1. The Hitchhiker's Guide to the Galaxy
“Probability factor of one to one...we have normality, I repeat we have normality. Anything you still can't cope with is therefore your own problem.”
The Hitchhiker's Guide to the Galaxy

2. Paper Topics
The final paper is to be 8-10 pages (not counting references), double-spaced.
On Feb. 19. You need to provide me two possible topics, each one having 2 references.
It will be worth 20 points.

3. The Boltzmann equation for excitation states (atomic levels)
And the Saha equation for ionization states:

4. Width of the line
The width of a spectral line is determined by 3 things:
1) Natural broadening (damping)
2) Doppler broadening
3) Pressure broadening (damping)

5. Pressure (and collisional) broadening
The bound-bound orbitals that we've already discussed can be perturbed in a collision with a neutral atom, or by a close encounter with the electric field of an ion. The results of individual collisions are called collisional broadening and the statistical effects of the electric fields of large number of closely passing ions can be termed pressure broadening.
In both cases, the broadening depends on the time between collisions or encounters with atoms or ions.

6. Pressure (and collisional) broadening
Calculating the precise shape is very complicated, but the general shape is similar to natural broadening- both of which contribute to what is called the damping profile.

7. Pressure (and collisional) broadening
Dto is the average time between collisions and is approximately equal to the mean free path (l=1/ns where s is a collision cross section) between collisions divided by the average speed of the atoms-
This leads to an equation for Dto
Where (as usual) n is the number density of atoms and m is the mass of the atoms. Then the width is

8. Pressure (and collisional) broadening
Putting in numbers for H atoms in the Sun (at T=5770 K) this is
Dl~2.36x10-5nm. Similar to natural broadening.

9. Pressure (and collisional) broadening
Now the luminosity classes become apparent:
Narrower lines in giants and supergiants (which happen to be luminous because of their large size) result from lower number densities. Pressure broadening ( which is proportional to n) broadens the lines formed in denser atmospheres.

10. The width of a spectral line is determined by 3 things:
Width of the line
The width of a spectral line is determined by 3 things:
1) Natural broadening (damping)
2) Doppler broadening
3) Pressure broadening (damping)

11. Voigt profiles: The total line profile is called the Voigt profile and has Doppler and damping components. Doppler dominates near the core and damping in the wings.

12. Voigt profiles: Since Doppler and damping components originate from different processes, the Voigt profile tells us the physical nature of the photosphere.

13. Quantities of atoms
The picture is a blackbody photosphere base that emits a continuous spectrum. The atoms above that absorb photons.
Let N be the number of atoms of a certain element lying above a unit area of photosphere. Then N is a column density and Na would be the number of atoms with electrons in the proper orbital for absorbing a photon at the wavelength of the spectral line (Na is the # of absorbers).

14. Voigt profiles showing the effect of changing the number of absorbers, Na

15. But eventually the line becomes optically thick and bottoms out
But eventually the line becomes optically thick and bottoms out. Adding more atoms no longer increases W at the same rate.
Initially, W is proportional to Na so adding twice as many atoms increases W by 2.

16. But soon adding atoms increases
the pressure broadening, and then
Initially the wings grow as

17. The goals is to find the number of absorbing atoms per unit area, Na,
Quantities of atoms
The goals is to find the number of absorbing atoms per unit area, Na,
Use a curve of growth technique which makes use of calculated and observed line profiles.
Basically it compares the equivalent width with fNa (oscillator strength multiplied by the Number of available absorbers- these are in the lower energy state).

18. Oscillator strengths: f
Not all transitions are equally likely (in fact no two are). The relative probabilities of an electron making a transition is called the oscillator strength, or f- value.
For H, f(Ha)=0.637 while f(Hb)=0.119, so the Ha line is about 5 times more likely to occur.

19. Oscillator strengths: f
Not all transitions are equally likely (in fact no two are). The relative probabilities of an electron making a transition is called the oscillator strength, or f- value.
For H, f(Ha)=0.637 while f(Hb)=0.119, so the Ha line is about 5 times more likely to occur.
These are measured in labs.

