BGK

Size: px
Start display at page:

Download "BGK"

Transcription

1 BGK

2

3 i X AGN Eckart [] Landau [2] [3] Israel Stewart Grad moment [4] BGK BGK

4

5 iii Boltzmann H Chapman-Enskog moment Boltzmann H Eckart Landau-Lifshitz Chapman-Enskog moment Boltzmann

6 iv 3.. BGK Result Discussion continuous spectrum Boltzmann Marle Marle BGK model Marle BGK Result Discussion Boltzmann Anderson-Witting Anderson-Witting BGK model Result Discussion Israel-Stewart A 5 B moment 23 C 25 D Fredholm 27

7 v E 29 E. BGK E.2 Marle BGK E.3 Anderson-Witting BGK F 39 F. BGK F.2 Anderson-Witting BGK F.3 Marle BGK

8

9 20 ev 20 X X AGN X 60 80

10 2 MRI saturation rate reconnection Navier-Stokes Navie-Stokes [2] [22] [23] Navier-Stokes Boltzmann

11 3 Chapman Enskog 0 Knudsen [24] 0 Euler Navier-Stokes Burnett [25] [26] Knudsen Chapman-Enskog Knudsen Knudsen Burnett Chen [27] [28] BGK [8] Navier-Stokes Knudsen Knudsen Grad [8] [9] moment method Knudsen Hilbert moment moment 3 moment expansion leading order Navier-Stokes 5 moment moment 5 Knudsen moment [29] [30] [3] [32] Knudsen Grad moment Levermore [33] [34] Eckart [] Landau [2]

12 4 [3] Chapman-Enskog Chapman-Enskog Israel Stewart Grad moment [4] moment B moment BGK BGK BGK [2] [8] BGK moment [9] BGK BGK review BGK BGK BGK Israel-Stewart

13 5

14

15 7 2 review Boltzmann review r N /3 N : d Nd 3 (d/ r) 3 k B = 2. review review - [5] Mihalas [6]

16 f(t, q, p) q q p dµ = dqdp fdµ. f(t, q, p) q q p fdµ = fdqdp f q r p p 2. f(t, r, p)d 3 p = n(t, r) (2.) ndv dv Boltzmann dv 3. dv d 3 dv L 3 d L 2

17 2. 9 d r L r L n r L n n n 2..2 p dp dp dp dp p p, p p, p p, p p, p w(p, p ; p, p ) fdp f dp dp dp (2.2) w(p, p ; p, p ) p, p p, p fdp w dp f f(t, r, p) f f(t, r, p ) w 4. dσ dσ = w(p, p ; p, p ) dp dp (2.3) v v w w(p, p ; p, p )dp dp = w(p, p ; p, p )dp dp (2.4) Ŝ

18 0 2 S in S nk = δ ik (2.5) n i = k S ni 2 = (2.6) n S ni 2 i n S in S nk = δ ik (2.7) n i = k S in 2 = (2.8) n 2 n = i S ni 2 = S in 2 (2.9) n i n i w 2..3 Boltzmann f f f t + f t + q (f q) + (fṗ) = 0 (2.0) p ( q f q + ṗ f p ) + f ( q q + ṗ ) = 0 (2.) p

19 2. Hamilton q = H p, ṗ = H q (2.2) q q = 2 H q p = ṗ p (2.3) (2.) 0 D Dt f = f ( t + q f ) q + ṗ f = 0 (2.4) p D/Dt f t ( ) f f + v f + ṗ p = t coll (2.5) ( f/ t) coll f f (x, v) 2 p p p, p p, p p p p dv dv dp w(p, p ; p, p )ff d 3 p d 3 p d 3 p (2.6) p, p p, p dv dp w(p, p ; p, p )f f d 3 p d 3 p d 3 p (2.7) dv dp (w f f wff )d 3 p d 3 p d 3 p (2.8)

20 2 2 w w(p, p ; p, p ), w w(p, p ; p, p ) (2.9) ( ) f = (w f f wff )d 3 p d 3 p d 3 p (2.20) t coll 2 (2.4) ( ) f = w (f f ff )d 3 p d 3 p d 3 p (2.2) t coll 0 (ṗ = 0) f t + v f = w (f f ff )d 3 p d 3 p d 3 p (2.22) (2.3) ( ) f = v rel (f f ff )dσd 3 p (2.23) t coll d 2..4 H f S = f ln e f dv d3 p (2.24) ( ds dt = f ln e ) dv d 3 p = ln f f t f t dv d3 p (2.25)

21 2. 3 (2.25) f t ( ) f f = v f F p + t coll (2.26) (2.25) ln f [ v f F f ] [ dv d 3 p = v + F p ] ( f ln f ) dv d 3 p (2.27) p e f = 0 0 ( ) ds f dt = ln f dv d 3 p (2.28) t 2.. q p ( ) f φ(p) dp (2.29) t φ p (2.20) ( ) f φ(p) dp = φ w(p, p ; p, p t )f f d 4 p φ w(p, p ; p, p )ff d 4 p (2.30) coll d 4 p = dpdp dp dp 4 p 2 p p p p ( ) f φ(p) dp = t coll coll coll (φ φ )w f f d 4 p (2.3) 2 p p p p /2 w ( ) f φ(p) dp = (φ + φ φ φ t coll 2 )w f f d 4 p (2.32) φ m p (v u) 2 /2 0 φ = ( ) f dp = t w (f f ff )d 4 p = 0 (2.33) coll

22 4 2 (2.32) (2.28) ds dt = 2 w f f ln f f d 4 pdv = ff 2 w ff x ln x d 4 pdv (2.34) d 4 p x = f f /ff 0 (2.33) /2 ds dt = w ff (x ln x x + ) d 4 pdv (2.35) 2 x > 0 x = 0 w f f H ds dt 0 (2.36) (2.35) d 4 p (2.2) molecular chaos ff v rel dσd 3 p d 3 pdv dt [5] 2..5 (2.22) p f(p) f(p) f(p )f(p ) f(p)f(p ) = 0 (2.37) log ln f(p ) + ln f(p ) = ln f(p) + ln f(p ) (2.38)

23 2. 5 ln f(p) 2 ln f(p) ln f(p) = α + βɛ (2.39) α β ɛ f (2.) β ρ f 0 = [ m(2πrt ) exp ɛ ] 3/2 T (2.40) R = m (2.4) m ρ = mn T Maxwell-Boltzmann f Maxwell-Boltzmann Boltzmann (2.35) 0 review - [7] 2..6 Boltzmann Boltzmann 5. l l nσ (2.42)

24 6 2 d σ d 2 n r 3 l r d l r ( r ) 2 l r (2.43) d L l L l d 3 vvf(t, r, v) = u : (2.44) d 3 v 2 (v u)2 f(t, r, v) = 3 ρrt ɛ : (2.45) 2 n u ɛ Boltzmann ( ) f f t + (vf) = t coll (2.46) m p (v u) 2 /2 d 3 v (2.32) φ m p (v u) 2 /2 0 Boltzmann m p (v u) 2 /2 p ρ + (ρu) = 0 t (2.47) t ρu α + Π αβ x β = 0 (2.48) n ɛ + q = 0 t (2.49) Π αβ q Π αβ = mv α v β fd 3 p (2.50) q = 2 (v u)2 vfd 3 p (2.5)

25 2. 7 f Π αβ q l L l L L/ v 0 f 0 (t, x) [ ] ρ(t, x) f 0 (t, x, v) = exp (v u(t, x))2 m(2πrt (t, x)) 3/2 2RT (t, x) (2.52) ρ u T f Boltzmann 0 0 Boltzmann Π αβ q u K v v = v + u Π αβ Π αβ = ρ v α v β = ρ (u α + v α)(u β + v β) = ρ(u α u β + v αv β ) (2.53) K v αv β = 3 v 2 δ αβ = T m δ αβ (2.54) v 2 = 3T/m Π αβ nt = P Π αβ Π αβ = ρu α u β + δ αβ P (2.55) P Boltzmann q K v 2 mv2 = 2 mv 2 + mu v + 2 mu2 (2.56) q [ mu 2 q = nu + m ] ( ) ρu v 2 + ɛ = u 2 + h (2.57)

26 8 2 h = P + n ɛ Navier-Stokes review [2] 2..7 Chapman-Enskog Navier-Stokes f f 0 f l/l f f 0 ( ) 2 l l f = f 0 + f L + f 2 + (2.58) L (2.52) ρ u T 0 L f f 0 f 0 matching condition f 0 m(f f 0 )d 3 p = 0, v α (f f 0 )d 3 p = 0, 2 mv2 (f f 0 )d 3 p = 0 (2.59) T Maxwell-Boltzmann matching condition f Boltzmann l/l Boltzmann l l/l (2.23) f v rel dσ /l l/l f

27 2. 9 l D Dt f n ( ) f0 0 = t l D ( ) Dt f f 0 = t ( ) n ( ) l fn = L t D/Dt coll coll coll (2.60) l (2.6) L ( ) n l (2.62) L matching condition f 0 l/l f n f 0 f n f 0 ρ u T Boltzmann Chapman-Enskog f n Chapman-Enskog f f p f = f 0 + f, f = f 0 ɛ χ(p) = T f 0χ (2.63) χ matching condition mχd 3 p = 0, v α χd 3 p = 0, 2 mv2 χd 3 p = 0 (2.64) collision f 0 matching condition f 0 f 0 ρ u T f 0 Chapman-Enskog f 0 ( ) f0 0 = (2.65) t (2.52) f 0 u = 0 { f 0 ɛ cp T v T + T T [ vα v β R coll δ αβ ] } ɛ V αβ c v f 0 I(χ) (2.66) T

28 20 2 ɛ v2 2R, V αβ 2 ( uα + u ) β x β x α (2.67) I(χ) ( ) f = f 0 I(χ) (2.68) t coll T I(χ) = w f 0 (χ + χ χ χ )dp dp dp (2.69) T V ρ + (ρu) = 0 t (2.70) u + u u = P t ρ (2.7) s + u s = 0 t (2.72) c v dt P dρ = T ds ρ2 (2.73) P = ρrt (2.74) (2.66) l/l f f 2 f Navier-Stokes 2..8 (2.66) ɛ c p T T v T = I(χ) (2.75) χ = g T (2.76) g χ (2.75) T ɛ c p T T v = I(g) (2.77)

29 2. 2 matching condition(2.64) f 0 gdp f 0 ɛgdp 0 matching condition f 0 vgdp = 0 (2.78) χ (2.5) f = f 0 + δf q = T 2 (v u)2 vf 0 χdp = T 2 (v u)2 vf 0 (g T )dp (2.79) T q α = κ αβ (2.80) x β κ αβ = T 2 (v u)2 f 0 v α g β dp (2.8) κ αβ δ αβ κ αβ = κδ αβ, κ = κ αα 3 (2.82) q = κ T (2.83) κ = 3T 2 (v u)2 f 0 v gdp (2.84) H f g κ Chapman-Enskog l ( ) f v t l (f f 0) = vf 0 l T χ (2.85) coll (2.66) (2.68) I v l χ (2.86)

30 22 2 (2.75) g l (2.87) (2.66) f f 0 l/l Chapman-Enskog g κ κ cnl v (2.88) c l l 2..9 (2.66) Π αβ = P δ αβ + ρu α u β σ αβ (2.89) ( σ αβ = 2η V αβ ) 3 δ αβ u + ζδ αβ u (2.90) (2.90) shear viscosity bulk viscosity (2.66) mv α v β (V αβ ) ( mv 2 3 δ αβ u + 3 ɛ ) u = I(χ) (2.9) c v shear viscosity (2.9) u = 0 m (v α v β 3 ) δ αβv 2 V αβ = I(χ) (2.92) trace 0 χ = g αβ V αβ (2.93) g αβ V αβ trace 0 δ αβ χ g αβ trace 0 χ

31 2. 23 (2.92) V αβ m (v α v β 3 ) δ αβv 2 = I(g αβ ) (2.94) g αβ matching condition g αβ (2.50) σ αβ = m v α v β f 0 χdp = η αβγδ V γδ (2.95) T η αβγδ = m v α v β f 0 g γδ dp (2.96) T η αβγδ {α, β} {γ, δ} {γ, δ} 0 δ αβ η αβγδ = η [δ αγ δ βδ + δ αδ δ βγ 23 ] δ αβδ γδ (2.97) η = m v α v β f 0 g αβ dp (2.98) 0T σ αβ = 2ηV αβ η η η m vnl (2.99) κ nc η mn vl (2.00) bulk viscosity (2.9) 0 ( mv 2 χ 3 ɛ ) u = I(χ) (2.0) c v χ = g u (2.02)

32 24 2 shear viscosity (2.0) u mv 2 3 ɛ c v = I(g) (2.03) σ αβ shear viscosity ζδ αβ u ζ = m v 2 f 0 gdp (2.04) 3T ɛ = mv 2 /2 c v = 3/2 (2.0) 0 (2.03) I(g) = 0 g = 0 (2.04) ζ = 0 bulk viscosity 0 review [5] Mihalas [6] 2..0 moment Chapman-Enskog f l L l/l l/l l/l l/l f f Hermite polynomials Grad [8] [9] (2.52) [ ρ(t, x) c 2 ] f 0 (t, x, v) = exp m(2πrt (t, x)) 3/2 2RT (t, x) (2.05) c v u (2.06) matching condition f f(t, x, v) f 0 (t, x, v) ρ(t, x) u(t, x) T (t, x) a (n) (t, x) ρ u T a (n) moment ρ

33 2. 25 u T a (n) moment a (m) a (n) = 0 moment moment ξ = c RT (2.07) g = (RT )3/2 f (2.08) n N u T g gdξ = (2.09) ξgdξ = 0 (2.0) ξ 2 gdξ = 3 (2.) ] g 0 = exp [ ξ2 (2π) 3/2 2 (2.2) g Hermite polynomials g(ξ, x, t) = g 0 (ξ) = g 0 (ξ) f n! a(n) i n=0 (x, t)h (n) i (ξ) (2.3) ( a (0) i H (0) + a () i H () i + 2! a(2) ij H(2) ij + f(ξ, x, t) = f 0 (ξ) n! a(n) i n=0 (x, t)h (n) i (ξ) (2.4) Hermite polynomials H (n) i ξ n a (n) i n i = (i i n ) Cartesian Hermite polynomials a (n) i = gh (n) i d 3 ξ = fh (n) i d 3 v (2.5) n )

34 26 2 H (n) i H (n) i H (0) = (2.6) H () i H (2) ij = ξ i = ξ i ξ j δ ij H (3) ijk = ξ iξ j ξ k (ξ i δ jk + ξ j δ ik + ξ k δ ij ) H (4) ijkl = ξ iξ j ξ k ξ l (ξ i ξ j δ kl + ξ i ξ k δ jl + ξ i ξ l δ jk + ξ j ξ k δ il + ξ j ξ l δ ik + ξ k ξ l δ ij ) +(δ ij δ kl + δ ik δ jl + δ il δ jk ) H (m) j n = m j i ξ i ± a (n) i a (0) = (2.7) a () i = 0 a (2) ij a (3) ijk = = p ij p S ijk p RT a (4) ijkl = Q ijkl prt p (p ijδ kl + p ik δ jl + p il δ jk + p jk δ il + p jl δ ik + p kl δ ij ) +(δ ij δ kl + δ ik δ jl + δ il δ jk ) a (0) f = f 0 0 a (3) a (3) i = a (3) irr = 2q i p RT (2.8) Hermite moment p ij q i a (n) i (2.3) a (n) i

35 2. 27 a (2) ij a (2) ij t a (2) ij + u r + a (2) ir x r u j + a (2) u i rj + (a (2) ij + δ ij ) x r x r RT DRT Dt (2.9) + RT a(3) ijr x r + RT n a(3) ijr N + 3 x r 2RT a(3) ijr RT x r + u i + u j = J (2) ij x j x i D/Dt J (n) ij J (n) i = N 2 r,s=0 β (nrs) ijk a (r) j a (s) k (2.20) i = (i i n ), j = (j j r ), k = (k k s ) β (nrs) ijk = ( r!s! m B θ, ξ ) ξ g 0 (ξ)g 0 (ξ )H (r) j (ξ)h (s) k (ξ )[H (n) i (ξ)]dθ dɛ dξ dξ RT (2.2) [φ] = φ + φ φ φ (2.22) B (2.20) J (n) a (n) i (2.9) a (2) i a (n) i a (2) i a (n) i U r 4 (2.20) Maxwell f a (n) i moment 3 moment 3 moment expansion bulk viscosity 4 moment f { f = f 0 + p ij 2pRT v iv j q )} i prt v i ( v2 5RT (2.23)

36 28 2 p ij q i p ij t + (u r p ij ) + 2 ( qi + q j 2 ) x r 5 x j x i 3 δ q r ij x r (2.24) u j u i + p ir + p jr 2 x r x r 3 δ u r ijp rs x ( s ui + p + u j 2 ) x j x i 3 δ u r ij + βnp ij = 0 x r q i t + (u r q i ) + 7 x r 5 q u r r + 2 x j 5 q u r r + 2 x i 5 q u r i (2.25) x r + RT p ir x r p ir RT x r p RT x i βnq i = 0 p ir N P rs x s P ij β β(rt ) = 2 2π x 6 e x2 /2 5 0 { } m B(θ, x 2RT ) sin 2 θ cos 2 θ dθ dx (2.26) β (2.24) (2.25) p ij + βnp ij = 0 (2.27) t q i t βnq i = 0 (2.28) p ij q i /βn β (2.20) β v rel σ /βn (2.24) (2.25) Chapman-Enskog 2..7 f ( ui p + u j 2 ) x j x i 3 δ u r ij + βnp () ij = 0 (2.29) x r 5 2 p RT x i βnq() i = 0 (2.30) f p () ij q() i Chapman-Enskog

