Spectroscopy – Continuous Opacities I. Introduction: Atomic Absorption Coefficents II. Corrections for Stimulated Emission III. Hydrogen IV. Negative Hydrogen Ion V. Negative Helium Ion VI. Metals VII. Electron Scattering VIII. Others IX. Summary

In order to calculate the transer of radiation through a model stellar atmosphere, we need to know the continuous absorption coefficient, . 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 computer a theoretical spectrum, you need to compute an atmospheric model, and this also depends on .

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  but in cool stars line density is so great it affects the continuum

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I. The atomic absorption coefficent The atomic absorption coefficient, , has units of area per absorber. The wavelength versus frequency question does not arise for  :  =   is not a distribution like I and I, but the power subtracted from I in interval d is  d. This is a distribution and has units of erg/(s cm 2 rad 2 Hz)  d =(c/ 2 )  d

II. Corrections for Stimulated Emission Recall that the stimulated emission (negative absorption) reduces the absorption:   = N ℓ B ℓu h – N u B uℓ h = N ℓ ℓ u h (1– N u B uℓ /N ℓ B ℓu ) = N ℓ B ℓu h [1– exp(–h /kT)] B ul B lu NuNu NℓNℓ ℓ

T cm K T 1 – exp(h /kT) % decrease T eff (10% reduction) Ang m2m

· · · Lyman n=1 n=2 n=3 n=4 n= ∞ BalmerPachen Brackett  = 13.6(1–1/n 2 ) eV ½ mv 2 = h –hRc/n 2 R = ×10 5 cm –1 III. Atomic Absorption Coefficient for Hydrogen HH 

Balmer Paschen Brackett Pfund At the ionization limit v=0, =Rc/n 2 ½ mv 2 = h –hRc/n 2 R = ×10 5 cm –1 n Ang Name Lyman Absorption edges:

III. Neutral Hydrogen: Bound-Free Original derivation is from Kramers (1923) and modified by Gaunt (1930):  n = 6.16 e6e6 h R n 5 3 gn∕gn∕  n = 6.16 e6e6 h3c3h3c3 R gn∕gn∕ 3 n5n5 = 00 gn∕gn∕  n5n5 a 0 = 1.044×10 –26 for in angstroms e = electron charge = 4.803×10 –10 esu gn∕gn∕ = Gaunt factor needed to make Kramer´s result in agreement with quantum mechanical results Per neutral H atom

 0, 10× –17 cm 2 /H atom Wavelength (Angstroms) n =1 n =2 n =1 n =2 n =3 n =1 n =2 n =3 n = 4 n =1 n =2 n =3 n = 4 n = 5 n =1 n =2 n =3 n = 4 n = 5 n = 6 n =1 n =2 n =3 n = 4 n = 5 n = 6 n = 7 n =1 n =2 n =3 n = 4 n = 5 n = 6 n = 7 ~ 3 /n 5 ~ 3

The sum of absorbers in each level times  n is what is needed. Recall: NnNn N = gngn u 0 (T) ( kT ) – exp  g n =2n 2  =I – hRc/n 2 = 13.6(1-1/n 2 ) eV u 0 (T) = 2 III. Neutral Hydrogen: Bound-Free

The absorption coefficient in square centimeters per neutral hydrogen atom for all continua starting at n 0  (H bf ) = Σ n0n0 ∞ nNnnNn N =  0 Σ n0n0 ∞ 3 n3n3 gn∕gn∕ ( kT ) – exp  =  0 Σ n0n0 ∞ 3 n3n3 gn∕gn∕ 10 –  III. Neutral Hydrogen: Bound-Free  = 5040/T,  in electron volts

Unsöld showed that the small contributions due to terms higher than n 0 +2 can be replaced by an integral: Σ n 0 +3 ∞ 1 n3n3 ( kT ) – exp  = ½ ∫ n 0 +3 ∞ ( kT ) – exp  d(1/n 2 )  =I – hRc/n 2 => d  = –Id(1/n 2 ) Σ n 0 +3 ∞ 1 n3n3 ( kT ) – exp  = ½ ∫ 33 I ( kT ) – exp  dd I  3 = I [ 1– 1 (n 0 +3) 2 ] III. Neutral Hydrogen: Bound-Free = kT I [ ( – exp   ) ( kT – exp  ) ] –

We can neglect the n dependence on g n ∕ and the final answer is: This is the bound free absorption coefficient for neutral hydrogen  (H bf ) =  0 3 [ Σ n0n0 n 0 +2 gn∕gn∕ n3n3 10 –  + log e 2I2I (10 –  3  – 10 –I  ) ] III. Neutral Hydrogen: Bound-Free

 bf (>3647)  bf (<3647) =  bf (n=3) +...  bf (n=2) +  bf (n=3) +... ≈  bf (n=3)  bf (n=2) 8 = 27 exp [ [ –(  3 –  2 )/kT = at 5000 K and at K

edge (Ang) T 3000 T 5000 T T Lyman  9×10 –19 6×10 –12 9×10 – Balmer Paschen Brackett Pfund III. Neutral Hydrogen: Bound-Free  (red side)/  (blue side)

III. Neutral Hydrogen: Bound-Free

III. Optical Depth and Height of Formation Wavelength Flux Recall:  =   dx  ~ 2/3 for Grey atmosphere As  increases,  increases => dx decreases You are looking higher in the atmosphere continuum Across a jump your are seeing very different heights in the atmosphere

<3647 A  =2/3 >3647 A  =2/3 z=0 Temperature profile of photosphere z=0 Temperature z z dx 1 dx 2  (  (>3647) => dx 2 > dx 1

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

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 absorbs the photon and its energy increases. The strength of this absorption depends on the velocity of the electron

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

III. Neutral Hydrogen: Free-Free Absorption According to Kramers the atomic coefficient is: d  ff = h  e 2 R m3m3 1 3 v dv 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.

