Prof. dr. A. Achterberg, Astronomical Dept., IMAPP, Radboud Universiteit

Transcription

1 Prof. dr. A. Achterberg, Astronomical Dept., IMAPP, Radboud Universiteit

2 Rough breakdown of MHD shocks

3 Jump conditions: flux in = flux out mass flux: ρv n magnetic flux: B n Normal momentum flux: ρv n B Bn + P+ 8π 4π Tangential momentum flux: ρvv n t BB n t 4π Energy flux: ρv n ( V B) + + ( γ 1) ρ 4πρ 4π V γ P B Bn

4 MHD Induction equation in conservative form: Tensor form: Component form: B t B t i = B V V B j ( ) = x ( BV V B) j i j i

5 Jump conditions: flux in = flux out + induction eqn mass flux: ρv n magnetic flux + Ind.: B n ce t = ( V B VB n t t n ) Normal momentum flux: ρv n + P+ B 8π Bn 4π Tangential momentum flux: ρv n V t BB n t 4π Energy flux: ρv n ( V B) V γ P B B + + ( γ 1) ρ 4πρ 4π n

6 There are just enough jump conditions to calculate the Downstream State (subscript ) from the Upstream State (subscript 1)

7 Case 1: parallel shock B = B B B n1 n n t1 = Bt = 0 magnetic field does NOT change! V t1 = V = 0 t

8 Case 1: parallel shock B = B B B n1 n n t1 = Bt = 0 magnetic field does NOT change! V t1 = V = t 0 Consequence: magnetic field completely drops out of the jump conditions! ( γ ) ( ) ( γ ) ( ) ρ s V1 1 = with s ρ1 γ 1 s + γ 1 Cs1 = 1

9 Case : perpendicular shock B V n1 t1 = Bn = 0 V1B 1 = VB = cet Vt 0 = =

10 Case : perpendicular shock B V n1 t1 = Bn = 0 V1B 1 = VB = cet Vt 0 = = Other jump conditions: ρv = ρ V J 1 1 B B ρv + P+ ρv P 8π = + + 8π 1 V γp B V γp B ρv + + = ρv + + ( γ 1) ρ 4πρ ( γ 1) ρ 4πρ 1

11 Interpretation of the V t B n -V n B t rule Tangential field due to MHD condition: E V = B c E y VB VB VB VB = = c c x z z x n t t n

12 Interpretation of the V t B n -V n B t rule Induction Equation: 1 c t 1 c B t = E ( d ) O B = dl E ( ) E E y = t1 t 0

13 Case : perpendicular shock Solution of the jump conditions: Define: 1 compression ratio; r β ρ = = ρ 1 pl B1 A 1 8π P = V = V 1 A1 = = plasma-beta parameter Alfven Mach Number ( γ) r γ γβ ( γ ) r ( γ ) + + pl + 1 A + 1 A = 0

14 Case : perpendicular shock Solution of the jump conditions: Define: 1 compression ratio; r β ρ = = ρ 1 pl B1 A 1 8π P = V = V 1 A1 = = plasma-beta parameter Alfven Mach Number Solutions with r 1 exist when: ( γ) r γ γβ ( γ ) r ( γ ) + + pl + 1 A + 1 A = 0 V C + V = C 1 s1 A1 f1

15 Compression ratio parallel & perpendicular shocks Parallel shock: Supersonic flow Perpendicular shock: Super-fast flow

16 Astrophysical Applications Solar physics: explosions (Solar Flares) Space Weather

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18 The Heliosphere around our Solar System Solar Wind

19 Hot Spots and knots in Jets from Quasars/AGN Supernova Remnants and the production of Galactic Cosmic Rays

20 Not all equilibrium solutions of the MHD equations are stable! Instability means: system moves away from equilibrium state at an accelerated rate (typically: exponential in time!) (In)stability can be determined by two methods: - Energy method - Wave method

21 Try to find the second-order potential energy of a small perturbation: x x + x,t eq eq ( ) ( ) ( ) Ueq Ueq + U1 + U vanishes for an equilibrium! formally of order ξ In MHD: 3 P B U = d x0 + + ρφ γ -1 8 π Thermal energy Magnetic energy Gravitational potential energy

22 3 P B U = d x0 + + ρφ < γ -1 8 π Gravitational Thermal energy Magnetic energy potential energy 0 Instability occurs if you can find a displacement such that it lowers the potential energy of the system!

