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

Transcription

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

2 Shocks occur in supersonic flows; Shocks are sudden jumps in velocity, density and pressure; Shocks satisfy flux in = flux out principle for - mass flux - momentum flux - energy flux

3 Three conservation laws means three fluxes for flux in = flux out! ( ρv) ( ρv) = 1 Mass flux ( ) ( ρv P ρv P) + = + 1 Momentum flux V γp V γp ρv + = ρv + ( γ 1) ρ ( γ 1) ρ 1 Energy flux Three equations for three unknowns: post-shock state () is uniquely determined by pre-shock state (1)!

4 = 1 V γ P + = = ( 1) γ ρ V γp V γp ρv ρv + = + ( γ 1) ρ ( γ 1) ρ ( ρv) ( ρv) 1 constant

5 1D case: Mach Number s pre-shock flow speed V = = pre-shock sound speed C s 1 Shocks can only exist if s >1! Weak shocks: s =1+e with e<< 1; Strong shocks: s >> 1.

6 V 1 S = = Cs 1 upstream flow speed upstream sound speed Shocks all have S > 1 r ρ ρ V V 1 = = = 1 γ + 1 ( ) γ 1 + ( ) Compression ratio: density contrast P P 1 γ = 1+ 1 γ + 1 ( ) Pressure jump

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8 ( ρv ) ( ρv ) n = 1 n ( ) ( ρv P ρv P) + = + n 1 n n V = V t1 t ( ρvv) ( ρvv) = n t 1 n t t V γp V γp ρvn + = ρvn + ( γ 1) ρ ( γ 1) ρ 1

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10 All relations remain the same if one makes the replacement: V V = V cos θ, = V / C = cosθ 1 n1 1 1 S n n1 s1 S 1 θ is the angle between upstream velocity and normal on shock surface Tangential velocity along shock surface is unchanged V = V sinθ = V = V sinθ t1 1 1 t

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13 Bell X1 Rocket Plane

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15 Diamond shocks in Jet Simulation

16 Fundamental parameter of shock physics: Mach Number normal shock speed n = sound speed V C n1 s1 Rankine-Hugoniot jump conditions: ρ ( γ + ) = = ρ ( ) 1 γ 1 + γ 1 P P 1 1 n γ + 1 ρ 1 n γ ( 1) n γ = P = ρ V γ + 1 γ + 1 ρ 1 n1 Strong shock limit

17 Trinity nuclear test explosion, New Mexico, 1945 Supernova remnant Cassiopeia A

18 Tycho s Remnant (SN 157AD)

19 Assumptions: 1. Explosion takes place in uniform medium with density ρ;. spherical expanding fireball! 3. Total available energy: E. Point explosion + uniform medium: no EXTERNAL scale imposed on the problem!

20 Dimensional analysis: [ E] [ m][ ] [ t] [ ρ] =, = [ m] [ ] 3 Et ρ 1/5 R S is a length! Sedov: fireball radius ~ Sedov radius R S rt () R() t t S /5

21 Steps: 1. Photo dissociation of Iron in hot nucleus star: loss of (radiation) pressure!. Collapse of core under its own weight formation of proto-neutron star when ρ ~ g/cm 3 3. Gravitational binding energy becomes more negative: positive amount of energy is lost from the system! 4. Core Bounce shock formation and ejection envelope

22 Evolution of a massive star (5 solar masses) Core collapse: t ~ 0. s (!) Collapse onset: photo-dissociation of iron

23 Processes around collapsed core

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25 Gravitational binding energy: E gr GM core mass potential Mcore = R GM R core

26 1 1 GM core 46 Egr GM core 10 J Rinit Rfinal Rfinal M 1 M ~ 10 kg, R ~ 10 km =10 m core 30 4 final

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28 neutronization core: 99% into neutrinos p+ e n+ ν e E sn = E = gr J 1% into explosion E snr 44 = 0.01Esn = 10 J

