Aerothermodynamics of high speed ows - Ltas-aea ::Welcome · 2019-05-13 · The Dawn of the quantum theory The de Broglie waves Exercise 1 Calculate the de Broglie wavelength of an

Aerothermodynamics of high speed flowsAERO 0033–1

Lecture 10: Quantum Mechanics and Statistical Physics (introduction)

Thierry Magin, Greg Dimitriadis, and Johan [email protected]

Aeronautics and Aerospace Departmentvon Karman Institute for Fluid Dynamics

Aerospace and Mechanical Engineering DepartmentFaculty of Applied Sciences, University of Liege

Room B52 +1/433Wednesday 9am – 12:00pm

February – May 2019

Magin (AERO 0033–1) Aerothermodynamics 2018-2019 1 / 80

Outline I

1 The Dawn of the quantum theoryThe photoelectric effectThe Hydrogen atomic spectrumThe de Broglie wavesThe Bohr theory of the hydrogen atomHeisenberg’s uncertainty principle

2 The Schrodinger equation and a particle in a boxThe classical wave equationThe Schrodinger equationA particle in a boxPostulates of quantum mechanics

3 Rovibrational energy levels of diatomic moleculesHarmonic oscillatorRigid rotator


Outline II

4 Statistical physicsPure perfect gasMixture of inert perfect gasesMixture of reacting perfect gases


First Solvay conference in 1911 at the Hotel Metropole (Brussels)

Seated (L-R): W. Nernst, M. Brillouin, E. Solvay, H. Lorentz, E. Warburg, J. Perrin, W. Wien,M. Curie, and H. Poincare, standing (L-R): R. Goldschmidt, M. Planck, H. Rubens, A.

Sommerfeld, F. Lindemann, M. de Broglie, M. Knudsen, F. Hasenohrl, G. Hostelet, E. Herzen,J.H. Jeans, E. Rutherford, H. Kamerlingh Onnes, A. Einstein and P. Langevin


The Dawn of the quantum theory

Outline

1 The Dawn of the quantum theoryThe photoelectric effectThe Hydrogen atomic spectrumThe de Broglie wavesThe Bohr theory of the hydrogen atomHeisenberg’s uncertainty principle

2 The Schrodinger equation and a particle in a box

3 Rovibrational energy levels of diatomic molecules

4 Statistical physics


The Dawn of the quantum theory The photoelectric effect

The photoelectric effect

Hertz’s experiment (1886)UV light causes electrons to be emitted from a metallic surface:

Electron KE independent of incident radiation intensityThreshold frequency ν0 below which no electrons are ejectedAbove ν0, electron KE varies linearly with frequency ν

Einstein’s interpretation (1905)Planck’s hypothesis for the energy of electrons in the constituents ofradiating blackbody: E = hν, with h = 6.626 10−34 J sEinstein: the photoelectric effect is caused by absorption of quanta oflight (photons)



The photoelectric effect

Kinetic Energy of ejected electron

KE = hν − φ

energy of incident photon - minimum energy required to remove anelectron from the surface of the metal

φ = hν0

minimum frequency ν0 that will eject an electron: frequency requiredto overcome the work function of the metal

Einstein obtained a value of h in close agreement with Planck’s valuededuced from the blackbody radiation formula



Exercise

When lithium is irradiated with light, the kinetic energy of the ejectedelectrons is 2.935 ×10−19 J for λ = 300 nm and 1.280 ×10−19 J forλ = 400 nm. Calculate

1 The Planck constant,

2 The threshold frequency,

3 The work function of lithium,

for these data. The speed of light is 2.998× 108 m/s.


The Dawn of the quantum theory The Hydrogen atomic spectrum

The Hydrogen atomic spectrum

Every atom, when subjected to high temperatures or electricaldischarge, emits electromagnetic radiation of characteristic frequencies

Characteristic emission spectrum of an atom: certain discretefrequencies called linesHydrogen atomic spectrum: Lyman (1906), Balmer (1885), andPaschen (1908) series

The Balmer and Rydberg formulae for the frequency of the lines(1/λ = ν/c , RH = 109 677.581 cm−1)

νBalmer = 8.2202×1014

(1− 4

n2

)Hz ;

1

λ= RH

(1

n21

− 1

n22

)cm−1


The Dawn of the quantum theory The de Broglie waves

The de Broglie waves

de Broglie (1924): wave-particle duality of light and matterEinstein’s eq. for photons: λ = h

p , with p = mvde Broglie: both light and matter obey this equation

Experimental observationWhen a beam of X rays / electrons is directed at a crystalline substance, the beamis scattered in a definite manner characteristic of the atomic structure of the crystalDiffraction occurs because the interatomic spacings in the crystal are about thesame as the wavelength of the X-rays / electrons


The Dawn of the quantum theory The de Broglie waves

Exercise

1 Calculate the de Broglie wavelength of an electron traveling at 1% ofthe speed of light. Data:

The mass of an electron is 9.109× 10−31 kg,The speed of light is 2.998× 108 m/s.

2 Calculate the de Broglie wavelength of a ball of 45 g traveling at avelocity of 40 m/s.


The Dawn of the quantum theory The Bohr theory of the hydrogen atom

The Bohr theory of the hydrogen atom

Bohr (1911)Hydrogen atom: rather massive nucleus with one associated electronNucleus assumed to be fixed with electron revolving about it

Classical physics:Because the electron is constantly accelerated, it should emit electromagneticradiation and lose energyIt will spiral into the nucleus and a stable orbit is classically forbidden

Two nonclassical assumptions:Existence of stationary electron orbitsThe de Broglie waves of the orbiting electron must be in phase as the electronmakes one complete revolution as shown in case (a), as opposed to cases (b) and (c)



The Bohr theory of the hydrogen atom

Bohr orbits are quantized

2πr = nλ, n = 1, 2, 3, . . .