20. Oscillator strengths: f
Multiplying Na by f gives the number of atoms lying above each unit area of the photosphere that are actively involved in producing a given spectral line.

21. The graph is called a curve of growth.
graph showing how equivalent width grows with Na
This can be translated to a
The graph is called a curve of growth.

22. A general curve of growth for the Sun

23. Curve of growth
Using the curve of growth and a measured equivalent width, we can determine the number of absorbing atoms per unit area (the graph was made in cgs!). Then the Boltzmann and Saha equations convert this into the total number of atoms lying above the base of the photosphere.

24. Let's try one! (sorry, in angstroms) This is for sodium in the Sun.
l(A)
W(A) f
log(W/l)
log[f(l/5000A)]
0.088
0.0214
-4.58
-1.85
0.730
0.645
-3.90
-0.12
Both are transitions from the ground state to the 1st and 2nd excited states of the Na I atom.
Use T=5,800K and Pe=10 dynes/cm2 (1Pa).

25. Na I at T=5800K, Pe=10 dynes/cm2(1Pa)
l(A)
W(A) f
log(W/l)
log[f(l/5000A)]
0.088
0.0214
-4.58
-1.85
0.730
0.645
-3.90
-0.12
Column 3 has units to match our CoG graph.
Then read off log Nf[l/5000]

26. Reminder: this graph was made in cgs units.
14.83
13.20
Reminder: this graph was made in cgs units.

27. Na I at T=5800K, Pe=10 dynes/cm2
l(A)
W(A) f
log(W/l)
log[f(l/5000A)]
0.088
0.0214
-4.58
-1.85
0.730
0.645
-3.90
-0.12
13.20 and (I got these from a table). But these are for log Nf and we have to correct them to log Na (which is the ground state). To do this, we use the log(f) provided:
Log Na(3302)=13.20-(-1.85)= 15.05
Log Na(5890)=14.83-(-0.12)= 14.95
These average to 15.0, which means there are 1015 Na I atoms in the ground state per cm2 of photosphere.

28. Now we have to find how many atoms are in other states.
Na I at T=5800K, Pe=10 dynes/cm2
l(A)
W(A) f
log(W/l)
log[f(l/5000A)]
0.088
0.0214
-4.58
-1.85
0.730
0.645
-3.90
-0.12
These average to 15.0, which means there are 1015 Na I atoms in the ground state per cm2 of photosphere.
Now we have to find how many atoms are in other states.

29. Na I at T=5800K, Pe=10dynes/cm2
Apply the Boltzmann equation to determine how many are in the excited state compared to the ground state. Use the wavelength to determine the difference in energies:
For the ground state, g1=2 and for the first excited state, g5890=6 g3302=2, k=8.6174x10-5eV/K, hc=12,400eV/Ang.

30. Na I at T=5800K, Pe=10dynes/cm2
Apply the Boltzmann equation to determine how many are in the excited state compared to the ground state. Use the wavelength to determine the difference in energies:
For our 2 lines, this results in 5.45x10-4 (3302) and (5890) so nearly all the Na I atoms are in the ground state (to ~5%).

31. Na I at T=5800K, Pe=10 dynes/cm2 Now the Saha equation:
With ZI=2.4, ZII=1.0, cI=5.14eV, k=8.6174x10-5eV/K,
h=6.626x10-27erg.s, me=9.109x10-28g,
k=1.38x10-16erg/K.
NOTE: constants in cgs units.

32. So for each Na from our curve of growth, there are 2,430 Na II atoms.
Na I at T=5800K, Pe=10 dynes/cm2
Now the Saha equation:
NII/NI=2,430.
So for each Na from our curve of growth, there are 2,430 Na II atoms.
Note that cII=47.3eV is high enough not to worry about Na III or higher ionizations.

