Magnetohydrodynamics of pulsar winds and plerions Yuri Lyubarsky

Transcription

1 Magnetohydrodynamics of pulsar winds and plerions Yuri Lyubarsky Ben-Gurion University, Israel

2 Pulsar magneto sphere e Pulsar wind +, e, (ions?), electro-magnetic fields Pulsar wind nebula

3 j j v B v v j v v j B v j v j v

4 The hoop stress F j B Magnetic hoop stress F = 1 c j B Total force 1 F E + j B = ρ e c 1 E + c v B = 0 As v c the electric and magnetic forces nearly cancel each other

5 In the far zone, the magnetic field is nearly azimuthal

6 Obliquely rotating magnetosphere B B cp equatorial plane B B

7 How the electromagnetic energy is transformed into the plasma energy? Non-oscillating fields: no energy release mechanism Waves: various dissipation mechanisms (Usov 1975; Michel 1982, 1994; Coroniti 1990; Melatos & Melrose 1996; Lyubarsky & Kirk 2001; Lyubarsky 2003; Kirk & Skjaeraasen 2003; Melatos & Skjeraasen 2004) B 1 r n 1 2 r v B λ j 1 current r j n r

8 Observations suggest that the energy flows from the pulsar predominantly within the equatorial belt What theory says about the angular distribution of the energy flux in the pulsar wind?

9 Split monopole solution Michel (1973) aligned rotator Bogovalov (1999) oblique rotator f w = r f (sin θ + ), 0 σ

10 B B equatorial plane magnetosphere B current sheet B In the equatorial belt, most of the energy is transferred by alternating electro-magnetic field

11 The fate of the alternating field 1. Dissipation in the wind: very difficult, extreme assumption (Lyubarsky&Kirk 2001; Kirk&Skjaeraasen 2003) 2.Dissipation at the termination shock: driven reconnection (Lyubarsky, in progress)

12 The shock in a striped wind (1.5D PIC simulations) γ = 7 σ =2.5 N = 2 400,000

13

14 B 2 /8π

15 MHD flow beyond the termination shock is determined only by the total energy flux and the mean magnetic field in the wind

16 The mean field=0 at the equator and at the axis

17 Origin of the get-torus structure (Lyubarsky 2002) magnetic collimation disk disk termination shock

18 MHD simulations of the pulsar wind nebula Komissarov & Lyubarsky 2003 f w = r f 2 (sin θ + σ ); 4πf ξ 2 0 θ B = sinθ 1 ; ξ 1 c r π σ = 0.1ξ 2

19 Pulsar plasma fills in the cavity within the expanded cold envelope magnetic field and velocity expansion velocity 5000 km/s ξ = σ =

20 Gas pressure and velocity field around the termination shock sprite towards the observer rim shock bright arch --DISK--- fluff termination shock Mach belt c

21 Magnetic pressure/gas pressure

22 Simulated images σ=0.004 σ=0.009 σ=0.025

23 Simulated image, σ=0.009, with magnetic field at the axis sprite sprite fluff bright arch jet rim shocks

24 Chandra image of the Crab Nebula Bright Arch fluff Counterjet Polar Jet

25 Particle acceleration at the termination shock in a striped wind

26 N(E) electron energy spectrum E -1.6 injection spectrum E -1.6 E E, GeV Radio emitting electrons are accelerated now in the same region as the ones responsible for optical to X-ray emission (Gallant & Tuffs; Bietenholtz, Frail & Hester)

27 Difference image at 4615 MHz (1998 Aug 9-Oct 13)

28 1.6 N( E) = KE from 100 E Emin MeV to 1 TeV br N K = N( E) de = 0.6E 0.6 min ε = EN E = 0.4 ( ) de K 0.4E br Particle acceleration in the standard (kinetic energy dominated) shock Γ shock N(E) Γmc 2 Fermi acceleration at ultra-relativistic shocks: N( E) (Bednarz & Ostrowski 1998; Gallant & Achterberg 1999; Kirk, Guthman, Gallant & Achterberg 2000) E = KE ( )

29 1. Gallant, van der Swalluw, Kirk, Achterberg (2002): Ion dominated wind 1 Γ 100; N & 10 s ; & ± p N e s N & GJ = s 1 2. Lyubarsky (2003): Dissipation of the Poynting flux at the termination shock

30 Conclusions 1. Most of the energy is transferred in the equatorial belt by alternating magnetic fields 2. Magnetization of the postshock flow is determined only by the mean magnetic field (=0 at the equator) 3. Termination shock is highly non-spherical 4. The jet is formed beyond the termination shock

31 Unsolved problems 1. Azimuthal symmetry of the internal ring. 2. Wisps 3. Flat spectrum of the radio emitting electrons

32 rl = ε eb δ ε eb max 0 E = B ε c v

33 Crab s jet Chandra HST difference image