37 2. 29 λ = p βn = T nl v β (2.3) κ = 5 pr 4 βn = 5 RT 4 β cnl v (2.32) β λ η m (2.24) (2.25) /βn (2.24) (2.25) p ij + A ij + βnp ij = 0 (2.33) t q i t + B i βnq i = 0 (2.34) A ij B i /βn A ij B i source term (2.27) (2.28) p ij = A ij βn q i = 3 B i 2 βn (2.35) (2.36) p ij q i /βn A ij B i Chapman-Enskog /βn Chapman-Enskog (2.24) q i u i Landau- Lifshitz moment 2..7 Chapman-Enskog Navier-Stokes moment (2.9)

38 30 2 (2.24) (2.25) moment B moment moment moment Chapman-Enskog l/l moment moment 3-moment moment moment Hermite (2.3) moment moment free streaming moment free streaming moment moment moment 3 δt ρr + δq = 0 2 t (2.37) τ δ q + δq = κ δt t (2.38) τ κ exp[ nt + ikx] δt δq n 2 τ n + Kk2 = 0 (2.39) K = κ/τc v n n = 2τ [ ± 4τKk 2 ] (2.40)

39 n k 2. K = τ = k = 0 n = 0 n = /τ couple couple BGK moment B

40 metric (+ ) c review de Groot [0] C. Cercignani [] 2.2. f(x, p) f f fd 3 xd 3 p f d 3 xd 3 p p d 4 p δ D (p µ p µ m 2 ) d 4 p δ D (p µ p µ m 2 ) (2.4) d 4 p δ D (p µ p µ m 2 ) = d 4 p 2E δ D(p 0 E) (2.42) p 0 >0, E= p 2 +m 2 d 3 p = 2p 0 p0 = p 2 +m 2 δ D (φ(x)) = i φ (a i ) δ D(x a i ) (2.43) (2.42) d 3 p p 0 = dp xdp y dp z p 0 (2.44) p 0 = p 2 +m 2

41 K K V K ( d d 3 xd 3 p = γd 3 x p 0 3 p ) = d 3 x d 3 p (2.45) d 3 xd 3 p f f N µ 7. N µ = (n, j) p 0 n j 8. T µν d 3 p p 0 pµ f : (2.46) d 3 p p 0 pµ p ν f : (2.47) 9. d S µ 3 p p 0 pµ f(ln f ) : (2.48) Boltzmann f dµ = d 3 xd 3 p N(t) N = f(t + t, x + v t, p + F t)dµ(t + t) f(t, x, p)dµ(t) = 0 (2.49) v = p/p 0 F t dµ

42 34 2 dµ(t + t) = J dµ(t) (2.50) J = ( x (t + t), x 2 (t + t),, p 3 (t + t) ) (x (t), x 2 (t),, p 3 (2.5) (t)) t J J = + F i p i t + O [ ( t) 2] (2.52) f (2.49) [ N f t = f + vi t x i + F i f p i + f F i p i = [ f t f + vi x i + (ff i ) p i ] dµ(t) = 0 ] dµ(t) (2.53) t s [ DN f Ds = γ t [ = p0 f m t [ p ν = m f + vi x i + (ff i ) p i f + vi x i + (ff i ) p i f x ν + p0 m (ff i ) p i ] dµ(t) (2.54) ] dµ(t) ] dµ(t) = 0 ( p F K µ = m, p 0 ) m F (2.55) p µ K µ = 0 (2.56) (2.54) / p x p 0 p p0 p + p = p p 0 p 0 + p (2.54) p 0 (ff i [ ( ) ) m p i = p 0 fk p p 0 p 0 p 0 + ] p 0 p (fk) = fk0 p 0 = fkµ p µ + p (fk) (2.57) (2.58)

43 [ ] DN p ν Ds = f m x ν + (fkν ) p ν dµ(t) = 0 (2.59) 0 p (2.23) v rel dσ n n 2 σ = dσ v rel dv dt dν dν = σv rel n n 2dV dt (2.60) dν dν dν = An n 2 dv dt (2.6) A σv rel dv dt dν An n 2 n n 0 ndv n = γn 0 = p0 m n 0 (2.62) An n 2 Ap 0 p 0 2 Ap 0 p 0 2 p0 σv rel A p0 p 0 2 p µ p µ 2 p 0 = A p 0 2 p 0 p0 2 p (2.63) p 2

44 36 2 p 0 2 = m 2 p 2 = 0 σv rel A = σv rel A A = σv rel p µ p µ 2 p 0 p0 2 (2.64) v rel p µ p µ 2 p µ p µ 2 = m v 2 rel m 2 (2.65) v rel = m2 m2 2 (p µ p µ 2 )2 (2.66) p p 2 dν n = d 3 p f ( ) f t coll dµ(t) = dµ(t) (f f ff ) p µp µ p 0 p 0 = m ( ) f p 0 dµ(t) s coll v rel σd 3 p (2.67) p ν f x ν + ) m (fkν p ν = p 0 (f f ff ) p µp µ p 0 p 0 v rel σd 3 p (2.68) p 0 f t + pi f p 0 x i + (ff i ) p i = (f f ff ) p µp µ p 0 p 0 v rel σd 3 p (2.69) H (2.68) H

45 µ S µ d µ S µ 3 p = p 0 pµ µ [f(ln f )] (2.70) d 3 p = ln f pµ µ f (2.68) p 0 d 3 p m p 0 ln f (fkµ ) d 3 p p µ p 0 ln f(f f ff )p µ p µ v relσ d3 p p 0 (2.7) d 3 p m p 0 ln f (fkµ ) p µ = m d 4 p p µ δ D(p 2 m 2 )θ(p 0 )f(ln f )K µ (2.72) θ(p 0 ) 0 φ(f f ff )p µ p µ v relσ d3 p d 3 p p 0 p 0 = (φ + φ φ φ 4 )(f f ff )p µ p µ v relσ d3 p d 3 p p 0 p 0 (2.73) µ S µ = ln ff ( 4 f f ff ) f f f f p µ p µ v relσ d3 p d 3 p p 0 p 0 (2.74) ( x) ln x x > 0 x = 0 µ S µ 0 (2.75) (2.68)

46 (2.68) p f 0 f 0f 0 f 0 f 0 = 0 (2.76) ln ln f 0 + ln f 0 ln f 0 ln f 0 = 0 (2.77) ln f 0 ln f 0 = α + β µ p µ (2.78) f 0 n f 0 = [ 4πm 2 T K 2 (ζ) exp u µp µ T ] (2.79) ζ = m T n T u µ u µ u µ u µ = T Maxwell-Jüttner (2.74) 0 ζ u µ

47 f f f v / c 2.2 ζ = v / c 2.3 ζ = v / c 2.4 ζ = 0.0 c ζ = 00 Maxwell ζ =

48 40 2 ζ = 0.0 N µ 2.2. d N µ 3 p n d 3 p = p 0 pµ f 0 = 4πm 2 T K 2 (ζ) p 0 pµ exp [ u µp µ T u µ ] (2.80) N µ = n u µ (2.8) u µ p = p d 3 p = p 2 dpdω = pp 0 dp 0 dω (2.82) z = p0 T (2.83) (2.80) (2.8) u µ u µ n n d 3 p = 4πm 2 T K 2 (ζ) p 0 p0 exp [ u µp µ ] (2.84) T n = ζ 2 dzz z K 2 (ζ) 2 ζ 2 e z (2.85) K n (ζ) = 2n n! (2n)! ζ = 2n (n )! (2n 2)! ζ n dz(z 2 ζ 2 ) n /2 e z (2.86) ζ ζ n dzz(z 2 ζ 2 ) n 3/2 e z (2.87) ζ (2.85) n = n N µ = nu µ 2.2. T µν T µν = d 3 p p 0 pµ p ν f 0 = n d 3 p 4πm 2 T K 2 (ζ) p 0 pµ p ν exp [ u µp µ ] T (2.88) u µ u ν η µν T µν = neu µ u ν P γ µν (2.89)

49 2.2 4 γ µν = η µν u µ u ν (2.89) e P f 0 d ne = T µν 3 p u µ u ν = p 0 (pµ u µ ) 2 f 0 (2.90) P = 3 T µν γ µν = d 3 p 3 p 0 pµ p ν γ µν f 0 (2.9) N µ P = nt (2.92) e = m K (ζ) + 3T K 2 (ζ) (2.93) (2.92) recurrence relation K n+ (ζ) = 2n K n(ζ) ζ + K n (ζ) (2.94) (2.93) e = m K 3(ζ) K 2 (ζ) T (2.95) (2.92) (2.95) h = e + P n = mk 3(ζ) K 2 (ζ) (2.96) c P = ( ) ( ) h e, c V = T P T V (2.97) c P = Γ Γ = d K 3 (ζ) dζ K 2 (ζ) (2.98) c V = c P (2.99) Γ recurrence relation d K n (ζ) dζ ζ n = K n+(ζ) ζ n (2.200)

50 42 2 (2.94) c P c P = Γ Γ = ζ2 + 5h T ( ) 2 h (2.20) T s d ns = S µ 3 p u µ = p 0 pµ u µ f(ln f ) (2.202) f 0 [ µ p µ ] u µ f 0 = exp (2.203) T (2.202) (2.85) (2.90) ns = n (e µ) + n (2.204) T µ = e + P n T s = h T s (2.205) µ T ζ ζ ζ massless K n (ζ) [ π 2ζ e ζ + 4n2 + (4n2 )(4n 2 ] 9) 8ζ 2!(8ζ 2 + ) (ζ ) (2.206) (2.95) (2.96) (2.20) e m T h m T + 5 Γ T 2 m + (2.207) T 2 m + (2.208) 8 T m + (2.209)

51 lim ζ 0 ζn K n (ζ) = 2 n (n )! (2.20) e = 3T (2.2) h = 4T (2.22) Γ = 4 3 (2.23) P = nt (2.2) P = ne 3 (2.24) trt µν = 0 (2.25) Eckart Landau-Lifshitz f time like u µ 0. u µ : (2.26) u µ u µ = (2.27). γ µν : (2.28) γ µν = g µν u µ u ν (2.29) g Minkowski η µν 2.2.

52 44 2 T µν = η µρ η νσ T ρσ (2.220) = (γ µρ + u µ u ρ )(γ νσ + u ν u σ )T ρσ = (T ρσ u ρ u σ )u µ u ν + (u ρ T ρσ γ σµ )u ν + (u ρ T ρσ γ σν )u µ + {(γ µρ γ νσ 3 ) γµν γ ρσ + 3 } γµν γ ρσ T ρσ f 0 T µν = neu µ u ν (2.22) + {(q µ + h E γ µν N ν )u ν + u µ (q ν + h E γ νµ N ν )} + p µν (P + Π)γ µν N µ N µ = nu µ + ν µ (2.222) ν µ u µ u µ ν µ = 0 (2.223) 2. n N µ u µ : (2.224) p µν (γ µρ γ νσ 3 ) γµν γ ρσ T ρσ : (2.225) P + Π 3 γµν γ ρσ T ρσ : + bulk viscosity (2.226) q µ u ρ T ρσ γ σµ h E γ µν N ν (2.227) = u ρ T ρσ γ σµ h E ν µ : e n T ρσu ρ u σ : (2.228) p µν traceless T 00 T µν u µ u ν

53 u µ u µ (2.22) (2.222) T µν T 0µ N µ Eckart [] 3. u µ E : Eckart velocity (2.229) N µ = nu µ E (2.230) T µν = neu µ E uν E (2.23) + (q µ u ν E + u µ E qν ) + p µν (P + Π)γ µν N µ u µ u µ E f u µ E N µ N µ = d 3 p N µ n p 0 pµ f (2.232) Landau-Lifshitz [2] 4. u µ L : Landau Lifshitz velocity (2.233) N µ = n L u µ L + νµ (2.234) T µν = n L e L u µ L uν L (2.235) + p (µν) L (P L + Π L )γ µν T µν u µ u µ L f u µ L T µν u Lν ulρ T ρσ T στ u τ L = u Lν ulρ T ρσ T στ u τ L d 3 p p 0 pµ p ν f (2.236) Eckart Landau-Lifshitz T µν u µ E uµ L u µ L = uµ E + U µ (2.237)

54 46 2 U µ u Lµ u µ L = = (uµ E + U µ )(u Eµ + U µ ) + 2U µ u Eµ (2.238) U µ u µ E u Lµ u µ E = (2.239) N µ = nu µ E = n Lu µ L + νµ (2.240) u µ L (2.239) n = n L (2.24) Eckart Landau- Lifshitz (2.237) (2.240) U µ = νµ n (2.242) T µν = neu µ E uν E + (q µ u ν E + u µ E qν ) + p µν (P + Π)γ µν (2.243) = n L e L u µ L uν L + p (µν) L (P L + Π L )γ µν (2.237) P L = P, e L = e, Π L = Π, p (µν) L = p µν, ν µ = h E q µ (2.244) h E = e + P/n Eckart Landau-Lifshitz u µ E = uµ L nh E q µ (2.245) q µ Eckart Landau-Lifshitz

55 j = ρv (2.246) (2.207) (2.22) (2.222) (2.68) p µ µ f = ( ) f s coll (2.247) p ν d 3 p/p 0 (2.73) µ N µ = 0 (2.248) µ T µν = 0 (2.249) (2.22) (2.222) Eckart Landau-Lifshitz µ u µ µ = η µν ν = u µ u ν ν + γ µν ν u µ D + µ (2.250) D u µ µ = t µ γ µν ν (2.25) = x i (2.252)

56 48 2 = u µ D u µ Lagrange µ (2.248) Dn = n µ u µ µ ν µ + ν µ Du ν (2.253) Eckart ν µ = 0 Dn = n µ u µ (2.254) Landau-Lifshitz ν µ = q µ /h E Dn = n µ u µ + µ q µ h E q µ h E Du ν (2.255) (2.22) γ ρ ν W µ u ν T νρ γ ρµ = q µ + h E ν µ (2.256) T µν (2.22) T µν = neu µ u ν + {W µ u ν + u µ W ν } + p µν (P + Π)γ µν (2.257) (2.256) nh E Du µ = µ P + µ Π γ ν µ σ p νσ (2.258) + (p µν Du ν γ ν µ DW ν W µ ν u ν W ν ν u µ ) u µ Eckart W µ = q µ nh E Du µ = µ P + µ Π γ ν µ σ p νσ (2.259) + (p µν Du ν γ ν µ Dq ν q µ ν u ν q ν ν u µ ) Landau-Lifshitz W µ = 0 nh E Du µ = µ P + µ Π γ ν µ σ p νσ (2.260) +p µν Du ν (2.22) u ν D(ne) = nh E µ u µ + p µν ν u µ µ W µ + 2W µ Du µ (2.26)

57 (2.255) Dn nde = P µ u µ Π µ u µ + p µν ν u µ µ W µ + e µ ν µ (2.262) +(2W µ eν µ )Du µ (2.26) (2.262) Eckart D(ne) = nh E µ u µ + p µν ν u µ µ q µ + 2q µ Du µ (2.263) nde = P µ u µ Π µ u µ + p µν ν u µ µ q µ + 2q µ Du µ (2.264) Landu-Lifshitz D(ne) = nh E µ u µ + p µν ν u µ (2.265) nde = P µ u µ Π µ u µ + p µν q µ ν u µ e µ + e q µ Du µ h E h E (2.266) P dn De + P Dn = (2.262) (2.253) De Dn n(de+p Dn ) = Π µ u µ +p µν ν u µ µ q µ ν µ µ h E +(2q µ +h E ν µ )Du µ (2.267) 0 Eckart n(de + P Dn ) = Π µ u µ + p µν ν u µ µ q µ + 2q µ Du µ (2.268) Landau-Lifshitz n(de + P Dn ) = Π µ u µ + p µν ν u µ + qµ h E µ h E q µ Du µ (2.269) q µ p µν Π (2.253) (2.258) (2.262) Dn = n µ u µ (2.270) nh E Du µ = µ P (2.27) nde = P µ u µ (2.272)

58 50 2 u µ exp( ik µ x µ ) = exp( iωt + ikx) (2.273) δu i = δu x ω kn 0 0 δn 0 ωnh E 0 k δu x 0 kp ωnc V 0 δt T 0 n δp iωδn = iknδu x (2.274) iωnh E δu x = ikδp (2.275) iωnδe = ikp δu x (2.276) δe = c V δt (2.277) δp = nδt + T δn (2.278) = (2.279) n 2 ω ( h E c V ω 2 c P T k 2) = 0 (2.280) ω = ±C s k (2.28) Γ C s = T (2.282) h E m h E (2.208) (2.209) h E m Γ 5 3 5T/3m (2.22) (2.23) h E 4T Γ 4 3 c/ 3

