Ok, although this question arises out of the global warming debate, this is a question purely for physicists and not intended to branch into that particular debate.

We are told that LWIR from the Earth to the atmosphere is absorbed by greenhouse gasses and then re-radiated (in the form of backradiation) partially to the surface of the planet, "...thus warming the surface...".

My specific question is this:

How exactly can the re-radiated LWIR from a cooler atmosphere warm the original source of its absorbed energy?

Edit: I am editing this question as several people have stated this is a duplicate question. I disagree. The other question quoted three alternatives and some answers stated the 2nd Law of Thermodynamics was not compromised. Please re-read my question: can the backradiated energy possibly warm a surface which was the original source of that energy? In other words, how can the surface heat itself further when there is no other energy source? I hope this clarifies, and thanks to all that have answered.

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    $\begingroup$ What is LWIR? Please, do not use unexplained abbreviations. I do not want to guess. $\endgroup$ – Bernhard Apr 21 '14 at 16:10
  • $\begingroup$ Long Wave InfraRed $\endgroup$ – DavePhD Apr 21 '14 at 16:35

Firstly, not all parts of the atmosphere are cooler than the surface. For example, the Thermosphere can be much hotter than Earth's surface.

More importantly, even for a cooler region of the atmosphere, yes some thermal radiation emitted from a cooler region is absorbed by a warmer region, it is just that more radition goes from warmer to cooler than cooler to warmer.

For example see this Earth-Atmosphere Energy Balance. For every 98 units of energy the atmosphere emits to the Earth's surface, the Earth emits 104 units to the atmosphere (not including direct emission through the atmosphere to outer space and non-radiative transfer of heat to the atmosphere).

  • $\begingroup$ Thank you, DavePhD, for answering but the diagram you linked too is actually part of the problem I have leading to my question. The diagram asserts that 98 'units of energy' emitted by the atmosphere (and I think we can discount the thermosphere here) is "absorbed by the surface". This is exactly what my question asked - exactly <how> does the surface absorb this radiation? Does it absorb for energy gain or is the radiation absorbed and instantly re-emitted (i.e. reflected)? $\endgroup$ – Arfur B Apr 21 '14 at 20:46
  • $\begingroup$ It is absorbed and not necessarily re-emitted. Emission depends upon temperature. If there is a hotter object and a cooler object, each will emit thermal radiation that the other absorbs, but more will be emitted by the hot object and absorbed by the cool object. $\endgroup$ – DavePhD Apr 21 '14 at 21:18
  • $\begingroup$ Thank you DavePhD. How can the receiving surface gain energy by absorbing radiation at wavelengths it is already transmitting at? Isn't this - conceptually - the same as adding two glasses of 5 deg C water together and expecting 10 degrees? $\endgroup$ – Arfur B Apr 22 '14 at 4:10
  • $\begingroup$ There is net loss for the hot surface and net gain for the cool surface. Think of how Earth constantly receives radiation from the cosmic microwave background, even though CMB corresponds to a 2.7K black body. Or think of chemical equilibrium reactions, where there is a forward and backward reaction at the same time. With the glass of water example, there is violation of conservation of energy. The main thing to consider is ALL materials above absolute zero emit radiation, proportion to the 4th power of absolute temperature. The wavelength distribution is different for different temps. $\endgroup$ – DavePhD Apr 22 '14 at 10:38

Back radiation is not a good physics term. When one has many different bodies at different temperatures, they each radiate according to the black body formula for their type and temperature. In a thermally isolated system ( within a dewar for example) they will come to the same temperature in the end.

The earth is not a closed system, and finally the black body radiation at the boundaries with space will escape. If it were not for the replenishment of energy by the radiation from the sun the earth would come at thermal equilibrium at the temperature of the space around it. ( well, forgetting the magma heat)

The presence in the atmosphere of so called ( bad terminology again) "green house gases" and some types of clouds, just delay the loss of radiation by reflecting some of it back . The same effect as when one covers a hot water bottle with a woolen cover: the woolen cover reflects a bit of the infrared heat it receives and delays the cooling of the bottle, even though it is cooler than the water in the bottle. The exact physical process of reflection depends on the material, in the case of gases it happens with absorption and re-radiation.

So in a sense, it is not heating, but delayed cooling that is the phenomenon.

  • $\begingroup$ Thank you, anna v. However, the woollen cover reduces heat loss by insulating the kettle, does it not? I do not think the kettle is warmed by the cover. At no stage does the 'backradiation' warm the kettle because the kettle is the only heat source and something hot cannot heat itself further without an additional source of heat. Or am I wrong? Backradiation is not the same argument as insulation. Therefore, 'delayed cooling' is not the same as 'warming'. My question specifically addresses the 'how' of the 'energy gain by absorption of backradiation' - if in fact that happens. $\endgroup$ – Arfur B Apr 21 '14 at 21:02
  • $\begingroup$ "At no stage does the 'backradiation' warm the kettle because the kettle is the only heat source" the same is true for the earth. The extra heat source from the sun does not enter this argument, as the extra heat source from the radiator or the fireplace in the room does not enter the woolen cover/water-bottle argument. They are calling it warming because of fascination with "warm". It is delayed cooling for the atmosphere too. Think. The moon with no atmosphere and earth with atmosphere. The atmosphere is an insulation. Water vapor and CO2 and Methane help in this because they are not $\endgroup$ – anna v Apr 22 '14 at 3:43
  • $\begingroup$ transparent to infrared the way most of Oxygen and Nitrogen are. So they absorb it lasp.colorado.edu/~bagenal/3720/CLASS5/5Spectroscopy.html (this is from the sun side, but it is the same from bottom up too). Absorption means the molecules get into an excited state, and then de-excite releasing a photon isotropically, and the effect is the same as reflection from a mirror for the percentage of the spherical angle covering the earth underneath, for each molecule. It builds up. The effect on the total system is a delayed cooling, the same as the woolen cover. In the woolen cover $\endgroup$ – anna v Apr 22 '14 at 3:50
  • $\begingroup$ it is the lattice of molecules making up the wool that get excited by the infrared and reflects it back . The statement about smaller temperature heat source not heating a higher temperature one is a classical thermodynamic one. Once photons enter the framework is quantum mechanical and the two frames coincide at the limit . The earth does not get warmer than it was because H2O and CO2 exist in the atmosphere, it just gets less cool. This would be explained in classical thermodynamics by the heat capacity of the air changing with CO2, not by calling the air a heat source. $\endgroup$ – anna v Apr 22 '14 at 3:55
  • $\begingroup$ anna v, thank you very much for your comprehensive answer. It confirms my assessment. Just one more piece of clarification, if I may: If the Earth was getting heated by a separate source, would I be right in thinking that the Earth would still not 'warm' as there would be no mechanism for it do so? Thank you for the explanation re the heat capacity. Thank you again. Will respond later if you comment again. $\endgroup$ – Arfur B Apr 22 '14 at 4:00

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