Does CO2 emit infrared photons at the same frequency in which it absorbs them?

When CO2 emits, it isn't acting as a greenhouse gas and no energy is absorbed by the atmosphere. It is in fact acting to cool the atmosphere or it has no effect at all. I suppose alarmists imagine this photon getting bounced around until it strikes the earth again. That is not how it works. The greenhouse effect doesn't work by CO2 molecules absorbing radiation and reemitting it like some sort of reflector. If that were the mechanism, the air would NEVER warm. The extra energy absorbed is translated into vibrational or momentum when the molecule of CO2 bumps into other molecules. That extra energy is heat. Warm bodies radiate heat in what is known as black body radiation. That heat is what is reflected back to the earth or space. Any emissions have virtually zero effect as the concentration is increased since it is already absorbed to extinction in relatively short distances in the primary bands except for tiny amounts of energy on the shoulders. When you shine an infrared source through a cylinder partially filled with CO2, it blocks the transmission of light. It warms the gas. It also only absorbs about 8 percent so the reemission of heat by the air isn't going to have much of the original heat reabsorbed by CO2. Less than one percent, 8 percent of 8 percent, will be reabsorbed. CO2 is a very inefficient greenhouse gas because of its limited absorption bands. All matter can emit heat. If CO2 absorbed heat from gas molecules around it and that energy was stored in the bonds and released, that would be cooling the atmosphere which is the opposite effect of the greenhouse effect. I imagine that happens but I doubt that changes in concentration has much effect at all in the net energy balance from that mechanism. I heard one guy claim it cools the upper atmosphere.

The main effect is just shortening the distance that the CO2 is absorbed to extinction. The greenhouse effect is CO2 absorbing the energy in narrow bands of infrared light and translating that to heat energy where the atmosphere warms. All the molecules of the atmosphere then emit IR radiation in wide bands, only a small amount of which is absorbed again by CO2 at least. That is talking about the heat that is generated from the ground. Some heat is also transferred higher into the atmosphere by convection, mainly because of water condensing in thunderstorms. That isn't greenhouse warming when that warmed air is insulated by greenhouse gasses but it obviously still affects it in the same way.

They generally emit and absorb at the same frequency. There may be slight changes due to rotational energy levels. d/dx+d/dy+d/dz is a spectroscopist and could probably give more details.

There are other processes that occur other than emitting a photon. Energy can be transferred to other molecules through collisions. For example, a collision could go to the translational energy of another molecule or promote a vibrational mode that does not relax via emitting an IR photon such as N2 vibrations.

I think the simple answer to why clouds are important is that they are simply a high density of water, a green house gas.

Edit: BZ said

" The greenhouse effect doesn't work by CO2 molecules absorbing radiation and reemitting it like some sort of reflector. If that were the mechanism, the air would NEVER warm. The extra energy absorbed is translated into vibrational or momentum when the molecule of CO2 bumps into other molecules.That extra energy is heat. "

Finally got something right

"Warm bodies radiate heat in what is known as black body radiation. That heat is what is reflected back to the earth or space."

The atmosphere is not a black body. An isolated molecule in its ground state will never emit black body radiation no matter what its translational energy is.

"When you shine an infrared source through a cylinder partially filled with CO2, it blocks the transmission of light. It warms the gas. It also only absorbs about 8 percent so the reemission of heat by the air isn't going to have much of the original heat reabsorbed by CO2. "

This is really a nonsensical statement since the amount of absorption is completely dependent on light source. A laser tuned to a vibrational band of CO2 will absorb 100% until extinction.

"Less than one percent, 8 percent of 8 percent, will be reabsorbed. "

What? It does not work that way. It does not re-emit as black body radiation, see above. It will emit in the frequencies that it absorbs.

Edit:

NW Jack

The two graphs are two different things. The top graph is a transmitted spectrum. The bottom graphs are absorption spectrums. One graph shows what frequencies are getting blocked, the other shows what is making it through. If you look carefully, top figure is nearly a mirror image of the Total Absorption graph. The colored lines in the transmission spectrum represent the expected black body radiation from the earth over a range of temperatures. There is no great mystery here, the areas of the spectrum that the atmosphere cannot absorb gets transmitted. Those areas that get absorbed do not show up on the transmitted spectrum. That is exactly what these figures are showing. It does not show a significant shift in the IR radiation from absorption and re-emmittance.

Also water in liquid and solid forms can absorb IR. I am not sure how the reflective properties of water in the visible region matter at night. Can you explain further since there is little visible light being transmitted upward at night (except by humans but that is minuscule).

Edit: You are really jumping all over the place. I questioned why clouds interaction with visible light matters (at night), then you give an explanation about longwave IR. Completely two different things.

"99.9% absorption at 15 micron in 10 meters = 50% absorption in 1 meter."

This is incorrect. Beer's Law says that absorption is linear with the distance through a medium.

Edit: Your point seems to change every time you write something so I have no idea what your point is. First you say reflectivity is important in the visible range, then you talk about IR absorption, then you say reflectivity is not important. You jump back and forth between interactions in the visible range and IR range. Just because something is a good reflector in the visible range, does not necessarily mean it be a good reflector in IR. This may help:

http://www.newton.dep.anl.gov/askasci/phy00/phy008...

In regards to Beer's law, you are confusing transmission or intensity with absorbance.