59 Chapman-Enskog Chapman-Enskog Chapman-Enskog Eckart Landau-Lifshitz Eckart f f(t, x, p) f 0 (t, x, p) + f (t, x, p) = f 0 (t, x, p)[ + φ(t, x, p)] (2.283) f l L l/l f 0 (t, x, p) n f 0 (t, x, p) = [ 4πm 2 T K 2 (ζ) exp u µp µ T ] (2.284) n T u i matching condition Eckart d 3 p p 0 pµ f 0 φ = 0 (2.285) d 3 p p 0 (pµ u µ ) 2 f 0 φ = 0 (2.286) (2.285) f 0 n u µ (2.286) e Landau-Lifshitz u µ d 3 p p 0 pµ f 0 φ = 0 (2.287) u µ d 3 p p 0 pµ p ν f 0 φ = 0 (2.288) 0 f 0 f

60 52 2 p µ µ f 0 = I[φ] (2.289) I[φ] (2.270) (2.27) (2.272) f 0 { T [ ( u c V 3 ζ2 c V (G 2 ζ 2 4Gζ ζ 2 µ ) p µ ) T ( u 3 (ζ2 + 5Gζ ζ 2 G 2 µ ) ] 2 p µ 4) ν u ν p µp ν T T µ u ν + p ( µ T 2 (p νu ν h E ) µ T T )} µ P = I[φ] nh E G (2.290) G = K 3(ζ) K 2 (ζ) = h E m (2.29) φ φ = [ a 0 + a u µ p µ + a 2 (u µ p µ ) 2] ν u ν (2.292) [ + p µ (a 3 + a 4 p ν u ν ) µ T T ] µ P + a 5 p µ p ν µ u ν nh E φ (2.290) matching condition(2.285) (2.286) a 0 a a 2 a 3 a 4 a 5 φ bulk viscosityπ q µ p µν traceless (2.292) Π = µ µ u µ (2.293) [ q µ = κ µ T T ] µ P (2.294) nh E p µν = 2η µ u ν (2.295) bulk viscosity e = c V T m T

61 P ( µ ) ( ) he d = T nt 2 dt + dp nt (2.296) µ q µ = κ nt 2 ( µ µ ) h E T (2.297) Y (P, T ) = 0 dv j [ e (E j+p V )/T = exp µ ] T (2.298) a i [0] [] µ = P 2 T m 2 I (20G + 3ζ 3G 2 ζ 2Gζ 2 + 2G 3 ζ 2 ) 2 ( 5Gζ ζ 2 + G 2 ζ 2 ) 2 (2.299) κ = 3P 2 m 2 (ζ + 5G G 2 ζ) 2 I I 2 (2.300) η = 30P 2 T m 2 G 2 2I 6I 2 + 3I 3 (2.30) I I 2 I 3 αr n n µ = T (20G + 3ζ 3G 2 ζ 2Gζ 2 + 2G 3 ζ 2 ) 2 ζ 4 K 2 (ζ) 2 64π σ ( 5Gζ ζ 2 + G 2 ζ 2 ) 2 (2K 2 (2ζ) + ζk 3 (2ζ)) κ = 3 (ζ + 5G G 2 ζ) 2 ζ 4 K 2 (ζ) 2 64πσ (ζ 2 + 2)K 2 (2ζ) + 5ζK 3 (2ζ) η = 5 T ζ 4 K 3 (ζ) 2 64π σ (2 + 5ζ 2 )K 2 (2ζ) + (3ζ ζ)K 3 (2ζ) (2.302) (2.303) (2.304) bulk viscosity

62 54 2 (2.302) (2.303) (2.304) ζ µ 25 mt 256σ π ζ 2 κ 75 mt 256mσ π η 5 mt 64σ π ( 83 6 ( ) ζ + ) ζ + ) ( ζ + (2.305) (2.306) (2.307) bulk viscosity ζ µ [ ( T π σ ζ ln ζ ) 2 + 6γ κ ( ) 2πσ 4 ζ2 + η 3 ( T + ) 0π σ 20 ζ2 + ] + bulk viscosity (2.308) (2.309) (2.30) moment Chapman-Enskog moment moment bulk viscosity 3 moment Grad 3-moment expansion bulk viscosity 4 moment Eckart Cercignani [] moment ψ =, p µ, p µ p ν d 3 p/p 0 µ N µ = 0 (2.3) µ T µν = 0 (2.32) µ T µνρ = P νρ (2.33)

63 (2.33) d T µνρ 3 p p 0 pµ p ν p ρ f (2.34) P νρ d 3 p d 3 p 2 p 0 p 0 (p ν p ρ + p ν p ρ pν p ρ p ν p ρ )ff p µ p µ v relσ (2.35) n, u µ, T, Π, p µν, q µ 4 moment T µνρ 4-moment moment moment f 4-moment f T µνρ u µ s 2.2. f s uµ n 4-moment 4 constraint d 3 p p 0 p µf(ln f ) (2.36) N µ u µ = u µ d 3 p p 0 pµ f (2.37) T µν u µ = u µ d 3 p p 0 pµ p ν f (2.38) T µν ρ u ρ = u ρ d 3 p p 0 p µ p ν p ρ f (2.39) N µ T µν moment constraint constraint F = uµ d 3 p n p 0 p µf(ln f ) + λ ( d + λ ν T µν 3 p u µ u µ p 0 pµ p ν f ( N µ u µ u µ d 3 p ) p 0 pµ f ) + λ µν ( T µν ρ u ρ u ρ d 3 p p 0 p µ p ν p ρ f (2.320) ) f 4-moment moment f Euler-Lagrange f [ ] f = exp n(λ + λ µ p µ + λ µν p µ p ν ) (2.32) 4 λ λ µ λ µν 4 moment λ = λ E + λ NE, λ µ = λ E µ + λ NE µ, λ µν = λ NE µν (2.322)

64 56 2 f f = exp [ n(λ E + λ E µ p µ ) ] exp [ ] n(λ NE + λ NE µ p µ + λ NE µν p µ p ν ) (2.323) f 0 moment f { } f = f 0 n(λ NE + λ NE µ p µ + λ NE µν p µ p ν ) (2.324) λ NE µ λne µν uµ λ NE µ = λu µ + λ ν γµ ν (2.325) λ NE µν = Λu µu ν + ( 2 Λ ρ(γµu ρ ν + γνu ρ µ ) + Λ ρσ γµγ ρ ν σ ) 3 γρσ γ µν (2.326) f f = f 0 { nλ NE n( λu µ + λ ν γµ)p ν µ (2.327) n [Λu µ u ν + 2 Λ ρ(γ ρµu ν + γ ρνu µ ) + Λ ρσ (γ ρµγ σν 3 )] } γρσ γ µν p µ p ν ) f Πq µ p µν f d N µ 3 p d 3 p = p 0 pµ f = p 0 pµ f 0 (2.328) N µ u µ γ ν µ ( λ NE + λm G ) ( + Λm G ) = 0 (2.329) ζ ζ λ µ γ µν + mgλ µ γ µν = 0 (2.327) T µν T µν p µν = 2n 2 m 2 T G ζ Λ µν (2.330) [ ( Π = n 2 T λ NE + Gm λ G ) ] m 2 Λ ζ [ ( q µ = n 2 T Gmγ µν λν G ) ] m 2 γ µν Λ ν ζ ( λ NE G ) + (3 ζ λm Gζ ) + + Λm (5 2 Gζ 2 + 2ζ ) + G = 0

65 (2.329) (2.330) 4 f { Π f = f 0 + α 0 (α + α 2 u µ p µ + ζm ) P 2 u µu ν p µ p ν (2.33) ( q µ G + α 3 P m pµ ) m 2 u νp µ p ν + p } µν ζ P 2Gm 2 pµ p ν α 0 = 5Gζ ζ 2 + G 2 ζ 2 20G + 3ζ 3G 2 ζ 2Gζ 2 + 2G 3 ζ 2 (2.332) α = 5G + 2ζ 6G2 ζ + 5Gζ 2 + ζ 3 G 2 ζ 3 5Gζ ζ 2 + G 2 ζ 2 α 2 = 3ζ 6G + ζ G 2 ζ m 5Gζ ζ 2 + G 2 ζ 2 ζ α 3 = ζ + 5G G 2 ζ f T µνρ P µν T µνρ = (nc + C 2 Π)u µ u ν u ρ + 6 (nm2 nc C 2 Π)(η µν u ρ + η µρ u ν + η νρ u µ ) +C 3 (η µν q ρ + η µρ q ν + η νρ q µ ) 6C 3 (u µ u ν q ρ + u µ u ρ q ν + u ν u ρ q µ ) +C 4 (p µν u ρ + p µρ u ν + p νρ u µ ) (2.333) P µν = B (η µν 4u µ u ν ) Π + B 2 (u µ q ν + u ν q µ ) + B 3 p µν (2.334) C i C = m2 (ζ + 6G) (2.335) ζ C 2 = 6m 2ζ 3 5ζ + (9ζ 2 30)G (2ζ 3 45ζ)G 2 9ζ 2 G 3 ζ 20G + 3ζ 3G 2 ζ 2ζ 2 G + 2ζ 2 G 3 C 3 = m ζ + 6G G 2 ζ ζ ζ + 5G G 2 ζ C 4 = m (ζ + 6G) Gζ

66 58 2 B i I i B = ζ ( ζ 2 5ζG + ζ 2 G 2 )I 3m 2 P 20G + 3ζ 3G 2 ζ 2ζ 2 G + 2ζ 2 G 3 (2.336) B 2 = ζ I I 2 3m 2 P ζ + 5G ζg 2 ζ 2I 6I 2 + 3I 3 B 3 = 30m 2 P G (2.33) C 2 2 DΠ + 2 (m2 + C )Dn ζ 2T nc DT 5C 3 µ q µ (2.337) + 6 (nm2 + 5nC ) µ u µ = 3B Π 5C 3 Dq µ [ (m 2 C ) µ n + ζ ] 6 T nc µ T C 2 µ Π C 4 ν p µν (2.338) 6 (nm2 + 5nC )Du µ = B 2 q µ C 4 Dp µν + 2C 3 µ q ν + 3 (nm2 nc ) µ u ν = B 3 p µν (2.339) ζ (2.337) (2.338) Dn Du µ DT Dn + n µ u µ = 0 (2.340) nh E Du µ = µ (P + Π) ν p µν Dq µ nc V DT = P µ u µ µ q µ (2.337) (2.338) C 2 2 DΠ n 2 (m2 + C ) µ u µ + ζ 2T C ζ 2 + 5ζG ζ 2 G 2 ( µ q µ + P µ u µ ) (2.34) 5C 3 µ q µ + 6 (nm2 + 5nC ) µ u µ = 3B Π 5C 3 Dq µ [ (m 2 C ) µ n + ζ ] 6 T nc µ T C 2 µ Π C 4 ν p µν (2.342) [ (nm 2 + 5nC ) µ (P + Π) ν p µν Dq µ] = B 2 q µ 6nh E (2.339) (2.340) (2.34) (2.342) 4-moment (2.339) (2.34) (2.342) p µν Π q µ (2.339) (2.34) (2.342) p µν Π q µ

67 Chapman-Enskog Chapman-Enskog Grad moment Chapman-Enskog Navier-Stokes Chapman-Enskog (2.83) Chapman-Enskog Navier-Stokes c P T t + ( κ T ) (2.343) c P κ T t = χ T (2.344) χ κ/c P T = 0 T (t = 0, x) = const δ D (x) T (t, x) = const ) ( 2 πχt exp x2 4χt (2.345) t x = 0 Chapman-Enskog Chapman-Enskog

68 60 2 [3] causal cone causal cone Lorentz causal cone Chapman-Enskog 2.5 causal cone Lorentz [2] Grad moment 2..0 moment Chapman-Enskog Chapman-Enskog moment

69 moment Chapman-Enskog Chapman-Enskog BGK BGK Navier-Stokes BGK BGK BGK Eckart Marle [3] BGK

70 62 3 Landau-Lifshitz Anderson-Witting [4] BGK 3. Boltzmann 3.. BGK (2.22) A k Bhatnagar Gross Krook [8] BGK ( ) f = f(t, x, v) f 0(t, x, v) (3.) t coll τ f 0 (t, x, v) = ρ (v u) 2 e 2RT (3.2) m(2πrt ) d/2 f 0 (t, x, v) f local matching condition (f 0 f) ψ µ d 3 v = 0 (3.3) ψ µ = m (, v, 2 ) (v u)2 BGK f t = f(t, x, v) f 0(t, x, v) τ (3.4) f f 0 e t/τ τ f 0 τ mean flight time f f 0 v [9] A τ BGK

71 3. Boltzmann 63 f0 f ψ µ d 3 v = 0 (3.5) τ ψ µ = m (, v, 2 ) (v u)2 matching condition(3.3) (3.5) τ v matching condition mean flight time τ v v C s 3..2 BGK E ( ) ( ) Df(t, x, v) f = Dt t + v f(t, x, v) = t coll (3.6) BGK ( ) f t coll = f(t, x, v) f 0(t, x, v) τ (3.7) f 0 (t, x, v) = ρ(t, x) (v u(t,x)) 2 e 2RT (3.8) m(2πrt (t, x)) 3/2 f 0 source term f matching condition f 0 f (f 0 f) ψ µ d 3 v = 0 (3.9) ψ µ = m (, v, 2 ) (v u)2 (3.0) f 0 (v) δf = f f 0, δf 0 = f 0 f 0, δτ = τ τ ( τ f 0 ) (3.)

72 64 3 f 0 ( ) t + v δf = δf δf 0 τ (3.2) τ τ τ δf = δ fe iω(t t 0)+ik x (3.3) (3.2) ( ) iω + ik v δf = τ τ δf 0 (3.4) δf 0 [ δf 0 = f δρ 0 + v δu + ρ 0 RT 0 ρ 0 f 0 = e v m(2πrt 0 ) 3/2 f 0 ( v 2 2RT ) ] δt 2T 0 (3.5) 2 /2RT 0 (3.6) δρ δu δt matching condition δf ( ) iω + ik v δf(v) (3.7) τ = d 3 v f [ 0 (v) m + v ( v v 2 + m τ ρ 0 RT 0 2RT0 2 3 ) ( v 2 3 )] δf(v ) 2T 0 2RT 0 2 ωτ ω, 2RT0 τk k, v 2RT0 v (3.8) δf(v) = d 3 v K(v, v )δf(v ) (3.9) [ f K(v, v 0 (v) m ) + 2v v + m ( v 2 3 ) ( v 2 3 )] (3.20) iω + ik v ρ 0 T iω + ikv 0 iω + ikv = 0 ω Fredholm 2 K(v, v )

73 3. Boltzmann 65 D v v 2 3/2 ω k E iω ( b δρ+ δρ + 2ib ) { k k δu b } πe b 2 Erfc(b) { ( b b 2 + } πe b 2) 2 Erfc(b) δt = 0 b b = iω k (3.2) (3.22) k v { ( k δu + ib b b 2 + } πe b 2) 2 Erfc(b) δt (3.23) ( i δρ + 2ib ) ( k k δu b πe Erfc(b)) b2 = 0 k v kv y { k } πe b2 Erfc(b) δu y = 0 (3.24) v 2 /2 3/2 (3.2) [ { 3k 2b( b 2 ) + (2b 4 b 2 + ) }] πe b2 Erfc(b) δt 2 + [ { 3k 2 b (b 2 ) }] πe b2 Erfc(b) δρ 2 [ { 2b b (b 2 ) } ] k δu πe b2 Erfc(b) 3 = 0, (3.25) k 3.2 δu δρ δt { ( b 2 b b 2 + } πe b 2) 2 Erfc(b) { ( 3 k 2b + 2 ) ( + 2 k { ( ( kb) + + 2b 2 2b = 0, { 3k + 2b 2 2b k ) ( b πe b2 Erfc(b)) k ( 2b( b 2 ) + (2b 4 b 2 + ) πe b2 Erfc(b) (3.26) ) ( b + ( b 2 ) πe Erfc(b)) } b2 } )}

74 66 3 k πe b2 Erfc(b) = 0 (3.27) ω δρ δu δt δf δf(v) = n C n f0 (v) iω n + ik v C n [ ( δρωn + 2v δu ωn + v 2 3 ) ] δtωn e i(ω nt+k x) ρ 0 2 T 0 (3.28) Erfc(z) arg(z) < π/4 ω (3.28) ψ µ (3.28) matching condition matching condition (3.2) b (3.23) i ωδρ + k δu = 0 (3.29) iω + ikv = 0 kv continuous mode (3.7) 0 = d 3 v f 0 (v) τ = δf 0 [ m ρ 0 + v ( v v 2 + m RT 0 2RT0 2 3 ) ( v 2 3 )] δf(v ) (3.30) 2T 0 2RT 0 2 ρ u T f moment (3.30) Result

75 3. Boltzmann (ω τ) k τ C s (ω τ) k τ C s (ω τ) k τ C s 3.3 shear flow

76 68 3 ω ω k 2 Navier-Stokes Navier-Stokes shear flow 00 0 (ω τ) k τ C s 3.4 shear flow 00 0 (ω τ) k τ C s 3.5

77 3. Boltzmann (ω τ) k τ C s Im ω 0.8 Re (k/ ω) / τ ω 3.7

78 Im (k/ ω) / τ ω 3.8 Bhatnagar [9] Sirovich [35] (3.2) (3.23) (3.25) δρ 0 δ ρ δ u δ T k τ C s 3.9

79 3. Boltzmann 7 0 δ ρ δ u δ T k τ C s δ ρ δ u δ T k τ C s δ ρ δ u δ T k τ C s 3.2 2

