Shocks in Heliophysics. Dana Longcope LWS Heliophysics Summer School & Montana State University

Transcription

1 Shocks in Heliophysics Dana Longcope LWS Heliophysics Summer School & Montana State University

2 Shocks in Heliophysics Vasyliunas, v.1 ch.10

3 Shocks in Heliophysics Tsuneta and Naito 1998 Forbes, T.G., and Malherbe 1986 (v.2 ch. 6)

4 Shocks in Heliophysics shock Gold 1962 Cane, Reames, and von Rosenvinge 1988 (v. 2 ch. 7)

5 Shocks in Heliophysics Forbes 2000 (v. 2, ch. 6)

6 Shocks in Heliophysics Crooker, Gosling, et al (v3, ch. 8)

7 Shocks in Heliophysics Opher v2, ch. 7

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9 Shocks on Earth

10 Shocks are Caused by a collision between 2 different plasmas A hydrodynamic surprise: one plasma is not ``expecting the collision

11 Shocks in Hydrodynamics mass momentum energy ρ t (ρv) t t = [ ρv] [ ] = ρvv + p I + σ Viscosity, exchanges thermal & kinekc energy 1 2 ρv 2 + p 1 = 2 ρv 2 v + γ γ 1 γ 1 pv + σ v kinekc thermal viscous stress tensor σ ij = µ v i + v j x j x i 2 3δ ij v

12 mass momentum energy ρ = [ ρv] t (ρv) t t [ ] = ρvv + p I + σ σ ij = µ v i + v j x j x i 1 2 ρv 2 + p 1 = 2 ρv 2 v + γ γ 1 γ 1 pv + σ v 2 3δ ij v v s v i collision v i

13 mass momentum energy ρ = [ ρv] t (ρv) t t [ ] = ρvv + p I + σ σ ij = µ v i + v j x j x i 1 2 ρv 2 + p 1 = 2 ρv 2 v + γ γ 1 γ 1 pv + σ v 2 3δ ij v pre- shock (1) - v s v 1 v 2 v 1 v 2 post- shock (2) v i - v s transform to ref. frame co- moving w/ shock - shock frame Assume 1. SoluKon is stakonary in shock frame 2. Regions 1 & 2 are uniform: ρ 1 v 1 p 1 ρ 2 v 2 p 2

14 0 = [ ρv] [ ] 0 = ρvv + p I + σ jump [[ f ]] = f 2 f 1 0 = [[ρv n ]] 0 = [[ρv n v + pˆ n ]] 1 0 = 2 ρv 2 v + γ γ 1 pv + σ v n ˆ pre- shock (1) post- shock (2) 0 = [[ 1 2 ρv n v 2 + γ γ 1 pv n ]] Jump condi4ons (Rankine- Hugoniot) v 1 v 2 v 1 v 2 ρ 1 ρ 2 2. Regions 1 & 2 are uniform: σ ij = µ v i + v j x j x i 2 3 δ ij v = 0

15 0 = [[ρv n ]] 0 = [[ρv n v + pˆ n ]] 0 = [[ 1 2 ρv n v 2 + γ γ 1 pv n ]] ρv n = const. ρv n [[v]] +[[p]]ˆ n = 0 ρv n [[ 1 2 v 2 + γ γ 1 ( p /ρ)]] = 0 Case 1: ρv n = 0 - no flow across interface p 2 =p 1 pressure balance

16 0 = [[ρv n ]] 0 = [[ρv n v + pˆ n ]] 0 = [[ 1 2 ρv n v 2 + γ γ 1 pv n ]] ρv n = const. ρv n [[v]] +[[p]]ˆ n = 0 ρv n [[ 1 2 v 2 + γ γ 1 ( p /ρ)]] = 0 Case 1: ρv n = 0 - no flow across interface p 2 =p 1 pressure balance v 1 HOT p 1 p 2 COLD v 2 Contact discon4nuity T 1 T 2 ρ 1 ρ 2 related to linear waves w/ ω=0 entropy mode: [[ρ]] 0 shear mode: [[v t ]] 0

