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Spectroscopy – Continuous Opacities

Spectroscopy – Continuous Opacities. Introduction: Atomic Absorption Coefficents Corrections for Stimulated Emission Hydrogen Negative Hydrogen Ion Negative Helium Ion Metals Electron Scattering Others Summary.

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Spectroscopy – Continuous Opacities

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  1. Spectroscopy – Continuous Opacities • Introduction: Atomic Absorption Coefficents • Corrections for Stimulated Emission • Hydrogen • Negative Hydrogen Ion • Negative Helium Ion • Metals • Electron Scattering • Others • Summary

  2. In order to calculate the transer of radiation through a model stellar atmosphere, we need to know the continuous absorption coefficient, kn. This shapes the continous spectrum → more absorption, less light. It also influences the strength of stellar lines → more continous absorption means a thinner photosphere with few atoms to make spectral lines. Also, before we compute a theoretical spectrum, you need to compute an atmospheric model, and this also depends on kn.

  3. dIn = –In +jn/kn = –In + Sn dtn Recall the radiative transfer equation: To solve this you need to know the opacity. You can have a nice solution, but it will not reproduce observations. The grey atmosphere has a simple opacity, but no bearing with reality. The problem is that there are a lot of opacity sources which are temperature dependent

  4. B1I V F0 V G2 V

  5. I. The atomic absorption coefficent • The total continuous absorption coefficient is the sum of absorption resulting from many physical processes. These are in two categories: • bound-free transition: ionization • free-free transition: acceleration of a charge when passing another charge bound-bound transitions result in a spectral line and are not included in knbut in cool stars line density is so great it affects the continuum

  6. I. The atomic absorption coefficent The atomic absorption coefficient, a, has units of area per absorber. The wavelength versus frequency question does not arise for a: an = al a is not a distribution like In and Il, but the power subtracted from In in interval dn is adn. This is a distribution and has units of erg/(s cm2 rad2 Hz) a dn =(c/l2) a dl

  7. Nu Blu Bul Nℓ II. Corrections for Stimulated Emission Recall that the stimulated emission (negative absorption) reduces the absorption: knr = NℓBℓuhn – NuBuℓhn = Nℓ Bℓuhn(1– NuBuℓ/NℓBℓu) = NℓBℓuhn[1– exp(–hn/kT)]

  8. ½ mv2 = hn–hRc/n2 R = 1.0968×105 cm–1 · · · Ha b III. Atomic Absorption Coefficient for Hydrogen Brackett Lyman Pachen Balmer 13.60 n=∞ 12.75 n=4 12.08 n=3 10.20 n=2 0.00 n=1 c = 13.6(1–1/n2) eV

  9. nlAng Name Absorption edges: 1 912 Lyman ½ mv2 = hn–hRc/n2 R = 1.0968×105 cm–1 At the ionization limit v=0, n=Rc/n2 2 3647 Balmer 3 8206 Paschen 4 14588 Brackett 5 22790 Pfund

  10. pe6 R gn∕ an = 6.16 h n5n3 gn∕ a0 l3 p2e6 l3 gn∕ = an = 6.16 R h3c3 n5 n5 e = electron charge = 4.803×10–10 esu gn∕ = Gaunt factor needed to make Kramer´s result in agreement with quantum mechanical results III. Neutral Hydrogen: Bound-Free Original derivation is from Kramers (1923) and modified by Gaunt (1930): Per neutral H atom a0 = 1.044×10–26 for l in angstroms

  11. n =2 n =2 n =2 n =2 n =2 n =2 n =2 n =1 n =1 n =1 n =1 n =1 n =1 n =1 n = 7 n = 6 n = 6 n = 5 n = 5 n = 5 n = 4 n = 4 n = 4 n = 4 n =3 n =3 n =3 n =3 n =3 ~l3 ~l3/n5 n = 7 n = 6 n = 5 n = 4 a0, 10×–17 cm2/H atom n =3 n =1 Wavelength (Angstroms)

  12. Recall: Nn gn ( c ) exp – = u0(T) kT N III. Neutral Hydrogen: Bound-Free The sum of absorbers in each level times an is what is needed. gn=2n2 c =I – hRc/n2 = 13.6(1-1/n2) eV u0(T) = 2