33. Na I at T=5800K, Pe=10 dynes/cm2 Now the Saha equation: NII/NI=2,430.
So for each Na from our curve of growth, there are 2,430 Na II atoms.
We found 1015 Na atoms/cm2 in our curve-of-growth analysis.
So the total number of Na atoms is N~2.43x1018/cm2.

34. The total number of Na atoms is N~2.43x1018/cm2.
Na I at T=5800K, Pe=10 dynes/cm2
Now the Saha equation:
The total number of Na atoms is N~2.43x1018/cm2.
The mass of a sodium atom is 3.82x10-23kg, so the mass of sodium atoms, per square centimeter (cgs, remember) is 9.3x10-5g/cm2

35. Na I at T=5800K, Pe=10 dynes/cm2 Mass of Na is 9.3x10-5g/cm2
For comparison:
H is 1.1
He is 0.43
O is 0.015
C is
Ne is
N is
Fe is

36. At this point, you have the ability to determine abundances in stellar atmospheres!

37. However….. curve-of-growth analysis is rarely used now.
At this point, you have the ability to determine abundances in stellar atmospheres!
However….. curve-of-growth analysis is rarely used now.
If used across different ions of the same element, it can deduce electron pressure
From May 2002

38. Spectral synthesis
This is comparing a model atmosphere to a spectrum. This is now the more common way to derive stellar abundances.

39. Then you would have a program that interpolates between models to fit actual spectra.

42. Nomenclature
It is more typical to use a comparative nomenclature. This is usually related to solar abundance and/or compared to H in the Sun.
The 12 scale compares the number of atoms to that of H in the Sun (the log of which is 12).
For the Sun, nH=6.6x1023 atoms/cm2 so in our example of Na for the 12 scale-
log(2.43x1018/6.6x1023 )+12= ?

43. This would be called 6.6 dex.
Nomenclature
It is more typical to use a comparative nomenclature. This is usually related to solar abundance and/or compared to H in the Sun.
For the Sun, nH=6.6x1023 atoms/cm2 so in our example of Na for the 12 scale-
log(2.43x1018/6.6x1023 )+12= 6.6
This would be called 6.6 dex.

44. Nomenclature
Another way. This gain uses solar H for the comparison, but also includes H for the target star.
This is what I've more commonly seen. It's also a log scale, so abundances are quoted as dex, but it compares the ratio of an element to H in both the star and the Sun.
So it's a direct comparison to solar abundances for any atom, not just H.

45. Nomenclature log[N(Fe)/N(H)] for the Sun is -4.33.
A direct comparison to solar abundances.
log[N(Fe)/N(H)] for the Sun is
So if a star has a solar abundance of iron, its [Fe/H]=0.
If it were down by a factor of 10 (1 dex), then [Fe/H]= -1

46. Back to the Book- Chapter 1: Thermal sources
Low pressure gases can still resemble continuous (blackbody) sources, so long as they're thermalized.
This means that the individual atoms have had time to interact fully such that as an ensemble, they have a roughly Gaussian distribution with most particles near a given temperature.
In this case, on average, each photon would have energy E~kT where k is Botzmann's constant.

47. Thermal sources
For the corona this isn't quite true, however it indicates that it is true for the mechanism that is heating the gas.

48. E~kT where k is Botzmann's constant .
Thermal sources
So if the corona is at a temperature of 1 million kelvin, what is the peak wavelength of light emitted?
E~kT where k is Botzmann's constant .
k= 8.617x10-5eV/K

49. E~kT where k is Botzmann's constant .
Thermal sources
So if the corona is at a temperature of 1 million kelvin, what is the peak wavelength of light emitted?
E~kT where k is Botzmann's constant .
k= 8.617x10-5eV/K
E = 86 eV
l=hc/E=1240/86 = 14.4nm Far-UV

50. Solar flares can produce hard x-rays and g-ray emission
Solar flares can produce hard x-rays and g-ray emission. This can make the Sun's output lethal to us Earthly inhabitants.

53. High energy space
Prior to satellites, we could not observe the Sun (or any star) in UV, X-rays or g-rays.