23 You have to calculate all physical quantities to second order in displacement vector!!!

24 Advantages: - You only need quantities linear in ξ(x,t) - You get the growth rate σ=im(ω) > 0 directly exp( iωt) exp( σt)

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26 Example 1: Kelvin-Helmholz Instability of fluids in relative motion Astrophysical significance: - Dynamics of MHD jets - Planetary Atmospheres (e.g. Jupiter)

27 B B

28 B B δ B ρ V (,) t δp x = + Π t x 4πρ x 4π Inertial force with comoving derivative Magnetic tension force Pressure force (kinetic + magnetic) ( x,) t = 0 Incompressibility condition

29 (,) B V t t x x ρ πρ = = Π Π= x

30 Write displacement and pressure perturbation as: ( x,) t = a() z exp( ikx iωt) Π( x,) t =Π () z exp( ikx iωt) Blue = plane wave part!

31 Π Π d Π Π= + = 0 k Π= 0 x z dz Equation of motion: ax ikπ ( ) ρ ω± kv k V A = Π a z z

32 Solution for the pressure should: 1. Stay finite for all z ;. Be continuous across contact discontinuity at z=0. d Π dz Π exp( k z) for z > 0 0 k Π= 0 Π ( z) = Π 0 k z z < exp( ) for 0

33 ax ikπ ρ ( ω± kv ) k V A = Π a z z Π ( z) exp( ± kz) for k > 0 ikπ ( z) ax ( z) = (+ sign: z < 0, -sign: z > 0) ρ ( ω± kv ) k V A ± kπ ( z) az ( z) = (+ sign: z < 0, -sign: z > 0) ρ ( ω± kv ) k V A

34 You can not draw vacuum bubbles or have the two fluids interpenetrate! kπ 0 kπ 0 az( z = 0 ) = = a ( 0 ) z z = + = ρ ( ω+ kv ) k V A ρ ( ω kv ) k V A

35 ( ) ( ) ρ ω kv k V A + ρ ω+ kv k V A = 0 ω = kv kv = A just the Alfven wave if V = 0 (trivially stable) positive if V < V A (stable case) negative if V > V (KH unstable case) A

36 UNSTABLE if centrifugal force > magnetic tension force ρv R curv > B 4π R curv V B 4πρ = > VA

37 ω ν = kc s wave frequency sound frequency Gray area: Range of unstable velocities

38 1. Instability needs a driver: flow in the case of the KH Instability;. Boundary conditions play a role: You have to solve differential equations! 3. Physical jump conditions at a contact discontinuity give the dispersion relation!

39 Astrophysical Background: Need for a source of turbulence in accretion disks

40 Force Balance in a Geometrically Thin Disk

41 Force balance in vertical (Z-)direction: Hydrostatic Equilibrium Basic assumption: temperature varies only with radius P T( R) ρ GM Z = = ρ 3 Z µ Z R Use Ideal Gas Law Vertical component of gravity Z ρ( RZ, ) = ρ( R,0)exp, ( R) T ( RR ) GM µ 3

42 Force Balance in the radial direction V dv P GM = + Ω RR dr R R ρ ρ ( ) ρ Order-of-magnitude estimates: ρv R P ρv θ = R R R Inertia term Pressure gradient Centrifugal term

43 3 R V R V V R R R V V R GM P P R R R R G G R R M M R θ θ θ ρ ρ ρ ρ ρ ρ ρ = = Ω = Inertia term: Pressure gradient:

44 Conclusion: balance between gravity and centrifugal force in R-direction! GM dθ GM Ω ( RR ) Ω ( R) = = ΩK( R) R dt R ρ ρ 3 Disk rotates with Keplerian circular angular velocity!