  13. l3 anNn ( c ) ∞ ∞ gn∕ Σ Σ exp – = a0 k(Hbf) = n3 kT N n0 n0 l3 ∞ gn∕ Σ = a0 10–qc n3 n0 III. Neutral Hydrogen: Bound-Free The absorption coefficient in square centimeters per neutral hydrogen atom for all continua starting at n0 q = 5040/T, c in electron volts

  14. ∞ ( c ) ∫ = ½ d(1/n2) exp – kT ( ( c c ) ) ∞ ∞ n0+3 Σ Σ exp exp – – kT kT n0+3 n0+3 1 1 I dc ( c ) n3 n3 ∫ = ½ exp – I kT 1 c3 [ ] ( ) c3 kT ( ) [ I ] c3 = I 1– exp – – = (n0 +3)2 exp – kT I kT III. Neutral Hydrogen: Bound-Free Unsöld showed that the small contributions due to terms higher than n0+2 can be replaced by an integral: c =I – hRc/n2 => dc = –Id(1/n2)

  15. log e gn∕ n0+2 [ Σ ] (10–c3q – 10 –Iq) + a0l3 k(Hbf) = 10–qc 2qI n3 n0 III. Neutral Hydrogen: Bound-Free We can neglect the n dependence on gn∕ and the final answer is: This is the bound free absorption coefficient for neutral hydrogen

  16. n =2 n =1 n = 7 n = 6 n = 5 n = 4 n =3

  17. kbf(n=3) + ... kbf(>3647) = kbf(<3647) kbf(n=2) + kbf(n=3) + ... kbf(n=3) ≈ kbf(n=2) 8 [ exp [ –(c3 – c2)/kT = 27 III. Neutral Hydrogen: Bound-Free = 0.0037 at 5000 K and 0.033 at 10000 K

  18. III. Neutral Hydrogen: Bound-Free k(red side)/k(blue side)

  19. III. Neutral Hydrogen: Bound-Free

  20. Across a jump your are seeing very different heights in the atmosphere continuum 912 3647 8602 Flux Wavelength III. Optical Depth and Height of Formation Recall: tn = knrdx tn ~ 2/3 for Grey atmosphere As an increases, kn increases => dx decreases You are looking higher in the atmosphere

  21. Temperature profile of photosphere 10000 8000 z Temperature 6000 4000 l<3647 A l>3647 A z=0 dx1 dx2 t=2/3 t=2/3 k(<3647) > k(>3647) => dx2 > dx1 z z=0

  22. B4 V

  23. But wait, I just said that the Balmer jump should be larger for cooler stars. Why is this not the case? For cooler stars other sources of opacity start to dominate, namely H–

  24. Maximum of black body = lT = 0.5099 cm K But peak implies T=13400 K But peak implies T=13400 K The stronger opacity of on the blue side of the Balmer jump distorts the Planck curve. One cannot use the peak of the intensity, but must fit the full spectral energy distribution

  25. Photometric Amplitude of rapidly oscillating Ap stars: Amplitude (mmag) Wavelength (Ang) Different wavelengths probe different heights in atmosphere

  26. III. Neutral Hydrogen: Free-Free The free-free absorption of hydrogen is much smaller. When the free electron has a collision with a proton its unbound orbit is altered. The electron can absorbs a photon and its energy increases. The strength of this absorption depends on the velocity of the electron

  27. III. Neutral Hydrogen: Free-Free proton e– Orbit is altered The absorption of the photon is during the interaction

  28. 1 h2e2 R daff = 0.385 dv n3 v pm3 III. Neutral Hydrogen: Free-Free Absorption According to Kramers the atomic coefficient is: This is the cross section in square cm per H atom for the fraction of the electrons in the velocity interval v to v + dv. To get complete f-f absorption must integrate over v.

