p)
kb T
ε(x,p)
− k T
b
x pe
and show that the average of ε is equal to kb T .
dxd p
USTH
Thermodynamics for physics and chemistry
1. Consider a mirror suspended from a torsion wire and placed in an enclosure thermostat at
278.1 K. It is found that the mirror makes a mean square angle ✓¯2 = 4.178.106 U.S.I with its
equilibrium position. Knowing that the spring constant of the wire is 9.428.10 16 kg.m2 s 2
estimate the Boltzmann constant by assuming that the system obeys the distribution of
the same name.
We recall that
r
Z +1
⇡
↵x2
e
=
↵
1
2. We consider a perfect gas of quantum particles (fermions or bosons) confined in an enclosure
of volume V . The particles have mass m and spin S.
(a) Show that by noting g = 2S + 1, with the approximation the density of states gs (✏) of
the system for a energy ✏ takes the form
gV m3/2 p
gs = p 2 3 ✏
2⇡ h̄
(b) Show that taking into account the quantum statistical effects the number of particles
having an energy between ✏ and ✏ + d✏ is
p
gV m3/2
✏d✏
dN = p 2 3 ✏ µ
2⇡ h̄ e kB T ± 1
where µ is the chemical potential of the system, kB the Boltzmann constant. Specify
the sign for the case of fermions.
(c) The number of particles is N . Explain how in principle one could after the change of
variable z = kB✏ T link the density of the system and its temperature to its chemical
potential.
(d) Express the grand potential ⌦ of the system and show that after passing to the continuum (by using the density of states) we obtain
✏ µ
kB gV T m3/2 Z 1 p
⌦=⌥ p 2 3
✏ ln(1 ± e kB T )d✏
0
2⇡ h̄
(e) Express in the same way the total energy E of the system.
(f) Using integration by parts, knowing that ⌦ =
P V with P the pressure show that
2
PV = E
3
(g) Comment on this result by comparing it to the results of classical statistical thermodynamics for a perfect gas.
1
USTH - Master AMSN
Thermodynamics in physics and chemistry
Consider a non-interacting classical two-dimensional (2D) gas of N non-relativistic identical particles confined in a harmonic trap with hard walls at r = R
1
mω 2 r2 for r < R
2
Vtrap (r) =
,
∞
for r ≥ R
where m is the particles mass, ω is the trap angular frequency and r is the radial coordinate.
(a) (2 points) Show that the density distribution n(r) of this gas in thermal equilibrium
at temperatureR T is n(r) = Ce−βVtrap (r) , with β −1 = kB T . Compute C such that n(r) is
normalized as n(r)d2 r = N . Sketch n(r)/n(0) as a function of r/R.
Next, this gas is set to rotate at a fixed angular velocity Ω around the z axis, defined as
perpendicular to the gas plane and intersecting it at the center of the trap r = 0. The thermodynamics of a non-interacting gas in equilibrium in the rotating frame can be calculated
by replacing the single-particle Hamiltonian in the lab frame hlab = p2 /(2m) + Vtrap (r) by
hrot = hlab − Ω · L, where Ω = Ω ẑ (ẑ is the unit vector along the z axis) and L = r × p
is the angular momentum of the particle (note that L is collinear with ẑ here).
(b) (2 points) Show that hrot can be written in the form hrot = 12m (p − mΩ × r)2 + Veff (r),
where Veff (r) = 12 m(ω 2 − Ω2 )r2 for r < R (and ∞ otherwise), and interpret this result.
(c) (2.5 points) Suppose the rotating gas is in equilibrium at a temperature T . Show that its
density distribution is nΩ (r) = De−βVeff (r) , and calculate D to normalize this distribution
2
2)
as in (a). For simplicity, express your result in terms of the parameter α = m(ω2kB−Ω
R2 .
T
Sketch nΩ (r)/nΩ (0) as a function of r/R for slow (Ω ω), fast (Ω = ω), and ultrafast
rotation (Ω ω).
(d) (1.5 points) Show that the canonical partition function of the classical gas has the form
Z = Z0(1 − e−α)N where Z0 is a proportionality factor independent of R (you do not need to
calculate Z0).
