Recycling of Heat in the Atmosphere is Impossible: A Note from Nasif S. Nahle

Introduction

Key diagrams on the Earth’s energy budget depicts an exchange of energy between the surface and the atmosphere and their subsystems considering each system as if they were blackbodies with emissivities and absorptivities of 100% 1, 2.

This kind of analyses shows a strange “multiplication” of the heat transferred from the surface to the atmosphere and from the atmosphere to the surface which is unexplainable from a scientific viewpoint. The authors of those diagrams adduce that such increase of energy in the atmosphere obeys to a “recycling” of the heat coming from the surface by the atmosphere 1, 2, as if the atmosphere-surface were a furnace or a thermos and the heat was a substance.

Such “recycling” of heat by the atmosphere does not occur in the real world for the reasons that I will expose later in this note.

Few authors have avoided plotting such unreal recycling of heat and only show the percentages related to the flow of energy among systems and subsystems of the Earth 3, 4.

We do know that serious science makes a clear distinction between heat and internal energy. However, we will not touch this abnormal definition of heat from those erroneous diagrams1, 2 on the annual Earth’s energy budget.

In addition to the wrong concept of heat that the authors let glimpse in their articles 1, 2, the recycling of heat by the atmosphere does not and cannot occur in the real world. There are many physical factors, already proven experimentally and observationally5, that nullify the ideas of the recycling of heat by the atmosphere.

The principal physical factor that inhibits the recycling of heat in the atmosphere is the degradation of the energy each time it is absorbed and emitted by any system10. This degradation of energy is well described by the second law of thermodynamics6, whose fundamental formulation is as follows:

The energy is always dispersed or diffused from an energy field with lesser available microstates towards an energy field with higher available microstates5.

In other words, the energy is always dispersed or diffused from the system with a higher energy density towards the system with a lower energy density5, 10.

The purpose of this essay is to demonstrate that some evaluations 1, 2 on the Earth’s annual energy budget are not considering the laws of basic physics and thermodynamics, that the “recycling” of heat in the atmosphere is unphysical and that the carbon dioxide works like a coolant of the surface, rather than like a warmer.

Analysis

The Earth and all its subsystems are gray-bodies3; consequently, any calculations made on the basis of blackbodies greatly differ from the real world, but only provide us an idea about what could be happening in such or that physical situation5. This means that they cannot absorb all the energy that they receive from a source and that, equally, they cannot emit the whole amount of such absorbed energy in the form of energy capable to do work on other systems, but rather that the main part of that energy is no longer accessible for making work and it is lost irremediably into the natural heat sinks.

All the spontaneous processes occurring in nature are irreversible processes 7, 8. Absolutely-reversible processes do not exist in the natural world 9, while absolutely-irreversible processes do exist in the natural world.

Shift to Red of Dispersed Quantum/Waves and Emitted by the Atmosphere Quantum/Waves

There are three bands of absorption of IR radiation by the carbon dioxide, i.e. 2.6 µm, 4.3 µm and 14.77 µm.

In this assessment, we will analyze the absorption of the energy of quantum/waves with wavelengths of 4.3 µm and 14.77 µm

The energy of an IR quantum/wave with a wavelength of 4.3 µm, emitted from the Earth’s surface is 4.62 x 10^-20 J. From this energy, a molecule of CO2 absorbs 9.24 x 10^-23 J.

4.61 x 10^-20 J are dispersed to other systems, except to the molecules that dispersed it. This amount of energy corresponds to a wavelength of 4.31 µm. The wavelength has been lengthened (redshift) by 0.01 µm.

A quantum/wave with wavelength = 14.77 µm –the band at which the carbon dioxide exhibits its maximum absorption potential- has an energy density of 1.345 x 10^-20 J. If it hits a molecule of CO2, the carbon dioxide molecule absorbs only 2.7 x 10^-23 J, while the energy carried by the dispersed quantum/wave is 1.3423 x 10^-20 J.

