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Spectroscopy – Lecture 2

Spectroscopy – Lecture 2. Atomic excitation and ionization Radiation Terms Absorption and emission coefficients Einstein coefficients Black Body radiation. ·. ·. ·. ·. ·. ·. I. c. I. Atomic excitation and ionization. E >0. E =0. ∞. qualitative energy level diagram. n.

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Spectroscopy – Lecture 2

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  1. Spectroscopy – Lecture 2 • Atomic excitation and ionization • Radiation Terms • Absorption and emission coefficients • Einstein coefficients • Black Body radiation

  2. · · · · · · I c I. Atomic excitation and ionization E>0 E=0 ∞ qualitative energy level diagram n 3 2 • Mechanisms for populating and depopulating the levels in stellar atmospheres: • radiative • collisional • spontaneous transitions E=–I 1

  3. H, He relatively hard to ionize → hot stars you see absorption lines of Hydrogen Metals relatively easy for first ionization

  4. statistical weight Boltzmann factor I. Atomic excitation and ionization The fraction of atoms (or ions) excited to the nth level is: Nn = constant gnexp(–cn/kT) Statistical weight is 2J+1 where J is the inner quantum number (Moore 1945)1. For hydrogen gn=2n2 1 Moore, C.E. 1945, A Multiplet Table of Astrophysical Interest, National Bureau of Standards

  5. Nn gn ( Dc ) exp – = gm kT Nm I. Atomic excitation and ionization Ratio of populations in two levels m and n : Dc = cn – cm

  6. ... + + + gn Nn Nn ( ( ( ( ( c3 ci c2 cn cn ) ) ) ) ) gn gn exp exp exp exp exp – – – – – g3 g2 kT kT kT kT kT = N N u(T) u(T) u(T) = S gi 10 –qcn = I. Atomic excitation and ionization The number of atoms in level n as fraction of all atoms of the same species: = g1 Partition Function q= log e/kT = 5040/T

  7. From Allen‘s Astrophysical Quantities Q = 5040/T Y = stage of ionization. Y = 1 is neutral, Y = 2 is first ion.

  8. Nr N1 I. Atomic excitation and ionization If we are comparing the population of the rth level with the ground level: gr –5040 log c + log = g1 T

  9. I. Atomic excitation and ionization Example: Compare relative populations between ground state and n=2 for Hydrogen g1 = 2, g2=2n2 =8 Temp. (K)q=5040/T N2/N1 6000 0.840 0.00000001 8000 0.630 0.0000016 10000 0.504 0.00031 15000 0.336 0.00155 20000 0.252 0.01100 40000 0.126 0.209

  10. I. Atomic excitation and ionization N2/N1 10000 20000 40000 60000

  11. ) ) ( ( kT 2pm 2u1(T) ( I ) Pe exp – kT u0(T) h3 N1 N1 u1 3 5 = Ratio of ions to neutrals N u0 N 2 2 Ratio of ionic to neutral partition function = I. Atomic excitation and ionization : Saha Eq. For collisionally dominated gas: = m = mass of electron, h = Planck´s constant, Pe = electron pressure

  12. –5040 u1 log Pe I + 2.5 log T + log 0.1762 – = u0 T N1 N1 F(T) 5 or N N 2 = Pe u1 u0 F(T) = 0.65 T 10–5040I/kT I. Saha Equation Numerically:

  13. Stellar Parameters: T = 10000 K Pe = 300 dynes cm–2 Atomic Parameters: Ca I = 6.11 ev log 2u1/uo = 0.18 I. Saha Equation Example: What fraction of calcium atoms are singly ionized in Sirius? log N1/N0 = 4.14 no neutral Ca

  14. N2/N1 = 6.6 N1/(N1+N2) = 0.13 I. Saha Equation Maybe it is doubly ionized: Second ionization potential for Ca = 11.87 ev u1 = 1.0 log 2u2/u1 = –0.25 log N2/N1 = 0.82 In Sirius 13% of the Ca is singly ionized and the remainder is doubly ionized because of the low ionization potential of Ca.