The angular momentum of the electron about the proton is quantized

λ = hmev⇒

mevr = n~, n = 1, 2, 3, . . . , with ~ =h

2π

Bohr radius: a0 = 4πε0~2

mee2 = 0.529× 10−10m

Coulomb attraction force = centrifugal force

e2

4πε0r2=

mev2

r⇒ r = a0n

2



Application: derivation of Rydberg’s formula

Total energy of the electron in a hydrogen atom

En = KE + V (r)

=1

2mev

2 − e2

4πε0r

=1

2

(e2

4πε0r

)− e2

4πε0r

= − e2

8πε0r

= − mee4

8πε20h

2

1

n2= −13.598 eV

1

n2

Observed spectrum of the hydrogen atom (RH = mee4

8πε20h

2 )

∆E = En2 − En1 = RH

(1

n21

− 1

n22

)= hν




The Dawn of the quantum theory Heisenberg’s uncertainty principle

Heisenberg’s uncertainty principle

The position and the momentum of a particle cannot be specifiedsimultaneously with unlimited precision: ∆x∆p & h

e.g ., to measure the position of an electron within a distance ∆x , we can use light with a

wavelength λ ∼ δx . For the electron to be “seen” a photon must interact or collide in

some way with the electron. During the collision, some of the photon momentum

p = h/λ will be transferred to the electron, changing its momentum. A smaller

wavelength leads to a smaller ∆x and a larger ∆p.

If we wish to locate any particle within a distance ∆x , then weintroduce atomically an uncertainty in its momentum ∆pThis uncertainty does not stem from poor measurement orexperimental technique, but it is a fundamental property of the act ofmeasurement itselfThe Heisenberg uncertainty principle is of no consequence formacroscopic bodies but it has very important consequences in dealingwith atomic and subatomic particles


The Dawn of the quantum theory Heisenberg’s uncertainty principle

Exercise

1 Calculate the uncertainty in velocity if we wish to locate an electronwithin an atom so that ∆x is approximately 0.5× 10−10m.

The mass of an electron is 9.109× 10−31 kg.

2 Calculate the uncertainty in the position of a baseball thrown at 40m/s velocity if we measure its momentum with a relative error of10−8.


The Schrodinger equation and a particle in a box

Outline

1 The Dawn of the quantum theory

2 The Schrodinger equation and a particle in a boxThe classical wave equationThe Schrodinger equationA particle in a boxPostulates of quantum mechanics




The Schrodinger equation and a particle in a box The classical wave equation

The classical wave equation

The one dimensional wave equation describes the motion of avibrating string

Newton’s law (T=tension of the string and µ its linear mass density)

µ dx∂2u

∂t2= T2 sinβ − T1 sinα

∼(T∂u

∂x

)x+dx

−(T∂u

∂x

)x

⇒ ∂2u(x , t)

∂t2= v2∂

2u(x , t)

∂x2with the propagation speed v =

√T

µMagin (AERO 0033–1) Aerothermodynamics 2018-2019 17 / 80

The Schrodinger equation and a particle in a box The Schrodinger equation

The Schrodinger equation (1925)

The Schrodinger eq. is the fundamental equation of quantummechanics

This eq. cannot be derived, just as Newton’s laws are fundamentalpostulates of classical mechanicsIt gives a description of a quantum system evolving with time, such asatoms, molecules, and subatomic particlesThe Schrodinger equation was developed principally from the deBroglie hypothesisThe solutions to this eq. are called wave functions

The time-independent Schrodinger eq. for finding the wave functionof a particle

Even though we cannot derive the Schrodinger eq., we can at leastshow that it is plausible

⇒ Finding the wave function of a particle


The Schrodinger equation and a particle in a box A particle in a box

A particle in a box

Classical 1D wave equation

u(x , t): amplitude of displacement of the string from its equilibriumhorizontal position

∂2u(x , t)

∂x2=

1

v2

∂2u(x , t)

∂t2

Time-independent eq.: separation of variables

u(x , t) = ψ(x) cosωt

⇒ ∂2ψ

∂x2+ω2

v2ψ(x) = 0

Seeing that ω = 2πν and v = νλ

∂2ψ

∂x2+

(2π

λ

)2

ψ(x) = 0



A particle in a box

Time-independent Schrodinger eq.

Idea of de Broglie: matter waves

λ =h

p

Total energy of the particle: kinetic energy + potential energy

E =p2

2m+ V (x)

Momentum: p = mv ⇒ λ = hp

= h√2m[E−V (x)]

Substituting λ the wave eq., one obtains

− ~2

2m

∂2ψ

∂x2+ V (x)ψ(x) = Eψ(x)



Exercise

Calculate the wavefunction and energy of a particle in 1D box of size a.

The particle is free in the box (V (x) = 0),

It experiences no potential energy in the region 0 ≤ x ≤ a,

The wave function vanishes at the box boundary:ψ(x = 0) = 0 and ψ(x = a) = 0.



Geometry of theproblem

Schrodinger eq. for 0 ≤ x ≤ a:

d2ψ

dx2+

2mE

~2ψ(x) = 0

with ψ(x = 0) = 0 and ψ(x = a) = 0

General solution

ψ(x) = A cos kx + B sin kx

with k =√

2mE~

Boundary conditionsA = 0 and B sin ka = 0

⇒ ka = nπ, n = 1, 2, . . .

Energy and wave function

En =h2n2

8ma2, ψn(x) = B sin

nπx

a, n = 1, 2, . . .



Interpretation of the wave function

Born’s interpretation of the wave functionψ∗(x)ψ(x)dx is the probability that the particle is located between x and x + dxThe probability that the particle is found outside of 0 ≤ x ≤ a is zero: ψ(x) = 0ψ(x) is assumed to be a continuous function ⇒ ψ(x = 0) = 0 and ψ(x = a) = 0

The wave functions must be normalizedThe particle is certain to be found in the region 0 ≤ x ≤ a:∫ a

0ψ∗(x)ψ(x)dx = 1

|B|2∫ a

0sin2 kx dx = 1

|B|2[

12x −

sin 2kx

4k

]a0

= 1 (2 sin2 kx = 1− cos2kx)

⇒ B =

(2

a

)1/2

ψn(x) =(

2a

)1/2sin nπx

a, n = 1, 2, . . .

Probability of finding a particle between x1 and x2Prob(x1 ≤ x ≤ x2) =

∫ x2x1ψ∗(x)ψ(x)dx



Quantization of the energy

Quantized energy: En = h2n2

8ma2 n = 1, 2, . . .

The energy of the particle is quantized and the integer n is called aquantum numberQuantization arises naturally from the boundary conditions, beyond thestage of Bohr and Planck where the are introduced in an ad hocmanner

Schrodinger, Annalen der Physik 79, 361 (1926)“In this communication, I wish to show that the usual rules of quantization can bereplaced by another postulate (the Schrodinger equation) in which there occurs nomention of whole number. Instead, the introduction of integers arises in the samenatural way as, for example, in a vibrating string, for which the number of nodes isintegral. The new conception can be generalized, and I believe that it penetratesdeeply into the true nature of the quantum rules.”



The energy levels, wavefunctions (a), and

probability densities (b)for the particle in a box

Application: excitation of πelectrons in butadiene

H2C = CHCH = CH2 is assumedto be a linear moleculeLength a= 578 pm with 2 C=Cbonds (2×135 pm), 1 C-C bond(154 pm), 2 C radii (2×77 pm)Pauli exclusion principle: eachenergy state can hold only twoelectrons (with opposite spins)The four π electrons fill the firsttwo energy levelsFirst excited state (3rd level):

∆E = h2

8mea2 (32 − 22)Wavenumber:∆Ehc = 1

λ = 4.54× 104cm−1

1/λ ∼ 4.61× 104cm−1, butadieneabsorption band



Exercise

For a particle moving in a 1D box of size a, show that:

The mean value of x is < x >= a/2, and the mean square deviation isσ2x =< x2 > − < x >2= (a2/12)[1− 6/(π2n2)].

⇒ As n becomes large, this value agrees with the classical value. Theclassical probability distribution is uniform.

Correspondance principle: quantum mechanical results and classicalmechanical results tend to agree in the limit of large quantumnumbers.

Show that Heisenberg uncertainty principle is satisfied:

σxσp =~2

(π2n2

3− 2

) 12

≥ ~2.



A particle in a 3D box

Time-independent Schrodinger eq. for 0 ≤ x ≤ a, 0 ≤ y ≤ b,0 ≤ z ≤ c :

− ~2

2m∇2ψ = Eψ(x , y , z)

with the Laplacian operator ∇2ψ = ∂2

∂x2 + ∂2

∂y2 + ∂2

∂z2

Boundary conditions

ψ(x = 0, y , z) = ψ(x = a, y , z) = 0, for all y and z

ψ(x , y = 0, z) = ψ(x , y = b, z) = 0, for all x and z

ψ(x , y , z = 0) = ψ(x , y , z = c) = 0, for all x and y




Wavefunction for 0 ≤ x ≤ a, 0 ≤ y ≤ b, 0 ≤ z ≤ c :

ψnxnynz (x , y , z) =

(8

abc

)1/2

sinnxπx

asin

nyπy

bsin

nzπz

c

with nx , ny , nz = 1, 2, . . .

Energy levels

En =h2

8m(n2x

a2+

n2y

b2+

n2z

c2), nx , ny , nz = 1, 2, . . .

Particular case: cube → a = b = c

En =h2

8ma2(n2

x + n2y + n2

z), nx , ny , nz = 1, 2, . . .




The energy levels for a particle in a cube (a=b=c)

The energy levels of a particle in a cube are degenerate (same energyvalue for different index triplets) because of the symmetry introducedwhen a general rectangular box becomes a cube


The Schrodinger equation and a particle in a box Postulates of quantum mechanics

Postulates

The development of quantum mechanics is now sufficiently complete that we can reduce

the theory to a set of postulates. MacQuarrie and Simon have reviewed the following

postulates in their book Physical Chemistry, a molecular approach

Postulate 1. The state of a quantum-mechanical system is completelyspecified by a function ψ(x) that depends on the coordinate of theparticle. All possible information about the system can be derivedfrom ψ(x). This function, called the wave function or the statefunction, has the important property that ψ∗(x)ψdx is the probabilitythat the particle lies in the interval dx , located at the position x .

For simplicity of notation, we have assumed, that only one coordinate (x) is needed to

specify the position of one particle, as in the case of a particle in a one-dimensional box

e.g., particle in 1D box: ψn(x) =(

2a

)1/2sin nπx

a , n = 1, 2, . . .



Postulates

Postulate 2’. To every observable in classical mechanics corresponds alinear Hermitian operator A in quantum mechanics.

e.g., particle in 1D box: total energy operator (Hamiltonian)

H = − ~2

2m

d2

dx2+ V (x)

Definition: A is Hermitian if and only if∫all space

f ∗(x) A g(x) dx =

[∫all space

g∗(x) A f (x) dx

]∗for f (x) and g(x) any two state functions.



Postulates

Postulate 3. In any measurement of the observable associated withthe operator A, the only values that will ever be observed are theeigenvalues an, which satisfy an eigenvalue equation

Aψn = anψn.

Property: The eigenvalues of Hermitian operators are real and theireigen functions are orthonormal.

e.g., particle in 1D box: Hψn(x) = Enψn(x), with energy

En =h2n2

8ma2, n = 1, 2, . . .



Postulates

Postulate 4. If a system in a state described by a normalized wavefunction ψ, then the average value of the observable corresponding toA is given by

< A >=

∫all space

ψ∗Aψ dx .

e.g., particle in 1D box: < x >=∫ a

0 ψ∗n x ψn dx = a/2



Application: commutators

Definition The commutator of two operators is [A, B] = AB − BA.These operators commute if [A, B] = 0.

Schrodinger’s uncertainty relation Consider two operators A andB. Their standard deviations are related by

σaσb ≥ 12

∣∣∣∣∫ ψ∗(x)[A, B]ψ(x) dx

∣∣∣∣Proof Using the relation AB = 1

2(AB + BA) + 1

2[A, B], one obtains the inequality

| < AB > |2 ≥ 14| < [A, B] > |2. Schwarz’s inequality < A2 >< B2 > ≥ | < AB > |2

yields to < A2 >< B2 > ≥ 14| < [A, B] > |2. The proof follows from the definition

σ2c =< (C− < C >)2 > and the following identity: [A− < A >, B− < B >] = [A, B].

There is an intimate connection between commuting operators and the UncertaintyPrinciple. If two operators A and B commute, then their eigen values a and b can bemeasured simultaneously to any precision. If two operators A and B do not commute,then their eigen values a and b cannot be measured simultaneously to any precision.



Exercise

Considering the simultaneous measurement of the momentum andposition of a particle, so that Px = −i~ d

dx and X = x , show

[Px , X ] = −i~I .Based on the previous relation, show that the Heisenberg uncertaintyprinciple is satisfied:

σxσp ≥~2.



Postulates

Postulate 5. The wavefunction, or state function, of a system evolvesin time according to the time-dependent Schrodinger equation

i~∂

∂tΨ(x , t) = HΨ(x , t).

For most of the cases, the solution to this equation is

Ψ(x , t) = ψ(x) exp

(−iE~

t

),

where quantity ψ(x) is a solution to the following time-independent Schrodinger eq.

Hψ(x) = Eψ(x).

Notice that the probability density is then entirely given by the stationary state wave eigenfunctions:

|Ψn(x , t)|2 = |ψn(x)|2.


Rovibrational energy levels of diatomic molecules

Outline



3 Rovibrational energy levels of diatomic moleculesHarmonic oscillatorRigid rotator



Rovibrational energy levels of diatomic molecules Harmonic oscillator

Harmonic oscillator

Two masses connected by aspring: model used to describe

the vibrational motion of adiatomic molecule

Observable: total energy E

Operator: Hamiltonian

H = − ~2

2µ∂2

∂x2 + V (x)

Total energy conservation (classicalNewton law)

1

2µx2 +

1

2kx2 = E

Relative coordinate:x = x2 − x1 − l0Undistorted length: l0Force constant (Hooke’s law): kReduced mass: µ

Schrodinger eq.

− ~2

2µ

∂2ψ

∂x2+

1

2kx2ψ(x) = Eψ(x)

(solution based on Hermitepolynomials)


Rovibrational energy levels of diatomic molecules Harmonic oscillator

Harmonic oscillator

Left: harmonic-oscillator wavefunctions for the first four

energy levels, Right:probability densities

Vibrational frequency

ω =

(k

µ

)1/2

Vibrational energy levels (nodegeneracy)

Ev = ~ω(v +1

2)

Vibrational quantum number:v = 0, 1, 2, . . .

Zero-point energy (uncertaintyprinciple)

E0 =1

2~ω

Constant energy spacing:Ev+1 − Ev = ~ω


Rovibrational energy levels of diatomic molecules Rigid rotator

Rigid rotator

Two masses rotating about theircenter of mass: model used to

describe the rotational motion ofa diatomic molecule

Observable: rotational kineticenergy E

Operator: Hamiltonian

H = L2

2I= − ~2

2Ir2∇2

Rotational kinetic energyconservation (classical Newton law)

L2

2I= E

Angular momentum: L = IωMoment of inertia: I = µr2

Reduced mass: µFixed separation of two masses:r = r1 + r2

Angular speed: ω

Schrodinger eq.

−~2

2I(r2∇2Y ) = EY (θ, φ)

(solution based on Legendrepolynomials)

Rigid-rotator wave function: YSpherical coordinates angles: θ, φ



Rigid rotator

Rotational energy levels:

EJ =~2

2IJ(J + 1)

Rotational quantum number: J = 0, 1, 2, . . .Degeneracy of the energy level:

gJ = 2J + 1

Energy spacing:

EJ+1 − EJ =~2

I(J + 1) = hν, J = 0, 1, 2, . . .

Rotational frequency:ν = 2B (J + 1)

Rotational constant:

B =h

8π2I



Limits of validity of rigid rotor and harmonic oscillatorassumption for the nitrogen molecule

Leroy’s rovibrational potential can be used for the ground electronicstate of the nitrogen molecule

Potential curve of N2 for J=0,20,40,. . .

The 9390 (v,J) rovibrational energy levels for N2 are obtainedtogether by solving the Schrodinger equation

These energy levels are either bound (below the dissociation energy)or quasi-bound (above the centrifugal barrier)



Limits of validity of rigid rotor and harmonic oscillatorassumption for the nitrogen molecule

Rotational energy spacing

0 50 100 150 200 250 300Rotational quantum number J [ - ]

0.0

2.5x10-21

5.0x10-21

7.5x10-21

10.0x10-21

12.5x10-21

15.0x10-21

E(v

,J+

1) -

E(v

,J)

[ J

]

12v = 0

Vibrational energy spacing (J=0)

0 10 20 30 40 50 60Vibrational quantum number v [ - ]

0

1x10-20

2x10-20

3x10-20

4x10-20

5x10-20

E(v

+1,

0) -

E(v

,0)

[ J

]

Non rigid rotor: E (J + 1, v)− E (J, v) is not ∝ JNon harmonic oscillator: E (v + 1, 0)− E (v , 0) 6= constant

[paper AIAA 2009-3837]


Statistical physics

Outline




4 Statistical physicsPure perfect gasMixture of inert perfect gasesMixture of reacting perfect gases


Statistical physics Pure perfect gas

Microscopic description of a nitrogen molecule gas

Potential energy diagram for N2 [Gilmore, 1964]

Translational energy spacing

ET2,1,1 − ET

1,1,1 = 3h2

8mN2a2 =

kBθTN2

, with θTN2= 10−15 K

(a = 1cm)

Internal energy spacing

RotationERJ=1 − ER

J=0 = ~2

IN2= 2kBθ

RN2

,

with θRN2= 2.88 K

VibrationEVv=1 − EV

v=0 = hνN2 = kBθVN2

,

with θVN2= 3393 K

ElectronicEEN2(A) − EE

N2(X) = kBθEN2

, with

θEN2= 72 293 K



Gas composed of independent particles (perfect gas) thatare identical

Let us assume that the total energy of level I for a molecule is givenby its translational, rotational, vibrational and electroniccontributions: εI = εTI + εRI + εVI + εEI , I = 1, 2, . . .

Consider the following system of independent and identical particlesNumber of molecules:

∑I NI = N

Total energy:∑

I NI εI = EVolume: V

A macrostate NI is a population distribution N1,N2, . . . over theenergy levels

At some given instant, the molecules are distributed over the energy levels in adistinct way.In the next instant, due to molecular collisions, the populations of some levels maychange, creating a different set of NI ’s, and hence a different macrostate.

Different states can be associated to the same energy level εI when itis degenerate. The occupancy of an energy level with degeneracy aicompletely defines a microstate for a given macrostate.



Macrostate and microstate

Let us examine some possible macrostates for∑

I NI = 5 particleswith total energy

∑I NI εI = 20 [au] for a system with 8 energy levels

Index I 1 2 3 4 5 6 7 8energy εI [au] 0 1 2 3 5 7 9 11degeneracy ai 2 1 3 2 4 1 3 2

macrostate 1: NI 0 1 1 0 2 1 0 0macrostate 2: NI 1 0 2 0 0 1 1 0macrostate 3: NI 1 1 0 1 1 0 0 1

Example of three possible microstates for macrostate 1

X

I = 3I = 2 I = 5 I = 6

X XX X X

X X

XX

X

X

X X

microstate 1

microstate 3 (bosons)

microstate 2

X



Exercise

Consider particles weakly interacting in a box (cube of volumeV = a3). Introducing the notation n2 = n2

x + n2y + n2

z , withnx , ny , nz = 1, 2, . . ., their translational energy is given byεT = h2/(8mV 2/3)n2

Show that the number of states with an energy less than ε isΓ = 4

3πV /h3(8mε)3/2

Compute the number of nitrogen molecules N0 in a volume of 1 cm3 attemperature T = 0 C and pressure p = 1 atm.Compute Γ assuming that ε = 10 < ε >, where < ε >= 3

2 kBT is theaverage kinetic energy of these particles.Show that in this case the great majority of states are emptyΓ/N0 1.

As consequence, it is possible to group the translational energy levels of similar energiesand to divide the energy scale into successive regions containing a range of energy valuessuch that the following inequality holds ai NI



Counting the number of microstatesTwo types of particles

Molecules and atoms with an odd number of elementary particles arecalled fermions and obey a Fermi-Dirac statistics.Molecules and atoms with an even number of elementary particles arecalled bosons and obey a Bose-Einstein statistics.

Quantum mechanical property: indistinguishability of particles

The number of microstates W in a given macrostate N1,N2, . . .follows Pauli’s exclusion principle

For fermions, only one molecule may be in any given degenerate state at anyinstant (C(ai ,NI ) combination of ai states taken NI times without repetition)

W (NI ) =∏I

ai !

(ai − NI )!NI !

For bosons, the number of molecules that can be in any given degenerate state atany instant is unlimited (C(ai + NI − 1,NI ) combination of ai states taken NI timeswith repetition)

W (NI ) =∏I

(NI + ai − 1)!

(ai − 1)!NI !Magin (AERO 0033–1) Aerothermodynamics 2018-2019 47 / 80


The most probable macrostate

The total number of microstates of the system is defined as

Ω =∑

over all sets of NIsuch that

∑I NI=N

and∑

I NI εI =E

W (NI )

The most probable macrostate NI is that macrostate which has themaximum number of microstates. It is derived to be

NI =ai

eαeβεI ± 1

where quantities α and β are Lagrange multipliers associated with theconstraints (+1 for fermions and -1 for bosons)

That constitutes the thermodynamic equilibrium of the system.For N large, it can be shown that only the largest term in the summakes any effective contribution to Ω and one has:

Ω = Wmax



The most probable macrostate

ProofFor mathematical convenience, one assumes that NI 1 and ai 1,and for fermions, that ai NI

The finiteness of the total energy requires that NI → 0 as εI →∞, however the

high-lying energy levels for which NI is small contribute little to the value of lnW .

For lower levels, we recall that suitable values of NI and ai can be formally obtained

by grouping the translational energy levels of similar energies and dividing the

energy scale into successive regions containing a range of energy values.

Applying Stirling’s formula ln a! = a ln a− a for very large values of a,

one gets, lnW =∑

I

[NI ln

(aiNI± 1)± ai ln

(1± NI

ai

)], where the

upper sign stands for bosons and -1 for fermionsThe constrained optimization problem is:∑

I

[ln(

aiNI± 1)]

dNI = 0,∑

I dNI = 0,∑

I εIdNI = 0

Introducing Lagrange multipliers, the thesis is obtained by cancelingthe terms between brackets in the relation∑

I

[ln(

aiNI± 1)− α− βεI

]dNI = 0



The limiting case: Boltzmann distribution

The values of α and β can be obtained from the constraints∑I NI = N and

∑I NI εI = E

We investigate the limiting case for which |eαeβεI | 1, the physicalmeaning of this assumption will be discussed later. The Boltzmanndistribution is derived as NI = aie

−αe−βεI .Notice that this result is based on indistinguishability of particles as opposed to a purely

classical derivation based on Boltzmann statistics.

Introducing the partition function Q =∑

I aie−βεI , quantity α is

given by α = exp(Q/N) and the following equation is obtained fromthe conservation of particles

NI

N=

1

Qaie−βεI

The energy can be obtained from the relation

E =N

Q

∑I

aiεI e−βεI



Boltzmann’s relationUsing the sparse population property ai NI shown in a previousexercise and the formula ln(1 + x) ∼ x for |x | 1, one obtainslnW =

∑I NI (ln ai

NI+ 1). The number of microstates is given by

ln Ω = lnWmax = N(ln QN + 1) + βE

All possible microstates of a system corresponding to given values of N, E , and the εI ’sare a priori equally probable. Since Ω = Wmax , in the Boltzmann limit, this system spendmost of its time in the macrostate corresponding to the Boltzmann distribution.

Boltzmann postulated a functionalrelationship between the entropy of asystem and the number of microstates

S = kB ln Ω

= kBN(ln QN + 1) + kBβE

where kB is Boltzmann’s constant

The entropy is additive: S1 + S2 =kB ln Ω1 + kB ln Ω2 = kB ln(Ω1Ω2)



Link with Gibbs’ relation

Writing Gibbs’ relation for a single-component gas

dS =1

TdE +

p

TdV − µ

TdN

Using the definition of temperature ( ∂S∂E )V ,N = 1T , and the relations

β = β(E ,N) andQ = Q[V , β(E ,N)], one obtains

β =1

kBT

The energy can be conveniently be derived from the partition functionas follows

E = NkBT2

(∂ lnQ

∂T

)V ,N

The pressure is given by the relation

p = T (∂S

∂V)T ,N − (

∂E

∂V)T ,N = T (

∂S

∂V)T ,N = NkBT (

∂ lnQ

∂V)T ,N



Exercise

In the Boltzmann limit, let us examine the number of ways W inwhich a distribution can occur for values NI = N∗I + ∆NI around theBoltzmann distribution N∗I .

