Contents 1 Jeans (

Contents 1 Jeans 2 1.1....................................... 2 1.2................................. 2 1.3............................... 3 2 3 2.1 ( )................................ 4 2.2 WKB........................ 4 2.3.................. 5 3 7 3.1 2........................... 7 3.2.......................................... 8 3.3....................................... 8 3.4 active disk.................................... 9 4 9 4.1...................................... 9 4.2.................................... 11 5 12 5.1.................................. 12 5.2.................................. 12 5.3.................................... 13 6 14 6.1 2................................. 14 6.2................................ 15 6.2.1................................. 15 6.2.2 2 (dynamical friction)............. 15 6.2.3 2.......................... 17 6.3.............................. 18 6.3.1.................................. 18 6.3.2 3...................... 18 6.3.3......................... 19 7 21 7.1.............................. 21 7.2................................ 21 8 24 9 24 1

1 1 Jeans 1.1, Eq. of continuity, Euler s eq. ρ + (ρv) = 0. (1) t v t + (v )v = 1 p Φ. (2) ρ, Poisson s eq. Φ = 4πGρ. (3) 1.2 (ρ 0 = p 0 = const., v 0 = Φ 0 = 0.) (a) ρ = ρ 0 + ρ 1, p = p 0 + p 1, v = v 1, Φ = Φ 1. (b) ( 2 ) Eq. of continuity Euler s eq. Poisson s eq. Eq. of state v 1 t ρ 1 t + ρ 0 v 1 = 0. (4) p 1 = = 1 ρ 0 p 1 Φ 1. (5) Φ 1 = 4πGρ 1. (6) ( ) p ρ 1 = c 2 s ρ 1 (7) ρ s (c) v 1, Φ 1 [ t Eq. (4) ρ 0 Eq. (5)] Eq. (6) 2 t 2 ρ 1 c 2 s ρ 1 4πGρ 0 ρ 1 = 0 (8) 2

(d) ρ 1 = C exp(ik x iωt). ω ω 2 = c 2 sk 2 4πGρ 0. (9) Jeans length k < k J λ J = 2π πc = 2 s k J Gρ 0 4πGρ0 c s (10) ( λ J c s 1 Gρ0 ) Jeans mass M J 4π 3 ρ 0(λ J /2) 3 1.3 (Molecular Cloud) Orion[1500 ], Taurus[450 ] 100-10 4 /cm 3 10 22-10 21 g/cm 3. : -. 10 4-10 7 M (M = 2 10 33 g). 10-30K ( 200-300m/sec). Jeans length Jeans mass λ J 10 19 cm 10. M J 10M λ J, M J ρ 1/2 ( ). 2 (accretion disk) passive disk ( active disk) 3

2.1 ( ) Σ = ρdz, P = pdz; v z = 0, v/ z 0 Eq. of continuity Σ t + 1 R R (RΣv R) + 1 R ϕ (Σv ϕ) = 0 (11) Euler s eq. v R t + v v R R R + v ϕ v R R ϕ v2 ϕ R = 1 Σ P R GM R 2 Φ D R (12) v ϕ t + v v ϕ R R + v ϕ v ϕ R ϕ + v Rv ϕ R = 1 P RΣ ϕ 1 Φ D R ϕ (13) Poisson s eq. Φ D = 4πGΣδ(z) (14) 2.2 WKB (a) Σ = Σ 0 + Σ 1, P = P 0 + P 1, P 1 = c 2 sσ 1 ; v R = v R,1, v ϕ = RΩ(R) + v ϕ,1 ; Φ D = Φ D,0 + Φ D,1 (b) WKB R 1 R ϕ, 1 R ; exp(ikr + imϕ iωt) (c) Poissson s eq. ( 2 R 2 + 2 z 2 )Φ D,1 = 4πGΣ 1 δ(z) ( WKB ) (15) -0 +0 z ( ) ( ) ΦD,1 ΦD,1 = z z z=+0 z 0 ( 2 R 2 + 2 z 2 )Φ D,1 = 0 Eqs. (16), (17) z=0 z= 0 = 2πGΣ 1. (16) Φ D,1 e k z exp(ikr + imϕ iωt). (17) Φ D,1 = 2πGΣ 1 /k. (18) (d) Eq. of continuity i(mω ω)σ 1 + ikσ 0 v R,1 + imσ 0 R v ϕ,1 = 0. (19) 4

