7 Stellar Winds & Supernovae Remnants

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7 Stellar Winds & Supernovae Remnants 36 Physics 426 Notes Spring 2009 7 Stellar Winds & Supernovae Remnants Whether or not a particular flow satisfies this condition depends on the starting conditions, such as with what Now let’s look at some supersonic flow situations velocity and temperature it left the stellar surface, and which involve shocks. You remember that we worked also what the boundary conditions at large distances with smooth transonic flows last term – for instance are. If it does not start in such a way to satisfy this the solar wind. These are rare; it’s very easy to form condition, it either stays subsonic (corresponding to fi- shocks when supersonic flows are decelerated or bent nite pressure at infinity), or cannot establish a steady by their surroundings. In this chapter we’ll work with flow. 2-1/2 examples: stellar wind shocks, and two types of Further, the solution beyond the sonic point depends supernova remnants. • on the temperature structure of the wind. The only so- 7.1 Stellar winds and the surrounding ISM lutions with dv/dr > 0 for r > rs are those for which 2 cs(r) drops off more slowly than 1/r; it is only these We worked with smooth, transonic stellar wind flow for which the right-hand side stays positive. In the case last term ... and found a solution in which the outer 2 of an isothermal wind, with cs = constant, (7.3) can be regions of the wind flow are supersonic. But this can’t solved in the limit r rs: carry on forever. At some point the wind flow must run " into the ambient ISM. We expect the deceration to lead v2(r) 4c2 ln r + constant (7.5) # s to an outer shock in the wind; what are the details? By way of review, I’ll repeat the basics of the inner-wind Thus, the wind will be supersonic, by a factor of a few, as r . The question of how the solar wind man- solution here. → ∞ ages to stay nearly isothermal is not solved; it is prob- 7.1.1 The basic solution ably due to energy transport by some sort of waves (MHD or plasma waves, for instance) which are gen- Mass conservation in a steady, spherical flow is • erated in the photosphere and damped somewhere far ρvr2 = constant; or, out in the wind. 1 dρ 1 dv 2 + + = 0 (7.1) ρ dr v dr r 7.1.2 The outer shock while the momentum equation becomes in this case The pressure in this supersonic wind is dropping with (noting that gravity from the central star is important), radius (since ρ 1/vr2, with v being only slowly ∝ varying; thus p ρT 1/r2, approximately, in dv dp GM ∝ ∝ ρv + = ρ (7.2) an isothermal wind). Therefore, when the wind pres- dr dr − r2 sure is close to the ISM pressure, the wind must slow 2 Writing dp/dr = csdρ/dr, these two equations com- down. At this outer boundary, we expect some sort of bine to give the basic wind equation, shock transition, since the wind is supersonic. Past this c2 dv 2c2 GM shock, the hot, shocked wind-gas will expand into the v s = s (7.3) ISM (at about its own sound speed, to start); as long − v dr r − r2 ! " as this expansion is supersonic relative to the ISM, the This does not have analytic solutions over the whole expanding hot gas will drive a “snowplowed” shell of range of r. However, we can learn quite a bit about the ISM, and a second shock, out into the ISM. nature of the solutions simply by inspection of (7.3), as A cartoon of this region, at some point in time, would follows. be that in Figure 7.1. Let region “a” be the wind; S1 First, the left hand side contains a zero, at v2 = c2. be the inner shock; region “b” be the wind-gas which • s If we want to consider well-behaved flows, that is to has been through the shock; C be the contact sur- say those in which the derivative dv/dr does not blow face between the wind and the ISM; region “c” be the up, then the right hand side of (7.3) must go to zero at shocked ISM; and S2 be the outer shock (moving into the same point. This defines the condition that must be the ISM). We expect S1 to be an adiabatic shock (since met at the sonic point: the