1. Spectroscopy – Lecture 2 I.Atomic excitation and ionization II. Radiation Terms III. Absorption and emission coefficients IV. Einstein coefficients V. Black Body radiation

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

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

5. I. Atomic excitation and ionization Ratio of populations in two levels m and n : NnNn NmNm = gngn gmgm ( kT ) – exp   =  n –  m

6. I. Atomic excitation and ionization The number of atoms in level n as fraction of all atoms of the same species: NnNn N = g1g1 gngn ( kT ) – exp nn + g2g2 ( kT ) – exp 22 + g3g3 ( kT ) – exp 33... + = gngn u(T) ( kT ) – exp nn u(T) =  ( kT ) – exp ii gigi Partition Function NnNn N = gngn u(T) 10 –  n  = log e/kT = 5040/T

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

8. I. Atomic excitation and ionization If we are comparing the population of the rth level with the ground level: NrNr N1N1 = T g r g 1 log –5040  + log

9. I. Atomic excitation and ionization Example: Compare relative populations between ground state and n=2 for Hydrogen g 1 = 2, g 2 =2n 2 =8 Temp. (K)  =5040/T N 2 /N 1 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 20000100004000060000 N 1 /N 2

11. I. Atomic excitation and ionization : Saha Eq. For collisionally dominated gas: = N1N1 N PePe h3h3 ( 2m2m ) 2 3 ( kT ) 2 5 2u 1 (T) u 0 (T) ( kT ) – exp I N1N1 N = Ratio of ions to neutrals u1u1 u0u0 = Ratio of ionic to neutral partition function m = mass of electron, h = Planck´s constant, P e = electron pressure

12. I. Saha Equation Numerically: N1N1 N = T PePe u 1 u 0 log –5040 I+2.5 log T+log – 0.1762or N1N1 N =  (T) PePe  (T) = 0.65 u 1 u 0 T 2 5 10 –5040I/kT

13. I. Saha Equation Example: What fraction of calcium atoms are singly ionized in Sirius? log N 1 /N 0 = 4.14no neutral Ca T = 10000 K P e = 300 dynes cm –2 Stellar Parameters: Ca I = 6.11 ev log 2u 1 /u o = 0.18 Atomic Parameters:

14. I. Saha Equation Maybe it is doubly ionized: Second ionization potential for Ca = 11.87 ev u 1 = 1.0 log 2u 2 /u 1 = –0.25 log N 2 /N 1 = 0.82 N 2 /N 1 = 6.6 N 1 /(N 1 +N 2 ) = 0.13 In Sirius 13% of the Ca is singly ionized and the remainder is doubly ionized

15. 250001000063004200 T 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 (H  ) Ionization theory thus explains the behavior of the Balmer lines along the spectral sequence.

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

18. II. Radiation Terms: Specific intensity Normal Observer AA  I =  E cos AA tt  lim I = dE cos  dAdA dd dt d  Consider a radiating surface :

19. II. Specific intensity Can also use wavelength interval: I d = I d Note: the two spectral distributions (, ) have different shape for the same spectrum For solar spectrum: I = max at 4500 Ang I = max at 8000 Ang c= d = –(c/ 2 ) d Equal intervals in  correspond to different intervals in. With increasing, a constant d corresponds to a smaller and smaller d

20. Circle indicates the integration is done over whole unit sphere on the point of interest II. Radiation Terms: Mean intensity I. The mean intensity is the directional average of the specific intensity: J = 1 44 ∫ I  d 

21. II. Radiation Terms: Flux Flux is a measure of the net energy across an area  A, in time  t and in spectral range  Flux has directional information: -F +F AA

22. II. Radiation Terms: Flux F  = lim   E  A  t  dE = ∫ d A d t d ∫ I  cos  d  F=F= = ∫ I = dE cos  dAdA dd dt d Recall:

23. II. Radiation Terms: Flux F = dd Looking at a point on the boundary of a radiating sphere ∫ 0 22 ∫ 0  I  cos  sin  d  = dd ∫ 0 22 ∫ 0  I  cos  sin  d  + dd ∫ 0 22 ∫  /2  I  cos  sin  d  = 0 For stars flux is positive Outgoing flux Incoming flux

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

25. II. Radiation Terms: Flux F = 22 If there is no azimuthal (  ) dependence ∫ 0 I  sin  cos  d   Simple case: if I is independent of direction: F =  I (∫ sin  cos  d  = 1/2 ) Note: I is independent of distance, but F obeys the standard inverse square law

26. r F Energy received ~ I  A 1 /r 2 Source image Source Detector element A1A1

27. 2r F Energy received ~ I  A 2 /4r 2 but  A 2 =4  A 1 = I  A  /r 2 A2A2

28. 10r F Energy received ~ I  A 3 /100r 2 but  A 3 = area of source Since the image source size is smaller than our detector element, we are now measuring the flux Detector element Source image A3A3

29. II. K-integral and radiation pressure = d  = ∫ 0  ∫∫ 0  sin  d  d  ∫ –1  22 d   = cos  K = ∫ I  cos   d  1 44 K = 1 2 dd ∫ –1 

30. 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 Pressure= 2 c d E cos  dt dA  component of momentum normal to wall per unit area per time = pressure

31. II. K-integral and radiation pressure P d d  =  cos 2  d d  2I c P d = ∫ I cos 2  d d  /c  P = 44 ∫ 0  I (  )  2  d  = 22 ∫ I (  )  2  d  c c 44 c K P =

32. II. K-integral and radiation pressure Special Case: I is indepedent of direction 3c3c P = 44 I Total radiation pressure: P = ∫ 0  I  d 3c3c 44 = 3c3c 44 T4T4 For Blackbody radiation P = 22 ∫  I (  )  2  d  c

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

35. II. Moments of radiation J = 1 44 ∫ I  d  J = 1 ∫ –1–1  I (  )  d  2 K = 1 2 I    d  ∫ –1  H = 1 2 I  d  ∫ –1  Mean intensity Flux = 4  H Radiation pressure

36. III. The absorption coefficient I I  + dI dx    is the absorption coefficient/unit mass [ ] = cm 2 /gm.  comes from true absorption (photon destroyed) or from scattering (removed from solid angle) dI  = –   I dx

37. III. Optical depth I I  + dI  The radiation sees neither   or dx, but a the combination of the two over some path length L.   = ∫ o L   dx Optical depth L Units: cm 2 gm cm 3 cm

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

39. III. Simple solution to radiative transfer equation I I  + dI dx  dI  = – I d  I  = I  e –  Optically thin e –  = 1-  I = I o (1-  ) dI  = –   I dx

40. III. The emission coefficient I I  + dI dx j dI  = j  I dx j  is the emission coefficient/unit mass [ ] = erg/(s rad 2 Hz gm) j comes from real emission (photon created) or from scattering of photons into the direction considered.

41. III. The Source Function The ratio of the emission to absorption coefficients have units of I. This is commonly referred to as the source function: S = j /  The physics of calculating the source function S can be complicated. Let´s consider the simple cases of scattering and absorption

42. III. The Source Function: Pure isotropic scattering isotropic scattering dd dj to observer The scattered radiation to the observer is the sum of all contributions from all increments of the solid angle like d  Radiation is scattered in all directions, but only a fraction of the photons reach the observer

43. III. The Source Function: Pure isotropic scattering The contribution to the emission from the solid angle d  is proportional to d  and the absorbed energy  I. This is isotropically re-radiated: dj =  I d  /4  ∫ j =  I d  /4  S = j  = ∫ I d  /4  = J The source function is the mean intensity

44. 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: S = 2h 3 c2c2 1 exp(h /kT) – 1

45. III. The Source Function: Scattering + Pure absorption j =  S I +  A B (T) S = j /  where  =  S +  A S =  S  S +  A J  A  S +  A B   + Sum of two source functions weighted according to the relative strength of the absorption and scattering

46. 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 h. The probability that an atom will emit its quantum energy in a time dt, solid angle d  is A ul. A ul is the Einstein probability coefficient for spontaneous emission.

47. IV. Einstein Coefficients If there are N u excited atoms per unit volume the contribution to the spontaneous emission is: j  = N u A ul h 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.

48. IV. Einstein Coefficients The probability for stimulated emission producing a quantum in a time dt, solid angle d  is B ul I dt d  B ul is the Einstein probability coefficient for stimulated emission. True absorption is defined in the same way and the proportionality constant denoted B lu.   I = N l B lu I h – N u B ul I h The amount of reduction in absorption due to the second term is only a few percent in the visible spectrum.

