Kinetic theory of gases

Toan T. Nguyen1,2

Penn State University

Graduate Student seminar, PSUJan 19th, 2017

Fall 2017, I teach a graduate topics course: same topics !

1Homepage: http://math.psu.edu/nguyen2Math blog: https://sites.psu.edu/nguyen

Toan T. Nguyen (Penn State) Kinetic theory of gases PSU, Jan 19th, 2017 1 / 20

Figure: 4 states of matter, plus Bose-Einstein condensate!

• James Clerk Maxwell, in 1859, gave birth to kinetic theory: use thestatistical approach to describe the dynamics of a (rarefied) gas.

Figure: 4 states of matter, plus Bose-Einstein condensate!

• James Clerk Maxwell, in 1859, gave birth to kinetic theory: use thestatistical approach to describe the dynamics of a (rarefied) gas.

(x , p)

Figure: Kinetic theory of gases: phase space!

Gas molecules are identical.

One-particle phase space: position x ∈ Ω ⊂ Rd , momentum p ∈ Rd

f (t, x , p) the probability density distribution. Mass: f (t, x , p) dxdp.

Interest: the dynamics of f (t, x , p) ? Non-equilibrium theory.

(x , p)

Collisionless kinetic theory

I. Collisionless Kinetic Theory

Hamilton 1830’s classical mechanics3: one-particle Hamiltonian H(x , p)(induced from Lagrangian):

x = ∇pH, p = −∇xH

Examples:

Free particles:

H(x , p) =1

2m|p|2

Particles in a potential:

H(x , p) =1

2m|p|2 + Vpot(x)

3In truth, it should be an N-particle Hamiltonian dynamics, then mean field limit orBBGKY hierarchy to kinetic theory. S1: Presentation on its formal derivation?

x = ∇pH, p = −∇xH

Examples:

Free particles:

H(x , p) =1

2m|p|2

H(x , p) =1

2m|p|2 + Vpot(x)

x = ∇pH, p = −∇xH

Examples:

Free particles:

H(x , p) =1

2m|p|2

H(x , p) =1

2m|p|2 + Vpot(x)

• Also for plasmas: charged particles in electromagnetic fields

H(x , p) =1

2m|p − qA(x)|2 + φ(x)

with electric and magnetic potentials φ,A.

• Force acting on particles:

F = −∇xH = E +q

m(p × B)

which is the Lorentz force, with E = −∇xφ and B = ∇x × A.

• Also for plasmas: charged particles in electromagnetic fields

H(x , p) =1

2m|p − qA(x)|2 + φ(x)

with electric and magnetic potentials φ,A.

• Force acting on particles:

F = −∇xH = E +q

m(p × B)

which is the Lorentz force, with E = −∇xφ and B = ∇x × A.

(x0, p0)

(xt , pt)t > 0

Figure: Liouville’s Theorem: volume in phase space remains constant (Exercise:prove this).

This yields the kinetic equation: ∂t f = H, f (Poisson bracket), explicitly

dtf (t, x(t), p(t)) = ∂t f +∇pH · ∇x f −∇xH · ∇pf

= ∂t f + v · ∇x f + Fforce · ∇v f

(x0, p0)

(xt , pt)t > 0

Figure: Liouville’s Theorem: volume in phase space remains constant (Exercise:prove this).

This yields the kinetic equation: ∂t f = H, f (Poisson bracket), explicitly

dtf (t, x(t), p(t)) = ∂t f +∇pH · ∇x f −∇xH · ∇pf

= ∂t f + v · ∇x f + Fforce · ∇v f

Three examples of kinetic equations (remain very active):

Vlasov dynamics (transport in phase space):

∂t f + v · ∇x f + F · ∇v f = 0

Vlasov-Poisson:

F = −∇xφ, −∆xφ = σ

f (t, x , v) dv ,

with σ = 1 (plasma physics: charged particles) or σ = −1(gravitational case: stars in a galaxy).

Vlasov-Maxwell (plasma): Lorentz force F = E + v × B with theelectromagnetic fields E ,B solving Maxwell (generated by charge andcurrent density).

∂t f + v · ∇x f + F · ∇v f = 0

Vlasov-Poisson:

F = −∇xφ, −∆xφ = σ

f (t, x , v) dv ,

∂t f + v · ∇x f + F · ∇v f = 0

Vlasov-Poisson:

F = −∇xφ, −∆xφ = σ

f (t, x , v) dv ,

∂t f + v · ∇x f + F · ∇v f = 0

Vlasov-Poisson:

F = −∇xφ, −∆xφ = σ

f (t, x , v) dv ,

• Can you solve Vlasov, the Cauchy problem?

In fact, it’s simply ODEs:

x = v , v = F (t, x , v)

Then, Vlasov f (t, x , v) = f0(x−t , v−t).

• One needs Lipschitz bounds on F (t, x , v).

