5. Since the range of the radius is small, R dτ. dr >> 1, atmosphere is eectively plane parallel.


 Λέανδρος Βικελίδης
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1 Stellar Atmospheres 1. Until now, we've worried mostly about the interiors of stars where the mean free path of a photon is small compared to the distance to the surface. Photons nearly isotropic. 2. As you get close to the surface, a large fraction of then photons are directed outward, and some actually escape. (a) Gray calls it the transition between the interior and the interstellar medium (ISM). (b) Shu calls it the transition from walking to ying. 3. While radius is a suitable variable for the interior of a star. Optical depth is more appropriate for the atmosphere. or R τ = κ ρdr r = κ ρdr 4. There's no formal denition of what constitutes atmosphere, but it's τ O(1), not τ = 1 3 or τ = 1 3. Here I'm using the Rosseland mean, which makes sense for atmosperes as well as for interiors. 5. Since the range of the radius is small, R dr >> 1, atmosphere is eectively plane parallel. The star has a radius R and a luminosity L, and therefore there exists T eff such that L = 4πR 2 σteff 4. Can treat g as constant and just as on earth! Two parameters: T eff and g. 6. Our fundamental equation of radiative transfer was cos θ I r I sin θ + I κ p j p = θ r The second term is roughly zero because 7. All atmospheres are radiative! d ln I dr >> 1 R. τ T r Radiative skin. 1
2 8. Let's rst ask how T varies as a function of τ. We will need to make four simplifying approximations: (a) Gray atmosphere (b) Local thermodynamic equilibrium (c) The Eddington approximation (d) Radiative equilibrium 9. In the course of deriving our radiative transport equation, we derived an intermediate result, which we called the energy ux equation, by taking the rst moment of the Fundamental Equation of Radiative Transport (FERT) and dω. c dp κ ρ dr + F = 1. Let's rst assume that κ is independent of. This is called the gray atmosphere assumption, and gives = κρdr c dp rad + F d = Integrating over frequency, we get L 4πR 2 = σt 4 eff c dp rad + σt 4 eff = 11. Temptation #1: let P rad = at 4 3 and we'd be done. (a) No; no longer isotropic; not in thermodynamic equilibrium. (b) Divide FERT by κ ρ: { dωd cos θ I + I + j } = τ κ j (Here, κ S, which is the source function.) It is a measure of the ratio of emission to absorption, and for thermodynamic equilibrium it is exactly equal to B. This is not true, however, if there's scattering. It's possible for j κ = B even when the radiation eld is not blackbody. This is called local thermodynamic equilibrium (refers to frequency). Take second [word]: { d dω cos 3 θ I } τ I cos 2 θ + B cos 2 θ = Now we make the Eddington approximation. I (τ, cos θ) = I (τ) + I 1 (τ) cos θ +... Only terms even in cos n will be nonzero when we integrate over dω. ( ddω cos 3 θ I ) τ I cos 2 θ + B cos 2 θ = 2
3 But 4π 5 ddω cos 4 θ I 1 τ cp rad + c 3 at 4 = d di 1 F = dω cos 2 θi 1 = 4π 3 I 1 ( F τ 3 F 5 τ cp rad + c 3 at 4 = goes to because ux does not change with τ.) This is what we wanted but could not justify. 12. Substituting for P rad we get c d 4 σ 3 c T 4 + σteff 4 = where we have used the relation between the StefanBoltzmann constant σ and the radiation constant a, a = 4σ c. T 4 T 4 eff = 3 (τ + q) 4 The righthand term is determined from the boundary condition F d = σteff 4. But what is the constant of integration q? We must do another calculation before we can determine it. 13. Derive a result for I in nongray case. Consequences for spectral lines. I = I (τ, θ) cos θ I τ I + S = Multiply by e τ sec θ ; e τ sec θ 1 I e τ sec θ I + S e τ sec θ = sec θ τ Keep θ constant. Let s = τ sec θ. We can then say e τ sec θ 1 I e τ sec θ I = d sec θ τ ds e s I = S e s Integrate with respect to s from to : I (, θ) = S is equal to B if LTE. S e τ sec θ sec θ 3
4 14. Now, expand B [T (τ)] around some value τ : B (τ ) = B (τ ) + (τ τ ) db +... τ Substitute this into the integral. I (, θ) = B (τ ) + db (τ τ )e τ sec θ sec θ Choose τ = cos θ. = B (τ ) + db τ (cos θ τ ) I (, θ) B [T (τ = cos θ)] This is the EddingtonBarbier relation. τ = 1 τ = cosθ θ Makes sense. limb darkening B (τ =cosθ) θ Sun B (τ =1) Now let us integrate over all outgoing angles. F = dω {}}{ 2π d cos θ cos θi (, θ) 4
5 If we let τ = 2 3 then the db nu term integrates to zero. We are left with F = πb (τ = 2 3 ). For a gray atmospere we have: F = σt 4 (τ = 2 3 ) (by the Eddington approximation). Therefore, Teff 4 = T 4 (τ = 2 3 ). But we had T 4 = 3 Teff 4 4 (τ + q). We can solve for our constant of integration and nd q = 2 3. Had we not made the Eddington approximation we would have found a q(τ) that varied between.577 and.71 as τ increases from to. 5