Model Photospheres

1. Model Photospheres • What is a photosphere? • Hydrostatic Equilibruium • Temperature Distribution in the Photosphere • The Pg-Pe-T relationship • Properties of the Models • Models for cool stars

2. I. What is a Stellar Atmosphere? • Transition between the „inside“ and „outside“ of the star • Boundary between the stellar interior and the interstellar medium • All energy generated in the core has to pass through the atmosphere • Atmosphere does not produce any energy • Two basic parameters: • Effective temperature. Not a real temperature but the temperature needed to produce the observed flux via 4pR2T4 • Surface Gravity – log g (although g is not a dimensionless number). Log g in stars range from 8 for a white dwarf to 0.1 for a supergiant. The sun has log g = 4.44

3. What is a Photosphere? • It is the surface you „see“ when you look at a star • It is where most of the spectral lines are formed What is a Model Photosphere? • It is a table of numbers giving the source function and the pressure as a function of optical depth. One might also list the density, electron pressure, magnetic field, velocity field etc. • The model photosphere or stellar atmosphere is what is used by spectral synthesis codes to generate a synthetic spectrum of a star

4. Real Stars: • Spherical • Can pulsate • Granulation, starspots, velocity fields • Magnetic fields • Winds and mass loss

5. Our model: • Plane parallel geometry • Hydrostatic equilibrium and no mass loss • Granulation, spots, and velocity fields are represented by mean values • No magnetic fields

6. P +dP A dr dm P Gravity II. Hydrostatic equilibrium dA P + dP r + dr P r M(r) The gravity in a thin shell should be balanced by the outward gas pressure in the cell

7. Fp = PdA –(P + dP)dA = –dP dA Pressure Force dM FG = –GM(r) Gravitational Force r2 P +dP A r dr M(r) = ∫ r(r) 4pr2dr 0 dm dM = r dA dr P Gravity FP + FG = 0 Both forces must balance: r(r)M(r) –G –dP dA + dA dr = 0 r2

8. = • Radiation pressure: 2.52 ×10–15 T4 dyne/cm2 2. Magnetic pressure: PM = B2 4s T4 PR = 8p 3c 3. Turbulent pressure: ~ ½rv2v is the root mean square velocity of turbulent elements The pressure in this equation is the total pressure supporting the small volume element. In most stars the gas pressure accounts for most of this. There are cases where other sources of pressure can be significant when compared to Pg. Other sources of pressure:

9. B2 8p Footnote: Magnetic pressure is what is behind the emergence of magnetic „flux ropes“ in the Sun Pphot Ptube + In the magnetic flux tube the magnetic field provides partial pressure support. Since the total pressure in the flux tube is the same as in the surrounding gas Ptube < Pphot. Thus rtube < rphot and the flux tube rises due to buoyancy force.

10. Pressures B is for magnetic pressure = Pg v is velocity that generates pressure equal to Pg according to ½rv2

11. We are ignoring magnetic fields in generating the photospheric models. But recall that peculiar A-type stars can have huge global magnetic fields of several kilogauss in strength. In these atmospheric models one has to treat the magnetic pressure as well.

12. r(r)M(r) –G = dP dr r2 dP = grdx The weight in the narrow column is just the density × volume × gravity g dP kn = dtn In our atmosphere F GM/r2 = g (acceleration of gravity) x increases inward so no negative sign P dA dx r F + dF P + dP dtn = knrdx gravity

13. Pg½ dPg = Pg½ dk0 k0 is a reference wavelength (5000 Å) g 3 2 3 2 2 3 k0 3/2 ⅔ dt0 Pg(t0) = Pg½ gPg½ to to ( dt0 ( Pg(t0) = ∫ ∫ g g k0 k0 0 0 ⅔ t0½ log to Pg½ ( dlog t0 ( ∫ Pg(t0) = k0 log e –∞ One way to integrate the hydrostatic equation Integrating on a logarithmic optical depth scale gives better precision

14. Numerical Procedure • Guess the function Pg(t0) and perform the numerical integration • New value of Pg(t0) is used in the next iteration until convergence is obtained • A good guess takes 2-3 iterations Problem: we must now k0 as a function of t0 since k0 appears in the integrand. kn is dependent on temperature and electron pressure. Thus we need to know how T and Pe depend on t0.