The carbon dioxide molecule emits a quantum/wave with energy = 5.4 x 10^-26 J, which corresponds to a wavelength λ of 3.75 m. The quantum/wave emitted by the carbon dioxide is not an IR quantum wave, but a Radio quantum/wave; therefore, its energy cannot be absorbed as heat neither by the surface neither by molecules of carbon dioxide.

Notice that 1.32145 x 10^-20 J is dispersed towards another energy field with more available microstates that resides in other systems; for example, the outer space, water vapor molecules, or dust. The wavelength of the dispersed quantum/waves has been elongated up to 14.8 µm (redshift); this elongation puts the IR quantum/wave out of the range of absorptivity of carbon dioxide by the specificity and sensitivity of absorption and emission potentials; consequently, the energy of these quantum/waves cannot be reabsorbed by molecules of carbon dioxide.

The following calculation over the resulting quantum/wave with wavelength of 14.8 µm absorbed by the carbon dioxide does not happen in nature; however, I decided to include it for readers take notice of the impossibility that heat can be “recycled” in the atmosphere.

Assuming that the absorption of that quantum/wave is still possible and another molecule of carbon dioxide could absorb it, we would have that:

For a wavelength 14.8 µm, the energy absorbed by the molecule of CO2 would be 1.3215 x 10^-20 J.

The energy of a quantum/wave emitted by that molecule of carbon dioxide would be 2.643x 10^-23 J, which would correspond to a wavelength of 0.7515 cm. This magnitude would match with the band of microwaves in the EM spectrum (microwaves’ wavelength = 0.01 to 20 cm). It still contains usable energy, but this energy can no longer be absorbed by molecules of carbon dioxide and it is lost into any of the energy sinks.

At this point, let us remember that the longer the wavelength is, the lower the energy density of that quantum/wave is.

The energy required to excite an electron for it shifts from a lower quantum microstate to the next higher quantum microstate is 5.4468 x 10^-19 J. Ref. 5

Therefore, the percentage of energy absorbed by a molecule of carbon dioxide with a wavelength of 14.77 µm represents 0.2% of the total energy required to excite an electron of the atoms of a molecule of carbon dioxide.

In that case, for an electron in the carbon dioxide molecule becomes excited and changes its energy configuration, a contribution of energy, supplied by 20554 IR quantum/waves, is required. Consequently, the carbon dioxide in the Earth’s atmosphere is in an energy field with higher number of available microstates.

This is the reason by which the flow of power is always transferred on a very specific directionality, i.e. from higher to lower and never the opposite.

How many molecules of carbon dioxide would be needed to get 249 Joules of energy in the total volume of carbon dioxide in the atmosphere?

~4.6 x 10^20 molecules of carbon dioxide are needed to get a volume of air absorbing 249 Joules of energy within the wavelength 14.77 µm.

There are ~2.61 x 10^9 molecules of carbon dioxide in one cubic meter of air; therefore, we need 1.76 x 10^11 m^3 of air for the molecules of carbon dioxide can absorb, simultaneously, 249 J.

The total volume of the Earth’s air is 4.2 × 10^18 m^3. There are ~1.1 x 10^28 molecules of carbon dioxide in the whole volume of air on Earth; consequently, almost the whole volume of molecules of carbon dioxide in the Earth’s atmosphere would absorb 249 J.

Therefore, there are 6.23 x 10^26 probabilities that the total amount of carbon dioxide in the Earth’s atmosphere absorbs the whole load of energy of 249 J; however, each molecule of carbon dioxide would absorb only 2.3 x 10^-26 J.

The molecules of carbon dioxide which had absorbed 2.3 x 10^-26 J of energy would emit quantum/waves with wavelength = 4.3 km, which correspond to the spectrum of vertical gravity waves (buoyancy). Therefore, those waves are lost in the gravity field.

As a result, the carbon dioxide is a coolant, rather than a warmer, of the Earth.

In conclusion

Key diagrams that purport to show the annual energy budget of Earth show a recycling in the atmosphere of the heat emitted by the surface.  But they are wrong.