80 72 3 δt δu 0 δu δρ δu δt 2 ω 3..4 Discussion (3.26) (3.27) shear (3.27) ω = 0 ( iω)/k b Erfc(x) Erfc(x) = 2 π e x2 2 π x e t2 dt (3.3) ( 2x ) (2n )!! 2 2 ( )n x4 2 n x 2n (3.32) (3.27) k b + ( ) 2b 3 + O b 4 = 0 (3.33) b 4 b = ( iω)/k k 2( + iω) 3 {k2 2( + iω)iω} = 0 (3.34) iω k 3 iω = k2 2 + O(k4 ) (3.35)

81 3. Boltzmann 73 shear flow k = 0 iω = (3.26) b b (3.26) ω 4 ( ) ( 3k 8 b + 5 kb b 3 2 2kb 4 2b b ( 4 kb + 2 kb 2 b kb b 4 ) ) ( 3k 3 b + 5 3b 3 ) = 0 (3.36) b = ( iω)/k ω 4 k k 2 ( k 2 )(iω) 4 (2 + 22k 2 )(iω) k 2 (iω) 2 0k 2 (iω) + 5k 4 = 0 (3.37) iω k 2 k = 0 ω = 0 3 iω = k2 2 + O(k3 ), k ± i 6 k + O(k3 ) (3.38) A (A.45) (A.46) A BGK λ = /τ (A.45) (A.46) (A.45) (A.46) k / 2 k 2RT 0 A RT 0 k Im ω Re ω k b (3.34) b 3..3 k = 0 ω = 0 shear flow k = 0 ω = i

82 74 3 A shear flow k = 0 ω = i A BGK ω = i (3.26) (3.2) b 0 /b (3.26) b = 0 k ω = i ω = iτ Im ω < F. [35] 3..5 collision n collision time n δρ δu k n collision BGK τ v k collision collisionless k collision δe Chapman-Enskog k

83 3. Boltzmann 75 collision n momentum density pressure viscosity Eigenvalue k τ C s momentum density pressure viscosity Eigenvalue k τ C s 3.4

84 momentum density pressure viscosity Eigenvalue k τ C s 00 momentum density pressure viscosity Eigenvalue k τ C s 3.5 2

85 3. Boltzmann continuous spectrum 3..2 iω + ikv = 0 continuous spectrum v y v z (3.30) C. Cercignani [40] v 0 δf δf = δf + Dδ D (v v 0 ) (3.39) δ D Dirac D δf (3.7) δf ( ) ( iω + ik v δf + τ τ iω + ik v 0 = d 3 v f [ 0 (v) m + v ( v v 2 + m τ ρ 0 RT 0 2RT0 2 3 ) ( v 2 3 2T 0 2RT f [ 0 (v) m + v ( v 2 v 0 + m 3 ) ( v )] D τ ρ 0 RT 0 2T 0 2RT 0 2 2RT 2 0 continuous spectrum ) D (3.40) )] δf (v ) ω = i τ + k v 0 (3.4) ω d 3 v [ f0 (v) m δf = + v ( v v 2 + m iτk (v v 0 ) ρ 0 RT 0 2RT0 2 3 ) ( v 2 3 )] δf (v ) 2T 0 2RT [ f 0 (v) m + iτk (v v 0 ) ρ 0 v ( v 2 v 0 + m RT 0 2RT0 2 3 ) ( v )] D 2T 0 2RT 0 2 (3.42) Fredholm shear flow v y { k } πe b2 Erfc(b) δu y = πe b2 Erfc(b) D v 0y (3.43) b = ik v 0 k = iv 0x (3.44)

86 78 3 continuous spectrum δu y k [ ] k δu y = πe v0xerfc( iv 2 0x ) D v 0y (3.45) δρ δu δt continuous spectrum v (3.30) continuous spectrum δf continuous spectrum A Maxwell BGK [7] r n n > 2 continuous mode f L 2

87 3.2 Boltzmann Marle Boltzmann Marle BGK BGK Eckart Marle BGK 3.2. Marle BGK model Marle [3] BGK ( ) f = m t coll τ (f(t, x, p) f 0(t, x, p)) (3.46) τ p µ f 0 [ n(t, x) f 0 (t, x, p) = 4πm 2 T (t, x)k 2 (ζ(t, x)) exp p µu µ ] (t, x) T (t, x) ζ = m T (3.47) (3.48) p µ Maxwell-Jüttner n u µ T x µ u µ u 2 = ( ) p µ µ f = p 0 t + v f = m τ (f(t, x, v) f 0(t, x, v)) (3.49) p µ µ N µ = m τ µ T µν = m τ d 3 p p 0 (f f 0) (3.50) d 3 p p 0 pν (f f 0 ) (3.5) 0 n u µ T 0

88 80 3 d 3 p p 0 (f f 0) = p 0 nk (ζ) mk 2 (ζ) = 0 (3.52) d 3 p p 0 pµ (f f 0 ) = N µ N µ 0 = 0 (3.53) matching condition f 0 matching condition n u e Marle BGK model n matching condition /e matching condition e 0 f 0 e e 0 Marle BGK e e 0 0 bulk viscosity (3.53) Eckart Marle BGK BGK [] ζ = m/t Marle BGK /ζ τ (3.46) BGK τ τ ζ ζ τ BGK τ BGK ( ) t + v f = τ (f(t, x, v) f 0(t, x, v)) (3.54) f (3.54) [ f(t) = f(0) + t ] e t/τ f 0 (t )dt e t/τ (3.55) τ 0

89 3.2 Boltzmann Marle 8 τ τ Marle BGK (3.49) t ] f(t) = [f(0) + τ e t /τ f 0 (t )dt e t/τ (3.56) 0 τ = p0 m τ (3.57) τ τ τ f(t, x, p) Marle BGK (3.49) p µ µ f(t, x, p) = p 0 ( t + v ) f(t, x, p) = m τ (f(t, x, p) f 0(t, x, p)) (3.58) p = 0 t f(t, x, 0) = m τ (f(t, x, 0) f 0(t, x, 0)) (3.59) BGK τ Marle BGK τ τ τ BGK τ τ τ Marle BGK τ τ τ M τ M m τ M p 0 τ (3.60) τ M 3.2. (3.52) τ M = m n d 3 p p 0 f 0 τ = K (ζ) τ (3.6) K 2 (ζ)

90 82 3 K n n K (ζ)/k 2 (ζ) ζ ζ 0 ζ/2 τ (3.6) Marle BGK Marle BGK ( ) D Ds f = pµ µ f = p 0 t + v f = m (f f 0 ) (3.62) τ M f 0 n f 0 = [ 4πm 2 T K 2 (ζ) exp p µu µ T ζ = m T ] (3.63) (3.64) f 0 source term f f 0 f (f 0 f) ψ µ d3 p p 0 = 0 (3.65) ψ µ = (, p µ ) (3.66) (3.65) f 0 f matching condition ( 3.2. f 0 (p µ ) δf = f f 0, δf 0 = f 0 f 0, δτ M = τ M τ M ( τ M f 0 ) (3.67) f 0 ( ) t + v δf = δf δf 0 (3.68) τ M τ M = p0 m τ M (3.69)

91 3.2 Boltzmann Marle 83 τ M τ M τ M δf = δ fe ik µx µ = δ fe iω(t t 0)+ik x (3.70) (3.68) ( ) iω + ik v δf = δf 0 (3.7) τ M τ M u µ δf 0 [ δf 0 = f δn 0 + ( + p0 + K 2 n 0 T 0 f 0 = n 0 4πm 2 T 0 K 2 (ζ 0 ) exp [ p0 ζ 0 K 2 ] T 0 ) δt p δu ] T 0 T 0 (3.72) (3.73) f 0 δu 3.2. Eckart K n K n ζ δn δu δt matching condition δf ( ) iω + ik v δf(p) (3.74) τ M d 3 p [ ) f0 (p) p 0 = p 0 + ( + p0 + K 2 A(ζ) p 0 ζ 0 p ] p δf(p ) τ M n 0 T 0 K 2 T 0 T 0 δt A(ζ) ω τ ω, τk k, p 0 T 0 z (3.75) d 3 p δf(p) = p 0 K(p, p )δf(p ) (3.76) [ ( ) f K(p, p 0 (p) p 0 ) ( ) + + z + K 2 A(ζ) p 0 ζ 0 p ] p iω + ik p K z n 0 K 2 T 0 T 0 p 0 K 2 ζ (3.77) ( iω + ik p ) K z p 0 K 2 ζ 0 (3.78)

92 84 3 (3.78) ω Fredholm 2 K(p, p ) p p 0 d 3 p/p 0 ω k E.2 [ K 2 k ζk 2 P (0) πk 2 e c] δn (3.79) [( + 3 K ) (kk 2 ζ K 2 ( πk 2 e c c 2 K + [ ik 2ζ k πk 2 e c i k 2 { ( ζ K2 k P (n) b ζk 2 P (0) ) + ζk 2 (kk ζp ()) )] δt ζ K 2 ) } P (0) iωp () K K { iω( + c) + K 2 ζ K }] k δu = 0 P (n) = dy e ζy y n arctan ζ y 2 b b = ζ ( ) K2 iωy k K (3.80) (3.8) c = K 2ζ K Im(ω) ζ c = iωk 2 ω 2 /k 2 k 2 + K ( K2 kk c F.3 ) 2 + ω2 k 2 if Im(ω) < K 2 K (3.82) if Im(ω) > K 2 K

93 3.2 Boltzmann Marle 85 k p [ ( ζ ik + i K ) 2ζ ωζ P (0) + K k k P () + πk { e c [ ( ) + iζ 3K K2 + i K 2ζ 2 + ωζ2 k + π k e c [ω ( 3P () K ζ [ { ζ 2 + k 2 K 2 ( iω) K 2ζ K k { + π k e c k 2 ω( + c) + i K 2 K ζ }] δn (3.83) ( ζ + ζk 2 3P (0) K ) ζ P (0) + ζp () K 2 K k K 2 ) P () + ζp (2) K 2 { } ( + c) K 2 ζ + c 2 + c + + i K ( )]] 2 ζ c + K 2 ζ δt K 2 K K 2 ( ) K2 P (0) iωp () + iωζ ( ) } K2 P () iωp (2) ζk K k K ω 2 (c 2 + 2c + 2) + 2iω K ( ) }] 2 2 K2 ζ ζ( + c) k δu = 0 K k p kp y [ [ 2kK dy e ζy bζ y 2 + { b 2 + ζ 2 (y 2 ) } ] arctan ζ y 2 b (3.84) π ) {(c ζ e c 2 + 2c + 2) ( ω2 k 2 2iK ( 2 K 2 k 2 ζω( + c) + 2 ζ 2 )}] K k 2 K 2 ζ 2 δu y = 0 p 0 [ ζp () K k + π ] ζ e c ( + c) δn (3.85) [ ( + ζ 3P () K ) ζp () + ζp (2) + πζ { }] K e c ( + c) K 2 ζ + c 2 + c + δt 2 K 2 [ + iζ { K 2 ζ ( )} K2 P () iωp (2) k k K + π i { ζ e c iω(c 2 + 2c + 2) + K }] 2 ζ( + c) k δu = 0 k K F.3 matching condition Marle BGK matching condition δn δu 3 (3.79) iζ/k k (3.85) K

94 86 3 ωζ/k (3.83) iωδn + ik δu = 0 δρ δu δt ω δρ δu δt δf δf(v) = n C n f0 (v) ( iω + ik p [ δnωn + n 0 p 0 ) K z K 2 ζ ( + p0 T 0 + K 2 K 2 ζ 0 ) δtωn T 0 p δu ω n T 0 ] e i(ω nt+k x) (3.86) C n (3.78) ω p x continuous mode Cercignani [40] 3..6 δf = δf + Dδ D ( ( iω + ik p p 0 ) ) K z K 2 ζ (3.74) δf (3.87) Result ζ = 00 ζ = ζ = 0.0

95 3.2 Boltzmann Marle damping rate n τ e-05 e-06 e wave number k τ c 3.6 ζ = damping rate n τ e wave number k τ c 3.7 ζ = 00 0 damping rate n τ e wave number k τ c 3.8 ζ = 0.0

96 88 3 Chapman-Enskog k 2 k 2 Chapman-Enskog ζ = Marle BGK BGK k ζ = 00 k ζ ζ = 0.0 ζ =

97 3.2 Boltzmann Marle Re (ω τ) wave number k τ c 3.9 ζ = 00 0 damping rate n τ wave number k τ c 3.20 ζ = 0 Re (ω τ) wave number k τ c 3.2 ζ = 0.0

98 damping rate n τ e-05 e-06 e wave number k τ c 3.22 ζ = Re (ω τ) e wave number k τ c 3.23 ζ = 0 damping rate n τ e wave number k τ c 3.24 ζ = 0.0

99 3.2 Boltzmann Marle 9 ω ζ ζ = ζ = ζ = 0.0 k = 0 C BGK shear flow ζ =

100 damping rate n τ e-05 e-06 e wave number k τ c 3.25 ζ = 00 shear flow damping rate n τ e wave number k τ c 3.26 ζ = shear flow damping rate n τ e wave number k τ c 3.27 ζ = 0.0 shear flow

101 3.2 Boltzmann Marle 93 shear flow shear flow ζ ζ = BGK Discussion Marle BGK Eckart Chapman-Enskog k 2 BGK Anderson-Witting Marle BGK ζ BGK ζ = 00 BGK Marle BGK ζ BGK ζ moment shear flow moment k = 0 n 0 Anderson-Witting BGK shear flow Israel-Stewart [4] moment Israel-Stewart BGK k couple n /τ

102 94 3 couple n /τ k n Israel-Stewart couple F (3.70) /τ Cercignani [4] Anderson-Witting BGK Marle BGK n BGK Sirovich [35] BGK

103 3.3 Boltzmann Anderson-Witting Boltzmann Anderson-Witting Anderson-Witting BGK [4] Marle BGK Eckart Landau-Lifshitz 3.3. Anderson-Witting BGK model Anderson Witting [4] BGK ( ) f = p ν f(t, x, p) f 0 (t, x, p) u ν t τ coll (3.88) τ p µ u µ f 0 f 0 (t, x, p) = ζ = m T [ n(t, x) 4πm 2 T (t, x)k 2 (ζ(t, x)) exp p µu µ ] (t, x) T (t, x) (3.89) (3.90) p µ Maxwell-Jüttner n u µ T x µ u µ u 2 = p µ µ f = p 0 ( t + v ) f = p ν u ν f(t, x, p) f 0 (t, x, p) τ (3.9) p µ µ N µ = u ν τ µ T µρ = u ν τ d 3 p p 0 pν (f f 0 ) (3.92) d 3 p p 0 pν p ρ (f f 0 ) (3.93) 0 n u µ T 0

104 96 3 u ν d 3 p p 0 pν (f f 0 ) = u ν N ν u ν N ν 0 = 0 (3.94) u ν d 3 p p 0 pν p ρ (f f 0 ) = u ν T νρ u ν T νρ 0 = 0 (3.95) matching condition Anderson-Witting BGK (3.94) f 0 n f (3.95) 0 f 0 e f BGK matching condition (3.95) matching condition (3.95) u µ T 0µ f 0 T 0µ 0 T 0µ 0 matching condition u µ Landau-Lifshitz Anderson-Witting [4] BGK ( ) D Ds f = pµ µ f = p 0 t + v f = u ν p ν f f 0 τ (3.96) f 0 n f 0 = [ 4πm 2 T K 2 (ζ) exp p µu µ T ζ = m T ] (3.97) (3.98) K 2 BGK Landau-Lifshitz matching condition f 0 n e f ( 3.3. ) f 0 source term f f 0 f u ν d 3 p p 0 pν ψ µ (f 0 f) = 0 (3.99)

105 3.3 Boltzmann Anderson-Witting 97 ψ µ = (, p µ ) (3.00) (3.99) f 0 f matching condition f 0 (p µ ) δf = f f 0, δf 0 = f 0 f 0, δτ = τ τ ( τ f 0 ) (3.0) f 0 ( ) t + v δf = δf δf 0 τ (3.02) τ τ τ δf = δ fe ik µx µ = δ fe iω(t t 0)+ik x (3.03) (3.02) ( ) iω + ik v δf = τ τ δf 0 (3.04) u µ δf 0 [ δf 0 = f δn 0 + ( + p0 + K 2 n 0 T 0 f 0 = n 0 4πm 2 T 0 K 2 (ζ 0 ) exp [ p0 ζ 0 K 2 ] T 0 ) δt p δu ] T 0 T 0 (3.05) (3.06) f 0 δu 3.3. Landau-Lifshitz K n K n ζ δn δu δt matching condition δf ( ) iω + ik v δf(p) (3.07) τ = d 3 p f [ ) 0 (p) + ( + p0 + K 2 A(ζ) p 0 ζ 0 p ] p δf(p ) τ n 0 T 0 K 2 T 0 T 0