Euler s eq. ( i(mω ω) 2Ω 2B i(mω ω) ) ( vr,1 v ϕ,1 ) = ( ik im/r ) ( ) c 2 s Σ 1 + Φ D,1 Σ 0 (20) ( vr,1 v ϕ,1 ) = 1 ( (mω ω)k i2bk ) ( ) c 2 s Σ 1 + Φ D,1 Σ 0 B = 1 d(r 2 Ω) (Oort s B constant), 2R dr κ 2 = 4BΩ (epicycle frequency), = κ 2 (mω ω) 2. (21) (22) Eqs. (18), (19), (21) ( 1 + c2 sk 2 ) 2πGΣ 0 k Σ 1 = 0. (23) (mω ω) 2 = c 2 sk 2 2πGΣ 0 k + κ 2. (24) (e) ω k., κ 2 > 0, =0 2. j (= R 2 Ω) R Rayleigh Toomre Q c sκ πgσ > 1 (25) Q Toomre s Q value Q=1 λ crit = 2π/k crit = 2πc s /κ. 2.3. (passive disk[ ] ). πd 2 L /(4πR 2 ) = 4πd 2 σt 4 (26) (L = 3.9 10 33 erg/sec) T = 280(R/1AU) 1/2 K. (27) c s = 1.2(R/1AU) 1/4 km/sec. (28) 5

Σ gas = 1700(R/1AU) 3/2 g/cm 2. (29) Σ dust = ( ) R 3/2 7 g/cm 2 (R < 2.7AU, ) 1AU ( ) R 3/2 28 g/cm 2 (R > 2.7AU, ) 1AU (30) M D = 100AU 0 Σ 2πRdR 0.03M 2 10 4 M Toomre Q Q = 50(R/1AU) 1/4 (31) Euler s eq z 1 p ρ z = GM z R 3 (32) ρ = Σ gas 2πh e z2 /2h 2. (33) h = c s /Ω. h/r 1/30 (1AU). : Binney and Tremaine (1987) Galactic Dynamics, Chap. 5 & 6, Princeton Univ. Press. (1997) 12 2, 1: 1. 100 /cm 3, T = 10K Jeans length Jeans mass 2.3 (27), (28) (29) Toomre Q 2. (21)-(23) 3. R Roche 2.4R M (ρ M /ρ m ) 1/3 (34) (R M : ρ M : ρ m : Toomre. κ = Ω. Σ/h. 6

3 2 (Lynden-Bell and Pringle 1974) α ν t = αc s h. αcs αh α 0.01 : : (MRI ) 3.1 2 -. - jet. - ϕ ( ) Eq. of continuity Σ t + 1 R R (RΣv R) = 0. (35) (cf. Eq. [11]) Navier-Stokes eq. ( ) ϕ t (Σv ϕ) + 1 ( R 2 R 2 Σv R v ϕ R 2 ) Π Rϕ = 0. (36) R j = Rv ϕ t (Σj) + 1 ( RΣjvR R 2 ) Π Rϕ = 0. (37) R R Π Rϕ R, ϕ ( vϕ Π Rϕ = Σν t R v ) ϕ = Σν t R dω R dr. (38) Eqs. (37), (38) t (Σj) + 1 ( R R RΣjv R } {{ } flux R 3 dω Σν t dr }{{} flux ) = 0. j t = 0, Eqs.(35), (39) v R ṀD flux) Ṁ D 2πRΣ v R = 2π ( ) R 3 dω Σ ν t. (40) (dj/dr) R dr Eq. (35) Σ 1 Σ t + 1 [ 1 R R (dj/dr) R (39) ( )] R 3 dω Σν t = 0. (41) dr 7

3.2 Σ t = 0 Eqs.(41), (39), Eq. (43) Ṁ D =, (42) J D jṁd + 2πR 3 dω Σν t =. dr (43), J D = ṀD j(r in ) (> 0) ν t Σ = jṁd J D 2πR 3 (dω/dr). (44) Ṁ D Ω R 3/2 (Kepler ) ( ν t Σ = 2πR 3 (dω/dr) [j(r) j(r in)]. (45) ṀD Σ = 3πν t 1 Rin R ). (46) 3.3 Ω R β Kepler β = 3/2 ν t = ν 1 (R/R 1 ) γ R Kepler γ = 1). (41) (Lynden-Bell & Pringle 1974; Hartmann et al. 1998) Σ(R, t) = M D (0) 2 γ 2πR 2 1 ( R R 1 M D (0) t = 0 R 1, R R in. ) γ 2 β 1 T 2 γ exp [ (R/R 1) 2 γ T ]. (47) T = t t 1 + 1, t 1 = (2 β)r2 1 (2 γ) 2 βν 1. (48) M D (t) M D (t) = M D (0) T 2 β 2 γ. (49) R flux ṀD(R) Ṁ D (R) = M [ D(t) 2 β T t 1 2 γ (R/R 1) 2 γ ] exp [ (R/R 1) 2 γ ]. (50) T T t disk R2 D 3ν t (R D ) 1 3αΩ(R D ) ( RD h ) 2. (51) α = 10 3 100AU 2 8