The lengthening of the wavelength of quantum/waves emitted by the absorber systems and the decrease of their frequency inhibit any possibility of re-absorption of the absorbed energy -in the form of infrared radiation- by the same absorber once it has been emitted out from the absorber system.

Additionally, this assessment confirms that the second law of thermodynamics is applicable to molecular and quantum levels.

The carbon dioxide does not act like a warmer of the Earth’s surface, but rather like a coolant of the Earth’s surface.

References

1.  Kiehl, J. T. and Trenberth, K. E. (1997). Earth’s Annual Global Mean Energy Budget. Bulletin of the American Meteorological Association. No. 78. Pp. 197-208.

 2. http://www.globalwarmingart.com/wiki/Image:Greenhouse_Effect.png

 3. Peixoto, José P., Oort, Abraham H. 1992. Physics of Climate. Springer-Verlag New York Inc. New York. Page 366.

 4. http://eosweb.larc.nasa.gov/EDDOCS/images/Erb/components2.gif

 5. Castellan, Gilbert W. Physical Chemistry-3rd Edition. Addison-Wesley-Longman Publishing Company, Inc. 1998.

 6. Van Ness, H. C. Understanding Thermodynamics. 1969. McGraw-Hill, New York.

 7. http://www.brighthub.com/engineering/mechanical/articles/4616.aspx

 8. http://pubs.acs.org/doi/abs/10.1021/ed010p45

 9. http://www.ecourses.ou.edu/cgi-bin/eBook.cgi?doc=&topic=th&chap_sec=05.3&page=theory

 10.  http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node49.html

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To read more from Nasif S. Nahle scroll and click here: http://jennifermarohasy.com/blog/author/nasif-s-nahle/

40 Responses to Recycling of Heat in the Atmosphere is Impossible: A Note from Nasif S. Nahle

  1. Mack March 12, 2011 at 11:19 am #

    Nasif Nahle,
    Thankyou for your excellent plain concise real world analysis of the physics. I was always of the contention that the CO2 Theory had extremely dubious flimsy tenuous physics,but now your article finally confirms to me that this is just plain junk quack science. And whatsmore CO2 may even have a cooling effect!
    Thankyou, well done.

  2. Luke March 12, 2011 at 3:32 pm #

    Well Mack’s easily pleased. I just love how empirical evidence just hits this nonsense out of the ball park. And don’t you love how Nobel prize winning AGW defeating material just always seems to never get published.

    I wonder where the images goes http://www.youtube.com/watch?v=SeYfl45X1wo

    Magic !

  3. cementafriend March 12, 2011 at 4:46 pm #

    Thanks Nahle for the interesting post.
    With regard to CO2 cooling the atmosphere that is what Dr Noor Van Andel put in his presentation to KNMI see here http://climategate.nl/wp-content/uploads/2011/02/CO2_and_climate_v7.pdf
    In case there are those that do not know KNMI, it stands for Royal Dutch Meteorological Institute which holds one of the three world wide climate databases (and the only one not subject to manipulation). The following will get you to the site http://www.knmi.nl/index_en.html then follow the link to data to get to climate explorer

  4. Nasif Nahle March 12, 2011 at 5:21 pm #

    @Mack… Thank you very much for your comment.

    @Luke… Thanks for your strange comment.

    I saw the short video you posted in your message. He says that carbon dioxide absorbs energy. Yep… Carbon dioxide absorbs energy. I say it many times in my article; for example, here:

    “There are three bands of absorption of IR radiation by the carbon dioxide, i.e. 2.6 µm, 4.3 µm and 14.77 µm.”

    And:

    “A quantum/wave with wavelength = 14.77 µm –the band at which the carbon dioxide exhibits its maximum absorption potential- has an energy density of 1.345 x 10^-20 J.”

    “The energy of a quantum/wave emitted by that molecule of carbon dioxide would be 2.643x 10^-23 J, which would correspond to a wavelength of 0.7515 cm. This magnitude would match with the band of microwaves in the EM spectrum (microwaves’ wavelength = 0.01 to 20 cm).”