106 98 3 δt A(ζ) ω τ ω, τk k, p 0 T 0 z (3.08) δf(p) = d 3 p K(p, p )δf(p ) (3.09) [ ( ) f K(p, p 0 (p) ) iω + ik p + + z + K 2 A(ζ) p 0 ζ 0 p ] p p n 0 0 K 2 T 0 T 0 (3.0) iω + ik p p 0 0 (3.) (3.) ω Fredholm 2 K(p, p ) p 0 p 0 p (p 0 ) 2 d 3 p/p 0 ω k E.3 p 0 [( (ζq(2) K 2 k) δn + 3 K ) ] ζ ζq(2) + ζ 2 Q(3) δt (3.2) K 2 i ( 3K2 + ζk bζ 2 Q(3) ) k δu = 0 k Q(n) b Q(n) = b = iω k dy e ζy y n arctan y2 b y (3.3) (3.4) p 0 k p ( 3K2 + ζk bζ 2 Q(3) ) δn (3.5) [ + K ζ i [ ζk 3 b k ( 3 + K ζ K 2 ) + (3 + ζ 2 )K 2 bζ 2 ( 3Q(3) K { (2 + ζ 2 )K 2 + 3ζK bζ 3 Q(4) }] k δu = 0 )] ζq(3) + ζq(4) K 2 δt

107 3.3 Boltzmann Anderson-Witting 99 p 0 k p kp 0 p y [ 2kζK3 ζ 3 { (b 2 + )Q(4) Q(2) } + b { (2 + ζ 2 )K 2 + 3ζK 2 }] δuy = 0 (3.6) (p 0 ) 2 [ k(3k2 + ζk ) ζ 2 Q(3) ] δn i [ bζ 3 Q(4) (ζ 2 ] + 2)K 2 3ζK k δu (3.7) [ ( k + k {(3 + ζ 2 )K 2 ζk 3 + K )} ( ζ ζ 2 3Q(3) K )] Q(3)ζ + ζq(4) δt = 0 K 2 K 2 matching condition Anderson- Witting BGK Landau-Lifshitz iωδn + ik δu = 0 (3.95) µ T µν = δ(ne) + (neδu) (3.8) t = iω n [ ( 0T 0 (3K 2 + ζk )δn + {(3 + ζ 2 )K 2 ζk 3 + K )} ] ζ δt K 2 K 2 + i ζk 3 K 2 n 0 T 0 k δu =0 matching condition δu matching condition (3.5) δt matching condition (3.7) (3.5) (3.7) b (3.8) δρ δu δt ω δρ δu δt δf δf(v) = n C n C n f0 (v) iω + ik p (3.9) p [ 0 ) δnωn + ( + p0 + K 2 δtωn ζ 0 p δu ] ω n e i(ω nt+k x) n 0 T 0 K 2 T 0 T 0 (3.) ω p x continuous mode Cercignani [40]

108 ( δf = δf + Dδ D iω + ik p ) p 0 (3.20) (3.07) δf Result Marle BGK ζ = 00 ζ = ζ = 0.0

109 3.3 Boltzmann Anderson-Witting damping rate n τ e-05 e-06 e wave number k τ c 3.28 ζ = 00 0 damping rate n τ e wave number k τ c 3.29 ζ = 0 damping rate n τ e wave number k τ c 3.30 ζ = 0.0

110 02 3 Chapman-Enskog k 2 k 2 Chapman-Enskog ζ = 0.0 ζ = Anderson-Witting BGK BGK k ζ = 00 k ζ ζ =

111 3.3 Boltzmann Anderson-Witting Re (ω τ) wave number k τ c 3.3 ζ = Re (ω τ) wave number k τ c 3.32 ζ = 00 0 Re (ω τ) wave number k τ c 3.33 ζ = 0.0

112 damping rate n τ e-05 e-06 e wave number k τ c 3.34 ζ = 00 0 damping rate n τ e wave number k τ c 3.35 ζ = 0 damping rate n τ e wave number k τ c 3.36 ζ = 0.0

113 3.3 Boltzmann Anderson-Witting 05 ω ζ ζ = ζ = ζ = 0.0 k = 0 C BGK shear flow ζ =

114 damping rate n τ e-05 e-06 e wave number k τ c 3.37 ζ = 00 shear flow damping rate n τ e wave number k τ c 3.38 ζ = shear flow damping rate n τ e wave number k τ c 3.39 ζ = 0.0 shear flow

115 3.3 Boltzmann Anderson-Witting 07 shear flow shear flow ζ Marle BGK ζ = BGK shear flow 00 damping rate n τ wave number k τ c 3.40 ζ = 00 shear flow 00 0 Re (ω τ) wave number k τ c 3.4 ζ = 0.0 shear flow moment Discussion Anderson-Witting BGK Landau-Lifshitz 3.3.2

116 Chapman-Enskog k 2 BGK Marle Anderson-Witting BGK ζ BGK ζ = 00 BGK Anderson-Witting BGK ζ BGK ζ moment shear flow moment k = 0 n 0 shear flow Israel-Stewart [4] moment Israel-Stewart Landau-Lifshitz moment couple n /τ Marle BGK couple n /τ k n Israel-Stewart Landau-Lifshitz couple

117 3.3 Boltzmann Anderson-Witting 09 Marle BGK Marle BGK Anderson-Witting BGK Landau-Lifshitz Marle BGK Eckart ζ = BGK C BGK

118 Israel-Stewart Discussion Israel-Stewart [4] Israel-Setewart review moment (2.339) (2.340) (2.34) (2.342) (2.339) (2.34) (2.342) Π = µ( ν u ν + β 0 Π α0 ν q ν ) (3.2) ( ) q µ = κγ µν T νt + u ν + β q ν α 0 ν Π α ρ p ρ ν (3.22) p µν = 2η( µ u ν + β 2 ṗ µν α µ q ν ) (3.23) α 0 α β 0 β β Israel-Stewart [5] ( ) n > 0 (3.24) Θ T ( ) ne > 0 (3.25) T Θ β 0, β, β 2 > 0 (3.26) β (ne + P ) > 0 (3.27) ( ) ne > 0 (3.28) T Θ = (ρ + P nst )/T n /T Isreal-Stewart moment l L l/l l/l Israel-Stewart moment (2.339) (2.34) (2.342) Chapman-Enskog n

119 3.4 Israel-Stewart B moment Israel-Stewart 4 4 Longitudinal-mode [] 0 damping rate n τ wave number k τ C s 3.42 ζ = 00 Israel-Stewart damping rate n τ wave number k τ C s 3.43 ζ = 0.0 Israel-Stewart

120 2 3 BGK Israel-Stewart ζ cτk = Chapman-Enskog c C s C s τk = k 2 moment moment moment [36] Israel-Stewart Israel-Stewart Chapman-Enskog Israel-Stewart

121 3 4 BGK Cauchy BGK BGK Marle BGK Anderson-Witting BGK Cauchy Israel-Stewart Israel-Stewart Isreal-Stewart [5]

122 4 4

123 5 A P. Résibois [6] f 0 f = f 0 + δf, δf f 0 ( ) t + v δf = ncδf (A.) (A.2) C collision operator Cδf d 3 v dωσ(χ; v rel )v rel [δf ϕ eq + ϕ eq δf δfϕ eq ϕeq δf ] (A.3) ϕ eq = f 0 N ϕ eq ϕ eq = ϕ eq ϕ eq Cδf = d 3 v [ ( ) ( ) δf δf ϕ eq + ϕ eq dωσ(χ; v rel )v rel ϕ eq ϕ eq (A.4) ( ) ( ) ] δf δf ϕ eq ϕ eq (A.5) collision operator 2..4 k h d 3 vϕ eq (v) k(v) Ch(v) = d 3 vd 3 v dωσ(χ; v rel )v rel ϕ eq ϕ eq (A.6) 4 [ ( ) ( ) ( ) ( ) ] k k k k + ϕ eq [ ( h ϕ eq ) + ϕ eq ) ( h ϕ eq ϕ eq ( h ϕ eq ) ϕ eq ( h ϕ eq ) ]

124 6 A collision operator Hilbert k h = d 3 vϕ eq (v) k(v) h(v) (A.7) collision operator C (A.6) k C h = h C k (A.8) λ 0 j nc φ 0 j = λ 0 j φ 0 j (A.9) (A.6) h C h 0 (A.0) λ 0 j = φ0 j nc φ0 j φ 0 j φ0 j 0 (A.) (A.6) φ 0 j /ϕeq collision operator C 0 φ 0 j Hilbert φ 0 j(v) φ 0 j(v ) = δ(v v )ϕ eq (v) (A.2) j φ 0 j φ 0 j = (A.3) j t δf = ncδf (A.4) δf δf(t, v) = c 0 jφ 0 j(v)e λ0 j t j c 0 j = φ 0 j δf(t = 0) = d 3 vϕ eq (v) φ 0 j(v) δf(0, v) (A.5) (A.6)

125 7 t λ 0 j lim δf(t, v) = t α= c 0 αφ 0 α(v) (A.7) δρ δu δt collision operator C φ 0 j(v) φ 0 rlm(v) = φ rl (v)y lm (θ v, Φ v ) (A.8) Y lm U r 4 Maxwell σ(χ, v rel )v rel F (χ) collision operator C Cδf = d 3 v dωf (χ)ϕ eq ϕ eq [ ( ) ( ) ( ) ( ) ] δf δf δf δf ϕ eq + ϕ eq ϕ eq ϕ eq (A.9) (A.20) v rel {v, v } {v, v } δf = ϕ eq h n (v) (A.2) Wang Chang Uhlenbeck [39] ( ) ( ) φ 0 rlm = A rl ϕ eq (v)s (r) v 2 l v l+/2 Y lm (θ v, Φ v ) 2RT 2RT (A.22) S r l Laguerre π [ ( λ 0 rl = 2πn dχ sin χf (χ) cos 2r+ χ ) Pl 0 (cos χ ) ( + sin 2r+l χ ) ( χ ) ] Pl 0 δ r,0 δ l,0 2 2 (A.23)

126 8 A δ r,l L. Sirovich [36] ( ) t + ξ x g = L(g) = x m [g] = g + g g g f 0 = ρ 0 exp (2πRT 0 ) 3/2 ( ξ2 ) = 2RT 0 f 0 [g] B(θ) dɛ dθ dξ ρ 0 (RT 0 ) 3/2 ω (A.24) (A.25) (A.26) ξ ξ g f = f 0 ( + g) (A.27) f ν = /τ νt = t, xν (RT 0 ) /2 = x, ξ (RT 0 ) /2 = ξ, ρ 0 Bν m = B (A.28) ( ) t + ξ x g = L(g) = x ω [g] B(θ) dɛ dθ dξ (A.29) L shear ψ rl = Sr l+/2 (ξ2 /2)ξ l P l (cos θ) 2 l+ Γ(r+l+3/2) π /2 r!(2l+) (A.30) Sl r P l Laguerre Legendre ωψ rl ψ r l dξ = δ rr δ ll (A.3) ψ } 2(r + l + 3/2) ξ x ψ rl = (l + ) {ψ r,l+ (2l + )(2l ) ψ 2r r,l+ (2l + 3)(2l + ) } 2(r + l + /2) + l {ψ r,l (2l + )(2l ) ψ 2(r + ) r+,l (2l + )(2l ) (A.32)

127 9 L { λ rl = 2π dθb(θ) cos 2r+l θ ( 2 P l cos θ ) + sin 2r+l θ 2 2 P l ( sin θ ) } ( + δ r0 δ l0 ) 2 (A.33) λ r0 = λ r, (A.34) λ rl [36] [39] λ 00 = λ 0 = λ 0 = 0 (A.35) λ [36] ψ rl g = b rl ψ rl, b rl = ω g ψ rl dξ (A.36) r,l b µν 3 b 00 = ρ, b 0 = u, b 0 = 2 T, b 02 = p, b = 5 S (A.37) ( ) t λ µν b µν + x ( 2(µ + ν + 3/2) ν {b µ,ν (2ν + )(2ν ) b µ+,ν 2µ + 2 (2ν )(2ν + ) }) 2(µ + ν + 3/2) +(ν + ) {b µ,ν+ (2ν + 3)(2ν + ) b 2µ µ,ν+ = 0 (2ν + 3)(2ν + ) (A.38) 3 (A.38) } ρ t + u x = 0 u t + p x + x (ρ + T ) = 0 T t + 2 S 3 x + 2 u 3 x = 0 (A.39) (A.40) (A.4)

128 20 A b rl = ˆb rl e σt ikx (A.42) ˆbrl ( } 2(µ + ν + 3/2) (σ λ µν )ˆb µν ik ν {ˆbµ,ν (2ν + )(2ν ) ˆb 2µ + 2 µ+,ν (2ν )(2ν + ) (A.43) +(ν + ) {ˆbµ,ν+ 2(µ + ν + 3/2) (2ν + 3)(2ν + ) ˆb µ,ν+ 2µ (2ν + 3)(2ν + ) 0 D(σ, k) = [36] k = 0 (A.43) }) = 0 σ = λ rl (A.44) λ rl σ k = 0 k = 0 λ rl k k = 0 σ = 0 3 [36] σ = k2 + O(k 4 ) λ 5 σ = ±i 3 [k + O(k3 )] + k 2 (A.45) ( + 2 ) + O(k 4 ) (A.46) 3λ 3λ 02 Maxwell Maxwell n > 2 U r n Pao [7] BGK g (A.36) ψ rl L ( ) t + ξ x f 0 g = L(g) = f 0 λ rl b rl ψ rl x r,l (A.47)

129 2 λ 00 = λ 0 = λ 0 = 0 λ rl λ ( ) t + ξ x f 0 g = f 0 λ b rl ψ rl (A.48) x r,l )} 32 = λ f 0 {g (ψ 00 ρ + ψ 0 u + ψ 0 T (A.49) λ = /τ BGK (3.2) (A.46) λ 02 BGK λ 02 λ BGK

130

131 23 B moment moment moment moment [0] A Boltzmann ˆLb = [ iωi + iak Λ] b = 0 (B.) I b A Λ b j = φ 0 j δf, A ij = φ 0 i v φ 0 j, Λ ij = φ 0 i C φ 0 j (B.2) C collision operator A Maxwell (A.43) [36] (B.) Λ source term ˆP ˆP = iωi + iak (B.3) Q = det ˆP = det( iωi + iak) = 0 (B.4) (B.4) k ω [20] A weakly hyperbolic

132 24 B moment moment (B.) b (b, b) iω(b, b) + ik(b, Ab) (b, Λb) = 0 (B.5) A (b, Λb) A (b, Ab) ω = ω R in (B.5) n(b, b) (b, Λb) = 0 iω R (b, b) + ik(b, Ab) = 0 (B.6) (B.7) (B.6) n n = (b, Λb) (b, b) (B.8) (b, Λb) moment λ (B.8) n 0 n λ (B.9) f moment moment n k λ collisionless moment moment (B.9) moment λ moment moment collisionless moment consistent

133 25 C Cercignani [42] p µ µ f = Q(f, f) (C.) Q p µ f f = f 0 ( + h) (C.2) f 0 Maxwell-Jüttner h p µ µ h = ˆLh (C.3) ˆL collision operator H = L 2 (f 0 dω) (g, h) = ḡhf 0 dω (C.4) (h, ˆLh) [43] h = g(p µ ) exp( ik ν x ν ) (C.5) k µ = (ω, k)

134 26 C g (C.3) ik µ p µ g = ˆLg (C.6) Cauchy k ω = k(a + ib) a (C.6) ikp 0 (a + ib)g + ikp x g = ˆLg (C.7) g H (C.7) g ik(a + ib)(g, p 0 g) + ik(g, p x g) = (g, ˆLg) (C.8) (C.8) a(g, p 0 g) + (g, p x g) = 0 (C.9) a = (g, px g) (g, p 0 g) < (C.0) p x < p 0 g H [42]

TOP URL 1

TOP URL   1 TOP URL http://amonphys.web.fc.com/ 1 19 3 19.1................... 3 19.............................. 4 19.3............................... 6 19.4.............................. 8 19.5.............................

More information

1 (Berry,1975) 2-6 p (S πr 2 )p πr 2 p 2πRγ p p = 2γ R (2.5).1-1 : : : : ( ).2 α, β α, β () X S = X X α X β (.1) 1 2

1 (Berry,1975) 2-6 p (S πr 2 )p πr 2 p 2πRγ p p = 2γ R (2.5).1-1 : : : : ( ).2 α, β α, β () X S = X X α X β (.1) 1 2 2005 9/8-11 2 2.2 ( 2-5) γ ( ) γ cos θ 2πr πρhr 2 g h = 2γ cos θ ρgr (2.1) γ = ρgrh (2.2) 2 cos θ θ cos θ = 1 (2.2) γ = 1 ρgrh (2.) 2 2. p p ρgh p ( ) p p = p ρgh (2.) h p p = 2γ r 1 1 (Berry,1975) 2-6

More information

.2 ρ dv dt = ρk grad p + 3 η grad (divv) + η 2 v.3 divh = 0, rote + c H t = 0 dive = ρ, H = 0, E = ρ, roth c E t = c ρv E + H c t = 0 H c E t = c ρv T

.2 ρ dv dt = ρk grad p + 3 η grad (divv) + η 2 v.3 divh = 0, rote + c H t = 0 dive = ρ, H = 0, E = ρ, roth c E t = c ρv E + H c t = 0 H c E t = c ρv T NHK 204 2 0 203 2 24 ( ) 7 00 7 50 203 2 25 ( ) 7 00 7 50 203 2 26 ( ) 7 00 7 50 203 2 27 ( ) 7 00 7 50 I. ( ν R n 2 ) m 2 n m, R = e 2 8πε 0 hca B =.09737 0 7 m ( ν = ) λ a B = 4πε 0ħ 2 m e e 2 = 5.2977

More information

TOP URL 1

TOP URL   1 TOP URL http://amonphys.web.fc2.com/ 1 30 3 30.1.............. 3 30.2........................... 4 30.3...................... 5 30.4........................ 6 30.5.................................. 8 30.6...............................