  15. From Lawrence Allen‘s The Atmospheres of the Sun and Stars T 25000 10000 6300 4200 The number of hydrogen atoms in the second level capable of producing Balmer lines reaches its maximum at Teff ≈ 10000 K

  16. Behavior of the Balmer lines (Hb) Ionization theory thus explains the behavior of the Balmer lines along the spectral sequence.

  17. How can a T=40000 star ionize Hydrogen? I = 13.6 ev = 2.2 x 10–11 ergs E = kT → T = 160.000 K So a star has to have an effective temperature to of 160.000 K to ionize hydrogen? Answer later.

  18. Predicted behavior according to Ionization Theory Ionization theory‘s achievement was the intepretation of the spectral sequence as a temperature sequence Observed behavior according to Ionization Theory

  19. Consider a radiating surface: q DA Dw DEn In = lim cos q DA Dw Dt Dn dEn In = cos q dA dw dt dn II. Radiation Terms: Specific intensity Normal Observer

  20. II. Specific intensity Can also use wavelength interval: Indn = Ildl Note: the two spectral distributions (n,l) have different shape for the same spectrum For solar spectrum: Il= max at 4500 Ang In = max at 8000 Ang c=ln dn = –(c/l2) dl Equal intervals in l correspond to different intervals in n. With increasing l, a constant dl corresponds to a smaller and smaller dn

  21. 1 ∫ 4p II. Radiation Terms: Mean intensity I. The mean intensity is the directional average of the specific intensity: Jn = Indw Circle indicates the integration is done over whole unit sphere on the point of interest

  22. +Fn -Fn DA II. Radiation Terms: Flux Flux is a measure of the net energy across an area DA, in time Dt and in spectral range Dn Flux has directional information:

  23. S DEn Fn = lim DA Dt Dn ∫ dEn = Incos q dw = ∫ dA dt dn Recall: ∫ Incos q dw Fn = dEn In = cos q dA dw dt dn II. Radiation Terms: Flux

  24. 2p 2p p 2p ∫ ∫ ∫ ∫ 0 0 0 0 p/2 = 0 ∫ = df In cos q sin q dq 0 p ∫ + df In cos q sin q dq p/2 II. Radiation Terms: Flux Looking at a point on the boundary of a radiating sphere Fn = df In cos q sin q dq Outgoing flux Incoming flux For stars flux is positive

  25. Hot Spot Cool star (K0IV) Hot star (DAQ3) II. Radiation Astronomical Example of Negative Flux: Close Binary system:

  26. Fn p In = II. Radiation Terms: Flux If there is no azimuthal (f) dependence p/2 ∫ Fn = 2p In sin q cos q dq 0 Simple case: if In is independent of direction: (∫ sin q cos q dq = 1/2 ) Note: In is independent of distance, but Fn obeys the standard inverse square law

  27. Flux radiating through a sphere of radius d is just F = L/4pd2 L = 4pR2In (=sT4) d

  28. Detector element Source F r DA1 Source image Energy received ~ InDA1/r2

  29. Energy received ~ InDA2/4r2 but DA2 =4DA1 = In DA1/r2 F 2r DA2

  30. Energy received ~ InDA3/100r2 but DA3 = area of source Source image F DA3 Detector element 10r Since the image source size is smaller than our detector element, we are now measuring the flux The Sun is the only star for which we measure the specific intensity

  31. Kn = In cos2q dw 1 ∫ 2p p ∫ ∫ ∫ +1 ∫ dw = sin q dq df = 2p dm 4p 0 0 –1 m = cos q +1 1 ∫ m2dm Kn = 2 –1 II. K-integral and radiation pressure

  32. d En cosq q Pressure= dt dA 2 c II. K-integral and radiation pressure This intergral is related to the radiation pressure. Radiation has momentum = energy/c. Consider photons hitting a solid wall component of momentum normal to wall per unit area per time = pressure

  33. 2In Pn dn dw = cos2q dn dw c ∫ Pndn = In cos2q dn dw/c +1 +1 ∫ ∫ 2p 4p In (m) m2dm = In (m) m2dm Pn = c c -1 0 4p Kn Pn = c II. K-integral and radiation pressure