    As you can see, I’m not saying that the carbon dioxide doesn’t absorb energy. The problem, that many people have not understood, is that the carbon dioxide emits energy with other wavelength that cannot be reabsorbed again by the carbon dioxide, and that this energy is lost into any of the “heat” sinks.

    You say that I just love how empirical evidence just hits this nonsense out of the ball park.

    Well, Luke… I never thought you didn’t like reading bibliography and references. You’re a bit lost with your strange commentary:

    Peter Debye and Arthur Holly Compton discovered the strengthening of quantum/waves in 1922. Compton’s experiments have been repeated hundreds of times with different wavelengths and all of them have confirmed Debye’s and Compton’s discovery.

    Please, don’t say that Debye’s and Compton’s experiments and formulas are nonsense. And please, don’t say that the experiments and observations of hundreds of modern physicists that have confirmed Compton’s experiments are non-sense.

    By the way, Compton’s experiments have been repeated since 1922 using radiation from all bands of the EM spectrum. The results are the same.

    The problem with AGW is that they expose whatever idea comes to their minds, without corroborating it with well proven scientific knowledge. This thing about “recycling of heat in the atmosphere” is not true, as I have demonstrated through my last two articles.

    :)

  5. Nasif Nahle March 12, 2011 at 5:39 pm #

    @Cementafriend…

    Wow! Thank you for providing those links to Dr. Noor Van Andel’s work. Let me tell you that I have plotted the changes of energy density with altitude in the atmosphere and our conclusions are the same. This is the way how science works. Thanks again!

    :)

  6. Luke March 12, 2011 at 6:57 pm #

    Well Nasif our previous discussion on direct measurement of the “non-existent” radiation says it all. Publish or perish.

  7. Nasif Nahle March 13, 2011 at 12:27 am #

    @Luke…

    Sorry… It’s not me who will publish it. Please, be patient.

    In the meantime, you could spend your valuable time on applying the following formula to know a little more about those “non-existent” measurements of 300 W/m^2 of “backradiation”:

    Δλ = [h/(me*c)] * (1-Cos θ)

    Where h is for Planck’s constant, me is the mass of the electron, and c is the speed of light.

    Do you remember what I told you about the angle of incidence of quantum/waves and the energy density of them? This formula is very simple. Good Luck, Luke!

    :)

  8. Schiller Thurkettle March 13, 2011 at 12:35 am #

    Nasif,

    Thanks for that wonderful post. Since the CAGWers don’t have their physics right, the rest of their stuff is totally bogus.

    A partial proof of this is Luke’s responses — all he can do is say things like, who got published. Shows he can’t and won’t show up on the same playing field.

    Luke is annoying, but he’s an effective reminder of how little the CAGWers have going for them.

  9. Mack March 13, 2011 at 9:12 am #

    All,
    As a layman what I can’t understand is why the CO2 molecule is not heated in the first place by incoming radiation. ie have a “shielding” effect. as demonstrated by Lukes U tube experiment. Am I to understand that if the infrared camera was taken out into space you wouldn’t get an image of the sun if it was pointed at it.???!!!!

  10. Nasif Nahle March 13, 2011 at 12:07 pm #

    @ Shiller Thurkettle…

    Indeed. Thanks for your comment.

    @Mack…

    Thanks a lot for your message.

    The “experiment” shows a great bias, starting from the fact that he saturated the space inside the tube with carbon dioxide. If he was intending to compare his tube with the Earth’s atmosphere, he would have filled the tube with air. First trick.

    The camera is at one end of the tube and the candle at the opposed end. In the video, we see a very tenuous figure of the candle flame after he fills the tube with carbon dioxide. The surroundings of the candle image, taken by the IR camera, remain without changes.

    It is because the man tuned the IR camera to capture only a determined range of IR radiation from the candle so the camera cannot capture other wavelengths; that’s the second trick. If he would have widened the range, we would not have seen any change.