49. IV. Einstein Coefficients NuNu NlNl B lu I True absorption, dependent on I Principle of detailed balance: N u [A lu + B ul I ] = N l B ul I A ul Spontaneous emission, independent of I Negative absorption, dependent on I +B ul I h

50. V. Black body radiation Light enters a box that is a perfect absorber. If the container is heated walls will emit photons that are reabsorbed (thermodynamic equilibrium). A small fraction of the photons will escape through the hole. Detector

51. V. Black body radiation: observed quantities I = c4c4 5 F(c/ T) I = 3 F(  T) F is a function that is tabulated by measurements. This scaling relation was discovered by Wien in 1893 I = 2kT 2 c2c2 2  ckT 4 Rayleigh-Jeans approximation for low frequencies I =

52. V. Black body radiation: The Classical (Wrong) approach Lord Rayleigh and Jeans suggested that one could calculate the number of degrees of freedom of electromagnetic waves in a box at temperature T assuming each degree of freedom had a kinetic energy kT and potential energy 2kT 2 c2c2 I = but as  ∞, I  This is the „ultraviolet“ catastrophe of classical physics Radiation energy density = number of degrees of freedom×energy per degree of freedom per unit volume.

53. V. Black body radiation: Planck´s Radiation Law Derive using a two level atom: NnNn NmNm = gngn gmgm ( kT ) – exp h Number of spontaneous emissions: N u A ul Rate of stimulated emission: N u B ul I Absorption: N l B ul I

54. V. Black body radiation: Planck´s Radiation Law In radiative equilibrium collisionally induced transitions cancel (as many up as down) N u A ul + N u B ul I = N l B lu I I = A ul B lu (N l /N u ) – B ul I = A ul (g l /g u )B lu exp(h /kT) – B ul

55. V. Black body radiation: Planck´s Radiation Law This must revert to Raleigh-Jeans relation for small I ≈ A ul (g l /g u )B lu – B ul + (g l /g u )B lu h /kT Expand the exponential for small values  e x = 1+x) h /kT << 1 this can only equal 2kT 2 /c 2 if B ul = B lu g l /g u A ul = 2h 3 c2c2 B ul Note: if you know one Einstein coeffiecient you know them all

56. V. Black body radiation: Planck´s Radiation Law I = 1 (exp(h /kT) – 1) 2h 3 c2c2 I = 1 (exp(hc/ kT) – 1) 2hc 2 5

57. Maximum I = T = 0.5099 cm K Maximum I = 5.8789×10 10 Hz K V. Black body radiation: Planck´s Radiation Law I  = 2kT  2 c2c2 Rayleigh-Jeans approximation → 0 2  ckT 4 I = I =I = 2h32h3 c2c2 e – h /kT I =I = 2  hc 2 5 e – h c /k T Wien approximation → ∞

58. V. Black body radiation: Stefan Boltzman Law In our black body chamber escaping radiation is isotropic and no significant radiation is entering the hole, therefore F =  I F d =  ∫ 0 ∞ ∫ 0 ∞ 1 (exp(h /kT) – 1) 2h 3 c2c2 d Integral =  4 /15 4 x3x3 = 22 c2c2 ( kT h ) ∫ 0 ∞ e x –1 dxdx ∞ 0 25k425k4 F ∫ d = 15h 3 c 2 T4T4 =  T 4 x=h  kT

59. V. Note on Einstein Coefficients and BB radiation In the spectral region where h /kT >> 1 spontaneous emissions are more important than induced emissions In the ultraviolet region of the spectrum replace I by Wien´s law: B ul I = B ul 2h 3 c2c2 e –h /kT = A ul e –h /kT << A ul Induced emissions can be neglected in comparison to spontaneous emissions

60. V. Note on Einstein Coefficients and BB radiation In the spectral region where h /kT << 1 negative absorption (induced emissions) are more important than spontaneous emissions In the far infrared region of the spectrum replace I  by Rayleigh-Jeans law: B ul I = B ul 2 kT c2c2 = A ul c2c2 2h 3 2 kT c2c2 = A ul kT h >> A ul The number of negative absorptions is greater than the spontaneous emissions

61. IUBJK log I ~ –4 log log I ~ –5 log  – 1/ 40000 20000 10000 5000 3000 1500 1000 500 750 T (K)

62. T = 6000 K I I

63. V. Black body radiation: Photon Distribution Law N = 1 (exp(h /kT) – 1) 2h 2 c2c2 N = 1 (exp(hc/ kT) – 1) 2hc 2 4 Detectors detect N, not I !