1 Fluid Equations for Rarefied Gases Jean-Luc Thiffeault Department of Applied Physics and Applied Mathematics Columbia University http://plasma.ap.columbia.edu/~jeanluc 21 May 2001 with E. A. Spiegel
2 Fluid Equations Equations for macroscopic variables: ρ, u, T. Navier Stokes is the most well-known. Discovered empirically. Hilbert and Chapman & Enskog derived N S as an expansion of the Boltzmann equation in the Knudsen number ɛ, the mean-free-path over macroscopic length scales. Domain of validity is limited: does poorly with rarefied gases (ɛ 1). Higher order expansions (Burnett) do not seem to do much better, and can actually do worse.
3 Asymptotic but not Convergent Grad (1963) expressed it best: The apparent ineffectiveness of the Burnett equations... is evidently a consequence of the fact that the Chapman Enskog expansion is asymptotic rather than convergent. The series must always be truncated. For rough answers,... one should use the Navier Stokes equations (or give up the Chapman Enskog approach entirely). Propose a modification of Chapman Enskog in an attempt to extend its domain of validity. [BGK: Chen, Rao, & Spiegel (2000), Spiegel & Thiffeault (2001); Fokker Planck: Spiegel & Thiffeault (2001)]
4 The Boltzmann Equation Liouville operator: Df = 1 ɛ C[f] D t + v i x i + K i v i Collision operator satisfies ψ α C[f] d 3 v = 0, α = 1,..., 5, where the ψ α are the collisional invariants, ψ α = m (1, v, 1 2 v2 ) There are no other collisional invariants.
5 Macroscopic Variables Can define slow variables, ρ mf d 3 v, u 1 ρ where the peculiar velocity is mvf d 3 v, T m 3R ρ c 2 f d 3 v, c(x, v, t) v u(x, t). There are no other slow variables.
6 Fluid Equations From the collisional invariants, can derive fluid equations for the slow variables, t ρ + (ρ u) = 0, (1) t u + u u = ρ 1 P + K, (2) 3 2 ρr ( tt + u T ) = P : u Q. (3) where P m c cf d 3 v, Q 1 2 m c2 cf d 3 v, (4) are the pressure tensor and the heat flux. Everything is still exact. But what are P and Q?
7 Mean-free-path Expansion Hilbert and Chapman & Enskog suggested an expansion of f in the mean-free-path, as expressed by the Knudsen number, ɛ. The equilibrium state is defined by C[f 0 ] = 0, and is given by the Maxwellian distribution, ) n f 0 (x, v, t) = exp ( c2. (2πRT ) 3/2 2R T At order ɛ 0, we have f (0) = f 0, and we recover the Euler equations, for which P = p I, Q = 0, where p = ρr T is the scalar pressure (ideal gas law).
8 BGK Operator Need to specify collision operator C. Use BGK (Bhatnager, Gross, & Krook, 1954), C[f] 1 τ (f 0 f) also known as the relaxation model. Phenomenological model: force f to relax to an equilibrium f 0. Conserves collisional invariants ψ α if the so-called matching conditions are satisfied: ψ α f 0 d 3 v = ψ α f d 3 v, α = 1,..., 5, Absorb the relaxation time τ in ɛ. BGK is very simple, and there is a trick to recover the Boltzmann result.
9 First Order For BGK, the first order distribution function f (1) is given directly from f (1) = Df 0 as a polynomial in the peculiar velocity c = v u multiplying a Maxwellian f 0, f (1) = ( α (3) ijk ci c j c k + α (2) ij ci c j + α (1) i c i + α (0)) f 0 f (1) f 0,
10 Pressure Tensor and Heat Flux After solving for f (1) and evaluating the first-order pressure tensor and heat flux, find [ P = p I ɛ ( u + u ) ( ) ] D ln T 23 u I ɛ + 2 Dt 3 u I, Q = 5 2 pr ɛ [ ( ) ] Du T + σ Dt K + T ln p, where the convective derivative is D Dt t + u How do we eliminate these convective derivatives?
11 The Chapman Enskog Result Chapman & Enskog use the Euler equations: P p I = 1 2 p ɛ E, Q = 5 6 p ɛ R T, where the traceless rate-of-strain tensor is E u + u 2 3 u I This gives the Navier Stokes equations.
12 But why use Euler? If instead we simply insert P and Q into the full fluid equations, we obtain t u + u u = 1 ) (p ρ 1 ɛ DS 2 1 C v Dt 2 ρ 1 (p ɛ E ), and a similarly modified equation for T. We have introduced the specific entropy, ( ) p S = C v ln. ρ 5/3 There is a bulk viscosity in the equation, given by the rate of entropy production. Implicit differential equation containing terms of all orders in ɛ.
13 Example: Ultrasonic Sound Waves Navier Stokes is notoriously bad at describing sound waves when the frequency of oscillations is comparable to the free flight time of particles between collisions. Linearize equations around equilibrium: u = 0, T = T 0, ρ = ρ 0. Find dispersion relation. The natural choice of Knudsen number is the frequency of the wave over the free flight time τ.
14 1.0 0.9 0.8 Inverse of Phase speed 0.7 0.6 0.5 0.4 0.3 This work Sirovich and Thurber NS 0.2 0.1 ET (16215 moments) Experiment 0.0 10-2 10-1 10 0 10 1 Inverse of Knudsen number [Chen, Rao, and Spiegel, Phys. Lett. A (2000)]
15 0.2 This work Woods NS Bimodal distribution ET(20 moments) 1/δ 0.1 Monte Carlo Experiment (Ar) Experiment (Xe) 0.0 1.0 1.1 1.2 1.3 1.4 1.5 1.6 M [Chen, Rao, and Spiegel, Phys. Lett. A (2000)]
16 Linearized pressure tensor: Second Order P (2) = p ɛ 2 2 t E + 2R T σt T + 5R T 3σT 2 T I + t Σ I 2R T σ Ψ 5R T 3σ Ψ I Linearized heat flux: Q (2) = 1 2 p R T τ 2 σ 1 5 σt t T + 14 E 5 σ tψ + 10 Σ Σ D ln T Dt + 2 3 u, Ψ 1 R T ( ) Du Dt K ln p
17 1 0.8 eplacements Inverse of phase speed 0.6 0.4 0.2-1.5-1 -0.5 0 0.5 1 Inverse of Knudsen number Blue: Chen, Rao, Spiegel; Green: 1st order; Red: 2nd order
18 Conclusions and Future Work Want fluid equations that are valid over a wide range of Knudsen numbers. Modification of Chapman Enskog method: does not use lower-order equations (Euler). Get implicit fluid equations: contain all orders in ɛ. Second-order theory: agreement seems to gets better Fokker Planck collision operator: application to plasmas, astrophysics, etc. Compare with more experiments.