3.4 active disk ϵ (e.g. ) ( ϵ = ν t R dω ) 2. (52) dr R R in T s = ( 2σTs 4 = Σν t R dω ) 2. (53) dr ( ) 1/4 1 dω2 ṀDR Ω 1/2. (54) 8πσ dr α = 10 3 1AU T s 100K. : z.,., d ( 16σT 3 ) dt = 1 ( dz 3κρ dz 2 ν tσ R dω ) 2 δ(z). (55) dr, τ z = z κρdz, Eq. (55) 4 d (σt 4 ) = RΩ dω 3 dτ z 4π drṁd. (56) τ z = 2/3 T = T s ( 3 T (z) = 4 τ z + 1 ) 1/4 T s. (57) 2 4 5nm-0.2µ (Mathis et al. 1977) (Goldreich and Ward 1973) 4.1 (a) F drag = ma(m)ρ g v. (58) A 2 9

Stokes law (a > 3l/2, l(= 1/[n H2 σ H2 ]) : gas mean free path) 8 3c s l A = π 2ρ solid a 2. (59) Epstein s law (a < 3l/2) t stop = A = 8 c s π ρ solid a. (60) A = 3C D v 8ρ solid a. (61) m v F drag = 1. (62) Aρ g (b) v = (v R, v ϕ = RΩ + v ϕ,1, v z ) (Ω = Ω K GM c /R 3 ) (20) ( 0 2Ω 2B 0 ) ( vr v ϕ,1 ) = 1 ρ g p R + ρ da(v R v R ) ρ d A(V ϕ,1 v ϕ,1 ). (63) V = (V R, V ϕ = RΩ + V ϕ,1, V z ) ( ) ( 0 2Ω VR 2B 0 V ϕ,1 (c) ) ( ρg A(V R v R ) = ρ g A(V ϕ,1 v ϕ,1 ) ). (64) v R = v z = 0, v ϕ = (1 η)rω, (65) η = 1 p ( 2RΩ 2 ρ g R = 1 cs ) 2 ln p 2 RΩ ln R. (66) 2t stopω V R = ηrω, (67) 1 + (t stop Ω) 2 V ϕ = RΩ 1 ηrω, (68) 1 + (t stop Ω) 2 V z = Ωz (t stop Ω). (69) 10

4.2 (a) dm dt = ρ dσ col v. (70) σ col 4πa 2, v V z t grow = a/ da dt = 3m/dm dt = 3mρ ga 4πa 2 ρ d zω 2 Σ g Σ d Ω 1. (71) z Epstein ( 100AU) (b) t grow h/ V z (h) 10t grow (c) h d ν t = αc s h h 2 d ν t h d V z (h d ). (72) α h d = h. (73) t stop Ω (71) (d) V R (67) t stop Ω K 1 V R ηrω 50m/sec. (74) (t stop Ω K 1) t drift = R/ V R 1 100 (at 1AU). (75) ηω 2: 1. (47) (41) (39) R in R R 1 T 1/(2 γ) (45) 2. Epstein s law (60) factor 3. (65)-(68) 11

3 5 5.1 (Goldreich and Ward 1973) Σ d /h d ( Ω 2 /G Toomre Q c s h d Ω [34 ]) Q h dω 2 πgσ d 1. (76) h d,crit πgσ d Ω 2 = πσ dr 3 M c. (77) 1AU h d,crit /R 2 10 6. h d,crit Ω 7cm/sec. k crit 1/h d,crit. (78) m crit π(2π/k crit ) 2 Σ d 4π 5 R6 Σ 3 d Mc 2. (79) 1AU m crit 10 18 g 5.2 (a) (ρ d ρ g ) v ϕ ηrω. (80) (Weidenschilling 1980) ( 1 GM c 2 R 3 z2 1 2 (ηrω)2. (81) z c2 RΩ 2 h2 R. (82) z/h 1/30 ( 1AU) ρ d < ρ g decouple 12