29 Main properties: 1. Strong shock propagating through the Interstellar Medium; (or through the wind of the progenitor star). Different expansion stages: - Free expansion stage (t < 1000 yr) R t - Sedov-Taylor stage (1000 yr < t < 10,000 yr) R t /5 - Pressure-driven snowplow (10,000 yr < t < 50,000 yr) R t 3/10

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31 Energy budget: E GM 3 c 46 grav = 5 10 J Rc 99% into neutrino's 1% into mechanical energy Expansion speed: V 1/ -1/ ESNR ESNR M ej exp = 3000 km/s 44 Mej 10 J 10 M

32 - Expansion decelerates due to swept-up mass; - Interior of the bubble is reheated due to reverse shock; - Hot bubble is preceded in ISM by strong shock: - the supernova blast wave.

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34 Shock relations for strong (high-mach number) shocks: ( γ ) ( ) ρ + 1 s γ + 1 = ρ1 γ 1 s + γ 1 P P 1 V1 ρ1v1 s cs 1 γ P1 γ ( 1) as s γ γ = = s γ + 1 γ+1 P= ρ1v1 γ + 1

35 P γ P = ρ V γ + 1 γ + 1 s 1 ism s Pressure behind strong shock (blast wave) P i = γ 1 e γ 1 E ( ) ( ) SNR i 4π 3 R 3 S Pressure in hot SNR interior

36 At contact discontinuity: equal pressure on both sides! ρ γ + 1 ism V γ 1 ( ) E SNR s 4π 3 Rs 3 This procedure is allowed because of high sound speeds in hot interior and in shell of hot, shocked ISM: No large pressure differences are possible!

37 At contact discontinuity: equal pressure on both sides! ρ γ + 1 ism V γ 1 ( ) E SNR s 4π 3 Rs 3 V s dr 8π s snr = dt 3 1 ρism ( γ ) E 1/ R 3/ s Relation between velocity and radius gives expansion law!

38 1/ 3/ 8π E snr Rs drs dt ( 3 γ 1) ρism Step 1: write the relation as difference equation

39 1/ 3/ 8π E snr Rs drs dt ( 3 γ 1) ρism ( 5/ ) 8π E snr d Rs ( 5 3 γ 1) ρism 1/ dt Step : write as total differentials and

40 1/ 3/ 8π E snr Rs drs dt ( 3 γ 1) ρism ( 5/ ) 8π E snr d Rs ( 5 3 γ 1) ρism 1/ dt integrate to find the Sedov-Taylor solution 1/5 R t C E t C snr /5 s( ) γ, ρism γ /5 5 8π = ( 3 γ 1) 1/5 1.96

41 1 MsnrVs = E snr V = 4π 3 Msnr = Mej + 3 ρismr s shock speed = expansion speed E snr s 4π 3 Mej + 3 ρismrs Deceleration radius R d : R 1/3 1/3 1/3 3Mej Mej nism d = -3 4πρism M 1 cm pc

42 V s 1/ 1/ E 1 1 snr = = V M ej 1 + ( R/ Rd) 1 + ( R/ Rd) /5 t R 3/ t

43 1. Energy is put in gradually: E(t)=L wind t L wind = mechanical luminosity 1 MV w M = r rv r = 4 π ρw( ) w( ) mass loss = constant

44 1. Energy is put in gradually: E(t)=L wind t L wind = mechanical luminosity 1 MV w M = r rv r = 4 π ρw( ) w( ) mass loss = constant. Dimensional analysis: R () t S 1/5 3 1/5 Ett () Lwindt = = ρ ρ t 3/5

45 View from rest frame FW Shock for V w >> V S Towards Star

46 ρ V = ρ V ism S w w V S Sedov: 1/5 dr 3 L 3R = = t = dt 5 ρism 5t S wind /5 S ρ () r Wind properties: M = = L wind w 3 4πrVw πrvw

47 ρ V = ρ V ism S w w r = RTS t /5 ρ ism V Sedov: 9ρ L = /5 ism wind 4/5 S 5 ρism t Wind properties: ρ () rv = L wind w w π rvw

48 Ring Nebula

49 Helix Nebula Eskimo Nebula

50 Eta Carinae Hourglass Nebula

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