(e) (2 points) Using the partition function in (d), calculate the force F = kBT ∂ log Z
∂R
F
exerted by the gas on the disk walls, and thus the pressure P = 2πR
. Express your results
P πR2
for N
as a function of α.
kB T
Find P for fast rotation (Ω = ω) and its asymptotic form in the limit of ultrafast
rotation (Ω ω), and interpret your results.
Useful reminders:
The canonical partition function of a classical gas of N identical particles is
R 2 2 −βh(r,p) N
Z = h2N1 N !
d rd p e
, where h(r, p) is the single-particle Hamiltonian.
RR
0
2
1−e−λR
2λ
2
pπ
re−λr dr =
R∞
−∞
e−λu du =
2
(recall the Jacobian in polar coordinates is dxdy = rdrdθ)
λ
A · (B × C) = B · (C × A) = C · (A × B)
A × (B × C) = (A · C)B − (A · B)C
Problem 2
Consider non-interacting spin-1/2 fermions in two dimensions (2D) with a linear dispersion relation
± (~k) = ±~v|~k| .
Positive energy states (with energy + ) define the conduction band and negative energy
states (with energy − ) define the valence band. Assume that the allowed wavevectors
~k = {kx , ky } correspond to periodic boundary conditions over a square region of area A.
At temperature T = 0 the valence band is completely filled and the conduction band is
completely empty. At finite T , excitations above this ground state correspond to adding
particles (occupied states) in the conduction band or holes (unoccupied states) in the
valence band.
(a) (2.5 points) Find the single-particle density of states D() as a function of the energy
in terms of ~, v, A. Sketch D() over both the negative and positive energy region.
In the next two parts we will argue that the chemical potential µ(T ) = 0 at any
temperature T (which can be assumed so for the rest of the problem).
(b) (2 points) Using the Fermi-Dirac distribution, show that if µ(T ) = 0 then the probability of finding a particle at energy is equal to the probability of finding a hole at
energy −.
(c) (1.5 points) Particle number conservation requires that the number of particles in the
conduction band must equal the number of holes in the valence band, Np = Nh . Show
that if µ(T ) = 0 then this condition holds at all T . (Do not worry if the integrals are
formally divergent – they will be cut off in any physical system.)
(d) (2.5 points) Find the total internal energy of the excitations above the T = 0 state,
U (T ) − U (0), expressed in terms of A, v, ~, kB . Note that since we are subtracting U (0),
in the valence band you only need to count the energy associated with holes.
Useful integral:
1
n!
where ζ(n + 1) =
P∞
k=1
Z
0
∞
xn
= (1 − 2−n )ζ(n + 1) ,
ex + 1
k −n−1 is the Riemann zeta function.
(e) (15 points) Use your answer in (d) to find the heat capacity at constant area
CA (T ) ∝ T α . What is the exponent α?
Problem inspired by Yale university online material
USTH
Master in Nanomaterials
Thermodynamics for physics and chemistry
December 14th 2018 (Durée : 2 h )
1. A gas of N hydrogen molecules (H2 ) without interactions is at thermal equilibrium at temperature T .
(a) Let us suppose that each molecule can be considered as a simple quantum
mechanical oscillator with frequency ω = 4111 cm −1 . Give an expression of
the corresponding partition function Zvib for this gas.
(b) Let us assume that each molecule is a quantum molecular rotator with energies
EJ = BJ(J + 1)
where J is an integer
B=
h
8π 2 cI
and
m 2
d
2
is the inertia matrix diagonal value, for which we assume that the molecule is
rigid m being the mass of the hydrogen atom and d the size of the molecule
(74.14 pm).
Let us remind that in quantum mechanics each rotation state is characterized
by a quantum mechanical number m going from −J to J.
Write the partition function zrot for this gas and approximate it for high
temperatures.
I=
(c) Give the translation partition function if the gas is confined in a volume V .
(d) Compute heat capacity at low temperatures for the gas.
2. Now we approximate the state equation
P = kB T (n + B2 (T )n2 )
.
with n = (N/V ) the density and B2 (T ) a coefficient.
