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

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

4. I. Atomic excitation and ionization
The fraction of atoms (or ions) excited to the nth level is:
Nn = constant gn exp(–cn/kT)
statistical weight
Boltzmann factor
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. ( ) I. Atomic excitation and ionization
Ratio of populations in two levels m and n : Nn Nm = gn gm ( kT ) –
exp Dc
Dc = cn – cm

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

8. I. Atomic excitation and ionization
If we are comparing the population of the rth level with the ground level: gr Nr N1
–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

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

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

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

13. I. Saha Equation
Maybe it is doubly ionized:
Second ionization potential for Ca = ev
u1 = 1.0 log 2u2/u1 = –0.25
N2/N1 = 6.6
N1/(N1+N2) = 0.13
log N2/N1 = 0.82
13% Ca II
87% Ca III

14. T
25000
10000
6300
4200

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

17. Can also use wavelength interval:
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

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

19. 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:
+Fn DA
-Fn

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

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

22. Astronomical Example of Negative Flux: Close Binary system:
II. Radiation
Astronomical Example of Negative Flux: Close Binary system:
Hot Spot
Cool star (K0IV)
Hot star (DAQ3)
Negative and postive flux through this area

23. ∫ (∫ sin q cos q dq = 1/2 ) II. Radiation Terms: Flux
If there is no azimuthal (f) dependence
p/2 ∫ Fn = 2p
In sin q cos q dq
Simple case: if In is independent of direction: Fn = p In
(∫ sin q cos q dq = 1/2 )

24. Energy received ~ InDA1/r2
Detector element
Source F
DA1 area covered by solid angle q2
Detector subtends angle q on the sky r
Source image

25. Energy received ~ InDA2/4r2 but DA2 =4DA1 = In DA1/r2F 2r
DA2

26. 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

27. ∫ ∫ ∫ ∫ II. K-integral and radiation pressure Kn = In cos2q dw 1 4p =2p p
sin q dq df ∫ –1 +1 2p dm
m = cos q
Kn = 1 2
m2 dm ∫ –1 +1

28. 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 En cosq
dt dA q
component of momentum normal to wall per unit area per time = pressure

29. ∫ ∫ II. K-integral and radiation pressure Pn dn dw = cos2q dn dw 2In c
In cos2q dn dw/c +1
Pn = 4p ∫
In (m) m2 dm = 2p -1 c 4p c Kn
Pn =

30. ∫ ∫ II. K-integral and radiation pressure
Special Case: In is indepedent of direction Pn = 2p ∫ -1 +1
In (m) m2 dm c 3c Pn = 4p In
Total radiation pressure: Pn = ∫ 
In dn 3c 4p 4s T4
For Blackbody radiation

33. ∫ ∫ ∫ ∫ II. Moments of radiation Jn = 1 4p In dw Jn = 1 In (m) dm 2–1 +1
In (m) dm 2
Mean intensity
Hn = 1 2
I(m) m dm ∫ –1 +1
Flux = 4pH
Kn = 1 2
I(m) m2 dm ∫ –1 +1
Radiation pressure

34. III. The absorption coefficient
In + dIn In dx kn
kn is 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

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

36. 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

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

38. III. The emission coefficient
In + dIn In dx jn
dIn = jnr In dx
jn is 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.

39. 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

40. • III. The Source Function: Pure isotropic scatteringdw
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

41. ∫ ∫ 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 ∫
jn =
knIn dw/4p
Sn = jn kn = ∫
In dw/4p
= Jn
The source function is the mean intensity

42. 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

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

44. 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. Aul is the Einstein probability coefficient for spontaneous emission.

45. IV. Einstein Coefficients
If there are Nu excited atoms per unit volume the contribution to the spontaneous emission is:
jnr = Nu Aul hn
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.

46. IV. Einstein Coefficients
The probability for stimulated emission producing a quantum in a time dt, solid angle dw is Bul In dt dw. Bul is the Einstein probability coefficient for stimulated emission.
True absorption is defined in the same way and the proportionality constant denoted Blu.
knrIn = NlBluInhn – NuBulInhn
The amount of reduction in absorption due to the second term is only a few percent in the visible spectrum.