(b) t stop Ω 1 50m/sec 100 (at 1AU) (71) (c) ( ) 50m/sec ( 180km) 5.3 <0.01eV, ), 0.1eV, ) γ [J/m 2 ] E stick = γ ( ). (83) R 2 ( ) a JKR ( ) 14γR 2 1/3 a. (84) E E [Pa = N/m 2 ] E elastic ( ) du 1 dx dv 2 E ( ) du 2 dv Eaδ 2 Ea 5 /R 2. (85) dx u δ δ = a 2 /R E stick + E elastic a ( γ 5 R 4 ) 1/3 E bond = E stick + E elastic 23. (86) m v crit E 4 v crit ( Ebond m ) { 1/2 3(R/0.1µm) 5/6 m/sec ( ), 0.3(R/0.1µm) 5/6 m/sec ( ). (87) 1 γ [J/m 2 ] E [GPa] (SiO 2 ) 0.03 50 0.1 7 13

4 6 6.1 2 2 M m (µ = Mm M + m ) r r r θ 2 = α r 2, (88) θ r θ + 2ṙ θ = 0, (89) α = G(M + m). (90) L = r 2 θ. (91) E = 1 2ṙ2 + L2 2r 2 α r. (92) e < 1 (x + ae) 2 y 2 a 2 + a 2 (1 e 2 = 1. (93) ) r = a(1 e2 ) 1 + e cos θ. (94) a e E(< 0) L a = α 2E, e = 1 + 2EL2 α 2. (95) y Y y a a 1- e 2 ae r θ x, X y=( e 2 1) 1/2 (x+ae ) x 14

e > 1 (93)-(95) a < 0 e > 1 v 0 b E = 1 G(M + m) 2 v2 0, L = bv 0, a = v0 2, e = 1 + b2 a 2. (96) ( ) 1 φ = 2 sin 1 = 2 sin 1 1 e 1 + [ bv0 2/G(M + m)] 2. (97) 6.2 6.2.1 (a) (b) 2 ( 1 6.2.2 2 (dynamical friction) (dynamical friction) (m Vm 2 = M VM 2 ) (a) 1 ( M) M : V M : m : n 15

1 2 v = V m V M. v 0 = V M. v v v 0 / v 0 = v 0 (cos φ 1) = 2v 0 sin 2 φ 2 = 2v 0 1 + [ bv0 2/G(M + m)] 2. (98) v = 2v [ 0 bv 2 0 /G(M + m) ] 1 + [ bv0 2/G(M + m)] 2. (99) M V M = m M + m v. (100) 2 V M = 0. V M = V M v 0 v 0. M dv M dt = V M = nv 0 2πbdb V M m 2v 0 = n v 0 2πbdb M + m 1 + [ V M bv0 2/G(M + m)] 2 V M = 2π [ ] G(M + m) 2 ln(λ 2 + 1)ρ m V M V M. (101) M + m V 2 M ρ m = mn Λ = b maxv 2 M G(M + m). (102) M dynamical friction [ ] G(M + m) 2 σ DF 2π ln(λ 2 + 1). (103) b max V 2 M (101) b b b max Λ > 100 b max factor 2 ln(λ 2 + 1) b max dynamical friction 2 16

(b) 1 0 Maxwell f( V m ) = 1 (2π Vm ) 2 exp 3/2 ( V m 2 2 V 2 m ). (104) M v 0 = V m V M dv M = d 3 V m f(v m ) m n v 0 2v 0 2πbdb dt M + m 1 + [ v 0 bv0 2/G(M + m)] 2 v 0 = 2πρ m G 2 (M + m) d 3 V m f(v m ) V m V M V m V M 3 ln(λ2 + 1). (105) (105) ln(λ 2 + 1) F (r 0 ) = d 3 rgρ(r) r r ( 3 r0 0 = G ρ(r)4πr 2 dr r r 0 0 dynamical friction (Chandrasekhar 1943) dv M dt = 2πρ m G 2 (M + m) ln(λ 2 + 1) = 2πρ m G 2 (M + m) ln(λ 2 + 1) VM 0 ) r0 f(v m )4πV 2 mdv m [ erf(x) 2X π e X2 ] r 0 3 (106) V M V M 3 VM V M 3. (107) 2 (104) X = V M / 2 V 2 m erf(x) erf(x) = 2 π X 0 e x2 dx Λ Λ = b max(vm 2 + V m ) 2. (108) G(M + m) ( V 2 m V 2 M ) (107) erf( ) = 1 (a) (101) ( V 2 m V 2 M ) (107) [ ] 4X3 /(3 π) dv M dt = 2 2πρm G 3 2 (M + m) ln(λ 2 + 1) Vm 2 3/2 (for Vm 2 VM). 2 (109) V M 6.2.3 2 dynamical friction V M t relax t relax = V M / dv M dt (VM 2 + V m ) 2 3/2 2πρ m G 2 (M + m) ln(λ 2 (110) + 1) t relax = V 2 M/ V M 2 (111) 17