  29. 1 1 ∞ ( ) ( ) ) m 1 ( mv2 h2e2 R 2 ∫ 2 2 v dv exp – aff = 0.385 kT 2kT n3 p pm3 0 3 2 ( ) h2e2 R 1 2m aff = 0.385 pkT n3 pm3 III. Neutral Hydrogen: Free-Free Absorption Using the Maxwell-Boltzmann distribution for v Quantum mechanical derivation by Gaunt is modified by f-f Gaunt factor gf

  30. afgfNiNe k(Hff) = N0 Density of neutral H Ni 5 3 N 2 2 ) ) ( ( kT 2pm ( I ) 2u1(T) exp – Pe = kT u0(T) h3 Pe = NekT III. Neutral Hydrogen: Free-Free Absorption The absorption coefficient in square cm per neutral H atom is proportional to the number density of electrons, Ne and protons Ni: Recall the Saha Equation:

  31. ) ( 2pmkT ( I ) exp – afgf k(Hff) = kT h3 log e 3 a0l3gf 10–qI k(Hff) = 2 2qI R=2p2me4/h3c I=hcR Using: q=log e/kT = 5040/T for eV III. Neutral Hydrogen: Free-Free Absorption

  32. III. Total Absorption Coefficient for Hydrogen total bound-free free-free

  33. IV. The Negative Hydrogen Ion The hydrogen atom is capable of holding a second electron in a bound state. The ionization of the extra electron requires 0.754 eV All photons with l < 16444 Ang have sufficient energy to ionize H – back to neutral H Very important opacity for Teff < 6000 K Where does this extra electron come from? Metals!

  34. IV. The Negative Hydrogen Ion For Teff > 6000 K, H– too frequently ionized to be an effective absorber For Teff < 6000 K, H– very important For Teff < 4000 K, no longer effective because there are no more free electrons

  35. a0 = 1.99654 a1 = –1.18267 × 10–5 a2 = 2.64243 × 10–6 a3 = –4.40524 × 10–10 a4 = 3.23992× 10–14 a5 = –1.39568 × 10–10 a6 = 2.78701 × 10–23 l is in Angstroms IV. The Negative Hydrogen Ion The bound free absorption coefficient can be expressed by the following polynomial abf = a0 + a1l + a2l2 + a3l3 + a4l4 + a5l5 + a6l6

  36. IV. The Negative Hydrogen Ion: Bound-Free abf, 10–18 cm2 per H– ion Wavelength (angstroms)

  37. N(H) 5040 – –log Pe = + 2.5 log T + 0.1248 I log T N(H –) in eV 5 2 k(Hbf–) = 4.158 × 10–10abfPeq 100.754q IV. The Negative Hydrogen Ion: bound-free The H – ionization is given by the Saha equation u0(T) = 1, u1(T) = 2

  38. IV. The Negative Hydrogen Ion: free-free f0+f1logq+f2log2q k(Hff–) = Peaff = 10–26× Pe 10 f0 = –2.2763–1.685 logl+0.766 log2l–0.0533464 log3l f1 = 15.2827–9.2846 logl+1.99381 log2l–0.142631 log3l f3 = –197.789+190.266 logl–67.9775 log2l+10.6913 log3l–0.625151 log4l

  39. bound-free free-free IV. The Negative Hydrogen Ion: Total

  40. V. The Negative Helium ion The bound–free absorption is neglible, but free-free can be important in the atmospheres of cool stars and at longer wavelengths

  41. VI. Metals • In the visible a minor opacity source because they are not many around • Contribute indirectly by providing electrons • In the visible kn(metals) ~ 1% kn(Hbf–) • A different story in the ultraviolet where the opacity is dominated by metals

  42. VI. Metals The absorption coefficient for metals dominate in the ultraviolet

  43. VII. Electron (Thompson) Scattering • Important in hot stars where H is ionized • Only true „grey“ opacity source since it does not depend on wavelength • Phase function for scattering ~ 1 + cos q • Stellar atmosphere people assume average phase ~ 0

  44. 2 ( e2 8p ( ae = = 0.6648 x 10–24 cm2/electron 3 mc2 aeNe aePe k(e) = = NH PH VII. Electron (Thompson) Scattering The absorption coefficient is wavelength independent: The absorption per hydrogen atom: PH = Partial pressure of Hydrogen

  45. S S N = NH Nj + Ne = Aj + Ne Solving for NH N–Ne Pg–Pe NH PH = = S S Aj Aj VII. Electron (Thompson) Scattering PH is related to the gas and electron pressure as follows: Nj particles of the jth element per cubic cm and Aj = Nj/NH

  46. S Independent of pressure ae k(e) = Aj VII. Electron (Thompson) Scattering S aePe Aj k(e) = Pg– Pe Electron scattering is important in O and Early B stars If hydrogen dominates their composition Pe = 0.5Pg

  47. 0.5 PTot = Pe + Pg Pe /PTot Teff

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