15. Limb darkening is due to the decrease of the continuum source function outwards ∞ In(0) = ∫Sne–tnsec q sec q dtn 0 The exponential extinction varies as tnsec q, so the position of the unit optical depth along the line of sight moves upwards, i.e. to smaller t. III. Temperature Distribution in the Solar Photosphere • Two probes of depth: • Limb Darkening • Wavelength dependence of the absorption coefficient The increment of path length along the line of sight is ds = dx sec q q ds

16. Temperature profile of photosphere Bottom of photosphere 10000 8000 z Temperature 6000 4000 q2 z=0 q1 dz tn =1 surface Top of photosphere Limb Darkening The path length dz is approximately the same at all viewing angles, but at larger the optical depth of t=1 is reached higher in the atmosphere

18. Solar limb darkening as a function of wavelength in Angstroms Solar limb darkening as a function of position on disk

19. z Temperature profile of photosphere and chromosphere 10000 In radio waves one is looking so high up in the atmosphere that one is in the chromosphere where the temperature is increasing with heigth 8000 chromosophere Temperature 6000 4000 z=0 At 1.3 mm the solar atmosphere exhibits limb brightening Horne et al. 1981

20. Limb darkening in other stars Use transiting planets No limb darkening transit shape At the limb the star has less flux than is expected, thus the planet blocks less light

21. The depth of the light curve gives you the Rplanet/Rstar, but the „radius“ of the star depends on the limb darkening, which depends on the wavelength you are looking at To get an accurate measurement of the planet radius you need to model the limb darkening appropriately q If you define the radius at which the intensity is 0.9 the full intensity: At l=10000 Å, cos q=0.6, q=67o, projected disk radius = sin q = 0.91 At l=4000 Å, cos q=0.85, q=32o, projected disk radius = sin q = 0.52 → disk is only 57% of the „apparent“ size at the longer wavelength

22. The transit duration depends on the radius of the star but the „radius“ depends on the limb darkening. The duration also depends on the orbital inclination When using different data sets to look for changes in the transit duration due to changes in the orbital inclination one has to be very careful how you treat the limb darkening.

23. Possible inclination changes in TrEs-2? Evidence that transit duration has decreased by 3.2 minutes. This might be caused by inclination changes induced by a third body But the Kepler Spacecraft does not show this effect.

24. One possible explanation is that this study had to combine different data sets taken at different wavelength band passes (filters). But the limb darkening depends on wavelength. At shorter wavelengths the star „looks“ smaller. The only star for which the limb darkening is well known is the Sun

25. Using: a + b cosq In(0) = This is the Eddington-Barbier relation which says that at cos q = tn the specific intensity on the surface at position equals the source function at a depth tn ∞ In(0) = ∫Sne–tnsec q sec q dtn 0 In the grey case had a linear source function: Sn = a + btn Limb darkening laws usually of the form: Ic(0) continuum intensity at disk center Ic = Ic(0) (1 – e + e cos q) e ≈ 0.6 for the solar case, 0 for A-type stars

26. d log tn log e Rewriting on a log scale: ∞ In(0) = ∫Sne–tnsec qtn sec q 0 Contribution function Sample solar contribution functions No light comes from the highest and lowest layers, and on average the surface intensity originates higher in the atmosphere for positions close to the limb.