• Vlasov-Poisson: F = −∇xφ and −∆φ = ±∫

f (t, x , v) dv . Roughlyspeaking: ∇xF = ∇2

x(−∆x)−1ρ ≈ ρ, up to a log of derivatives of f .

• Global unique solutions (1991): Pfaffelmoser, Lions-Perthame, Schaeffer,Glassey’s book.

• Vlasov-Maxwell: Outstanding open problem! Glassey Strauss’s theorem.

• Can you solve Vlasov, the Cauchy problem? In fact, it’s simply ODEs:

x = v , v = F (t, x , v)

Several mathematical issues: ∂t f = H, f (focus on previous examples).

Infinitely many conserved quantities, and infinitely many equilibria:

dtϕ(f ) = 0, f∗ = ϕ(H(x , p))

for arbitrary ϕ(·). Which one is dynamically stable ? and shouldn’t itbe just Gaussian ? Maxwell-Boltzmann distribution (next section) ?

Hence, large time dynamics and stability of equilibria are verydelicate!

dtϕ(f ) = 0, f∗ = ϕ(H(x , p))

VP: ∂t f + v · ∇x f −∇xφ · ∇v f = 0, −∆φ = ρ− 1.

For instance, homogenous equilibria

µ = µ(e), e :=1

2|v |2, φ0 = 0.

Arnold: Look at casimir’s invariant

A[f ] =

∫∫ [1

2|v |2f + ϕ(f )

]dvdx +

∫|∇φ|2 dx .

µ is a critical point iff ϕ′(µ) = −12 |v |

2 or ϕ′ = −µ−1. Convexity ofA[f ] requires

µe < 0.

µe < 0 implies Arnold’s nonlinear stability. But, what’s long-timedynamics?

VP: ∂t f + v · ∇x f −∇xφ · ∇v f = 0, −∆φ = ρ− 1.

µ = µ(e), e :=1

2|v |2, φ0 = 0.

A[f ] =

∫∫ [1

2|v |2f + ϕ(f )

]dvdx +

∫|∇φ|2 dx .

µe < 0.

VP: ∂t f + v · ∇x f −∇xφ · ∇v f = 0, −∆φ = ρ− 1.

µ = µ(e), e :=1

2|v |2, φ0 = 0.

A[f ] =

∫∫ [1

2|v |2f + ϕ(f )

]dvdx +

∫|∇φ|2 dx .

µe < 0.

VP: ∂t f + v · ∇x f −∇xφ · ∇v f = 0, −∆φ = ρ− 1.

µ = µ(e), e :=1

2|v |2, φ0 = 0.

A[f ] =

∫∫ [1

2|v |2f + ϕ(f )

]dvdx +

∫|∇φ|2 dx .

µe < 0.

vmin vvmin

Figure: Stable vs unstable equilibria. Beautiful Penrose’s iff criteria:

linearlyunstable iff vmin exists and

µ′(v)dv

v − vmin> 0.

S2: Presentation on Penrose’s criteria?

vmin vvmin

Figure: Stable vs unstable equilibria. Beautiful Penrose’s iff criteria: linearlyunstable iff vmin exists and

µ′(v)dv

v − vmin> 0.

S2: Presentation on Penrose’s criteria?

• Mouhot-Villani: nonlinear Landau damping of Penrose stablehomogenous equilibria of VP (earning Villani a Fields medal).• S3: presentation on linearized case?

• Open: Are there stable inhomogenous equlibria (VP)? f∗ = ϕ(H(x , p)).

• Some of my works (available on my homepage):

Stability of inhomogenous equilibria of VM (w/ Strauss).

Various asymptotic limits of Vlasov (w/ Han-Kwan).

Non-relativistic limits of VM (w/ Han-Kwan and Rousset).

Collisional kinetic theory

II. Collisional Kinetic Theory

v ′∗

v vv ′

v∗ = 0 v ′∗v∗

Figure: Elastic collisions: momentum and energy are exchanged between particles,however conserved, after collision:

v + v∗ = v ′ + v ′∗

|v |2 + |v∗|2 = |v ′|2 + |v ′∗|2

v ′∗v+v∗2

Sphere Svv∗ :

v ′∗

v vv ′

v∗ = 0 v ′∗v∗

Figure: Elastic collisions: momentum and energy are exchanged between particles,however conserved, after collision:

v + v∗ = v ′ + v ′∗

|v |2 + |v∗|2 = |v ′|2 + |v ′∗|2

v ′∗v+v∗2

Sphere Svv∗ :

One has the Boltzmann equation:

∂t f − H, f = Q[f ]

in which collision operator Q[f ]:

Q[f ](x , v) =

∫∫∫R9

ω(v , v∗|v ′, v ′∗)︸︷︷︸Scattering

[f2(x , v ′, v ′∗)︸︷︷︸

− f2(x , v , v∗)︸︷︷︸Loss

]dv∗dv ′dv ′∗

with f2(x , v , v∗) the probability distribution of finding two particles withrespective momenta.• Molecular chaos assumption:

f2(x , v , v∗) = f (x , v)f (x , v∗)

asserting velocities are uncorrelated before the collision (unlikeHamiltonian, here it introduces a notion of time arrow!).