47. IV. Einstein CoefficientsNu hn
BluIn
True absorption, dependent on In
Aul
Spontaneous emission, independent of In
Negative absorption, dependent on In
+BulIn Nl
Principle of detailed balance:
Nu[Aul + BulIn] = NlBluIn

48. V. Black body radiation
Detector
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.

49. V. Black body radiation: observed quantities
Il =
F(c/lT) l5
F is a function that is tabulated by measurements
In = n3
F(n/T)
I n=
2kTn2 c2
2pckT l4
Rayleigh-Jeans approximation for low frequencies
I l=

50. 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
Radiation energy density = number of degrees of freedom×energy per degree of freedom per unit volume.
2kTn2 c2
I n =
but as n
∞, In 
This is the „ultraviolet“ catastrophe of classical physics

51. ( ) V. Black body radiation: Planck´s Radiation Law
Derive using a two level atom: Nn Nm = gn gm ( kT ) –
exp hn
Number of spontaneous emissions: NuAul
Rate of stimulated emission: NuBulIn
Absorption: NlBluIn

52. NuAul + NuBulIn = NlBluIn
V. Black body radiation: Planck´s Radiation Law
In radiative equilibrium collisionally induced transitions cancel (as many up as down)
NuAul + NuBulIn = NlBluIn
In =
Aul
Blu(Nl/Nu) – Bul
In =
Aul
(gl/gu)Bluexp(hn/kT) – Bul

53. V. Black body radiation: Planck´s Radiation Law
This must revert to Raleigh-Jeans relation for small n
Expand the exponential for small values (ex = 1+x)
In ≈
Aul
(gl/gu)Blu – Bul + (gl/gu)Bluhn/kT
hn/kT << 1 this can only equal 2kTn2/c2 if
Bul = Blugl/gu
Aul =
2hn3 c2
Bul
Note: if you know one Einstein coeffiecient you know them all

54. V. Black body radiation: Planck´s Radiation Law
In = 1
(exp(hn/kT) – 1)
2hn3 c2
Il = 1
(exp(hc/lkT) – 1)
2hc2 l5

55. V. Black body radiation: Planck´s Radiation Law
Maximum Il = lT = cm K
Maximum In = ×1010 Hz K
I n =
2kTn 2 c2
Rayleigh-Jeans approximation n→ 0
2pckT l4
I l=
I n =
2phn 3 c2
e–hn/kT
I l =
2phc2 l5
e–hc/klT
Wien approximation
n→∞

56. ∫ ∫ ∫ ( ) V. Black body radiation: Stefan-Boltzmann Law
In our black body chamber escaping radiation is isotropic and no significant radiation is entering the hole, therefore Fn = pIn Fn
dn = p ∫ ∞ 1
(exp(hn/kT) – 1)
2hn3 c2 dn 4 x3 = 2p c2 ( kT h ) ∫ ∞
ex–1 dx
x=hn/kT ∞
2p5k4 Fn ∫
dn =
15h3c2 T4
= sT4
Integral = p4/15

57. V. Note on Einstein Coefficients and BB radiation
In the spectral region where hn/kT >> 1 spontaneous emissions are more important than induced emissions
In the ultraviolet region of the spectrum replace In by Wien´s law:
BulIn = Bul
2hn3 c2
e–hn/kT
= Aul
e–hn/kT
<< Aul
Induced emissions can be neglected in comparison to spontaneous emissions

58. V. Note on Einstein Coefficients and BB radiation
In the spectral region where hn/kT << 1 negative absorption (induced emissions) is more important than spontaneous emissions
In the far infrared region of the spectrum replace In by Rayleigh-Jeans law:
BulIn = Bul
2nkT c2 =
Aul c2
2hn3
2nkT =
Aul kT hn
>>Aul
The number of negative absorptions is greater than the spontaneous emissions

59. log I ~ –4 log l log I ~ –5 log l – 1/l I U B J K T (K) 40000 20000
10000
5000
log I ~ –5 log l – 1/l
3000
1500
1000
750
500

60. T = 6000 K Il In

61. V. Black body radiation: Photon Distribution Law
Nn = 1
(exp(hn/kT) – 1)
2hn2 c2
Nl = 1
(exp(hc/lkT) – 1)
2hc2 l4
Detectors detect N, not I !

62. Il Nl