27. Wavelength Variation of the Absorption Coefficient • Since the absorption coefficient depends on the wavelength you look into different depths of the atmosphere. For the Sun: • See into the deepest layers at 1.6mm • Towards shorter wavelengths kn increases until at l = 2000 Å it reaches a maximum. This corresponds to a depth of formation at the temperature minimum (before the increase in the chromosphere)

28. Solar Temperature distributions Best agreements are deeper in the atmosphere where log t0 = –1 to 0.5 Poor aggreement is higher up in the atmosphere

29. ¼ [t + q(t)] Teff ¾ T(t) = Temperature Distribution in other Stars The simplest method of obtaining the temperature distribution in other stars is to scale to a standard temperature distribution, for example the solar one. T(t0) = S0Tסּ(t0) In the grey case: Teff In the grey case the scaling factor is the ratio of effective temperatures T(t) = Tסּ(t) Tסּeff

30. Scaled solar models agree well (within a few percent) to calculations using radiative equilibrium. They also agree well when applying to giant stars. Numerically it was easier to use scaled solar models in the past. Now, one just uses a grid of models calculated using radiative equilibrium

31. Fj(T) N1j = Pe N0j u1 u0 IV. The Pg–Pe–T Relation When solving the hydrostatic pressure equation we start with an initial guess for Pg(t0). We then require that the electron pressure Pe(t0) = Pe(Pg) in order to find k0(t0) = k0(T,Pe) for the integrand. The electron pressure depends on the temperature and chemical composition. N1j = number of ions per unit volume of the jth element See Saha equation from 2nd lecture N0j =number of neutrals 5/2 F(T) = 0.65 T 10–5040I/kT

32. Nej Nej = = Fj(T) N0j Nj –Nej Pe Fj(T)/Pe Nej = Nj 1 + Fj(T)/Pe Neglect double ionization. N1j = Nej, the number of electrons per unit volume that are contributed by the jth element. The total number of jth element particles is Nj = N1j + N0j. Solving for Nej The pressures are: Pe = SNejkT j Pg = S (Nej +Nj)kT j

33. SNejkT Pe = S (Nej +Nj)kT Pg Fj(T)/Pe Using the number abundance Aj = Nj/NH NH = number of hydrogen SNj Pe 1 + Fj(T)/Pe = Pg Fj(T)/Pe SNj 1 + 1 + Fj(T)/Pe Fj(T)/Pe SAj 1 + Fj(T)/Pe Pg Pe = Fj(T)/Pe SAj 1 + 1 + Fj(T)/Pe Taking ratios:

34. This is a transcendental equation in Pe that has to be solved iteratively. F(T) are constants for such an iteration. Pe and Pg are functions of t0. This equation is solved at each depth using the first guess of Pg(t0). For the cooler models the temperature sensitivity of the electron pressure is very large with d log Pe/d log T ≈ 12 since the absorption coefficient is largely due to the negative hydrogen ion log t 1 0 –1 –2 The absorption coefficient kn is largely due to the negative hydrogen ion which is proportional to Peso the opacity increases very rapidly with depth. –3 –4

35. F(T)/Pe Pe = Pg 1 + 2 F(T)/Pe 2 Pe = F(T)Pg– 2F(T)Pe = F(T)(Pg– 2Pe) 2 Pe ≈ F(T)Pg Hydrogen dominates at high temperatures and when it is fully ionized Pg ≈ 2Pe At cooler temperatures Pe ~ Pg½ Where does the later come from? Assume the photosphere is made of single element this simplifies things: Pg >> Pe in cool stars

36. 3 2 g ⅔ t0½ log to Pg½ ( dlog t0 ( ∫ Pg(t0) = k0 log e –∞ Completing the model • We can now can compute this • Take T(t0) and our guess for Pg(t0) • Compute Pe(t0) and k0(t0) • Above equation gives new Pg(t0) • Iterate until you get convergence (≈ 1%) • Can now calculate geometrical depth and surface flux

37. t0 1 ∫ dt0 x(t0) = k0(t0)r(t0) 0 Pg–Pe N–Ne NH = = kT S Aj S Aj The Geometric Depth We are often interested in the geometric depth scale (i.e. where the continuum is formed). This can be computed from dx = dt0/k0r The density can be calculated from the pressure ( P = (r/m)KT ) r = NH (hydrogen particles per cm3) x SAjmj grams/H particle) where mj is the atomic weight of the jth element

