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> Any wavelengths which were gonna heat the air by absorption, were absorbed on the way in and didn't reach the roof in the first place. What is the basis for

> Any wavelengths which were gonna heat the air by absorption, were absorbed on the way in and didn't reach the roof in the first place. What is the basis for this? Intuitively (I am no expert) it seems more likely that the amount of thermal energy produced would be a function of the distance light travels through a medium and the amount of energy carried by the light in the first place. So I would expect that the "inbound" light would produce more thermal absorption energy than the reflected "outbound" light, but it wouldn't arbitrarily all be absorbed on the way in. Reality seems to be more complicated: https://www.sciencedirect.com/science/article/abs/pii/0960168693901047 https://www.sciencedirect.com/science/article/abs/pii/096016...

myself248 4y ago
It was a bit of an oversimplification on my part, since for a given wavelength, the atmosphere isn't binary, completely transparent or completely opaque (i.e. the radiation would go a very very short distance before being absorbed), but for _most_ wavelengths it's _very close_ to binary. Either it's absorbed or it isn't, to a first approximation. Yes, some wavelengths are absorbed very weakly, and some are scattered rather than absorbed. But these represent a tiny fraction of the spectrum, the nuance would've overcomplicated the post, and they don't change the result much. For wavelengths that are not absorbed, they can be reflected back out just as easily as they came in. For wavelengths that're strongly absorbed, they heat the first parts of the atmosphere they encounter, and this is why the thermosphere is called that. For those few wavelengths that're weakly absorbed, yes they produce some heating on the way in, and if you reflect them back out, they do produce some heating on the way out. But as you've said, because the intensity of the radiation is decreasing along the whole path, the "outbound" contribution is weak indeed. It turns out to be basically a rounding-error in the thermal contribution, which is overwhelmingly dominated by blackbody absorption and re-emission, which often does produce wavelengths that're strongly absorbed. On a perfectly-white planet (or snow-covered region), you might have to start caring about weakly-absorbed wavelengths, but for most types of ground cover, other effects swamp it.

myself248 4y ago

It was a bit of an oversimplification on my part, since for a given wavelength, the atmosphere isn't binary, completely transparent or completely opaque (i.e. the radiation would go a very very short distance before being absorbed), but for _most_ wavelengths it's _very close_ to binary. Either it's absorbed or it isn't, to a first approximation. Yes, some wavelengths are absorbed very weakly, and some are scattered rather than absorbed. But these represent a tiny fraction of the spectrum, the nuance would've overcomplicated the post, and they don't change the result much. For wavelengths that are not absorbed, they can be reflected back out just as easily as they came in. For wavelengths that're strongly absorbed, they heat the first parts of the atmosphere they encounter, and this is why the thermosphere is called that. For those few wavelengths that're weakly absorbed, yes they produce some heating on the way in, and if you reflect them back out, they do produce some heating on the way out. But as you've said, because the intensity of the radiation is decreasing along the whole path, the "outbound" contribution is weak indeed. It turns out to be basically a rounding-error in the thermal contribution, which is overwhelmingly dominated by blackbody absorption and re-emission, which often does produce wavelengths that're strongly absorbed. On a perfectly-white planet (or snow-covered region), you might have to start caring about weakly-absorbed wavelengths, but for most types of ground cover, other effects swamp it.