Stellar Atmospheres: Hydrostatic Equilibrium Hydrostatic Equilibrium Particle conservation 1 Stellar Atmospheres: Hydrostatic Equilibrium Ideal gas dr dA P+dP P r P= pressure k PT=⋅=ρρ mass density Am H A = atomic weight forces acting on volume element: dV== dAdr dmρ dV GM dm GM ρ dF=−rr =− dAdr g rr22 buoyancy: =− (pressure difference * area) dFP dPdA 2 1 Stellar Atmospheres: Hydrostatic Equilibrium Ideal gas In stellar atmospheres: = MMr ∗ mass of atmosphere negligible rR=<<∗ thickness of atmosphere stellar radius ρ →=−GM∗ =ρ dFg 2 dAdrg dAdr R∗ = GM ∗ with g : 2 surface gravity R∗ Type log g usually written as log(g / cm s-2 ) Main sequence star 4.0 .... 4.5 log g is besides Teff the 2nd Sun 4.44 fundamental parameter of Supergiants 0 .... 1 static stellar atmospheres White dwarfs ~8 Neutron stars ~15 Earth 3.0 3 Stellar Atmospheres: Hydrostatic Equilibrium Hydrostatic equilibrium, ideal gas buoyancy = gravitational force: += dFPg dF 0 −−dPdA gρ dAdr =0 dP =−gρ()r dr dP Arm() eliminate ρ()rPr with ideal gas equation: =−g H ( ) dr kT (r) example: Tr( )== T const , Ar( ) == A const (i.e., no ionization or dissociation) dPAm1 dP Am =−gPrHH()⇒ =− g dr kTPdrkT solution: −− = ()rr0 gAmH / kT P(r) P(r0 )e −− = ()/r r0 H P(r)P(r)0 e kT H := pressure scale height 4 gAmH 2 Stellar Atmospheres: Hydrostatic Equilibrium Atmospheric pressure scale heights ≈ Earth: A 28 (N2 ) kT T ≈ 300 K H = 9 km H = gAm H log g = 3 Sun: A = 1 (H) T ≈ 6000 K H = 180 km log g = 4.44 = + + White dwarf: A 0.5 (H ne ) T = 15000 K H = 0.25 km = log g 8 = + + Neutron star: A 0.5 (H ne ) 6 T =10 K H =1.6 mm ! = log g 15 5 Stellar Atmospheres: Hydrostatic Equilibrium Effect of radiation pressure π = 4 2nd moment of intensity PR (v) Kv ' c 1st moment of transfer equation (plane-parallel case) ' dK v = H dvτ () v dP 4π R ==Hdvvdr with τκ( ) ( ) dvτ () c v dP 4π R = κ ()vH dr c v integration over frequencies: ∞ dP 4π R = κ ()vHdv ∫ v dr c 0 6 3 Stellar Atmospheres: Hydrostatic Equilibrium Effect of radiation pressure Extended hydrostatic equation ∞ dP dP 4π =−grρρ()R =− gr () κ()v Hdv ∫ v dr dr c 0 = ρ greff () () r definition: effective gravity ∞ 41π g (r):()=−gvHκ dv =g − g (depth dependent!) effρ ∫ v rad c (r) 0 In the outer layers of many stars: ∞ 41π gg<=0 i.e. κ( vHdvg ) > eff rad ρ ∫ v cr()0 Atmosphere is no longer static, hydrodynamical equation Expanding stellar atmospheres, radiation-driven winds 7 Stellar Atmospheres: Hydrostatic Equilibrium The Eddington limit Estimate radiative acceleration Consider only (Thomson) electron scattering as opacity σ = σ (v) e (Thomson cross-section) q = number of free electrons per atomic mass unit Pure hydrogen atmosphere, completely ionized q = 1 Pure helium atmosphere, completely ionized q = 2/ 4 = 0.5 ∞ ∞ 4π 1 4π q 4π qσ g e = σ n H dv = σ H dv = e H rad c n m / q ∫ e e v c m ∫ e v c m e H 0 σ H 0 H Flux conservation: HT= 4 4π eff ' g e 41πσqMσ qRTσπσ1 4 24 Γ=rad =e TG4 = e eff eeffππ2 gm4cRcGMm4H H qLLσ L − / Γ=e =10 4.51 q e π 8 4mcGMH MM / 4 Stellar Atmospheres: Hydrostatic Equilibrium The Eddington limit Consequence: for given stellar mass there exists a maximum luminosity. No stable stars exist above this luminosity limit. =⋅⋅−4.51 Lmax LqMM10 1 Γ << Sun: e 1 Main sequence stars (central H-burning) ≈→=()3 Mass luminosity relation: LL/ M / M Mmax 180 M Gives a mass limit for main sequence stars Eddington limit written with effective temperature Γ = −15.12 4 = and gravity e 10 qTeff / g 1 − + + − = 15.12 log q 4logTeff log g 0 Straight line in (log Teff,log g)-diagram 9 Stellar Atmospheres: Hydrostatic Equilibrium The Eddington limit Positions of analyzed central stars of planetary nebulae and theoretical stellar evolutionary tracks (mass labeled in solar masses) 10 5 Stellar Atmospheres: Hydrostatic Equilibrium Computation of electron density At a given temperature, the hydrostatic equation gives the gas pressure at any depth, or the total particle density N: = PNkTgas NN=++=+ N n N n atoms ions e N e NN massive particle density The Saha equation yields for given (ne,T) the ion- and atomic densities NN. The Boltzmann equation then yields for given (NN,T) the population densities of all atomic levels: ni. Now, how to get ne? α We have k different species with abundances k Particle density of species k: K ==−αα() = NkkNNnN keN , and it is ∑ NN k k1= 11 Stellar Atmospheres: Hydrostatic Equilibrium Charge conservation Stellar atmosphere is electrically neutral Charge conservation electron density=ion density * charge K jk =⋅ = njNNe ∑∑ jk , jk density of j-th ionization stage of species k kj==11 jk-1 Combine with Saha equation (LTE) ∏ neΦlk(T) N = by the use of ionization fractions: f = jk = l j jk N jk jk-1 k + 1 ∑∏ neΦlk(T) == We write the charge conservation as m 1 l m K jk K jk = ⋅ = α − ⋅ ne ∑∑ j Nk f jk (ne ,T ) ∑∑k (N ne ) j f jk (ne ,T) k ==1 j 1 k ==11j K jk = − α ⋅ = ne (N ne )∑∑k j f jk (ne ,T ) F(ne ) k ==11j Non-linear equation, iterative solution, i.e., determine zeros of − = F(ne ) ne 0 use Newton-Raphson, converges after 2-4 iterations; yields ne and fij, and with Boltzmann all level populations 12 6 Stellar Atmospheres: Hydrostatic Equilibrium Summary: Hydrostatic Equilibrium 13 Stellar Atmospheres: Hydrostatic Equilibrium Summary: Hydrostatic Equilibrium Hydrostatic equation including radiation pressure ∞ dP dP 4π =−=−grρρκ()R gr () () vHdv ∫ v dr dr c 0 Photon pressure: Eddington Limit Hydrostatic equation → N → Combined charge equation + ionization fraction ne → Population numbers nijk (LTE) with Saha and Boltzmann equations 14 7 Stellar Atmospheres: Hydrostatic Equilibrium 3 hours Stellar …per day… …is too much!!! Atmospheres… 15 8.