Energy build- up & release erup3ons. MHD instabili3es Lecture 2 Jan. 23, 2017

Transcription

1 Energy build- up & release erup3ons MHD instabili3es Lecture 2 Jan. 23, 2017

2 Our mission: explain what has happened here Our tool: magnetohydrodynamics (MHD)

3 MHD describes Plasma as a single fluid (combining e - s & ions) Fluid fully described by densi3es* Mass density ρ(x,t) Momentum density ρ(x,t) u(x,t) u(x,t) is C.M. velocity NOT par3cle velocity Pressure p(x,t) = p e + p i thermal energy density = (3/2)p Slow dynamics ( τ 10-3 sec for Sun) Good conductor: E = E + u B 0 (no charge density! quasi- neutrality ) * instead of par3cle informa3on o\en par3cle veloci3es have Maxwellian dist n w/ moments specified by ρ, u, and p

4 MHD equa3ons Dynamical evolu3on of fluid densi3es & B ρ t = ρ u ( ) $ ρ& % t + u ' )u = p + 1 c ( J B + ρg p t + u p = 3 5 p u + 2 η 3 e J 2 + B t = cη ej u B ( ) J = c 4π B J is NOT related to u mass con3nuity momentum energy induc3on (Faraday+Ohm) Ampere cgs

5 mass accelera3on (of CM) MHD equa3ons Dynamical evolu3on of fluid densi3es & B ρ t = ρ u $ ρ& % p t + u p = 5 3 p u η e J 2 + B t = cη ej u B ( ) J = c 4π B ( ) fluid pressure magne3c (Lorentz) t + u ' )u = p + 1 c ( J B + ρg gravity

6 ρ MHD equa3ons Dynamical evolu3on of fluid densi3es & B t = ( ρ u ) $ ρ t + u ' & )u = p + 1 % ( J B + ρg c p t + u p = 3 5 p u + 2 η 3 e J 2 + B t = cη ej u B ( ) J = c 4π B Faraday ce = ce' u B = cη e J u B Ohm

7 MHD equa3ons Dynamical evolu3on of fluid densi3es & B ρ t = ρ u $ ρ& % ( ) t + u ' )u = p + 1 ( 4π ( B ) B + ρg p t + u p = 5 3 p u + η 6π B 2 + B t = u B ( ) +η 2 B Use Ampere to eliminate J from dynamical eqs. closed eqs. for densi3es & B alone magne3c diffusivity η= c2 4π η e [ cm 2 s - 1 ]

8 Integrated densi3es mass M = V ρ(x)d 3 x V P = momentum V ρ(x)u(x) d 3 x

9 Integrated densi3es Total energy E = " 1 ρ u p + 8π 1 B 2 + ρ Ψ$ # % d 3 x V thermal magne3c bulk kine3c (i.e. KE of fluid NOT of par3cles) g = Ψ gravita3onal

10 Conserva3on of MHD energy 1 E = " 2 ρ u p + 8π 1 B 2 # + ρ Ψ$ % d 3 x V Take d/dt, bring into integral, use MHD equa3ons, integrate by parts. de dt =! $ 1 2 ρ u 2 % u pu + 4π 1 B ( u B ) + ρ Ψu& ' da V enthalpy flux Poyn3ng flux All fluxes vanish if u = 0 on bndry V

11 B Pos. work stressing field from bndry ^ n u de dt =! $ 1 2 ρ u 2 % u pu + 4π 1 B ( u B ) + ρ Ψu& ' da V enthalpy flux Poyn3ng flux All fluxes vanish if u = 0 on bndry V Tangen3al bndry mo3on only: u.da = u n da = 0 de dt = 1 4π! V ( u B )B n da Work done by stressing field from bndry

12 Integrated densi3es Total energy E = " 1 ρ u p + 8π 1 B 2 + ρ Ψ$ # % d 3 x V thermal magne3c bulk kine3c (i.e. KE of fluid NOT of par3cles) g = Ψ gravita3onal thermal magne3c ~ p B 2 / 8π β ~ 10-3 in corona neglect thermal (& grav.) focus on magne3c & kine3c

13 L ρ = g/cm 3 (n e = 10 9 cm - 3 ) B = 100 G p = 0.1 erg/cm 3 L M = V ρ(x)d 3 x (T = 10 6 K) = ρ L 3 = g E = " 1 ρ u p + 8π 1 B 2 + ρ Ψ$ # % d 3 x V E T ~ pl 3 = erg E grav ~ MgL/2 = erg E mag ~ B 2 L 3 /8π = erg u = 10 8 cm/s : E kin ~ Mu 2 /2 = erg

14 CME model ingredients Current = twist in flux rope Increasing twist (current) instability/loe Erup3on converts energy: magne3c kine3c

15 Simple example* arcade field w/ flux rope line current *based on van Tend & Kuperus 1978 and Forbes & Isenberg 1991

16 Simple example arcade field J = B = 0 fix B z (x,0)

17 arcade field J = B = 0 Simple example line dipole: ~ z - 2 z fix B z (x,0) B x