More information

. ev=,604k m 3 Debye ɛ 0 kt e λ D = n e n e Ze 4 ln Λ ν ei = 5.6π / ɛ 0 m/ e kt e /3 ν ei v e H + +e H ev Saha x x = 3/ πme kt g i g e n

. ev=,604k m 3 Debye ɛ 0 kt e λ D = n e n e Ze 4 ln Λ ν ei = 5.6π / ɛ 0 m/ e kt e /3 ν ei v e H + +e H ev Saha x x = 3/ πme kt g i g e n 003...............................3 Debye................. 3.4................ 3 3 3 3. Larmor Cyclotron... 3 3................ 4 3.3.......... 4 3.3............ 4 3.3...... 4 3.3.3............ 5 3.4.........

More information

30

30 3 ............................................2 2...........................................2....................................2.2...................................2.3..............................

More information

July 28, H H 0 H int = H H 0 H int = H int (x)d 3 x Schrödinger Picture Ψ(t) S =e iht Ψ H O S Heisenberg Picture Ψ H O H (t) =e iht O S e i

July 28, H H 0 H int = H H 0 H int = H int (x)d 3 x Schrödinger Picture Ψ(t) S =e iht Ψ H O S Heisenberg Picture Ψ H O H (t) =e iht O S e i July 8, 4. H H H int H H H int H int (x)d 3 x Schrödinger Picture Ψ(t) S e iht Ψ H O S Heisenberg Picture Ψ H O H (t) e iht O S e iht Interaction Picture Ψ(t) D e iht Ψ(t) S O D (t) e iht O S e ih t (Dirac

More information

tomocci ,. :,,,, Lie,,,, Einstein, Newton. 1 M n C. s, M p. M f, p d ds f = dxµ p ds µ f p, X p = X µ µ p = dxµ ds µ p. µ, X µ.,. p,. T M p.

tomocci ,. :,,,, Lie,,,, Einstein, Newton. 1 M n C. s, M p. M f, p d ds f = dxµ p ds µ f p, X p = X µ µ p = dxµ ds µ p. µ, X µ.,. p,. T M p. tomocci 18 7 5...,. :,,,, Lie,,,, Einstein, Newton. 1 M n C. s, M p. M f, p d ds f = dxµ p ds µ f p, X p = X µ µ p = dxµ ds µ p. µ, X µ.,. p,. T M p. M F (M), X(F (M)).. T M p e i = e µ i µ. a a = a i

More information

/ Christopher Essex Radiation and the Violation of Bilinearity in the Thermodynamics of Irreversible Processes, Planet.Space Sci.32 (1984) 1035 Radiat

/ Christopher Essex Radiation and the Violation of Bilinearity in the Thermodynamics of Irreversible Processes, Planet.Space Sci.32 (1984) 1035 Radiat / Christopher Essex Radiation and the Violation of Bilinearity in the Thermodynamics of Irreversible Processes, Planet.Space Sci.32 (1984) 1035 Radiation and the Continuing Failure of the Bilinear Formalism,

More information

Part () () Γ Part ,

Part () () Γ Part , Contents a 6 6 6 6 6 6 6 7 7. 8.. 8.. 8.3. 8 Part. 9. 9.. 9.. 3. 3.. 3.. 3 4. 5 4.. 5 4.. 9 4.3. 3 Part. 6 5. () 6 5.. () 7 5.. 9 5.3. Γ 3 6. 3 6.. 3 6.. 3 6.3. 33 Part 3. 34 7. 34 7.. 34 7.. 34 8. 35

More information

25 7 18 1 1 1.1 v.s............................. 1 1.1.1.................................. 1 1.1.2................................. 1 1.1.3.................................. 3 1.2................... 3

More information

Dirac 38 5 Dirac 4 4 γ µ p µ p µ + m 2 = ( p µ γ µ + m)(p ν γ ν + m) (5.1) γ = p µ p ν γ µ γ ν p µ γ µ m + mp ν γ ν + m 2 = 1 2 p µp ν {γ µ, γ ν } + m

Dirac 38 5 Dirac 4 4 γ µ p µ p µ + m 2 = ( p µ γ µ + m)(p ν γ ν + m) (5.1) γ = p µ p ν γ µ γ ν p µ γ µ m + mp ν γ ν + m 2 = 1 2 p µp ν {γ µ, γ ν } + m Dirac 38 5 Dirac 4 4 γ µ p µ p µ + m 2 p µ γ µ + mp ν γ ν + m 5.1 γ p µ p ν γ µ γ ν p µ γ µ m + mp ν γ ν + m 2 1 2 p µp ν {γ µ, γ ν } + m 2 5.2 p m p p µ γ µ {, } 10 γ {γ µ, γ ν } 2η µν 5.3 p µ γ µ + mp

More information

18 I ( ) (1) I-1,I-2,I-3 (2) (3) I-1 ( ) (100 ) θ ϕ θ ϕ m m l l θ ϕ θ ϕ 2 g (1) (2) 0 (3) θ ϕ (4) (3) θ(t) = A 1 cos(ω 1 t + α 1 ) + A 2 cos(ω 2 t + α

18 I ( ) (1) I-1,I-2,I-3 (2) (3) I-1 ( ) (100 ) θ ϕ θ ϕ m m l l θ ϕ θ ϕ 2 g (1) (2) 0 (3) θ ϕ (4) (3) θ(t) = A 1 cos(ω 1 t + α 1 ) + A 2 cos(ω 2 t + α 18 I ( ) (1) I-1,I-2,I-3 (2) (3) I-1 ( ) (100 ) θ ϕ θ ϕ m m l l θ ϕ θ ϕ 2 g (1) (2) 0 (3) θ ϕ (4) (3) θ(t) = A 1 cos(ω 1 t + α 1 ) + A 2 cos(ω 2 t + α 2 ), ϕ(t) = B 1 cos(ω 1 t + α 1 ) + B 2 cos(ω 2 t

More information

v v = v 1 v 2 v 3 (1) R = (R ij ) (2) R (R 1 ) ij = R ji (3) 3 R ij R ik = δ jk (4) i=1 δ ij Kronecker δ ij = { 1 (i = j) 0 (i

v v = v 1 v 2 v 3 (1) R = (R ij ) (2) R (R 1 ) ij = R ji (3) 3 R ij R ik = δ jk (4) i=1 δ ij Kronecker δ ij = { 1 (i = j) 0 (i 1. 1 1.1 1.1.1 1.1.1.1 v v = v 1 v 2 v 3 (1) R = (R ij ) (2) R (R 1 ) ij = R ji (3) R ij R ik = δ jk (4) δ ij Kronecker δ ij = { 1 (i = j) 0 (i j) (5) 1 1.1. v1.1 2011/04/10 1. 1 2 v i = R ij v j (6) [

More information

: 2005 ( ρ t +dv j =0 r m m r = e E( r +e r B( r T 208 T = d E j 207 ρ t = = = e t δ( r r (t e r r δ( r r (t e r ( r δ( r r (t dv j =

: 2005 ( ρ t +dv j =0 r m m r = e E( r +e r B( r T 208 T = d E j 207 ρ t = = = e t δ( r r (t e r r δ( r r (t e r ( r δ( r r (t dv j = 72 Maxwell. Maxwell e r ( =,,N Maxwell rot E + B t = 0 rot H D t = j dv D = ρ dv B = 0 D = ɛ 0 E H = μ 0 B ρ( r = j( r = N e δ( r r = N e r δ( r r = : 2005 ( 2006.8.22 73 207 ρ t +dv j =0 r m m r = e E(

More information

TOP URL 1

TOP URL   1 TOP URL http://amonphys.web.fc2.com/ 1 6 3 6.1................................ 3 6.2.............................. 4 6.3................................ 5 6.4.......................... 6 6.5......................

More information

構造と連続体の力学基礎

構造と連続体の力学基礎 II 37 Wabash Avenue Bridge, Illinois 州 Winnipeg にある歩道橋 Esplanade Riel 橋6 6 斜張橋である必要は多分無いと思われる すぐ横に道路用桁橋有り しかも塔基部のレストランは 8 年には営業していなかった 9 9. 9.. () 97 [3] [5] k 9. m w(t) f (t) = f (t) + mg k w(t) Newton

More information

1 1.1 / Fik Γ= D n x / Newton Γ= µ vx y / Fouie Q = κ T x 1. fx, tdx t x x + dx f t = D f x 1 fx, t = 1 exp x 4πDt 4Dt lim fx, t =δx 3 t + dxfx, t = 1

1 1.1 / Fik Γ= D n x / Newton Γ= µ vx y / Fouie Q = κ T x 1. fx, tdx t x x + dx f t = D f x 1 fx, t = 1 exp x 4πDt 4Dt lim fx, t =δx 3 t + dxfx, t = 1 1 1.1......... 1............. 1.3... 1.4......... 1.5.............. 1.6................ Bownian Motion.1.......... Einstein.............. 3.3 Einstein........ 3.4..... 3.5 Langevin Eq.... 3.6................

More information

I ( ) 1 de Broglie 1 (de Broglie) p λ k h Planck ( Js) p = h λ = k (1) h 2π : Dirac k B Boltzmann ( J/K) T U = 3 2 k BT

I ( ) 1 de Broglie 1 (de Broglie) p λ k h Planck ( Js) p = h λ = k (1) h 2π : Dirac k B Boltzmann ( J/K) T U = 3 2 k BT I (008 4 0 de Broglie (de Broglie p λ k h Planck ( 6.63 0 34 Js p = h λ = k ( h π : Dirac k B Boltzmann (.38 0 3 J/K T U = 3 k BT ( = λ m k B T h m = 0.067m 0 m 0 = 9. 0 3 kg GaAs( a T = 300 K 3 fg 07345

More information

II No.01 [n/2] [1]H n (x) H n (x) = ( 1) r n! r!(n 2r)! (2x)n 2r. r=0 [2]H n (x) n,, H n ( x) = ( 1) n H n (x). [3] H n (x) = ( 1) n dn x2 e dx n e x2

II No.01 [n/2] [1]H n (x) H n (x) = ( 1) r n! r!(n 2r)! (2x)n 2r. r=0 [2]H n (x) n,, H n ( x) = ( 1) n H n (x). [3] H n (x) = ( 1) n dn x2 e dx n e x2 II No.1 [n/] [1]H n x) H n x) = 1) r n! r!n r)! x)n r r= []H n x) n,, H n x) = 1) n H n x) [3] H n x) = 1) n dn x e dx n e x [4] H n+1 x) = xh n x) nh n 1 x) ) d dx x H n x) = H n+1 x) d dx H nx) = nh

More information

量子力学 問題

量子力学 問題 3 : 203 : 0. H = 0 0 2 6 0 () = 6, 2 = 2, 3 = 3 3 H 6 2 3 ϵ,2,3 (2) ψ = (, 2, 3 ) ψ Hψ H (3) P i = i i P P 2 = P 2 P 3 = P 3 P = O, P 2 i = P i (4) P + P 2 + P 3 = E 3 (5) i ϵ ip i H 0 0 (6) R = 0 0 [H,

More information

No δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i x j δx j (5) δs 2

No δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i x j δx j (5) δs 2 No.2 1 2 2 δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i δx j (5) δs 2 = δx i δx i + 2 u i δx i δx j = δs 2 + 2s ij δx i δx j

More information

Z: Q: R: C: sin 6 5 ζ a, b

Z: Q: R: C: sin 6 5 ζ a, b Z: Q: R: C: 3 3 7 4 sin 6 5 ζ 9 6 6............................... 6............................... 6.3......................... 4 7 6 8 8 9 3 33 a, b a bc c b a a b 5 3 5 3 5 5 3 a a a a p > p p p, 3,

More information

K E N Z U 2012 7 16 HP M. 1 1 4 1.1 3.......................... 4 1.2................................... 4 1.2.1..................................... 4 1.2.2.................................... 5................................

More information

1 1.1 H = µc i c i + c i t ijc j + 1 c i c j V ijklc k c l (1) V ijkl = V jikl = V ijlk = V jilk () t ij = t ji, V ijkl = V lkji (3) (1) V 0 H mf = µc

1 1.1 H = µc i c i + c i t ijc j + 1 c i c j V ijklc k c l (1) V ijkl = V jikl = V ijlk = V jilk () t ij = t ji, V ijkl = V lkji (3) (1) V 0 H mf = µc 013 6 30 BCS 1 1.1........................ 1................................ 3 1.3............................ 3 1.4............................... 5 1.5.................................... 5 6 3 7 4 8

More information

S I. dy fx x fx y fx + C 3 C dy fx 4 x, y dy v C xt y C v e kt k > xt yt gt [ v dt dt v e kt xt v e kt + C k x v + C C k xt v k 3 r r + dr e kt S dt d

S I. dy fx x fx y fx + C 3 C dy fx 4 x, y dy v C xt y C v e kt k > xt yt gt [ v dt dt v e kt xt v e kt + C k x v + C C k xt v k 3 r r + dr e kt S dt d S I.. http://ayapin.film.s.dendai.ac.jp/~matuda /TeX/lecture.html PDF PS.................................... 3.3.................... 9.4................5.............. 3 5. Laplace................. 5....

More information

IA

IA IA 31 4 11 1 1 4 1.1 Planck.............................. 4 1. Bohr.................................... 5 1.3..................................... 6 8.1................................... 8....................................

More information

4. ϵ(ν, T ) = c 4 u(ν, T ) ϵ(ν, T ) T ν π4 Planck dx = 0 e x 1 15 U(T ) x 3 U(T ) = σt 4 Stefan-Boltzmann σ 2π5 k 4 15c 2 h 3 = W m 2 K 4 5.

4. ϵ(ν, T ) = c 4 u(ν, T ) ϵ(ν, T ) T ν π4 Planck dx = 0 e x 1 15 U(T ) x 3 U(T ) = σt 4 Stefan-Boltzmann σ 2π5 k 4 15c 2 h 3 = W m 2 K 4 5. A 1. Boltzmann Planck u(ν, T )dν = 8πh ν 3 c 3 kt 1 dν h 6.63 10 34 J s Planck k 1.38 10 23 J K 1 Boltzmann u(ν, T ) T ν e hν c = 3 10 8 m s 1 2. Planck λ = c/ν Rayleigh-Jeans u(ν, T )dν = 8πν2 kt dν c

More information

n (1.6) i j=1 1 n a ij x j = b i (1.7) (1.7) (1.4) (1.5) (1.4) (1.7) u, v, w ε x, ε y, ε x, γ yz, γ zx, γ xy (1.8) ε x = u x ε y = v y ε z = w z γ yz

n (1.6) i j=1 1 n a ij x j = b i (1.7) (1.7) (1.4) (1.5) (1.4) (1.7) u, v, w ε x, ε y, ε x, γ yz, γ zx, γ xy (1.8) ε x = u x ε y = v y ε z = w z γ yz 1 2 (a 1, a 2, a n ) (b 1, b 2, b n ) A (1.1) A = a 1 b 1 + a 2 b 2 + + a n b n (1.1) n A = a i b i (1.2) i=1 n i 1 n i=1 a i b i n i=1 A = a i b i (1.3) (1.3) (1.3) (1.1) (ummation convention) a 11 x

More information

: , 2.0, 3.0, 2.0, (%) ( 2.

: , 2.0, 3.0, 2.0, (%) ( 2. 2017 1 2 1.1...................................... 2 1.2......................................... 4 1.3........................................... 10 1.4................................. 14 1.5..........................................

More information

( ) ( 40 )+( 60 ) Schrödinger 3. (a) (b) (c) yoshioka/education-09.html pdf 1

( ) ( 40 )+( 60 ) Schrödinger 3. (a) (b) (c)   yoshioka/education-09.html pdf 1 2009 1 ( ) ( 40 )+( 60 ) 1 1. 2. Schrödinger 3. (a) (b) (c) http://goofy.phys.nara-wu.ac.jp/ yoshioka/education-09.html pdf 1 1. ( photon) ν λ = c ν (c = 3.0 108 /m : ) ɛ = hν (1) p = hν/c = h/λ (2) h

More information

TOP URL 1

TOP URL   1 TOP URL http://amonphys.web.fc.com/ 3.............................. 3.............................. 4.3 4................... 5.4........................ 6.5........................ 8.6...........................7

More information

20 4 20 i 1 1 1.1............................ 1 1.2............................ 4 2 11 2.1................... 11 2.2......................... 11 2.3....................... 19 3 25 3.1.............................