Assuming that |∆NI | N∗I , expand ln(N∗I + ∆NI ) neglecting thirdorder terms.Show that the number of microstate satisfies the following relation

ln WWmax

∼ − 12

∑I

(∆NI

N∗I

)2

N∗I

Consider 1 cm3 of nitrogen gas at 1atm pressure and 0 C. Supposethat the average deviation |∆NI |/N∗I is 0.1%, compute the value of theratio W /Wmax .

We note that the right-hand side is negative showing that the extremum is a maximum.

In a system with a large number of particles, even a small deviation in the distribution

numbers from the Boltzmann values N∗I leads to an enormous reduction in the number of

microstates corresponding to a given macrostate.



Exercise

Show that the Lagrange multiplier is given in the Boltzmann limit bythe following expression

α = ln

[VN

(2πmkBT

h2

)3/2]

+ ln(Q int

).

The internal partition function Q int is not small. This equation shows a posteriori thatthe condition |eαeβεI | 1 for the Boltzmann limit reduces to

V

N

(2πmkBT

h2

)3/2

1.

This requirement would be violated and the gas degenerate, for example, when theparticle mass is very small and the number density is sufficiently large, as in the case of anelectron gas in metals. It is also violated for normal molecular values of the mass m andnumber density N/V when the temperature T is very low. In this situation the particlestend to crowd together in the lower energy states.


Statistical physics Mixture of inert perfect gases

Mixture of inert perfect gases

Consider a mixture of species i ∈ S = H + e (heavy species + freeelectrons), with H = Ha + Hp (atoms + polyatomic species)

The total number of particles of each species Ni is kept constantindependently: it is assumed that species are inert and do not react(frozen gas)

The energy level I of species i is εi ,IThe system is assumed to be constrained as follows

Number of particles of species i :∑

I Ni,I = Ni

Total energy:∑

i∈S

∑I Ni,I (εi,I + mih

Fi ) = E

Volume: V

where hFi is the formation enthalpy

The total number of particles is N =∑

i∈S Ni



Energy of chemical bonds

Formation enthalpy

Energy is released in the gas or absorbed by the gas when chemicalreactions occurA common level from which all the energies are measured can beestablished by means of a formation enthalpy hFiNotice that assigning a formation enthalpy to species is an arbitraryconvention, solely a difference of energy can be measured in a chemicalreactor (1st law of thermodynamics)This formation enthalpy can be fixed for instance at 0 K

Formation entropy

There is no need for a formation entropyThe entropy of a perfect crystal at 0 K is exactly equal to zero (3rd lawof thermodynamics)Notice that the semi-classical expression of the entropy presented inthis introduction does not satisfy this property



Boltzmann distribution for a mixture of perfect gases

The most probable macrostate of gas mixture can be obtained in theBoltzmann limiting case following a method similar to the onedeveloped for the single species gas

Ni ,I = ai ,I e−αi−βmih

Fi e−βεi,I

Boltzmann distribution for equilibrium population Ni ,I of energy levelsI of species i Ni ,I

Ni=

1

Qiai ,I exp(

−εi ,IkBT

)

Energy level I of species i : εi,IDegeneracy of level εi,I of species i : ai,IPartition function of species i : Qi =

∑I ai,I exp(

−εi,IkBT

)



Evaluation of the thermodynamic properties in terms of

the partition function Qi =∑

I ai ,I exp(−εi,IkBT

)

For a pure gas of Ni particles of species i in equilibrium in a volumeV , the energy per unit mass is given by

ei =kB

miT 2

(∂ lnQi

∂T

)V

+ hFi

The partial pressure is given by pi = NikBT (∂ lnQi∂V )T

The enthalpy per unit mass is expressed as

hi = ei +piρi

The entropy per unit mass is given by

si =kB

mi

[T

(∂ lnQi

∂T

)V

+ 1

]+

kB

miln

Qi

Ni



Translational energy of a particle in a box

Translational partition function for species i

QTi (V ,T ) ∼

∫ ∞0

exp[

−h2P

8mia2kBT(n2

x + n2y + n2

z)]

dnxdnydnz

= V(

2πmikBTh2

P

) 32

(Hint:∫∞

0 e−αx2dx =

√π

4α , α > 0)

Translational thermodynamic properties for species i ∈ S

eTi (T ) = 32

kBTmi

pi = NiV kBT = nikBT (perfect gas)

hTi (T ) = eTi (T ) + kBTmi

= 52

kBTmi

sTi (pi ,T ) =hTi (T )

T + kBmi

ln

[kBTpi

(2πmikBT

h2P

) 32

]Magin (AERO 0033–1) Aerothermodynamics 2018-2019 59 / 80


Rotational energy of rigid rotorRotational partition function for diatomic molecule i ∈ Hp

QRdiai (T ) ∼

∫ ∞0

(2J + 1) exp(− θRi

T J(J + 1))

dJ

= − TθRi

[exp

(− θRi

T J(J + 1))]∞

0= T

θRi

θRi : characteristic rotational temperature

Polyatomic molecule

QRi (T ) = 1

σi

(TθRi

)L/2

σi : steric factor: symmetry and indistinguishability propertiesσi = 1 for heteronuclear diatomic molecules (CO, NO, NO+, . . .)σi = 2 for CO2 and homonuclear diatomic molecules (N2, O2, N+

2 , . . .)