6.3 6.3.1 (a) e, i 1 RΩ v (e + i)rω RΩ (b) 3 + particle-in-a-box 2 (viscous stirring) 2 or 3 ν (er) 2 /t relax 6.3.2 3 (a) 2 e, i 1 (x, y, z) x = R R 0, y = R 0 (Θ Θ 0 Ω 0 t), z = Z. Ω 0 = Ω(R 0 ). x = a R 0 er 0 cos[ω 0 (t τ)], y = y 0 3 2 Ω 0(a R 0 )t + 2eR 0 sin[ω 0 (t τ)], z = ir 0 sin[ω 0 (t ω)]. (112) (113) or (b) e, i, (a R 0 )/R 0 2 2 3x dr dt = 2Ω 0 v + Ω 2 0 0 G(M + m)r r z 3. (114) 18

x : 3 ( ) M + m 1/3 r H = R 0 (115) 3M star r = r/r H, t = Ω 0 t, v = v/(r H Ω 0 ) r H R 0 /100 dr dt = 2e z v + 3x 0 z Fig.1. 3r r 3 (116) 2 particle-in-a-box e r H /R 0 2 6.3.3 (a) 4 0 e 2 1/2 ) v = e 2 1/2 RΩ 2 dv dt v t relax. (117) 2 t relax (110) t relax = (2v 2 ) 3/2 4πρ m G 2 m ln(λ 2 + 1). (118) ρ m Σ d ρ m Σ d /(v/ω) (119) t relax v 4 e 2 1/2 t 1/4 y 40 20 0 20 40 4 Figure 1: e = i = 0 x = 1 4 2 19

i e 2 1/2 / i 2 1/2 2 (e.g., Ohtsuki et al. 2002) N Fig. 2 (c) v t stop mv t stop = 1 2 C Dπrmρ 2 g v. (120) 2 r m m km C D = 1 t relax = t stop (33) v [ ] π Σ d c 1/5 ln(λ 2 + 1) v esc v esc 1 3 Σ g ( ) Σd /Σ 1/5 ( g c ) 1/5 ( rm 1/250 1km s 1 r Earth ) 1/5 v esc. (121) v esc = 2Gm/r m m 4/5. dynamical friction ) (121) (121) v esc Σ d r m. Figure 2: N (Ohtsuki et al. 2002, Icarus 7 ) Analytic 20

5 7 7.1 dm dt = ρ m σ col v m (122) 1 2 v2 m = 1 2 v 2 m GM/r M b v m = r M v m 2 σ col = πr 2 M M (119) ρ m Σ d Ω/v m t grow = M dm dt M πσ d ΩrM 2 ( 1 + 2GM ) r M vm 2. (123) ( 1 + 2GM ) 1 r M vm 2 (124) (M = m, v m v esc /3) ( ) M 1/3 ( ) R 3 8 10 6 yr (R < 2.7AU) M Earth 1AU t grow = ( ) M 1/3 ( ) R 3 2 10 8 yr (R > 2.7AU) M Earth 5AU (125) 10 4 10 8, 10 11 (10 7 7.2 { (a) (runaway growth) 2 m ( M ( m v m v esc, M M 21

(124) t grow v 2 mm 1/3 (126) N Fig. 3-2.5 M (v m M 1/6 ) t grow M 0 (b) (oligarchic growth) N (Kokubo et al. 1998) (Fig. 4a) R 10r H ( M 1/3 ) t grow M 1/3 v m v esc,m /6 (125) 1 2 N(>m) m/m0 Figure 3: N (Kobayashi et al. 2010, Icarus 1 ) N (Kokubo et al. 2000) 3 (c) 22

(isolation mass) isolation mass (c) M iso = Σ d 2πR R. (127) R = b r H ( ) b 3/2 ( ) R 3/4 (2πbR M iso = 2 Σ d ) 0.1 M Earth (R < 2.7AU) 3 10 1AU = 3M star ( ) b 3/2 ( ) R 3/4 3 M Earth (R > 2.7AU) 10 5AU (128) 1AU 30r H (Chambers et al. 1996) R (Fig. 4b) 1 R = 8r H 10 8 e T a R r Figure 4: (a) N (Kokubo et al. 2000, Icarus 7 ) - 10r H (b) (Chambers et al. 1996, Icarus 1 3 R 3 1/3 23

8 (a) c 2 < GM r M v 2 esc (129) 300K (10 25 g) (b), (Mizuno ), (Kelvin-Helmholtz opacity opacity 5-15 (c) 1 ( R 10r H ) 2 gap ( r H > c/ω) gap? Kanagawa et al. 2015, Tanigawa & Tanaka 2016) 3: (photoevapolation) 9 (a) (b) gap 24