wind is probably hot and low density, and thus will have a long cooling time); region “b” will contain hot, 2 2 GM 2 v = c at r = rs = (7.4) 3 mvwind 7 s 2c2 shocked wind, with Tb several 10 K s ∼ 16 kB ∼ × Physics 426 Notes Spring 2009 37 1 (noting that m = 2 mp is the mean mass per particle if bit of algebra tells us that α = 3/5 is the allowed solu- region “b” is fully ionized). The outer shock will prob- tion, and we can also find an expression for A. Thus, ably be isothermal, since the ISM is denser and cooler the solution is than the wind. Thus, the shocked ISM will be in a 1/5 E˙ thin shell, containing all of the original ISM that lay R(t) = 0.76 wind t3/5 (7.9) between S2 and the star. % ρo & S 2 and C ˙ 1/5 dR Ewind −2/5 S 1 v(t) = = 0.46 t (7.10) dt % ρo & a Thus, the shell decelerates with time, as it ought to if R(t) b it is picking up more and more ISM. One can then use c d these, and (7.6), to work out the pressure acting on the outer shell: Figure 7.1 Cartoon of the structure of a stellar wind, and 7 2 −4/5 its interaction with the ISM. Left: the shock structure within pb(t) = A ρot (7.11) the wind. Right: The outer shell of dense, snowplowed ISM. 25 From Dyson & Williams figures 7.3 and 7.4. so that the outer pressure drops with time. Finally, one can also work out the kinetic energy of the shell, which To start, consider the position of this shell, R(t), as a must give the energy input to the general ISM from the function of time. We start with a force equation. The wind. This turns out to be mass is the ISM shell is the mass that was originally 2 2π dR within R(t): 4π ρ R3(t). The force acting on the shell R3ρ 0.2E˙ t (7.12) 3 o 3 o dt # wind is just the pressure behind it, which drives it outward: ! " so that about 20% of the wind energy goes to the ISM. d 4π 3 dR 2 (The rest goes to heating the bubble, and to “pdV” R ρo = 4πR pb (7.6) dt 3 dt work). ! " if pb is the pressure, in the outer part of region “b”, 7.1.3 What about the inner shock? which acts on the shell. Next, we need an equation The location of the inner shock, S1, is determined by a for pb(t). The energy within region “b” – which we will take to be nearly all of the volume within R(t) combination of the jump conditions, applied at S1, and – is 2πp R3(t); the rate of change of this energy the pressure at the outside of the region, pb, as follows. # b is given by the difference between the input (from the The jump conditions, at S1, are ρsb = 4ρsa, if “sb” • wind) and the “pdV” work done by the expansion: and “sa” subscripts refer to postshock (region “b”) and preshock (region “a”), respectively. Also, vsb = vsa/4 d 3 ˙ d 4π 3 – here we assume 1, the strong shock limit, 2πR pb = Ewind pb R (7.7) M " dt − dt 3 and also an adiabatic shock. At the shock, momentum ! " # $ conservation1 tells us that ρv2 + p is conserved; thus, if E˙ wind is the energy input rate from the wind, as- 3 2 3 2 sumed constant. These two equations can be com- p = ρsav + psa ρsav sb 4 sa # 4 sa bined, for instance by eliminating pb, to get where we have used the fact that v c in the wind sa " s 3 2 3 ˙ region. 4 d R 3 dR d R 2 dR 3 Ewind aR 3 + bR 2 + cR = dt dt dt dt 2π ρo In region “b”, where we can ignore gravity (com- ! " • 3 (7.8) pared to the internal energy, 2 p), the momentum equa- where a, b, c are numerical constants (that you’ll find in tion is dv dp the homework). Noting that each term on the left hand ρv + 0 side has the same dependence on R and t ( R5t−3), dr dr # ∼ we can try a power law solution, R(t) = Atα. A small 1Check back to chapter 4 from last term, P425; equation 4.4. 38 Physics 426 Notes Spring 2009 and, since v c in most of region “b”, ρ con- very high temperature and pressure, driving an expan- ) s # stant, and we have 1 ρv2 + p constant. Now – sion. This expansion will be very supersonic, setting 2 # we know that v drops from vsb = vsa/4, at S1, to up a spherical shock wave moving into the surround- v t−2/5 c at S (from the shock conditions). ings and sweeping up gas as it goes.
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