18 Simple example z add line current I h I line current: B x = I h z B x

19 Simple example z ~ z - 1 I Image current required to maintain B z (x,0) I ++++ B x = I h + z B x

20 Simple example z B x = B x Sum all contribu3ons

21 Forces on coronal line current F mag = I B Simple example I F ic line dipole: ~ z - 2 z ~ z - 1 F ar h B x -I

22 Forces on coronal line current F mag = I B Simple example I F ic line dipole: ~ z - 2 z ~ z - 1 F ar h F z F ic - F ar equilibrium height for flux rope B x F net = F ic + F ar ~ z - 1 ~ z - 2 h

23 E mag E mag (h) = E + F net ( h ") d h" h work to place line current E : energy of arcade alone F z F ic - F ar equilibrium height for flux rope h F net = F ic + F ar ~ z - 1 ~ z - 2 h

24 equilibrium unstable to torus instability : E mag E : energy of arcade alone Restoring force from bipole falls off faster (~z - 2 ) than repulsive force from Image current (~z - 1 ) F z - F ar F ic F net = F ic + F ar ~ z - 1 ~ z - 2 h

25 E mag No instability if AR field falls off ~ z - p w/ decay index p < 1 (in cylindrical goem. cri3cal index is p < 1.5) F z - F ar F ic ~ z - p F net = F ic + F ar ~ z - 1 h

26 E mag Q: How would system ever find itself in unstable equilibrium? E : energy of arcade alone F z h F ic h

27 E mag A: it probably wouldn t E : energy of arcade alone F z Increase current I (i.e. twist flux rope) h F ic Flux rube rises h

28 E mag ΔE mag E : energy of arcade alone F z h F ic Flux rube rises h

29 E mag ΔE mag E : energy of arcade alone F z F ic Loss of equilibrium h cri3cal I Flux rube rises h

30 I. Equilibrium From Toy to Reality equilibrium β << 1 ρ β << 1 $ & % t + u ' )u = p + 4π 1 ( ( B ) B + ρg ( B 0 ) B 0 = J 0 B 0 = 0 Force free field B 0 = α(x) B 0 B 0 is parallel to B 0 Special case: α = 0 B 0 = J 0 = 0 B 0 = χ, B 0 = 2 χ = 0 potencal field

31 Equilibria Example 1: constant- α equilbria vector Helmholtz eq. B 0 = α B 0 ( B ) 0 = 2 B 0 = α 2 B 0 Linear eq. constant- α a.k.a. Linear force- free field (LFFF)

32 Equilibria Example 1: constant- α equilbria vector Helmholtz eq. B 0 = α B 0 ( B ) 0 = 2 B 0 = α 2 B 0 Example 1a: cylindrical sym. = Lundquist flux rope 2 B z = α 2 B z (r) B 0 (r) = B 0! " J 0 (αr)ẑ + J 1 (αr) ˆφ # $ B & J B z (r) = B 0 J 0 (αr) Bessel Func. α > 0 right- handed helices 2π/α z

33 Kusano et al Equilibria Example 1b: Cartesian sym. = periodic arcade 2 B x = α 2 B x (y, z) B x (y, z) ~cos(ky) e z k2 α 2 π/k x B 0 = B 0 k 2 α 2 cos(ky) ŷ k +sin(ky)ẑ}e z k2 α 2! α " # k cos(ky)ˆx π/k PIL: B z = 0

34 B(x) = B 0 (x)+! ξ B 0 From Toy to Reality II. Perturb magne3c field B t = u B ( ) +η 2 B η << is very small (Rm >>1) ideal Make small displacement! ξ (x) = u(x) δt ( ) + 1! 2 ξ (! ξ B ) 0 # $ % & { } +! Perturbed field no field lines broken

35 From Toy to Reality III. Perturba3on to Energy E m = 1 8π introduce B(x) 2 d 3 x B(x) = B 0 (x)+! ξ B 0 Expand, integrate by parts, E m = 1 8π 2 B 0 d 3 x 1 4π ( ) + 1! 2 ξ (! ξ B ) 0 % &( B 0 ) B 0 ' (! ξ d 3 x # $ % & { } +! { } d 3 x + 1 (! ξ B ) 2 0 (! ξ B 0 ) (! ξ B ) 0 +" 8π

36 From Toy to Reality III. Perturba3on to Energy First varia3on: vanishes if B 0 (x) is equilibrium E m E m (h 0 ) = 0 ( B 0 ) B 0 = 0 h E m = 1 8π 2 B 0 d 3 x 1 4π % &( B 0 ) B 0 ' (! ξ d 3 x { } d 3 x + 1 (! ξ B ) 2 0 (! ξ B 0 ) (! ξ B ) 0 +" 8π

37 From Toy to Reality III. Perturba3on to energy Energy second varia3on: equilibrium is unstable if any perturba3on ξ(x) can make δ 2 W < 0 E m E m (h 0 ) < 0 h E m = 1 8π 2 B 0 d 3 x 1 4π % &( B 0 ) B 0 ' (! ξ d 3 x { } d 3 x + 1 (! ξ B ) 2 0 (! ξ B 0 ) (! ξ B ) 0 +" 8π δ 2 W{ ξ(x) }