∂t f − H, f = Q[f ]

Q[f ](x , v) =

∫∫∫R9

[f2(x , v ′, v ′∗)︸︷︷︸

− f2(x , v , v∗)︸︷︷︸Loss

with f2(x , v , v∗) the probability distribution of finding two particles withrespective momenta.

• Molecular chaos assumption:

f2(x , v , v∗) = f (x , v)f (x , v∗)

∂t f − H, f = Q[f ]

Q[f ](x , v) =

∫∫∫R9

[f2(x , v ′, v ′∗)︸︷︷︸

− f2(x , v , v∗)︸︷︷︸Loss

with f2(x , v , v∗) the probability distribution of finding two particles withrespective momenta.• Molecular chaos assumption:

f2(x , v , v∗) = f (x , v)f (x , v∗)

• Conservation laws: v ′, v ′∗ ∈ Svv∗ and hence

dv ′dv ′∗ = dσ(Svv∗) = |v − v∗|dσ(n), n ∈ S2.

• (symmetric) Scattering: Elastic hard-sphere collisions:

ω(v , v∗|v ′, v ′∗) =|n · (v − v∗)||v − v∗|

(angle of collision)

• The Boltzmann equation (1872) for hard-sphere gases:

∂t f − H, f =

|n · (v − v∗)|[f ′f ′∗ − ff∗

]dndv∗.

Is it solvable ? Di Perna-Lions ’89, Guo 2010. There remain manyoutstanding open problems.

ω(v , v∗|v ′, v ′∗) =|n · (v − v∗)||v − v∗|

∂t f − H, f =

|n · (v − v∗)|[f ′f ′∗ − ff∗

]dndv∗.

ω(v , v∗|v ′, v ′∗) =|n · (v − v∗)||v − v∗|

∂t f − H, f =

|n · (v − v∗)|[f ′f ′∗ − ff∗

]dndv∗.

ω(v , v∗|v ′, v ′∗) =|n · (v − v∗)||v − v∗|

∂t f − H, f =

|n · (v − v∗)|[f ′f ′∗ − ff∗

]dndv∗.

• Boltzmann’s H-Theorem: Look at the entropy:

H(t) =

∫∫R6

f log f dxdv .

• One has

dtH(t) =

|n · (v − v∗)|[f ′f ′∗ − ff∗

]log f dndxdvdv∗

|n · (v − v∗)|[f ′f ′∗ − ff∗

ff∗f ′f ′∗

dndxdvdv∗

≤ 0.

• Entropy is decreasing ! (unlike Hamiltonian, there is a time arrow!).

H(t) =

∫∫R6

f log f dxdv .

• One has

dtH(t) =

|n · (v − v∗)|[f ′f ′∗ − ff∗

]log f dndxdvdv∗

|n · (v − v∗)|[f ′f ′∗ − ff∗

ff∗f ′f ′∗

dndxdvdv∗

≤ 0.

H(t) =

∫∫R6

f log f dxdv .

• One has

dtH(t) =

|n · (v − v∗)|[f ′f ′∗ − ff∗

]log f dndxdvdv∗

|n · (v − v∗)|[f ′f ′∗ − ff∗

ff∗f ′f ′∗

dndxdvdv∗

≤ 0.

• Boltzmann’s H-Theorem: There is more: equilibria iff

ff∗ = f ′f ′∗ , v ′, v ′∗ ∈ Svv∗

and so iff (presentation?):

Maxwellian distribution

f (x , v) = Ce−β|v−u|2

=ρ(x)

(2πθ(x))3/2e− |v−u(x)|2

2θ(x)

in which (ρ, u, θ) denotes macroscopic density, velocity, and temperatureof gases.

• Boltzmann’s H-Theorem: There is more: equilibria iff

ff∗ = f ′f ′∗ , v ′, v ′∗ ∈ Svv∗

and so iff (presentation?): Maxwellian distribution

f (x , v) = Ce−β|v−u|2

=ρ(x)

(2πθ(x))3/2e− |v−u(x)|2

2θ(x)

in which (ρ, u, θ) denotes macroscopic density, velocity, and temperatureof gases.

• Hydrodynamics limits (if time permits).

• Mathematical difficulty of the Cauchy problem and convergence toequilibrium.

On justification of Maxwell-Boltzmann distribution (w/ Bardos,Golse, Sentis)

on Quantum Boltzmann, interaction between fermions, bosons, andBose-Einstein condenstate (w/ Tran)

Kinetic theory of gases - personal.psu.edu

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