17 0 = [[ρv n ]] 0 = [[ρv n v + pˆ n ]] 0 = [[ 1 2 ρv n v 2 + γ γ 1 pv n ]] (ρv 2 + p) 2 ( ) = γ 1 2(ρv) 1 2 ρv 3 + γ γ 1 pv 0 = [[ρv n 2 + p]] Case 2: [ ] = F(M) (γm 2 +1) 2 γ 2 M 2 (γ 1)M ρv n = const. 0 ρv n [[v ]] = 0 v,1 = v,2 = v transform away M = v γp/ρ Mach number F(M 1 ) = F(M 2 ) M 2 = F 1 ( F(M 1 ))

18 pre- shock (1) post- shock (2) super- sonic sub- sonic v 1 v 2 M 2 = F 1 ( F(M 1 )) M 2 = (γ 1)M γM 2 1 γ +1 v 1 c s,1 c s,2 v 2 v 2 = (γ 1)M v 1 (γ +1)M 1 ρ 1 ρ 2 ρ 2 = (γ +1)M 1 ρ 1 (γ 1)M p 1 p 2 p 2 = 2γM 2 1 γ +1 p 1 γ +1 s 1 s 2 s 2 s 1 = ln p 2 γ ln ρ 2 p 1 ρ 1

19 Hypersonic limit M 1 >> 1 γ = 5/3 M 2 = (γ 1)M γM 2 1 γ +1 M 2 (γ 1) 2γ = 1 5 v 2 = (γ 1)M v 1 (γ +1)M 1 ρ 2 = (γ +1)M 1 ρ 1 (γ 1)M p 2 = 2γM 2 1 γ +1 p 1 γ +1 2 v 2 (γ 1) v 1 (γ +1) ρ 2 (γ +1) ρ 1 (γ 1) p 2 2γ p 1 γ +1 M 2 1 = 1 4 = 4 p ρ 1 v 1 2

20 Entropy s tale s 2 s 1 pre- shock must be supersonic fluid goes from super- to sub- sonic " " " fluid slows down crossing shock fluid compresses crossing shock t viscosity = 0 s 2 = s ρv 2 + p 1 = 2 ρv 2 v + γ γ 1 γ 1 pv + σ v viscosity does not appear in jump condikons (conservakon 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? µ v x const.

21 A nearby shock asin(1/m ) M=1 Mach cone M =8 γ = 5/3 M M= o M =8.0 Spreiter, Summers & Alksne 1966

22 Farther afield (the picture before May) T 1 = 7000 K c s1 =14 km/s v 1 = 25 km/s shock contact disconknuity shock Axford & Suess 1994 p ρ ˆ n ˆ n ISM Opher v2, ch. 7

23 Hundhausen 1973 (v.3, ch.8) ˆ n ˆ n High- speed stream colliding with slower wind creates 2 shocks: Forward shock Reverse shock Both move outward (v s >0) Q: which way is ˆ for each? n

24 Shocks in MHD ρ t (ρv) t t = [ ρv ] [ ] ( 8π B2 ) = ρvv - 1 BB + p + 1 4π ( 1 2 ρv p + 1 γ 1 8π B2 ) 1 = kinekc thermal magnekc B t = v B [ ] 0 = B I + σ 2 ρv 2 v + γ pv [ 1 γ 1 4π (v B) B + σ v]

25 0 = [ ρv] [ ] ( 8π B 2 ) 0 = ρvv - 1 4π BB + p + 1 I + σ [ ] 1 0 = 2 ρv 2 v + γ γ 1 pv 1 4π (v B) B + σ v 0 = [ v B] 0 = B v 2 B 1 B 2 v 1

26 0 = [ ρv] 0 = B 0 = [[ρv n ]] 0 = [[B n ]] [ ] ( 8π B2 ) 0 = ρvv - 1 BB + p + 1 4π I + σ ρv n = const. B n = const. ρv n [[v]] B n [[B]]/4π +[[p + B 2 /8π]]ˆ n = 0 ρv n [[v ]] = B n [[B ]]/4π v 2 B 1 B 2 v 1

27 0 = [ ρv] 0 = B [ ] ( 8π B2 ) 0 = ρvv - 1 BB + p + 1 4π I + σ ρv n = const. B n = const. ρv n [[v ]] = B n [[B ]]/4π Case 1: ρv n = 0 Case 2: B n = 0 v 2 0 = B n [[B]] Case 1a: B n = 0 Tangen4al Discon4nuity B 1 v 1 B 2 Case 1b: [[ B ]] = 0 Contact Discon4nuity

28 Case 1a: Tangen4al Disc. v n = 0, B n = 0 [ ] ( 8π B2 ) 0 = ρvv - 1 BB + p + 1 4π I + σ 0 = [ ρv] 0 = B ρv n = const. B n = const. [[p + B 2 /8π]]ˆ n = 0 J v 1 B 1 v 2 B 2 An equilibrium current sheet separakng uniform B fields w/ different angles. total pressure balance possible shear B 12 /8π p 1 p 2 B 22 /8π

29 Case 1b: Contact Disc. v n = 0, [[ B ]] = 0 [ ] ( 8π B2 ) 0 = ρvv - 1 BB + p + 1 4π I + σ 0 = [ ρv] 0 = B ρv n = const. B n = const. [[p]]ˆ n = 0 0 = [ v B] [[v ]] = 0 HOT B COLD p 1 p 2 T 1 T 2 ρ 1 ρ 2 related to linear waves w/ ω=0 entropy mode: [[ρ]] 0 shear mode: [[v t ]] 0

30 Case 2: perpendicular shock B n = 0, v n = [ ρvv - 4π BB + p + 1 I + σ ] ( B 2 ) 8π [ ] 1 0 = 2 ρv 2 v + γ γ 1 pv 1 4π (v B) B + σ v ρv n [[v ]] = B n [[B ]]/4π v,1 = v,2 = v transform away [[ρv 2 + p + B 2 /8π]] = 0 [[ 1 2 ρv n v 2 + γ γ 1 pv n + 1 4π B 2 v n ]] = 0 0 = [ v B] [[Bv n ]] = 0 v 1 v 2 B 1 B 2 HD shock w/ 2 ``fluids plasma: γ=5/3 mag. field: γ=2 B 1 B 2 p 1 p 2

31 Gabriel 1976 B n = 0 B n 0 hot v n = 0 TangenKal disconknuity v n 0 Perpendicular shock Contact disconknuity General MHD shock cold

32 Case 3: General MHD shock B n 0, v n 0 0 = [ ρv] 0 = [ v B] ρv n = const. 0 0 = B B n = const. 0 0 = [ ρvv - 1 BB + ( p + 1 4π 8π B2 ) I + σ ] 1 0 = 2 ρv 2 v + γ γ 1 pv 1 4π (v B) B + σ v B 1 v 1 [ ] v 2 B 2 ρv n [[v ]] = B n [[B ]]/4π v,1 v,2 cannot transform away v can transform to dehoffman- Teller frame v dht = v,1 v n,1 B n,1 B,1 In dh- T frame: v 1 B 1 E 1 = v 1 X B 1 = 0 E 2 = v 2 X B 2 = 0 v 2 B 2 flow is parallel to field lines on both sides

33 Case 3: General MHD shock B n 0, v n 0 0 = [ ρv] 0 = [ v B] ρv n = const. 0 0 = B B n = const. 0 0 = [ ρvv - 1 BB + ( p + 1 4π 8π B2 ) I + σ ] 1 0 = 2 ρv 2 v + γ γ 1 pv 1 4π (v B) B + σ v [ ] v 1 B 1 v 2 B 2 ρv n [[v ]] = B n [[B ]]/4π v,1 v,2 cannot transform away v can transform to dehoffman- Teller frame v dht = v,1 v n,1 B n,1 B,1 In dh- T frame: v 1 B 1 E 1 = v 1 X B 1 = 0 E 2 = v 2 X B 2 = 0 v 2 B 2 flow is parallel to field lines on both sides

34 Jump condi4ons in de Hoffman- Teller frame ρv n [[v ]] B n [[B ]]/4π = 0 [[ρv n 2 + p + B 2 /8π]] = 0 [[ 1 2 ρv n v 2 + γ γ 1 pv n ]] = 0 v 1 B 1 θ 1 v 2 θ 2 B 2 β 1 β 2 Shock characterized by 3 dimensionless pre- shock parameters: M 1A = v 1 B 1 / 4πρ 1 = θ 1 = tan 1 β 1 = B 1 B n1 p 1 B 1 2 /8π v 1n B 1n / 4πρ 1 = tan 1 v 1 v n1 Use 3 jump. conds to solve for downstream parameters M A2 θ 2 β 2

35 ρv n [[v ]] B n [[B ]]/4π = 0 [[(M A 2 1)tanθ]] = 0 eliminate θ 2 in terms of θ 1, M A1 2 and M A2 2 etc. unkl v 1 B 2 B 1 v 2 β 1 β 2 θ 2 θ 1

36 θ 1 =30 0 β 1 =0.05 NB: ρ 2 /ρ 1 = M A1 2 / M A2 2 region M A2 > M A1 (ρ 2 < ρ 1 ) disallowed by 2 nd law thermo. M A2 = M A1 reps. 3 linear waves of MHD each shock: flow faster (slower) than lin. wave in region 1 (2)

37 3 MHD shocks fast: 1 < M A2 < M A1 intermediate: M A2 < 1 < M A1 slow: M A2 < M A1 < 1

38 3 MHD shocks fast: 1 < M A2 < M A1 intermediate: M A2 < 1 < M A1 slow: M A2 < M A1 < 1 [[(M A 2 1)tanθ]] = 0 M A1 2 M A = tanθ 2 tanθ 1 = B 2 B 1 fast slow intermediate B 1 B 2 B 1 B 2 B 1 B 2 θ 1 θ 2 θ θ 1 2 θ 2 > θ 1 > 0 B θ 1 > θ 2 > 0 1 < B B 2 1 > B 2 θ 1 θ 1 > 0 > θ 2 θ 2

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41 Hypersonic limit 4 MagneKc pressure: only x16 Plasma pressure, no limit

42 Fast Shocks Observed Voyager r=1.6 AU (v esc = 34 km/s) H- sphere: θ 1 =84 o, β 1 =15 T N R 305 km/s 282 km/s cm cm - 3 M A1 = 8.0/cos(θ 1 ) Vinas & Scudder 1986

43 ISEE- 1 crosses bow shock M A1 = 8.1/cos(θ 1 ) θ 1 =74 o, β 1 =6 Vinas & Scudder 1986

44 2 FM shocks 2 (1?) TDs r=3.5 AU Gosling et al. 1994

45 Solar Fast Shocks Type II Liu et al n e =10 7 cm - 3 n e =10 6 cm - 3 Band spliung reveals both n e1 & n e2 Type III n e2 /n e1 =1.59 M A1 ~

46 Solar Fast Shocks Type II Liu et al ground n e =10 7 cm n - 3 e =3 x 10 6 cm - 3 n e =10 6 cm - 3 Type III Space (STEREO) Band spliung reveals both n e1 & n e2 n e2 /n e1 =1.59 M A1 ~

47 Science May 10, 2012 v fms = 26.8 km/s Breaking news from IBEX v fms = 21.4 km/s

48 Slow shocks: reconnection Lin & Lee 1999 SS p

49 Model reconnection Tsuneta and Naito 1998 Forbes, T.G., and Malherbe 1986 (v.2 ch. 6)

50 ISEE- 1 at the magnetopause: SS from reconneckon Walthour et al. 1994

51 Intermediate Shocks ReconnecKon between skewed fields B z Lin & Lee 1999 B y Petschek & Thorne 1967

52 Entropy s Tale the sequel A shock converts kinetic to thermal energy How? What is the µ-physics? Collisions (i.e. viscosity) high enough density Otherwise (story not finished yet) Instabilities, chaotic particle motion, Related to acceleration of non-thermal particles Easier if particles gyrate near shock: quasi-perp. thermalizes easier than quasi-parallel

53 Summary

54 v s rarefackon v i v i - v s v i v 1 v 2 - v s co- moving frame v 1 v 2 saksfies conservakon but violates 2 nd law TD ρ 1 ρ 2 What happens instead?

55 c s rarefackon v i v i v i t v i c s v i

56 Forward shock Voyager 2 at r=2.17 AU θ 1 =23 o, β 1 =1.7, M A1 =2.4 upstream downstream Scudder et al. 1984