38. log t0 S AjkT(t0)t0 ∫ x(t0) = k0(t0)S Ajmj[Pg(t0)t0 – Pe(t0)] –∞ A more interesting form is to integrate on a Pg scale with dPg = rgdx d log t0 d log e Pg S AjkT(p) ∫ x(t0) = S Ajmj –∞ 1 g dp This makes physical sense if you recall the scale height of the atmosphere: Scale height H = kT/mg p The thickness of the atmosphere is inversely proportional to the surface gravity since T(Pg) depends weakly on gravity

40. ∞ ∫ It is customary to integrate on a log t scale 0 ∞ ∫ kn(t0)t0 d log t0 Fn(0)= 2p Sn(tn)E2(tn) k0(t0) d log e –∞ Flux contribution function Computation of the Spectrum The spectrum Fn(0)= 2p Sn(tn)E2(tn)dtn

41. Flux Contribution Functions as a Function of Wavelength But cross the Balmer jump and the flux dramatically increases. This is because there is a sharp decrease in the opacity across the Balmer jump. Flux at 8000 Å originates higher up in the atmosphere than flux at 5000 or 3646 Å

42. Flux Contribution Functions as a Function of Effective Temperature T= 10400 K T= 8090 K T= 4620 K A hotter star produces more flux, but this originates higher up in the atmosphere

43. ∞ ∫ ∞ d Sn(tn) ∫ E3(tn)dtn Fn(0)= pSn(0) + 0 dtn 0 The flux arises from the gradient of the source function. Depths where dS/dt is larger contribute more to the flux Computation of the Spectrum There are other techniques for computing the flux → Different integrals. Integrating flux equation by parts: Fn(0)= 2p Sn(tn)E2(tn)dtn

44. Cannot scale T(Pg), unlike T(t0)!!! V. Properties of Models: Pressure d log T/dlog Pg = 0.4 Convection gradient Temperature Relationship between pressure and temperature for models of effective temperatures 3500 to 50000 K. The dashed line marks where the slope exceeds 1–1/g ≈ 0.4 and implies instability to convection

45. = = 0.85 0.62 dlogPg dlogPg dlog g dlog g Teff = 8750 K Effects of gravity Increasing the gravity increases all pressures. For a given T the pressure increases with gravity

46. 3 2 g ⅔ t0½ log to Pg½ ( dlog t0 ( ∫ Pg(t0) = k0 log e –∞ Pg ≈ C(T) g ⅔ since pressure dependence in the integral is weak. So dlog P/dlog g ~ 0.67 In general Pg ~ gp In cool models p ranges from 0.64 to 0.54 in going from deep to shallow layers In hotter models p ranges from 0.85 to 0.53 in going from deep to shallow layers Recall Pe ≈ Pg½ in cool stars → Pe ≈ constant g⅓ Pe ≈ 0.5 Pg in hot stars → Pe ≈ constant g⅔

47. Properties of Models: Chemical Composition Gas pressure Electron pressure • In hot models hydrogen takes over as electron donor and the pressures are indepedent of chemical composition • In cool models increasing metals → increasing number of electrons → larger continous absorption → shorter geometrical penetration in the line of sight → gas pressure at a given depth decreases with increasing metal content

48. Pe Pe Pe NH kT kT Pg Pg Pg S N j = ≈ Pe PH S N jkT S N jkT Qualitatively: Using SAj for the sum of the metal abundances 1 = S A j Since PH, the partial pressure of Hydrogen dominates the gas pressure SNj = S(N1 + N0)j, the number of element particles is the sum of ions and neutrals and Pe=NekT = SN1jkT for single ionizations

49. g 1 1 S N 1j Pe Pe k0 ≈ S A j S A j S (N 1 + N0)j Pg Pg ≈ 1 g dk0 g = = Pek0/Pe PgSAjk0/Pe dk0 dPg = In the solar case metals are ionized SN1j >> N0j k0 is dominated by the negative hydrogen ion, so k0/Pe is independent of Pe

50. t0 1 g ∫ 2 dt0 ½Pg = k0/Pe S Aj g and T are constants 0 For metals being neutral: SN1J << S(N1 + N0)j can show –⅓ –½ ½ ⅓ Pe =c0 Pg =c0 Pg =c0 Pe =c0 (S Aj) (S Aj) (S Aj) (S Aj) Integrating: