1. Continuous Opacity Sources

2. Continuous Opacity 2 Continuous Opacity Sources Principal Sources: –Bound-Bound Transitions –Bound-Free –Free-Free (Bremstralung) –Electron Scattering (Thompson & Compton) –Molecular Transitions We consider only H or H-Like cases –We will do this classically (and correct to the QM result) –Stars are mostly H (in one form or another).

3. Continuous Opacity 3 Dominant Opacity Sources Type of Star SpeciesType Cool Stars (G- M) of Normal Composition H-H-H-H-b-fDominant Warmer A-F stars H-H-H-H-f-fSecondary Hb-fDominant Hf-fSecondary Interiors and hottest stars Hf-f

4. Continuous Opacity 4 Caveats and Details At High Temperatures: (1-e -hν/kT ) → 0 so all of the bb, bf, and ff sources go to 0! Electron scattering takes over (is always there and may be important). –Free-Free is not the same as electron scattering: Conservation of momentum says a photon cannot be absorbed by a free particle! In principal we start with a QM description of the photon - electron interaction which yields the cross section for absorption/scattering of the photon of energy hν, call the cross section a i (ν).

5. Continuous Opacity 5 The Opacity Is The opacity (g/cm 2 ) for the process is Κ i (ν) = n i a i (ν)/ρ n i is the number density (#/cm 3 ) of the operant particles The subscript i denotes a process/opacity The total opacity is Κ total = ∑Κ i (ν) Κ total = ∑Κ i (ν)

6. Continuous Opacity 6 Compton Scattering This process is important only for high energy photons as the maximum change is 0.024Å. Reference: Eisberg -- Fundamentals of Modern Physics p. 81 ff.

7. Continuous Opacity 7 Electron Scattering Conditions Collision of an electron and a photon –Energy and Momentum Must be conserved In stellar atmospheres during photon-electron collisions the wavelength of the photon is increased (assume the E electron < Photon (hν)) –At 4000Å: hν = 4.966(10 -12 ) ergs –½ mv 2 v v < 10 8 cm/s –At 5000K v RMS = 6.7(10 5 ) cm/s –At 100000K v RMS = 3(10 6 ) cm/s

8. Continuous Opacity 8 Thompson Scattering Reference: Marion - Classical Electromagnetic Radiation p. 272 ff Low Energy Process: v << c Energy Absorbed from the EM field is a = acceleration of the electron: a = e E /m e and E is the magnitude of the electric field. Classical Electron Scattering

9. Continuous Opacity 9 Thompson Scattering Field Energy Density = (Time Average) Energy Flux Per Electron = c Energy Flux Per Electron = c Now Take the Time Average of (#): But that has to be the energy flux per electron times the cross section which is σ T c.

10. Continuous Opacity 10 The Thompson Cross Section At High Temperatures this breaks down: T ≥ 10 9 K

11. Continuous Opacity 11 Electron Scattering Opacity Κ e = σ T N e /ρ There is no frequency dependence! Scattering off other ions is unimportant: Cross Section Goes as (1/m) 2 so for ions (1/Am H ) 2 while for electrons it goes as (1/m e ) 2 Ratio: Ions/Electrons = (m e /Am H ) 2 = (1/A(1840)) 2 < 10 -6

12. Continuous Opacity 12 Rayleigh Scattering: σ R Scattering of a low energy photon by a bound electron. Classically: Rayleigh scattering occurs when a photon of energy less than the atomic energy spacing is absorbed. –The electron then oscillates about the unperturbed energy level (harmonically). –The electron reradiates the same photon but remains in the same energy state.

13. Continuous Opacity 13 The Rayleigh Cross-Section The cross section is: σ = σ T /(1-(ν 0 /ν) 2 ) 2 –Where hν is the photon energy –hν 0 is the restoring force for the oscillator When ν << ν 0 : σ R = σ T (λ 0 /λ) 4 –Now since ν << ν 0 we have  E << kT –Which implies T ~ 1000 K for this process.

