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tu144
2011-May-24, 01:01 PM
I'm trying to read and understand this page on Interpreting Emission Line Spectra (http://www.astr.ua.edu/keel/galaxies/emission.html)
In part named "Temperature and Density Determinations" one can read:

For temperature determinations, we want to measure the ratio of electrons with different energies - so we look for pairs of lines of the same species with very different excitation energies. The most-used example is [O III] with lines at 4363 and 4959+5007 Angstrems.
The ratio being:
http://www.astr.ua.edu/keel/galaxies/emisseqn8.gif
4959+5007 Angstrems means we look for intensity of 4959line and add it with intensity of 5007 line?
Why can't we use only one line?

Ok, now something more fundamental - the temperature.
I was always taught that temperature is kinetic energy of atoms moving around and colliding with each other.
Now, when we get these 4959A and 5007A lines, from black body spectrum we can see that this corresponds to temperatures 5845K and 5788K.
When atoms begin to move so fast that their kinetic energy corresponds to 5788K the double ionized oxygen OIII starts to emit 5007A photons. If they start to move even faster (corresponding temperature 5845K) OIII starts to emit 4959A photons.
Is that correct?

Now, if it is correct, how do we know that OIII is being excited to these energy levels by the temperature(kinetic energy of atoms) and not by other photons incoming from nearby star or other light source?

Amber Robot
2011-May-24, 01:27 PM
Now, when we get these 4959A and 5007A lines, from black body spectrum we can see that this corresponds to temperatures 5845K and 5788K.

As I tried to point out in another thread, a single photon does not have a temperature, so you can't assign a wavelength to a temperature like this. Am I misunderstanding what you are suggesting here?

Ken G
2011-May-24, 01:52 PM
Now, if it is correct, how do we know that OIII is being excited to these energy levels by the temperature(kinetic energy of atoms) and not by other photons incoming from nearby star or other light source?We have to assume certain things about how the atom is excited in order for the formula to apply-- it is not a general result.

ngc3314
2011-May-24, 05:42 PM
I'm trying to read and understand this page on Interpreting Emission Line Spectra (http://www.astr.ua.edu/keel/galaxies/emission.html)
In part named "Temperature and Density Determinations" one can read:

The ratio being:
http://www.astr.ua.edu/keel/galaxies/emisseqn8.gif
4959+5007 Angstrems means we look for intensity of 4959line and add it with intensity of 5007 line?
Why can't we use only one line?

We can, it's just the convention to use the sum in that ratio. Because of statistical weights, decays in the 5007 line have exactly 3x the number of photons in the 4959 line.


Ok, now something more fundamental - the temperature.
I was always taught that temperature is kinetic energy of atoms moving around and colliding with each other.
Now, when we get these 4959A and 5007A lines, from black body spectrum we can see that this corresponds to temperatures 5845K and 5788K.
When atoms begin to move so fast that their kinetic energy corresponds to 5788K the double ionized oxygen OIII starts to emit 5007A photons. If they start to move even faster (corresponding temperature 5845K) OIII starts to emit 4959A photons.
Is that correct?

No. Blackbody relations don't take the "ordinary" meaning when dealing with emission lines, because they tell you about what happens at energies that may be much different from the energy of their particular photons. (Example - any recombination line of hydrogen tells you what happens at energies >13.6 eV even if it's a high-n IR or radio line) Emission line spectra reflect excitations in whatever the upper state is, and the decay in a particular line does not generally go all the way down in one transition. For example, [O III] 4363 starts in a higher energy level than the 4959/5007 lines; the transition from that upper level to the ground state is down around 2300 A.

Excitation by photons will be important only for resonance transitions - those starting from the ground state); excited states generally have radiative lifetimes too short to have many ions available for absorbing photons before they decay. (Exceptions - there are a few transitions with such fine energy splitting that they can be pumped upward on absorption of the plentiful microwave background photons. In hindsight, Colin McKellar found evidence for the CMB in interstellar absorption lines in the 1940s). In nebular conditions, collisions, mostly with electrons, are frequent enough to do the job, plus being able to scramble ionic states in ways that ignore the radiative selection rules. On top of that, one electron liberated by photoionization can collisionally excite many ions before losing too much energy to do more.

tu144
2011-May-25, 10:07 PM
As I tried to point out in another thread, a single photon does not have a temperature, so you can't assign a wavelength to a temperature like this. Am I misunderstanding what you are suggesting here?
Nope. It's just, the more I am trying to study this stuff the more confused I become, so I might throw the same stupid questions again from time to time.


We can, it's just the convention to use the sum in that ratio. Because of statistical weights, decays in the 5007 line have exactly 3x the number of photons in the 4959 line.



No. Blackbody relations don't take the "ordinary" meaning when dealing with emission lines, because they tell you about what happens at energies that may be much different from the energy of their particular photons. (Example - any recombination line of hydrogen tells you what happens at energies >13.6 eV even if it's a high-n IR or radio line) Emission line spectra reflect excitations in whatever the upper state is, and the decay in a particular line does not generally go all the way down in one transition. For example, [O III] 4363 starts in a higher energy level than the 4959/5007 lines; the transition from that upper level to the ground state is down around 2300 A.

Excitation by photons will be important only for resonance transitions - those starting from the ground state); excited states generally have radiative lifetimes too short to have many ions available for absorbing photons before they decay. (Exceptions - there are a few transitions with such fine energy splitting that they can be pumped upward on absorption of the plentiful microwave background photons. In hindsight, Colin McKellar found evidence for the CMB in interstellar absorption lines in the 1940s). In nebular conditions, collisions, mostly with electrons, are frequent enough to do the job, plus being able to scramble ionic states in ways that ignore the radiative selection rules. On top of that, one electron liberated by photoionization can collisionally excite many ions before losing too much energy to do more.
More or less understood.
Thanks!