Shocks in Heliophysics Dana Longcope LWS Heliophysics Summer School & Montana State University Shocks in Heliophysics Vasyliunas, v.1 ch.10 Shocks in Heliophysics Tsuneta and Naito 1998 Forbes, T.G., and Malherbe 1986 (v.2 ch. 6) Shocks in Heliophysics shock Gold 1962 Cane, Reames, and von Rosenvinge 1988 (v. 2 ch. 7) Shocks in Heliophysics Forbes 2000 (v. 2, ch. 6) Shocks in Heliophysics Crooker, Gosling, et al. 1999 (v3, ch. 8) Shocks in Heliophysics Opher v2, ch. 7 Shocks in the air Shocks are… • Collision between 2 different plasmas • A hydrodynamic surprise Shocks in Hydrodynamics Viscosity, exchanges ∂ρ mass = −∇⋅ [ρv] thermal & kineKc energy ∂t ∂(ρv) momentum = −∇⋅ [ρvv + pI +σ ] ∂t € ⎛ ⎞ ⎡ ⎤ ∂ 1 2 p 1 2 γ energy ⎜ 2 ρv + ⎟ = −∇⋅ ⎢ 2 ρv v + pv +σ ⋅ v⎥ ∂t ⎝ γ −1⎠ ⎣ γ −1 ⎦ € kinec thermal € ⎛ ⎞ ∂vi ∂v j 2 viscous stress tensor σij = µ⎜ + − 3 δij∇⋅ v⎟ ⎝ ∂x j ∂xi ⎠ € ∂ρ mass = −∇⋅ [ρv] ∂t ⎛ v ⎞ ∂(ρv) ∂vi ∂ j 2 momentum σij = µ⎜ + − 3 δij∇⋅ v⎟ = −∇⋅ [ρvv + pI +σ ] ⎜ ∂x ∂x ⎟ ∂t ⎝ j i ⎠ € ⎛ ⎞ ⎡ ⎤ ∂ 1 2 p 1 2 γ energy ⎜ 2 ρv + ⎟ = −∇⋅ ⎢ 2 ρv v + pv +σ ⋅ v⎥ ∂t ⎝ γ −1⎠ ⎣ γ −1 ⎦ € € vs € collision vi vi ∂ρ mass = −∇⋅ [ρv] ∂t ⎛ v ⎞ ∂(ρv) ∂vi ∂ j 2 momentum σij = µ⎜ + − 3 δij∇⋅ v⎟ = −∇⋅ [ρvv + pI +σ ] ⎜ ∂x ∂x ⎟ ∂t ⎝ j i ⎠ € ⎛ ⎞ ⎡ ⎤ ∂ 1 2 p 1 2 γ energy ⎜ 2 ρv + ⎟ = −∇⋅ ⎢ 2 ρv v + pv +σ ⋅ v⎥ ∂t ⎝ γ −1⎠ ⎣ γ −1 ⎦ € € pre-shock (1) post-shock (2) Assume € vi -vs -vs 1. Soluon is staonary in v1 v2 shock frame v 1 v2 2. Regions 1 & 2 are uniform: transform to ref. frame co-moving ρ1 v1 p1 ρ2 v2 p2 w/ shock - shock frame 0 = −∇⋅ [ρv] jump 0 = [[ρvn ]] [[ f ]] = f 2 − f1 0 = −∇⋅ ρvv + pI +σ ˆ [ ] 0 = [[ρvn v + pn]] € ⎡ ⎤ € 1 2 γ 1 2 γ 0 = −∇⋅ ⎢ 2 ρv v + pv +σ ⋅ v⎥ 0 = [[ ρv v + pv ]] € 1 2 n γ −1 n ⎣ γ − ⎦ € nˆ € Jump condions pre-shock (1) post-shock (2) (Rankine-Hugoniot) € € € v1 v2 2. Regions 1 & 2 are v 1 v2 uniform: ⎛ ⎞ ∂vi ∂v j 2 σij = µ⎜ + − 3 δij∇⋅ v⎟ = 0 ρ1 ρ2 ⎝ ∂x j ∂xi ⎠ € 0 = [[ρvn ]] ρvn = const. ˆ 0 = [[ρvn v + pnˆ ]] ρvn [[v]]+[[p]]n = 0 1 2 γ v [[ 1 v 2 γ (p/ )]] 0 0 = [[ 2 ρvnv + γ −1 pvn ]] ρ n 2 + γ −1 ρ = € € € € Case 1: ρvn = 0 - no flow across interface € p2=p1 pressure balance € 0 = [[ρvn ]] ρvn = const. ˆ 0 = [[ρvn v + pnˆ ]] ρvn [[v]]+[[p]]n = 0 1 2 γ v [[ 1 v 2 γ (p/ )]] 0 0 = [[ 2 ρvnv + γ −1 pvn ]] ρ n 2 + γ −1 ρ = € € € € Case 1: ρvn = 0 - no flow across interface € p2=p1 pressure balance € v2 Contact HOT COLD disconnuity v1 p p 1 2 related to linear T1 T2 waves w/ ω=0 entropy mode: [[ρ]]≠0 ρ1 ρ2 shear mode: [[vt]] ≠0 0 = [[ρvn ]] Case 2: ρvn = const.≠ 0 ρv [[v ]] = 0 0 = [[ρvn v + pnˆ ]] n ⊥ 1 2 γ ⇒ v⊥,1 = v⊥,2 = v⊥ 0 = [[ 2 ρvnv + γ −1 pvn ]] € € transform away € (ρv 2 + p)2 γ −1 (γM 2 +1€)2 = = F(M) 2( v) 1 v 3 γ pv γ 2 M 2 ( 1)M€2 2 € ρ ( 2 ρ + γ −1 ) [ γ − + ] v M = Mach € γp/ρ number F(M1) = F(M2 ) € −1 M2 = F (F(M1)) € € pre-shock (1) post-shock (2) −1 M2 = F (F(M1)) super-sonic 2 sub-sonic (γ −1)M + 2 M = 1 2 2γM 2 −γ +1 v1 v2 € 1 c v1 s,2 2 v2 (γ −1)M1 + 2 v2 = 2 cs,1 € v1 (γ +1)M1 ρ (γ +1)M 2 ρ1 ρ2 2 1 = 2 ρ1 (γ −1)M1 + 2 € p 2 2 p 2γM − γ +1 2 = 1 p1 p γ +1 € 1 ⎛ ⎞ ⎛ ⎞ p2 ρ2 s1 s2 − s1 = ln⎜ ⎟ −γ ln⎜ ⎟ s ⎝ p1 ⎠ ⎝ ρ1 ⎠ 2 € € Hypersonic limit M1 >> 1 γ = 5/3 2 (γ −1)M1 + 2 (γ −1) 1 M2 = 2 M2 → = 2γM1 −γ +1 2γ 5 2 v (γ −1)M + 2 v2 (γ −1) 1 2 = 1 → = 2 v (γ +1) 4 € v1 (γ +1)M1€ 1 € ρ (γ +1)M 2 ρ (γ +1) 2 = 1 2 → = 4 ρ (γ −1)M 2 + 2 ρ (γ −1) 1 1 1 € € € 2 p 2γ 2 p2 2γM1 − γ +1 2 → M 3 1 2 = € 1 p2 → 2 × 2 ρ1v1 p γ +1 p1 γ +1 € 1 € € € € Entropy’s tale s2 ≥ s1 pre-shock must be supersonic fluid goes from super- to sub-sonic " # # ! fluid slows down crossing shock ! fluid compresses crossing shock viscosity = 0 ! s2 = s1 ⎛ ⎞ ⎡ ⎤ ∂ 1 2 p 1 2 γ ⎜ 2 ρv + ⎟ = −∇⋅ ⎢ 2 ρv v + pv +σ ⋅ v⎥ t 1 1 ∂ ⎝ γ − ⎠ ⎣ γ − ⎦ viscosity does not appear in jump condiKons (conservaon laws) but it is necessary for transiKon from 1 to 2 € Q: what goes wrong if we set µ = 0 ? Q: is that diff. from taking µ $ 0 ? vs rarefacKon vi vi v -v i s -vs co-moving frame v1 v2 sasfies conservaon v1 v 2 but violates 2nd law TD What happens instead? ρ2 ρ1 cs rarefacKon vi vi vi v t i cs vi A nearby shock M=1 asin(1/M∞) Mach cone M∞ M∞=8 γ = 5/3 M=5.2 49o 8.0 M∞=8.0 Spreiter, Summers & Alksne 1966 Farther afield… shock contact disconKnuity shock Opher v2, ch. 7 1973 (v.3, ch.8) High speed stream colliding with slower wind creates Hundhausen 2 shocks: Forward shock Reverse shock Both move outward (vs>0) Q: which way is for each? nˆ € Shocks in MHD ∂ρ = −∇⋅ [ρv] ∂t ∂(ρv) 1 1 2 = −∇⋅ ρvv - 4π BB + (p + 8π B )I +σ ∂t [ ] € ∂ 1 2 1 1 2 1 2 γ 1 ( 2 ρv + γ −1 p + 8π B ) = −∇⋅ [ 2 ρv v + γ −1 pv − 4π (v × B) × B +σ ⋅ v] ∂ t € kinec thermal magneKc € ∂B = −∇ × [v × B] ∂t 0 = ∇⋅ B € € 0 = −∇⋅ [ρv] 0 = −∇⋅ ρvv - 1 BB + p + 1 B2 I +σ [ 4π ( 8π ) ] € 0 = −∇⋅ 1 ρv 2v + γ pv − 1 (v × B) × B +σ ⋅ v [ 2 γ −1 4π ] € 0 = −∇ × [v × B] € 0 = ∇⋅ B € € v2 B1 B2 v1 0 = −∇⋅ [ρv] ρvn = const. B const. 0 = ∇⋅ B n = 1 1 2 0 = −∇⋅ ρvv - 4π BB + (p + 8π B )I +σ € [ € ] € € 2 ρvn [[v]] − Bn[[B]]/4π +[[p + B /8π]]nˆ = 0 € ρvn [[v⊥]] = Bn[[B⊥]]/4π € v2 € B1 B2 v1 0 = −∇⋅ [ρv] ρvn = const. B const. 0 = ∇⋅ B n = 0 = −∇⋅ ρvv - 1 BB + p + 1 B2 I +σ € [ 4π ( €8π ) ] € € ρvn [[v⊥]] = Bn[[B⊥]]/4π 0 = Bn[[B]] € Case 1: ρvn = 0 Case 1a: Bn = 0 € Case 2: Bn = 0 € v2 Tangen4al Discon4nuity B1 B2 Case 1B: [[ B ]] = 0 v1 Contact Disconnuity Case 1a: Tangenal Disc. 0 = −∇⋅ [ρv] ρvn = const. vn = 0, Bn = 0 0 = ∇⋅ B Bn = const. 0 = −∇⋅ ρvv - 1 BB + p + 1 B€2 I +σ € 2 [ 4π ( 8π ) ] [[p + B /8π]]nˆ = 0 € € J € € An equilibrium current B1 v2 sheet separang uniform B fields w/ different angles. v 1 B2 • total pressure balance • possible shear 2 B1 /8π p2 p 2 1 B2 /8π Case 1B: Contact Disc. 0 = −∇⋅ [ρv] ρvn = const. vn = 0, [[ B ]] = 0 0 = ∇⋅ B Bn = const. 0 = −∇⋅ ρvv - 1 BB + p + 1 B2 I +σ ˆ [ 4π ( 8π € ) ] € [[p]]n = 0 € € 0 = −∇ × [v × B] [[v⊥]] = 0 € € B € HOT COLD € p p2 1 related to linear T1 T2 waves w/ ω=0 entropy mode: [[ρ]]≠0 ρ1 ρ2 shear mode: [[vt]] ≠0 Case 2: perpendicular shock ρvn [[v⊥]] = Bn[[B⊥]]/4π Bn = 0, vn ≠ 0 ⇒ v⊥,1 = v⊥,2 = v⊥ transform away 1 1 2 2 2 0 = −∇⋅ ρvv - 4π BB + (p + 8π B )I +σ € [[ρv + p + B /8π]] = 0 [ ] 1 2 γ 1 € 0 = −∇⋅ 2 ρv v + γ −1 pv − 4π (v × B) × B +σ ⋅ v [ ] 1 2 γ 1 2 € [[ 2 ρvnv + 1 pvn + 4 B vn ]] = 0 € γ − π € 0 = −∇ × [v × B] [[Bvn ]] = 0 v v 1 €2 € HD shock w/ 2 ``fluids’’ B1 B 2 €• plasma: γ=5/3 • mag. field: γ=2 p B1 B2 2 p1 Gabriel 1976 hot Bn = 0 Bn ≠ 0 vn = 0 TangenKal Contact disconKnuity disconKnuity cold vn ≠ 0 Perpendicular General shock MHD shock Case 3: General MHD shock ρvn [[v⊥]] = Bn[[B⊥]]/4π Bn ≠ 0, vn ≠ 0 ⇒ v⊥,1 ≠ v⊥,2 ρvn = const.≠ 0 0 = −∇⋅ [ρv] cannot transform away v⊥ € 0 = ∇⋅ B Bn = const.≠ 0 can transform to 1 1 2 € deHoffman-Teller frame 0 = −∇⋅ ρvv - 4π BB + (p + 8π B )I +σ [ € ] € vn,1 1 2 γ 1 v €= v − B € 0 = −∇⋅ 2 ρv v + γ −1 pv − 4π (v × B) × B +σ ⋅ v dHT ⊥,1 ⊥,1 [ € ] Bn,1 0 = −∇ × [v × B] € In dH-T frame: • v || B € 1 1 B2 € • E1 = v1 X B1 = 0 v1 € • E 2 = v2 X B2 = 0 B v2 1 • v2 || B2 • flow is parallel to field lines on both sides Shock characterized by 3 dimensionless Jump condions in pre-shock parameters: de Hoffman-Teller frame v1 v1n ρvn [[v⊥]] − Bn[[B⊥]]/4π = 0 M = = 1A B / 4πρ B / 4πρ 2 2 1 1 1n 1 [[ρvn + p + B⊥ /8π]] = 0 ⎛ B ⎞ 1 2 γ −1 ⊥1 θ1 = tan ⎜ ⎟ [[ 2 ρvnv + γ −1 pvn ]] = 0 € ⎝ Bn1 ⎠ € p1 € β1 = 2 B1 /8π B2 € v1 € Use 3 jump. conds B v2 β β1 1 2 to solve for € θ2 downstream parameters θ1 MA2 θ2 β2 2 2 ρvn [[v⊥]] − Bn[[B⊥]]/4π = 0 [[(MA −1)tan θ]] = 0 2 2 eliminate θ2 in terms of θ1, MA1 and MA2 etc.