14. Continuous Opacity 14 Free-Free Opacities Bremstralung –Electron moving in the field of an ion of charge Ze emits or absorbs a photon: Acceleration in field produces a photon of hν De-acceleration in field consumes a photon of hν Consider the Emission Process –Initial Electron Velocity: v′ –Final Electron Velocity: v Conservation of Energy Yields ½ m e v 2 + hν = ½ m e v′ 2 ½ m e v 2 + hν = ½ m e v′ 2 Absorption Events

15. Continuous Opacity 15 Energy Considerations Energy Absorbed: dE/dt = 2/3 (e 2 /c 3 ) a 2 where a = e E /m e (a is the acceleration) Most energy is absorbed during the time t  b / v′ when the electron is close to the ion –b is called the impact parameter –b is the distance of closest approach Acceleration is ~ Ze 2 /m e b 2 E abs  2/3 (e 2 /c 3 ) (Ze 2 /m e b 2 ) 2 (b / v′)  2/3 ((Z 2 e 6 )/(m e 2 c 3 b 3 v′))  2/3 ((Z 2 e 6 )/(m e 2 c 3 b 3 v′))

16. Continuous Opacity 16 Frequency Dependence Greatest contribution is when 2πνt  1 during time t  b / v′ –Therefore 2πν b / v′ = 1 or 2πν = v′ / b The energy emitted per electron per ion in the frequency range d is –dq ν = 2πb db E abs Why? 2πb db is an area! –dq ν = energy emitted per electron per ion per unit frequency Expand dE/dT in a Fourier Series

17. Continuous Opacity 17 Bremstralung Energy Total Energy Emitted: n i n e v′f(v′)dv′ dq ν –n i = Ion number density –n e = electron number density (note that the electron flux is n e v′f(v′)dv′) The reverse process defines the Bremstralung absorption coefficient a ν giving the absorption per ion per electron of velocity v from the radiation field. In TE: Photon Energy Density = U νp = (4π/c) B ν (T)

18. Continuous Opacity 18 The Absorption Coefficient Net energy absorbed must be the product of the photon flux (photon energy = hν) cU νp dν and n i n e f(v)dva ν and (1-e -hν/kT ) or –cU νp dν n i n e f(v)dva ν (1-e -hν/kT ) But in TE that must be equal to the emission: –cU νp dν n i n e f(v)dva ν (1-e -hν/kT ) = n i n e v′f(v′)dv′ dq ν a ν = π/3 (Z 2 e 6 )/(hc m e 2 ν 3 v) This is off by 4/  3 from the exact classical result.

19. Continuous Opacity 19 The Bremstralung Opacity V is v in the previous equations This reduces to This reduces to κ ff (ν) = 4/3 n i n e (2π/3m e kT) 1/2 ((Z 2 e 6 )/(hc m e ν 3 )) g ff (ν) κ ff (ν) = 4/3 n i n e (2π/3m e kT) 1/2 ((Z 2 e 6 )/(hc m e ν 3 )) g ff (ν) g ff (ν) is the mean Gaunt factor and the result has been corrected to the exact classical result.

20. Continuous Opacity 20 Bound Free Opacities This process differs from the free-free case due to the discrete nature of one of the states N th Discrete State: Then the electron capture/ionization process must satisfy: ½ m e v 2 - E n = hν ½ m e v 2 - E n = hν Transition from a Bound State to Continuum or Visa Versa

21. Continuous Opacity 21 Bound Free V is the velocity of the ejected or absorbed electron. Semiclassical treatment of electron capture –Electron Initial Energy: ½ m e v 2 and is positive –The energy decreases in the electric field as it accelerates seeing ion of charge Ze –Q: Why does it loose energy as it accelerates? –A: It radiates it away The energy loss per captured electron may be estimated as: –dq ν = 2πbdbE abs = (8π 2 /3) ((Z 2 e 6 )/(m e 2 c 3 v′ 2 dν))