It's just a part of the electromagnetic radiation spectrum that is emitted by warm objects. The heat of the object is exciting the atoms who are releasing photons of energy to remain in their sleep state. Radio waves, visible light, xrays, infrared etc are all just made up of waves of massless particles called photons. Depending on how much energy is being emitted, the type of radiation will vary between all. If it's a lot of energy, it will release gamma rays but if it's only a small amount of energy to be released, radio waves will be released. Anything between will come under an other name, beit light, heat etc. They are all the same thing essentially only differing on the energy to be emitted. That will decide the wavelength and frequency of the waves of energy,

<Does CO2 emit infrared photons at the same frequency in which it absorbs them?> You need to understand a bit of quantum mechanics and statistical mechanics for this one. CO2 has 3 atoms, each can move in x,y and z directions giving 9 degrees of freedom. 3 degrees of freedom are taken by translation of the molecule as a unit and 3 degrees of freedom are taken by rotations, but two rotations are indistinguishable (degenerate) since CO2 is a linear molecule. This leaves 3 internal vibrational modes: a degenerate bending mode, a symmetric stretching mode and an assymetric stretching mode. Each of these has a quantum number ranging from 0 (ground state) to a large integer. The maximum is limited by molecular disintigration. The bending and assymetric stretching modes are infrared active: transitions from one quantum number to another in either mode involves the absorption or emission of and infrared photon. The symmetric stretching mode is not infrared active, but still participates via the Raman effect and combination bands. In addition to the 3 modes noted above, sums and differences of the fundamentals are allowed (for example nu1 + 2 nu2 or nu1 - nu3). There is a large number of permutations. A symmetry analysis is needed to determine the infrared activity of each one. If, as in the case of CO2, a combination band is close to a fundamental mode, there is an interaction called a Fermi resonance that perturbs both. So, there are a lot of absorptions possible just considering vibrational modes, but we are just getting started. There are rotational quantum numbers +/-j superimposed on each vibrational mode. Each band in the low resolution spectra given in the Wiki is actually composed of thousands of narrow lines, each with its own infrared absorption characteristic. CO2 does absorb and emit within the same set of frequencies, but an individual molecule does not necessarily re-emit at the same frequency as it absorbed: interactions with other molecules or with other excited states in the same molecule may cause emission at a different frequency or conversion of the energy to kinetic energy (heat). Enter statistical mechanics. An ensemble of molecules in thermal equilibrium will absorb and emit at precisely the same frequencies. While some energy may be lost by one molecule to translation, another molecule will gain the same energy in a collision to emit at the same frequency the first molecule absorbed at. The energy of the ensemble will be distributed among all of the available modes according to the partition function Z = sum (exp(-Ei/kT)), where Ei is the energy of the ith state. The probability that state i is populated is exp(-Ei/kT)/Z. The notion that IR is absorbed to extinction is utter nonsense in the context of atmospheric physics. Most of the IR from a beam is absorbed at room temperature because you are comparing a source at 1300 K with a gas at 300 K. The IR energy exiting is the spectral pattern for 300 K, not zero. The Beer Lambert Law is an approximation that works when the temperature of the source is much higher than the temperature of the sample. Further, the exiting radiation is scattered over 4pi radians whereas the input beam subtends a small solid angle. If the gas was at 1300 K, the IR leaving would be the same as the IR entering, albeit over a larger solid angle. The loss of directionality - call it homogenization, is a key concept for understanding the greenhouse effect. On average, half of the re-radiated energy is re-radiated in the direction of the ground and is available to warm layers closer to the surface.

Clouds. Clouds are full of water vapor, which absorbs IR and micron scale droplets that scatter IR strongly in all directions. Try shining a bright light through your flesh. The light loses directionality and decays exponentially with distance from the source. The same happens in clouds albeit at a slightly different length scale.

Edit: Please look at the blurb on statistical mechanics again. There is no need for a change in frequency for the energy to be transmitted through the atmosphere, albeit in small steps. At thermal equilibrium, a good analogy would be water flowing through a pipe. The molecule of H2O entering at the bottom is not the same as the molecule leaving at the top, but the numbers must be equal to maintain the same mass in the pipe.

The outgoing infrared radiation from Earth has a wavelength of 0.3 to 0.4 micrometers, this is the same vibrational frequency as that of the greenhouse gases, of which carbon dioxide is one of many.

This similarity allows for the partial absorption and subsequent retransmission of same wavelength thermal energy.

Not all of this outgoing energy is absorbed by the greenhouse gases. An infrared photo may be transmitted from Earth and travel out into space unimpeded or it may be absorbed by a molecule of greenhouse gas then retransmitted. The direction of retransmission determines whether this photon would contribute to global warming or not.

Once retransmitted the energy may go out into space or it may return to Earth, whereupon the cycle begins again. Feasibly it’s possible that a single photon could pass through the absorption/transmission cycle several times before finally being lost to space.

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RE: YOUR ADDED COMMENTS

Your comments are somewhat lacking in coherence and seem to be going of at numerous tangents. Your maths is still wrong (arithmetically and procedurally).

Fortunately d/dx+d/dy+d/dz has answered your question, this is one of his areas of expertise and he is far more knowledgeable about this subject than I am (and probably any other user). To compliment d/dx’s answer you may wish to refer to the diagrams and equations provided here http://brneurosci.org/co2.html

In my opinion, this is THE most important question. Starred.

My understanding is yes!

Indeed, this is the defining property of a greenhouse gas, if my (non expert) understanding is correct.

Can't really help with the clouds part of the question though ...

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Edit to bravozuzu: Just to be clear, you are saying that CO2 is not a greenhouse gas then? because what you are describing wouldn't produce a greenhouse effect!

I would be interested to know your view on the greenhouse effect produced by water vapour. Unless something extremely different is going on, then water vapour is not a greenhouse gas either!