More information

meiji_resume_1.PDF

meiji_resume_1.PDF β β β (q 1,q,..., q n ; p 1, p,..., p n ) H(q 1,q,..., q n ; p 1, p,..., p n ) Hψ = εψ ε k = k +1/ ε k = k(k 1) (x, y, z; p x, p y, p z ) (r; p r ), (θ; p θ ), (ϕ; p ϕ ) ε k = 1/ k p i dq i E total = E

More information

II ( ) (7/31) II ( [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Re

II ( ) (7/31) II (  [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Re II 29 7 29-7-27 ( ) (7/31) II (http://www.damp.tottori-u.ac.jp/~ooshida/edu/fluid/) [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Reynolds [ (4.6), (45.8)] [ p.186] Navier Stokes I Euler Navier

More information

24 I ( ) 1. R 3 (i) C : x 2 + y 2 1 = 0 (ii) C : y = ± 1 x 2 ( 1 x 1) (iii) C : x = cos t, y = sin t (0 t 2π) 1.1. γ : [a, b] R n ; t γ(t) = (x

24 I ( ) 1. R 3 (i) C : x 2 + y 2 1 = 0 (ii) C : y = ± 1 x 2 ( 1 x 1) (iii) C : x = cos t, y = sin t (0 t 2π) 1.1. γ : [a, b] R n ; t γ(t) = (x 24 I 1.1.. ( ) 1. R 3 (i) C : x 2 + y 2 1 = 0 (ii) C : y = ± 1 x 2 ( 1 x 1) (iii) C : x = cos t, y = sin t (0 t 2π) 1.1. γ : [a, b] R n ; t γ(t) = (x 1 (t), x 2 (t),, x n (t)) ( ) ( ), γ : (i) x 1 (t),

More information

n ξ n,i, i = 1,, n S n ξ n,i n 0 R 1,.. σ 1 σ i .10.14.15 0 1 0 1 1 3.14 3.18 3.19 3.14 3.14,. ii 1 1 1.1..................................... 1 1............................... 3 1.3.........................

More information

~nabe/lecture/index.html 2

~nabe/lecture/index.html 2 2001 12 13 1 http://www.sml.k.u-tokyo.ac.jp/ ~nabe/lecture/index.html nabe@sml.k.u-tokyo.ac.jp 2 1. 10/ 4 2. 10/11 3. 10/18 1 4. 10/25 2 5. 11/ 1 6. 11/ 8 7. 11/15 8. 11/22 9. 11/29 10. 12/ 6 1 11. 12/13

More information

微分積分 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. このサンプルページの内容は, 初版 1 刷発行時のものです.

微分積分 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます.   このサンプルページの内容は, 初版 1 刷発行時のものです. 微分積分 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. ttp://www.morikita.co.jp/books/mid/00571 このサンプルページの内容は, 初版 1 刷発行時のものです. i ii 014 10 iii [note] 1 3 iv 4 5 3 6 4 x 0 sin x x 1 5 6 z = f(x, y) 1 y = f(x)

More information

医系の統計入門第 2 版 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. このサンプルページの内容は, 第 2 版 1 刷発行時のものです.

医系の統計入門第 2 版 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます.   このサンプルページの内容は, 第 2 版 1 刷発行時のものです. 医系の統計入門第 2 版 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. http://www.morikita.co.jp/books/mid/009192 このサンプルページの内容は, 第 2 版 1 刷発行時のものです. i 2 t 1. 2. 3 2 3. 6 4. 7 5. n 2 ν 6. 2 7. 2003 ii 2 2013 10 iii 1987

More information

OHP.dvi

OHP.dvi 7 2010 11 22 1 7 http://www.sml.k.u-tokyo.ac.jp/members/nabe/lecture2010 nabe@sml.k.u-tokyo.ac.jp 2 1. 10/ 4 2. 10/18 3. 10/25 2, 3 4. 11/ 1 5. 11/ 8 6. 11/15 7. 11/22 8. 11/29 9. 12/ 6 skyline 10. 12/13

More information

untitled

untitled 0. =. =. (999). 3(983). (980). (985). (966). 3. := :=. A A. A A. := := 4 5 A B A B A B. A = B A B A B B A. A B A B, A B, B. AP { A, P } = { : A, P } = { A P }. A = {0, }, A, {0, }, {0}, {}, A {0}, {}.

More information

Untitled

Untitled II 14 14-7-8 8/4 II (http://www.damp.tottori-u.ac.jp/~ooshida/edu/fluid/) [ (3.4)] Navier Stokes [ 6/ ] Navier Stokes 3 [ ] Reynolds [ (4.6), (45.8)] [ p.186] Navier Stokes I 1 balance law t (ρv i )+ j

More information

5 1.2, 2, d a V a = M (1.2.1), M, a,,,,, Ω, V a V, V a = V + Ω r. (1.2.2), r i 1, i 2, i 3, i 1, i 2, i 3, A 2, A = 3 A n i n = n=1 da = 3 = n=1 3 n=1

5 1.2, 2, d a V a = M (1.2.1), M, a,,,,, Ω, V a V, V a = V + Ω r. (1.2.2), r i 1, i 2, i 3, i 1, i 2, i 3, A 2, A = 3 A n i n = n=1 da = 3 = n=1 3 n=1 4 1 1.1 ( ) 5 1.2, 2, d a V a = M (1.2.1), M, a,,,,, Ω, V a V, V a = V + Ω r. (1.2.2), r i 1, i 2, i 3, i 1, i 2, i 3, A 2, A = 3 A n i n = n=1 da = 3 = n=1 3 n=1 da n i n da n i n + 3 A ni n n=1 3 n=1

More information

0406_total.pdf

0406_total.pdf 59 7 7.1 σ-ω σ-ω σ ω σ = σ(r), ω µ = δ µ,0 ω(r) (6-4) (iγ µ µ m U(r) γ 0 V (r))ψ(x) = 0 (7-1) U(r) = g σ σ(r), V (r) = g ω ω(r) σ(r) ω(r) (6-3) ( 2 + m 2 σ)σ(r) = g σ ψψ (7-2) ( 2 + m 2 ω)ω(r) = g ω ψγ

More information

i 18 2H 2 + O 2 2H 2 + ( ) 3K

i 18 2H 2 + O 2 2H 2 + ( ) 3K i 18 2H 2 + O 2 2H 2 + ( ) 3K ii 1 1 1.1.................................. 1 1.2........................................ 3 1.3......................................... 3 1.4....................................

More information

(1.2) T D = 0 T = D = 30 kn 1.2 (1.4) 2F W = 0 F = W/2 = 300 kn/2 = 150 kn 1.3 (1.9) R = W 1 + W 2 = = 1100 N. (1.9) W 2 b W 1 a = 0

(1.2) T D = 0 T = D = 30 kn 1.2 (1.4) 2F W = 0 F = W/2 = 300 kn/2 = 150 kn 1.3 (1.9) R = W 1 + W 2 = = 1100 N. (1.9) W 2 b W 1 a = 0 1 1 1.1 1.) T D = T = D = kn 1. 1.4) F W = F = W/ = kn/ = 15 kn 1. 1.9) R = W 1 + W = 6 + 5 = 11 N. 1.9) W b W 1 a = a = W /W 1 )b = 5/6) = 5 cm 1.4 AB AC P 1, P x, y x, y y x 1.4.) P sin 6 + P 1 sin 45

More information

( )

( ) 7..-8..8.......................................................................... 4.................................... 3...................................... 3..3.................................. 4.3....................................

More information

S I. dy fx x fx y fx + C 3 C vt dy fx 4 x, y dy yt gt + Ct + C dt v e kt xt v e kt + C k x v k + C C xt v k 3 r r + dr e kt S Sr πr dt d v } dt k e kt

S I. dy fx x fx y fx + C 3 C vt dy fx 4 x, y dy yt gt + Ct + C dt v e kt xt v e kt + C k x v k + C C xt v k 3 r r + dr e kt S Sr πr dt d v } dt k e kt S I. x yx y y, y,. F x, y, y, y,, y n http://ayapin.film.s.dendai.ac.jp/~matuda n /TeX/lecture.html PDF PS yx.................................... 3.3.................... 9.4................5..............

More information

N/m f x x L dl U 1 du = T ds pdv + fdl (2.1)

N/m f x x L dl U 1 du = T ds pdv + fdl (2.1) 23 2 2.1 10 5 6 N/m 2 2.1.1 f x x L dl U 1 du = T ds pdv + fdl (2.1) 24 2 dv = 0 dl ( ) U f = T L p,t ( ) S L p,t (2.2) 2 ( ) ( ) S f = L T p,t p,l (2.3) ( ) U f = L p,t + T ( ) f T p,l (2.4) 1 f e ( U/

More information

(Bessel) (Legendre).. (Hankel). (Laplace) V = (x, y, z) n (r, θ, ϕ) r n f n (θ, ϕ). f n (θ, ϕ) n f n (θ, ϕ) z = cos θ z θ ϕ n ν. P ν (z), Q ν (z) (Fou

(Bessel) (Legendre).. (Hankel). (Laplace) V = (x, y, z) n (r, θ, ϕ) r n f n (θ, ϕ). f n (θ, ϕ) n f n (θ, ϕ) z = cos θ z θ ϕ n ν. P ν (z), Q ν (z) (Fou (Bessel) (Legendre).. (Hankel). (Laplace) V = (x, y, z) n (r, θ, ϕ) r n f n (θ, ϕ). f n (θ, ϕ) n f n (θ, ϕ) z = cos θ z θ ϕ n ν. P ν (z), Q ν (z) (Fourier) (Fourier Bessel).. V ρ(x, y, z) V = 4πGρ G :.

More information

b3e2003.dvi

b3e2003.dvi 15 II 5 5.1 (1) p, q p = (x + 2y, xy, 1), q = (x 2 + 3y 2, xyz, ) (i) p rotq (ii) p gradq D (2) a, b rot(a b) div [11, p.75] (3) (i) f f grad f = 1 2 grad( f 2) (ii) f f gradf 1 2 grad ( f 2) rotf 5.2

More information

all.dvi

all.dvi 29 4 Green-Lagrange,,.,,,,,,.,,,,,,,,,, E, σ, ε σ = Eε,,.. 4.1? l, l 1 (l 1 l) ε ε = l 1 l l (4.1) F l l 1 F 30 4 Green-Lagrange Δz Δδ γ = Δδ (4.2) Δz π/2 φ γ = π 2 φ (4.3) γ tan γ γ,sin γ γ ( π ) γ tan

More information

D v D F v/d F v D F η v D (3.2) (a) F=0 (b) v=const. D F v Newtonian fluid σ ė σ = ηė (2.2) ė kl σ ij = D ijkl ė kl D ijkl (2.14) ė ij (3.3) µ η visco

D v D F v/d F v D F η v D (3.2) (a) F=0 (b) v=const. D F v Newtonian fluid σ ė σ = ηė (2.2) ė kl σ ij = D ijkl ė kl D ijkl (2.14) ė ij (3.3) µ η visco post glacial rebound 3.1 Viscosity and Newtonian fluid f i = kx i σ ij e kl ideal fluid (1.9) irreversible process e ij u k strain rate tensor (3.1) v i u i / t e ij v F 23 D v D F v/d F v D F η v D (3.2)

More information

( ) ,

( ) , II 2007 4 0. 0 1 0 2 ( ) 0 3 1 2 3 4, - 5 6 7 1 1 1 1 1) 2) 3) 4) ( ) () H 2.79 10 10 He 2.72 10 9 C 1.01 10 7 N 3.13 10 6 O 2.38 10 7 Ne 3.44 10 6 Mg 1.076 10 6 Si 1 10 6 S 5.15 10 5 Ar 1.01 10 5 Fe 9.00

More information

Einstein 1905 Lorentz Maxwell c E p E 2 (pc) 2 = m 2 c 4 (7.1) m E ( ) E p µ =(p 0,p 1,p 2,p 3 )=(p 0, p )= c, p (7.2) x µ =(x 0,x 1,x 2,x

Einstein 1905 Lorentz Maxwell c E p E 2 (pc) 2 = m 2 c 4 (7.1) m E ( ) E p µ =(p 0,p 1,p 2,p 3 )=(p 0, p )= c, p (7.2) x µ =(x 0,x 1,x 2,x 7 7.1 7.1.1 Einstein 1905 Lorentz Maxwell c E p E 2 (pc) 2 = m 2 c 4 (7.1) m E ( ) E p µ =(p 0,p 1,p 2,p 3 )=(p 0, p )= c, p (7.2) x µ =(x 0,x 1,x 2,x 3 )=(x 0, x )=(ct, x ) (7.3) E/c ct K = E mc 2 (7.4)

More information

all.dvi

all.dvi 72 9 Hooke,,,. Hooke. 9.1 Hooke 1 Hooke. 1, 1 Hooke. σ, ε, Young. σ ε (9.1), Young. τ γ G τ Gγ (9.2) X 1, X 2. Poisson, Poisson ν. ν ε 22 (9.) ε 11 F F X 2 X 1 9.1: Poisson 9.1. Hooke 7 Young Poisson G

More information

gr09.dvi

gr09.dvi .1, θ, ϕ d = A, t dt + B, t dtd + C, t d + D, t dθ +in θdϕ.1.1 t { = f1,t t = f,t { D, t = B, t =.1. t A, tdt e φ,t dt, C, td e λ,t d.1.3,t, t d = e φ,t dt + e λ,t d + dθ +in θdϕ.1.4 { = f1,t t = f,t {

More information

( ; ) C. H. Scholz, The Mechanics of Earthquakes and Faulting : - ( ) σ = σ t sin 2π(r a) λ dσ d(r a) =

( ; ) C. H. Scholz, The Mechanics of Earthquakes and Faulting : - ( ) σ = σ t sin 2π(r a) λ dσ d(r a) = 1 9 8 1 1 1 ; 1 11 16 C. H. Scholz, The Mechanics of Earthquakes and Faulting 1. 1.1 1.1.1 : - σ = σ t sin πr a λ dσ dr a = E a = π λ σ πr a t cos λ 1 r a/λ 1 cos 1 E: σ t = Eλ πa a λ E/π γ : λ/ 3 γ =

More information

II A A441 : October 02, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka )

II A A441 : October 02, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka ) II 214-1 : October 2, 214 Version : 1.1 Kawahira, Tomoki TA (Kondo, Hirotaka ) http://www.math.nagoya-u.ac.jp/~kawahira/courses/14w-biseki.html pdf 1 2 1 9 1 16 1 23 1 3 11 6 11 13 11 2 11 27 12 4 12 11

More information

all.dvi

all.dvi I 1 Density Matrix 1.1 ( (Observable) Ô :ensemble ensemble average) Ô en =Tr ˆρ en Ô ˆρ en Tr  n, n =, 1,, Tr  = n n  n Tr  I w j j ( j =, 1,, ) ˆρ en j w j j ˆρ en = j w j j j Ô en = j w j j Ô j emsemble

More information

SFGÇÃÉXÉyÉNÉgÉãå`.pdf

SFGÇÃÉXÉyÉNÉgÉãå`.pdf SFG 1 SFG SFG I SFG (ω) χ SFG (ω). SFG χ χ SFG (ω) = χ NR e iϕ +. ω ω + iγ SFG φ = ±π/, χ φ = ±π 3 χ SFG χ SFG = χ NR + χ (ω ω ) + Γ + χ NR χ (ω ω ) (ω ω ) + Γ cosϕ χ NR χ Γ (ω ω ) + Γ sinϕ. 3 (θ) 180

More information

1.1 foliation M foliation M 0 t Σ t M M = t R Σ t (12) Σ t t Σ t x i Σ t A(t, x i ) Σ t n µ Σ t+ t B(t + t, x i ) AB () tα tαn µ Σ t+ t C(t + t,

1.1 foliation M foliation M 0 t Σ t M M = t R Σ t (12) Σ t t Σ t x i Σ t A(t, x i ) Σ t n µ Σ t+ t B(t + t, x i ) AB () tα tαn µ Σ t+ t C(t + t, 1 Gourgoulhon BSSN BSSN ϕ = 1 6 ( D i β i αk) (1) γ ij = 2αĀij 2 3 D k β k γ ij (2) K = e 4ϕ ( Di Di α + 2 D i ϕ D i α ) + α ] [4π(E + S) + ĀijĀij + K2 3 (3) Ā ij = 2 3Āij D k β k 2αĀikĀk j + αāijk +e

More information

20 6 4 1 4 1.1 1.................................... 4 1.1.1.................................... 4 1.1.2 1................................ 5 1.2................................... 7 1.2.1....................................

More information

I A A441 : April 21, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka ) Google

I A A441 : April 21, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka ) Google I4 - : April, 4 Version :. Kwhir, Tomoki TA (Kondo, Hirotk) Google http://www.mth.ngoy-u.c.jp/~kwhir/courses/4s-biseki.html pdf 4 4 4 4 8 e 5 5 9 etc. 5 6 6 6 9 n etc. 6 6 6 3 6 3 7 7 etc 7 4 7 7 8 5 59

More information

Note.tex 2008/09/19( )

Note.tex 2008/09/19( ) 1 20 9 19 2 1 5 1.1........................ 5 1.2............................. 8 2 9 2.1............................. 9 2.2.............................. 10 3 13 3.1.............................. 13 3.2..................................