(a) One shows that
1
B2 (T ) =
2
Z
f (r)d3 r
with
f (r) = e−βu(r) − 1
and u(r) the intermolecular interaction potential. Expand within first order
as a function of βu and compute B2 .
1
(b) Besides
P
N
=1−
nkB T
6kB T V
Z
g(r)r
du(r) 3
d r
dr
Supposing that g(r) ≈ e−βu(r) prove the previous result with an interaction
by parts and an approximation.
(c) Now we assume
−12
r
u(r) = 4(
σ
−6
−
r
σ
)
Find the minimum of u(r), plot the function, and conclude about phase
transitions.
3. Suppose that we have N bosons with density of states
g(E) = 0
for E < 0 and
g(E) =
N0
E0
else.
(a) Express N and the chemical potential µ. Show that
1
µ = − (1 − e−βE0 (N/N0 ) )
β
(b) Can the system exhibit Bose-Einstein condensation ?
2
USTH
Master 1 AMSN
Thermodynamics for physics and chemistry
January 11th 2016 (Durée : 2 h )
1. A gas of N diatomic molecules (X2 ) without interactions is in thermal equilibrium at temperature T .
(a) Let us assume that, for the calculation of the vibrational properties, each molecule X2 can be considered as
a simple harmonic harmonic quantum mechanical oscillator of frequency w in cm 1 . Give an expression
for the partition function Zvib of this gas.
(b) Let us suppose that, for the calculation of the rotational properties, the molecule can be treated as a rigid
mechanical quantum rotator. rigid. Thus, the quantum states of rotation have energies of the form
EJ = BJ(J + 1)
where J is the rotation quantum number and
B=
h
8p2 cI
with
M 2
d
2
the moment of inertia, for which we assume a a rigid dumbbell model, M being the mass of the atom of the
atom X and d the size of the molecule
Recall that in quantum mechanics, in addition to the quantum number J, each spin energy state is also
characterized by a quantum number a quantum number m, which varies from J to J.
Write a formal expression for zrot as a function of rotational partition of this gas. Evaluate it in the limit of
high temperature.
I=
(c) Also in the high temperature limit, calculate the total average energy, including translational including
translation, vibration and rotation.
(d) Calculate the specific heat at low temperature, assuming that the temperature is still high enough for the
molecules remain in gaseous form
(e) Sketch the variation of the specific heat as a function of temperature with an approximate numerical application ; what really happens at high What really happens at high temperatures ?
(f) Comment on the following graph giving the heat capacity as a function of the logarithm of the logarithm
of the temperature and give the scales of the the ordinate axis
1
2. A classical monoatomic, non-ideal gas with N particles confined in volume V is in thermal equilibrium at
temperature T .
An approximation of the equation of state of this gas is
P = kB T (n + B2 (T )n2 )
.
Here, n = (N/V ) is the density and B2 (T ) is called the second coefficient of the virial .
(a) Derive an expression for the average energy E of this gas.
(b) Derive an expression for the heat capacity at constant volume, Cv Cv of this gas.
(c) Derive an expression for the entropy S of this gas.
(d) Deduce an expression for the chemical potential µ of this gas.
(e) It can be shown that
B2 (T ) =
1
2
with
Z
f (r) = e
f (r)d 3 r
bu(r)
1
and u(r) the two-body interaction potential, assumed to describe the system. Develop to first order in bu
and recalculate B2 .
(f) On the other hand, we have the virial equation
P
=1
nkB T
Assuming g(r) ⇡ e
bu(r)
N
6kB TV
Z
g(r)r
du(r) 3
d r
dr
find the previous result by an integration by parts and a physical argument.
(g) We take as interaction
u(r) = 4e(
r
s
12
r
s
6
)
Find the minimum of u(r) and plot the shape of this function. A such a system exhibit one or more phase
transitions when the temperature when the temperature is decreased ?
2
3. White dwarfs
The white dwarfs are the stage of end of life of the stars of solar type stars. These stars are small : their
typical radius R is of the Earth, about 5000 km, compared to the 700 000 km radius of the Sun. However,
they have a mass M almost equal to that of the Sun Ms = 2.1030 kg. The central regions regions of a white
dwarf are composed mainly of C and O (isotopes 12 and 16 respectively), elements that the star too little
massive can not burn (there are indeed stars in which the physical stars in which the physical conditions
allow to start nuclear reactions with to start nuclear reactions with C and O). The temperature of the central
regions is estimated at 107 K.