    Besides, the Earth is not a crystal tube filled with frozen compresed carbon dioxide, therefore, there is no point of comparison and the conclusion of the man is completely invalid. Third trick

    Carbon dioxide is transparent to the incoming shortwave IR radiation, i.e. the carbon dioxide does not absorb solar shortwave IR radiation.

    Carbon dioxide absorbs longwave IR radiation, especifically, wavelengths corresponding to 2.6 µm, 4.3 µm and 14.77 µm.

    If you would take the same camera, tuned to capture the same range of IR radiation of the candle, out to the space, you would see the Sun emitting at the same range on which your camera is tuned. Most digital cameras take photos of the Sun with an obscure point in the center of the solar disk. It happens because the camera is adjusted to not absorb the solar radiation out of the ranges of visible light; otherwise, the camera would be damaged by the incident solar IR radiation concentrated on the camera’s silicon sensor plate that captures the image. Pure modern technology.

    :)

  11. Luke March 13, 2011 at 12:29 pm #

    Come off it Nahle – this is sheer nonsense as the many papers on radiation and spectroscopic measurement at the Earth’s surface well show.

    You can now also rewrite the hundreds of surface energy balance papers with your new theory.

    So poppycock and balderdash !

    Again – if it were true it would be Nobel prize winning stuff that a biologist totally overthrew the entire knowledge of radiative atmospheric physics – so I look forward to your urgent rush-to-publication hold-the-presses paper in Nature or Science.

    When may we expect it’s publication !

  12. Nasif Nahle March 13, 2011 at 12:41 pm #

    @Luke…

    Heh! Please, tell me how you reach to the conclusion that it is my theory; you’re honoring me in excess.

    Tell me also, if your mood is properly balanced, in scientific terms, that the theory, which you say is mine, is poppycock and balderdash.

    :)

  13. cohenite March 13, 2011 at 1:11 pm #

    luke; it’s to the penalty box for you; Nasif is describing the operation of MEP as it works at the quantum level in respect to the energy functioning of one molecule type, CO2. Being a bureacrat I don’t suprise you’ve come across work, which is what MEP is, before!

    The only other paper I’ve seen which posits a cooling effect from increased CO2 is this:

    http://www.informaworld.com/smpp/content~db=all?content=10.1080/15567030701568727

    Unfortunately I don’t have the full copy but Chilingar describes a different mechanism based on reduced pressure and Henry’s Law to cause the cooling effect.

  14. Alan Siddons March 13, 2011 at 2:01 pm #

    “…the energy of these quantum/waves cannot be reabsorbed by molecules of carbon dioxide.”

    Critics, be cautious because professor Nahle is articulating and quantifying what is already known about so-called greenhouse gases. A US Department of Energy document says the same, although much more informally.

    “What happens after the GHG molecules absorb infrared radiation? The hot molecules release their energy, usually at lower energy (longer wavelength) radiation than the energy previously absorbed. The molecules cannot absorb energy emitted by other molecules of their own kind.” http://www.eia.doe.gov/cneaf/alternate/page/environment/appd_a.html

    So much for recycling…

  15. cementafriend March 13, 2011 at 3:04 pm #

    Cohenite, maybe you overlooked my comment at #3 above (please look at the link) and Nasif’s reply at #5.
    Here is the Chilingar et al paper http://www.google.com.au/#q=chilingar&hl=en&ei=exmSTOSQGdO6ccSmyPAG&start=30&sa=N&fp=4ea4a308ac696def . Some AGW people have tried to criticise Chilingar et al but there are other peer reviewed papers that make the same point.
    Here is one which has been published behind a paywall http://arxiv.org/abs/1003.1508

  16. Mack March 13, 2011 at 4:14 pm #

    Thanks Nasif,
    I was fooled by the old “camera trick”
    These AGWers will stop at nothing! :) :) :) :)

