It looks like the 5.xxln(C/C,sub>o) is a fitted eq. above a baseline of 278ppmv in 1750. I'll have to look into that.
what does 1750 data have to do with doing a curve fit against spectral data? You use a database of spectoscopy data to provide the information needed. We are talking about a Line-by-Line analysis of the CO2 & H2O absorption spectrums as a function of concentration and blackbody radiance, a spectroscopic analysis, not a fit to a how concentration varies with time.
The eq.
FCO2 = 5.35ln(C/Co) = watts absorbed above the 278ppmv baseline from 1750.
That's my understanding of it. Note that at C=Co, the watts = 0. The eq is an equivalent to Beer's law:
I=Io(1-e- α *c*x) , where α is the absorbtion coefficient, c is the baseline concentration, and x is the pathlength. The eq. is then used to find FCO2 above 1750's value.
FCO2 = Inow - I1750 = I1750(e- α *xc1750 - e- α *xcnow).
Each molecule can absorb at many wavelengths, and each has it's own transition probability. So, the exponential terms are of the form Σn e- αn *xc.
FCO2 = I1750( Σne- αn *xc1750 - Σne- αn *xcnow).
You can see the constants in there. I assume Myhre considered this as a change in intensity due to the probability of more excited states with a change in concentration, dEexcited = A*dc/C. I= A*ln(C/Co), where A represents some sum of absorbtion constants, as above. Fitting obtains, 5.35ln(C/C1750). Then FCO2=I.
At any particular instant, there will be a population of excited molecules. They will decay and reradiate in another direction, they will scatter radiation, and they will decay by transfer of kinetic energy to the surounding gas. The caption to the graph you showed me, then provided a link for says, "The graph at left shows the percentage of energy absorbed in a clear tropical sky by water vapor (green) and carbon dioxide (brown)." In order to absorb 100% of the radiation, there must be unexcited molecules with states available.
"There are more CO2 molecules than there are photons of IR at the proper wavelengths to be absorbed in a given path length of atmosphere.
If that were true, then the caption on that figure could never say 100%. It would have to say something less.
" Increasing concentration effectively broadens the spectral line allowing more absorption by weak lines and the skirts either side of the central spectral line which are not saturated.
Increasing concentration provides more unexcited molecules to absorb radiation. More States are available. The width of those lines around a vibration is due to rotational coupling with the vibration and dopler spreading. The wavelength must match for any particular combo of vibration, rotation and molecular velocity. It is less probable that those vibrations with other angular momentum will be excited. The angular momnetum is coupled, so the vibrational E changes and has max values at J=0 and in the p and r branches, about 4 Δ Js away from there. That's why their peaks are successively lower as the wavelength moves from the base vibrational wavelength. The pressure broadening should be constant.