Using the Maxwell-Boltzmann distribution for v  ff = h  e 2 R m3m3 1 3 ( m kT ) 2 3 exp ( mv2mv2 2kT ) – v ( 2  ) 2 1 ∫ 0 ∞ dv ( 2m  kT )  ff = h  e 2 R m3m Quantum mechanical derivation by Gaunt is modified by f-f Gaunt factor g f III. Neutral Hydrogen: Free-Free Absorption

The absorption coefficient in square cm per neutral H atom is proportional to the number density of electrons, N e and protons N i :  (H ff ) = fgfNiNefgfNiNe N0N0 Density of neutral H Recall the Saha Equation: ( = NiNi N PePe h3h3 2m2m ) ( kT ) 2u 1 (T) u 0 (T) ( kT ) – exp I P e = N e kT III. Neutral Hydrogen: Free-Free Absorption

 (H ff ) = fgffgf 2 3 ( 2  mkT ) h3h3 ( kT ) – exp I  I  (H ff ) =  0 3 g f log e 10 –  I Using: I=hcR R=2  2 me 4 /h 3 c  =log e/kT = 5040/T for eV III. Neutral Hydrogen: Free-Free Absorption

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

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 eV All photons with  < Ang have sufficient energy to ionize H – back to neutral H Very important opacity for T eff < 6000 K Where does this extra electron come from? Metals!

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

IV. The Negative Hydrogen Ion The bound free absorption coefficient can be expressed by the following polynomial  bf = a 0 + a 1 + a a a a a 6 6 a 0 = a 1 = – × 10 –5 a 2 = × 10 –6 a 3 = – × 10 –10 a 4 = × 10 –14 a 5 = – × 10 –10 a 6 = × 10 –23 is in Angstroms

IV. The Negative Hydrogen Ion: Bound-Free Wavelength (angstroms)  bf, 10 –18 cm 2 per H – ion

IV. The Negative Hydrogen Ion: bound-free The H – ionization is given by the Saha equation log N(H) N(H – ) = –log P e – 5040 T I +2.5 log T in eV  (H bf – ) = × 10 –10  bf P e   u 0 (T) = 1, u 1 (T) = 2

IV. The Negative Hydrogen Ion: free-free  (H ff – ) = P e  ff = 10 –26 × P e 10 f 0 +f 1 log  +f 2 log2  f 0 = –2.2763–1.685 log log 2 – log 3 f 1 = – log log 2 – log 3 f 3 = – log – log log 3 – log 4

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

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

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

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

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  Stellar atmosphere people assume average phase ~ 0

The absorption coefficient is wavelength independent:  e = 88 3 ( ( e2e2 mc 2 2 = x 10 –24 cm 2 /electron The absorption per hydrogen atom:  (e) = eNeeNe NHNH ePeePe PHPH = P H = Partial pressure of Hydrogen VII. Electron (Thompson) Scattering

P H is related to the gas and electron pressure as follows: N =  N j + N e =  NHNH A j + N e N j particles of the jth element per cubic cm and A j = N j /N H Solving for N H NHNH = N–NeN–Ne  AjAj PHPH = Pg–PePg–Pe  AjAj VII. Electron (Thompson) Scattering

 (e) = Electron scattering is important in O and Early B stars P g – P e ePeePe  AjAj If hydrogen dominates their composition P e = 0.5P g  (e) = ee  AjAj Independent of pressure VII. Electron (Thompson) Scattering

P e /P Tot T eff P Tot = P e + P g 0.5

VIII. Other Sources of Opacity H 2 (neutral) has no significant absorption in the visible H 2 +, H 2 – do have significant absorption H 2 + (bf) important in the ultraviolet, in A-type stars it is ~ 10% of H – bound-free opacity Peaks in opacity around  ≈ 1100 Å, is dominated by the Balmer continuum below 3600 Å in most stars H 2 – (free-free) important in the infrared (cool stars) and fills the opacity minimum of H – at Å H 2 molecules

bound-free free-free

VIII. Other Sources of Opacity Important only in O and B-type stars He II (bound-free) is hydrogenic → multiply hydrogen cross sections by Z 4 or 16. He I (bound-free), He II (bound-free)

VIII. Other Sources of Opacity Important in cool stars Scattering by molecules and atoms Has a 1/ 4 dependence Rayleigh Scattering

VIII. Other Sources of Opacity Molecules and ions: CN –, C 2 –, H 2, He, N 2, O 2, TiO,.... Cool Stars Basically Cool Stars are a mess and only for the bravest theoretical astrophysicist

IX. Summary of Continuous Opacities Spectral Type Dominant opacity O–BElectron scattering, He I,II (b-f), H(f-f) B–AH I: b-f, f-f He II: b-f, some electron scattering A–Fequal contributions from H I (b-b) H I (f-f), H – (b-f, f-f) G–K H I (b-f), H – (b-f, f-f), Rayleigh Scattering off H I

IX. Summary of Continuous Opacities Spectral Type Dominant opacity K–Early M H – (b-f, f-f), Rayleigh scattering (UV) off H I and H 2, molecular opacities (line blanketing) M:Molecules and neutral atoms, H – (b-f, f-f), Rayleigh scattering off other molecules

He II/He III ionization zone Contraction During compression He II ionized to He III, He III has a higher opacity. This blocks radiation causing star to expand Cepheid Pulsations are due to an opacity effect: During expansion He zone cools, He III recombines, opacity decreases allowing photons to escape. Star then contracts under gravity. Expansion

Most pulsating stars can be explained by opacity effects