38 Q: can we find perturba3on ξ(x) to make this nega3ve? δ 2 W {! ξ (x)} = 1 8π { (! ξ B ) 2 0 (! ξ B 0 ) (! ξ B )} d 3 0 x

39 Q: can we find perturba3on ξ(x) to make this nega3ve? δ 2 W {! ξ (x)} = 1 8π { (! ξ B ) 2 0 (! ξ B 0 ) (! ξ B )} d 3 0 x always > 0 Not if B 0 = poten3al fields are stable Only if B 0 = J 0 is large enough Take- home message: To be unstable a force- free equilibrium must have current exceeding some threshold value J 0

40 Applica3on to constant- α equilibria: B 0 = α B 0 δ 2 W {! ξ (x)} = 1 8π { (! ξ B ) 2 0 α (! ξ B 0 ) (! ξ B )} d 3 0 x Find* ξ & µ s.t. (! ξ B ) 0 = µ (! ξ B ) 0 δ 2 W = µ(µ α) 8π! ξ B 0 2 d 3 x δ 2 W < 0 if we can find* 0 < µ < α * Will depend on equilibrium, boundary condi3ons, and α

41 Example 1a: Lundquist flux rope B 0 (r) = B 0! " J 0 (αr)ẑ + J 1 (αr) ˆφ # $ rigid conductor Take eg. L = 4a periodic a 2π/α z (! ξ B ) 0 = µ (! ξ B ) 0 µ = L Unstable when! ξ ~ cos(φ 2π z / L) α > L kink mode (m=1) αl 2π = 2.01 L turns

42 Linton et al Kink mode instability Kliem et al. 2004

43 Non- erup3on observed by TRACE on May turns Torok & Kliem 2005

44 Weaker field above full erup3on 2.5 turns Torok & Kliem 2005

45 Example 1b: Cartesian sym. = periodic arcade! ξ B 0 = L π z cos( π y L y )cos( πz L z )ŷ + L π y sin( π y L y )sin( πz L z )ẑ + µ 1 cos( π y L y )sin πz L z!! ( ξ B ) 0 = µ ( 1 ξ B0 ) L y π µ 1 = ( ) 2 Ly + ( L π ) 2 z ( ) ˆx Pitchfork bifurca3on L z α=0 0<α<µ 1 α>µ 1

46 Instability Unstable equilibrium δ 2 W < 0 (ideal instability: η=0) How does system end up in unstable equilib? Evolu3on can change stable unstable Termed a bifurca3on Pitchfork bif n: 2+ new stable equilibria are born system usually drops into new stable equilibrium Slow evolu3on small energy release

47 Loss of equilibrium Saddle- node bifurca3on Stable & unstable equilibria annihilate Energy release: indep t of evolu3on speed E M U E M S ΔE M

48 Instability in ac3on Mikic et al Reconnec3on produces island Expanding equilibrium α=0 Start w/ poten3al field (α=0) Increase current by imposing slow boom bndry Cross stability threshold (probably not LOE)

49 Instability in ac3on Mikic et al E M Energy build- up by bndry shear E K

50 Instability? η=0 Mikic & Linker 1994 E M η 0

51 Back to the toy E mag ΔE mag E : energy of arcade alone F z F ic Flux rube rises Loss of h cri3cal I h

52 Erup3on

53 Scenario requires E BUT dφ dt =! E dl = E (h) L 0 y y E = E' u B = E' 0 E' = ηj X- point Since η is small Rm = ul η ~ 1010 >>1

54 Erup3on w/ E =0

55 Current sheet B z J y (z) B z discon3nuous X- point CS Integral: total current in sheet Exerts downward force on line current Balances upward force from image: CS is new equilibrium Equivalent: tension from overlying field holds flux rope down

56 stable Lin & Forbes 2000 torus unstable bipole strength λ

57 stable E M Lin & Forbes 2000 torus unstable h bipole strength λ

58 stable E M Lin & Forbes 2000 torus unstable h bipole strength λ

59 E M Q: is erup3on ( h ) possible w/ E =0? h A: No. h produces open field r appears like monopole B mp ~ r - 1 E open = 1 8π B 2 r dr dφ ~ dr r Aly- Sturrock conjecture: E M < E open NB: spherical B mp ~ r - 2 E open <

60 Erup3on via reconnec3on Assume E CS (more later) Φ beneath CS increases Downward force decreases (reconnec3on reduces overlying flux) Flux rope rises Flare signatures produced by E

61 Erup3on via reconnec3on

62 CME model ingredients Current = twist in flux rope Increasing twist (current) instability/loe Erup3on converts energy: magne3c kine3c Frozen- flux leads to current sheet (CS) CS eliminates overlying flux Can lead to instability Can permit further erup3on Will produce flare signatures (later)

63 Amari et al. 2001

64 From Chen 2011 adapted from Fan & Gibson 2007

65 Van Driel- Gesztelyi & Culhane 2009 Forbes 2000

66 Summary Erup3on governed by MHD MHD instability (or LoE) releases magne3c energy kine3c energy LoE can lead to sig. energy release Reconnec3on prob. plays important role Next What is reconnec3on & what role does it play?