  17. cohenite March 13, 2011 at 4:46 pm #

    Thanks cementafriend, I am familiar with the Gerlich paper and the follow up paper and the Josh Halpern gerfuffle; I did have a link to the full copy of the Chilingar CO2 ‘cooling/convection greater than radiative transport of energy’ paper but it has closed. Chilingar was my number 9 in my best anti-AGW papers list:

    http://jennifermarohasy.com/blog/2008/09/ten-of-the-best-climate-research-papers-nine-peer-reviewed-a-note-from-cohenite/

  18. Luke March 13, 2011 at 5:50 pm #

    Mack – so will deniers (stop at nothing :-( :-( :-( …)

    You guys are like zombies – no matter how many times this drivel is refuted you line up for another helping. Talk about drongo-ism.

  19. Mack March 13, 2011 at 9:30 pm #

    Yes but Luke I absolutely love to line up to refute this drivel you espouse to.
    It’s great fun!
    AAaaahahahahahahahahahahaha.
    Another helping please!

  20. Otter March 13, 2011 at 9:45 pm #

    ‘each system as if they were blackbodies with emissivities and absorptivities of 100%’

    As a Geologist I can of course get the drift of most of the physics here, but not all. What I wish to know with the above line is this: IS that a correct statement, that black-body emmissivity and absorpties, are always 100%?

    I need to be certain of this in my own arguments, mind-numbed sheep like luke not included of course.

  21. cementafriend March 13, 2011 at 11:03 pm #

    Otter, the Planck spectral distribution of radiant energy flux is based on the premise of blackbody, “The charcteristic properties of a blackbody are that it absorbs all the radiation incident on its surface and that the quality and intensity of the radiation it emits are completely determined by its temperature”. The Stefan-Boltzman equation can be derived from the Planck equation for the distribution with the assumption of a hemispherical surface. Both Planck’s distribution and the Stefan-Boltzman equation assume a vacuum and that there is no other form of heat transfer.
    As I and others such as Nasif have indicated previously a) there are no blackbodies (including the sun) b) CO2 is a gas and not a surface c) CO2 only absorbs in very narrow wavelengths (and thus is far far from being a blackbody even if it was a solid with a surface)
    Engineers use an emissivity to relate the actual radiant absorption or emission to that of a blackbody. The emissivity varies with temperature. As Nasif has shown (and I have also put in my post http://jennifermarohasy.com/blog/index.php?s=cementafriend) the emissivity of CO2 is very small. Then, absorption is dependent of the amount present ( represented by partial pressure in Hoyt Hottel’s equation) The low emissivity and the low amount in the atmosphere leads to insignificant radiant energy absorption. CO2 of course can radiate to space at the top of the atmosphere. Nasif’s explanation fills in the details how this happens.

  22. williamcg March 14, 2011 at 1:32 am #

    Cohenite,

    The link that cementafriend provided for the Chilingar paper did not work for me. I have a stand alone copy of the PDF but I’m not sure how to get it to you.

  23. Kevan Hashemi March 14, 2011 at 2:28 am #

    You say, “If it hits a molecule of CO2, the carbon dioxide molecule absorbs only 2.7 x 10^-23 J, while the energy carried by the dispersed quantum/wave is 1.3423 x 10^-20 J.”

    Perhaps you can explain to your readers how a fraction of the energy of a photon can be absorbed by a molecule, leaving the photon to continue with the remainder of its energy. According to quantum theory, the process you describe is impossible. Either the CO2 molecule absorbs the entire photon, or it does not. There is no in-between of partial absorption.

    Furthermore, your suggestion that CO2 emits at a wavelength that it itself does not absorb is a violation of the Second Law of Thermodynamics, as described by Kirchoff’s thought experiment. See below for explanation.

    http://homeclimateanalysis.blogspot.com/2010/01/radiative-symmetry.html

  24. Schiller Thurkettle March 14, 2011 at 3:42 am #

    This information is difficult to reconcile with Luke’s claim that CO2 acts like a high-efficiency parabolic reflector which is oriented in the direction of the Earth’s center of mass.

