But I thought we were talking about the real impact on radiation. The difference is going to be quantitative: radiation coming from the earth’s surface is going to be a lot stronger than coming from a bit of gas.

So there isn’t much difference between a gas and a solid, conceptually. A mm thickness of gas probably has the same number of atoms as the first few microns of a solid surface.

But there are .
Solids (and plasmas) have free electrons what gives them very special characteristics while gazes don’t .
Solids present a sharply defined interface while gazes don’t .
Solids (and plasmas) exhibit reflection at their interface while gazes scatter .
Also solids (and dense plasmas) can be approximated by a black bodies while gazes can’t .
It is not the number of atoms that counts but their structure and presence of free electrons .
So conceptually the interaction of electromagnetical waves with solids and with gazes is treated in a quite different manner .
Solids can be treated like a phonon/photon interaction while gazes can’t .

Sure, that’s why I used the word “conceptually” in post #73:

So there isn’t much radiative difference between a gas and a solid, conceptually

I felt that posts 70 and 75 were assigning too big a conceptual difference between molecules in a gas and ‘molecules’ (e.g., unit-cells) in a solid, as far as IR radiative effects are concerned.

What I keep wondering is how does the 98 + percent O2 and N2 affect the emission/absorption of IR by the trace amounts of GHGs. Thermalization of all those molecules HAS to have a profound effect on the energy levels of the GHGs. I wonder just how much rocking and rolling the CO2 and HOH molecules are doing in the presence of N2 and O2.

The presence of the N2/O2 does indeed have a profound effect on the energy levels of the GHGs in the lower troposphere, in the case of CO2 molecules excited in the 15 micron band the vibrational energy increase is ~8kJ/mole, this is almost completely removed by collisions with the surrounding molecules at a timescale much shorter than the emission timescale.

It’s a question of degree & amount.

If you take the same kind of atoms (say, gold) and disperse them as a cloud of vapor vs packing them into a solid, 1 cm of travel through each of these cases will experience dramatically different total absorption.

Experiment and measurements “trump” theory.

http://junkscience.com/blog/2008/01/12/processing-33-years-of-ir-longwave-data/

But the population of the excited states is determined, at LTE, by the temperature of the surrounding gas through inelastic collisions (the Boltzmann distribution, IIRC). For high optical density, emission becomes limited by the value of the Planck function at the wavelength of interest. Stefan-Boltzmann only applies to grey or black bodies, so unless absorption/emission is saturated at all wavelengths (which can happen at high enough humidity and temperature), the temperature dependence is not well described by a T^4 dependence.

My understanding is that this is correct (but I’m certainly no thermo expert). What I keep wondering is how does the 98 + percent O2 and N2 affect the emission/absorption of IR by the trace amounts of GHGs. Thermalization of all those molecules HAS to have a profound effect on the energy levels of the GHGs. I wonder just how much rocking and rolling the CO2 and HOH molecules are doing in the presence of N2 and O2.

photons intersecting a solid HAVE to interact in some way.

It depends on what you mean by ‘interact’. Photons can certainly pass through a solid without absorption. If by ‘interact’ you mean that the photon is influenced by fields due to the atoms, photons in gases also HAVE to interact with atomic/molecular fields.

The difference between a gas and a solid is that photons intersecting a solid HAVE to interact in some way.

Whereas in passing through a thin gas, the photons can avoid interaction, and thus avoid becoming thermalized. When the Optical Depth gets up to about 1, this issue goes away; at that point, the intensity approximates the blackbody intensity for that frequency.