In special relativity, the total energy of a particle of mass mass m and momentum p is given by
E 2 = p2 c2 + m2 c4
For ultra-relativistic particles (p
mc ), we obtain
E ⇡ pc
(a) Density of a white dwarf.
i. Estimate the density of a white dwarf, in kg/m3 then in number of nuclei per m3 . Calculate the available
volume per nucleus.
ii. What can you say about the state of matter in the central regions of a of a white dwarf : neutral gas,
ions, dissociated plasma of electrons on one side and nuclei on the other ?
(b) Properties of the gas of nuclei
i. Estimate the degeneracy temperature of the nuclei, i.e., the temperature below which the nuclei the
temperature below which the quantum effects become important. We recall that the Bose temperature
TB is given by
2p h̄2
TB =
mkb
✓
N 1 1
V 2s + 1 I
◆2
3
where m denotes the mass of the bosons, N their total number, s their spin, V the volume and I the
integral :
2
I=p
p
Z • 1/2
x
0
ex
1
dx ⇡ 2.612
ii. Can we apply the Maxwell-Boltzmann statistics to the nuclei ?
iii. Estimate the order of magnitude of the pressure due to the nuclei.
iv. Properties of the electron gas
A. Calculate the value of the impulse pF corresponding to the Fermi level of the electrons assuming
that the temperature of the system is zero.
B. Deduce that the electrons at the Fermi level are relativistic, i.e. pF ⇡ me c (with me the mass of an
electron). Calculate the position of the Fermi level for zero temperature. We will take for zero of
the energies, the rest energy of the particles.
C. Justify a posteriori the validity of the approximation T = 0.. We will admit that the relativistic
corrections necessary according to the previous question do not modify the order of magnitude of
TF . What is the origin of the pressure due to the electrons ? Estimate its order of magnitude.
(c) Equilibrium of the white dwarf
We are now interested in the equilibrium of the white dwarf. The previous questions have shown that we
can consider the medium as a mixture of two unrelated gases : a gas of electrons and a gas of nuclei. a
gas of nuclei. The kinetic energy of the electrons is sufficiently large enough to neglect the kinetic energy
of the nuclei and the electrostatic interactions between the charged particles. The influence of the nuclei,
negligible from the kinetic point of view, is is however preponderant when we are interested in the gravitational forces. gravitational forces. In all the following, we place ourselves in the approximation where the
temperature of the the approximation where the temperature of the system is zero.
i. Density of energy
3
A. Give the relativistic expression of the kinetic energy density E/V in the form of an integral depending on the variable variable x = mpe c with p the impulse of the electrons of mass me and c the speed
of light. We will not try to to explain this expression in the general case.
B. What happens to these expressions in the ultra-relativistic case ultra-relativistic case, i.e. for x 1 ?
We will calculate the first two dominant terms of the limited development of E/V as a function of
x.
ii. Equilibrium radius of the white dwarf The gravitational pressure allows to counteract the effects of the
degeneracy pressure of the electrons. In the hypothesis of a homogeneous distribution distribution of
the mass in the white dwarf, the gravitational forces are manifested by an energy
EG =
3 GM 2
5 R
with G the gravitational constant.
A. How is the equilibrium condition of the white dwarf written ?
B. Determine the equilibrium radius R0 as a function of the mass M in the ultra-relativistic case. In the
non-relativistic case and and making the approximation that the mass of the neutron and that of the
proton are equal, we admit that we end up with :
4h̄2
R0 =
9pGme
✓
9p
4m p
◆5
3
1
1
M3
Discuss the results obtained. Show in particular the existence of of a limit mass Mc in the ultrarelativistic case. Compare Mc to the mass of the Sun.
4. Data
Mass of the proton m p = 1.67.10 27 kg
Mass of the electron me = 9.11.10 31 kg
h̄ = 1.05.10 34 USI
G = 6.6742.10 11 USI
c = 299792458 USI
kb = 1, 38.10 23 USI
4