Does CO2 emit infrared photons at the same frequency in which it absorbs them?
If infrared is captured every couple of meters and reemitted, then why are clouds so important for keeping the surface warm at night?
At the surface, the extinction distance for 15 micron light is only 10 meters. Since the light is not going directly up or down, an energy shell should be less than 7 meters high. "several" in this cause should mean many thousands of times according to your theory.
Clouds are visible because they consist of droplets of H2O in solid and liquid state. Thus, they act as reflectors rather than absorb/retransmit steam zones below them.
Since at least one of the molecules in a CO2 collision is CO2, do you suspect that a greater than 8% of the re-emitted radiation is going to be still in the CO2 absorption spectrum? Or does that have nothing to do with it?
Good question. I hope this helps:
At night, especially in the open desert, longwave blackbody radiation is emitted from the ground. http://en.wikipedia.org/wiki/Blackbody_r...
If there is no cloud cover, things cool off rapidly, and enormous temperature drops happen. It there is cloud cover, things do not cool off so quickly, and more heat is retained for the next day. http://ww2010.atmos.uiuc.edu/%28Gh%29/gu...
For a straight shot: 99.9% absorption at 15 micron in 10 meters = 50% absorption in 1 meter. Doing the RMS thing, that leaves 0.7 meters as the thickness of an absorption shell for present concentrations of CO2 at 15um at the surface. For a square centimeter of air, that is 70 ml.
Mass = 70 ml * 0.029 Kg / 22400 ml = .000906% of the atmosphere per absoption shell.
50% absorption Shells = 1/0.0000906 = 11000
For Trevor's theory: 2^11000 shells ~ 2 exp 1000 absorptions and transmitions on average to exit the atmosphere at 15 um, and the
I did make some arithmetic errors above, but not procedural errors, and nothing that would change the point:
For 15 um (CO2's main absorption contribution), and clouds an altitude where the pressure is 0.9 atm, the photon would go through 10^328 absorptions before reaching the cloud if released from the ground at 1 atm. Thus, the reflectivity of the particles in the cloud should make no measurable difference.
I am correct about the application of the Beer-Lambert Law: http://en.wikipedia.org/wiki/Beer%E2%80%...
"In regards to Beer's law, you are confusing transmission or intensity with absorbance."
1) Good point. Actually, I never used "absorbance" above, but:
"absorption" = 100% - Transmission
However, I see where this led to confusion, and will avoid that terminology in the future.
2) Indeed, d/dx's explanation is helpful, according to that, theoretically, absorption is always zero since everything gets transmitted eventually, but not always at the same frequency.
3) I made no point about UV light.
4) My point, after correcting my bad arithmetic:
- 707 ml, and don't miss count the zeros again...
is that an crazy high number of retransmissions in random directions occur before the IR at ~15 um reaches the clouds, so how can a bunch more possibly make a difference at that height, unless the re-emission is at a different frequency (that does not get captured so much), as Bravozulu and d/dx suggest? Transferring energy to other molecules via collision?