Art said:
Vanesch if I am following your correspondence with the Middlebury author correctly he seems to be advocating a model which equates to an atmosphere 1 molecule thick and so any non-CO2 molecules act as spacers (ignoring any possible absorption by other atmospheric components) an IR photon has only one chance to hit a CO2 molecule and if it misses it's lost to space?
The other extreme is the second viewpoint you reference which suggests the atmosphere is so deep that inevitably all IR photons of the right wave length/frequency will strike a CO2 molecule meaning we are already at saturation point so adding more CO2 is irrelevant?
If this summary is correct how would one go about proving which if any is correct?
In fact, both viewpoints can be correct. In order to understand this, what you need to know is an elementary concept in radiation/matter interaction, which is "interaction cross section".
If you have a beam of particles which transports N particles per second, and this beam hits a slab of material of a thickness d and a density rho (number of interacting molecules/atoms/... per cm^3), then it turns out that the number of reactions per second, M, is given by:
M = A x N x d x rho (at least, as long as M is much smaller than N, which means that the beam is not "depleted" throughout the slab).
In other words, the number of reactions is proportional to the thickness, it is proportional to the density of target molecules, and it is proportional to the number of particles per second in the beam.
A is a constant of proportionality, and if you work out its dimensions, it is a surface.
(indeed, N has no dimension,d is in cm, rho is in 1/cm^3, and M has no dimension).
A characterises the interaction between a particle in the beam, and a target molecule, and is a microscopic quantity that is only dependent on the incident particle and the target (and the collision energy). It can in some cases be calculated from first principles. It is the so-called "interaction cross section" for the specific reaction under study, and is independent of the macroscopic material presentation.
The interaction cross section has the dimension of a surface, so officially m^2, but one usually uses a smaller unit: 1 barn = 10^(-24) cm^2.
For instance, the incident particle can be a photon of this or that wavelength (or energy), the target molecule can be a CO2 molecule, and that's it.
One usually gives the interaction cross section (in barn) as a function of the "incident energy" - or for photons, the photon energy or wavelength. So interaction cross sections are usually plots or tables of the cross section as a function of wavelength.
This function can show "peaks", which means that for certain incident energies (or wavelengths for photons) the value of the cross section increases strongly, to fall back again at a bit higher value. Usually one specifies the center energy/wavelength of that peak, and calls it "an absorption line", but don't forget that it is actually a continuous rising and falling of the cross section.
Note that it is a pretty universal concept. You can have scattering cross sections for neutrons on nucleae, or absorption cross sections for photons on molecules, or fission cross sections for neutrons on uranium nucleae...
It all follows the same universal concept of a 2-particle interaction: you specify the two particles (beam and target) and you specify the kind of interaction (scattering, absorption, another reaction), and you can define a cross section for that.
Now, one can re-shuffle the formula:
M = A x N x (d x rho)
d x rho is the number of target molecules PROJECTED on a perpendicular surface to the beam in the entire slab. So this is the number of molecules one would find per unit of SURFACE if the entire slab were compressed into a thin sheet.
This was part of my argument: the number of interactions doesn't change as long as (d x rho) is the same, so whether or not you have a thick slab (a column) or a compressed sheet doesn't matter. It is also the "difference" between the two views.
Now, we can now calculate A x (d x rho). A was a (tiny) surface, the cross section. (d x rho) was the compressed number of molecules per unit of surface.
So it looks as if we can associate a tiny surface to each of these molecules, of size A, and A x (d x rho) is then the FRACTION of the initial unit surface which is now "covered" with "molecules"... at least, as long as this number is much smaller than 1, so that there is no "overlap". It gives you then "the probability" that a beam particle will "hit" a molecule.
Indeed, if you have N beam particles per second, and each of them has a probability Ax(dxrho) to "get hit", you have N x A x (d x rho) hits per second: our initial formula.
But let us not forget, that's only valid when this probability of hitting is small. If not, you have to "re-adjust" the beam as it depetes throughout the slab - which comes down to saying that the molecules "hide one behind the other" ; it can then be shown that you have:
M = N (1 - exp( - A x d x rho) ), the exponential absorption law, at least under the hypothesis that there is only one kind of interaction: the one under study, and that upon interaction, the beam particle disappears. If scattering happens, the problem becomes way more complicated, but the elementary concepts remain valid for thin slabs.
My main argument was, that in this formula M = A x N x (d x rho), at no point, the ratio of our target molecules (CO2) to any other species appears. It is not because you ADD another, neutral, species to the mixture, that the number of interactions diminishes - which was the principal argument in the first page.
Now, in the second argument, of course, if A x d x rho is a very big number (say, 2000 or so), then the number of interactions M ~ N. Indeed, exp(-2000) is a very small number, negligible in comparison to 1. If now, we double the density of CO2, we would obtain A x d x rho = 4000, and we STILL have M ~ N.
However, things are in reality more complicated. We don't have a monochromatic beam, and there are other interactions. The cross section is a function of the photon energy, in the reference frame of the target molecule. If these molecules move around, then in their reference system, the incoming photon has a bit more or less energy, and hence the cross section is different: this is what one calls "doppler broadening". There are many other small effects. Once you want to take all this into account, you need to write a computer simulation (such as MODTRANS). I would like, myself, to see a more detailed explanation of which effects are dominant, and what does what. It can seriously alter the balance of the radiation.
I've written (small) transport codes myself, in the nuclear domain. I can assure you that the results are sometimes surprising. That said, as long as you take into account all relevant interactions, and if you have accurate enough tables of cross sections (and if you don't make any silly bugs), then there is not much doubt about the result of such a computer calculation: the theory of radiation transport is rather well-understood.
You can compare it a bit to finite-element calculations in mechanical structures.
NOTE:
sometimes one defines cross section in a different (but equivalent) way:
M = A x N x d x rho
Let us consider the geometrical section of the incoming beam to be S (surface), and let us assume that the beam is homogeneous across this section.
Let's multiply by 1: S / S:
M = A x N x d x rho x S / S
Now, d x rho x S is rho x V is the number of target particles in all exposed to the beam.
M = A x (N/S) x (d x rho x S)
N/S is the number of incident particles in the beam per unit of time and per unit of surface: it is the flux density of the beam.
So we have:
M = A x (incoming beam flux density) x (total number of exposed target particles).