More information

I A A441 : April 15, 2013 Version : 1.1 I Kawahira, Tomoki TA (Shigehiro, Yoshida )

I A A441 : April 15, 2013 Version : 1.1 I   Kawahira, Tomoki TA (Shigehiro, Yoshida ) I013 00-1 : April 15, 013 Version : 1.1 I Kawahira, Tomoki TA (Shigehiro, Yoshida) http://www.math.nagoya-u.ac.jp/~kawahira/courses/13s-tenbou.html pdf * 4 15 4 5 13 e πi = 1 5 0 5 7 3 4 6 3 6 10 6 17

More information

4 19

4 19 I / 19 8 1 4 19 : : f(e, J), f(e) Phase mixing Landau Damping, violent relaxation : 2 2 : ( ) http://antwrp.gsfc.nasa.gov/apod/ap950917.html ( ) http://www-astro.physics.ox.ac.uk/~wjs/apm_grey.gif

More information

phs.dvi

phs.dvi 483F 3 6.........3... 6.4... 7 7.... 7.... 9.5 N (... 3.6 N (... 5.7... 5 3 6 3.... 6 3.... 7 3.3... 9 3.4... 3 4 7 4.... 7 4.... 9 4.3... 3 4.4... 34 4.4.... 34 4.4.... 35 4.5... 38 4.6... 39 5 4 5....

More information

(iii) 0 V, x V, x + 0 = x. 0. (iv) x V, y V, x + y = 0., y x, y = x. (v) 1x = x. (vii) (α + β)x = αx + βx. (viii) (αβ)x = α(βx)., V, C.,,., (1)

(iii) 0 V, x V, x + 0 = x. 0. (iv) x V, y V, x + y = 0., y x, y = x. (v) 1x = x. (vii) (α + β)x = αx + βx. (viii) (αβ)x = α(βx)., V, C.,,., (1) 1. 1.1...,. 1.1.1 V, V x, y, x y x + y x + y V,, V x α, αx αx V,, (i) (viii) : x, y, z V, α, β C, (i) x + y = y + x. (ii) (x + y) + z = x + (y + z). 1 (iii) 0 V, x V, x + 0 = x. 0. (iv) x V, y V, x + y

More information

( ) (ver )

( ) (ver ) ver.3.1 11 9 1 1. p1, 1.1 ψx, t,, E, p. = E, p ψx, t,. p, 1.8 p4, 1. t = t ρx, t = m [ψ ψ ψ ψ] ρx, t = mi [ψ ψ ψ ψ] p4, 1.1 = p6, 1.38 p6, 1.4 = fxδ ϵ x = fxδϵx = 1 π fxδ ϵ x dx = fxδ ϵ x dx = [ 1 fϵ π

More information

6 2 T γ T B (6.4) (6.1) [( d nm + 3 ] 2 nt B )a 3 + nt B da 3 = 0 (6.9) na 3 = T B V 3/2 = T B V γ 1 = const. or T B a 2 = const. (6.10) H 2 = 8π kc2

6 2 T γ T B (6.4) (6.1) [( d nm + 3 ] 2 nt B )a 3 + nt B da 3 = 0 (6.9) na 3 = T B V 3/2 = T B V γ 1 = const. or T B a 2 = const. (6.10) H 2 = 8π kc2 1 6 6.1 (??) (P = ρ rad /3) ρ rad T 4 d(ρv ) + PdV = 0 (6.1) dρ rad ρ rad + 4 da a = 0 (6.2) dt T + da a = 0 T 1 a (6.3) ( ) n ρ m = n (m + 12 ) m v2 = n (m + 32 ) T, P = nt (6.4) (6.1) d [(nm + 32 ] )a

More information

V(x) m e V 0 cos x π x π V(x) = x < π, x > π V 0 (i) x = 0 (V(x) V 0 (1 x 2 /2)) n n d 2 f dξ 2ξ d f 2 dξ + 2n f = 0 H n (ξ) (ii) H

V(x) m e V 0 cos x π x π V(x) = x < π, x > π V 0 (i) x = 0 (V(x) V 0 (1 x 2 /2)) n n d 2 f dξ 2ξ d f 2 dξ + 2n f = 0 H n (ξ) (ii) H 199 1 1 199 1 1. Vx) m e V cos x π x π Vx) = x < π, x > π V i) x = Vx) V 1 x /)) n n d f dξ ξ d f dξ + n f = H n ξ) ii) H n ξ) = 1) n expξ ) dn dξ n exp ξ )) H n ξ)h m ξ) exp ξ )dξ = π n n!δ n,m x = Vx)

More information

1 filename=mathformula tex 1 ax 2 + bx + c = 0, x = b ± b 2 4ac, (1.1) 2a x 1 + x 2 = b a, x 1x 2 = c a, (1.2) ax 2 + 2b x + c = 0, x = b ± b 2

1 filename=mathformula tex 1 ax 2 + bx + c = 0, x = b ± b 2 4ac, (1.1) 2a x 1 + x 2 = b a, x 1x 2 = c a, (1.2) ax 2 + 2b x + c = 0, x = b ± b 2 filename=mathformula58.tex ax + bx + c =, x = b ± b 4ac, (.) a x + x = b a, x x = c a, (.) ax + b x + c =, x = b ± b ac. a (.3). sin(a ± B) = sin A cos B ± cos A sin B, (.) cos(a ± B) = cos A cos B sin

More information

X G P G (X) G BG [X, BG] S 2 2 2 S 2 2 S 2 = { (x 1, x 2, x 3 ) R 3 x 2 1 + x 2 2 + x 2 3 = 1 } R 3 S 2 S 2 v x S 2 x x v(x) T x S 2 T x S 2 S 2 x T x S 2 = { ξ R 3 x ξ } R 3 T x S 2 S 2 x x T x S 2

More information

,. Black-Scholes u t t, x c u 0 t, x x u t t, x c u t, x x u t t, x + σ x u t, x + rx ut, x rux, t 0 x x,,.,. Step 3, 7,,, Step 6., Step 4,. Step 5,,.

,. Black-Scholes u t t, x c u 0 t, x x u t t, x c u t, x x u t t, x + σ x u t, x + rx ut, x rux, t 0 x x,,.,. Step 3, 7,,, Step 6., Step 4,. Step 5,,. 9 α ν β Ξ ξ Γ γ o δ Π π ε ρ ζ Σ σ η τ Θ θ Υ υ ι Φ φ κ χ Λ λ Ψ ψ µ Ω ω Def, Prop, Th, Lem, Note, Remark, Ex,, Proof, R, N, Q, C [a, b {x R : a x b} : a, b {x R : a < x < b} : [a, b {x R : a x < b} : a,

More information

2019 1 5 0 3 1 4 1.1.................... 4 1.1.1......................... 4 1.1.2........................ 5 1.1.3................... 5 1.1.4........................ 6 1.1.5......................... 6 1.2..........................

More information

(yx4) 1887-1945 741936 50 1995 1 31 http://kenboushoten.web.fc.com/ OCR TeX 50 yx4 e-mail: yx4.aydx5@gmail.com i Jacobi 1751 1 3 Euler Fagnano 187 9 0 Abel iii 1 1...................................

More information

Report98.dvi

Report98.dvi 1 4 1.1.......................... 4 1.1.1.......................... 7 1.1..................... 14 1.1.................. 1 1.1.4........................... 8 1.1.5........................... 6 1.1.6 n...........................

More information

数学Ⅱ演習(足助・09夏)

数学Ⅱ演習(足助・09夏) II I 9/4/4 9/4/2 z C z z z z, z 2 z, w C zw z w 3 z, w C z + w z + w 4 t R t C t t t t t z z z 2 z C re z z + z z z, im z 2 2 3 z C e z + z + 2 z2 + 3! z3 + z!, I 4 x R e x cos x + sin x 2 z, w C e z+w

More information

Cercignani Shen Kuščer

Cercignani Shen Kuščer 22 8 23 29 3 22 1 3 2 3 3 4 4 7 5 9 5.1................................ 9 5.2 Cercignani...................... 12 5.3 Shen Kuščer..................... 12 6 13 7 17 7.1 Maxwell..........................

More information

i

i 009 I 1 8 5 i 0 1 0.1..................................... 1 0.................................................. 1 0.3................................. 0.4........................................... 3

More information

ω 0 m(ẍ + γẋ + ω0x) 2 = ee (2.118) e iωt x = e 1 m ω0 2 E(ω). (2.119) ω2 iωγ Z N P(ω) = χ(ω)e = exzn (2.120) ϵ = ϵ 0 (1 + χ) ϵ(ω) ϵ 0 = 1 +

ω 0 m(ẍ + γẋ + ω0x) 2 = ee (2.118) e iωt x = e 1 m ω0 2 E(ω). (2.119) ω2 iωγ Z N P(ω) = χ(ω)e = exzn (2.120) ϵ = ϵ 0 (1 + χ) ϵ(ω) ϵ 0 = 1 + 2.6 2.6.1 ω 0 m(ẍ + γẋ + ω0x) 2 = ee (2.118) e iωt x = e 1 m ω0 2 E(ω). (2.119) ω2 iωγ Z N P(ω) = χ(ω)e = exzn (2.120) ϵ = ϵ 0 (1 + χ) ϵ(ω) ϵ 0 = 1 + Ne2 m j f j ω 2 j ω2 iωγ j (2.121) Z ω ω j γ j f j

More information

* 1 2014 7 8 *1 iii 1. Newton 1 1.1 Newton........................... 1 1.2............................. 4 1.3................................. 5 2. 9 2.1......................... 9 2.2........................

More information

H 0 H = H 0 + V (t), V (t) = gµ B S α qb e e iωt i t Ψ(t) = [H 0 + V (t)]ψ(t) Φ(t) Ψ(t) = e ih0t Φ(t) H 0 e ih0t Φ(t) + ie ih0t t Φ(t) = [

H 0 H = H 0 + V (t), V (t) = gµ B S α qb e e iωt i t Ψ(t) = [H 0 + V (t)]ψ(t) Φ(t) Ψ(t) = e ih0t Φ(t) H 0 e ih0t Φ(t) + ie ih0t t Φ(t) = [ 3 3. 3.. H H = H + V (t), V (t) = gµ B α B e e iωt i t Ψ(t) = [H + V (t)]ψ(t) Φ(t) Ψ(t) = e iht Φ(t) H e iht Φ(t) + ie iht t Φ(t) = [H + V (t)]e iht Φ(t) Φ(t) i t Φ(t) = V H(t)Φ(t), V H (t) = e iht V (t)e

More information

211 kotaro@math.titech.ac.jp 1 R *1 n n R n *2 R n = {(x 1,..., x n ) x 1,..., x n R}. R R 2 R 3 R n R n R n D D R n *3 ) (x 1,..., x n ) f(x 1,..., x n ) f D *4 n 2 n = 1 ( ) 1 f D R n f : D R 1.1. (x,

More information

OHP.dvi

OHP.dvi t 0, X X t x t 0 t u u = x X (1) t t 0 u X x O 1 1 t 0 =0 X X +dx t x(x,t) x(x +dx,t). dx dx = x(x +dx,t) x(x,t) (2) dx, dx = F dx (3). F (deformation gradient tensor) t F t 0 dx dx X x O 2 2 F. (det F

More information

21 2 26 i 1 1 1.1............................ 1 1.2............................ 3 2 9 2.1................... 9 2.2.......... 9 2.3................... 11 2.4....................... 12 3 15 3.1..........

More information

JKR Point loading of an elastic half-space 2 3 Pressure applied to a circular region Boussinesq, n =

JKR Point loading of an elastic half-space 2 3 Pressure applied to a circular region Boussinesq, n = JKR 17 9 15 1 Point loading of an elastic half-space Pressure applied to a circular region 4.1 Boussinesq, n = 1.............................. 4. Hertz, n = 1.................................. 6 4 Hertz

More information

Maxwell

Maxwell I 2018 12 13 0 4 1 6 1.1............................ 6 1.2 Maxwell......................... 8 1.3.......................... 9 1.4..................... 11 1.5..................... 12 2 13 2.1...................

More information

80 4 r ˆρ i (r, t) δ(r x i (t)) (4.1) x i (t) ρ i ˆρ i t = 0 i r 0 t(> 0) j r 0 + r < δ(r 0 x i (0))δ(r 0 + r x j (t)) > (4.2) r r 0 G i j (r, t) dr 0

80 4 r ˆρ i (r, t) δ(r x i (t)) (4.1) x i (t) ρ i ˆρ i t = 0 i r 0 t(> 0) j r 0 + r < δ(r 0 x i (0))δ(r 0 + r x j (t)) > (4.2) r r 0 G i j (r, t) dr 0 79 4 4.1 4.1.1 x i (t) x j (t) O O r 0 + r r r 0 x i (0) r 0 x i (0) 4.1 L. van. Hove 1954 space-time correlation function V N 4.1 ρ 0 = N/V i t 80 4 r ˆρ i (r, t) δ(r x i (t)) (4.1) x i (t) ρ i ˆρ i t

More information

2000年度『数学展望 I』講義録

2000年度『数学展望 I』講義録 2000 I I IV I II 2000 I I IV I-IV. i ii 3.10 (http://www.math.nagoya-u.ac.jp/ kanai/) 2000 A....1 B....4 C....10 D....13 E....17 Brouwer A....21 B....26 C....33 D....39 E. Sperner...45 F....48 A....53

More information

t = h x z z = h z = t (x, z) (v x (x, z, t), v z (x, z, t)) ρ v x x + v z z = 0 (1) 2-2. (v x, v z ) φ(x, z, t) v x = φ x, v z

t = h x z z = h z = t (x, z) (v x (x, z, t), v z (x, z, t)) ρ v x x + v z z = 0 (1) 2-2. (v x, v z ) φ(x, z, t) v x = φ x, v z I 1 m 2 l k 2 x = 0 x 1 x 1 2 x 2 g x x 2 x 1 m k m 1-1. L x 1, x 2, ẋ 1, ẋ 2 ẋ 1 x = 0 1-2. 2 Q = x 1 + x 2 2 q = x 2 x 1 l L Q, q, Q, q M = 2m µ = m 2 1-3. Q q 1-4. 2 x 2 = h 1 x 1 t = 0 2 1 t x 1 (t)

More information

2007 5 iii 1 1 1.1.................... 1 2 5 2.1 (shear stress) (shear strain)...... 5 2.1.1...................... 6 2.1.2.................... 6 2.2....................... 7 2.2.1........................

More information

201711grade1ouyou.pdf

201711grade1ouyou.pdf 2017 11 26 1 2 52 3 12 13 22 23 32 33 42 3 5 3 4 90 5 6 A 1 2 Web Web 3 4 1 2... 5 6 7 7 44 8 9 1 2 3 1 p p >2 2 A 1 2 0.6 0.4 0.52... (a) 0.6 0.4...... B 1 2 0.8-0.2 0.52..... (b) 0.6 0.52.... 1 A B 2

More information

() n C + n C + n C + + n C n n (3) n C + n C + n C 4 + n C + n C 3 + n C 5 + (5) (6 ) n C + nc + 3 nc n nc n (7 ) n C + nc + 3 nc n nc n (

() n C + n C + n C + + n C n n (3) n C + n C + n C 4 + n C + n C 3 + n C 5 + (5) (6 ) n C + nc + 3 nc n nc n (7 ) n C + nc + 3 nc n nc n ( 3 n nc k+ k + 3 () n C r n C n r nc r C r + C r ( r n ) () n C + n C + n C + + n C n n (3) n C + n C + n C 4 + n C + n C 3 + n C 5 + (4) n C n n C + n C + n C + + n C n (5) k k n C k n C k (6) n C + nc

More information

Hanbury-Brown Twiss (ver. 2.0) van Cittert - Zernike mutual coherence

Hanbury-Brown Twiss (ver. 2.0) van Cittert - Zernike mutual coherence Hanbury-Brown Twiss (ver. 2.) 25 4 4 1 2 2 2 2.1 van Cittert - Zernike..................................... 2 2.2 mutual coherence................................. 4 3 Hanbury-Brown Twiss ( ) 5 3.1............................................

More information

note1.dvi

note1.dvi (1) 1996 11 7 1 (1) 1. 1 dx dy d x τ xx x x, stress x + dx x τ xx x+dx dyd x x τ xx x dyd y τ xx x τ xx x+dx d dx y x dy 1. dx dy d x τ xy x τ x ρdxdyd x dx dy d ρdxdyd u x t = τ xx x+dx dyd τ xx x dyd

More information

ii p ϕ x, t = C ϕ xe i ħ E t +C ϕ xe i ħ E t ψ x,t ψ x,t p79 やは時間変化しないことに注意 振動 粒子はだいたい このあたりにいる 粒子はだいたい このあたりにいる p35 D.3 Aψ Cϕdx = aψ ψ C Aϕ dx

ii p ϕ x, t = C ϕ xe i ħ E t +C ϕ xe i ħ E t ψ x,t ψ x,t p79 やは時間変化しないことに注意 振動 粒子はだいたい このあたりにいる 粒子はだいたい このあたりにいる p35 D.3 Aψ Cϕdx = aψ ψ C Aϕ dx i B5 7.8. p89 4. ψ x, tψx, t = ψ R x, t iψ I x, t ψ R x, t + iψ I x, t = ψ R x, t + ψ I x, t p 5.8 π π π F e ix + F e ix + F 3 e 3ix F e ix + F e ix + F 3 e 3ix dx πψ x πψx p39 7. AX = X A [ a b c d x

More information

2002 11 21 1 http://www.sml.k.u-tokyo.ac.jp/members/nabe/lecture2002 http://www.sml.k.u-tokyo.ac.jp/members/nabe/lecture nabe@sml.k.u-tokyo.ac.jp 2 1. 10/10 2. 10/17 3. 10/24 4. 10/31 5. 11/ 7 6. 11/14

More information