  34. +1 ∫ 2p Pn In (m) m2dm = c -1 4p Pn In = 3c 4p  ∫ 4s For Blackbody radiation Pn Indn = = T4 3c 3c 0 II. K-integral and radiation pressure Special Case: In is indepedent of direction Total radiation pressure:

  35. Radiation pressure is a significant contribution to the total pressure only in very hot stars.

  36. Jn = Indw ∫ +1 1 Jn = 1 In (m) dm ∫ 2 4p –1 +1 1 ∫ I(m) m dm Hn = 2 –1 +1 ∫ 1 I(m) m2dm Kn = 2 –1 II. Moments of radiation Mean intensity Flux = 4pH Radiation pressure m = cos q

  37. In + dIn In III. The absorption coefficient dx kn knis the absorption coefficient/unit mass [ ] = cm2/gm. kn comes from true absorption (photon destroyed) or from scattering (removed from solid angle) dIn = –knr In dx

  38. L ∫ o gm cm2 gm cm3 III. Optical depth L In + dIn In kn The radiation sees neither knr or dx, but a the combination of the two over some path length L. tn= knr dx Optical depth Units: cm

  39. III. Optical depth Optically thick case: t >> 1 => a photon does not travel far before it gets absorbed Optically thin case: t << 1 => a photon can travel a long distance before it gets absorbed

  40. Optically thin t < 1 t ≈1 Optically thick t > 1 LucaSebben

  41. III. Simple solution to radiative transfer equation In + dIn In dIn = –knr In dx dx kn dIn = – In dt In = Inoe–t Optically thin e–t = 1-t In = Ino(1-t)

  42. III. The emission coefficient In + dIn In dx jn dIn = jnr In dx jnis the emission coefficient/unit mass [ ] = erg/(s rad2 Hz gm) jn comes from real emission (photon created) or from scattering of photons into the direction considered.

  43. III. The Source Function The ratio of the emission to absorption coefficients have units of In. This is commonly referred to as the source function: Sn = jn/kn The physics of calculating the source function Sn can be complicated. Let´s consider the simple cases of scattering and absorption

  44. dw • III. The Source Function: Pure isotropic scattering djn to observer isotropic scattering The scattered radiation to the observer is the sum of all contributions from all increments of the solid angle like dw. Radiation is scattered in all directions, but only a fraction of the photons reach the observer

  45. ∫ ∫ jn = knIn dw/4p jn Sn = = In dw/4p = Jn kn III. The Source Function: Pure isotropic scattering The contribution to the emission from the solid angle dw is proportional to dw and the absorbed energy knIn. This is isotropically re-radiated: djn = knIn dw/4p The source function is the mean intensity

  46. III. The Source Function: Pure absorption All photons are destroyed and new ones created with a distribution governed by the physical state of the material. Emission of a gas in thermodynamic equilibrium is governed by a black body radiator: 2hn3 1 Sn = c2 exp(hn/kT) – 1

  47. knS knA Sn = Jn Bn (T) + knS+ knA knS+ knA III. The Source Function: Scattering + Pure absorption jn = knSIn +knABn(T) Sn = jn/kn where kn = knS+ knA Sum of two source functions weighted according to the relative strength of the absorption and scattering

  48. IV. Einstein Coefficients When dealing with spectral lines the probabilities for spontaneous emission can be described in terms of atomic constants Consider the spontaneous transition between an upper level u and lower level l, separated by energy hn. The probability that an atom will emit its quantum energy in a time dt, solid angle dw is Aul. Aulis the Einstein probability coefficient for spontaneous emission.

  49. IV. Einstein Coefficients If there are Nu excited atoms per unit volume the contribution to the spontaneous emission is: jnr = NuAulhn If a radiation field is present that has photons corresponding to the energy difference between levels l and u, then additional emission is induced. Each new photon shows phase coherence and a direction of propagation that is the same as the inducing photon. This process of stimulated emission is often called negative absorption.

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