45 Time scale for mass flow in a thin accretion disk: 3 tres = Re tff = Re R / GM = V R ν R R Observations: t res t Re ff

46 Angular momentum loss and the radial inflow velocity dλ ρv = with λ RV =Ω( R) R dr visc θ q-component Equation of motion GM Ω ( R) = Ω 3 K ( R) R Radial force balance Specific angular momentum is also Keplerian: λ = λ =Ω = R GM R K K

47 Height-integrated angular momentum equation: dλ ρv = dr visc + + dλ visc visc dz ρv = dz L dr dλ Σ ( RV ) ( R) = dr L visc

48 Height-integrated angular momentum eqn: ( RV ) ( R) d λ Σ = L dr visc Rotation is Keplerian! λ = λ =Ω = R GM R K K dλk Σ ( RV ) ( R) dr = L visc

49 Height-integrated angular momentum eqn: dλk Σ ( RV ) ( R) dr = L visc Rotation is Keplerian! dλk d GM 1 1 = GM R = = ΩK ( R) R dr dr R Σ ( RV ) ( R) visc L = Ω ( RR ) K

50 Conclusion: mass flow is proportional to angular momentum loss Σ ( RV ) ( R) visc L = Ω ( RR ) K visc 4π L M = π RΣ ( RV ) ( R) = Ω ( R) K

51 Viscous Torque Shear stress visc λ d ( R ( R) ) ηr Ω = = t R R R R dr Viscosity coefficient: σ η ρ = ρν 3

52 Finally: the angular momentum equation visc dλ λ 1 3 dω ρv = = ρν R dr t R R dr Integrate over thickness disk: dλ 1 dω dr R R dr 3 Σ V = Σν R

53 Finally: the angular momentum equation dλ 1 dω dr R R dr 3 Σ V = Σν R Use mass conservation: M = π RΣ V = constant d dr d M Ω λ R + π Σ νr = dr 3 ( ) 0 3 d M Ω λ( R) + π Σ νr = constant J dr

54 Radial flow speed and the Reynolds number M = π RΣ V = constant M νσ= 1 3π R i R V( R) 3ν R t res R R = V( R) 3ν

55 Radial flow speed and the Reynolds number t t res orbit R R V( R) = 3ν tres R RVθ Re Ω = = torbit 3πν 3πν 3π π Ω

56 t t res orbit RV Re 3πν 3πν 3π Ω R = θ = From measured disk temperature and accretion rate Re 10 t orbit mol days t res days 10 yr

57 t t res orbit RV Re 3πν 3πν 3π Ω R = θ = From measured disk temperature and accretion rate Re 10 t orbit mol days t res days 10 yr Oops!

58 ~ 5 days High M Visual magnitude Low M Julian day

59 Stability condition: Without magnetic field: d ( ΩR ) dλ dλ = λ = λ < 0 dr dr dr With magnetic field: dω R < 0 d R

60 Radial force without magnetic field: f R GM λ GM R R R Ω R = 3 Keplerian Disk: f R GM = 0 Ω=, λ= 3 R GM R Effect of small axisymmetric perturbation: df R R+ ξ, δ f = ξ = Ω ξ R R R R R dr λ = constant A restoring force! STABILITY!

61 Magnetic tension force keeps Ω constant!

62 f R GM λ GM R R R Ω R = 3 f R GM = 0 Ω=, λ= 3 R GM R df GM R R ξ, δ f ξ + = =Ω ξ + ξ =+Ω 3 ξ R R R R R 3 R R dr Ω= constant R Force now amplifies the perturbation: INSTABILITY!

63 Accretion Disk turbulence and the Magneto-rotational instability (Figure: James M. Stone, Princeton Univ.)