22. Continuous Opacity 22 Cross Section The cross section for emission of photons into frequency interval dν is defined by –hν dσ ν = dq ν = hν (dσ ν /dν) dν The final state is discrete so define σ cn as the cross section for capture into state n (energy E n ) characterized by n in the range (n, n+dn) so that dσ ν = σ cn dn. Then –hν (dσ ν /dν) = hν σ cn dn/dν –Solve for σ cn and use E n = -I H Z 2 /n 2 to get dn/dν Thus: σ cn = ((2I H Z 2 /n 2 )/(h 2 νn 3 ))(dq ν /dν) = (32/3) π 4 ((Z 2 e 10 )/(m e c 3 h 4 v 2 νn 3 )) = (32/3) π 4 ((Z 2 e 10 )/(m e c 3 h 4 v 2 νn 3 ))

23. Continuous Opacity 23 Photoionization Let σ νn = photoionization cross section The number of photons absorbed of energy hν with the emission of electrons of energy ½m e v 2 - E n from the n th atomic state is –(cU νp /hν) dν σ νn N n (1-e -hν/kT ) –N n is the number of atoms in state n The reverse process: the number of electrons with initial velocities v captured per second into state n is N i σ cn n e v f(v) dv The reverse process is related by detailed balance

24. Continuous Opacity 24 State Population The Boltzmann Equation for the system is: where g 1 = 2 and g n = 2n 2 (H-like ions) and the Saha equation is Detailed Balance says: (cU νp /hν)dνσ νn N n (1-e -hν/kT ) = N i σ cn n e vf(v)dv

25. Continuous Opacity 25 The Bound Free Coefficient We assume –Maxwellian distribution of speeds –Boltzmann and Saha Equations σ νn = (g i /g n ) (m e vc/hν) 2 σ cn We assume the Z-1 electrons in the atom do not participate, set g i = 1, and correct for QM then

26. Continuous Opacity 26 Limits and Conditions Photon must have hν > E n –σ = 0 for hν < En (= I H Z 2 /n 2 ) Recombination –Levels coupled to ν by bound free processes have E n for n > n* where n* = (I H Z 2 /hν) 1/2 Total Bound Free Opacity:

27. Continuous Opacity 27 Other Opacities SourceBFFFOther H I YesYesRayleigh H2-YesYes H-YesYes He I YesYesRayleigh He II YesYes He-Yes Low Temperature Atomic: T < 10000K C I YesYes Mg I YesYes Si I YesYes Al I YesYes Intermediate Atomic: 10000 < T < 20000 Mg II YesYes Si II YesYes Ca II YesYes N I YesYes O I YesYes Hot Atomic: T > 20000K C II – IV YesYes N II – V YesYes Ne I - VI YesYes

28. Continuous Opacity 28 The Rosseland Mean Opacity

29. Continuous Opacity 29 More Spectroscopic Notation Electron and proton have –Spin ½ –Angular momentum  /2 and each has a corresponding magnetic moment The magnetic moment of the electron interacts with both the orbital magnetic moment of the atom (spin-orbit) and the spin of the proton (spin-spin). –Spin-orbit: fine structure (multiplets) –Spin-spin: hyperfine structure H Like Only

30. Continuous Opacity 30 Notation Principal Quantum Number n Orbital Quantum Number l : l ≤ n - 1 – SPDF –l = 0123 Total angular momentum of a state specified by l = l [( l +1)] ½  Spin angular momentum (s=1/2): [s(s+1)] ½  =  3  /2 Orbital and spin angular momentum interacts to produce a total angular momentum quantum number j = l ± 1/2: S States have j = 1/2, P states have j = 1/2, 3/2; and D has j = 3/2, 5/2

31. Continuous Opacity 31 H Like Notation