L = 2 for linear molecules, L = 3 otherwise

Rotational thermodynamic properties

eRi (T ) = hRi (T ) = LkBT2mi

sRi (T ) =hRi (T )

T + kBmi

[L2 ln

(TθRi

)+ ln

(1σi

)]Magin (AERO 0033–1) Aerothermodynamics 2018-2019 60 / 80


Vibrational energy of harmonic oscillator

Vibrational modes of CO2

Number of vibrationalmodes for a moleculecomposed of N atoms

3N− 3− L

Diatomic molecule:3× 2− 3− 2 = 1CO2 molecule (linear):3× 3− 3− 2 = 4




Vibrational partition function for normal mode m of molecule i ∈ Hp

Shifting the zero-point energy to zero, one obtains

QVim(T ) =

∞∑v=0

exp(−θVimv

T

)=∞∑v=0

[exp(

−θVimT )

]v= 1

1−exp(−θVimT )

(Hint:∑n

v=0 xv = 1−xn+1

1−x →∑∞

v=0 xv = 1

1−x for |x | < 1)

θVim: characteristic vibrational temperature

Vibrational thermodynamic properties

eVi (T ) = hVi (T ) =kB

mi

∑m

θVim

exp(θVimT )− 1

sVi (T ) =hVi (T )

T− kB

mi

∑m

ln[1− exp(−θVimT )], i ∈ Hp




Low temperature limit

limT→ 0 K

eVi (T ) = 0

High temperature limit

limT→ ∞ K

eVi (T )

T=

kB

mi

∑m

θVim “ 00 ”

=kB

mi

∑m

θVim limT→ ∞ K

−1T 2

θVim exp(θVimT )−1

T 2

=kB

mi(3N− 3− L)

Example: diatomic molecule

eVi (T ) ∼ kB

miT forT → ∞ K



Electronic energy of an atom / molecule

Electronic partition function for species i ∈ H

QEi (T ) =

∑n

gEin exp(

−θEinT

)

Cut-off criterion: n = 1, 2, . . . , nmax with finite value of nmax to avoiddivergence of QE

i (T ) when nmax →∞Electronic thermodynamic properties

eEi (T ) = hEi (T ) =kB

mi

∑ng

Einθ

Ein exp(

−θEinT )

QEi

sEi (T ) =hEi (T )

T+

kB

milnQE

i

Spectroscopic data for electronic energy level nθEin: characteristic electronic temperaturegEin : degeneracy



Species thermodynamic properties

Species enthalpies for species i are obtained by summing over theenergy modes

Atoms: hi (T ) = hTi (T ) + hEi (T ) + hFi , i ∈ Ha

Molecules: hi (T ) = hTi (T ) + hEi (T ) + hRi (T ) + hVi (T ) + hFi , i ∈ Hp

Electrons: he(T ) = hTe (T ) + hFe

Species energies are obtained in a similar fashion

Species entropy for species i are given by

Atoms: si (pi ,T ) = sTi (pi ,T ) + sEi (T ), i ∈ Ha

Molecules: si (pi ,T ) = sTi (pi ,T ) + sEi (T ) + sRi (T ) + sVi (T ), i ∈ Hp

Electrons: se(pi ,T ) = sTe (pi ,T ) + kB

meln 2

⇒ The spin contribution is added to calculate the free electron entropy. Itcan be formally treated as a degeneracy



Thermodynamic properties

Mixture energies and enthalpies are obtained by summing over thespecies properties weighted by the mass fractions

e =nS∑i=1

yiei (T ) and h =nS∑i=1

yihi (T )

These quantities are linked by the expression e = h − p/ρ

The contribution of the entropy of mixing is added to evaluate themixture entropy

s(p,T ) =∑j∈S

yjsj(pj ,T )

=∑j∈S

yjsj(p,T ) +kB

ρ

∑j∈S

nj ln(1/xj)



Thermodynamic properties: asymptotic values

Equipartition theorem: energy per unit mass for fully excited energymode δ of species i

eδi (T ) = 12

kT

mi= 1

2

RT

Mi= 1

2RiT

When the mode δ is not excited then its energy vanishes

eδi (T ) = 0

Species energy for species i obtained by summing over the modes

ei (T ) =∑δ

eδi (T )

Atomic species with translational modes excited: ei (T ) = 32

kBTmi

Molecular species with translational and rotational modes excited: ei (T ) = 52

kBTmi



Exercise1 Compute the specific heat ratio for:

A monoatomic gas such as He,Air at ambient temperature.

Molecular mass data: MN = 14× 10−3kg /mol, MO = 16× 10−3kg /mol, and

MHe = 4× 10−3kg /mol.

2 Explain why your voice changes when you breath helium.

The speed of a sound wave through helium will then be much higher. So by inhaling

helium and using it as the source of the perceived sound, the frequency of the voice

changes but neither the pitch, since your vocal chords are vibrating at the same speed as

when you are using air, nor the configuration of the vocal tract. So while the base

frequency of the chords remains the same, the frequency of the sound heard by others is

increased.

3 Give bounds for the specific heat of a nitrogen gas at the exit of thenozzle of the Longshot facility assuming that the nitrogen moleculesare not dissociated and their vibrational energy mode partially excited.

Some high-temperature effects in nitrogen can be simulated by means of cold helium.


Statistical physics Mixture of reacting perfect gases

Local Thermodynamic Equilibrium (LTE) composition

8-species carbon dioxide mixture: CO, CO2, O2, C, O, C+, O+, e−

⇒ Equilibrium composition derived from

Conservation of charge (quasi-neutrality)

Ne− -NC+ -NO+ =0

Conservation of the number of elements

NCO+2NCO2+ 2NO2+NO+NO+ =NO

NCO+NCO2+NC+NC+ =NC

Equilibrium conditions of a minimal reaction set

CO2 ↔ C + 2OCO ↔ C + OO2 ↔ 2OC+ + e− ↔ CO+ + e− ↔ O



Mixture of reacting perfect gases

Gas mixture of species i ∈ S = R∪ EIndependent chemical components (elements / charge) i ∈ EDependent chemical components i ∈ R

A convenient set of chemical reactions, sufficient to compute theequilibrium composition, is obtained by writing formation of thedependent components in terms of elements

Xi ∑j∈E

ν ijXj , i ∈ R,

Notice that these reactions do not correspond to elementary processes

Stoichiometric matrix ν ij , i ∈ R, j ∈ E , for 8-species carbon dioxide

j\i C O e− CO CO2 O2 C+ O+

C 1 0 0 1 1 0 1 0O 0 1 0 1 2 2 0 1e− 0 0 1 0 0 0 -1 -1



Boltzmann distribution

The system is assumed to be constrained as follows

Number of Independent chemical components j ∈ E :∑i∈S ν

ij

∑I Ni,I = Nj

Total energy:∑

i∈S

∑I Ni,I (εi,I + mih

Fi ) = E

Volume: V

The number of macrostates in the Boltzmann limiting case islnW =

∑i∈S

∑I Ni ,I (ln

ai,INi,I

+ 1)