  25. Otter March 14, 2011 at 7:48 am #

    Thank you, Cementafriend!

  26. cohenite March 14, 2011 at 10:19 am #

    Williamcg; I already have a hard copy thanks; my concern was whether there was alink still out there; apparently there is not.

    Kevan; all surfaces emit at different, indeed varying, wavelengths compared with the wavelengths they receive; how could it not be otherwise with processes of absorption, work and reflection operating.

    With a CO2 molecule absorption transfers that energy into kinetic movement which expends energy; if the CO2 molecule does not collide with another molecule than the remitted photon must have less energy than the one received because of the vibrational work of the molecule when absorption took place.

    As I said earlier Nasif’s calculations are consistent with MEP and therefore the 2nd law of thermodynamics.

  27. Kevan Hashemi March 14, 2011 at 2:17 pm #

    Cohenite: By your argument, a CO2 molecule in isolation at 300 K will never emit a photon unless it first absorbs a photon of higher energy. But that is not the case. So your assertion cannot be true.

  28. cohenite March 14, 2011 at 3:18 pm #

    Kevan; if a CO2 molecule at any temperature absorbs a photon its energy will increase; it will expend that energy through vibration and reemission; by definition the reemission cannot be at the same energy level as the absorption because some of the absorbed energy has been expressed kinetically . So your rebuttal is incorrect.

  29. Johnathan Wilkes March 14, 2011 at 6:24 pm #

    Kevan Hashemi March 14th, 2011 at 2:17 pm

    Cohenite: By your argument, a CO2 molecule in isolation at 300 K will never emit a photon unless it first absorbs a photon of higher energy.
    ==========================================
    When something is at rest or equilibrium, at any temperature, why would it emit a photon or DO anything?
    Admittedly, this situation seldom happens, but you need an input to produce an output!

  30. Nasif Nahle March 15, 2011 at 7:21 am #

    @Kevan Hashemi…

    Thanks for your comment.

    You say:

    Cohenite: By your argument, a CO2 molecule in isolation at 300 K will never emit a photon unless it first absorbs a photon of higher energy. But that is not the case. So your assertion cannot be true.

    Unless the molecule fo carbon dioxide you refer collides into another molecule, which would generate randomness of its kinetic energy. However, you’re referring to an isolated molecule of CO2, so collisions don’t apply and the molecule is at rest. The molecule won’t emitt a quantum/wave until another quantum/wave hits on it. Such event would cause the electron moves backwards and emit a photon.

    Please, don’t forget the anti-bunching effect discovered by Kimble, Dagenais and Mandel, which has been confirmed on different materials by many other scientists.

    :)

  31. Nasif Nahle March 15, 2011 at 7:23 am #

    By the way, the anti-bunching effect is also dismissed by AGW hypothesis.

    :)

  32. Luke March 15, 2011 at 10:00 am #

    So much theoretical dross – pity basic energy and spectroscopic measurements at the earth’s surface fly in the face of all this. Has anyone rung Davos and told them their life’s work is invalid …. hahahahahahahaa….

  33. hunter March 15, 2011 at 12:22 pm #

    Luke is correct on this.
    AGW calamity belief does not need new physics to fail.
    So mark the calendar.
    I agree with Luke on this one.
    ;^)

  34. Nasif Nahle March 15, 2011 at 12:42 pm #

    @Luke…

    Thanks for your… laughs?

    Please, tell me where those ridiculous 24 W/m2 of “downwelling” radiation comes from. Also, tell me how you could demonstrate that the “downwelling” radiation is absorbed by the surface.

    Anti-bunching effect is real, Luke; it has been demonstrated by experimentation… No place to your “hahahahaha”:

    http://prl.aps.org/abstract/PRL/v69/i10/p1516_1

    http://www.optique-quantique.u-psud.fr/FichiersPDF/OL25-1294.pdf

    http://www.physics.queensu.ca/~shughes/papers/EmyPRB2010.pdf

    :)

  35. Nasif Nahle March 15, 2011 at 12:45 pm #

    @hunter…

    Indeed! Luke is right… AGW belief doesn’t need modern physics to fail.