Introducing Lagrange multipliers, the constrained optimizationproblem is written as:∑

i∈S

∑I

[ln

ai,INi,I−∑

j∈E νijαj − β(εi ,I + mih

Fi )]

dNi ,I = 0

The Boltzmann distribution is Ni ,I = ai ,I e−∑

j∈E νijαj−βmih

Fi e−βεi,I , for

species i ∈ S



Local Thermodynamic equilibrium

Summing of the energy levels, one obtains Ni = e−∑


Fi Qi

with the partition function Qi =∑

I ai ,I e−βεi,I , for i ∈ S

⇒ The Boltzmann distribution is found to beNi,I

Ni= 1

Qiai ,I e

−βεi,I , forspecies i ∈ S

⇒ The equilibrium of reaction Xi ∑

j∈E νijXj , i ∈ R, is expressed by

means of a generalized Saha equation

Πk∈E

(NkQk

)ν ikNiQi

=e−

∑j,k∈E ν

ikν

kj αj−β

∑k∈E ν

ikmkh

Fk

e−∑


Fi

= e−β(∑

k∈E νikmkh

Fk−mih

Fi )

since νkj = δjk for j , k ∈ EIt can be shown from the Gibbs’ equation that β = 1

kBT



Exercise

Show that the composition of a gas mixture satisfies van ’t Hoff’sequation ∑

j∈Eν ij ln xj − ln xi = lnKi , i ∈ R,

With the equilibrium constant

Ki (p,T ) = exp[(migi (p,T )−

∑j∈E ν

ijmjgj(p,T ))

kBT]

The species Gibbs free energy is defined by the relation

gi (p,T ) = hi (T )− Tsi (p,T )

= hFi −kBT

miln

Qi

N



A library for high enthalpy and plasma flows at VKI

Quantities relevant to engineering design for hypersonic flows

Heat flux & shear stress to the surface of a vehicleTheir prediction strongly relies on completeness and accuracy of thenumerical methods & physico-chemical models

Why a library for physico-chemical models?

Implementation common to several CFD codesNonequilibrium models, not satisfactory today, are regularly improvedBasic data are constantly updated(spectroscopic constants, cross-sections,. . .)

Constraints for the library implementationHigh accuracy of the physical models

Laws of thermodynamics must be satisfiedValidation based on experimental data

Low computational costUser-friendly interface



MUTATION++ library

MUTATION++ library: MUlticomponent Transport AndThermodynamic properties / chemistry for IONized gases written inC++

⇒ Breeding of the following libraries: Pegase (VKI), Chemkin(Livermore), and EGlib (Ecole Polytechnique)

Modules1 Thermodynamic properties2 Chemistry3 Transport properties4 Energy / chemistry exchange terms

New features versus the fortran Mutation libraryFunctionality rebuilt into a modern, object oriented, extensible framework (J.B.Scoggins)Improved methodology when solving equilibrium compositionsEfficient calculation of costly model parameters using lookup tables defined a priorior during initialization based on error constraints



Definition of the mixture speciesEarth atmosphere: 79%N2 and 21%O2

Ablation of carbon phenolic composites

Mars atmosphere: 95.55%CO2, 2.7%N2, 1.6%Ar, and 0.15%O2

Titan atmosphere: 98.6%N2, 1.4%CH4

⇒ e.g. multitemperature models

11-species air: N2, NO, O2, N, O, N+2 , NO+, N+, O+

2 , O+, e−

8-species carbon dioxide: CO, CO2, O2, C, O, C+, O+, e−

⇒ e.g. electronic collisional-radiative (CR) models

17-species+7 pseudo-species: methane-nitrogen mixture+ CN(X), CN(A), CN(B), N2(X), N2(A), N2(B), N2(C)9-species+86 pseudo-species: air + N(1-46), O(1-40)

⇒ e.g. rovibrational CR models

1-species+9390 pseudo-species: N +9390 (v,J) levels of N2



Chemical reactions: LTE composition

0 2500 5000 7500 10000 12500 15000

Temperature [ K ]

0

0.2

0.4

0.6

0.8

1.0

x i [ -

]

8-species carbon dioxide mixtureat 1 atm pressure

O

CO

CO2

O2

Ce

-

O+

C+

0 2500 5000 7500 10000 12500 15000

Temperature [ K ]

0

0.2

0.4

0.6

0.8

x i [ -

]

11-species air mixture at 1 atm pressure

O

NO

N2

O2

N

e-

O+

N+



Chemical reactions: LTE composition

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000Temperature [ K ]

0.0001

0.001

0.01

0.1

1

Spec

ies

mol

e fr

actio

n (L

TE

com

posi

tion)

Carbon phenolic at 0.1atm pressureC:H:O=0.2293:0.6606:0.1101 (Mutation library)

CO C

H

O

e-

O+

H+

C2H

2

C3

H2

C2

CH

C+

OH

C2H

CH4

C3H

CO2

OH

C2H

4

C6H

6

LTE composition



Mixture enthalpy

0 2500 5000 7500 10000 12500 15000

Temperature [ K ]

0

25

50

75

100

125

h [

MJ

kg-1

]


Total

TranslationalRotationalVibrational

Electronic

Form

atio

n

0 2500 5000 7500 10000 12500 15000

Temperature [ K ]

0

1

2

3

4

5

h [

MJ

kg-1

]


Tot

alT

rans

latio

nal

RotationalVibrational

Electronic

Form

atio

nMagin (AERO 0033–1) Aerothermodynamics 2018-2019 79 / 80


Specific heat at constant pressure

0 2500 5000 7500 10000 12500 15000

Temperature [ K ]

0

5

10

15

20

c p [ k

J kg

-1 K

-1 ]

8-species carbon dioxide mixtureat 1 atm pressure

Frozen

Equilibrium

0 2500 5000 7500 10000 12500 15000

Temperature [ K ]

0

5

10

15

20

25

c p [ k

J kg

-1 K

-1 ]

11-species air mixtureat 1 atm pressure

Frozen

Equilibrium

Equilibrium specific heat: cp = (∂Th)p, frozen specific heat: c fp =∑

i yi (∂Thi )p


Documents

Aerothermodynamics of high speed ows - Ltas-aea ::Welcome · 2019-05-13 · The Dawn of the quantum theory The de Broglie waves Exercise 1 Calculate the de Broglie wavelength of an