    :)

  36. jennifer March 15, 2011 at 3:12 pm #

    from Cementafriend…

    ****************

    Cohenite, seems my comment got lost in the ether (cold space) maybe it got surrounded by CO2 as I was typing.

    Firstly, this link to the Chilingar paper seems to work

    http://www.boatdesign.net/…/46771d1282714505-what-do-we-think-about-climate-change-chilingar-cooling-due-co2.pdf

    Google must have pulled the other one.

    Secondly, you and some others refer to photons. What if photons do not exist -look at this paper from real experience and measurements

    http://www.worldsci.org/pdf/abstracts/abstracts_5711.pdf

    Do you really think there are billions and billions of little photons, emitted by every molecule, having a specific narrow frequency (would that be one at 14.77777777777 micron and another at 14.7777777778 micron?) and related to some specific emission temperature?

    Plancks distribution is a continium which takes in X-rays, UV, visible light, IR, microwaves, radio waves. Note the word and description “waves”. Waves which can combine, and cancel.”

    *******
    sorry about comments not getting through, and I am travelling with limited internet access to check. Jen

  37. cohenite March 15, 2011 at 4:36 pm #

    Thanks Jennifer; unfortunately cementafriend’s link goes to some very nice cruising motor yachts; no matter, I do have a hard copy of the paper; I’ll keep searching for a link.

    No photons, eh? I will spend some more time on cementafriend’s linked paper but my intital impression is there appears to be a notable omission from it, namely Thomas Young and his double slit experiment.

  38. cementafriend March 15, 2011 at 10:29 pm #

    Sorry cohenite and others I checked the link just before writing the comment last night. It has now disappeared- maybe a co-incidence but maybe someone is trying to suppress access to free versions. It is behind a paywall here http://www.informaworld.com/smpp/content~content=a788582859~db=all~order=page
    Further to photons from an article by W.E. Lamb, Jr. Appl. Phys. B 60, 77-84 (1995) (I did not save link) -Summary “It is high time to give up the use of the word “photon”, and
    of a bad concept which will shortly be a century old. Radiation does not consist of particles, and the classical, i.e., non-quantum, limit of QTR is described by Maxwell’s equations for the electromagnetic fields, which do not involve particles. Talking about radiation in terms of
    particles is like using such ubiquitous phrases as “You know” or “I mean” which are very much to be heard in some cultures. For a friend of Charlie Brown, it might serve as a kind of security blanket.”

  39. mkelly March 16, 2011 at 4:26 am #

    How many molecules of carbon dioxide would be needed to get 249 Joules of energy in the total volume of carbon dioxide in the atmosphere?

    I am curious about how did you obtain the 249 joule figure.

    All gases dissipate heat. We heat our homes, cool our cars, dry our hair with that understanding. Gases don’t add heat.

  40. Nasif Nahle March 16, 2011 at 5:51 am #

    @mkelly…

    Thanks for your comment.

    Exactly! Gases don’t add heat. Let’s say it more properly: Gases don’t add energy, but disperse it.

    Regarding the 249 J figure, it is the load of energy transferred from the surface to the atmosphere. Let me tell you that you could see more than a dozen quantities assigned to the energy transferred from the surface to the air, depending on the impact the author wants to cause on his readers. ;)

    The real formula is: P = e (A) (σ) (Tw4 – Tcold4); for example, let’s say the surface is at 338.15 K and the air is at 303.15 K, the power transferred from the surface to the air would be 249.34 W, or 249.34 J/s. As it happens in one second, the energy would be 249.34 J/s * s = 249.34 J.

    Your last phrase says it all: All gases dissipate heat. We heat our homes, cool our cars, dry our hair with that understanding. Gases don’t add heat. In other words, the transfer of energy is always from the higher energy density to the lower energy density.

    :)

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